# Thread: [Complex Analysis] Integral of a multi-valued function

1. ## [Complex Analysis] Integral of a multi-valued function

Hey! I need some help to prove the following identity:

$\int_{0}^{\infty} \frac{x^a}{x^2+b^2}dx\ =\ \frac{\pi b^{a-1}}{\cos(\pi a/2)} , \ 1>a>-1, \ b>0$

I'm supposed to use the formula

$\int_{0}^{\infty} x^{a}F(x)dx= \frac{2\pi i}{1-e^{2\pi ia}}\sum_{z\neq 0} Res(z^{a}F(z))$

So, in this case I get the following:

$\int_{0}^{\infty} \frac{x^a}{x^2+b^2}dx\ =\ \frac{2\pi i}{1-e^{2\pi ia}}(Res_{ib}\frac{x^a}{x^2+b^2}+Res_{-ib}\frac{x^a}{x^2+b^2})$

$Res_{ib}\frac{x^a}{x^2+b^2}=\ \frac{(ib)^{a-1}}{2}$

and

$Res_{-ib}\frac{x^a}{x^2+b^2}=\ \frac{(-ib)^{a-1}}{2}$

$\int_{0}^{\infty} \frac{x^a}{x^2+b^2}dx\ =\ \frac{\pi i}{1-e^{2\pi ia}}((ib)^{a-1}+(-ib)^{a-1})$

Now

$(ib)^{a-1}+(-ib)^{a-1}=\ b^{a-1}[(i)^{a-1}+(-i)^{a-1}]=\ b^{a-1}\sin(\pi a/2)$

So

$\int_{0}^{\infty} \frac{x^a}{x^2+b^2}dx\ =\ \frac{\pi i}{1-e^{2\pi ia}}b^{a-1}\sin(\pi a/2)$

And I'm stuck... What do I do next?

Any help will be appreciated.

2. ## Re: [Complex Analysis] Integral of a multi-valued function

$i\sin(\pi a/2) = e^{i\pi a/2} - \cos(\pi a/2)$.

$\dfrac{i\pi b^{a-1} \sin(\pi a/2)}{1-e^{2i \pi a}} = \pi b^{a-1} \dfrac{e^{i\pi a/2} - \cos(\pi a/2)}{1 - e^{2\pi i a}} = -\dfrac{1}{4}e^{-i \pi a}\dfrac{\pi b^{a-1}}{\cos(\pi a/2)}$.

So, you may have done something wrong. You get close to what you want...

3. ## Re: [Complex Analysis] Integral of a multi-valued function

Originally Posted by SlipEternal
$i\sin(\pi a/2) = e^{i\pi a/2} - \cos(\pi a/2)$.

$\dfrac{i\pi b^{a-1} \sin(\pi a/2)}{1-e^{2i \pi a}} = \pi b^{a-1} \dfrac{e^{i\pi a/2} - \cos(\pi a/2)}{1 - e^{2\pi i a}} = -\dfrac{1}{4}e^{-i \pi a}\dfrac{\pi b^{a-1}}{\cos(\pi a/2)}$.

So, you may have done something wrong. You get close to what you want...
Thanks. I noticed one mistake:i^(a-1)+(-i)^(a-1) =/= sin(pi*a/2)

i^(a-1)+(-i)^(a-1)= 2sin(pi*a/2)

But that just replaces that 1/4 by 1/2, doesn't it?. There must be another mistake I can't find.

4. ## Re: [Complex Analysis] Integral of a multi-valued function

Wolframalpha evaluates that integral to $\dfrac{1}{2}\dfrac{b^{a-1}\pi}{\cos(\pi a/2)}$, so it is getting half of your "expected" identity result.

5. ## Re: [Complex Analysis] Integral of a multi-valued function

Originally Posted by SlipEternal
Wolframalpha evaluates that integral to $\dfrac{1}{2}\dfrac{b^{a-1}\pi}{\cos(\pi a/2)}$, so it is getting half of your "expected" identity result.
Oh, you're right. So I'm getting closer to the result now. I'll start from zero to see where's the mistake. Thanks for your help!