In one problem I am stuck with an integral that can be mapped to the following form:
$$I = \int_0^\infty dx \ \dfrac{x^a}{(1+x^b)^c}$$
where it can be assumed that $b>0$ and $c>0$. I know that the special case of $a>0$ and $c=1$ can be solved with the help of the residue theorem (see e.g. Show that $\int_0^ \infty \frac{1}{1+x^n} dx= \frac{ \pi /n}{\sin(\pi /n)}$ , where $n$ is a positive integer.), but am not sure how to treat the branch cut that appears for a rational $c$.
As Maxim pointed out the integral is most easily "solved" by mapping it to the beta function.
$$ \begin{align} I &= \int_0^\infty \frac{x^a}{(1+x^b)^c} d x \\ &= \frac{1}{b}\int_0^1 \frac{d u}{u(1-u)} \left(\frac{u}{1-u}\right)^{\frac{1}{b}} \frac{\left(\frac{u}{1-u}\right)^{\frac{a}{b}}}{\left(1 + \frac{u}{1-u}\right)^c} \\ &= \frac{1}{b}\int_0^1 d u \ u^{\frac{1+a}{b}-1} \ (1-u)^{1-\frac{1+a}{b}-c} \\ &\equiv \frac{1}{b} B\left(\frac{1+a}{b},2-\frac{1+a}{b}-c\right) \end{align} $$
where in the first step we substituted $x=\left(\frac{u}{1-u}\right)^{\frac{1}{b}} \ \ \rightarrow \ \ d x = \frac{d u}{bu(1-u)} \left(\frac{u}{1-u}\right)^{\frac{1}{b}} $. Note that in the case of $c=1$ we can use the reflection formula for gamma functions to turn the beta function in a simple $\csc$.