I am trying to show that $I = \int_0^{+\infty} \frac{x^a}{x^2 + 1} = \frac{\pi / 2} {\cos \frac{\pi a}{2}}$ provided $-1 < a < 1$.
So I consider $I = \int_{C_R}\frac{z^a}{z^2 + 1}$ where $C_R$ is the positively oriented contour consisting of the real segment [-R,R] and the upper semi circle of radius R.
Now $C_R$ only contains one sigularity of the above integrand, $\exp(\frac{i \pi}{2})$.
So I take the residue at this point, multiply by $2\pi i $ and get $\pi \left( \cos \frac{a \pi}{2} + i \sin \frac{a \pi}{2} \right)$
Now am I supposed to take the real part of this? And even then, why do I not get the desired result? Note that I should take half of the answer since the complex integral is from negative to plus infinity. However I am aware that there might be an issue here, since $a$ can be negative and not necessarily an integer, thus possibly screwing up the parity of I.
Any help would be appreciated.
Because of the branch point, you need to deform your contour so as to avoid the branch point. To do this, introduce a little circular bump above the branch point. Thus, consider
$$\oint_C dz \frac{z^a}{z^2+1}$$
where $C$ is the above-described deformed semicircle. The contour integral is then equal to
$$\int_{-R}^{-\epsilon} dx \frac{x^a}{x^2+1} + i \epsilon \int_{\pi}^0 d\phi \, e^{i \phi} \frac{\epsilon^a e^{i a \phi}}{\epsilon^2 e^{i 2 \phi}+1} \\ + \int_{\epsilon}^R dx \frac{x^a}{x^2+1} + i R \int_{0}^{\pi} d\theta \, e^{i \theta} \frac{R^a e^{i a \theta}}{R^2 e^{i 2 \theta}+1}$$
We consider the limits as $R \to \infty$ and $\epsilon \to 0$; in each of these limits, the fourth and second integrals vanish, respectively, because $a \in (-1,1)$. Therefore, the contour integral is the sum of the first and third integrals. Note that, in the first integral, $-1=e^{i \pi}$. Therefore, sub $x \mapsto e^{i \pi} x$ and get
$$\left (1+e^{i \pi a} \right ) \int_0^{\infty} dx \frac{x^a}{x^2+1} $$
The contour integral, on the other hand, is equal to $i 2 \pi$ times the residue of the pole inside $C$, i.e., $z = e^{i \pi/2}$; we therefore have
$$\int_0^{\infty} dx \frac{x^a}{x^2+1} = i 2 \pi \frac{1}{2 i} \frac{e^{i \pi a/2}}{1+e^{i \pi a}} = \frac{\pi}{2 \cos{(\pi a/2)}}$$