The beta integral can be evaluated elegantly using the convolution theorem for Laplace transforms.

Start with the following Laplace transform: $$ s^{-\alpha} = {\cal L} \left[ {t^{\alpha - 1} \over \Gamma(\alpha)} \right] = \int_0^\infty e^{-st} {t^{\alpha - 1} \over \Gamma(\alpha)} dt $$

Since $s^{-q} s^{-p} = s^{-q - p}$ , the convolution theorem imples that $$ {t^{q - 1} \over \Gamma(q)} * {t^{p - 1} \over \Gamma(p)} = {t^{q + p - 1} \over \Gamma(q + p)} $$

Writing out the definition of convolution, this becomes $$ \int_0^t {(t-s)^{q - 1} \over \Gamma(q)} {s^{p - 1} \over \Gamma(p)} ds = {t^{q + p - 1} \over \Gamma(p + q)} $$

Setting $t=1$ and simplifying, we conclude that $$ \int_0^1 x^{p - 1} (1-x)^{q - 1} \,dx= {\Gamma(p) \Gamma(q) \over \Gamma(p + q)} $$