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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$\ds{\int_{-\infty}^{\infty}{\dd t \over \pars{1 + \ic t}^{x}\pars{1 - \ic t}^{y}}
     =2^{2 - x - y}\,\pi\,
     {\Gamma\pars{x + y - 1} \over \Gamma\pars{x}\Gamma\pars{y}}:\ {\large ?}}$

\begin{align}&\color{#c00000}{%
\int_{-\infty}^{\infty}{\dd t \over \pars{1 + \ic t}^{x}\pars{1 - \ic t}^{y}}}
=-\ic\int_{-\infty\ic}^{\infty\ic}{\dd t \over \pars{1 + t}^{x}\pars{1 - t}^{y}}
\\[3mm]&=\ic\int_{-\infty}^{-1}
{\dd t \over \verts{1 + t}^{x}\expo{\ic\pi x}\pars{1 - t}^{y}}
+\ic\int_{-1}^{-\infty}
{\dd t \over \verts{1 + t}^{x}\expo{-\ic\pi x}\pars{1 - t}^{y}}
\\[3mm]&=\ic\expo{-\ic\pi x}\int_{1}^{\infty}
{\dd t \over \verts{1 - t}^{x}\pars{1 + t}^{y}}
-\ic\expo{\ic\pi x}\int_{1}^{\infty}{\dd t \over \verts{1 - t}^{x}\pars{1 + t}^{y}}
\\[3mm]&=-\ic\pars{\expo{\ic\pi x} - \expo{-\ic\pi x}}
\int_{1}^{\infty}{\pars{t - 1}^{-x} \over \pars{1 + t}^{y}}\,\dd t
=2^{1 - y}\sin\pars{\pi x}\int_{0}^{\infty}
{t^{-x} \over \pars{1 + t/2}^{y}}\,\dd t
\\[3mm]&=\color{#c00000}{%
2^{2 - x - y}\sin\pars{\pi x}\int_{0}^{\infty}
{t^{-x} \over \pars{1 + t}^{y}}\,\dd t}
\end{align}

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
  $$
\Re\pars{x} < 1\qquad\mbox{and}\qquad\Re\pars{x + y} > 1
$$

