# Tricky contour integral

1. Apr 24, 2007

Contour integral

How would you deal with this?

$$\int \frac{\rho \sin{\theta} d \rho d \theta}{\cos{\theta}} \frac{K^2}{K^2 + \rho^2} e^{i \rho \cos{\theta} f(\mathbf{x})}$$

if the cos(theta) were'nt on the bottom I'd have no problem; I'd simply substitute for cos(theta) and the sin(theta) would cancel...

but as it stands.. I'm stumped.

Help/hints would be much appreciated. Thanks for reading.

Last edited: Apr 24, 2007
2. Apr 24, 2007

### Dick

Are you sure there should be a cos(theta) in the denominator? Where did that expression come from?

3. Apr 24, 2007

$$\int \frac{d^3 \mathbf{q}}{\mathbf{q_z}} \frac{2K^2 + \mathbf{q}^2 - (\mathbf{q} \cdot \hat{x})^2}{2(K^2 + \mathbf{q}^2)}e^{i \mathbf{q_{\perp}} \cdot \mathbf{x}}e^{i \mathbf{q_z} f(\mathbf{x})}$$

I made q_z parallel to x, (which means the first exponential disappears), converted into spherical coords, integrated over \phi from 0 to 2pi, and thus...

Last edited: Apr 24, 2007
4. Apr 24, 2007

### Dick

I guess I'm stumped as well. Anybody else seen something like that?

5. Apr 24, 2007