Relation between Beta and Gamma functions How do I demonstrate the relationship between the Beta and the Gamma function, in the cleanest way possible? I am thinking one (or two) substitution of variables is necessary, but when and how is the question.
Here is the beta function: $B(\alpha,\beta)=\int_0^1x^{\alpha-1}(1-x)^{\beta-1}dx$.
Here is the gamma function $\Gamma(\alpha)=\int_0^{\infty}t^{\alpha-1}e^{-t}dt$.
Here is the relationship between the Beta and Gamma functions: $B(\alpha,\beta)=\frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)}$.
Thanks for any help!
 A: Compute the double integral
$$
\Gamma(\alpha)\Gamma(\beta)=\iint t^{\alpha-1}\mathrm e^{-t}s^{\beta-1}\mathrm e^{-s}\mathrm ds\mathrm dt,
$$
using the change of variable $x=t/(t+s)$, $y=t+s$.
A: $\Gamma(\alpha)=\int_0^{\infty}t^{\alpha-1}e^{-t}dt$. Let $t = x^2$, $dt = 2xdx$. Thus,
$$
\Gamma(\alpha)=\int_0^{\infty}x^{2\alpha-2}e^{-x^2}2xdx = 2\int_{0}^{\infty}x^{2\alpha - 1}e^{-x^2}dx \ \text{and} \ 
\Gamma(\beta)= 2\int_{0}^{\infty}y^{2\beta - 1}e^{-y^2}dy
$$
$$
\Gamma(\alpha)\Gamma(\beta) = 4\int_{0}^{\infty}\int_{0}^{\infty}x^{2\alpha - 1}y^{2\beta - 1}e^{-(x^2 + y^2)}dxdy
$$
Let $x = r\cos \theta$, $y = r\sin \theta$. Thus, $dxdy = rdrd\theta$ and
$$
\begin{split}
\Gamma(\alpha)\Gamma(\beta) &= 4\int_{0}^{\pi/2}\int_{0}^{\infty}r^{2\alpha + 2\beta - 1}\cos^{2\alpha - 1}\theta\sin^{2\beta - 1}\theta drd\theta\\
&= \biggl[2\int_{0}^{\pi/2}\cos^{2\alpha - 1}\theta\sin^{2\beta - 1}\theta d\theta \biggr]\biggl[2\int_{0}^{\infty}r^{2(\alpha + \beta) - 1}e^{-r^2}dr \biggr]\\
&=B(\alpha,\beta)\Gamma(\alpha + \beta)
\end{split}
$$
