Part C · the analytic toolkit · spine

The Fourier transform, complex analysis, and why causality is a statement about poles

A linear system that cannot answer before it is asked has a frequency response with no poles in one half of the complex plane — and that one fact ties refraction to absorption, turns unit 06's step function into a theorem, and evaluates integrals nothing else can.

needs 06~3½ h slides + 7 h problemsproblems: ./mr problems 20

Unit 06 solved one oscillator completely and left four promissory notes: the delta function it avoided, the Fourier transform it computed without naming, the causality slide that said poles sit on one side of the real axis without saying why that matters, and the fundamental theorem of algebra, cited. Unit 09 left three more: the Laplace limit $0.6627\ldots$, the constants in the large-$x$ form of $J_n$, and how fast $I_n(ne)$ grows with $n$. This unit pays all seven.

It does so by building three pieces of mathematics that belong together. The Fourier transform diagonalises every linear time-invariant system, turning the differential equation into division. Distributions make the delta function and the Green's function honest. Complex analysis — holomorphy, Cauchy's theorem, residues — is what makes the transform invertible in practice, and it is also where causality turns out to live: a response function that vanishes before $t = 0$ is exactly one whose transform is holomorphic in the upper half-plane. Then asymptotics of integrals, which is contour deformation used for estimation rather than exact evaluation.

What you need, and what you do not

The dependency is unit 06 only, so nothing from units 03–05 is assumed: partial derivatives are defined here in the one line Cauchy–Riemann needs, and integrals along curves are defined here from scratch, as ordinary one-variable integrals of a parametrised path. No Green's theorem is used — Cauchy's theorem is proved by Goursat's subdivision argument, which needs only the triangle inequality and compactness. Two facts about double integrals are cited where they are used: that an absolutely integrable integrand may be integrated in either order (Fubini), and that a limit may pass under an integral sign when the integrands are dominated by one integrable function (dominated convergence). Both are proved in Stein & Shakarchi, Real Analysis, ch. 1–2.

How to read this deck

One object runs through it: unit 06's galvanometer, $\omega_0 = 8\ \mathrm{s^{-1}}$, $\gamma = 0.4\ \mathrm{s^{-1}}$, and its susceptibility $\chi(\omega)$. Section 1 shows $\chi$ is the Fourier transform of the Green's function; section 2 proves $LG = \delta$; section 3 finds $\chi$ holomorphic off two points and reads its Taylor radius off them; section 4 computes its residues; section 5 inverts it by closing contours and proves the causality theorem; section 6 reads its high-frequency tail by Watson's lemma. Every number quoted was computed. Proofs are on the slides where they are short and in these notes under Full proof where they are not; anything imported is labelled cited. N opens the notes, G lists the glossary terms.

20 · overview

One oscillator, inverted, and six sections

Unit 06 found the response of the galvanometer to a drive at one frequency, $\chi(\omega)$, and to one kick, $G(t)$. This unit proves they are one object seen twice — and inverts $\chi$ by closing a contour: up for $t < 0$, where there are no poles, so $G$ vanishes; down for $t > 0$, collecting two residues.

1 · the transform

Inversion, Plancherel, convolution, uncertainty.

2 · distributions

$\delta$ defined; $LG = \delta$ proved.

3 · holomorphy

Cauchy–Riemann, Goursat, Cauchy, Liouville.

4 · residues

Laurent, residues, Jordan, cuts, Rouché.

5 · causality

Poles, Kramers–Kronig, glass, Bromwich.

6 · asymptotics

Watson, Laplace, stationary phase, saddles, packets.

getting around

A dotted underline is a glossary term — click it, or press G. A solid underline like the Green's function is the running example: hover it. N shows the notes, where the long proofs are.

The problem sheet follows the same order. P1–P3 are transforms and distributions; P4–P6 are holomorphy and contour integrals, including Fresnel's integrals by an octant; P7 is Kramers–Kronig in full; P8–P9 are branch cuts and the Bromwich contour; P10 is the Laplace limit, finally proved; P11–P12 are stationary phase and steepest descent. All twelve end in a number; P3, P7 and P10 are also proofs.

Why this example. The galvanometer's $\chi$ is a rational function with two simple poles. That makes every step of the causality story computable by hand — which is the point: once the mechanism is visible on a function you can write down, the theorem that generalises it is a statement about something you have already seen. Section 5's theorem is about any causal response; section 5's worked example is about this one.

A change of convention from unit 06, made once and kept for the rest of the course: the drive is written $e^{-i\omega t}$, not $e^{i\omega t}$. The first slide of section 1 says why and what it changes.

section 1

The Fourier transform

A drive that is not periodic is still a superposition of frequencies — continuously many of them.

20 · transform

From one frequency to all of them

$f:\mathbb R\to\mathbb C$ is absolutely integrable if $\int_{-\infty}^\infty|f(t)|\,\mathrm dt < \infty$ (an improper Riemann integral). For such $f$:

the Fourier transform $$\hat f(\omega) = \int_{-\infty}^{\infty} f(t)\,e^{i\omega t}\,\mathrm dt,\qquad \omega\in\mathbb R .$$

It converges absolutely, since $|e^{i\omega t}| = 1$, and $|\hat f(\omega)| \le \int|f|$.

Riemann–Lebesgue lemma

$\hat f$ is continuous and $\hat f(\omega)\to0$ as $|\omega|\to\infty$.

Shift, modulation and scaling, each one substitution: $f(t-a)\mapsto e^{i\omega a}\hat f$, $\ e^{-i\nu t}f\mapsto\hat f(\omega-\nu)$, $\ f(t/a)\mapsto |a|\hat f(a\omega)$.

a convention, changed from unit 06

Unit 06 drove with $e^{i\omega t}$ and got $\chi_{06} = 1/(\omega_0^2 - \omega^2 + i\gamma\omega)$. From here on a single frequency is $e^{-i\omega t}$ — the convention of every later unit and of the physics literature — and

$$\chi(\omega) = \frac{1}{\omega_0^2 - \omega^2 - i\gamma\omega} = \chi_{06}(-\omega).$$

For real $\omega$ that is the complex conjugate: same modulus, lag $\varphi$ now read as $\arg\chi$. The poles move to the lower half-plane.

Why a transform at all: the galvanometer's response to $e^{-i\omega t}$ is $\chi(\omega)e^{-i\omega t}$. If a drive is a superposition of such exponentials, linearity gives the response as the same superposition, each term multiplied by $\chi$. The transform finds the weights.

Where the normalisation goes. Conventions differ in three places: the sign in the exponent, the placement of $2\pi$, and whether $\omega$ or $\nu = \omega/2\pi$ is used. This course puts the whole $1/2\pi$ in the inverse and uses $+i\omega t$ in the forward transform, so that a signal $e^{-i\omega_1 t}$ has its spectrum at $+\omega_1$. With $e^{-i\omega t}$ as the unit of oscillation, a wave travelling towards $+x$ is $e^{i(kx - \omega t)}$ with $k, \omega > 0$: that is the reason for the change, and it is what unit 24 and every later unit use.

Full proof of the Riemann–Lebesgue lemma

Continuity. $|\hat f(\omega) - \hat f(\nu)| \le \int|f(t)||e^{i(\omega-\nu)t} - 1|\,\mathrm dt$. Given $\varepsilon$, choose $T$ with $\int_{|t|>T}|f| < \varepsilon$; on $|t|\le T$, $|e^{i(\omega-\nu)t} - 1|\le|\omega-\nu|T$. So the difference is at most $2\varepsilon + |\omega-\nu|T\int|f|$, which is $<3\varepsilon$ once $|\omega - \nu|$ is small.

Decay. For the indicator of an interval, $\int_a^be^{i\omega t}\mathrm dt = (e^{i\omega b} - e^{i\omega a})/i\omega$, of modulus at most $2/|\omega|$. So step functions (finite combinations of indicators) have transforms tending to zero. A Riemann-integrable function on $[-T, T]$ is within $\varepsilon$ in $\int|\cdot|$ of a step function $s$ (lower Darboux sums), and the tails beyond $T$ are within $\varepsilon$ too; then $|\hat f(\omega)| \le |\hat s(\omega)| + 2\varepsilon$, and the first term is below $\varepsilon$ for large $|\omega|$. $\square$

The three rules. $\int f(t-a)e^{i\omega t}\mathrm dt$ with $u = t - a$ is $e^{i\omega a}\hat f(\omega)$. $\int e^{-i\nu t}f(t)e^{i\omega t}\mathrm dt = \hat f(\omega - \nu)$ by inspection. $\int f(t/a)e^{i\omega t}\mathrm dt$ with $u = t/a$ is $|a|\int f(u)e^{ia\omega u}\mathrm du$, the modulus because a negative $a$ reverses the limits.

Dimensions. If $t$ is a time and $f$ has dimension $[f]$, then $\hat f$ has dimension $[f]\,\mathrm T$ and its argument $\omega$ is an inverse time. Every formula in this section can be checked against that.

20 · transform

Differentiation becomes multiplication

the dictionary

If $f$ is $C^1$ and $f, f'$ are absolutely integrable, then $\widehat{f'}(\omega) = -i\omega\hat f(\omega)$. If $f$ and $tf$ are absolutely integrable, then $\hat f$ is $C^1$ and $\widehat{tf} = -i\,\hat f\,{}'$.

Proof of the first. Integrate by parts on $[-T, T]$: $\int f'e^{i\omega t} = [fe^{i\omega t}]_{-T}^{T} - i\omega\int fe^{i\omega t}$, and $f(\pm T)\to0$ because $f$ has limits at $\pm\infty$ ($f'$ integrable) which must be $0$ ($f$ integrable). $\square$

Apply it to the galvanometer, $\ddot x + \gamma\dot x + \omega_0^2x = f$:

$$(\omega_0^2 - \omega^2 - i\gamma\omega)\,\hat x = \hat f,\quad\text{so}\quad\hat x = \chi\,\hat f .$$

A differential equation has become division. Dimensions: $[\chi] = \mathrm T^2$, $[\hat f] = [x]\mathrm T^{-2}\cdot\mathrm T$, so $[\hat x] = [x]\mathrm T$ ✓.

worked example 1 — the Gaussian, by an ODE

$f(t) = e^{-t^2/2}$. It satisfies $f' = -tf$. Transform both sides with the dictionary:

$$-i\omega\hat f = i\hat f\,{}',\quad\text{so}\quad\hat f\,{}' = -\omega\hat f .$$

Separable (unit 00), unique through its initial value (unit 01), and $\hat f(0) = \int e^{-t^2/2}\mathrm dt = \sqrt{2\pi}$ (notes):

$$\hat f(\omega) = \sqrt{2\pi}\,e^{-\omega^2/2}.$$

A Gaussian is its own transform, up to $\sqrt{2\pi}$. By scaling, $e^{-t^2/2\sigma^2}\mapsto\sigma\sqrt{2\pi}\,e^{-\sigma^2\omega^2/2}$: width $\sigma$ in time, $1/\sigma$ in frequency.

Check: $\sigma\to\infty$, a constant signal, piles the spectrum up at $\omega = 0$; both sides carry $\mathrm T$ ✓.

Full proof of the second rule

$\frac{\hat f(\omega + h) - \hat f(\omega)}{h} = \int f(t)e^{i\omega t}\frac{e^{iht} - 1}{h}\,\mathrm dt$, and $|e^{iht} - 1|/|h| \le |t|$. The integrand is dominated by $|tf(t)|$, which is integrable, and converges pointwise to $itf(t)e^{i\omega t}$; dominated convergence (cited, Stein–Shakarchi, Real Analysis, ch. 2 Thm 1.13) gives $\hat f\,{}' = \widehat{itf}$, which is the rule. Continuity of $\hat f\,{}'$ is Riemann–Lebesgue applied to $tf$. $\square$

Full proof that $\int e^{-t^2/2}\mathrm dt = \sqrt{2\pi}$, without polar coordinates

Polar coordinates are unit 05, which this unit does not assume, so here is a one-variable proof. Let $A(x) = \big(\int_0^xe^{-s^2}\mathrm ds\big)^2$ and $B(x) = \int_0^1\frac{e^{-x^2(1+u^2)}}{1+u^2}\mathrm du$. Then $A'(x) = 2e^{-x^2}\int_0^xe^{-s^2}\mathrm ds$, and differentiating under the integral sign (the integrand and its $x$-derivative are continuous on the compact $[0,1]$, so this is the elementary Leibniz rule) gives $B'(x) = -2x\int_0^1e^{-x^2(1+u^2)}\mathrm du = -2e^{-x^2}\int_0^xe^{-s^2}\mathrm ds$ with $s = xu$. So $A + B$ is constant, equal to $A(0) + B(0) = \int_0^1\frac{\mathrm du}{1+u^2} = \frac\pi4$. As $x\to\infty$, $B(x)\le e^{-x^2}\to0$, so $\big(\int_0^\infty e^{-s^2}\big)^2 = \pi/4$. Hence $\int_{\mathbb R}e^{-s^2} = \sqrt\pi$, and $s = t/\sqrt2$ gives $\sqrt{2\pi}$. $\square$

Why the ODE route matters. It computes the transform without evaluating a single oscillatory integral. Section 3 extends the answer to a complex width — a step that needs analytic continuation to be legitimate — and section 6 uses that extension for a spreading wave packet.

The oscillator equation, transformed. The dictionary applies twice ($\widehat{\ddot x} = (-i\omega)^2\hat x$) provided $x, \dot x, \ddot x$ are integrable — true for the response to an integrable drive once the transient has died, and exactly true for $x = G * f$.

20 · transform

Inversion

the Fourier inversion theorem

If $f$ is continuous and both $f$ and $\hat f$ are absolutely integrable, then for every $t$

$$f(t) = \frac{1}{2\pi}\int_{-\infty}^{\infty}\hat f(\omega)\,e^{-i\omega t}\,\mathrm d\omega .$$

The kernels $K_\varepsilon$ of the proof: every one has area $1$, and as $\varepsilon\to0$ the area piles up at $u = 0$. Averaging $f$ against a kernel this narrow returns $f(t)$.

The idea of the proof. Insert a Gaussian $e^{-\varepsilon\omega^2/2}$ to make the double integral absolutely convergent, swap the order, and recognise the result:

$$\frac1{2\pi}\!\int\!\hat f e^{-i\omega t}e^{-\varepsilon\omega^2/2}\mathrm d\omega = \!\int\! f(s)K_\varepsilon(t - s)\mathrm ds,$$

where $K_\varepsilon(u) = e^{-u^2/2\varepsilon}/\sqrt{2\pi\varepsilon}$ is worked example 1 read backwards: positive, total integral $1$, concentrating at $u = 0$ as $\varepsilon\to0$. So the right side tends to $f(t)$, and the left side to the inversion integral.

On the running example: $G$ is continuous and integrable and $|\chi|\le C/(1+\omega^2)$ is integrable, so $G$ is recovered from $\chi$ by this formula. Section 5 evaluates it.

In particular $\hat f = 0$ forces $f = 0$: the transform is injective, so an identity between transforms is an identity between functions.

Full proof

Step 1 — the swap. $\int\!\!\int|f(s)|e^{-\varepsilon\omega^2/2}\,\mathrm ds\,\mathrm d\omega < \infty$, so Fubini (cited) allows $\frac1{2\pi}\int\big(\int f(s)e^{i\omega s}\mathrm ds\big)e^{-i\omega t}e^{-\varepsilon\omega^2/2}\mathrm d\omega = \int f(s)\Big(\frac1{2\pi}\int e^{-i\omega(t-s)}e^{-\varepsilon\omega^2/2}\mathrm d\omega\Big)\mathrm ds$. The inner integral is the transform of a Gaussian of width $\sigma = 1/\sqrt\varepsilon$, evaluated at $-(t-s)$: by worked example 1 and scaling, $\frac1{2\pi}\sqrt{2\pi/\varepsilon}\,e^{-(t-s)^2/2\varepsilon} = K_\varepsilon(t - s)$.

Step 2 — the right side tends to $f(t)$. $\int f(s)K_\varepsilon(t-s)\,\mathrm ds - f(t) = \int\big(f(t - u) - f(t)\big)K_\varepsilon(u)\,\mathrm du$. Given $\eta > 0$ choose $\delta$ with $|f(t-u) - f(t)| < \eta$ for $|u| < \delta$ (continuity). The part $|u| < \delta$ is below $\eta$. The part $|u|\ge\delta$ is at most $\sup_{|u|\ge\delta}K_\varepsilon(u)\int|f| + |f(t)|\int_{|u|\ge\delta}K_\varepsilon$, and both terms tend to $0$ with $\varepsilon$ because $K_\varepsilon(\delta)\to0$ and the Gaussian's mass concentrates.

Step 3 — the left side tends to the inversion integral. $|\hat f(\omega)e^{-i\omega t}(e^{-\varepsilon\omega^2/2} - 1)|\le2|\hat f(\omega)|$, integrable by hypothesis, and pointwise the difference tends to $0$: dominated convergence. $\square$

Reading the figure. $K_\varepsilon$ is a Gaussian of variance $\varepsilon$, so its width is $\sqrt\varepsilon$ and its height $1/\sqrt{2\pi\varepsilon}$; the three curves are $\varepsilon = 0.5$, $0.1$, $0.02$, heights $0.56$, $1.26$, $2.82$. Step 2 of the proof is the statement that a continuous $f$ is nearly constant over the width of the tallest one.

History. Fourier wrote the inversion formula in 1807–1822 for heat flow on an infinite rod, without proof. Cauchy and Poisson both had versions; the approximate-identity argument above is essentially Weierstrass's (1885). Inserting a convergence factor and removing it at the end is the single most useful trick in this subject, and it returns in section 2 and section 5.

20 · transform

Inversion at a jump: the box

the hypothesis on $\hat f$ is not decoration

The box $f = \mathbf 1_{[-1,1]}$ has $\hat f(\omega) = 2\sin\omega/\omega$, which is not absolutely integrable — because $f$ jumps. At the jump the symmetric limit $\lim_{W\to\infty}\frac1{2\pi}\int_{-W}^{W}$ still converges, to the midpoint $\frac12$ (notes).

The truncated inversion $f_W(t) = \frac1{2\pi}\int_{-W}^{W}\hat f(\omega)e^{-i\omega t}\,\mathrm d\omega$ can be written down: with $\operatorname{Si}(x) = \int_0^x\frac{\sin u}{u}\,\mathrm du$,

$$f_W(t) = \tfrac1\pi\big[\operatorname{Si}\big(W(1+t)\big) + \operatorname{Si}\big(W(1-t)\big)\big].$$

At $t = 1$ it is $\operatorname{Si}(2W)/\pi$, which tends to $\frac12$ because $\int_0^\infty\frac{\sin u}{u}\mathrm du = \frac\pi2$ (worked example 2, section 4). A slowly decaying transform is the signature of a jump.

Top: $\hat f$ falls off only like $1/\omega$. Bottom: the truncated inversions for $W = 6$ and $W = 30$. Away from the jumps they close in on the box; at $t = \pm1$ they pass near the midpoint. Both overshoot by about $9\%$ beside each jump — the larger $W$ narrower, not lower.

The truncated integral. $\frac1{2\pi}\int_{-W}^{W}\frac{2\sin\omega}{\omega}e^{-i\omega t}\mathrm d\omega$: the odd part $-i\sin\omega t$ integrates to zero against the even $\sin\omega/\omega$, leaving $\frac1\pi\int_0^W\frac{2\sin\omega\cos\omega t}{\omega}\mathrm d\omega = \frac1\pi\int_0^W\frac{\sin\omega(1+t) + \sin\omega(1-t)}{\omega}\mathrm d\omega$ (product to sum), and $u = \omega(1\pm t)$ turns each piece into $\operatorname{Si}$. The curves in the figure are this formula evaluated; at $t = 1$ they give $0.479$ ($W = 6$) and $0.505$ ($W = 30$). The overshoot beside the jump is a computed observation about the two curves drawn — its limit, $\operatorname{Si}(\pi)/\pi - \frac12 = 0.0895$, is the Gibbs phenomenon, cited (Körner, Fourier Analysis, the chapters on Gibbs's phenomenon) — and nothing later depends on it.

The box. $\int_{-1}^1e^{i\omega t}\mathrm dt = (e^{i\omega} - e^{-i\omega})/i\omega = 2\sin\omega/\omega$. Not absolutely integrable: over $[k\pi, (k+1)\pi]$ the integral of $|\sin\omega|/\omega$ is at least $2/((k+1)\pi)$, and the harmonic series diverges. The midpoint, for the box. At $t = 1$, $\frac1{2\pi}\int_{-W}^{W}\frac{2\sin\omega}{\omega}e^{-i\omega}\,\mathrm d\omega = \frac1{2\pi}\int_{-W}^{W}\frac{2\sin\omega\cos\omega}{\omega}\,\mathrm d\omega = \frac1{2\pi}\int_{-W}^{W}\frac{\sin2\omega}{\omega}\,\mathrm d\omega$ (the odd part integrates to zero), which tends to $\frac1{2\pi}\cdot\pi = \frac12$ by worked example 2: halfway between $f(1^-) = 1$ and $f(1^+) = 0$. The general statement — the symmetric limit gives the midpoint at every jump of a piecewise-$C^1$ integrable $f$ — is cited (Titchmarsh, Introduction to the Theory of Fourier Integrals, ch. 1) and is not needed later in this course.

20 · transform

Plancherel and convolution

Plancherel's theorem

For $f, g$ as in the inversion theorem,

$$\int f\,\bar g\,\mathrm dt = \frac1{2\pi}\int\hat f\,\bar{\hat g}\,\mathrm d\omega,$$

and in particular $\int|f|^2\mathrm dt = \frac1{2\pi}\int|\hat f|^2\mathrm d\omega$.

Proof, one line. Write $\bar g$ by inversion, swap (Fubini): $\int f(t)\frac{1}{2\pi}\int\bar{\hat g}(\omega)e^{i\omega t}\mathrm d\omega\,\mathrm dt = \frac{1}{2\pi}\int\bar{\hat g}\hat f\,\mathrm d\omega$. $\square$

Energy in time equals energy in frequency: the transform is, up to $\sqrt{2\pi}$, an isometry. It extends to all square-integrable functions (cited: Stein–Shakarchi, Real Analysis, ch. 5 §1).

the convolution theorem

If $f, g$ are absolutely integrable and bounded, $(f*g)(t) = \int f(s)g(t-s)\,\mathrm ds$ satisfies

$$\widehat{f*g} = \hat f\,\hat g .$$

Proof: swap the two integrals and substitute $u = t - s$, which splits $e^{i\omega t} = e^{i\omega s}e^{i\omega u}$. $\square$ So unit 06's $x = G*f$ becomes $\hat x = \hat G\hat f$ — and comparing with the dictionary slide,

$$\hat G(\omega) = \chi(\omega).$$

checked on the running example

$\int_0^\infty G^2\,\mathrm dt = \dfrac{1}{2\gamma\omega_0^2} = 0.01953125\ \mathrm{s^3}$, and $\dfrac1{2\pi}\displaystyle\int|\chi|^2\mathrm d\omega = \dfrac{0.1227185}{2\pi} = 0.01953125$ ✓.

