Show $\int_{0}^{\infty} \frac{x^{-z}}{(1 + x)^{2}} ~ \mathrm{d}{x} = \frac{\pi z}{sin(\pi z)}$ I need to solve the following integral:
$$
I = \int_{0}^{\infty} \frac{x^{-z}}{(1 + x)^{2}} ~ \mathrm{d}{x}.
$$

Wolfram Alpha gives the answer as $ \frac{\pi z}{sin(\pi z)}$, or equivalently, $\pi z csc(\pi z)$
My ultimate goal is to demonstrate that $z!(-z)! = \frac{\pi z}{sin(\pi z)}$
So far, I arrived at this integral by gamma and beta functions:
$$
z!(-z)! = \Gamma(z+1)\Gamma(-z+1) = \Gamma(m)\Gamma(n)\\
= B(z+1,-z+1)\Gamma(m+n) = B(z+1,-z+1)\Gamma(2)\\
\\
=B(z+1,-z+1) = \int_{0}^{\infty} \frac{x^{-z+1-1}}{(1 + x)^{2}} ~ \mathrm{d}{x}\\ 
= \int_{0}^{\infty} \frac{x^{-z}}{(1 + x)^{2}} ~ \mathrm{d}{x} 
$$
This integral is effectively the answer. Can you solve this with a contour integral? Also, is there a special name for this integral?
 A: By considering the contour integral
$$\oint_C d\zeta \frac{\zeta^{-z}}{(1+\zeta)^2} $$
about a keyhole contour and using the residue theorem, we may derive the relation
$$\left (1-e^{-i 2 \pi z} \right) \int_0^{\infty} dx \frac{x^{-z}}{(1+x)^2} = i 2 \pi \left [\frac{d}{d\zeta} e^{-z \log{\zeta}} \right ]_{\zeta=e^{i \pi}} = i 2 \pi (-z e^{-i \pi} ) e^{-i \pi z}$$
The result follows after a little algebra.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
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\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{\infty}
{x^{-z} \over \pars{1 + x}^{2}}\,\dd x} =
\int_{0}^{\infty}
x^{-z}\bracks{\sum_{k = 0}^{\infty}{-2 \choose k}x^{k}}\,\dd x
\\[5mm] = &\
\int_{0}^{\infty}
x^{-z}\bracks{\sum_{k = 0}^{\infty}{k + 1 \choose k}
\pars{-1}^{k}x^{k}}\,\dd x
\\[5mm] = &\
\int_{0}^{\infty}
x^{\pars{\color{red}{1 - z}} - 1}\bracks{\sum_{k = 0}^{\infty}\color{red}{\Gamma\pars{2 + k}}
{\pars{-x}^{k} \over k!}}\,\dd x
\\[5mm] = &\
\Gamma\pars{\color{red}{1 - z}}
\Gamma\pars{2 - \bracks{\color{red}{1 - z}}} =
\Gamma\pars{1 - z}\Gamma\pars{z}z\quad
\pars{\substack{\mbox{Ramanujan's}\\[0.5mm] \mbox{Master}
\\[1mm] \mbox{Theorem}}}
\\[5mm] = &\
\bbx{\pi z \over \sin\pars{\pi z}} \\ &
\end{align}
