Integral (using complex analysis)

1. Oct 25, 2009

Texxor22

1. The problem statement, all variables and given/known data

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

2. Relevant equations

3. The attempt at a solution

I integrated $$z^a/(z^b+1)$$ 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 $$0<\arg z<2\pi$$...so I got my total contour integral as

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

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

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