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A tricky countour integral (branch cuts)

  1. Dec 25, 2007 #1
    Hi all. I need to take the following integral by contour integration (it's actually a simplified version of another integral that I'm working on):

    [tex]\int_{-\pi}^\pi \frac{e^{i\omega}}{\sqrt{e^{i\omega}-1/2}\sqrt{e^{i\omega}-3}}[/tex].

    I am transforming it using [tex]z=e^{i\omega}[/tex] into
    [tex]\frac{1}{i}\int \frac{dz}{\sqrt{z-1/2}\sqrt{z-3}}[/tex], where the integral is around the unit contour around the origin.

    Now, it looks to me that there are two branch points: z=1/2 and z=3, and I'm choosing a branch cut from 1/2 to -infinity and from 3 to infinity. This is illustrated in the picture here
    http://img509.imageshack.us/my.php?image=clipboardqx7.jpg

    Along the branch cut from 1/2 to -infinity, the sqrt(z-1/2) is multivalued. The illustration that I linked shows how this can be used to solve the integral. However, I don't know how to account for the second branch cut.

    To clarify: if, for instance, I use [tex]z-3[/tex] instead of [tex]\sqrt{z-3}[/tex], the integral works out. However, with the square root it doesn't work -- I don't think the second branch cut should matter because I'm not integrating anywhere in its vicinity, but apparently it does. I don't have anyone to ask about it... so I'm hoping there's a contour integration maven that reads this and explains things to me...

    Edit: I wanted to clarify the linked diagram (link doesn't click for some reason so I have to copy/paste the URL).
    The green line is the branch cut, the yellow contour is the integral that I seek (i.e. int dw from -pi to pi is the same as int of dz over the yellow contour); the red contour does not enclose singularities, so it's zero by the residue theorem...
     
    Last edited: Dec 25, 2007
  2. jcsd
  3. Dec 25, 2007 #2

    Hurkyl

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    Your original integral: is it supposed to be w.r.t. omega?

    Anyways, I note that your original integrand is multivalued; what branch cut is it supposed to be using? I'm going to assume the branch cut required for the original integrand is the same one you decided to use for the integrand to your contour integral.

    Could you clarify in what way your approach doesn't work?
     
    Last edited: Dec 25, 2007
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