- #1
mathmadx
- 17
- 0
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
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
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..
Last edited by a moderator: