# Evaluating integrals using the residue theorem

1. Nov 13, 2007

### sschmidt

Hi,

I have trouble evaluating simple integrals like

$$\int_{-\infty}^{\infty} \frac{dx}{\sqrt{x^2 + 2}(x^2+1)} = \frac{\pi}{2}$$
I'd like to calculate the integral closing the integration loop in the upper half-plane enclosing the pole at +i. The residue is - i / 2 and hence 2 \pi i ( - i \ 2 ) is \pi, but the integral should evaluate to \pi / 2. I just don't seem to see what is happening here. The other pole at - i shouldn't interfere and the residue at +i is not evaluated twice, so where does the missing factor 1/2 come from? Is this related to the Riemann sphere and its branch cuts? Does the contour with semicircle cross branch cuts and change the leaf?
What are the branch cuts here anyway?

The same problem occurs with

$$\int_{-\infty}^{\infty} \frac{dx}{\sqrt{x^2 + 4}(x^2+1)} = \frac{2}{3} \pi i (-\frac{i}{\sqrt{3}})$$

Thank you very much and kind regards, Sven

Last edited: Nov 13, 2007