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Hi guys,

I need to show that:

[tex] \int_{0}^{\infty } \frac{x^{a}}{(x+1)^2} \dx = \frac{\pi a}{\sin(\pi a)} [/tex]

, where -1<a<1.

The problem is, that although a hint is given,the path of integrating it, I have difficulty what they really mean with "cut line, branch points, multivalued functions" etc.

In the hint , the path of integration is the following, which I don't understand why they have chosen it like this, could some explain their reasoning? For example, why do they exclude z=-1?

http://img143.imageshack.us/img143/151/acf1.jpg [Broken]

EDIT:

I have now calculated this integral using my own , simpler, contour, which is basically the same as this one, except it includes the point z=-1 ( Which I still don't understand why my book excludes..).

By the way; the whole idea of branch cuts etc basically means that if you wounded one time around the origin, you have to use z=r*exp(i(theta+2pi)); is this right? I calculated using this and I got the correct answer..

I need to show that:

[tex] \int_{0}^{\infty } \frac{x^{a}}{(x+1)^2} \dx = \frac{\pi a}{\sin(\pi a)} [/tex]

, where -1<a<1.

The problem is, that although a hint is given,the path of integrating it, I have difficulty what they really mean with "cut line, branch points, multivalued functions" etc.

In the hint , the path of integration is the following, which I don't understand why they have chosen it like this, could some explain their reasoning? For example, why do they exclude z=-1?

http://img143.imageshack.us/img143/151/acf1.jpg [Broken]

EDIT:

I have now calculated this integral using my own , simpler, contour, which is basically the same as this one, except it includes the point z=-1 ( Which I still don't understand why my book excludes..).

By the way; the whole idea of branch cuts etc basically means that if you wounded one time around the origin, you have to use z=r*exp(i(theta+2pi)); is this right? I calculated using this and I got the correct answer..

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