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Integral (using complex analysis)

  1. Oct 25, 2009 #1
    1. The problem statement, all variables and given/known data

    I'm trying to prove that [tex]\int_0^{\infty}\frac{t^a}{t^b+1}dt=\frac{\pi}{b}\csc\left[\frac{\pi}{b}(a+1)\right][/tex] for -1<Re(a)<Re(b)-1.


    2. Relevant equations



    3. The attempt at a solution

    I integrated [tex]z^a/(z^b+1)[/tex] along a positively oriented keyhole contour C. As I took the outer circle's radius to infinity and the inner circle's radius to 0 I got the correct bounds, -1<Re(a)<Re(b)-1. The branch I'm working with has [tex]0<\arg z<2\pi[/tex]...so I got my total contour integral as

    [tex]\oint_{C}\frac{z^a}{z^b+1}dz=\int_0^{\infty}\frac{z^a}{z^b+1}dz-\int_0^{\infty}\frac{(e^{2\pi i}z)^a}{(e^{2\pi i}z)^b+1}dz=[1-e^{2\pi ia}(e^{2\pi ib})^{-(a+1)/b}]\int_0^{\infty}\frac{z^a}{z^b+1}dz.[/tex] I substituted there to get rid of that exponential in the denominator.

    Then I figured that [tex]z^b+1[/tex] has order 1 poles (because it can be factored) at [tex]e^{i\pi(2k-1)/b},\;\;k=0,1,2...b-1.[/tex] I calculated the residues at these points as [tex]\frac{1}{b}e^{i\pi a(2k-1)/b}(e^{i\pi(2k-1)/b})^{1-b}.[/tex]

    Then I summed the residues from k=0->b-1, multiplied by 2 pi i, equated with the contour integral before, but when I do the algebra it doesn't reduce to a cosecant, though it comes really close. I think I messed up with my branch cut contributions on the contour integral. Any ideas? Thanks
     
    Last edited: Oct 25, 2009
  2. jcsd
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