- #1
MadMax
- 99
- 0
How would you tackle this?
[tex]\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}[/tex]
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 canceled 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.
[tex]\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}[/tex]
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 canceled 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.
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