$\hat G = \chi$, directly. $\int_0^\infty e^{-\gamma t/2}\frac{\sin\omega_\mathrm dt}{\omega_\mathrm d}e^{i\omega t}\mathrm dt$: write $\sin = (e^{i\omega_\mathrm dt} - e^{-i\omega_\mathrm dt})/2i$ and integrate the exponentials, $\int_0^\infty e^{-at}\mathrm dt = 1/a$ for $\operatorname{Re}a>0$, with $a_\pm = \gamma/2 - i\omega \mp i\omega_\mathrm d$. Then $\hat G = \frac{1}{2i\omega_\mathrm d}\big(\frac1{a_+} - \frac1{a_-}\big) = \frac{1}{2i\omega_\mathrm d}\cdot\frac{a_- - a_+}{a_+a_-} = \frac{1}{a_+a_-}$, and $a_+a_- = (\gamma/2 - i\omega)^2 + \omega_\mathrm d^2 = \omega_0^2 - \omega^2 - i\gamma\omega$ ✓. This is unit 06's worked example 3 with the new sign of $\omega$.

The Plancherel numbers. $\int_0^\infty e^{-\gamma t}\sin^2(\omega_\mathrm dt)\,\mathrm dt = \frac12\big(\frac1\gamma - \frac{\gamma}{\gamma^2 + 4\omega_\mathrm d^2}\big) = \frac{2\omega_\mathrm d^2}{\gamma\cdot4\omega_0^2}$ using $\gamma^2 + 4\omega_\mathrm d^2 = 4\omega_0^2$; divided by $\omega_\mathrm d^2$ that is $1/(2\gamma\omega_0^2) = 1/51.2$. The frequency side, $\int\frac{\mathrm d\omega}{(\omega_0^2-\omega^2)^2 + \gamma^2\omega^2} = \frac{\pi}{\gamma\omega_0^2}$, is problem P5, by residues; numerically $0.12271846$. Dimensions: $[G] = \mathrm T$, so $\int G^2\mathrm dt$ is $\mathrm T^3$; $[\chi]^2[\omega] = \mathrm T^4\mathrm T^{-1}$ ✓. Limit: as $\gamma\to0$ both sides diverge like $1/\gamma$ — an undamped oscillator rings forever and its response is concentrated on two infinitely sharp lines.

Full proof of the convolution theorem

$f*g$ is bounded by $\sup|g|\int|f|$, and $\int|f*g|\le\int|f|\int|g|$ by Tonelli (the nonnegative case of Fubini, cited with it). Then $\widehat{f*g}(\omega) = \int\!\!\int f(s)g(t-s)e^{i\omega t}\mathrm ds\,\mathrm dt$; the double integral converges absolutely, so Fubini lets the $t$-integral go inside, and with $u = t - s$ it is $\int f(s)e^{i\omega s}\big(\int g(u)e^{i\omega u}\mathrm du\big)\mathrm ds = \hat f\hat g$. $\square$

The structural point. Every exponential $e^{-i\omega t}$ is an eigenfunction of every time-invariant linear operator — time translation acts on it by a scalar, and anything commuting with all translations must too. The Fourier transform is the change of basis to those eigenfunctions, and in that basis every such operator is diagonal: multiplication by its eigenvalue $\chi(\omega)$. That is why one function of frequency is the whole system.

20 · transform

The uncertainty relation

For $f$ with $\int|f|^2 = 1$ define the spreads about the means $\bar t$, $\bar\omega$:

$$\begin{aligned}\Delta t^2 &= \textstyle\int(t-\bar t)^2|f|^2\,\mathrm dt,\\ \Delta\omega^2 &= \textstyle\frac1{2\pi}\int(\omega - \bar\omega)^2|\hat f|^2\,\mathrm d\omega .\end{aligned}$$

the uncertainty relation

If $f$ is $C^1$ with $f, tf, f'$ square-integrable and $\sqrt{|t|}f\to0$ at $\pm\infty$,

$$\Delta t\,\Delta\omega \ge \tfrac12,$$

with equality exactly for Gaussians $f = Ce^{-a(t - \bar t)^2 - i\bar\omega t}$, $a > 0$.

Physics: a short pulse cannot be monochromatic. A 10 fs laser pulse carries a spread of frequencies of order $10^{14}\ \mathrm{s^{-1}}$, whatever the laser.

proof, by Cauchy–Schwarz

Translate and modulate so $\bar t = \bar\omega = 0$ (neither changes $\Delta t$, $\Delta\omega$). Integrate by parts:

$$1 = -\!\int t\,(|f|^2)'\,\mathrm dt = -2\operatorname{Re}\!\int t\bar ff'\,\mathrm dt .$$

Cauchy–Schwarz: $1\le2\,\|tf\|\,\|f'\|$. By the dictionary and Plancherel, $\|f'\|^2 = \frac1{2\pi}\int\omega^2|\hat f|^2 = \Delta\omega^2$, and $\|tf\|^2 = \Delta t^2$. So $1\le 2\Delta t\,\Delta\omega$. $\square$

checks

Equality: $e^{-t^2/2}/\pi^{1/4}$ has $\Delta t^2 = \Delta\omega^2 = \frac12$ ✓. Dimensions: $\mathrm T\cdot\mathrm T^{-1}$ ✓. The running example: $G$ gives $\Delta t\,\Delta\omega = \sqrt{Q^2 - \frac12 + \frac1{4Q^2}} = 19.9875$ — far from minimal, and P2 says why.

Full proof, the steps the slide compresses

The integration by parts. $(|f|^2)' = (f\bar f)' = f'\bar f + f\bar f' = 2\operatorname{Re}(\bar ff')$. On $[-T, T]$, $\int t(|f|^2)' = [t|f|^2]_{-T}^T - \int|f|^2$, and the boundary term vanishes as $T\to\infty$ by the hypothesis $\sqrt{|t|}f\to0$. Cauchy–Schwarz in the inner product $\langle u, v\rangle = \int u\bar v$: $|\int(t f)\overline{f'}| \le \|tf\|\|f'\|$, and $|\operatorname{Re}z|\le|z|$. Plancherel for $f'$ with $\widehat{f'} = -i\omega\hat f$ gives $\int|f'|^2 = \frac1{2\pi}\int\omega^2|\hat f|^2$.

Removing the means. $g(t) = e^{i\bar\omega t}f(t + \bar t)$ has $|g(t)| = |f(t+\bar t)|$, so its time spread is the same about mean $0$; its transform is $e^{-i(\omega+\bar\omega)\bar t}\,\hat f(\omega + \bar\omega)$ (shift and modulation rules), whose modulus is $|\hat f(\omega+\bar\omega)|$, so its frequency spread is the same about mean $0$.

The equality case. Cauchy–Schwarz is an equality iff $f' = ctf$ for a constant $c$, and the real part must equal the modulus, which forces $c$ real; integrability forces $c < 0$. The ODE $f' = ctf$ has the unique solution $f = f(0)e^{ct^2/2}$ (unit 01). Undoing the translation and modulation gives the family on the slide. P2 proves the converse.

The 10 fs pulse. For a Gaussian pulse of intensity full width $10$ fs, $\Delta t = 10/(2\sqrt{2\ln2}) = 4.25$ fs in the $|f|^2$ sense, so $\Delta\omega\ge1/(2\Delta t) = 1.18\times10^{14}\ \mathrm{s^{-1}}$. The optical carrier at $800$ nm is $\omega = 2.35\times10^{15}\ \mathrm{s^{-1}}$, so the pulse spans about $\pm5\%$ of its own frequency: it is white-ish, and section 6 shows what glass does to it.

Quantum mechanics, which this course does not reach, reads this theorem with $t\to x$, $\omega\to p/\hbar$, and calls it Heisenberg's; the mathematics is the slide, unchanged. Weyl gave this proof in 1928.

The galvanometer's value, derived in P2: $\Delta\omega = \omega_0$ exactly and $\Delta t^2 = \gamma^{-2} - \frac{1}{2\omega_0^2} + \frac{\gamma^2}{4\omega_0^4}$, product $\sqrt{400 - 0.5 + 0.000625} = 19.9875$ at $Q = 20$. The spread in frequency is measured about $\bar\omega = 0$, and a real signal has two spectral peaks at $\pm\omega_\mathrm d$: most of $\Delta\omega$ is the distance between the peaks, not their width.

section 2

Distributions

A kick too short to resolve is not a function of time. It is a rule for what happens to every smooth test of it.

20 · distributions

Test functions and distributions

test functions

$\mathcal D = C_c^\infty(\mathbb R)$: smooth functions vanishing outside a bounded set. $\varphi_n\to\varphi$ in $\mathcal D$ means: all supports in one bounded set, and $\varphi_n^{(k)}\to\varphi^{(k)}$ uniformly for every $k$.

They exist: $\psi(t) = e^{-1/(1-t^2)}$ for $|t|<1$, $0$ otherwise, is $C^\infty$.

a distribution

A linear map $T:\mathcal D\to\mathbb C$, written $\varphi\mapsto\langle T,\varphi\rangle$, that is continuous: $\varphi_n\to\varphi$ implies $\langle T,\varphi_n\rangle\to\langle T,\varphi\rangle$.

Physics first, here too: a ballistic galvanometer reads $\int i\,\mathrm dt$ and nothing else about the pulse (unit 06). An idealised kick is exactly that: defined by what it does to smooth responses, not by its values.

Three examples.

Functions. A locally integrable $f$ gives $\langle T_f,\varphi\rangle = \int f\varphi$. Different continuous $f$ give different $T_f$, so functions sit inside distributions.
The Dirac delta: $\langle\delta,\varphi\rangle = \varphi(0)$.
The principal value: $\big\langle\mathrm{PV}\tfrac1t,\varphi\big\rangle = \displaystyle\lim_{\varepsilon\to0^+}\int_{|t|>\varepsilon}\frac{\varphi(t)}{t}\,\mathrm dt = \int_0^\infty\frac{\varphi(t) - \varphi(-t)}{t}\,\mathrm dt$.

The next slide draws the delta as a limit of pulses, and shows it is not a function.

Full proof that $\psi$ is smooth

It suffices that $h(x) = e^{-1/x}$ for $x>0$, $h(x) = 0$ for $x\le0$, is $C^\infty$, since $\psi(t) = h(1 - t^2)$. For $x > 0$, induction gives $h^{(k)}(x) = p_k(1/x)e^{-1/x}$ with $p_k$ a polynomial, and $p_k(1/x)e^{-1/x}\to0$ as $x\to0^+$ because $y^me^{-y}\to0$ as $y\to\infty$. So every one-sided derivative at $0$ exists and is $0$, matching the left side. $\square$ Such a function is not analytic at $0$ — its Taylor series there is $0$ — and section 3 will show that a holomorphic function can never do this. That is the first sign that complex differentiability is a much stronger condition than real.

The principal value is a distribution. $\int_{|t|>\varepsilon}\varphi/t$: the odd part of $1/t$ pairs the region $t>\varepsilon$ with $t<-\varepsilon$, giving $\int_\varepsilon^\infty(\varphi(t) - \varphi(-t))/t\,\mathrm dt$. By the mean value theorem $|\varphi(t) - \varphi(-t)|\le2t\sup|\varphi'|$, so the integrand is bounded near $0$ and the limit exists. If $\varphi$ is supported in $[-A, A]$, then $|\langle\mathrm{PV}\tfrac1t,\varphi\rangle|\le2A\sup|\varphi'|$, which gives continuity.

History. Dirac used $\delta$ freely from 1927; Sobolev (1936) and Schwartz (1945–1950) made it a theorem-bearing object, and Schwartz's Théorie des distributions is the source of the definitions here. Heaviside had been using its integral, the step function, since the 1890s.

20 · distributions

Pulses that become δ

A sequence $T_n\to T$ in distributions if $\langle T_n,\varphi\rangle\to\langle T,\varphi\rangle$ for each $\varphi$. Unit 06's square pulses of unit area, $p_\varepsilon = \varepsilon^{-1}\mathbf 1_{[0,\varepsilon]}$, have

$$\langle p_\varepsilon,\varphi\rangle = \frac1\varepsilon\int_0^\varepsilon\varphi\,\mathrm dt\ \to\ \varphi(0) = \langle\delta,\varphi\rangle,$$

so $p_\varepsilon\to\delta$ — whatever the pulse shape, provided its area is $1$.

$\delta$ is not a function

If $\int f\varphi = \varphi(0)$ for all $\varphi$, then $\int f\varphi = 0$ for every $\varphi$ vanishing near $0$, so $f = 0$ away from $0$, so $\int f\varphi = 0$ always — contradicting $\varphi(0)\ne0$.

The pulses converge; their pointwise limit (infinite at $0$, zero elsewhere) is no function at all. The limit lives only in what it does to $\varphi$.

Three pulses of area $1$, $\varepsilon = 1, \frac12, \frac14$, against a test function $\varphi$ (a stretched copy of the bump $\psi$). Pairing a pulse with $\varphi$ averages $\varphi$ over $[0,\varepsilon]$, and those averages close in on $\varphi(0)$.

20 · distributions

Derivatives by duality, and $LG = \delta$

the distributional derivative $$\langle T',\varphi\rangle := -\langle T,\varphi'\rangle .$$

For $C^1$ functions this is integration by parts, so nothing old changes. Every distribution now has derivatives of all orders.

The step: $H = \mathbf 1_{t>0}$ has $\langle H',\varphi\rangle = -\int_0^\infty\varphi' = \varphi(0)$, so $H' = \delta$.

the jump formula

If $f$ is $C^1$ on $t<0$ and on $t>0$ with one-sided limits, jump $[f] = f(0^+) - f(0^-)$, then

$$f' = \{f'\} + [f]\,\delta,$$

$\{f'\}$ being the classical derivative off $0$. Proof: split $-\int f\varphi'$ at $0$ and integrate each half by parts. $\square$

the jump condition, as a theorem

Let $L = \frac{\mathrm d^2}{\mathrm dt^2} + \gamma\frac{\mathrm d}{\mathrm dt} + \omega_0^2$ and $G$ vanish for $t < 0$, $C^2$ on $t > 0$ with one-sided limits. Then $LG = \delta$ in distributions iff

$LG = 0$ on $t>0$, $\ G(0^+) = 0$, $\ \dot G(0^+) = 1$.

The galvanometer's $G$ was built in unit 06 to meet exactly these three conditions. So $LG = \delta$ — the defining property, now a statement about a well-defined object. The proof, and a picture of the three conditions, are on the next slide.

Full proof of the jump formula

$\langle f',\varphi\rangle = -\int_{-\infty}^0f\varphi' - \int_0^\infty f\varphi'$. On $(-\infty,0)$: $-[f\varphi]_{-\infty}^{0^-} + \int_{-\infty}^0f'\varphi = -f(0^-)\varphi(0) + \int_{-\infty}^0f'\varphi$. On $(0,\infty)$: $-[f\varphi]_{0^+}^\infty + \int_0^\infty f'\varphi = f(0^+)\varphi(0) + \int_0^\infty f'\varphi$. Sum: $\int\{f'\}\varphi + (f(0^+) - f(0^-))\varphi(0)$. $\square$ ($\varphi$ has bounded support, so the terms at $\pm\infty$ vanish.)

Uniqueness of the causal $G$. If $G_1, G_2$ both satisfy the theorem's hypotheses and $LG_i = \delta$, their difference $D$ vanishes for $t < 0$ and satisfies $LD = 0$ on $t > 0$ with $D(0^+) = \dot D(0^+) = 0$, so $D \equiv 0$ by uniqueness for the initial value problem (unit 01). The step function in $G$ is not an extra assumption on top of the equation; it is the choice of which of the many solutions of $LG = \delta$ you want, and section 5 shows it is equivalent to a choice of half-plane.

Unit 06's jump, recovered physically. Integrating $LG = \delta$ "across $0$" was the heuristic there; the theorem is its rigorous form. The dimensionless $1$ in $\dot G(0^+) = 1$ is the unit impulse per unit inertia: $[\delta] = \mathrm T^{-1}$ and $[LG] = [G]\mathrm T^{-2}$, so $[G] = \mathrm T$ ✓.

20 · distributions

The jump condition, proved

Proof. The jump formula once gives $\dot G = \{\dot G\} + [G]\,\delta$. Differentiate that, using the jump formula again on $\{\dot G\}$:

$$\ddot G = \{\ddot G\} + [\dot G]\,\delta + [G]\,\delta' .$$

Add $\gamma\dot G + \omega_0^2G$ ($G$ itself is a function, so $\omega_0^2G$ brings no delta):

$$LG = \{LG\} + \big([\dot G] + \gamma[G]\big)\delta + [G]\,\delta' .$$

$\delta$ and $\delta'$ are independent (test with $\varphi$ having $\varphi(0) = 0\ne\varphi'(0)$, and conversely), so $LG = \delta$ iff $\{LG\} = 0$, $[G] = 0$ and $[\dot G] = 1$. $\square$

The galvanometer's $G$ and $\dot G$ across the kick. $G$ is continuous with a corner, so $[G] = 0$ and no $\delta'$ appears; $\dot G$ jumps by exactly $1$, which is the single $\delta$ in $LG$.

The second application. $\dot G$ is itself piecewise $C^1$ with jump $[\dot G]$, so $\ddot G = \{\ddot G\} + [\dot G]\delta + ([G]\delta)'$ — the last term because the distributional derivative of $\{\dot G\} + [G]\delta$ is $\{\ddot G\} + [\dot G]\delta + [G]\delta'$. Then $LG = \ddot G + \gamma\dot G + \omega_0^2G$ collects to the displayed formula; $\omega_0^2G$ contributes no delta because $G$ itself is a function.

Independence of $\delta$ and $\delta'$. Suppose $a\delta + b\delta' = 0$. Then for every $\varphi$, $a\varphi(0) - b\varphi'(0) = 0$. Take $\varphi = \psi$ (the bump, even, $\psi'(0) = 0$, $\psi(0) = e^{-1}$): $a = 0$. Take $\varphi(t) = t\psi(t)$: $\varphi(0) = 0$, $\varphi'(0) = e^{-1}$, so $b = 0$.

Reading the figure. Both panels are the formulas of unit 06 evaluated: $G = e^{-\gamma t/2}\sin(\omega_\mathrm dt)/\omega_\mathrm d$ and $\dot G = e^{-\gamma t/2}\big(\cos\omega_\mathrm dt - \frac{\gamma}{2\omega_\mathrm d}\sin\omega_\mathrm dt\big)$ for $t > 0$, both zero before. Had $G$ itself jumped, the jump formula applied to $\ddot G$ would have produced a $\delta'$ — an impulse of an impulse, which no kick supplies.

20 · distributions

Tempered distributions and their transforms

$\mathcal S$: smooth $\varphi$ with $t^m\varphi^{(k)}$ bounded for all $m, k$ (the Schwartz class). The dictionary maps $\mathcal S$ to itself. A tempered distribution is a continuous linear map on $\mathcal S$; since $\int\hat f\varphi = \int f\hat\varphi$ (swap the integrals), define

the transform, by duality $$\langle\hat T,\varphi\rangle := \langle T,\hat\varphi\rangle .$$

$T$	$\hat T$	why
$\delta$	$1$	$\hat\varphi(0) = \int\varphi$
$1$	$2\pi\delta$	inversion at $t = 0$
$e^{-i\nu t}$	$2\pi\delta(\omega - \nu)$	modulation
$\operatorname{sgn}t$	$2i\,\mathrm{PV}\frac1\omega$	limit of $e^{-\varepsilon\|t\|}\operatorname{sgn}t$

Sokhotski–Plemelj

As distributions, $\displaystyle\lim_{\varepsilon\to0^+}\frac{1}{x \mp i\varepsilon} = \mathrm{PV}\frac1x \pm i\pi\,\delta$.

Proof. $\dfrac{1}{x - i\varepsilon} = \dfrac{x}{x^2 + \varepsilon^2} + \dfrac{i\varepsilon}{x^2 + \varepsilon^2}$. The second is $i\pi$ times a positive kernel of total mass $1$ concentrating at $0$, so tends to $i\pi\delta$; the first tends to $\mathrm{PV}\frac1x$ (notes). $\square$

the running example, undamped

As $\gamma\to0$, $\chi$ $\to\dfrac{1}{\omega_0^2 - (\omega + i0)^2}$ — shorthand for $\omega + i\varepsilon$ with the limit $\varepsilon\to0^+$ taken last, as in Sokhotski (here $\varepsilon = \gamma/2$; section 5 names it the $i0$ prescription) — and

$$\operatorname{Im}\chi\to\frac{\pi}{2\omega_0}\big[\delta(\omega - \omega_0) - \delta(\omega + \omega_0)\big].$$

A lossless oscillator absorbs only at resonance — as a spike. Weight check: $\int_0^\infty\omega\operatorname{Im}\chi\,\mathrm d\omega = \frac\pi2$ for every $\gamma$ (section 5), and the spike gives $\omega_0\cdot\frac{\pi}{2\omega_0}$ ✓.

Why "tempered". $\mathcal D$ is not mapped to itself by the transform — the transform of a compactly supported function is never compactly supported unless it is zero (section 3's tools prove it: such a transform is entire, by Morera exactly as for $\chi$ there, and an entire function vanishing on an interval vanishes identically, by the identity theorem). So the transform is defined on the larger test space $\mathcal S$, and the distributions that act continuously on it are those of at most polynomial growth. Every function of polynomial growth, $\delta$ and all its derivatives, and $\mathrm{PV}\frac1t$ are tempered.

The swap. $\int\hat f(\omega)\varphi(\omega)\mathrm d\omega = \int\!\!\int f(t)e^{i\omega t}\varphi(\omega)\mathrm dt\,\mathrm d\omega = \int f(t)\hat\varphi(t)\,\mathrm dt$ by Fubini, which is why duality is the right definition: it agrees with the old transform on functions.

