Show $\int_{0}^{\infty} \frac{x^{-z}}{(1 + x)^{2}} ~ \mathrm{d}{x} = \frac{\pi z}{sin(\pi z)}$

Question

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

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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?

Answer 1

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.

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}