We take an example of applying the Cauchy residue theorem in evaluating usual real improper integrals.

We shall calculate the integral $$I := \int_0^\infty\frac{\cos{kx}}{1+x^2}\,dx$$ where $k$ is any real number. One may prove that the integrand has no antiderivative among the elementary functions if $k \neq 0$ .

Since the integrand is an even and $x\mapsto\frac{\sin{kx}}{1+x^2}$ an odd function, we may write $$I = \frac{1}{2}\int_{-\infty}^\infty\frac{\cos{kx}+i\sin{kx}}{1+x^2}\,dx = \frac{1}{2}\int_{-\infty}^\infty\frac{e^{ikx}}{1+x^2}\,dx,$$ using also Euler's formula. Let's consider the contour integral $$J := \oint_\gamma\frac{e^{ikz}}{1+z^2}\,dz$$ where $\gamma$ is the perimeter of the semicircle consisting of the line segment from $(-R,\,0)$ to $(R,\,0)$ and the semi-circular arc $c$ connecting these points in the upper half-plane ($R > 1$ ). The integrand is analytic on and inside of $\gamma$ except in the point $z = i$ which is a simple pole. Because we have (cf. the coefficients of Laurent series) $$\operatorname{Res}\left(\frac{e^{ikz}}{1+z^2},\;i\right) = \lim_{z\to i}(z-i)\frac{e^{ikz}}{z^2+1} = \lim_{z\to i}\frac{e^{ikz}}{z+i} = \frac{e^{-k}}{2i},$$ the residue theorem yields $$J = 2\pi i\!\cdot\!\frac{e^{-k}}{2i} = \pi e^{-k}.$$ This does not depend on the radius $R$ of the circle.

As for the latter part of $J$ , we denote $z := x+iy$ ($x,\,y\in\mathbb{R}$ ); then on the arc $c$ , where $|z| = R$ and $y\geqq 0$ , we have $$\left|\frac{e^{ikz}}{1+z^2}\right| = \frac{|e^{-ky+ikx}|}{|1+z^2|} = \frac{e^{-ky}}{|1+z^2|} \leqq \frac{1}{R^2\!-\!1}.$$ Using this estimation of the integrand we get, according the integral estimating theorem, the inequality $$\left|\int_c\frac{e^{ikz}}{1+z^2}\,dz\right| \leqq \frac{1}{R^2\!-\!1}\!\cdot\!\pi R = \frac{\pi}{R\!-\!\frac{1}{R}}.$$ Since the right hand member tends to 0 as $R\to\infty$ , then also the left hand member.