KindiakMath

Cauchy’s Residue Theorem

We developed the notion of Laurent series in order to more easily compute integrals. In particular, the powerful technique of Cauchy’s residue theorem becomes accessible to our use.

Lemma 1. Suppose f : \mathbb C \to \mathbb C is holomorphic on \bar B(z_0, r) \backslash \{z_0\} with Laurent series

\displaystyle f(z) = \sum_{k=-\infty}^\infty c_k \cdot (z - z_0)^k.

Defining \Gamma_0 := \partial B(z_0, r) with anti-clockwise direction (i.e. parameterisation \mathbf r(t) = z_0 + re^{it} for t \in [0, 2\pi]),

\displaystyle \oint_{\Gamma_0} f(z)\, \mathrm dz = 2 \pi i \cdot c_{-1}.

Proof. By the uniqueness of Laurent coefficients,

\displaystyle c_{-1} = \frac 1{2\pi i} \oint_{\Gamma_0} \frac{f(z)}{ (z-z_0)^{-1+1} }\, \mathrm dz = \frac 1{2\pi i} \oint_{\Gamma_0} f(z)\, \mathrm dz.

Theorem 1 (Cauchy’s Residue Theorem). Using the notation in Lemma 1, define the residue of f at z_0 by \mathrm{Res}(f, z_0). Suppose f is holomorphic on an open set U containing a simple closed contour \Gamma = \partial D (directed anti-clockwise) and its interior D, except for isolated singularities at z_1,\dots, z_n. Then

\displaystyle \oint_{\Gamma} f(z)\, \mathrm dz = 2\pi i \cdot \sum_{k=1}^n \mathrm{Res}(f, z_k).

Note that if f has no singularities, then we recover the Cauchy-Goursat theorem.

Proof. Fix circles \Gamma_k := B(z_k, r) for sufficiently small r > 0, so that the hypotheses in Lemma 1 are satisfied, and that

\displaystyle \oint_{\Gamma_k} f(z)\, \mathrm dz = 2\pi i \cdot \mathrm{Res}(f, z_k).

By the Cauchy-Goursat theorem and Cauchy’s integral formula,

\begin{aligned} \oint_{\Gamma}f(z)\, \mathrm dz &= \sum_{k=1}^n \oint_{\Gamma_k} f(z)\, \mathrm dz = 2\pi i \cdot \sum_{k=1}^n \mathrm{Res}(f, z_k). \end{aligned}

That felt a little anti-climatic, no? Well, the key trick is this: if we have simple techniques to compute the right-hand side, then the left-hand side no longer requires the painful work of re-parametrisation, and becomes much easier to compute.

Lemma 2. Suppose f has a pole of order m at z_0. Then

\displaystyle \mathrm{Res}(f, z_0) = \lim_{z \to z_0} \left[ \frac{\mathrm d^{n-1}}{\mathrm dz^{m-1}} ( (z-z_0)^m f(z) ) \right].

Proof. Consider the Laurent series of f given by

\displaystyle f(z) = \sum_{k=-m}^\infty c_k (z- z_0)^k,

since z_0 is a pole of order m. Then

\displaystyle (z-z_0)^m f(z) = \sum_{k = 0}^\infty c_{k-m} (z - z_0)^{k}

has a removable singularity at z_0 and extends to a function g holomorphic at z_0. Taking (m-1) derivatives,

\displaystyle g^{(m-1)}(z) = \sum_{k = m-1}^\infty c_{k-m} \cdot \frac{k!}{(k-m+1)!} \cdot (z - z_0)^{k-(m-1)}

Setting z = z_0,

\begin{aligned} \lim_{z \to z_0} \frac{\mathrm d^{n-1}( (z-z_0)^m f(z) )}{\mathrm dz^{m-1}} &= g^{(m-1)}(z_0) \\ &= (m-1)! \cdot c_{-1} \\ &= (m-1)! \cdot \mathrm {Res}(f, z_0), \end{aligned}

as required.

I’m gonna be honest—that looks even more frightening than the basic Cauchy residue theorem. True. But something beautiful happens when we deal only with simple poles.

Corollary 1. Suppose f has a simple at z_0. Then

\displaystyle \mathrm{Res}(f, z_0) = \lim_{z \to z_0} [ (z-z_0) f(z) ].

Now that looks far more pleasant. Let’s test out this new machinery with a not-so-trivial example.

Example 1. Evaluate \displaystyle \lim_{R \to \infty} \int_{-R}^{R} \frac{\cos x + \sin x}{1 + x^2}\, \mathrm dx.

