EvalutaionOfBetaFunctionUsingLaplaceTransform 2004-09-22 21:53:22 2007-04-14 09:57:12 Derivation beta function % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here The beta integral can be evaluated elegantly using the \PMlinkname{convolution theorem}{LaplaceTransform} 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 \[ \int_0^1 x^{p - 1} (1-x)^{q - 1} \,dx= {\Gamma(p) \Gamma(q) \over \Gamma(p + q)} \] \rightline{QED}