The table. $\langle\hat1,\varphi\rangle = \int\hat\varphi = 2\pi\varphi(0)$ by inversion at $t = 0$. For the sign: $\int_{-\infty}^\infty\operatorname{sgn}(t)e^{-\varepsilon|t|}e^{i\omega t}\mathrm dt = \frac{1}{\varepsilon - i\omega} - \frac{1}{\varepsilon + i\omega} = \frac{2i\omega}{\omega^2 + \varepsilon^2}$, and the first half of the Sokhotski proof shows $\omega/(\omega^2+\varepsilon^2)\to\mathrm{PV}\frac1\omega$. Similarly $H(t)e^{-\varepsilon t}\mapsto\frac1{\varepsilon - i\omega} = \frac{i}{\omega + i\varepsilon}$, so $\hat H = i\,\mathrm{PV}\frac1\omega + \pi\delta$ — the transform of the step is where causality first meets $i0$.

Full proof of Sokhotski–Plemelj

The delta part. $P_\varepsilon(x) = \frac{\varepsilon}{\pi(x^2 + \varepsilon^2)}$ is positive with $\int P_\varepsilon = \frac1\pi[\arctan(x/\varepsilon)]_{-\infty}^\infty = 1$, and $\int_{|x|>\delta}P_\varepsilon\to0$. The argument of the inversion proof, step 2, gives $\int P_\varepsilon\varphi\to\varphi(0)$.

The principal-value part. $\int\frac{x\varphi(x)}{x^2+\varepsilon^2}\mathrm dx = \int_0^\infty\frac{x(\varphi(x) - \varphi(-x))}{x^2+\varepsilon^2}\mathrm dx$. The integrand is dominated by $|\varphi(x) - \varphi(-x)|/x$, which is bounded near $0$ (mean value theorem) and integrable at infinity for $\varphi\in\mathcal S$; pointwise it tends to $(\varphi(x) - \varphi(-x))/x$. Dominated convergence gives $\langle\mathrm{PV}\frac1x,\varphi\rangle$. $\square$

The undamped limit. $-\omega^2 - i\gamma\omega = -(\omega + i\gamma/2)^2 - \gamma^2/4$, so for small $\gamma$ the denominator is $\omega_0^2 - (\omega + i0)^2$. Near $\omega_0$, $\chi\approx-\frac{1}{2\omega_0}\cdot\frac{1}{\omega - \omega_0 + i0}$, and Sokhotski with the lower sign gives $\operatorname{Im}\frac{1}{\omega - \omega_0 + i0} = -\pi\delta(\omega - \omega_0)$, hence $+\frac{\pi}{2\omega_0}\delta(\omega - \omega_0)$. Near $-\omega_0$ the prefactor changes sign. The positive sign at $+\omega_0$ matters: $\operatorname{Im}\chi > 0$ for $\omega > 0$ is the statement that a passive oscillator absorbs energy from the drive (unit 06's power formula, with this unit's convention), and the limit keeps it.

section 3

Holomorphic functions

Ask for a derivative in the complex sense and get, for free, infinitely many derivatives, a convergent power series, and Cauchy's theorem.

20 · holomorphy

Complex differentiability, and Cauchy–Riemann

holomorphic

$f$ on an open $U\subseteq\mathbb C$ is complex-differentiable at $z$ if $\displaystyle f'(z) = \lim_{h\to0,\ h\in\mathbb C}\frac{f(z+h) - f(z)}{h}$ exists; holomorphic on $U$ if at every point. Entire: holomorphic on $\mathbb C$.

Sums, products, quotients and compositions obey the real rules, with the real proofs. So polynomials and rational functions are holomorphic off the zeros of their denominators — $\chi$ on $\mathbb C\setminus\{\omega_+,\omega_-\}$.

Physics: $\chi$ at complex $\omega = \omega_r + i\omega_i$ is the response to $e^{-i\omega t} = e^{\omega_it}e^{-i\omega_rt}$ — a drive growing in time when $\omega_i > 0$, switched on from nothing at $t = -\infty$.

Partial derivatives, all this unit needs of them: for $u(x, y)$, $u_x$ is the derivative of $x\mapsto u(x,y)$ with $y$ held fixed, and $u_y$ likewise.

the Cauchy–Riemann equations

If $f = u + iv$ is complex-differentiable at $z = x + iy$, then

$$u_x = v_y,\qquad u_y = -v_x,\qquad f' = u_x + iv_x .$$

Proof. Let $h\to0$ along the reals: $f' = u_x + iv_x$. Along $h = ik$: $f' = \frac1i(u_y + iv_y) = v_y - iu_y$. Equate. $\square$

most smooth functions are not holomorphic

$\bar z = x - iy$: $u_x = 1\ne-1 = v_y$ everywhere. $|z|^2$ is differentiable at $0$ only. Holomorphy is a strong condition, and everything in this section is its price.

The converse, with its full proof

Theorem. If $u, v$ have continuous partial derivatives on $U$ satisfying Cauchy–Riemann, then $f = u + iv$ is holomorphic on $U$.

Proof. Write $h = a + ib$. Split the increment of $u$ in two one-variable steps: $u(x+a, y+b) - u(x,y) = [u(x+a,y+b) - u(x, y+b)] + [u(x,y+b) - u(x,y)]$. By the one-variable mean value theorem this is $u_x(x + \theta a, y + b)a + u_y(x, y + \theta'b)b$ with $\theta,\theta'\in(0,1)$, and by continuity of the partials it is $u_xa + u_yb + o(|h|)$. Same for $v$. Then $f(z+h) - f(z) = (u_x a + u_y b) + i(v_x a + v_y b) + o(|h|)$, and Cauchy–Riemann turns the bracket into $(u_x + iv_x)(a + ib)$: indeed $(u_x + iv_x)(a + ib) = u_xa - v_xb + i(v_xa + u_xb) = u_xa + u_yb + i(v_xa + v_yb)$. So $(f(z+h) - f(z))/h\to u_x + iv_x$. $\square$

The exponential. $e^z = e^x(\cos y + i\sin y)$ (unit 06's definition): $u_x = e^x\cos y = v_y$, $u_y = -e^x\sin y = -v_x$, all continuous, so $e^z$ is entire with $(e^z)' = u_x + iv_x = e^z$. Then $\cos z = (e^{iz} + e^{-iz})/2$ and $\sin z$ are entire too, and $\sin(\pi/2 + ix) = \cosh x$, which P10 needs.

What CR says geometrically. The real derivative of $(x,y)\mapsto(u,v)$ is the matrix $\begin{pmatrix}u_x&u_y\\v_x&v_y\end{pmatrix} = \begin{pmatrix}a&-b\\b&a\end{pmatrix}$, which is multiplication by $a + ib$: a rotation and a dilation. Holomorphic maps are the ones that are infinitesimally rotations-and-scalings — they preserve angles wherever $f'\ne0$.

The physics of complex frequency is worth holding on to, because section 5 turns on it. A drive $e^{\omega_it}e^{-i\omega_rt}$ with $\omega_i > 0$ was vanishingly small in the distant past; a causal system's response to it is well defined, $\int_0^\infty G(\tau)e^{i\omega\tau}\mathrm d\tau$, because the weight $e^{-\omega_i\tau}$ kills the old history. With $\omega_i < 0$ the drive was enormous in the past and the response integral may diverge. The half-plane in which a response function is automatically defined is fixed by the direction of time.

20 · holomorphy

Integrals along curves

the contour integral

For a piecewise $C^1$ curve $\gamma:[a,b]\to U$ and continuous $f$,

$$\int_\gamma f(z)\,\mathrm dz := \int_a^b f(\gamma(s))\,\gamma'(s)\,\mathrm ds,$$

an ordinary integral of a complex-valued function of one real variable. Reparametrising does not change it (substitution); reversing the direction flips its sign.

the ML inequality $$\Big|\int_\gamma f\,\mathrm dz\Big|\le\max_\gamma|f|\cdot L(\gamma),$$

with $L(\gamma) = \int_a^b|\gamma'(s)|\,\mathrm ds$ the length.

Every arc estimate in this unit is this inequality, used carefully.

primitives kill closed integrals

If $F' = f$ on $U$ then $\int_\gamma f\,\mathrm dz = F(\gamma(b)) - F(\gamma(a))$, and $\oint_\gamma f\,\mathrm dz = 0$ round every closed curve.

Proof: $\frac{\mathrm d}{\mathrm ds}F(\gamma(s)) = F'(\gamma(s))\gamma'(s)$, from the definition of $F'$; then the fundamental theorem of calculus. $\square$

the one integral that is not zero

Round $\gamma(s) = re^{is}$, $0\le s\le2\pi$, the integrand is $ir^{n+1}e^{i(n+1)s}$, so

$$\oint z^n\,\mathrm dz = \begin{cases}2\pi i, & n = -1,\\ 0, & n\ne-1.\end{cases}$$

So $1/z$ has no primitive on $\mathbb C\setminus\{0\}$. Residue theory is this computation, organised.

Why no line-integral machinery is needed. Unit 04 (not a prerequisite here) builds line integrals of vector fields in the plane. A contour integral is the same object in disguise — $f\,\mathrm dz = (u + iv)(\mathrm dx + i\,\mathrm dy)$ is two real line integrals — but nothing in this unit uses that: the definition above is a one-variable integral, and every property needed follows from one-variable calculus.

Proof of the ML inequality. For a complex-valued $g$ on $[a,b]$, $|\int g|\le\int|g|$: write $\int g = |\int g|e^{i\theta}$; then $|\int g| = \operatorname{Re}\int e^{-i\theta}g = \int\operatorname{Re}(e^{-i\theta}g)\le\int|g|$. With $g = f(\gamma)\gamma'$ that gives $\le\int|f(\gamma)||\gamma'|\le\max|f|\,L$. $\square$

The chain rule step. $\frac{F(\gamma(s+\eta)) - F(\gamma(s))}{\eta} = \frac{F(\gamma(s+\eta)) - F(\gamma(s))}{\gamma(s+\eta) - \gamma(s)}\cdot\frac{\gamma(s+\eta) - \gamma(s)}{\eta}$ when $\gamma(s+\eta)\ne\gamma(s)$; the first factor tends to $F'(\gamma(s))$ and the second to $\gamma'(s)$. (Where $\gamma' (s) = 0$ write $F(w) - F(\gamma(s)) = (F'(\gamma(s)) + \epsilon(w))(w - \gamma(s))$ with $\epsilon\to0$ instead.)

The circle. $\gamma'(s) = ire^{is}$, so $\gamma^n\gamma' = ir^{n+1}e^{i(n+1)s}$, whose integral over a period is $0$ unless $n + 1 = 0$. For $n\ne-1$ the primitive $z^{n+1}/(n+1)$ gives the same answer. The failure at $n = -1$ is the failure of $\log z$ to be single-valued: going once round adds $2\pi i$ to its imaginary part, $\arg z$. Section 4 turns that into branch cuts.

20 · holomorphy

Goursat's theorem

If $f$ is holomorphic on $U$ and $T\subset U$ is a closed solid triangle, then $\displaystyle\oint_{\partial T}f\,\mathrm dz = 0$.

proof, by subdivision

Join the midpoints of the sides: four triangles, whose boundary integrals sum to $\oint_{\partial T}$ (inner edges cancel). One, $T_1$, has $|\oint_{\partial T_1}|\ge\frac14|\oint_{\partial T}|$. Repeat: nested $T_n$ of diameter $2^{-n}d$, perimeter $2^{-n}p$, and $|\oint_{\partial T_n}f|\ge4^{-n}|\oint_{\partial T}f|$.

Compactness: $\bigcap T_n = \{z_0\}$, and $f(z) = f(z_0) + f'(z_0)(z - z_0) + \epsilon(z)(z - z_0)$ with $\epsilon\to0$. The linear part has a primitive; ML on the rest:

$4^{-n}|\oint_{\partial T}f|\le\sup_{T_n}|\epsilon|\cdot2^{-n}d\cdot2^{-n}p$. Multiply by $4^n$; let $n\to\infty$. $\square$

One subdivision. The four boundaries, each run anticlockwise, add up to $\partial T$: every inner edge is run once each way and cancels. The shaded $T_1$ keeps at least a quarter of the integral; its own subdivisions (darker) close down on $z_0$. Schematic.

Full proof, the three details

The inner edges cancel because each is traversed once in each direction by the two triangles sharing it, and reversal flips the sign. Choosing $T_1$: if all four had $|\oint| < \frac14|\oint_{\partial T}|$ the triangle inequality would give $|\oint_{\partial T}| < |\oint_{\partial T}|$. Nested compact sets with diameters tending to $0$ meet in exactly one point (the learner's metric-space topology: pick $z_n\in T_n$, a Cauchy sequence, whose limit lies in every $T_n$), and $z_0\in T\subset U$ so $f$ is differentiable there. The remainder estimate: on $\partial T_n$, $|z - z_0|\le\operatorname{diam}T_n$, so $|\epsilon(z)(z - z_0)|\le\sup_{T_n}|\epsilon|\,2^{-n}d$, and the ML inequality multiplies by $L(\partial T_n) = 2^{-n}p$.

History. Cauchy's integral theorem appears in his 1825 memoir on integrals between imaginary limits, proved with what amounts to Green's theorem and so with $f'$ assumed continuous. Goursat's subdivision argument is from 1884, published in its final form in 1900; Pringsheim noticed triangles are simpler than Goursat's rectangles. The argument is the model for many later proofs: assume the quantity is not small, localise by bisection, and contradict a local property.

20 · holomorphy

Goursat with one exceptional point

what was used, and what was not

Only the existence of $f'$ at each point: not its continuity, not Green's theorem, not a double integral. Cauchy (1825) assumed $f'$ continuous; Goursat showed it superfluous — as it must be, for $f'$ to come out smooth on the next slides.

one exceptional point is allowed

The conclusion still holds if $f$ is continuous on $U$ and holomorphic on $U\setminus\{p\}$. This is what the integral formula will need.

Proof. If $p$ is a vertex, cut off a tiny triangle $T_\eta$ there, of perimeter $\le\eta$; the rest splits into triangles without $p$, whose integrals vanish, so $|\oint_{\partial T}f|\le\eta\max_T|f|\to0$. Otherwise split $T$ into triangles having $p$ as a vertex. $\square$

The two cases. Only the tiny shaded triangle can carry an integral, and its perimeter goes to zero while $f$ stays bounded.

For $\chi$: any triangle avoiding $\omega_\pm$ has $\oint\chi = 0$. A triangle enclosing $\omega_+$ need not — the next section computes what it gives.

Full proof of the exceptional-point version

If $p\notin T$ this is Goursat. If $p$ is a vertex of $T$, cut off a tiny triangle $T_\eta$ at that vertex, of perimeter $\le\eta$; the rest of $T$ is a union of triangles not containing $p$, whose integrals vanish, so $|\oint_{\partial T}f| = |\oint_{\partial T_\eta}f|\le\eta\max_T|f|\to0$ (continuity makes $f$ bounded on the compact $T$). If $p$ is on an edge or inside, split $T$ into two or three triangles having $p$ as a vertex. $\square$

Why the integral formula needs this. Its proof applies Goursat to $g(\zeta) = (f(\zeta) - f(z))/(\zeta - z)$, which is continuous at $\zeta = z$ (its value there is $f'(z)$) but is not yet known to be differentiable there. One exceptional point is exactly the room that argument needs.

20 · holomorphy

Cauchy's theorem and the integral formula

Cauchy's theorem (convex version)

If $f$ is holomorphic on a convex open set $U$, then $F(z) = \int_{[c,z]}f$ (straight segment from a fixed $c$) is a primitive of $f$, and so $\oint_\gamma f\,\mathrm dz = 0$ for every closed curve $\gamma$ in $U$.

Proof. Goursat on the triangle $c, z, z+h$: $F(z+h) - F(z) = \int_{[z,z+h]}f = hf(z) + \int_{[z,z+h]}(f - f(z))$, and the last term is at most $|h|\max_{[z,z+h]}|f - f(z)| = o(|h|)$. So $F' = f$; then the primitive rule. $\square$

contours that are not in a convex set — cited

Semicircles, rectangles and sectors are convex. For keyholes and indented lines the theorem holds too: cut the region into convex pieces (notes). The general form — $\oint_\gamma f = 0$ when $\gamma$ winds zero times round every point outside $U$ — is cited: Ahlfors, Complex Analysis, ch. 4 §4.

Cauchy's integral formula

If $f$ is holomorphic on an open set containing the closed disc $|\zeta - z_0|\le r$, then for $|z - z_0| < r$

$$f(z) = \frac{1}{2\pi i}\oint_{|\zeta - z_0| = r}\frac{f(\zeta)}{\zeta - z}\,\mathrm d\zeta .$$

Proof. $g(\zeta) = \frac{f(\zeta) - f(z)}{\zeta - z}$, with $g(z) = f'(z)$, is continuous, and holomorphic except at $z$; so $\oint g = 0$ (exceptional-point Goursat, convex version, on a slightly larger disc). Hence $\oint\frac{f(\zeta)}{\zeta - z} = f(z)\oint\frac{\mathrm d\zeta}{\zeta - z} = 2\pi i\,f(z)$ (notes). $\square$

The values of a holomorphic function inside a circle are determined by its values on the circle. Nothing like this is true of real $C^\infty$ functions — the bump function of section 2 is zero on a whole interval and not zero.

Full proof that $\oint\frac{\mathrm d\zeta}{\zeta - z} = 2\pi i$ for every $z$ inside the circle

Write $\frac{1}{\zeta - z} = \frac{1}{(\zeta - z_0) - (z - z_0)} = \sum_{n\ge0}\frac{(z - z_0)^n}{(\zeta - z_0)^{n+1}}$, a geometric series with ratio $|z - z_0|/r < 1$ uniformly on the circle, so it may be integrated term by term. By the circle computation of the previous slide (centred at $z_0$), only $n = 0$ survives, giving $2\pi i$. $\square$

Why the convex version reaches every contour this unit uses

A keyhole (P8) — a large circle and a small one joined along both edges of a cut — bounds a region that the two perpendicular lines through the centre cut into four pieces, each lying in a convex set (a slit sector of opening less than $\pi$) on which the integrand is holomorphic; the boundaries of the four pieces add up to the keyhole, the internal cuts cancelling. An indented line closed by a semicircle is the difference of a half-disc and a small half-disc, cut in the same way along the imaginary axis. The annulus needed for Laurent series is handled the same way. So in this unit the cited general theorem is a convenience: each application could be checked by cutting. Stein & Shakarchi, Complex Analysis, ch. 2, formalise this as "toy contours".

The general formula. For any closed curve $\gamma$ in $U$ and $z\notin\gamma$, the winding number $n(\gamma, z) = \frac{1}{2\pi i}\oint_\gamma\frac{\mathrm d\zeta}{\zeta - z}$ is an integer, and $n(\gamma, z)f(z) = \frac{1}{2\pi i}\oint_\gamma\frac{f(\zeta)}{\zeta - z}\mathrm d\zeta$ whenever $\gamma$ is null-homologous in $U$ (Ahlfors, ch. 4 §5.1, cited). Every contour here winds once, anticlockwise, round the points inside it.

History. Cauchy, 1831, in Turin, in the exile he chose rather than swear allegiance to the new king. The formula is the single most productive identity in analysis: the next slide takes six theorems from it.

20 · holomorphy

What the integral formula buys

regularity, and power series

A holomorphic $f$ is infinitely differentiable, with

$$f^{(n)}(z) = \frac{n!}{2\pi i}\oint\frac{f(\zeta)}{(\zeta - z)^{n+1}}\,\mathrm d\zeta,$$

and it is analytic: its Taylor series at $z_0$ converges to $f$ on the largest open disc about $z_0$ inside $U$.

Cauchy estimates: if $|f|\le M$ on the circle of radius $r$, then $|f^{(n)}(z_0)|\le n!M/r^n$ — ML on the formula.

Liouville's theorem

A bounded entire function is constant. Proof: $|f'(z_0)|\le M/r$ for every $r$. $\square$

the fundamental theorem of algebra

A nonconstant polynomial $p$ has a complex root. Proof: otherwise $1/p$ is entire, and $\to0$ as $|z|\to\infty$, so bounded, so constant. $\square$ Unit 06's debt, paid.

two more

Maximum modulus: if $|f|$ has a local maximum in a connected $U$, $f$ is constant. Morera: a continuous $f$ with $\oint_{\partial T}f = 0$ for every triangle is holomorphic.

Full proofs

Derivatives. Differentiate the integral formula under the integral sign in $z$: the integrand $f(\zeta)(\zeta - z)^{-1}$ has $z$-derivatives $n!f(\zeta)(\zeta - z)^{-n-1}$, and for $z$ in a smaller disc these difference quotients converge uniformly in $\zeta$ on the circle, so the limit passes under the (one-variable) integral. Induction gives every order.

Power series. In the integral formula expand $\frac1{\zeta - z} = \sum\frac{(z-z_0)^n}{(\zeta - z_0)^{n+1}}$ as before, uniformly convergent on the circle; integrate term by term: $f(z) = \sum a_n(z - z_0)^n$ with $a_n = \frac{1}{2\pi i}\oint\frac{f(\zeta)}{(\zeta - z_0)^{n+1}}\mathrm d\zeta = f^{(n)}(z_0)/n!$. This works for every $r$ less than the distance from $z_0$ to the complement of $U$, which gives the radius claim. Consequently the radius of convergence of the Taylor series is at least the distance to the nearest singularity — and exactly that distance when the function cannot be continued to any larger disc (a power series is holomorphic inside its disc of convergence). P10 uses precisely this.

Maximum modulus. The integral formula at the centre, $\zeta = z_0 + re^{is}$, is the mean value property $f(z_0) = \frac1{2\pi}\int_0^{2\pi}f(z_0 + re^{is})\,\mathrm ds$. If $|f(z_0)|$ is a local maximum, then $|f(z_0)|\le\frac1{2\pi}\int|f(z_0 + re^{is})|\mathrm ds\le|f(z_0)|$, forcing $|f| = |f(z_0)|$ on small circles; a holomorphic function of constant modulus on a disc is constant (Cauchy–Riemann applied to $u^2 + v^2 = c$), and the identity theorem (next slide) spreads this over connected $U$.

Morera. In any disc in $U$, $F(z) = \int_{[c,z]}f$ satisfies $F' = f$ by the argument of the convex Cauchy theorem, which used only the vanishing of triangle integrals. So $F$ is holomorphic; by the first result, so is $F' = f$. $\square$ Morera is how one proves that a function defined by an integral, such as $\int_0^\infty G(t)e^{i\omega t}\mathrm dt$, is holomorphic: swap the triangle integral with the $t$-integral (Fubini) and use Goursat on $e^{i\omega t}$.

