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.

Title	example of using residue theorem
Canonical name	ExampleOfUsingResidueTheorem
Date of creation	2013-03-22 15:19:30
Last modified on	2013-03-22 15:19:30
Owner	pahio (2872)
Last modified by	pahio (2872)
Numerical id	19
Author	pahio (2872)
Entry type	Example
Classification	msc 30E20
Related topic	ImproperIntegral
Related topic	Residue
Related topic	IntegralsOfEvenAndOddFunctions
Related topic	UsingResidueTheoremNearBranchPoint