Shyan said:
I want to calculate the integral \int_0^{\infty} \frac{x^a}{(1+x)^2}dx \ (-1<a<1) 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 \frac{\pi a}{\sin{\pi a}})What's the point?
Thanks
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, z^a. Let's look at that. We have:
\frac{z^a}{(1+z)^2}
and so that's a second-order pole at z=-1 so the residue is just the derivative of z^a at z=-1. But that's a multi-valued function for -1<a<1 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 z^{1/2}. Then we could integrate over the branch
z^{1/2}=e^{1/2(\ln(r)+i\theta)},\quad 0<\theta\leq 2\pi
Ok, then that's the expression we would use to compute the residue:
\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
Same dif for any a in that range if we use that particular determination of z^a:
\text{Res}\left(\frac{z^{a}}{(1+z)^2},-1\right)=ae^{(a-1)(\ln|z|+i\theta)},\quad 0<\theta\leq 2 \pi
and therefore the residue for that particular sheet of the multivalued function z^a at z=-1 is ae^{(a-1)\pi i} 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, z^a=e^{a(\ln(r)+2\pi i)} 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:
\left(1-e^{2\pi i a}\right) \int_0^{\infty}\frac{x^a}{(1+x)^2}dx=2\pi i r
where r is that residue above.