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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 $c$.

JohnColtraneisJC
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user6566791
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  • Special case got me... we all have special needs and special cases... can you isolate the branch your interested in and HOW your trying? – Jay Aug 14 '19 at 16:59
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    Can you show what work you've done so far? You seem to be stuck in a specific place. – Moya Aug 14 '19 at 16:59
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    Your integral is not elementary. Substituting $x = (u/(1- u))^{1/b}$ gives $C ! \int_0^1 u^\alpha (1 - u)^\beta du$, which is a beta function. – Maxim Aug 14 '19 at 21:56

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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 $\csc$.

user6566791
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