History. Liouville stated his theorem in lectures in 1847, for doubly periodic functions; Cauchy had published it in 1844. The proof of the fundamental theorem of algebra above is the shortest there is, and it is honest: all the analysis is in Liouville, and all of Liouville is in the integral formula.

20 · holomorphy

The Taylor series dies at resonance

the running example

$\chi$ is holomorphic on $|\omega| < |\omega_\pm| = \sqrt{\omega_\mathrm d^2 + \gamma^2/4} = \omega_0 = 8$, so its low-frequency series $\chi = \omega_0^{-2}\big(1 + i\gamma\omega/\omega_0^2 + \dots\big)$ converges exactly for $|\omega| < \omega_0$: the stiffness-dominated description fails at resonance, and the poles say so.

Read the plot: below $\omega_0$ the polynomials hug $\chi$, more closely the higher the degree; past $\omega_0$ they run away, and raising the degree only makes them run faster. On the real axis $\chi$ is smooth at $8$, merely large; the series fails there because of two poles $0.2$ below the axis.

$|\chi(\omega)|$ on the real axis (log scale) against its Taylor polynomials about $0$ of degree $10$ and $40$. Both follow $\chi$ inside the disc $|\omega| < 8$ and diverge outside it: the radius is the distance to the poles.

20 · holomorphy

Analytic continuation

the identity theorem

If $f, g$ are holomorphic on a connected open $U$ and agree on a set with a limit point in $U$ — a segment of the real axis, say — then $f = g$ on all of $U$.

Proof, in outline. Let $h = f - g$. The set of points where every derivative of $h$ vanishes is closed, and open (the Taylor series there is $0$ on a disc), and nonempty (at the limit point, the first nonzero Taylor coefficient would make the zeros isolated). Connectedness. $\square$

analytic continuation

A holomorphic $F$ on $U\supset V$ extending $f$ on $V$. By the identity theorem it is unique — so a formula valid on part of a domain is valid wherever both sides make sense.

a Gaussian with a complex width

For $a > 0$, $\int e^{-at^2/2}\mathrm dt = \sqrt{2\pi/a}$ (scaling). Both sides are holomorphic in $a$ on $\operatorname{Re}a > 0$ (left: Morera; right: the principal square root). They agree on the positive reals, so

$$\int_{-\infty}^\infty e^{-at^2/2}\,\mathrm dt = \sqrt{\frac{2\pi}{a}},\qquad\operatorname{Re}a > 0 .$$

Section 6 uses this with $a = \sigma_0^2 + i\beta t$: a spreading wave packet.

The running example. The integral $\int_0^\infty G(t)e^{i\omega t}\mathrm dt$ converges only for $\operatorname{Im}\omega > -\gamma/2$. The rational formula for $\chi$ is its continuation to $\mathbb C\setminus\{\omega_\pm\}$: the integral fails below the line through the poles; the function does not.

Full proof of the identity theorem

Let $h = f - g$ and $A = \{z\in U : h^{(n)}(z) = 0\ \forall n\}$. Closed in $U$: an intersection of zero sets of continuous functions. Open: if $z\in A$, the Taylor series of $h$ at $z$ is identically $0$ and converges to $h$ on a disc, so $h = 0$ there and the disc is in $A$. Nonempty: let $w$ be the limit point of the set $\{h = 0\}$; $h(w) = 0$ by continuity. If some derivative at $w$ were nonzero, let $m$ be the first: $h(z) = (z - w)^mk(z)$ with $k$ holomorphic and $k(w)\ne0$, so $k\ne0$ near $w$ and $h$ has no zeros in a punctured disc — contradicting that $w$ is a limit of zeros. So $w\in A$. A connected space has no nonempty proper clopen subset, so $A = U$. $\square$

The Gaussian, carefully. $I(a) = \int e^{-at^2/2}\mathrm dt$ converges absolutely for $\operatorname{Re}a > 0$, locally uniformly, and each truncated integral $\int_{-T}^Te^{-at^2/2}\mathrm dt$ is entire in $a$ (differentiate under a finite integral). A locally uniform limit of holomorphic functions is holomorphic — Morera: triangle integrals pass to the limit. And $\sqrt{2\pi/a}$ with the principal branch ($\arg a\in(-\pi/2,\pi/2)$, so $\arg\sqrt{\cdot}\in(-\pi/4,\pi/4)$) is holomorphic on the right half-plane. The half-plane is connected, the positive reals have limit points in it. $\square$ Setting $a = \varepsilon - i\lambda$ and letting $\varepsilon\to0$ gives Fresnel's integral — but the limit needs an argument, and that argument is P6.

Where continuation meets physics. Past the line $\operatorname{Im}\omega = -\gamma/2$ the susceptibility is no longer a response to any drive; it is a function that knows where the system's free modes are. Unit 06 called those the poles; in the physics of resonances they are called quasinormal modes, and the ringdown frequencies of a perturbed black hole are found exactly this way: as the poles of a continued response function.

Uniqueness can fail round a hole. Continue $\log z$ from $z = 1$ along the upper half of the unit circle and along the lower half: at $z = -1$ you arrive with $i\pi$ and $-i\pi$. The domain $\mathbb C\setminus\{0\}$ is connected, but no single-valued continuation exists on it. The identity theorem is not contradicted — there is no single holomorphic function on the whole domain to compare. Section 4 handles this with cuts.

section 4

Singularities and residues

Round a closed contour, only the singularities inside speak — and each says one number.

20 · residues

Laurent series and isolated singularities

Laurent series

If $f$ is holomorphic on the annulus $r < |z - z_0| < R$, then on it

$$f(z) = \sum_{n\in\mathbb Z}a_n(z - z_0)^n,$$

$a_n = \frac{1}{2\pi i}\oint f(\zeta)(\zeta - z_0)^{-n-1}\mathrm d\zeta$ round any circle in the annulus. The negative powers are the principal part.

An isolated singularity: $f$ holomorphic on $0 < |z - z_0| < R$. Three kinds:

removable — no principal part; happens iff $f$ is bounded near $z_0$ (Riemann);
pole of order $m$ — principal part ends at $a_{-m}\ne0$; iff $|f|\to\infty$;
essential — infinitely many, as $e^{1/z}$; $f$ then comes close to every value near $z_0$ (Casorati–Weierstrass).

the residue $$\operatorname*{Res}_{z_0}f = a_{-1} = \frac1{2\pi i}\oint_{|z - z_0| = \varepsilon}f(z)\,\mathrm dz .$$

It is the one Laurent coefficient that survives integration: every other power has a primitive on the punctured disc.

the running example

$\chi$ $= \dfrac{-1}{(\omega - \omega_+)(\omega - \omega_-)}$ has simple poles at $\omega_\pm = \pm\omega_\mathrm d - i\gamma/2$ and is holomorphic everywhere else, including at $\infty$ in the sense that $\chi(1/w)$ has a removable singularity at $w = 0$.

Physics: a pole is a free mode — its real part the ringing frequency $7.9975\ \mathrm{s^{-1}}$, its imaginary part the decay rate $0.2\ \mathrm{s^{-1}}$. Its residue will be that mode's amplitude in $G$.

Full proof of the Laurent expansion

Fix $z$ in the annulus and circles $C_1$ (radius $r_1$) and $C_2$ (radius $r_2$) with $r < r_1 < |z - z_0| < r_2 < R$. Cauchy's formula for the region between them (a toy contour, cut into convex pieces as described on the Cauchy slide) gives $f(z) = \frac1{2\pi i}\oint_{C_2}\frac{f(\zeta)}{\zeta - z}\mathrm d\zeta - \frac1{2\pi i}\oint_{C_1}\frac{f(\zeta)}{\zeta - z}\mathrm d\zeta$. On $C_2$ expand $\frac{1}{\zeta - z} = \sum_{n\ge0}\frac{(z - z_0)^n}{(\zeta - z_0)^{n+1}}$ (ratio $|z - z_0|/r_2 < 1$); on $C_1$ expand $-\frac{1}{\zeta - z} = \frac{1}{(z - z_0) - (\zeta - z_0)} = \sum_{m\ge1}\frac{(\zeta - z_0)^{m-1}}{(z - z_0)^m}$ (ratio $r_1/|z - z_0| < 1$). Both converge uniformly on their circles; integrate term by term. The coefficient formulas are independent of the circle because each integrand is holomorphic on the annulus and circles in it can be deformed into one another by the same cutting argument. $\square$

Riemann's removable singularity theorem. If $|f|\le M$ near $z_0$, then $|a_{-n}| = |\frac1{2\pi i}\oint f(\zeta)(\zeta - z_0)^{n-1}\mathrm d\zeta|\le M\varepsilon^n\to0$ for $n\ge1$, by ML on a circle of radius $\varepsilon\to0$. Poles. If the principal part stops at $a_{-m}\ne0$, then $f = (z - z_0)^{-m}h$ with $h(z_0) = a_{-m}\ne0$, so $|f|\to\infty$. Conversely if $|f|\to\infty$, $1/f$ has a removable singularity with a zero of some finite order $m$ (finite by the identity theorem), and $f$ has a pole of order $m$.

Casorati–Weierstrass. Suppose some disc $|w - c| < \delta$ is missed by $f$ near $z_0$. Then $g = 1/(f - c)$ is bounded by $1/\delta$, so it has a removable singularity; hence $f = c + 1/g$ has either a removable singularity or a pole — not essential. $\square$

The poles as modes. A pole at $\omega_+ = \omega_\mathrm d - i\gamma/2$ contributes to the inverse transform a term proportional to $e^{-i\omega_+t} = e^{-\gamma t/2}e^{-i\omega_\mathrm dt}$: a damped oscillation at the ringing frequency. That is why a physicist reads a spectrum by its poles, and why the ringdown of anything — a bell, a galvanometer, a black hole — is quoted as a complex frequency.

20 · residues

The residue theorem

the residue theorem

Let $\gamma$ be a circle, or a contour of the kinds on the Cauchy slide, traversed anticlockwise, and $f$ holomorphic on an open set containing $\gamma$ and its inside, except at finitely many points $z_1,\dots,z_k$ inside. Then

$$\oint_\gamma f\,\mathrm dz = 2\pi i\sum_{j=1}^k\operatorname*{Res}_{z_j}f .$$

Proof. Let $P_j$ be the principal part at $z_j$; it converges on $\mathbb C\setminus\{z_j\}$. Then $f - \sum P_j$ has only removable singularities, so its integral is $0$ by Cauchy. And $\oint P_j = 2\pi i\,a^{(j)}_{-1}$, term by term, by the circle computation. $\square$

computing residues

Simple pole, $f = g/h$ with $h(z_0) = 0\ne h'(z_0)$, $g$ holomorphic: $\operatorname{Res}f = g(z_0)/h'(z_0)$.

Pole of order $m$: $\operatorname{Res}f = \dfrac{1}{(m-1)!}\lim\limits_{z\to z_0}\dfrac{\mathrm d^{m-1}}{\mathrm dz^{m-1}}\big[(z - z_0)^mf(z)\big]$.

the running example's residues

$h(\omega) = \omega_0^2 - \omega^2 - i\gamma\omega$ has $h'(\omega) = -2\omega - i\gamma$, so $h'(\omega_\pm) = \mp2\omega_\mathrm d$ and

$$\operatorname*{Res}_{\omega_\pm}\chi(\omega)e^{-i\omega t} = \mp\frac{e^{-i\omega_\pm t}}{2\omega_\mathrm d} .$$

Dimensions: $\mathrm T^{-1}\cdot\mathrm T^{2}$ per pole = $\mathrm T$ — a residue of $\chi$ carries $[\chi][\omega]$ ✓.

Full proof of the two formulas

Simple pole. $h(z) = h'(z_0)(z - z_0) + O((z - z_0)^2)$, so $(z - z_0)f(z) = g(z)/\big(h'(z_0) + O(z - z_0)\big)\to g(z_0)/h'(z_0)$, and for a simple pole $(z - z_0)f\to a_{-1}$. Order $m$. $(z - z_0)^mf = a_{-m} + \dots + a_{-1}(z - z_0)^{m-1} + a_0(z - z_0)^m + \dots$, a power series; its $(m-1)$-th derivative at $z_0$ is $(m-1)!\,a_{-1}$.

The running example's arithmetic. $h'(\omega_+) = -2(\omega_\mathrm d - i\gamma/2) - i\gamma = -2\omega_\mathrm d$ and $h'(\omega_-) = -2(-\omega_\mathrm d - i\gamma/2) - i\gamma = 2\omega_\mathrm d$. With $g = e^{-i\omega t}$, entire in $\omega$ for each fixed $t$, the residues follow. With the numbers, $2\omega_\mathrm d = 15.9950\ \mathrm{s^{-1}}$.

Where the proof is thin, said plainly. "Cauchy" for the function $f - \sum P_j$ on the inside of $\gamma$ requires that inside to be convex or cut into convex pieces. For a circle, a semicircle, a rectangle and a sector it is convex; for the keyhole and the indented contours it is the cutting argument. The theorem for an arbitrary closed curve, with winding numbers in place of "inside", is cited: Ahlfors, ch. 4 §5.1.

History. Cauchy introduced the residue — résidu, what is left over — in 1826, as the tool that unified dozens of definite integrals Euler, Laplace and Legendre had evaluated one trick at a time. The theorem in this form is in his 1831 Turin memoir.

20 · residues

Closing the contour: Jordan's lemma

Jordan's lemma

Let $C_R$ be the upper semicircle $z = Re^{i\theta}$, $0\le\theta\le\pi$, and $M_R = \max_{C_R}|g|\to0$ as $R\to\infty$. Then for every $a > 0$

$$\Big|\int_{C_R}g(z)e^{iaz}\,\mathrm dz\Big|\le\frac{\pi M_R}{a}\to0 .$$

proof

$|e^{iaz}| = e^{-aR\sin\theta}$ on $C_R$, so the integral is at most $M_RR\int_0^\pi e^{-aR\sin\theta}\mathrm d\theta = 2M_RR\int_0^{\pi/2}e^{-aR\sin\theta}\mathrm d\theta$. Concavity gives Jordan's inequality $\sin\theta\ge2\theta/\pi$ on $[0,\pi/2]$, so this is at most $2M_RR\int_0^\infty e^{-2aR\theta/\pi}\mathrm d\theta = \pi M_R/a$. $\square$

why Jordan and not just ML

ML gives $\pi RM_R$: useless when $g\sim1/R$ — $\sin x/x$, Kramers–Kronig, any transform of a function with a jump. The exponential's decay over most of the arc is what saves you.

$|e^{-i\omega t}| = e^{t\operatorname{Im}\omega}$: it decays in the upper half-plane when $t < 0$, in the lower when $t > 0$.

For $e^{-i\omega t}$ the lemma applies with $a = -t$ above (if $t < 0$) or, reflected, with $a = t$ below (if $t > 0$). The sign of $t$ chooses the half-plane. For $\chi$, $M_R\sim1/R^2$.

Jordan's inequality. $\sin$ is concave on $[0,\pi/2]$, so its graph lies above the chord from $(0,0)$ to $(\pi/2, 1)$, which is $2\theta/\pi$. The symmetry step uses $\sin(\pi - \theta) = \sin\theta$.

The lower half-plane version. For the lower semicircle $z = Re^{-i\theta}$ and $a > 0$, $|e^{-iaz}| = e^{-aR\sin\theta}$, and the same bound holds for $\int g(z)e^{-iaz}\mathrm dz$. The lower semicircle traversed from $+R$ to $-R$ through $-iR$ closes the real axis clockwise, which costs a minus sign in the residue theorem: that minus sign is where most errors in this subject are made.

For the running example the lemma is more than needed: $|\chi(\omega)|\le1/(R^2 - \omega_0^2 - \gamma R)$ on $|\omega| = R$ once $R$ is large, so even ML gives $\pi R\cdot O(R^{-2})\to0$. Jordan's lemma is needed in section 5's theorem, where $\chi$ is only assumed to tend to zero, and for the Dirichlet integral on the next slide.

History. Camille Jordan's Cours d'analyse (second edition, 1894). The same Jordan as the Jordan normal form of unit 07 and the Jordan curve theorem.

20 · residues · example

Worked: the Dirichlet integral, with an indentation

worked example 2 — $\int_0^\infty\frac{\sin x}{x}\,\mathrm dx$

Step 1 — the function. Not $\sin z/z$: $\sin z$ grows like $e^{|\operatorname{Im}z|}$ in both half-planes. Take $f(z) = e^{iz}/z$: imaginary part $\sin x/x$ on the axis, decaying upwards.

Step 2 — the contour. $[-R,-\varepsilon]$, the small upper semicircle $c_\varepsilon$ clockwise round the pole at $0$, $[\varepsilon,R]$, the big semicircle $C_R$. Nothing inside, so the total is $0$.

Step 2's contour. The pole at $0$ is stepped round from above, so nothing is inside; schematic.

Step 3 — the straight parts. Pair $x$ with $-x$: $\int_{\varepsilon<|x|<R}\frac{e^{ix}}{x}\mathrm dx = \int_\varepsilon^R\frac{e^{ix} - e^{-ix}}{x}\mathrm dx = 2i\int_\varepsilon^R\frac{\sin x}{x}\mathrm dx$.

Step 4 — the big arc. Jordan with $g = 1/z$, $M_R = 1/R$, $a = 1$: at most $\pi/R\to0$.

Step 5 — the small arc. $e^{iz}/z = 1/z + h(z)$, $h$ bounded near $0$. $\int_{c_\varepsilon}\mathrm dz/z = \int_\pi^0i\,\mathrm d\theta = -\pi i$, and $|\int_{c_\varepsilon}h|\le\pi\varepsilon\max|h|\to0$.

Step 6 — assemble. $2i\int_0^\infty\frac{\sin x}{x}\mathrm dx - \pi i = 0$, so

$$\int_0^\infty\frac{\sin x}{x}\,\mathrm dx = \frac\pi2 = 1.5708 .$$

Checks. $\int_0^\infty\frac{\sin ax}{x}\mathrm dx = \frac\pi2\operatorname{sgn}a$, independent of $|a|$, as $u = ax$ demands; with $x$ a time, a pure number ✓.

20 · residues

Half a residue: the indentation lemma

the indentation lemma

If $f$ has a simple pole at $x_0$ and $c_\varepsilon$ is an anticlockwise arc of angle $\alpha$ of $|z - x_0| = \varepsilon$, then $\int_{c_\varepsilon}f\to i\alpha\operatorname*{Res}_{x_0}f$. A half-circle gives half a residue.

Proof. $f = \frac{c}{z - x_0} + h$ with $h$ bounded near $x_0$; on the arc $\int\frac{c\,\mathrm dz}{z - x_0} = ic\alpha$ exactly, and $|\int h|\le\alpha\varepsilon\max|h|\to0$. $\square$

Step 3 read differently: $\mathrm{PV}\!\int_{-\infty}^{\infty}\frac{e^{ix}}{x}\mathrm dx = i\pi$. A pole on the contour contributes $\pm i\pi\operatorname{Res}$, the sign set by which way you step round it — Sokhotski–Plemelj, seen from the complex plane.

Two ways round the pole of $e^{iz}/z$. Above, the arc is clockwise ($-\pi i$) and nothing is enclosed; below, it is anticlockwise ($+\pi i$) and the residue $1$ is enclosed. Both give $2i\int_0^\infty\frac{\sin x}{x}\mathrm dx = \pi i$.

Proof of the indentation lemma. $f(z) = \frac{c}{z - x_0} + h(z)$ with $c$ the residue and $h$ holomorphic near $x_0$, hence bounded. On $z = x_0 + \varepsilon e^{i\theta}$, $\int\frac{c\,\mathrm dz}{z - x_0} = c\int i\,\mathrm d\theta = i\alpha c$ exactly, and $|\int h|\le\alpha\varepsilon\max|h|\to0$. $\square$ For a pole of order two or more the arc integral generally diverges as $\varepsilon\to0$, and there is no principal value: the lemma is really about simple poles.

What changes if the indentation goes below the pole. Then the pole is inside, the residue theorem gives $2\pi i\cdot1$ on the right, and the small arc (now anticlockwise under the pole) gives $+\pi i$: $2i\int_0^\infty\frac{\sin x}{x} + \pi i = 2\pi i$ — the same answer. Where you step round a pole on the contour is a free choice; what is not free is being consistent about it. In section 5 physics makes that choice, and calls it causality.

20 · residues

Branch points and cuts

$\oint\mathrm dz/z = 2\pi i$ means no continuous logarithm exists on $\mathbb C\setminus\{0\}$: each turn round $0$ adds $2\pi i$. Remove a ray and one does:

branch cut

On $\mathbb C\setminus(-\infty,0]$, the principal logarithm $\operatorname{Log}z = \ln|z| + i\operatorname{Arg}z$, $\operatorname{Arg}\in(-\pi,\pi)$, is holomorphic with derivative $1/z$. Then $z^a := e^{a\log z}$ for any $a\in\mathbb C$.

$0$ is a branch point: continuing $z^a$ once round it multiplies it by $e^{2\pi ia}$. The cut is a choice; the branch point is not.

the rule that prevents every error here

Choose the cut and the branch before writing an integral, and state the value of $\arg z$ on each edge of the cut.

The keyhole: cut along $[0,\infty)$, $\arg\in(0,2\pi)$. The two edges of the cut are the same points with different values of $z^a$.

What it buys. For $\int_0^\infty x^{a-1}R(x)\,\mathrm dx$ the two edges give $(1 - e^{2\pi ia})\int_0^\infty$; the circles vanish; the residue theorem finishes. P8: $\displaystyle\int_0^\infty\frac{x^{a-1}}{1 + x}\mathrm dx = \frac{\pi}{\sin\pi a}$ for $0 < a < 1$.

Why $\operatorname{Log}$ is holomorphic. On the slit plane, $\operatorname{Arg}z$ is continuous (it is $2\arctan\big(y/(x + |z|)\big)$, continuous where $x + |z|\ne0$, i.e. off $(-\infty,0]$), and $e^{\operatorname{Log}z} = z$. A continuous inverse of a holomorphic function with nonzero derivative is holomorphic, with derivative the reciprocal: $\frac{\operatorname{Log}(z+h) - \operatorname{Log}z}{h} = \frac{w' - w}{e^{w'} - e^w}\to\frac1{e^w} = \frac1z$ where $w = \operatorname{Log}z$, $w' = \operatorname{Log}(z+h)\to w$ by continuity. Alternatively check Cauchy–Riemann on $u = \frac12\ln(x^2 + y^2)$, $v = \operatorname{Arg}$.

