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# evaluation of beta function using Laplace transform

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

$\int_{0}^{1}x^{{p-1}}(1-x)^{{q-1}}\,dx={\Gamma(p)\Gamma(q)\over\Gamma(p+q)}$ |

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