Applications of Contour Integration

Contour integration is a powerful tool for evaluating definite integrals.

Theorem 9.1. If $f(x) = P(x)/Q(x)$ where $\deg(Q) \geq \deg(P) + 2$ and $Q$ has no real roots, Then

$\int_{-\infty}^{\infty} f(x)\, dx = 2\pi i \sum_{\mathrm{Im}(z_k) > 0} \mathrm{Res}(f, z_k)$

Where the sum is over poles in the upper half-plane.

Proof. Integrate $f(z)$ over the semicircular contour $\gamma_R$ consisting of $[-R, R]$ on the Real axis and the semicircle $|z| = R$ in the upper half-plane. As $R \to \infty$The integral over The semicircle vanishes (since $|f(z)| \leq M/R^2$ and the length is $\pi R$). $\blacksquare$

Problem. Evaluate $\int_{-\infty}^{\infty} \frac{dx}{x^2 + 1}$.

Solution. $f(z) = \frac{1}{z^2 + 1}$ has simple poles at $z = \pm i$.

Only $z = i$ is in the upper half-plane.

$\mathrm{Res}\left(\frac{1}{z^2 + 1}, i\right) = \frac{1}{2z}\Big|_{z = i} = \frac{1}{2i}$.

$\int_{-\infty}^{\infty} \frac{dx}{x^2 + 1} = 2\pi i \cdot \frac{1}{2i} = \pi$. $\blacksquare$

For integrals of the form $\int_0^{2\pi} R(\cos\theta, \sin\theta)\, d\theta$Substitute $z = e^{i\theta}$So $dz = iz\, d\theta$, $\cos\theta = \frac{z + z^{-1}}{2}$ $\sin\theta = \frac{z - z^{-1}}{2i}$.

The integral becomes $\int_{|z|=1} f(z)\, dz$ where $f(z)$ is a rational function.

Problem. Evaluate $\int_0^{2\pi} \frac{d\theta}{2 + \cos\theta}$.

Solution. Substitute $z = e^{i\theta}$: $d\theta = \frac{dz}{iz}$ $\cos\theta = \frac{z + 1/z}{2}$.

$\int_{|z|=1} \frac{dz}{iz\left(2 + \frac{z + 1/z}{2}\right)} = \int_{|z|=1} \frac{2\, dz}{i(z^2 + 4z + 1)}$

Poles: $z^2 + 4z + 1 = 0 \Rightarrow z = -2 \pm \sqrt{3}$.

$|z_1| = |-2 + \sqrt{3}| = 2 - \sqrt{3} \lt 1$ (inside). $|z_2| = |-2 - \sqrt{3}| = 2 + \sqrt{3} \gt 1$ (outside).

$\mathrm{Res}\left(\frac{1}{z^2 + 4z + 1}, z_1\right) = \frac{1}{2\sqrt{3}}$.

$\int_0^{2\pi} \frac{d\theta}{2 + \cos\theta} = \frac{2}{i} \cdot 2\pi i \cdot \frac{1}{2\sqrt{3}} = \frac{2\pi}{\sqrt{3}}$. $\blacksquare$

Theorem 9.2 (Jordan’s Lemma). If $f(z) \to 0$ uniformly as $|z| \to \infty$ in the upper Half-plane and $a \gt 0$Then

$\lim_{R \to \infty} \int_{C_R} e^{iaz}f(z)\, dz = 0$

Where $C_R$ is the upper semicircle $|z| = R$, $\mathrm{Im}(z) \geq 0$.

This allows evaluation of integrals of the form $\int_{-\infty}^{\infty} f(x)\cos(ax)\, dx$ and $\int_{-\infty}^{\infty} f(x)\sin(ax)\, dx$.

For integrals where the integrand has poles on the real axis, we use the Cauchy principal value:

$\mathrm{PV}\!\int_{-\infty}^{\infty} f(x)\, dx = \lim_{\varepsilon \to 0^+} \left(\int_{-\infty}^{a-\varepsilon} f(x)\, dx + \int_{a+\varepsilon}^{\infty} f(x)\, dx\right)$