I have to show that $$ \int_{0}^{\infty}\frac{x^{p-1}}{1+x}dx = \frac{\pi}{\sin(p\pi)}; 0<p<1$$
The hint provided says that I should choose as my contour the region between two circles of radius $\epsilon$ and $R$ and integrate $f(\mathbb{z}) = \frac{\mathbb{z}^{p-1}}{1+\mathbb{z}}$
Why such a contour? Shouldn't I be trying to choose some sort of semi-circles to eliminate the integral as $R \rightarrow \infty$ ?
I wanted to understand HOW to decide what contour to use in this problem and others. Once that is decided, I can proceed. Is there a good strategy to arrive at the right contour? Or is it just something you get better at the more problems you see?
I gues, you have to use "Key-Hole contour" with branch cut on positive real axes. The next you are solving integrals and two of them will be your initial by simplifying. Two integral, which countours was seleceted on "circles $\epsilon$ and $R$" will disapper (proving by inequality).
Finally you have to use Cauchy Residue Theorem, in your function $f(z)$ you have simple pole $z=-1=e^{\pi i}$
For more information check this video: https://www.youtube.com/watch?v=9H3NS0D5gng
P.S.
The semi circle countour is just one of popular contour because Jordan's Lemma says to us, that integrals kind of $\lim\limits_{R\to \infty}\oint_{C_R} f(z) e^{iaz} \,dz=0$, where $a>0$ and $f(z)$ is continuous in this area formed by the contour.