Solution. Recall that \sin x + \cos x = \sqrt 2 \sin(x + \pi/4), so that we want to equivalently evaluate

\displaystyle \lim_{R \to \infty} \int_{-R}^{R} \frac{\sqrt 2 \sin(x + \pi/4)}{1 + x^2}\, \mathrm dx.

Define the holomorphic function f : \mathbb C \backslash \{\pm i\} \to \mathbb C by

\displaystyle f(z) = \frac{e^{i z }}{1 + z^2},

so that the desired integral is

\begin{aligned} \displaystyle \lim_{R \to \infty} \int_{-R}^{R} \frac{\sin(x + \pi/4)}{1 + x^2}\, \mathrm dx &= \mathrm{Im} \left(\sqrt 2 e^{i \pi/4} \cdot \lim_{R \to \infty} \int_{-R}^{R} \frac{e^{iz}}{1 + z^2}\, \mathrm dz\right) \\ &= \mathrm{Im} \left(\sqrt 2 e^{i \pi/4} \cdot \lim_{R \to \infty} \int_{-R}^{R} f(z)\, \mathrm dz\right). \end{aligned}

Consider the contour \Gamma_R : -R \to R \to -R directed anti-clockwise comprising of \Gamma_1 := [-R, R] and the anti-clockwise semicircle with centre 0 and radius R parameterised by \Gamma_2 := r(t) = Re^{it}, t \in [0, \pi]. By definition,

\displaystyle \int_{-R}^{R} f(z)\, \mathrm dz = \oint_{\Gamma_1} f(z)\, \mathrm dz.

For R > 2, since |z| = R so that

|1 + z^2| \geq |z|^2 - 1 = R^2 - 1 > 0,

|e^{iz}| \leq 1 on \Gamma_2, and \| \Gamma_2 \| = R \pi, by the ML-inequality,

\displaystyle \left| \oint_{\Gamma_2} f(z)\, \mathrm dz \right| = \left| \oint_{\Gamma_2} \frac{e^{iz}}{1 + z^2}\, \mathrm dz \right| \leq \frac 1{R^2 - 1} \cdot R \pi.

On the other hand, since the only residue of f in \Gamma is the simple pole i, by Cauchy’s residue theorem,

\begin{aligned} \oint_{\Gamma} f(z)\, \mathrm dz &= 2\pi i \cdot \mathrm{Res}(f, i) \\ \int_{\Gamma_1} f(z)\, \mathrm dz + \int_{\Gamma_2} f(z)\, \mathrm dz &= 2 \pi i \cdot \lim_{z \to i} [ (z-i) f(z) ] \\ \int_{-R}^{R} f(z)\, \mathrm dz + \int_{\Gamma_2} f(z)\, \mathrm dz &= 2 \pi i \cdot \lim_{z \to i} \left[ (z-i) \cdot \frac{e^{iz}}{z^2 + 1} \right] \\ &= 2 \pi i \cdot \lim_{z \to i} \left[ \frac{e^{iz}}{z + i} \right] \\ &= 2 \pi i \cdot \frac{e^{i \cdot i}}{i + i} = \frac{\pi}{e}. \end{aligned}

Then

\begin{aligned} \left| \int_{-R}^{R} f(z)\, \mathrm dz - \frac{\pi}{e} \right| &= \left| \int_{\Gamma_2} f(z)\, \mathrm dz \right| \leq \lim_{R \to \infty}\frac{R}{R^2 - 1}. \end{aligned}

Taking R \to \infty and using continuity,

\displaystyle \lim_{R \to \infty } \int_{-R}^{R} \frac{e^{iz}}{z^2+1}\, \mathrm dz = \frac{\pi}{e}.

Therefore,

\begin{aligned}\lim_{R \to \infty} \int_{-R}^{R} \frac{\sin(x + \pi/4)}{1 + x^2}\, \mathrm dx &= \mathrm{Im} \left(\sqrt 2 e^{i \pi/4} \cdot \frac{\pi}{e} \right) \\ &= \frac{\pi}{e} \cdot \sqrt 2 \sin(\pi/4) = \frac {\pi}{e}. \end{aligned}

A rather surprising integral! I’m convinced there’s no purely real-variable calculus method to compute this integral. The satisfying part of this example is its utility of the all of the topics we have discussed before: holomorphicity, contour integration, the ML-inequality, and Cauchy’s residue theorem that in turn required the massive complex-analytic theorems: the Cauchy-Goursat theorem, Cauchy’s integral formula, and Laurent series.

This technique is also often known as the calculus of residues, whose tools we finally have access to in studying Fourier transforms.

—Joel Kindiak, 25 Aug 25, 1445H

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