Compute the integral of x^a / (1+x^2) for x going from 0 to + infinity

[disregardthat]

Compute $$\int^{\infty}_0 \frac{x^{\alpha}}{1+x^2} dx$$ for some $$-1<\alpha<1$$.

EDIT: This was slightly wrong.

The hint given is that we can integrate from -p to p except for a small semi-circle around 0, and a large semicircle from p to -p, and choose a branch of $$z^{\alpha}$$. Wouldn't this in any case be the method for computing $$\int^{\infty}_{-\infty} \frac{x^{\alpha}}{1+x^2} dx$$?

I'd appreciate some help in choosing a proper contour. I'd know how to integrate from -infinity to infinity, but from 0 I have no idea. Also, can I choose my branch for the logarithm arbitrarily?

 

[snipez90]

$$\int^{\infty}_{-\infty} \frac{z^{\alpha}}{1+z^2} \, dz =<br /> \int^{\infty}_{0} (z^{\alpha} + (-z)^{\alpha})\frac{1}{1+z^2} \, dz$$

But
$$(-z)^{\alpha} = e^{\pi i \alpha}z^{\alpha}$$
so
$$\int^{\infty}_{-\infty} \frac{z^{\alpha}}{1+z^2} dz = \left(1 + e^{\pi i \alpha}\right)\int^{\infty}_0} \frac{z^{\alpha}}{1+z^2} dz$$