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Need method to solve contour integral

  1. Jul 7, 2011 #1
    Hello, I am working on a problem and I have to solve a nasty integral. The problem is that I am not sure if the method I am using is correct.

    The integral I need to solve is:

    [itex]\int[/itex][itex]^{\infty}_{-\infty}[/itex][itex]\frac{e^{ikx}dk}{\sqrt{k^2 + a} - (b\pm i\lambda)}[/itex]

    At this point, I have tried multiplying the top and bottom by [itex]\sqrt{k^2 + a} + (b\pm i\lambda)[/itex] and continuing forth by splitting into partial fractions and using the Cauchy residue theorem, but I am not sure if this is a valid way to solve this.

    If anyone has any insight I would be more than indebted! Thank you so much!
  2. jcsd
  3. Jul 7, 2011 #2


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    It can work, as long as you are taking into account the fact that your integrand is multiple valued. You have branch points at [itex]\pm i a[/itex]. The typical approach is to define a single valued "branch" of the integrand with a corresponding branch cut, then do the normal contour integral thing. But in this case, the path you use to close your contour has to contend with the branch cut - you will most likely end up with integrals along the two sides of the branch cut in addition to your residue. Now these new integrals may be easier than your original integral, or recognizable as some special function or something, or they may be beasts unto themselves.

    Most complex analysis books should show examples of this sort of thing.

    hope that helps ...

  4. Jul 8, 2011 #3
    Thanks for the reply!

    Actually, I am already familiar with those methods for contour integration. The main problem I am having is whether or not it is legal to multiply the integrand by a specific value of "1," namely,

    [itex]\frac{\sqrt{k^2 + a} + (b\pm i\lambda)}{ \sqrt{k^2 + a} + (b\pm i\lambda) }[/itex].

    This would allow me to rewrite the integrand as a sum over two first order polynomials in the denominator, which is then solvable by the Cauchy residue method. However, it is the multiplication by "1" which I am having trouble justifying. Hope that is clearer.

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