# Cauchy's Theorem

1. Jun 24, 2011

### bruno67

A book on which I am studying (Arfken: Mathematical Methods for Physicists), uses the following result in order to derive an asymptotic expansion:

$$\int_{0}^{-i\infty} \frac{e^{-xu}}{1+iu}du = \int_{0}^{\infty} \frac{e^{-xu}}{1+iu}du,$$
where the change of limits in the integral is justified by invoking Cauchy's theorem. I am familiar with Cauchy's theorem, but I am not sure why it justifies this passage. How does it work?

2. Jun 24, 2011

### nonequilibrium

The only singularity of the denominator is for u = i, so for any path that doesn't go around i, cauchy's theorem gives that the contour integral is zero. Now which path to take? Take a quarter of a circle centered around zero and with radius R; take the quarter that lies in the region where Re(u) > 0 and Im(u) < 0, in other words "at the bottom right". So we have a contour with 3 "sides": a straight line from 0 to R, one on a piece of a circle going from R to -iR and then a line from -iR to 0. If we can prove that the contribution of the circle segment goes to zero for R -> infinity, then we have the statement in your OP.

However, I do not know how to prove that the contribution goes to zero...

3. Jun 25, 2011

### bruno67

Thanks for your help. I'll try to prove the remaining part and will post here the results, if it works.