Evaluation of Real Integrals via Residues

One of the most powerful applications of the residue theorem is the evaluation of definite integrals that are difficult or impossible to compute by real-variable methods.

Trigonometric Integrals

Definition6.3Trigonometric integral method

To evaluate $\int_0^{2\pi} R(\cos\theta, \sin\theta)\,d\theta$ where $R$ is a rational function, substitute $z = e^{i\theta}$ :

$\cos\theta = \frac{z + z^{-1}}{2}, \quad \sin\theta = \frac{z - z^{-1}}{2i}, \quad d\theta = \frac{dz}{iz}.$

This converts the integral to a contour integral over $|z| = 1$ .

ExampleTrigonometric integral

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

With $z = e^{i\theta}$ : $I = \oint_{|z|=1} \frac{1}{2 + (z+z^{-1})/2}\cdot\frac{dz}{iz} = \oint_{|z|=1}\frac{2\,dz}{i(z^2 + 4z + 1)}$ .

The roots of $z^2 + 4z + 1 = 0$ are $z = -2 \pm \sqrt{3}$ . Only $z_0 = -2 + \sqrt{3}$ lies inside $|z|=1$ .

$\text{Res}\left(\frac{2}{i(z^2+4z+1)}, z_0\right) = \frac{2}{i \cdot 2z_0 + 4i} = \frac{2}{i \cdot 2\sqrt{3}} = \frac{1}{i\sqrt{3}}.$

Therefore $I = 2\pi i \cdot \frac{1}{i\sqrt{3}} = \frac{2\pi}{\sqrt{3}}$ .

Improper Integrals on the Real Line

Theorem6.2Semicircular contour method

If $f(z) = P(z)/Q(z)$ where $P, Q$ are polynomials with $\deg Q \geq \deg P + 2$ and $Q$ has no real zeros, then

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

The integral is evaluated by closing the contour with a large semicircle in the upper half-plane, whose contribution vanishes by the ML inequality.

ExampleIntegral of a rational function

Evaluate $\int_{-\infty}^\infty \frac{dx}{(x^2+1)(x^2+4)}$ .

Partial fractions or direct residue computation. Poles in upper half-plane: $z = i$ (simple) and $z = 2i$ (simple).

$\text{Res}(f, i) = \frac{1}{2i(i^2+4)} = \frac{1}{2i \cdot 3} = \frac{1}{6i}$

$\text{Res}(f, 2i) = \frac{1}{(4i^2+1)\cdot 4i} = \frac{1}{(-3)(4i)} = \frac{-1}{12i}$

$\int_{-\infty}^\infty f\,dx = 2\pi i\left(\frac{1}{6i} - \frac{1}{12i}\right) = 2\pi i \cdot \frac{1}{12i} = \frac{\pi}{6}.$

Integrals Involving Exponentials

Theorem6.3Jordan's lemma

If $f(z) \to 0$ uniformly as $|z| \to \infty$ on the upper half-plane and $a > 0$ , then

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

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

ExampleFourier-type integral

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

Consider $\int_{-\infty}^\infty \frac{xe^{ix}}{x^2+1}\,dx$ and take the imaginary part. The pole in the upper half-plane is $z = i$ :

$\text{Res}\left(\frac{ze^{iz}}{z^2+1}, i\right) = \frac{ie^{-1}}{2i} = \frac{1}{2e}.$

By Jordan's lemma: $\int_{-\infty}^\infty \frac{xe^{ix}}{x^2+1}\,dx = 2\pi i \cdot \frac{1}{2e} = \frac{\pi i}{e}$ .

Taking the imaginary part: $\int_{-\infty}^\infty \frac{x\sin x}{x^2+1}\,dx = \frac{\pi}{e}$ .

Integrals with Branch Cuts

RemarkKeyhole contour for branch cuts

For integrals like $\int_0^\infty \frac{x^{\alpha-1}}{1+x}\,dx$ ( $0 < \alpha < 1$ ), use a keyhole contour around the branch cut $[0, \infty)$ of $z^{\alpha-1}$ . The contributions from the two sides of the cut differ by a factor of $e^{2\pi i(\alpha-1)}$ , allowing extraction of the desired integral. The result is

$\int_0^\infty \frac{x^{\alpha-1}}{1+x}\,dx = \frac{\pi}{\sin(\pi\alpha)}.$