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Tricky contour integral

  1. Apr 24, 2007 #1
    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: Apr 24, 2007
  2. jcsd
  3. Apr 24, 2007 #2

    Dick

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    Are you sure there should be a cos(theta) in the denominator? Where did that expression come from?
     
  4. Apr 24, 2007 #3
    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...
     
    Last edited: Apr 24, 2007
  5. Apr 24, 2007 #4

    Dick

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    I guess I'm stumped as well. Anybody else seen something like that?
     
  6. Apr 24, 2007 #5
    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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