# example of using residue theorem

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

 $I\;:=\;\int_{0}^{\infty}\frac{\cos{kx}}{1+x^{2}}\,dx$

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 (http://planetmath.org/EvenoddFunction) 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 (http://planetmath.org/CauchyResidueTheorem) yields

 $J\;=\;2\pi i\!\cdot\!\frac{e^{-k}}{2i}\;=\;\pi e^{-k}.$

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  $R\to\infty$,  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

 $I\;=\;\int_{0}^{\infty}\frac{\cos{kx}}{1+x^{2}}\,dx\;=\;\frac{\pi}{2e^{k}}.$

As for the latter part of $J$, we denote  $z:=x+iy$ ($x,\,y\in\mathbb{R}$);  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.

Title example of using residue theorem ExampleOfUsingResidueTheorem 2013-03-22 15:19:30 2013-03-22 15:19:30 pahio (2872) pahio (2872) 19 pahio (2872) Example msc 30E20 ImproperIntegral Residue  IntegralsOfEvenAndOddFunctions UsingResidueTheoremNearBranchPoint