example of using residue theorem

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

We shall the integral

$$I:={\int }_0^{\mathrm{\infty }}\frac{\cos kx}{1+x^2}𝑑x$$

where $k$ is any real number.  One may prove that the integrand has no antiderivative among the elementary functions if  $k\ne 0$.

Since the integrand is an even (http://planetmath.org/EvenoddFunction) and  $x\mapsto \frac{\sin kx}{1+x^2}$  an odd function, we may write

$$I=\frac{1}{2}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{\cos kx+i\sin kx}{1+x^2}𝑑x=\frac{1}{2}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{e^{ikx}}{1+x^2}𝑑x,$$

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 $(-R,\mathrm{\hspace{0.17em}0})$ to $(R,\mathrm{\hspace{0.17em}0})$ 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)

$$\mathrm{Res}(\frac{e^{ikz}}{1+z^2},i)=\underset{z\to i}{lim}(z-i)\frac{e^{ikz}}{z^2+1}=\underset{z\to i}{lim}\frac{e^{ikz}}{z+i}=\frac{e^{-k}}{2i},$$

the residue theorem (http://planetmath.org/CauchyResidueTheorem) yields

$$J=\mathrm{\hspace{0.33em}2}\pi i\cdot \frac{e^{-k}}{2i}=\pi e^{-k}.$$

This does not depend on the radius $R$ of the circle.

We split the integral $J$ in two portions:  one along the diameter and the other along the circular arc $c$.  So we obtain

$${\int }_{-R}^R\frac{e^{ikx}}{1+x^2}𝑑x+{\int }_c\frac{e^{ikz}}{1+z^2}𝑑z=\pi e^{-k}.$$

When  $R\to \mathrm{\infty }$,  the former portion tends to the limit $2I$ and the latter — as we at once shall see — to the limit 0.  Hence we get the result

$$I={\int }_0^{\mathrm{\infty }}\frac{\cos kx}{1+x^2}𝑑x=\frac{\pi }{2e^k}.$$

As for the latter part of $J$, we denote  $z:=x+iy$ ($x,y\in ℝ$);  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  $R\to \mathrm{\infty }$, then also the left hand member.