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Mathematical Methods

The residue theorem and real integrals

Residues and how to compute them, the residue theorem, integrals on the line with a semicircle, Jordan's lemma, the Dirichlet integral, trigonometric integrals and the argument principle.

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The residue theorem is the most used result of complex analysis. It reduces the computation of an integral round a closed curve to the sum of a few numbers, the residues, which are found with derivatives or limits. And it lets you compute real integrals that would be impossible with the methods of Analysis 1: exactly those that appear in Fourier and Laplace transforms.

The residue

The residue of f at an isolated singularity z_0 is the coefficient a_{-1} of the Laurent series, the one of \frac{1}{z - z_0}:

    \[ \operatorname{Res}(f, z_0) = a_{-1} = \frac{1}{2\pi i}\oint_{|z - z_0| = r}f(z)\,dz \]

The equality follows by integrating the Laurent series term by term: from the lesson on Cauchy’s theorem, every term (z - z_0)^n integrates to zero except n = -1, which gives 2\pi i. The residue is “what is left” after integration.

The residue theorem

Theorem. If f is holomorphic inside and on a simple closed curve \gamma, traversed anticlockwise, except at finitely many interior singularities z_1, \dots, z_k, then

    \[ \oint_\gamma f(z)\,dz = 2\pi i\sum_{j=1}^{k}\operatorname{Res}(f, z_j) \]

Proof. By Cauchy’s theorem \gamma can be deformed into small circles, one around each singularity (as in the figure of the lesson on Cauchy’s theorem). The integral round each is 2\pi i times the corresponding residue.

Computing residues

  • Simple pole: \operatorname{Res}(f, z_0) = \lim_{z\to z_0}(z - z_0)f(z).
  • Simple pole of a ratio f = \frac{g}{h}, with g(z_0) \ne 0, h(z_0) = 0, h'(z_0) \ne 0: \operatorname{Res} = \frac{g(z_0)}{h'(z_0)}. The handiest formula.
  • Pole of order m:

    \[ \operatorname{Res}(f, z_0) = \frac{1}{(m - 1)!}\lim_{z\to z_0}\frac{d^{m-1}}{dz^{m-1}}\Big[(z - z_0)^m f(z)\Big] \]

  • Essential singularity: write the Laurent series and read off a_{-1}.

Examples. \frac{e^z}{z^2 + 1} at z = i: \frac{e^i}{2i}. \frac{\cos z}{z^3} at 0: triple pole, \frac{1}{2!}\frac{d^2}{dz^2}\cos z\big|_0 = -\frac12. e^{1/z} at 0: from the series 1 + \frac1z + \dots, the residue is 1.

Real integrals with a semicircle

We want \int_{-\infty}^{\infty}f(x)\,dx. The idea is to close the segment [-R, R] with a semicircle \Gamma_R in the upper half-plane, apply the residue theorem and show that the integral over the semicircle tends to zero.

Semicircular contour in the upper half-plane with one pole inside and one outside
The closed contour is made of the segment from −R to R on the real axis and the semicircle Γ. For R large enough it encloses the pole i, while the pole −i stays outside. If the integral over the semicircle tends to zero, the real integral is 2πi times the sum of the residues in the upper half-plane.

Example.

    \[ I = \int_{-\infty}^{\infty}\frac{dx}{1 + x^2} \]

The function \frac{1}{1 + z^2} has poles at \pm i; only i is inside the contour, with residue \frac{1}{2i}. On the semicircle |f| \le \frac{1}{R^2 - 1} and the length is \pi R, so the semicircle integral is at most \frac{\pi R}{R^2 - 1}\to 0. What remains is

    \[ I = 2\pi i\cdot\frac{1}{2i} = \pi \]

The same result comes from the arctangent, but the method also works when the antiderivative is unknown. General rule: if f = P/Q is rational, with no poles on the real axis and the degree of Q at least two more than the degree of P, the semicircle integral tends to zero and

    \[ \int_{-\infty}^{\infty}\frac{P(x)}{Q(x)}\,dx = 2\pi i\sum_{\operatorname{Im}z_j \gt 0}\operatorname{Res}\left(\frac PQ, z_j\right) \]

Jordan’s lemma and Fourier integrals

For integrals like \int_{-\infty}^{\infty}f(x)\,e^{i\omega x}dx, which are Fourier transforms, the condition is weaker: with \omega \gt 0 it is enough that f\to 0 on the upper semicircle (Jordan’s lemma), because |e^{i\omega z}| = e^{-\omega y} is small where y \gt 0.

Example. \int_{-\infty}^{\infty}\frac{\cos x}{1 + x^2}dx = \operatorname{Re}\int\frac{e^{ix}}{1 + x^2}dx. Residue of \frac{e^{iz}}{1 + z^2} at i: \frac{e^{-1}}{2i}. So the integral is 2\pi i\cdot\frac{e^{-1}}{2i} = \frac{\pi}{e} \approx 1.156.

The Dirichlet integral. With a small semicircle to go round the pole at the origin one gets

    \[ \int_{-\infty}^{\infty}\frac{\sin x}{x}\,dx = \pi \]

It is the area under the sinc function, which we will meet again as the Fourier transform of the rectangular pulse.

Trigonometric integrals

Integrals \int_0^{2\pi}R(\cos\theta, \sin\theta)\,d\theta become integrals round the unit circle with z = e^{i\theta}, \cos\theta = \frac{z + z^{-1}}{2}, \sin\theta = \frac{z - z^{-1}}{2i}, d\theta = \frac{dz}{iz}.

Example. \int_0^{2\pi}\frac{d\theta}{2 + \cos\theta} = \oint\frac{1}{2 + \frac{z + 1/z}{2}}\cdot\frac{dz}{iz} = \oint\frac{2\,dz}{i(z^2 + 4z + 1)}. The roots of the denominator are -2\pm\sqrt3; only z_1 = -2 + \sqrt3 is inside the unit circle. The residue of \frac{2}{i(z^2 + 4z + 1)} is \frac{2}{i\cdot 2\sqrt3}, so the integral is 2\pi i\cdot\frac{1}{i\sqrt3} = \frac{2\pi}{\sqrt3}.

The argument principle

One last application, essential in control engineering: for a function with zeros and poles, \frac{1}{2\pi i}\oint\frac{f'(z)}{f(z)}dz = Z - P, the number of zeros minus the number of poles inside the curve. Geometrically it is the number of times the image f(\gamma) winds round the origin. The Nyquist criterion uses exactly this to decide whether a feedback system is stable, looking only at the open-loop frequency response.

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