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Beta function
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== Relationship to the gamma function == To derive this relation, write the product of two factorials as integrals. Since they are integrals in two separate variables, we can combine them into an iterated integral: :<math>\begin{align} \Gamma(z_1)\Gamma(z_2) &= \int_{u=0}^\infty\ e^{-u} u^{z_1-1}\,du \cdot\int_{v=0}^\infty\ e^{-v} v^{z_2-1}\,dv \\[6pt] &=\int_{v=0}^\infty\int_{u=0}^\infty\ e^{-u-v} u^{z_1-1}v^{z_2-1}\, du \,dv. \end{align}</math> Changing variables by {{math|''u'' {{=}} ''st''}} and {{math|''v'' {{=}} ''s''(1 β ''t'')}}, because {{math|''u + v'' {{=}} ''s''}} and {{math| ''u'' / ''(u+v)'' {{=}} ''t''}}, we have that the limits of integrations for {{math| ''s''}} are 0 to β and the limits of integration for {{math| ''t''}} are 0 to 1. Thus produces :<math>\begin{align} \Gamma(z_1)\Gamma(z_2) &= \int_{s=0}^\infty\int_{t=0}^1 e^{-s} (st)^{z_1-1}(s(1-t))^{z_2-1}s\,dt \,ds \\[6pt] &= \int_{s=0}^\infty e^{-s}s^{z_1+z_2-1} \,ds\cdot\int_{t=0}^1 t^{z_1-1}(1-t)^{z_2-1}\,dt\\ &=\Gamma(z_1+z_2) \cdot \Beta(z_1,z_2). \end{align}</math> Dividing both sides by <math>\Gamma(z_1+z_2)</math> gives the desired result. The stated identity may be seen as a particular case of the identity for the [[convolution#Integration|integral of a convolution]]. Taking :<math>\begin{align}f(u)&:=e^{-u} u^{z_1-1} 1_{\R_+} \\ g(u)&:=e^{-u} u^{z_2-1} 1_{\R_+}, \end{align}</math> one has: :<math> \Gamma(z_1) \Gamma(z_2) = \int_{\R}f(u)\,du\cdot \int_{\R} g(u) \,du = \int_{\R}(f*g)(u)\,du =\Beta(z_1,z_2)\,\Gamma(z_1+z_2).</math> See ''The Gamma Function'', page 18β19<ref>{{citation|last1=Artin|first1=Emil|title=The Gamma Function|pages=18β19|url=http://www.plouffe.fr/simon/math/Artin%20E.%20The%20Gamma%20Function%20(1931)(23s).pdf|access-date=2016-11-11|archive-url=https://web.archive.org/web/20161112081854/http://www.plouffe.fr/simon/math/Artin%20E.%20The%20Gamma%20Function%20(1931)(23s).pdf|archive-date=2016-11-12|url-status=dead}}</ref> for a derivation of this relation.
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