# Definite integral using complex analysis

jncarter

## Homework Statement

Considering the appropriate complex integral along a semi-circular contour on the upper half plan of z, show that
$\int^{\infty}_{\infty} \frac{cos(ax)}{x^2 + b^2} dx = \frac{\pi}{b}e^{-ab} (a>0, b>0)$​

## Homework Equations

$\int_{C} = 0$ For C is a semi-circle of infinite radius in the upper half of the complex plane.
$\oint f(z)dz = 0$ for analytic functions
$\oint f(z)dz = 2\pi i \sum Res[f(z_{0})]$ Where z0 is a singularity and you sum over all the residues within the contour.
$z^2 + b^2 = (z+ib)(z-ib)$

## The Attempt at a Solution

I've been trying to solve this by using $\int \frac{e^{iaz}}{z^2 + b^2}dz$. Which must be solved using Cauchy's Formula (the residue thing above), since f(z) is not analytic over all space. There is a singularity of order one at $\pm ib$.
Is there a better contour integral I could use? Right now the algebra is getting messy.

## Answers and Replies

Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
Show us your work. You must be doing something wrong because that contour integral is pretty straightforward to evaluate.

jncarter
I tried using the contour integral $\int^{\infty}_{\infty} \frac{e^{-iaz}}{z^2+b^2}dz$ and found that it was equal to $\frac{\pi}{b}e^{-ab}$. Now I think I have to consider that a sum of different integrals. We have the top hemisphere, the x-axis and the circular contour about z0 and the lines leading to +ib. I'm finding that must of those are zero and I think I can pull out the cosine function from the integral along the x-axis...

I think before I was just forgetting how to factor...