Complicated vector integral?

1. Jun 6, 2007

How would you tackle this?

$$\int \frac{db^3}{(2 \pi)^3} \frac{2K^4 - a^2 b^2 + (a \cdot b)^2}{2K^2 b^2 + b^4 + b^2a^2 / 2 - (a \cdot b)^2 + A}$$

The way I tried was expressing the whole thing in spherical polar coords, setting a_z parallel to b, and then carrying out the angular parts of the integration. Unfortunately this involved quite a few partial fractions... This in turn led to complicated logs popping up all over the place. Also there were some obvious singularities which came out, (which would be cancelled out by singularities hidden in the log terms). Then would come the scary task of integrating that mess over the non angular dimension... In a word "yuck"...

Can't help thinking that I'm missing a trick, or that there is a better way to tackle it.

EDIT: Hmmm I've just had an idea. I'll try to tackle the angular integral not as a standard integral, but instead using contour integration... I'll keep you posted.

Still any ideas/hints/suggestions would be much appreciated. Thanks.

Last edited: Jun 6, 2007
2. Jun 7, 2007

hmm nevermind...

Last edited: Jun 7, 2007
3. Jun 7, 2007

esalihm

numerator and denominator might have some common facton if you go into binomial expressions

can't get it though...

4. Jun 7, 2007

OK I think I found a method in some text books by which it can be done: "The residue theorem".

An example relevant for the angular part of the integration I'm considering is

$$\int^{2 \pi} _{0} F(\cos{\theta}, \sin{\theta}) d \theta$$

We use the subtitution

$$z=e^{i \theta}$$

But why does the integral become a contour integral, where the contour is the unit circle centred around the origin?

Presumably if we were integrating from 0 to pi, the contour would be a cemi-circle? But which plane would we use? Upper or lower, or is it right hand or left hand? Also why is it of unit radius? Why would it be incorrect to have the radius of the contour larger or smaller?