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Help for a contour integral

  1. Oct 22, 2013 #1

    ShayanJ

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    I wanna calculate the integral [itex] \int_0^{\infty} \frac{x^a}{(1+x)^2}dx \ (-1<a<1) [/itex] via contour integration But it seems a little tricky.
    I tried to solve it like example4 in the page ( http://en.wikipedia.org/wiki/Contour_integral#Example_.28IV.29_.E2.80.93_branch_cuts ) but I arrived at zero which I know is wrong.(The answer is [itex] \frac{\pi a}{\sin{\pi a}} [/itex])What's the point?
    Thanks
     
  2. jcsd
  3. Oct 24, 2013 #2
    The integral isn't too hard via a key-hole contour with the slot along the positive real axis except you have to be careful to compute the residue of the multi-valued function, [itex]z^a[/itex]. Let's look at that. We have:

    [tex]\frac{z^a}{(1+z)^2}[/tex]

    and so that's a second-order pole at [itex]z=-1[/itex] so the residue is just the derivative of [itex]z^a[/itex] at z=-1. But that's a multi-valued function for [itex]-1<a<1[/itex] so that expression has a potentially infinite number of answers. Well, the particular residue depends over which sheet of the function we integrate over. Suppose it was just [itex]z^{1/2}[/itex]. Then we could integrate over the branch

    [tex]z^{1/2}=e^{1/2(\ln(r)+i\theta)},\quad 0<\theta\leq 2\pi[/tex]

    Ok, then that's the expression we would use to compute the residue:

    [tex]\text{Res}\left(\frac{z^{1/2}}{(1+z)^2},-1\right)=1/2 e^{-1/2(\ln(-1)+\pi i)}=1/2 e^{-\pi i}=-1/2[/tex]

    Same dif for any a in that range if we use that particular determination of [itex]z^a[/itex]:

    [tex]\text{Res}\left(\frac{z^{a}}{(1+z)^2},-1\right)=ae^{(a-1)(\ln|z|+i\theta)},\quad 0<\theta\leq 2 \pi[/tex]

    and therefore the residue for that particular sheet of the multivalued function [itex]z^a[/itex] at z=-1 is [itex]ae^{(a-1)\pi i}[/itex] right?

    Now it's easy to compute the integral over the various legs of that contour: On the top leg along the real axis it's just the real integral. But on the lower leg, [itex]z^a=e^{a(\ln(r)+2\pi i)}[/itex] and I'll leave it to you to verify the integrals over the circular large and small arcs of the contour go to zero. And we're left with:

    [tex]\left(1-e^{2\pi i a}\right) \int_0^{\infty}\frac{x^a}{(1+x)^2}dx=2\pi i r[/tex]
    where r is that residue above.
     
    Last edited: Oct 24, 2013
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