How do I evaluate $\int \frac{x^a}{(x + 1)^2}$ over $\mathbb{R}^+$

Question

I'm currently trying to evaluate $$\int_{0}^{\infty} \frac{x^a}{(1 + x)^2}dx$$ I'm not sure how this integral works because it can require an arbitrary amount of branch cuts. I have tried using a quarter-circle integral. I've tried using the power series around the integral circle and then on the line, $Re(z) = -1 $, and some other contours, but the branch cut is making my contours ill-defined.

Answer 1

$\newcommand{\d}{\,\mathrm{d}}\newcommand{\H}{\mathscr{H}}$One way to deal with branch cuts is to use them rather than avoid them!

Fix a noninteger $a\in(-1,1)$. For $\epsilon>0$ let $\H_\epsilon$ denote the partial Hankel contour, the one that runs $\epsilon^{-1}-i\epsilon\to-i\epsilon$ in a straight line, $-i\epsilon\to i\epsilon$ in a clockwise semicircle, $i\epsilon\to\epsilon^{-1}+i\epsilon$ in a straight line. Let $f:z\mapsto\exp(a\log z)(1+z)^{-2}$, $\Bbb C^\times\to\Bbb C$, where $\log$ is the logarithm induced by $0\le\arg<2\pi$.

$f$ is meromorphic on $\Bbb C\setminus[0,\infty)$. Integrating $f$ around $\H_\epsilon$ will equal $-2\pi i$ times the residue of $f$ at $-1$ in addition with an error term that is vanishing as $\epsilon\to0^+$, noting that $zf(z)\to0$ as $|z|\to\infty$. The residue is easily computed to be $a\exp(a\log(-1))=a\exp(\pi ia)$. Also observe that the integral of $f$ around the semicircular portion of $\H_\epsilon$ vanishes as $\epsilon\to0$ since $zf(z)\to0$ as $|z|\to0^+$.

Therefore: $$\begin{align}-2\pi ia\exp(\pi ia)&=\lim_{\epsilon\to0^+}\int_{\H_\epsilon}f(z)\d z\\&=\int_0^\infty[f(x^+)-f(x^-)]\d x\\&=(1-\exp(2\pi ia))\int_0^\infty\frac{x^a}{(1+x)^2}\d x\end{align}$$Where $x^{\pm}$ denotes taking the limit as $z=x\pm i\epsilon\to x,\epsilon\to0^+$. Because $a$ is noninteger, the "branch jump" factor $(1-\exp(2\pi ia))$ is nonzero and we can recover the value of the integral!

We find: $$\int_0^\infty\frac{x^a}{(1+x)^2}\d x=\frac{2\pi ia}{\exp(\pi ia)-\exp(-\pi ia)}=\frac{\pi a}{\sin\pi a}=\pi a\csc\pi a,\,a\in(-1,1)\setminus\{0\}$$

And we take the limit as $a\to0$ to recover the integral's value at zero.