Beta function in terms of the gamma function

Remarks

Write $\Gamma(x)\Gamma(y)$ as the double integral $\int_0^\infty \int_0^\infty u^{x-1} v^{y-1} e^{-u-v}\, du\, dv$. The change of variables $u = st$, $v = s(1-t)$ with $s > 0$ and $0 < t < 1$ has Jacobian $s$. It turns the double integral into $\int_0^\infty s^{x+y-1} e^{-s}\, ds \cdot \int_0^1 t^{x-1}(1-t)^{y-1}\, dt = \Gamma(x+y)\, B(x, y)$. Absolute convergence, which needs $\operatorname{Re} x > 0$ and $\operatorname{Re} y > 0$, justifies the change of order.

Example: $B(\frac12, \frac12) = \Gamma(\frac12)^2 / \Gamma(1) = (\sqrt{\pi})^2 = \pi$. For positive integers $m, n$ the formula gives $B(m, n) = \frac{(m-1)!\,(n-1)!}{(m+n-1)!}$.

Without the condition on the real parts the integral defining $B$ diverges, for example at $x = 0$. The right side still makes sense there, and it is used to extend $B$ to a meromorphic function of $x$ and $y$.

Going deeper

For a student

The formula is the bridge between the two Euler integrals. It shows at once that $B(x, y) = B(y, x)$ and gives the analytic continuation of $B$. With $t = \sin^2 \theta$ it computes integrals $\int_0^{\pi/2} \sin^{a}\theta \cos^{b}\theta\, d\theta$, and with $x = y = \frac12$ it gives $\Gamma(\frac12) = \sqrt{\pi}$. In probability it gives the normalising constant of the beta distribution.