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 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 at 15:27
The proof is given in @ShaneChern link on page 8. –  Nathaniel Jan 28 at 15:55
see also here: de.wikibooks.org/wiki/…, example 4.1 –  Cortizol Mar 17 at 22:30