evaluation of beta function using Laplace transform

The beta integral can be evaluated elegantly using the convolution theorem (http://planetmath.org/LaplaceTransform) for Laplace transforms.

Start with the following Laplace transform:

$$s^{-\alpha }=ℒ\left[\frac{t^{\alpha -1}}{\mathrm{\Gamma }(\alpha )}\right]={\int }_0^{\mathrm{\infty }}{e}^{-st}\frac{t^{\alpha -1}}{\mathrm{\Gamma }(\alpha )}𝑑t$$

Since $s^{-q}{s}^{-p}=s^{-q-p}$, the convolution theorem imples that

$$\frac{t^{q-1}}{\mathrm{\Gamma }(q)}*\frac{t^{p-1}}{\mathrm{\Gamma }(p)}=\frac{t^{q+p-1}}{\mathrm{\Gamma }(q+p)}$$

Writing out the definition of convolution, this becomes

$${\int }_0^t\frac{{(t-s)}^{q-1}}{\mathrm{\Gamma }(q)}\frac{s^{p-1}}{\mathrm{\Gamma }(p)}𝑑s=\frac{t^{q+p-1}}{\mathrm{\Gamma }(p+q)}$$

Setting $t=1$ and simplifying, we conclude that

$${\int }_0^1{x}^{p-1}{(1-x)}^{q-1}𝑑x=\frac{\mathrm{\Gamma }(p)\mathrm{\Gamma }(q)}{\mathrm{\Gamma }(p+q)}$$

QED