Tricky contour integral

  • Thread starter MadMax
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  • #1
MadMax
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Contour integral

How would you deal with this?

[tex]\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})}[/tex]

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:

Answers and Replies

  • #2
Dick
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Are you sure there should be a cos(theta) in the denominator? Where did that expression come from?
 
  • #3
MadMax
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well I had something like:

[tex]\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})}[/tex]

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...
 
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  • #4
Dick
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I guess I'm stumped as well. Anybody else seen something like that?
 
  • #5
MadMax
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I wasn't correct in some of the things I wrote in the two equations. The second exponential in particular...

I edited them so they're correct now. Not sure if it makes a difference but perhaps it helps? Might make sense how I got from one to the other now anyway...
 

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