prove $\Gamma(a)\Gamma(b) = \Gamma(a+b)B(a,b)$ using polar transformation Question: prove $\Gamma(a)\Gamma(b) = \Gamma(a+b)B(a,b)$ using polar transformation:
my attempt:
L.H.S = $$\Gamma(a)\Gamma(b) = \int_0^{\infty}\int_0^{\infty}e^{-(x+y)}x^{a-1}y^{b-1}dxdy$$
Let $x = r\cos(\theta),y = r\sin(\theta)$
then $$\Gamma(a)\Gamma(b) = \int_0^{\pi/2}\int_0^{\infty}re^{-r(\cos(\theta)+\sin(\theta))}r^{a+b-2}\cos^{a-1}(\theta)\sin^{b-1}(\theta)drd\theta$$
$$= \int_0^{\pi/2}\cos^{a-1}(\theta)\sin^{b-1}(\theta)d\theta\int_0^{\infty}e^{-r(\cos(\theta)+\sin(\theta))}r^{a+b-1}dr $$
the first integral is almost $ = B(a,b)$ and the second integral is almost $ = \Gamma(a+b)$
but I am stuck there and cannot separate $\cos(\theta) + \sin(\theta)$ from $e$
any help is much appreciated.
 A: Clearly for $\theta\in[0,\pi/2]$, $\cos(\theta)+\sin(\theta)>0$. Let $t=r(\cos(\theta)+\sin(\theta))$ and then $r=\frac{t}{\cos(\theta)+\sin(\theta)}$. So
\begin{eqnarray}
&&\int_0^{\infty}e^{-r(\cos(\theta)+\sin(\theta))}r^{a+b-1}dr\\
&=&\int_0^{\infty}e^{-t}\frac{t^{a+b-1}}{[\cos(\theta)+\sin(\theta)]^{a+b}}dt\\
&=&\frac{1}{[\cos(\theta)+\sin(\theta)]^{a+b}}\int_0^{\infty}e^{-t}t^{a+b-1}dt\\
&=&\frac{1}{[\cos(\theta)+\sin(\theta)]^{a+b}}\Gamma(a+b).
\end{eqnarray}
Therefore
\begin{eqnarray}
\Gamma(a)\Gamma(b) &=& \int_0^{\pi/2}\int_0^{\infty}re^{-r(\cos(\theta)+\sin(\theta))}r^{a+b-2}\cos^{a-1}(\theta)\sin^{b-1}(\theta)drd\theta\\
&=& \int_0^{\pi/2}\cos^{a-1}(\theta)\sin^{b-1}(\theta)d\theta\int_0^{\infty}e^{-r(\cos(\theta)+\sin(\theta))}r^{a+b-1}dr\\
&=& \Gamma(a+b) \int_0^{\pi/2}\frac{\cos^{a-1}(\theta)\sin^{b-1}(\theta)}{[\cos(\theta)+\sin(\theta)]^{a+b}}d\theta\\
&=& \Gamma(a+b) \int_0^{\pi/2}\frac{\tan^{b-1}(\theta)}{[1+\tan(\theta)]^{a+b}}d\tan(\theta)\\
&\stackrel{\tan(\theta)\to t}{=}& \Gamma(a+b) \int_0^{\infty}\frac{t^{b-1}}{(1+t)^{a+b}}dt\\
&=& \Gamma(a+b)B(a,b).
\end{eqnarray}