The keyhole in detail (this is P8's first rung). On the upper edge $z = x$ with $\arg z = 0$, $z^{a-1} = x^{a-1}$; on the lower edge, traversed from $R$ back to $\varepsilon$, $\arg z = 2\pi$ and $z^{a-1} = x^{a-1}e^{2\pi i(a-1)} = x^{a-1}e^{2\pi ia}$. So the two edges give $(1 - e^{2\pi ia})\int_\varepsilon^Rx^{a-1}R(x)\,\mathrm dx$. The large circle is $O(R^{a-1}\cdot R^{-1}\cdot R) = O(R^{a-1})\to0$ for $a < 1$ and $R(x)\sim1/x$; the small circle is $O(\varepsilon^a)\to0$ for $a > 0$. Those two conditions are exactly the convergence conditions of the real integral.

Where branch points meet physics. A branch point in a response function signals a continuum of modes rather than a discrete set — a string on an infinite line, a diffusing heat pulse, a radiating antenna. The transform of the diffusion kernel involves $\sqrt{-i\omega}$, and the wave equation's Green's function in even space dimensions has a cut, which is why a pulse in two dimensions leaves a wake while in three it does not (unit 24). The singularity in P10 — where Kepler's equation stops defining $E(e)$ — is a square-root branch point.

20 · residues

Counting zeros: the argument principle and Rouché

the argument principle

If $f$ is holomorphic inside and on a circle (or toy contour) $\gamma$ except for poles, with no zeros or poles on $\gamma$, then

$$\frac{1}{2\pi i}\oint_\gamma\frac{f'(z)}{f(z)}\,\mathrm dz = Z - P,$$

zeros minus poles inside, with multiplicity: the number of times $f(\gamma)$ winds round $0$.

Proof. Near a zero of order $m$, $f = (z - z_0)^mk$ with $k(z_0)\ne0$, so $f'/f = \frac{m}{z - z_0} + \frac{k'}{k}$: residue $m$. At a pole of order $m$, residue $-m$. Residue theorem. $\square$

Rouché's theorem

If $f, g$ are holomorphic inside and on $\gamma$ and $|g| < |f|$ on $\gamma$, then $f$ and $f + g$ have the same number of zeros inside $\gamma$.

Proof. For $s\in[0,1]$, $|f + sg|\ge|f| - |g| > 0$ on $\gamma$. So $N(s) = \frac1{2\pi i}\oint\frac{(f + sg)'}{f + sg}$ is defined, continuous in $s$, and an integer: constant. $N(0) = N(1)$. $\square$

Why $N(s)$ is continuous. The integrand $(f' + sg')/(f + sg)$ is jointly continuous in $(s, z)$ on the compact $[0,1]\times\gamma$, since the denominator is bounded away from $0$ there; so it is uniformly continuous and the integral depends continuously on $s$. A continuous integer-valued function on the connected $[0,1]$ is constant.

Why "winding number". $\frac{f'}{f}\mathrm dz = \mathrm d\log f$ locally; integrating round $\gamma$ adds up the increase of $\log|f|$ (zero, since the curve closes) and $i$ times the increase of $\arg f$. So the left side is $\frac{1}{2\pi}\times$(total change of $\arg f$ along $\gamma$).

Engineering. Nyquist's stability criterion (1932) is the argument principle applied to $1 + K(s)$ along the boundary of the right half-plane: it counts the unstable poles of a feedback loop from a plot of its measured frequency response, without solving for the poles. For the galvanometer the count is immediate — $\omega_0^2 - \omega^2 - i\gamma\omega$ has both zeros in the lower half-plane — but for a loop with delay there is no formula and the count is the only tool.

20 · residues

Where the roots of $z^5 + 3z + 1$ are

where the roots of $z^5 + 3z + 1$ are

On $|z| = 2$: $|3z + 1|\le7 < 32 = |z^5|$, so all five roots have $|z| < 2$. On $|z| = 1$: $|z^5 + 1|\le2 < 3 = |3z|$, so exactly one has $|z| < 1$. (Computed: $-0.33199$, and four of modulus $1.2629$ and $1.3743$ ✓.)

P10 uses Rouché to show Kepler's equation has exactly one root in a disc for every $|e| < 0.6627$ — and that is the Laplace limit.

Why it works: on a circle where one term dominates, $f$ cannot reach $0$, and its image winds round $0$ as often as the dominant term's does — $z^5$ five times on $|z| = 2$, $3z$ once on $|z| = 1$.

Left: the five roots, computed, with $|z| = 1$ and $|z| = 2$. Right: the image of $|z| = 1$ under $f(z) = z^5 + 3z + 1$. It never reaches $0$ and goes round it once — one zero inside, by the argument principle. (The image of $|z| = 2$, too large to draw here, goes round five times.)

section 5

Causality is a statement about poles

A system that cannot respond before it is driven has a frequency response holomorphic in the upper half-plane — and conversely.

20 · causality · example

Worked: the galvanometer's Green's function by residues

worked example 3 — invert $\chi$, and watch the step function appear

Step 0 — what is to be computed. By inversion, $G(t) = \dfrac1{2\pi}\displaystyle\int_{-\infty}^\infty\frac{e^{-i\omega t}\,\mathrm d\omega}{\omega_0^2 - \omega^2 - i\gamma\omega}$, the limit of $\int_{-R}^R$.

Step 1 — the poles. $\omega^2 + i\gamma\omega - \omega_0^2 = 0$ gives $\omega_\pm = \pm\omega_\mathrm d - i\gamma/2 = \pm7.9975 - 0.2\,i$. Both below the real axis.

Step 2 — the arcs. On $|\omega| = R$: $|\omega_0^2 - \omega^2 - i\gamma\omega|\ge R^2 - \gamma R - \omega_0^2$, so $|\chi|\le2/R^2$ once $R\ge12$; and $|e^{-i\omega t}| = e^{t\operatorname{Im}\omega}\le1$ on whichever semicircle has $t\operatorname{Im}\omega\le0$. ML: arc $\le\pi R\cdot2/R^2\to0$.

Step 3 — $t < 0$: close upwards. On the upper arc $t\operatorname{Im}\omega\le0$ ✓. Inside the half-disc there is no pole, so by Cauchy $\int_{-R}^R + \int_{\text{arc}} = 0$ for every $R$; the arc $\to0$. Therefore

$$G(t) = 0\quad\text{for } t < 0 \quad\text{— exactly.}$$

The two closures, poles drawn deeper than their true $0.2$ below the axis. Which one is allowed is decided by the sign of $t$ alone.

Nothing here was assumed about $t < 0$. Unit 06 built $G$ by imposing $G = 0$ before the kick and matching a jump. Here the only input is $\chi$ — which unit 06 found from steady sinusoidal driving, an experiment in which no kick and no "before" ever occur — and the vanishing for $t < 0$ comes out of the position of two poles. That is the sense in which the step function has been explained rather than imposed.

The arc bound, in full. For $|\omega| = R$, $|\omega^2 + i\gamma\omega - \omega_0^2|\ge|\omega|^2 - \gamma|\omega| - \omega_0^2 = R^2 - 0.4R - 64$, which exceeds $R^2/2$ once $R^2/2\ge0.4R + 64$, true for $R\ge12$. So $|\chi|\le2/R^2$ there, the arc has length $\pi R$, and $|e^{-i\omega t}|\le1$ on the chosen semicircle. Here plain ML suffices; Jordan's lemma is needed only when $|\chi|$ falls like $1/R$.

Reading the figure. The segment $[-R, R]$ is common to both contours. For $t < 0$ it is closed by the upper arc, anticlockwise, and the half-disc it bounds contains no pole; for $t > 0$ by the lower arc, which runs from $R$ back to $-R$ through $-iR$, so the whole contour is clockwise and holds both poles. In the figure $R$ is about $11.5$, enough to enclose the poles at $|\omega_\pm| = 8$.

20 · causality · example

Worked: closing downwards, and unit 06's $G$ returns

worked example 3, continued — $t \ge 0$

Step 4 — $t > 0$: close downwards, clockwise, enclosing both poles:

$$\int_{-R}^R + \int_{\text{arc}} = \frac{-2\pi i}{2\omega_\mathrm d}\big(e^{-i\omega_-t} - e^{-i\omega_+t}\big),$$

the residues from the residue slide. With $e^{-i\omega_\pm t} = e^{-\gamma t/2}e^{\mp i\omega_\mathrm dt}$ the bracket is $2i\,e^{-\gamma t/2}\sin\omega_\mathrm dt$. Let $R\to\infty$, divide by $2\pi$, and $(-i)(2i)/2\omega_\mathrm d = 1/\omega_\mathrm d$:

$$G(t) = e^{-\gamma t/2}\,\frac{\sin\omega_\mathrm dt}{\omega_\mathrm d} .$$

Step 5 — $t = 0$. No exponential; both closures are legitimate by ML, and the upper one gives $G(0) = 0$: continuous, as unit 06 required.

The result, plotted: exactly $0$ before the kick (Step 3), then a ring at $\omega_\mathrm d = 7.9975\ \mathrm{s^{-1}}$ inside the envelope $\pm e^{-\gamma t/2}/\omega_\mathrm d$ (dashed).

Checks. Unit 06's $G$, to the letter ✓. $[G] = \mathrm T$ ✓. $\gamma\to0$: $\sin(\omega_0t)/\omega_0$, the undamped kick ✓. Number: first maximum near $t = \pi/2\omega_\mathrm d = 0.1964$ s, $G = 0.1202$ s.

Reading the figure. Each pole contributes one complex exponential, $e^{-i\omega_\pm t} = e^{-\gamma t/2}e^{\mp i\omega_\mathrm dt}$; their difference is the sine, and the common factor $e^{-\gamma t/2}$ — the poles' depth below the axis — is the dashed envelope. A pole's real part sets the ringing, its imaginary part the decay.

The orientation. The lower contour runs $-R\to R$ along the axis and back through $-iR$: clockwise. The residue theorem for a clockwise curve carries $-2\pi i$. The two residues from the residue slide are $-e^{-i\omega_+t}/2\omega_\mathrm d$ and $+e^{-i\omega_-t}/2\omega_\mathrm d$.

The numbers. $\omega_\mathrm d = \sqrt{63.96} = 7.9974996$, so $\pi/(2\omega_\mathrm d) = 0.196411$ s and $G = e^{-0.2\times0.196411}/7.9974996 = 0.120222$ s. (The true maximum of $G$ is slightly earlier, where $\tan\omega_\mathrm dt = 2\omega_\mathrm d/\gamma$, at $t = 0.193285$ s with $G = 0.120260$ s; the difference is in the fourth figure.) Physically: a unit angular impulse per unit moment of inertia, $1\ \mathrm{rad\,s^{-1}}$ of initial angular velocity, gives a first swing of $0.12$ rad. Unit 06's charge pulse gave $0.2\ \mathrm{rad\,s^{-1}}$ and a first swing of $0.02405$ rad: the same shape scaled by $0.2$ — with the exact maximum, $0.2\times0.120260 = 0.024052$ ✓.

What the undamped limit does to the calculation. At $\gamma = 0$ the poles are on the real axis and the integral in Step 0 does not exist. The damped answer tells you which way to step round them: the poles came from below, so the contour passes above them. The next slide makes that the rule.

20 · causality

The step function, explained

The example used two facts only: $\chi\to0$ at infinity, and no poles in the upper half-plane. Any response with those two properties vanishes for $t < 0$, by Step 3 verbatim. And for $t > 0$

$$G(t) = -i\sum_{\text{poles } \omega_k}\operatorname*{Res}_{\omega_k}\chi(\omega)e^{-i\omega t},$$

a sum of damped modes, one per pole — when there are finitely many poles and $\chi$ decays.

the $i0$ prescription

With no damping, $\chi = 1/(\omega_0^2 - \omega^2)$ has poles on the axis and the inversion integral is undefined. Causality decides: replace $\omega$ by $\omega + i0$ — poles pushed below, contour passing above — and

$$G_\mathrm{ret}(t) = H(t)\,\frac{\sin\omega_0t}{\omega_0} .$$

In the damped case no choice is left: $\gamma > 0$ has already moved the poles down by $\gamma/2$, and the real-axis integral is the retarded one. Dissipation and causality point the same way.

Forward: unit 24 inverts the wave operator's $1/(\omega^2/c^2 - k^2)$ with exactly this prescription, and it is how light is emitted after, not before, the charge moves.

The mode sum, justified. For $t > 0$ close downwards: if $\chi$ is rational with numerator degree at most denominator degree minus one, $|\chi|\le C/R$ on large arcs, Jordan's lemma (lower version, $a = t$) kills the arc, and the residue theorem with its clockwise sign gives $2\pi G(t) = -2\pi i\sum\operatorname{Res}$. For $\chi$ with infinitely many poles — a string, a cavity — the same formula holds when the arcs can be chosen to pass between poles with $\chi$ still small on them, and that is a property to be checked, not assumed.

Dimensions and limits. $[G_\mathrm{ret}] = \mathrm T$ ✓. $\omega_0\to0$: $G_\mathrm{ret}\to tH(t)$, the response of a free mass to a unit kick — velocity $1$ from $t = 0$ on ✓.

History. The prescription was used by Lorentz and by the founders of quantum field theory without a name; Feynman made the $i\varepsilon$ explicit in 1949, where choosing it differently at positive and negative frequency gives his propagator. The idea that dispersion relations follow from causality alone is due to Kronig (1926) and Kramers (1927), and was made a theorem by Titchmarsh's work on Fourier integrals (1937) — the causality slide, two on.

20 · causality

Retarded, advanced, and time's arrow

$LG = \delta$ does not know about time's arrow

Pass below both poles and closing upwards for $t < 0$ collects them: $G_\mathrm{adv}(t) = -H(-t)\sin(\omega_0t)/\omega_0$, which also satisfies $LG = \delta$ (its $\dot G$ jumps from $-1$ to $0$). Principal values at both poles give $\frac12(G_\mathrm{ret} + G_\mathrm{adv})$. The equation admits all of them; the choice of half-plane is the physics.

The difference $G_\mathrm{ret} - G_\mathrm{adv} = \sin(\omega_0t)/\omega_0$ for all $t$: a free oscillation, which $L$ annihilates. Any two solutions of $LG = \delta$ differ by one, and the contour picks which.

Undamped poles at $\pm\omega_0$ on the axis, with $\omega_0 = 8$. Passing above them (poles pushed below, $\omega + i0$) gives the retarded $G$, silent before the kick. Passing below gives the advanced one, which rings before the kick and stops dead at it.

The retarded and advanced computations. With poles at $\pm\omega_0 - i0$: closing up for $t < 0$ encloses nothing, $G = 0$; closing down for $t > 0$ gives, from the running example with $\gamma\to0$, $\sin(\omega_0t)/\omega_0$. With poles at $\pm\omega_0 + i0$: closing down for $t > 0$ encloses nothing; closing up for $t < 0$, anticlockwise, gives $G = \frac{1}{2\pi}\cdot2\pi i\big(\frac{-e^{-i\omega_0t}}{2\omega_0} + \frac{e^{i\omega_0t}}{2\omega_0}\big) = i\cdot\frac{2i\sin\omega_0t}{2\omega_0} = -\frac{\sin\omega_0t}{\omega_0}$. Both satisfy the three conditions of the jump theorem, the advanced one with the roles of $0^+$ and $0^-$ exchanged: $G_\mathrm{adv}(0^-) = 0$, $\dot G_\mathrm{adv}(0^-) = -1$, $\dot G_\mathrm{adv}(0^+) = 0$, jump $+1$ ✓. Their difference is $\sin(\omega_0t)/\omega_0$ for all $t$, a free oscillation — an element of the two-dimensional kernel of $L$ (unit 06), as it must be.

Reading the figure. Both right-hand panels are $\mp\sin(8t)/8$ restricted to a half-line. The advanced function satisfies the same equation and the same jump of $\dot G$ by $+1$ — from $-1$ to $0$ — but it is the response of an oscillator that knows the kick is coming. Nothing in $L$ forbids it; only the choice of half-plane does.

20 · causality

Causality, analyticity, Kramers–Kronig: one theorem

the causality theorem

Let $G$ be continuous except possibly at $0$, with $G$, $tG$ and $\chi = \hat G$ absolutely integrable. The following are equivalent:

(a) causality: $G(t) = 0$ for all $t < 0$;

(b) analyticity: $\chi$ extends continuously to $\operatorname{Im}\omega\ge0$, holomorphic on $\operatorname{Im}\omega > 0$, with $\max_{|\omega| = R,\,\operatorname{Im}\omega\ge0}|\chi|\to0$;

(c) dispersion: for every real $\omega$,

$$\chi(\omega) = \frac{1}{i\pi}\,\mathrm{PV}\!\!\int_{-\infty}^{\infty}\frac{\chi(\omega')}{\omega' - \omega}\,\mathrm d\omega' .$$

The sharp form, for square-integrable $\chi$, is Titchmarsh's theorem (cited: Titchmarsh, Introduction to the Theory of Fourier Integrals, 1937, Thm 95).

(a) ⇒ (b)

$\chi(\omega) = \int_0^\infty G(t)e^{i\omega t}\mathrm dt$ and $|e^{i\omega t}| = e^{-t\operatorname{Im}\omega}\le1$ for $t\ge0$: absolutely convergent on the closed upper half-plane, continuous there, holomorphic inside (Morera). Decay: for $G$ the indicator of $[a,b]\subset[0,\infty)$, $|\chi|\le2/|\omega|$; approximate by steps. $\square$

(b) ⇒ (a)

For $t < 0$ close upwards: no singularity inside, and Jordan's lemma with $a = -t > 0$ kills the arc. So $\int_{-R}^R\chi e^{-i\omega t}\mathrm d\omega = -\int_\text{arc}\to0$, and by inversion $G(t) = 0$. $\square$

(a) ⇔ (c) is the next slide. On the running example: $G$, $tG$, $\chi$ are integrable; (a) is unit 06's construction, (b) the position of two poles, and (c) is checked with numbers two slides on.

Full proof of (a) ⇒ (b)

Continuity on the closed half-plane: for $\omega_n\to\omega$ with $\operatorname{Im}\omega_n\ge0$, the integrands $G(t)e^{i\omega_nt}$ are dominated by $|G|$ and converge pointwise: dominated convergence. Holomorphy inside: for a triangle $T$ in the open upper half-plane, $\oint_{\partial T}\chi = \int_0^\infty G(t)\big(\oint_{\partial T}e^{i\omega t}\mathrm d\omega\big)\mathrm dt = 0$ (Fubini, the double integral converging absolutely; Goursat for the entire function $e^{i\omega t}$); Morera. Decay: for $G = \mathbf 1_{[a,b]}$, $\chi = (e^{i\omega b} - e^{i\omega a})/i\omega$ and both exponentials have modulus $\le1$ when $\operatorname{Im}\omega\ge0$. Given $\varepsilon$, a step function $s$ supported in $[0,\infty)$ has $\int|G - s| < \varepsilon$, so $|\chi(\omega)|\le|\hat s(\omega)| + \varepsilon$ uniformly on the closed half-plane, and $\hat s = O(1/|\omega|)$ there. $\square$

Full proof of (b) ⇒ (a)

$\chi$ is only assumed continuous up to the real axis, so apply Cauchy to the half-disc raised by $\eta > 0$ (where $\chi$ is holomorphic on a neighbourhood) and let $\eta\to0$: on the compact closed half-disc $\chi$ is uniformly continuous, so the integrals converge. Jordan's lemma, with $g = \chi$ and $M_R\to0$ by hypothesis, disposes of the arc. Since $\chi$ is absolutely integrable, $\lim_R\int_{-R}^R$ is the inversion integral, which equals $2\pi G(t)$ at every $t\ne0$ (inversion theorem, whose proof is pointwise at continuity points). $\square$

Why the hypotheses. $G$ integrable makes $\chi$ exist; $tG$ integrable makes $\chi$ of class $C^1$, which the principal value in (c) needs; $\chi$ integrable makes the inversion integral converge. A response with a delta function at $t = 0$ (instantaneous response, $\chi\to$ const at infinity) is outside the hypotheses — apply the theorem to $\chi - \chi(\infty)$ instead, which is what optics does with $\varepsilon(\omega) - 1$.

What (b) says physically. A causal system responds finitely to every drive that was smaller in the past than it is now; the upper half-plane is the set of such drives, and holomorphy there is the statement that the response depends smoothly — in fact analytically — on the drive's growth rate.

20 · causality

The dispersion relations, proved

(a) ⇔ (c), in three lines

1. Causality says $G(t) = \operatorname{sgn}(t)\,G(t)$ for $t\ne0$. Transforms are injective, so (a) $\iff\chi = \widehat{\operatorname{sgn}\cdot G}$.

2. The transform of a product is $\frac1{2\pi}$ times the convolution of the transforms (the convolution theorem run backwards, notes), and $\widehat{\operatorname{sgn}} = 2i\,\mathrm{PV}\frac1\omega$:

$$\widehat{\operatorname{sgn}\cdot G}(\omega) = \frac{i}{\pi}\,\mathrm{PV}\!\!\int\frac{\chi(\omega')}{\omega - \omega'}\,\mathrm d\omega' .$$

3. That is the right side of (c). $\square$

Real and imaginary parts, $\chi = \chi' + i\chi''$:

$$\chi'(\omega) = \frac1\pi\mathrm{PV}\!\!\int\frac{\chi''(\omega')\,\mathrm d\omega'}{\omega' - \omega},$$ $$\chi''(\omega) = -\frac1\pi\mathrm{PV}\!\!\int\frac{\chi'(\omega')\,\mathrm d\omega'}{\omega' - \omega}.$$

the Kramers–Kronig relations, for real $G$

A real $G$ gives $\chi(-\omega) = \overline{\chi(\omega)}$: $\chi'$ even, $\chi''$ odd. Folding the integrals onto $\omega' > 0$:

$$\chi'(\omega) = \frac2\pi\,\mathrm{PV}\!\!\int_0^\infty\frac{\omega'\chi''(\omega')}{\omega'^2 - \omega^2}\,\mathrm d\omega',$$ $$\chi''(\omega) = -\frac{2\omega}{\pi}\,\mathrm{PV}\!\!\int_0^\infty\frac{\chi'(\omega')}{\omega'^2 - \omega^2}\,\mathrm d\omega' .$$

Read it: the in-phase (reactive, dispersive) response at one frequency is an integral of the out-of-phase (absorptive) response at all frequencies. Neither can be designed without the other.

Dimensions: $[\chi''][\omega]/[\omega] = [\chi]$ ✓. Limit: $\chi''\equiv0$ forces $\chi'\equiv0$ — the glass slide.

