$B(x,y)\Gamma(x+y)=\Gamma(x) \Gamma(y)$ for all $x,y \in \mathbb{C}_{Re>0}$

Question

Define $\Gamma: \mathbb{C}_{Re>0} \to \mathbb{C}$ with $\Gamma(x):=\int_0^\infty\gamma_x(t)dt=\int_0^\infty t^{t-1} \exp(-t)dt$ and $B: \mathbb{C}^2_{Re>0} \to \mathbb{C}$ with $B(x,y):=\int_0^1 \beta_{x,y}(t)dt=\int_0^1t^{x-1}(1-t)^{y-1}dt$.

How do I show that $B(x,y)\Gamma(x+y)=\Gamma(x) \Gamma(y)$ for all $x,y \in \mathbb{C}_{Re>0}$?

Answer 1

In $\operatorname{B}$, substitute $t=\cos^2\theta$; in the three $\Gamma$s, substitute $t=r^2$, $t=X^2$, $t=Y^2$. The claim becomes$$\begin{align}&2\int_0^{\pi/2}\cos^{2x-1}\theta\sin^{2y-1}\theta d\theta\cdot2\int_0^\infty r^{2x+2y-1}e^{-r^2}dr\\&=2\int_0^\infty X^{2x-1}e^{-X^2}dX\cdot2\int_0^\infty Y^{2y-1}e^{-Y^2}dY.\end{align}$$Rewrite both sides of a double integral, then use polar coordinates $r=X\cos\theta,\,Y=X\sin\theta$ over the first quadrant so $rdrd\theta=dXdY$.