I have to prove this using the residue theorem, but I don't know which contour I have to use. I also have problems finding which function I have to consider, but I suppose is something related with the exponential because of the sinus of the result. Any advice?
$$\int_{0}^{\infty}\frac{t^{a-1}}{1 + t}dt = \frac{\pi}{{sin}(\pi a)} \; \; \; \forall a \in (0, 1).$$
Thank you!
First note that by a change of variables we can write
$\displaystyle \int_0^\infty \frac{t^{a-1}}{1+t}dt = \int_{-\infty}^\infty \frac{e^{ax}}{1+e^{x}}dx$ by setting $t = e^x$.
Then integrate the function $\displaystyle f(z) = \frac{e^{az}}{1+e^z}$ along a rectangle, lying on the real axis ("centered" at the origin) going out a distance of $R$ to both sides, and with a height of $2\pi i$.
The only pole in this contour is at $\pi i$, and the residue of $f$ at $\pi i$ is $\displaystyle \lim_{z\to \pi i} (z-\pi i)\frac{e^{a\pi i}}{e^z - e^{\pi i}} = e^{a\pi i} \lim_{z \to \pi i} \frac{z-\pi i}{e^z - e^{\pi i}} = -e^{a\pi i}$. Do you see why? Note that I used the fact that $e^{i\pi}=-1$.
You can show that along the vertical sides go to $0$ as $R \to \infty$, and moreover the integral along the top segment of the rectangle is a complex multiple of the integral along the bottom segment (the one you want). Factor, and you'll have it.