Full proof of step 2

Put $s_\varepsilon(t) = \operatorname{sgn}(t)e^{-\varepsilon|t|}$, bounded and integrable, with $\hat s_\varepsilon(\nu) = \frac{2i\nu}{\nu^2 + \varepsilon^2}$ (tempered-distribution slide). For the product, write $G$ by inversion and swap (Fubini; $|\chi(\omega')||s_\varepsilon(t)|$ is integrable): $\widehat{s_\varepsilon G}(\omega) = \int s_\varepsilon(t)\frac1{2\pi}\int\chi(\omega')e^{-i\omega't}\mathrm d\omega'\,e^{i\omega t}\mathrm dt = \frac1{2\pi}\int\chi(\omega')\hat s_\varepsilon(\omega - \omega')\,\mathrm d\omega' = \frac{i}{\pi}\int\chi(\omega')\frac{\omega - \omega'}{(\omega - \omega')^2 + \varepsilon^2}\,\mathrm d\omega'$. As $\varepsilon\to0$ the left side tends to $\widehat{\operatorname{sgn}\cdot G}(\omega)$ by dominated convergence ($|s_\varepsilon G|\le|G|$). The right side tends to the principal value by the argument of the Sokhotski proof, which used only that the function paired with the kernel is $C^1$ near $\omega$ and integrable — true of $\chi$, because $tG$ is integrable. $\square$

Step 1, both directions. If (a), then $\operatorname{sgn}\cdot G$ and $G$ agree except at $t = 0$, so their transforms agree. If $\chi = \widehat{\operatorname{sgn}\cdot G}$, then $h = G - \operatorname{sgn}\cdot G = 2G\,\mathbf 1_{t<0}$ has $\hat h = 0$, so by inversion $h(t) = 0$ at each $t\ne0$: (a). $\square$

The real and imaginary parts. Multiply (c) by $i\pi$: $i\pi(\chi' + i\chi'') = \mathrm{PV}\int\frac{\chi' + i\chi''}{\omega' - \omega}$. Real parts: $-\pi\chi'' = \mathrm{PV}\int\frac{\chi'}{\omega' - \omega}$. Imaginary parts: $\pi\chi' = \mathrm{PV}\int\frac{\chi''}{\omega' - \omega}$. The folding: split at $0$, send $\omega'\to-\omega'$ on the negative half, use parity, and combine $\frac{1}{\omega' - \omega} + \frac{1}{\omega' + \omega} = \frac{2\omega'}{\omega'^2 - \omega^2}$ (for the odd $\chi''$) or $\frac{1}{\omega' - \omega} - \frac{1}{\omega' + \omega} = \frac{2\omega}{\omega'^2 - \omega^2}$ (for the even $\chi'$, with an overall sign from the odd reflection).

An operator reading. The map $\chi\mapsto\frac{1}{\pi}\mathrm{PV}\int\frac{\chi(\omega')}{\omega - \omega'}\mathrm d\omega'$ is the Hilbert transform $\mathcal H$, and step 2 says $\mathcal H\hat f = -i\,\widehat{\operatorname{sgn}f}$. Causality is then the eigenvalue statement $\mathcal H\chi = -i\chi$, and the real-imaginary form says $\chi'$ and $\chi''$ are a Hilbert-transform pair. Since $\operatorname{sgn}^2 = 1$, $\mathcal H^2 = -1$.

History. Kronig, 1926, on the dispersion of X-rays; Kramers, 1927, at the Como conference, for light in general. Both argued physically. The mathematical content — boundary values of functions holomorphic in a half-plane are Hilbert-transform pairs — was Plemelj's (1908) and Titchmarsh's.

20 · causality · example

Worked: Kramers–Kronig on the galvanometer

worked example 4 — the dispersion relations, checked three ways

Step 1 — the two parts. With $D(\omega) = (\omega_0^2 - \omega^2)^2 + \gamma^2\omega^2$,

$$\chi' = \frac{\omega_0^2 - \omega^2}{D},\qquad \chi'' = \frac{\gamma\omega}{D} .$$

$\chi''>0$ for $\omega>0$: the oscillator absorbs.

Step 2 — at $\omega = 0$, exactly. The first relation gives

$$\chi'(0) = \frac2\pi\int_0^\infty\frac{\chi''(\omega')}{\omega'}\mathrm d\omega' = \frac{2\gamma}{\pi}\int_0^\infty\frac{\mathrm d\omega'}{D},$$

and P5's integral makes this $\frac{2\gamma}{\pi}\cdot\frac{\pi}{2\gamma\omega_0^2} = \frac1{\omega_0^2}$. That is $\chi(0) = 1/64 = 0.015625\ \mathrm{s^2}$: the static stiffness ✓.

The two parts of $\chi$ on the real axis. $\chi''$ is a narrow absorption peak at $\omega_0$; $\chi'$ swings through zero there. Kramers–Kronig computes either curve from the other; the dots are Step 4's check at $\omega = 7$.

Step 1. Multiply numerator and denominator of $\chi$ by the conjugate $\omega_0^2 - \omega^2 + i\gamma\omega$.

Step 2, the integral. $\int_{-\infty}^\infty\frac{\mathrm d\omega}{D} = \frac{\pi}{\gamma\omega_0^2}$ is problem P5, by residues at the two poles in the upper half-plane ($D = |h|^2$ has four zeros, $\omega_\pm$ and their conjugates). Half of it, by evenness, is $\pi/(2\gamma\omega_0^2)$. Numerically $0.0613592$ against $\pi/(2\cdot0.4\cdot64) = 0.0613592$ ✓.

Reading the figure. The peak of $\chi''$ is $1/(\gamma\omega_0) = 0.3125$ and its half-width is about $\gamma/2 = 0.2$; $\chi'$ reaches $+0.160$ and $-0.152$ just either side of $\omega_0$ and is $\chi(0) = 0.015625$ at $\omega = 0$, too small to see at this scale — which is why Step 2 checks it with numbers rather than by eye.

20 · causality · example

Kramers–Kronig on the galvanometer: the sum rule

worked example 4, continued — high frequency, and an arbitrary point

Step 3 — at large $\omega$. For $\omega$ beyond the region where $\chi''$ lives, the first relation gives $\chi'(\omega)\approx-\frac{2}{\pi\omega^2}\int_0^\infty\omega'\chi''\,\mathrm d\omega'$. But directly $\chi'\approx-1/\omega^2$ — a free mass. So the sum rule

$$\int_0^\infty\omega\,\chi''(\omega)\,\mathrm d\omega = \frac\pi2$$

must hold, whatever $\gamma$ and $\omega_0$. Directly: $\gamma\int_0^\infty\omega^2/D = \gamma\cdot\frac{\pi}{2\gamma} = \frac\pi2$ ✓.

Step 4 — at $\omega = 7\ \mathrm{s^{-1}}$, off every special point: $\chi(7) = 0.0644219 + 0.0120254\,i$. The principal-value integrals, computed numerically, return $0.06442192$ and $0.01202542$ ✓.

Dimensions in Step 3: $[\omega\chi''\,\mathrm d\omega] = \mathrm T^{-1}\mathrm T^2\mathrm T^{-1} = 1$ ✓.

The sum rule, drawn: $\omega\chi''(\omega)$ for the galvanometer's $\gamma = 0.4$ and for $\gamma = 2$ (same $\omega_0$). Damping trades height for width; the area under each curve is $\pi/2$.

Step 3, carefully. For large $\omega$ write $\frac{\omega'}{\omega'^2 - \omega^2} = -\frac{\omega'}{\omega^2} - \frac{\omega'^3}{\omega^2(\omega^2 - \omega'^2)}$; since $\chi''$ decays like $\omega'^{-3}$ the second term's contribution is $o(\omega^{-2})$ (split the integral at $\omega/2$ and $2\omega$; near $\omega'\approx\omega$ the principal value of a function of size $O(\omega^{-3})$ is $O(\omega^{-3}\log\omega)$). So $\chi'(\omega) = -\frac{2}{\pi\omega^2}\int_0^\infty\omega'\chi''\,\mathrm d\omega' + o(\omega^{-2})$, and comparison with $\chi' = -\omega^{-2} + O(\omega^{-4})$ gives the rule. The same integral, $\int_{-\infty}^\infty\omega^2/D = \pi/\gamma$, is also a two-residue computation; numerically $\gamma\int_0^\infty\omega^2/D = 1.5707963$ ✓.

Why a sum rule is physics. The coefficient of $-1/\omega^2$ is $\dot G(0^+) = 1$ — the jump that made $LG = \delta$, the inertia of the coil (section 6 derives this from Watson's lemma). So the total absorption, weighted by frequency, is fixed by Newton's second law at short times and is independent of the spring and the damper. For electrons in matter this is the Thomas–Reiche–Kuhn or $f$-sum rule, and it counts electrons: the absorption spectrum of a material, integrated, tells you how many charges there are per unit volume.

Step 4. $\chi(7) = 1/(15 - 2.8i) = (15 + 2.8i)/232.84 = 0.06442192 + 0.01202542i$. The principal values were computed by folding each integral onto $u = \omega' - 7 > 0$, which removes the singularity, and integrating adaptively; both agree with the direct values to seven or more significant figures.

The figure's numbers. $\omega\chi'' = \gamma\omega^2/D$ peaks at $\omega_0$ with height $1/\gamma$: $2.5$ for $\gamma = 0.4$ and $0.5$ for $\gamma = 2$. Both areas, computed, are $1.5708 = \pi/2$.

20 · causality

Why glass separates colours

Physics, cited: in a dielectric made of bound charges $q$, mass $m$, density $N$, each a copy of the oscillator with its own $\omega_0$, light sees the relative permittivity $\varepsilon(\omega)$ (Jackson, Classical Electrodynamics, §7.5; not derived in this course). Here $n$ is the refractive index — light travels at $c/n$, $c$ its speed in vacuum — and $\varepsilon_0$ the vacuum permittivity, a constant; this $\varepsilon$ is not the small $\varepsilon$ of earlier slides.

$$n(\omega)^2 = \varepsilon(\omega) = 1 + \frac{Nq^2}{\varepsilon_0m}\,\chi(\omega) .$$

$\varepsilon - 1$ is a causal response, so Kramers–Kronig applies to it.

Sellmeier is Kramers–Kronig with each absorption line squeezed to a delta function — the undamped limit of the Sokhotski slide.

a medium that refracts must also absorb

If $\operatorname{Im}\varepsilon = 0$ at every frequency, then $\operatorname{Re}\varepsilon = 1$ at every frequency. Proof: the first dispersion relation for $\varepsilon - 1$. $\square$ So transparent glass with $n\ne1$ in the visible must absorb somewhere else — and $n(0)^2 - 1 = \frac2\pi\int_0^\infty\frac{\operatorname{Im}\varepsilon}{\omega}\mathrm d\omega > 0$.

borosilicate crown glass, BK7

Its Sellmeier formula $n^2 = 1 + \sum_{i=1}^3\frac{B_i\lambda^2}{\lambda^2 - C_i}$ is three undamped oscillators, $\frac{B_i\lambda^2}{\lambda^2 - C_i} = \frac{B_i\omega_i^2}{\omega_i^2 - \omega^2}$ with $\omega_i = 2\pi c/\sqrt{C_i}$ ($\lambda = 2\pi c/\omega$ the vacuum wavelength).

The resonances sit at $\sqrt{C_i} = 77.5$ nm, $141.5$ nm (ultraviolet) and $10.18\ \mu$m (infrared) — where glass is opaque.

What is cited and what is proved. The formula $\varepsilon = 1 + (Nq^2/\varepsilon_0m)\chi$ needs electrodynamics — the polarisation $P = Nqx$ of charges each driven by the light's field and obeying this unit's oscillator equation — and is cited from Jackson §7.5; this course does not derive it, and here it is only the physical motivation. What is proved here is the mathematics: whatever the microscopic model, if $\varepsilon - 1$ is the transform of a causal, integrable response it satisfies Kramers–Kronig, and then zero absorption everywhere forces no refraction anywhere. Dimensions: $Nq^2/\varepsilon_0m$ is $\mathrm{m^{-3}C^2}/(\mathrm{C^2N^{-1}m^{-2}}\cdot\mathrm{kg}) = \mathrm{s^{-2}}$, the square of the plasma frequency, and $\chi$ is $\mathrm{s^2}$: $\varepsilon$ is dimensionless ✓.

The static sum rule is Step 2 of the worked example, for $\varepsilon - 1$: the zero-frequency polarisability is an integral over the absorption spectrum weighted by $1/\omega$, so absorption at low frequency (infrared) counts for more than absorption in the ultraviolet. Positive because $\operatorname{Im}\varepsilon > 0$ for $\omega > 0$ in a passive medium.

20 · causality

BK7: why blue bends more

From the maker's Sellmeier coefficients:

blue, 486.1 nm	$n = 1.52238$
yellow, 587.6 nm	$n = 1.51680$
red, 656.3 nm	$n = 1.51432$

$n_\mathrm{blue} - n_\mathrm{red} = 0.00805$: blue bends more, so a prism fans white light out, blue deflected most.

Why that way round: the visible band lies between the ultraviolet and infrared resonances, on the rising flank of the ultraviolet ones. Below a resonance an oscillator's $\chi'$ is positive and grows as $\omega$ approaches it — the left flank of the galvanometer's curve two slides back — and $n$ grows with it: higher frequency, shorter $\lambda$, larger $n$.

$n(\lambda)$ for BK7 from its Sellmeier formula. Steep towards the ultraviolet resonances, nearly flat through the visible band (shaded), falling again towards the infrared one. The three dots are the table's blue, yellow and red lines.

The numbers. Schott's N-BK7 coefficients: $B = (1.03961212, 0.231792344, 1.01046945)$, $C = (0.00600069867, 0.0200179144, 103.560653)\ \mu\mathrm m^2$. At the Fraunhofer F, d and C lines ($486.1327$, $587.5618$, $656.2725$ nm) the formula gives $1.522376$, $1.516800$, $1.514322$; the difference $n_F - n_C = 0.008054$, and the Abbe number $(n_d - 1)/(n_F - n_C) = 64.17$, the value in the catalogue. The three $\sqrt{C_i}$ are $0.07746$, $0.14148$ and $10.1765\ \mu$m. They are effective resonances fitted over the transparent range, not individual molecular lines; real glass absorbs in broad bands in the far ultraviolet and the mid-infrared, and the fit places one delta function to stand in for each band.

The identity used. $\frac{\lambda^2}{\lambda^2 - C} = \frac{1}{1 - C/\lambda^2}$ and $C/\lambda^2 = \omega^2/\omega_i^2$ with $\omega = 2\pi c/\lambda$, $\omega_i = 2\pi c/\sqrt C$. So the Sellmeier term is $\omega_i^2/(\omega_i^2 - \omega^2)$, which is $\omega_i^2\chi(\omega)$ for an undamped oscillator at $\omega_i$: $B_i\omega_i^2$ plays the role of $Nq^2/\varepsilon_0m$.

Newton and the prism. Newton's 1672 letter on colours established that white light is a mixture and that each colour has its own refrangibility. Why the refrangibility depends on colour had to wait for the oscillator model (Sellmeier 1871, Helmholtz 1875, Lorentz) — and why it could not have been otherwise, for Kramers and Kronig.

20 · causality

The Laplace transform and the Bromwich contour

the Laplace transform $$F(s) = \mathcal L[f](s) = \int_0^\infty f(t)\,e^{-st}\,\mathrm dt,$$

for $f$ on $[0,\infty)$ with $|f|\le Ce^{\sigma t}$; it converges and is holomorphic on $\operatorname{Re}s > \sigma$. For real $c > \sigma$ — a number here, not the speed of light — it is the Fourier transform of $f(t)e^{-ct}H(t)$ at $\omega$, with $s = c - i\omega$.

the Bromwich integral

For $c > \sigma$, at points where $f$ is continuous and when $F(c - i\omega)$ is integrable in $\omega$,

$$f(t) = \frac{1}{2\pi i}\int_{c - i\infty}^{c + i\infty}F(s)\,e^{st}\,\mathrm ds .$$

Proof: Fourier inversion for $f(t)e^{-ct}H(t)$, then $s = c - i\omega$, $\mathrm ds = -i\,\mathrm d\omega$. $\square$

Why bother: it takes growing functions and initial data in its stride, $\mathcal L[\dot f] = sF - f(0)$, so it solves an initial value problem in one pass.

For the oscillator with data $x(0)$, $\dot x(0)$ and drive $u$, applying the rule twice:

$$(s^2 + \gamma s + \omega_0^2)X = U + (s + \gamma)x(0) + \dot x(0).$$

The initial conditions arrive as a polynomial on the right, and the transient comes from it.

P9 uses it for the galvanometer's response to a switched-on current: the needle overshoots to $1.9244$ times its final deflection.

The substitution in full. Fourier inversion for $g(t) = f(t)e^{-ct}H(t)$: $g(t) = \frac{1}{2\pi}\int\hat g(\omega)e^{-i\omega t}\mathrm d\omega$ with $\hat g(\omega) = \int_0^\infty f(t)e^{-(c - i\omega)t}\mathrm dt = F(c - i\omega)$. Multiply by $e^{ct}$: $f(t) = \frac{1}{2\pi}\int F(c - i\omega)e^{(c - i\omega)t}\mathrm d\omega$. With $s = c - i\omega$, as $\omega$ runs from $-\infty$ to $\infty$, $s$ runs from $c + i\infty$ down to $c - i\infty$, and $\mathrm d\omega = i\,\mathrm ds$; reversing the direction absorbs a sign, giving $\frac{1}{2\pi}\cdot(-i)\int_{c-i\infty}^{c+i\infty} = \frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}$ ✓. The result is independent of $c > \sigma$: two Bromwich lines bound a strip with no singularity, and the horizontal ends vanish when $F\to0$ there.

The derivative rule. $\int_0^\infty\dot fe^{-st}\mathrm dt = [fe^{-st}]_0^\infty + s\int_0^\infty fe^{-st} = sF - f(0)$ for $\operatorname{Re}s > \sigma$. So $\ddot x + \gamma\dot x + \omega_0^2x = u$ with data $x(0), \dot x(0)$ becomes $(s^2 + \gamma s + \omega_0^2)X = U + (s + \gamma)x(0) + \dot x(0)$ — the initial conditions enter as a polynomial on the right, which the transient then comes from.

The step response (P9). A current switched on at $t = 0$ gives $u = u_0H(t)$, $U = u_0/s$, so $X = u_0/[s(s^2 + \gamma s + \omega_0^2)]$: three poles, at $0$ and at $-\gamma/2\pm i\omega_\mathrm d$. Residues give $x(t) = \frac{u_0}{\omega_0^2}\big[1 - e^{-\gamma t/2}\big(\cos\omega_\mathrm dt + \frac{\gamma}{2\omega_\mathrm d}\sin\omega_\mathrm dt\big)\big]$, first maximum at $t = \pi/\omega_\mathrm d = 0.39282$ s, value $1 + e^{-\pi\gamma/2\omega_\mathrm d} = 1.92444$ times the final deflection. A $Q = 20$ galvanometer nearly doubles on the first swing, which is why unit 06 recommended damping it.

History. Laplace used the transform in probability (1812). Heaviside's operational calculus (1890s) solved circuit equations by treating $\mathrm d/\mathrm dt$ as an algebraic $p$, with spectacular success and no justification; Bromwich (1916) supplied the contour integral that made Heaviside's rules theorems.

20 · causality

The running example, turned a quarter-turn

the running example, rotated

$\mathcal L[G](s) = \dfrac{1}{s^2 + \gamma s + \omega_0^2}$, poles $s = -\gamma/2\pm i\omega_\mathrm d$ — in the left half-plane — and $\chi(\omega) = \mathcal L[G](-i\omega)$: the same picture turned a quarter-turn.

The Bromwich line runs to the right of every singularity. For $t > 0$, $e^{st}$ decays to the left: close left, collect the poles. For $t < 0$ close right: nothing there, $f = 0$. Causality, once more.

With $s = -i\omega$ the $s$-plane is the $\omega$-plane rotated: the $\omega$-plane's lower half-plane is the left half here, and the real $\omega$-axis is the imaginary $s$-axis.

The galvanometer in the $s$-plane: poles at $-0.2\pm7.9975i$ (drawn further left than to scale), the Bromwich line to their right, and the closure to the left that collects them for $t > 0$. For $t < 0$, $e^{st}$ decays to the right, where there is nothing.

section 6

Asymptotics of integrals

When an integral has no closed form, find the one small region that decides it — moving the contour until there is one.

20 · asymptotics

Watson's lemma

Notation (unit 21 develops it): $f = O(g)$ if $|f|\le C|g|$ eventually; $f\sim g$ if $f/g\to1$; and $f(\lambda)\sim\sum a_n\varphi_n(\lambda)$, an asymptotic expansion, if for every $N$ the error after $N$ terms is $O(\varphi_N)$ — convergent or not.

The Gamma function: $\Gamma(\alpha) = \int_0^\infty t^{\alpha-1}e^{-t}\mathrm dt$ for $\alpha>0$; $\Gamma(n+1) = n!$ by parts, $\Gamma(\frac12) = \sqrt\pi$ by $t = u^2$.

Watson's lemma

If $|g(t)|\le Ce^{bt}$ and $g(t)\sim\sum_{n\ge0}a_nt^{\alpha_n}$ as $t\to0^+$, with $-1 < \alpha_0 < \alpha_1 < \dots\to\infty$, then as $\lambda\to+\infty$

$$\int_0^\infty g(t)e^{-\lambda t}\,\mathrm dt\sim\sum_{n\ge0}\frac{a_n\,\Gamma(\alpha_n+1)}{\lambda^{\alpha_n+1}} .$$

Only $t\lesssim1/\lambda$ matters: expand $g$ there, integrate term by term to $\infty$.

the running example: what a coil does at high frequency

From the ODE, $G$ $= t - \frac\gamma2t^2 + \frac{\gamma^2 - \omega_0^2}{6}t^3 + \dots$ near $0^+$ ($G(0) = 0$, $\dot G = 1$, $\ddot G = -\gamma$). Watson:

$$\mathcal L[G](s)\sim\frac1{s^2} - \frac{\gamma}{s^3} + \frac{\gamma^2 - \omega_0^2}{s^4} .$$

Directly, $1/(s^2 + \gamma s + \omega_0^2)$ expands in powers of $1/s$ to the same thing ✓ — and that Laurent series converges for $|s| > \omega_0$, so the expansion holds on the imaginary axis too. With $s = -i\omega$:

$$\chi(\omega) = -\frac{1}{\omega^2} + \frac{i\gamma}{\omega^3} + O(\omega^{-4}) .$$

A free mass, $\ddot x = f$, is $-1/\omega^2$: at high frequency every oscillator is pure inertia, and the coefficient is the jump $\dot G(0^+) = 1$ — the sum rule's constant.

