How can I calculate this integral using contour integration?

[ShayanJ]

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

 

[jackmell]

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.