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

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

I'm supposed to use the formula

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

So, in this case I get the following:

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

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

and

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

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

Now

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

So

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

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

Any help will be appreciated.