Full proof of Watson's lemma

Fix $N$ and $\delta > 0$ small enough that $|g(t) - \sum_{n<N}a_nt^{\alpha_n}|\le Kt^{\alpha_N}$ on $(0,\delta]$. Split $\int_0^\infty = \int_0^\delta + \int_\delta^\infty$. Tail: $|\int_\delta^\infty ge^{-\lambda t}|\le C\int_\delta^\infty e^{-(\lambda - b)t} = Ce^{-(\lambda-b)\delta}/(\lambda - b)$, exponentially small, hence $O(\lambda^{-M})$ for every $M$. Head: $\int_0^\delta a_nt^{\alpha_n}e^{-\lambda t}\mathrm dt = a_n\Gamma(\alpha_n+1)\lambda^{-\alpha_n-1} - a_n\int_\delta^\infty t^{\alpha_n}e^{-\lambda t}$, the last term again exponentially small; and the remainder is at most $K\int_0^\infty t^{\alpha_N}e^{-\lambda t}\mathrm dt = K\Gamma(\alpha_N+1)\lambda^{-\alpha_N-1}$. So the error after $N$ terms is $O(\lambda^{-\alpha_N-1})$, which is the definition. $\square$ The substitution $u = \lambda t$ gives $\int_0^\infty t^\alpha e^{-\lambda t}\mathrm dt = \Gamma(\alpha+1)\lambda^{-\alpha-1}$.

The Taylor coefficients of $G$. $G(0^+) = 0$ and $\dot G(0^+) = 1$ are the jump conditions; the ODE gives $\ddot G = -\gamma\dot G - \omega_0^2G = -\gamma$ and $\dddot G = -\gamma\ddot G - \omega_0^2\dot G = \gamma^2 - \omega_0^2$ at $0^+$. Watson with $\alpha_n = n+1$: $1\cdot1!/s^2 - \frac\gamma2\cdot2!/s^3 + \frac{\gamma^2 - \omega_0^2}{6}\cdot3!/s^4$. Directly, $\frac{1}{s^2}\big(1 + \frac\gamma s + \frac{\omega_0^2}{s^2}\big)^{-1} = \frac1{s^2}\big(1 - \frac\gamma s - \frac{\omega_0^2}{s^2} + \frac{\gamma^2}{s^2} + O(s^{-3})\big)$ ✓.

Why the imaginary axis needs a word. Watson's lemma as proved is for real $\lambda\to\infty$, and $s = -i\omega$ is not real. Here no extension of the lemma is needed: $\mathcal L[G]$ is rational, holomorphic for $|s| > \omega_0$ including at $\infty$, so its Laurent series about $\infty$ converges there, and by the identity theorem it is the series Watson produced. For a general $G$ with a jump in $\dot G$, integration by parts twice gives $\chi = -\dot G(0^+)/\omega^2 + o(\omega^{-2})$ directly — the same statement.

Dimensions. $[\chi] = \mathrm T^2$ and $[1/\omega^2] = \mathrm T^2$ ✓; $[\gamma/\omega^3] = \mathrm T^2$ ✓. Limit. $\omega_0\to0$ and $\gamma\to0$ leave $\chi = -1/\omega^2$ exactly: the free mass.

History. G. N. Watson, 1918, in a paper on the harmonic functions of the parabolic cylinder. The lemma is the reason most asymptotic expansions of integrals are easy once the integral has been put in Laplace form; Laplace's method, stationary phase and steepest descent are ways of putting it there.

20 · asymptotics

Laplace's method

Let $h$ be $C^2$ on $[a,b]$ with a unique maximum at an interior $t_0$, $h''(t_0) < 0$, and $g$ continuous with $g(t_0)\ne0$. Then as $\lambda\to\infty$

$$\int_a^bg(t)e^{\lambda h(t)}\,\mathrm dt\sim g(t_0)\,e^{\lambda h(t_0)}\sqrt{\frac{2\pi}{\lambda|h''(t_0)|}} .$$

Why. $e^{\lambda h}$ is a spike of height $e^{\lambda h(t_0)}$ and width $\sim(\lambda|h''|)^{-1/2}$; inside the spike $h\approx h(t_0) - \frac12|h''|(t - t_0)^2$ and $g\approx g(t_0)$, and the Gaussian integral does the rest. Everything outside is exponentially smaller. Full proof in the notes.

Dimensions: with $t$ a time, $\lambda h$ dimensionless, $\sqrt{1/\lambda|h''|}$ is a time, and both sides carry $[g]\,\mathrm T$ ✓.

The spike, for $h(u) = \ln u - u$ (the next slide's integrand), scaled to height $1$. As $n$ grows it narrows like $n^{-1/2}$ and becomes the Gaussian that Laplace's method integrates (dashed): only the neighbourhood of the maximum survives.

Full proof of Laplace's method

Write $h_0 = h(t_0)$, $k = |h''(t_0)|$. Given $\eta\in(0,1)$, choose $\delta$ so that on $|t - t_0|\le\delta$: $|g(t) - g(t_0)|\le\eta|g(t_0)|$ and $-\frac12k(1+\eta)(t - t_0)^2\le h(t) - h_0\le-\frac12k(1-\eta)(t - t_0)^2$ (Taylor with the continuity of $h''$). Outside the $\delta$-interval, $h\le h_0 - \mu$ for some $\mu > 0$ (unique maximum, compactness), so that part is at most $(b-a)\max|g|\,e^{\lambda(h_0 - \mu)}$ — smaller than $e^{\lambda h_0}\lambda^{-1/2}$ by a factor $e^{-\lambda\mu}\sqrt\lambda\to0$. Inside, the integral lies between $(1\mp\eta)g(t_0)e^{\lambda h_0}\int_{-\delta}^{\delta}e^{-\frac12\lambda k(1\pm\eta)u^2}\mathrm du$ (for $g(t_0) > 0$; otherwise reverse), and $\int_{-\delta}^\delta e^{-\frac12\lambda cu^2}\mathrm du = \sqrt{2\pi/\lambda c}\,(1 + o(1))$. So the ratio of the integral to the claimed form lies between $(1-\eta)/\sqrt{1+\eta}$ and $(1+\eta)/\sqrt{1-\eta}$ for large $\lambda$; $\eta$ was arbitrary. $\square$

Reading the figure. Each curve is $e^{n(h(u) - h(1))} = (u\,e^{1-u})^n$, and each dashed curve $e^{-n(u-1)^2/2}$ — the second-order Taylor expansion of $h$ at the maximum, exponentiated. At $n = 2$ the spike is lopsided (the cubic term of $h$ still matters); by $n = 50$ the two are hard to tell apart. The proof's "$\delta$-interval" is any fixed neighbourhood of $u = 1$: the spike eventually fits inside it.

History. Laplace, 1774, Mémoire sur la probabilité des causes par les événements, evaluating integrals of $x^p(1-x)^q$ for large $p, q$ — the beta integrals of Bayesian inference. De Moivre had $n!\approx Cn^{n+1/2}e^{-n}$ in 1730 and Stirling found $C = \sqrt{2\pi}$ the same year.

20 · asymptotics

Stirling's formula, by Laplace's method

Stirling's formula

$n! = \Gamma(n+1) = \int_0^\infty t^ne^{-t}\mathrm dt$. Put $t = nu$:

$$n! = n^{n+1}\int_0^\infty e^{n(\ln u - u)}\,\mathrm du .$$

$h(u) = \ln u - u$: maximum at $u = 1$, $h(1) = -1$, $h''(1) = -1$. So

$$n!\sim n^{n+1}e^{-n}\sqrt{\frac{2\pi}{n}} = \sqrt{2\pi n}\,\Big(\frac ne\Big)^n .$$

Check at $n = 10$: $3\,598\,695.6$ against $10! = 3\,628\,800$, low by $0.83\%$. The next term multiplies by $1 + \frac1{12n}$ and gives $3\,628\,684.7$, low by $3\times10^{-5}$.

Real integrand, real maximum. Stationary phase and steepest descent are what to do when the exponent is imaginary, or when there is no real maximum at all.

The relative error of Stirling's formula against $n$, computed. The leading form errs by about $1/12n$ (slope $-1$); with the factor $1 + \frac1{12n}$ the error drops to about $1/288n^2$ (slope $-2$). Asymptotic, and good from the start.

The infinite interval. The theorem is stated on $[a,b]$; Stirling's integral runs over $[0,\infty)$. On $u\ge2$, $h'(u) = 1/u - 1\le-\frac12$, so $h(u)\le\ln2 - 2 - (u-2)/2$ and the tail is at most $\frac2ne^{n(\ln2 - 2)}$; since $\ln2 - 2 = -1.307 < -1 = h(1)$, it is exponentially smaller than the main term. On $(0,\frac12]$, $h$ is increasing, so $h\le h(\frac12) = -\ln2 - \frac12 = -1.193 < -1$ and that piece is exponentially smaller too. The theorem applies on $[\frac12, 2]$ with $g = 1$.

The Stirling numbers. $\sqrt{20\pi}\,(10/e)^{10} = 3\,598\,695.62$; ratio to $3\,628\,800$ is $0.991704$. Times $1 + 1/120$: $3\,628\,684.75$. The correction series $1 + \frac{1}{12n} + \frac{1}{288n^2} - \dots$ comes from carrying the Taylor expansion of $h$ and the Gaussian moments to higher order — a Watson's lemma computation after the substitution $\ln u - u + 1 = -\tau^2/2$; it is cited here (de Bruijn, Asymptotic Methods in Analysis, §4.5) and it is an example of a divergent asymptotic series, which is unit 21's subject.

20 · asymptotics

Stationary phase

the method of stationary phase

Let $g, \psi$ be smooth and real-valued ($\psi$) on $[a,b]$, with $\psi'\ne0$ except at one interior $t_0$, where $\psi''(t_0)\ne0$. Then

$$\begin{aligned}\int_a^bg\,e^{i\lambda\psi}\mathrm dt ={}& g(t_0)\sqrt{\frac{2\pi}{\lambda|\psi''(t_0)|}}\,e^{i\lambda\psi(t_0)\pm i\pi/4}\\ &+ O(1/\lambda),\end{aligned}$$

the sign being that of $\psi''(t_0)$.

Where $\psi'\ne0$ the integral is $O(1/\lambda)$, by one integration by parts:

$$\int ge^{i\lambda\psi} = \Big[\frac{ge^{i\lambda\psi}}{i\lambda\psi'}\Big] - \frac{1}{i\lambda}\!\int\!\Big(\frac{g}{\psi'}\Big)'e^{i\lambda\psi}.$$

Near $t_0$ the phase stands still, the contributions stop cancelling, and a Fresnel integral — $\int e^{i\lambda u^2/2}\mathrm du = \sqrt{2\pi/\lambda}\,e^{i\pi/4}$, problem P6 — gives the size.

Why the stationary point decides. Top: $\cos(\lambda t^2/2)$ oscillates ever faster away from $t = 0$, slowly in the shaded zone of width $\sim\lambda^{-1/2}$. Bottom: its running integral barely moves outside the zone and climbs to nearly the whole answer, $\sqrt{2\pi/\lambda}\cos\frac\pi4$ (dashed), inside it.

Full proof

Localise. Choose a smooth cutoff $\rho$ equal to $1$ near $t_0$ and supported where $|\psi''|\ge\frac12|\psi''(t_0)|$ (built from the bump function of section 2). On the support of $1 - \rho$, $\psi'\ne0$, and the integration by parts on the slide gives $O(1/\lambda)$ (boundary terms at $a, b$ included).

Straighten the phase. Near $t_0$ with $\psi''(t_0) > 0$, put $u = \operatorname{sgn}(t - t_0)\sqrt{2(\psi(t) - \psi(t_0))}$. Since $\psi - \psi(t_0) = (t - t_0)^2q(t)$ with $q$ smooth and $q(t_0) = \psi''(t_0)/2 > 0$, $u = (t - t_0)\sqrt{2q(t)}$ is smooth with $\mathrm du/\mathrm dt = \sqrt{\psi''(t_0)}$ at $t_0$: a smooth change of variable with $\psi = \psi(t_0) + u^2/2$. The localised integral becomes $e^{i\lambda\psi(t_0)}\int k(u)e^{i\lambda u^2/2}\mathrm du$, with $k = \rho g\,\mathrm dt/\mathrm du$ smooth, compactly supported, and $k(0) = g(t_0)/\sqrt{\psi''(t_0)}$.

Evaluate. $k(u) = k(0)\rho_1(u) + u\,m(u)$ with $\rho_1$ a cutoff equal to $1$ on the support of $k$ and $m$ smooth and compactly supported. Then (i) $\int u\,m\,e^{i\lambda u^2/2}\mathrm du = \frac{1}{i\lambda}\int m\,\mathrm d(e^{i\lambda u^2/2}) = -\frac{1}{i\lambda}\int m'e^{i\lambda u^2/2} = O(1/\lambda)$; (ii) $\int\rho_1e^{i\lambda u^2/2} = \int e^{i\lambda u^2/2} - \int(1 - \rho_1)e^{i\lambda u^2/2}$, and the second piece is $O(1/\lambda)$ by integrating by parts against $u$ (the integrand vanishes near $0$). The Fresnel integral (P6, with a scaling) is $\sqrt{2\pi/\lambda}\,e^{i\pi/4}$. Assemble: $e^{i\lambda\psi(t_0)}g(t_0)\sqrt{2\pi/\lambda\psi''(t_0)}\,e^{i\pi/4} + O(1/\lambda)$. For $\psi'' < 0$ take complex conjugates. $\square$

Dimensions. With $t$ a time and $\lambda\psi$ dimensionless, $\sqrt{1/\lambda|\psi''|}$ is a time: both sides of the theorem carry $[g]\,\mathrm T$ ✓.

Reading the figure. With $\psi = t^2/2$, $g = 1$, $\lambda = 40$ on $[-2, 2]$, the theorem predicts $\int\cos(\lambda t^2/2)\,\mathrm dt\approx\sqrt{2\pi/40}\cos(\pi/4) = 0.280$, the real part of $\sqrt{2\pi/\lambda}\,e^{i\pi/4}$. The running integral was computed by the trapezium rule; it ends at $0.255$, within the $O(1/\lambda)$ the theorem allows. The ripples at the left and right come from the endpoints, the boundary terms of the integration by parts.

History. Stokes (1850, for the Airy integral) and Kelvin (1887, for water waves). The Fresnel integrals are Fresnel's, 1818, from diffraction at an edge.

20 · asymptotics

Unit 09's debt: the constants in $J_n$

Bessel's integral, by stationary phase

Bessel's integral $J_n(x) = \frac1{2\pi}\int_{-\pi}^{\pi}e^{i(n\theta - x\sin\theta)}\mathrm d\theta$ has phase $-\sin\theta$, stationary at $\theta = \pm\frac\pi2$ with $\psi'' = \pm1$. Two contributions, complex conjugates (P11):

$$J_n(x)\approx\sqrt{\frac{2}{\pi x}}\cos\Big(x - \frac{n\pi}{2} - \frac\pi4\Big).$$

$J_0(10) = -0.245936$ against $-0.246761$; $J_0(20) = 0.167025$ against $0.166656$ ✓.

Physics: in a superposition of waves, what survives at large distance or time is the part whose phase is stationary — Kelvin's argument (1887) for the V of a ship's wake, and the wave-packet slide's argument for group velocity.

$J_0$ (solid, computed) against the stationary-phase form (dashed). Useless near $x = 0$, where it diverges; by $x = 5$ the two are hard to tell apart. The dots are the slide's checks at $x = 10$ and $20$.

20 · asymptotics

Steepest descent

For $I(\lambda) = \int_Cg(z)e^{\lambda\varphi(z)}\mathrm dz$ with $g, \varphi$ holomorphic, Cauchy lets you move $C$, paying for any arcs by estimates. Where to?

saddle point, and the path through it

$|e^{\lambda\varphi}| = e^{\lambda\operatorname{Re}\varphi}$ has no local maximum (maximum modulus). The best available is a saddle, $\varphi'(z_0) = 0$. Through it, the curve on which $\operatorname{Im}\varphi = \operatorname{Im}\varphi(z_0)$ is the path of steepest descent: along it there is no oscillation and $\operatorname{Re}\varphi$ falls as fast as possible.

the method of steepest descent

On that path the integral is a Laplace integral, and

$$I(\lambda)\sim g(z_0)\,e^{\lambda\varphi(z_0)}\sqrt{\frac{2\pi}{-\lambda\varphi''(z_0)}},$$

the root's branch set by the direction in which the path crosses $z_0$.

Why the level curves of $\operatorname{Im}\varphi$. With $\varphi = u + iv$, Cauchy–Riemann makes the vectors of partials orthogonal: $u_xv_x + u_yv_y = u_xv_x - v_xu_x = 0$. So a level curve of $v$ runs along the direction of fastest change of $u$. Near $z_0$, $\varphi\approx\varphi(z_0) + \frac12\varphi''(z_0)(z - z_0)^2$, and the descent direction $e^{i\theta}$ makes $\varphi''(z_0)e^{2i\theta}$ negative real.

one method, three faces

Laplace: real saddle, real path. Stationary phase: saddle on the real axis, path crossing it at $45^\circ$ — the $e^{\pm i\pi/4}$. Genuinely complex saddles: exponentially small answers that real methods see only as "$O(\lambda^{-N})$ for all $N$".

where the work is

The saddle is two lines. The proof is the arc estimates that justify the move — the checkpoint's second clause.

The saddle, geometrically. $e^{\lambda\varphi}$ is holomorphic, so by the maximum modulus theorem its modulus $e^{\lambda\operatorname{Re}\varphi}$ — and hence $\operatorname{Re}\varphi$ — has no interior local maximum. Near a point where $\varphi' = 0\ne\varphi''$, $\varphi - \varphi(z_0)\approx\frac12\varphi''(z_0)(z - z_0)^2$, and $\operatorname{Re}$ of that rises in two opposite directions and falls in the two perpendicular ones: a saddle. A contour that must get from one valley to another crosses the ridge; the lowest crossing is the saddle. That is why the method is also called the saddle-point method.

Proof of the formula, given the deformation. Parametrise the steepest path near $z_0$ by $\tau$ with $\varphi(z(\tau)) = \varphi(z_0) - \tau^2/2$, $\tau$ real; this is the phase-straightening of the stationary-phase proof done in the complex plane, legitimate because $\varphi''(z_0)\ne0$ (the holomorphic inverse function theorem, cited: Ahlfors, ch. 4 §3.3). Then $I = e^{\lambda\varphi(z_0)}\int g(z(\tau))z'(\tau)e^{-\lambda\tau^2/2}\mathrm d\tau$, a real Laplace integral, and $z'(0) = (-\varphi''(z_0))^{-1/2}$ with the branch pointing along the path. Laplace's method (or Watson's lemma, for the full series) finishes. $\square$

Practice is often easier than the theory. The exact steepest path is rarely needed: any path through the saddle along which $\operatorname{Re}\varphi$ has a strict nondegenerate maximum at $z_0$ works, because Laplace's method for a complex-valued exponent with a real-part maximum is proved by the same estimates. The next slide uses a straight line.

Unit 09's last debt: how fast $I_n(ne)$ grows

Unit 09 needed the growth in $n$ of $I_n(ne) = \sum_k\frac{(ne/2)^{n+2k}}{k!(n+k)!}$, the sum of the absolute values of the terms it rearranged, and of $J_n(ne)$. Take as given data the integral $I_n(x) = \frac1{2\pi}\int_{-\pi}^{\pi}e^{x\cos\theta + in\theta}\,\mathrm d\theta$ (expand $e^{x\cos\theta}$ in powers and integrate term by term to recover the series). With $x = ne$ the exponent is $n\varphi$, $\varphi(\theta) = e\cos\theta + i\theta$, and $\varphi' = -e\sin\theta + i = 0$ at $\theta = i\beta$ with $\sinh\beta = 1/e$. The move costs nothing: the integrand is $2\pi$-periodic, so on the rectangle $-\pi, \pi, \pi + i\beta, -\pi + i\beta$ the two vertical sides cancel exactly, and Cauchy moves $[-\pi,\pi]$ to $\theta = s + i\beta$. There $\operatorname{Re}\varphi = e\cosh\beta\cos s - \beta$, with its unique maximum on $[-\pi,\pi]$ at $s = 0$, and $\varphi''(i\beta) = -e\cosh\beta = -\sqrt{1 + e^2}$ is negative real, so Laplace's method applies along the line: $$I_n(ne)\sim\frac{r(e)^n}{\sqrt{2\pi n\sqrt{1 + e^2}}},\qquad r(e) = e^{\varphi(i\beta)} = \frac{e\,e^{\sqrt{1+e^2}}}{1 + \sqrt{1 + e^2}},$$ using $e\cosh\beta = \sqrt{1+e^2}$ and $e^{\beta} = (1 + \sqrt{1+e^2})/e$. So $\sum_nI_n(ne)$ converges iff $r(e) < 1$, and $r(e) = 1$ exactly at $e = 0.6627434$: with $x_* = \sqrt{1 + \lambda^2}$ it is P10's $x_*\tanh x_* = 1$ in another form. The same computation for $J_n(ne) = \frac1{2\pi}\int e^{in(\theta - e\sin\theta)}\mathrm d\theta$, with the saddle at $i\alpha$, $\cosh\alpha = 1/e$, gives $J_n(ne)\sim q(e)^n/\sqrt{2\pi n\sqrt{1 - e^2}}$ with $q(e) = e\,e^{\sqrt{1-e^2}}/(1 + \sqrt{1 - e^2})$ — the ratio unit 09 quoted, $0.9943$ for Halley and $0.127$ for Mars. Numerically the ratio of $I_n(ne)$ to the formula is $0.99924$ at $e = 0.5$, $n = 50$, and $0.99981$ at $n = 200$. Debye (1909) did exactly this.

History. Riemann used the idea in 1863 for hypergeometric functions; Debye (1909), for Bessel functions of large order, made it the method it is. It is how the Airy function's exponentially small tail, the large-order behaviour of Bessel functions that unit 09 left open, and the classical limit of wave mechanics are all computed.

20 · asymptotics · example

Worked: an exponentially small integral by steepest descent

worked example 5 — $\displaystyle I(\lambda) = \int_{-\infty}^\infty e^{i\lambda(t^3/3 + t)}\,\mathrm dt = 2\int_0^\infty\cos\big(\lambda(\tfrac{t^3}3 + t)\big)\mathrm dt$ as $\lambda\to\infty$

Step 1 — real methods fail. The phase $t^3/3 + t$ has derivative $t^2 + 1 > 0$: no stationary point. Integrating by parts repeatedly shows only $I = O(\lambda^{-N})$ for every $N$.

