Proof of Cauchy's Beta Integral $\int_{-\infty}^\infty \frac{dt}{(1+it)^x(1-it)^y}$

The Cauchy's Beta Integral is given by

$$\int_{-\infty}^\infty \frac{dt}{(1+it)^x(1-it)^y}=\frac{\pi 2^{2-x-y}\Gamma(x+y-1)}{\Gamma(x)\Gamma(y)}$$

I would like to know how it is proved.

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I'm not familiar with complex analysis, could you explain me how to interpret $(1+it)^x$ for $x=1/2$, $t=1$ (for example). It seems to me that this expression is multivalued, so your integral is not well defined. – Norbert Jan 27 '13 at 10:24
You may read this paper: www1.maths.leeds.ac.uk/~kisilv/courses/sp-funct.pdf section 1.4.2 – Shane Chern Jan 27 '13 at 15:27
The proof is given in @ShaneChern link on page 8. – nbubis Jan 28 '13 at 15:55
see also here: de.wikibooks.org/wiki/…, example 4.1 – Cortizol Mar 17 '13 at 22:30

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\int_{-\infty}^{\infty}{\dd t \over \pars{1 + \ic t}^{x}\pars{1 - \ic t}^{y}} =2^{2 - x - y}\,\pi\, {\Gamma\pars{x + y - 1} \over \Gamma\pars{x}\Gamma\pars{y}}:\ {\large ?}}$
With $\ds{\xi \equiv {1 \over 1 + t}\quad\imp\quad t = {1 \over \xi} - 1}$: \begin{align}&\color{#c00000}{% \int_{-\infty}^{\infty}{\dd t \over \pars{1 + \ic t}^{x}\pars{1 - \ic t}^{y}}} =2^{2 - x - y}\sin\pars{\pi x} \int_{1}^{0}\pars{{1 \over \xi} - 1}^{-x}\xi^{y}\,\pars{-\,{\dd\xi \over \xi^{2}}} \\[3mm]&=2^{2 - x - y}\sin\pars{\pi x} \int_{1}^{0}\xi^{x + y - 2}\pars{1 - \xi}^{-x}\,\dd\xi =2^{2 - x - y}\sin\pars{\pi x}\ \overbrace{{\rm B}\pars{x + y - 1,-x + 1}}^{\ds{\mbox{Beta Function}}} \\[3mm]&=2^{2 - x - y}\sin\pars{\pi x}\ \overbrace{{\Gamma\pars{x + y - 1}\Gamma\pars{-x + 1}\over \Gamma\pars{y}}} ^{\ds{\Gamma's:\ \mbox{Gamma Functions}}} \end{align}
With Euler Reflection Formula: $\ds{\Gamma\pars{-x + 1} = {\pi \over \sin\pars{\pi x}\Gamma\pars{x}}}$ such that $$\color{#44f}{\large% \int_{-\infty}^{\infty}{\dd t \over \pars{1 + \ic t}^{x}\pars{1 - \ic t}^{y}} =2^{2 - x - y}\,\pi\, {\Gamma\pars{x + y - 1} \over \Gamma\pars{x}\Gamma\pars{y}}}$$ where $$\Re\pars{x} < 1\qquad\mbox{and}\qquad\Re\pars{x + y} > 1$$