- #1

jncarter

- 49

- 0

## Homework Statement

Considering the appropriate complex integral along a semi-circular contour on the upper half plan of z, show that

[itex] \int^{\infty}_{\infty} \frac{cos(ax)}{x^2 + b^2} dx = \frac{\pi}{b}e^{-ab} (a>0, b>0) [/itex]

## Homework Equations

[itex] \int_{C} = 0 [/itex] For C is a semi-circle of infinite radius in the upper half of the complex plane.

[itex] \oint f(z)dz = 0 [/itex] for analytic functions

[itex]\oint f(z)dz = 2\pi i \sum Res[f(z_{0})] [/itex] Where z

_{0}is a singularity and you sum over all the residues within the contour.

[itex] z^2 + b^2 = (z+ib)(z-ib) [/itex]

## The Attempt at a Solution

I've been trying to solve this by using [itex] \int \frac{e^{iaz}}{z^2 + b^2}dz [/itex]. 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 [itex]\pm ib[/itex].

Is there a better contour integral I could use? Right now the algebra is getting messy.