Step 2 — the saddles. $\varphi(z) = i(z^3/3 + z)$, $\varphi' = i(z^2 + 1) = 0$ at $z = \pm i$, with $\varphi(i) = i(-\frac i3 + i) = -\frac23$ and $\varphi(-i) = +\frac23$. Go up to $z = i$, where the integrand is small.

Step 3 — the arcs. Rectangle $-R, R, R+i, -R+i$; the integrand is entire. On the side $z = \pm R + iy$, $0\le y\le1$:

$$\operatorname{Re}\,\lambda\varphi = \lambda\big(-R^2y + \tfrac{y^3}3 - y\big)\le-\lambda R^2y,$$

so each side is at most $\int_0^1e^{-\lambda R^2y}\mathrm dy\le\frac{1}{\lambda R^2}\to0$.

The landscape of $\operatorname{Re}\varphi$, $\varphi = i(z^3/3 + z)$: level curves cross at the saddles $\pm i$. The real-axis contour is moved up through $i$ to $z = s + i$, the short sides vanishing as $R\to\infty$. Above the axis the integrand is small; below, near $-i$, it is huge.

Step 3, the algebra. With $z = R + iy$: $z^3 = R^3 + 3iR^2y - 3Ry^2 - iy^3$, so $i z^3/3 = iR^3/3 - R^2y - iRy^2 + y^3/3$ and $iz = iR - y$. Real part of $i(z^3/3 + z)$: $-R^2y + y^3/3 - y$. For $0\le y\le1$, $y^3/3 - y\le0$. Same at $-R$ ($R^2$ is even). Each side has length $1$, so the ML-type bound is $\int_0^1e^{-\lambda R^2y}\mathrm dy = (1 - e^{-\lambda R^2})/\lambda R^2$. Cauchy's theorem for the rectangle (convex) then says: integral along the real axis from $-R$ to $R$ $=$ integral along the top from $-R+i$ to $R+i$ $+$ the two sides; let $R\to\infty$. The real integral converges as an improper integral (integrate by parts once against $t^2 + 1$), so the limit is $I(\lambda)$.

Why this line and not the exact steepest path. The steepest path through $i$ is the level curve $\operatorname{Im}\varphi = \operatorname{Re}(z^3/3 + z) = x(x^2/3 - y^2 + 1) = 0$, that is $y = \sqrt{1 + x^2/3}$ (dashed in the figure), tangent to the line $\operatorname{Im}z = 1$ at $i$. Along the straight line the phase is $\operatorname{Im}\varphi = s^3/3$: not zero, but only third-order small, and that cubic phase is what Step 5 expands. Why not $-i$: on $z = R - iy$ the real part of the exponent is $\lambda(R^2y - y^3/3 + y)$, which grows with $R$ — the sides cannot be discarded, and the saddle there has $e^{+2\lambda/3}$: it would be a huge number cancelling itself.

20 · asymptotics · example

Worked: on the line through the saddle

worked example 5, continued — the moved contour

Step 4 — on the line through the saddle, $z = s + i$: expanding, $i\big(\frac{(s+i)^3}3 + s + i\big) = -\frac23 - s^2 + \frac{is^3}{3}$. So, exactly,

$$I(\lambda) = e^{-2\lambda/3}\!\int e^{-\lambda s^2}\cos\tfrac{\lambda s^3}3\,\mathrm ds .$$

No cancellation left: a Gaussian times a slowly turning phase.

Step 5 — Laplace. $\cos x = 1 - \frac{x^2}2 + O(x^4)$ and $\int s^6e^{-\lambda s^2} = \frac{15}{8}\sqrt\pi\,\lambda^{-7/2}$:

$$I = \sqrt{\frac\pi\lambda}\,e^{-2\lambda/3}\Big(1 - \frac{5}{48\lambda} + O(\lambda^{-2})\Big).$$

Check at $\lambda = 5$: quadrature gives $0.0277624$; one term $0.0282775$ ($+1.9\%$); two terms $0.0276884$ ($-0.27\%$) ✓. Limit: exponentially small, consistent with Step 1.

The same integral, before and after the move, at $\lambda = 5$. On the real axis the integrand swings between $\pm1$ for ever and cancels down to $0.0278$. On the line through the saddle it is one positive bump of height $e^{-2\lambda/3} = 0.036$: nothing left to cancel.

Step 4, the algebra. $(s+i)^3 = s^3 + 3is^2 - 3s - i$; divide by $3$ and add $s + i$: $\frac{s^3}3 + is^2 + \frac{2i}3$. Times $i$: $\frac{is^3}3 - s^2 - \frac23$. The odd part $\sin(\lambda s^3/3)$ integrates to zero against the even Gaussian.

Step 5, with its error. $|\cos x - 1 + x^2/2|\le x^4/24$, and with $x = \lambda s^3/3$, $\int e^{-\lambda s^2}\frac{\lambda^4s^{12}}{24\cdot81}\mathrm ds = \frac{\lambda^4}{1944}\cdot\frac{10395\sqrt\pi}{64}\lambda^{-13/2} = O(\lambda^{-5/2})$, which is $O(\lambda^{-2})$ relative to $\sqrt{\pi/\lambda}$. The second term: $-\frac{\lambda^2}{18}\cdot\frac{15\sqrt\pi}{8}\lambda^{-7/2} = -\frac{5}{48\lambda}\sqrt{\frac\pi\lambda}$. All rigorous — no step was an approximation that was not bounded.

The numbers. The exact value is $2\pi\lambda^{-1/3}\mathrm{Ai}(\lambda^{2/3})$, where Ai is the Airy function — a name to meet later, not a fact used here; at $\lambda = 5$ the quadrature and this formula agree, $0.02776242$. $\sqrt{\pi/5}\,e^{-10/3} = 0.02827754$, and times $1 - 5/240$ it is $0.02768843$. At $\lambda = 2$ the two-term value is $0.313163$ against $0.317375$, error $-1.3\%$; the error falls like $\lambda^{-2}$, as the bound says.

20 · asymptotics

Wave packets: phase and group velocity

A wave in a dispersive medium obeys a dispersion relation $\omega = \omega(k)$, and a packet is a superposition

$$u(x,t) = \frac1{2\pi}\int A(k)\,e^{i(kx - \omega(k)t)}\,\mathrm dk .$$

Each component's crests move at the phase velocity $v_\mathrm p = \omega/k$. Where is the packet?

group velocity, by stationary phase

Look along $x = vt$ as $t\to\infty$: the phase is $t\,(kv - \omega(k))$, stationary where $\omega'(k) = v$. So

$$u(vt, t)\approx\frac{A(k_*)\,e^{it(k_*v - \omega(k_*))\mp i\pi/4}}{\sqrt{2\pi t|\omega''(k_*)|}},$$

where $\omega'(k_*) = v$.

If $A$ is concentrated near $k_0$, $u$ is small unless $x/t\approx\omega'(k_0)$: the packet moves at $v_\mathrm g = \mathrm d\omega/\mathrm dk$, and decays like $t^{-1/2}$ — it spreads.

light in BK7, at 587.6 nm

In glass $k = n(\omega)\omega/c$, so $1/v_\mathrm g = \mathrm dk/\mathrm d\omega$ gives $v_\mathrm g = c/n_\mathrm g$ with the group index $n_\mathrm g = n - \lambda\frac{\mathrm dn}{\mathrm d\lambda}$. From the Sellmeier formula:

$n = 1.51680$, $v_\mathrm p = 0.65928\,c$;
$n_\mathrm g = 1.54136$, $v_\mathrm g = 0.64878\,c$.

A pulse travels $1.6\%$ slower than its own crests; the crests slide forward through it. Normal dispersion, $\mathrm dn/\mathrm d\lambda < 0$, is what makes $n_\mathrm g > n$.

Checks. No dispersion, $n$ constant: $n_\mathrm g = n$, $v_\mathrm g = v_\mathrm p$ ✓. Dimensions: $\mathrm d\omega/\mathrm dk$ is a speed ✓.

The stationary-phase step. Apply the theorem to $\int A(k)e^{it\psi(k)}\mathrm dk$ with $\lambda = t$ and $\psi(k) = kv - \omega(k)$: $\psi' = v - \omega'(k)$, $\psi'' = -\omega''(k)$. The contribution is $A(k_*)\sqrt{2\pi/t|\omega''(k_*)|}\,e^{it\psi(k_*) - i\frac\pi4\operatorname{sgn}\omega''(k_*)}$, divided by $2\pi$. If no $k$ in the support of $A$ has $\omega'(k) = v$, the integral is $O(1/t)$ — faster decay — so the packet's bulk is on the rays $x/t\in\omega'(\operatorname{supp}A)$.

The group index. $k(\omega) = n(\omega)\omega/c$ gives $\frac{\mathrm dk}{\mathrm d\omega} = \frac{n + \omega\,\mathrm dn/\mathrm d\omega}{c}$, and with $\lambda = 2\pi c/\omega$, $\omega\frac{\mathrm dn}{\mathrm d\omega} = -\lambda\frac{\mathrm dn}{\mathrm d\lambda}$. The derivative of the Sellmeier $n(\lambda)$ at $0.5875618\ \mu$m is $-0.04180\ \mu\mathrm m^{-1}$, so $n_\mathrm g = 1.516800 + 0.024560 = 1.541360$; $1/n = 0.659283$ and $1/n_\mathrm g = 0.648778$.

Deep water has $\omega = \sqrt{gk}$ (a result of fluid mechanics, not derived in this course), so $v_\mathrm g = \frac12\sqrt{g/k} = \frac12v_\mathrm p$: a group of swells travels at half the speed of its crests, which appear at the back of the group, run through it and die at the front. Kelvin's ship-wake angle, $2\arcsin(1/3) = 38.94^\circ$, is stationary phase applied to this relation in two dimensions.

History. Hamilton (1839) and Stokes (1876) distinguished the two velocities; Rayleigh's Theory of Sound (1877) gave $v_\mathrm g = \mathrm d\omega/\mathrm dk$ its name and its beat-frequency derivation. Energy and information travel at $v_\mathrm g$ in normal dispersion; near an absorption line $v_\mathrm g$ can exceed $c$, and Sommerfeld and Brillouin (1914) showed — by exactly this unit's contour methods — that the front of a signal still travels at $c$, because $\chi\to0$ as $|\omega|\to\infty$ in the upper half-plane.

20 · asymptotics

Packets spread

Take a Gaussian packet, $A(k) = e^{-\sigma_0^2(k - k_0)^2/2}$, and expand the dispersion relation to second order: $\omega\approx\omega_0 + v_\mathrm g q + \frac12\beta q^2$, $q = k - k_0$. Then

$$u = \frac{e^{i(k_0x - \omega_0t)}}{2\pi}\int e^{-aq^2/2 + iq(x - v_\mathrm gt)}\,\mathrm dq,$$

a Gaussian integral with complex width $a = \sigma_0^2 + i\beta t$, $\operatorname{Re}a > 0$: by analytic continuation it is $\sqrt{2\pi/a}\,e^{-(x - v_\mathrm gt)^2/2a}$. Taking the modulus:

the spreading law $$|u|\propto e^{-(x - v_\mathrm gt)^2/2\sigma(t)^2},$$ $$\sigma(t) = \sigma_0\sqrt{1 + \big(\beta t/\sigma_0^2\big)^2} .$$

A Gaussian packet at $\beta t/\sigma_0^2 = 0$, $1$, $2.5$, moving at $v_\mathrm g$: thick, $|u|$; thin, $\operatorname{Re}u$. The envelope widens as $\sigma(t)$ and its height falls as $\sigma(t)^{-1/2}$; the carrier is squeezed at one end and stretched at the other — the chirp. Parameters chosen to be visible, not physical.

Checks. $\beta = 0$: no spreading ✓. Late times $\sigma\approx|\beta|t/\sigma_0$: the narrower the packet, the faster it spreads — the uncertainty relation, moving. $[\beta] = \mathrm L^2\mathrm T^{-1}$, so $\beta t/\sigma_0^2$ is a number ✓.

The integral with complex width. The analytic continuation slide gave $\int e^{-aq^2/2}\mathrm dq = \sqrt{2\pi/a}$ for $\operatorname{Re}a > 0$. With a linear term: for $a > 0$, $\int e^{-aq^2/2 + ibq}\mathrm dq = \sqrt{2\pi/a}\,e^{-b^2/2a}$ (worked example 1 with scaling). For fixed real $b$ both sides are holomorphic in $a$ on $\operatorname{Re}a > 0$ (Morera, as before), so the identity theorem extends it. Here $b = x - v_\mathrm gt$.

The modulus. $|e^{-b^2/2a}| = e^{-b^2\operatorname{Re}(1/a)/2}$ and $\operatorname{Re}\frac1a = \frac{\sigma_0^2}{\sigma_0^4 + \beta^2t^2}$, so $\sigma(t)^2 = \frac{\sigma_0^4 + \beta^2t^2}{\sigma_0^2}$. Also $|\sqrt{2\pi/a}| = \sqrt{2\pi}\,(\sigma_0^4 + \beta^2t^2)^{-1/4}$, decaying like $t^{-1/2}$ as the stationary-phase slide predicted, and $\int|u|^2\mathrm dx$ is constant, as it must be (Plancherel: the dispersion only changes the phases of the components).

Reading the figure. Units with $\sigma_0 = 1$, $\beta = 1$, $k_0 = 6$, $v_\mathrm g = 6$ and $\omega_0 = 18$; the three snapshots are the exact formula $u\propto a^{-1/2}e^{i(k_0x - \omega_0t)}e^{-(x - v_\mathrm gt)^2/2a}$ with $a = \sigma_0^2 + i\beta t$. The widths are $\sigma = 1$, $\sqrt2$, $\sqrt{7.25} = 2.69$, the peak heights $1$, $0.84$, $0.61$. The wavelength of the carrier changes along the packet because $1/a$ has an imaginary part: the phase is quadratic in $x - v_\mathrm gt$.

The same law elsewhere. With $\omega = \hbar k^2/2m$ this is the spreading of a free quantum wave packet, $\beta = \hbar/m$; with a time-dependent width it is the diffraction of a Gaussian laser beam, the Rayleigh range playing the role of $\sigma_0^2/\beta$. The mathematics is one Gaussian integral with a complex parameter, legitimised by the identity theorem.

20 · asymptotics

A 10 fs pulse through 1 cm of glass

a 10 fs pulse through 1 cm of glass

In time rather than space: $k(\omega)\approx k_0 + (\omega - \omega_0)/v_\mathrm g + \frac12k''(\omega - \omega_0)^2$, and the same algebra gives a duration $\tau(z) = \tau_0\sqrt{1 + (k''z/\tau_0^2)^2}$.

BK7 at 800 nm: $k'' = 44.65\ \mathrm{fs^2\,mm^{-1}}$ from the Sellmeier formula. A pulse of intensity full width $10$ fs has $\tau_0 = 6.006$ fs; after $z = 10$ mm, $k''z = 446.5\ \mathrm{fs^2}$ and

$$\tau/\tau_0 = 12.42:\quad10\ \mathrm{fs}\to124\ \mathrm{fs}.$$

A centimetre of ordinary glass stretches the pulse twelvefold. Ultrafast optics spends much of its ingenuity undoing exactly this.

The pulse's intensity before and after $1$ cm of BK7, from the formula. Twelve times longer and twelve times lower, the same area: the energy is all still there, spread over $124$ fs. The dashed curve is the output magnified tenfold.

20 · summary

The galvanometer, closed, and where this goes

one oscillator, every section

1 $\chi = \hat G$. 2 $LG = \delta$. 3 Two poles on $|\omega| = \omega_0$; the Taylor series dies at resonance. 4 Residues $\mp e^{-i\omega_\pm t}/2\omega_\mathrm d$. 5 Close up: $G = 0$ for $t < 0$; close down: unit 06's $G$; Kramers–Kronig to seven digits. 6 $\chi\sim-1/\omega^2$: inertia, and the sum rule.

unit	what it takes from here
24	The wave operator's retarded Green's function in three dimensions, by residues, with the $i0$ prescription made explicit; its two-dimensional cousin, which has a tail.
38	Gravitational waves, radiated through that same retarded Green's function.

three things to carry forward

1. A linear time-invariant system is one function of frequency, and its poles are its free modes.

2. Causality is holomorphy in a half-plane: it decides which way a contour closes, which Green's function is physical, and why dispersion and absorption come together.

3. An integral is decided by its critical points — the maximum, the stationary point, the saddle — once the contour has been moved to where there is no cancellation left.

Reading. For the complex analysis: Stein & Shakarchi, Complex Analysis, ch. 1–3 — the route here (Goursat, toy contours, residues) is theirs — and ch. 4 for the Fourier transform done with contours, including the Paley–Wiener theorem, the sharp form of "supported on a half-line iff holomorphic in a half-plane". Ahlfors, Complex Analysis, ch. 4 is the reference for the general Cauchy theorem cited here. For distributions: Strichartz, A Guide to Distribution Theory and Fourier Transforms, ch. 1–4, which is short and physical; it states results on $\mathbb R^n$, so read every statement with $n = 1$ and skip the passages on partial derivatives and several-variable integrals, which this course has not built; skip ch. 5 onwards. For causality: Nussenzveig, Causality and Dispersion Relations, ch. 1, and Toll's 1956 paper in Physical Review, which proves that causality alone implies the dispersion relations. For asymptotics: Bender & Orszag, Advanced Mathematical Methods for Scientists and Engineers, ch. 6 — the examples are superb, and the asymptotic notation it assumes is the one defined on the Watson slide. Jackson's Classical Electrodynamics §7.5 and §7.10 on dispersion and Kramers–Kronig will read easily after unit 24, and not before.

What was deliberately left out. The Fourier transform on $L^2$ beyond the statement of Plancherel's extension; the Paley–Wiener theorems (cited above); Riemann surfaces, beyond the remark that branch cuts are a choice; conformal mapping, which two-dimensional electrostatics can use but this course does not need; the Mellin transform; uniform asymptotics near coalescing saddles (Airy-type transitions); the Stokes phenomenon. Each is a real subject and none of them is needed to reach the field equations.

The debts, settled. Unit 06: the delta function (section 2), the transform (section 1), the causality slide (section 5), the fundamental theorem of algebra (section 3). Unit 09: the Laplace limit (P10), the Bessel constants (stationary phase slide and P11), and the growth rate of $I_n(ne)$ and $J_n(ne)$ in $n$ (steepest descent slide, notes). Lagrange's inversion formula by residues is part of P10.

20 · checkpoint

Done when

checkpoint

On paper and without notes, you can:

compute a retarded Green's function by contour integration from its frequency response — locate the poles, compute the residues — and prove it vanishes for $t < 0$ by closing the contour in the half-plane with no poles and bounding the arc, not by inspecting the answer;
evaluate a real integral by steepest descent: find the saddle, move the contour through it, and justify every arc and connecting segment by an explicit estimate;
settle unit 09's Laplace limit: prove that Lagrange's series for Kepler's equation converges for every $M$ when $e < 0.6627434$ and diverges at $M = \pi/2$ when $e > 0.6627434$.

Failing it is unambiguous

A Green's function nonzero before the kick; a contour closed where the exponential grows; an arc dropped without a bound; the wrong saddle or a lost $e^{i\pi/4}$; a Laplace limit asserted, not proved.

21 divergent series · 24 Maxwell, where this contour is closed in four dimensions.

Try the checkpoint on something not in the deck before opening the sheet. A good one: the critically damped galvanometer, $\gamma = 2\omega_0 = 16\ \mathrm{s^{-1}}$, where $\chi = -1/(\omega + i\omega_0)^2$ has a double pole at $-8i$. Closing upwards for $t < 0$ gives $0$ (no pole above, $|\chi|\le1/(R - 8)^2$ on the arc). Closing downwards for $t > 0$, clockwise, the double-pole formula gives $\operatorname{Res}\,(-e^{-i\omega t})/(\omega + 8i)^2 = -\frac{\mathrm d}{\mathrm d\omega}e^{-i\omega t}\big|_{-8i} = ite^{-8t}$, so $G(t) = \frac{1}{2\pi}(-2\pi i)(ite^{-8t}) = te^{-8t}$ — unit 06's critically damped kick, whose maximum $1/(8e) = 0.04598$ s occurs at $t = 1/8$ s. Then do P12 for the steepest-descent clause and P10 for the third.

Every number on this slide and in this deck was computed. If yours disagree, the arithmetic is the thing to re-read first.

The Fourier transform, complex analysis, and why causality is a statement about poles

One oscillator, inverted, and six sections

The Fourier transform

From one frequency to all of them

Differentiation becomes multiplication

Inversion

Inversion at a jump: the box

Plancherel and convolution

The uncertainty relation

Distributions

Test functions and distributions

Pulses that become δ

Derivatives by duality, and $LG = \delta$

The jump condition, proved

Tempered distributions and their transforms

Holomorphic functions

Complex differentiability, and Cauchy–Riemann

Integrals along curves

Goursat's theorem

Goursat with one exceptional point

Cauchy's theorem and the integral formula

What the integral formula buys

The Taylor series dies at resonance

Analytic continuation

Singularities and residues

Laurent series and isolated singularities

The residue theorem

Closing the contour: Jordan's lemma

Worked: the Dirichlet integral, with an indentation

Half a residue: the indentation lemma

Branch points and cuts

Counting zeros: the argument principle and Rouché

Where the roots of $z^5 + 3z + 1$ are

Causality is a statement about poles

Worked: the galvanometer's Green's function by residues

Worked: closing downwards, and unit 06's $G$ returns

The step function, explained

Retarded, advanced, and time's arrow

Causality, analyticity, Kramers–Kronig: one theorem

The dispersion relations, proved

Worked: Kramers–Kronig on the galvanometer

Kramers–Kronig on the galvanometer: the sum rule

Why glass separates colours

BK7: why blue bends more

The Laplace transform and the Bromwich contour

The running example, turned a quarter-turn

Asymptotics of integrals

Watson's lemma

Laplace's method

Stirling's formula, by Laplace's method

Stationary phase

Unit 09's debt: the constants in $J_n$

Steepest descent

Worked: an exponentially small integral by steepest descent

Worked: on the line through the saddle

Wave packets: phase and group velocity

Packets spread

A 10 fs pulse through 1 cm of glass

The galvanometer, closed, and where this goes

Done when

Failing it is unambiguous

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