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How do I demonstrate the relationship between the Beta and the Gamma function, in the cleanest way possible? I am thinking one (or two) substitution of variables is necessary, but when and how is the question.

Here is the beta function: $B(\alpha,\beta)=\int_0^1x^{\alpha-1}(1-x)^{\beta-1}dx$.

Here is the gamma function $\Gamma(\alpha)=\int_0^{\infty}t^{\alpha-1}e^{-t}dt$.

Here is the relationship between the Beta and Gamma functions: $B(\alpha,\beta)=\frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)}$.

Thanks for any help!

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  • $\begingroup$ Yes! I was able to solve it. Thanks. $\endgroup$ – user85362 Oct 4 '13 at 22:18
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Compute the double integral $$ \Gamma(\alpha)\Gamma(\beta)=\iint t^{\alpha-1}\mathrm e^{-t}s^{\beta-1}\mathrm e^{-s}\mathrm ds\mathrm dt, $$ using the change of variable $x=t/(t+s)$, $y=t+s$.

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    $\begingroup$ Can you be more specific and explain more please $\endgroup$ – user416990 Mar 27 '17 at 21:13
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    $\begingroup$ @user416990 Certainly not for such a vague query, no. $\endgroup$ – Did Mar 27 '17 at 21:17
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$\Gamma(\alpha)=\int_0^{\infty}t^{\alpha-1}e^{-t}dt$. Let $t = x^2$, $dt = 2xdx$. Thus, $$ \Gamma(\alpha)=\int_0^{\infty}x^{2\alpha-2}e^{-x^2}2xdx = 2\int_{0}^{\infty}x^{2\alpha - 1}e^{-x^2}dx \ \text{and} \ \Gamma(\beta)= 2\int_{0}^{\infty}y^{2\beta - 1}e^{-y^2}dy $$ $$ \Gamma(\alpha)\Gamma(\beta) = 4\int_{0}^{\infty}\int_{0}^{\infty}x^{2\alpha - 1}y^{2\beta - 1}e^{-(x^2 + y^2)}dxdy $$ Let $x = r\cos \theta$, $y = r\sin \theta$. Thus, $dxdy = rdrd\theta$ and $$ \begin{split} \Gamma(\alpha)\Gamma(\beta) &= 4\int_{0}^{\pi/2}\int_{0}^{\infty}r^{2\alpha + 2\beta - 1}\cos^{2\alpha - 1}\theta\sin^{2\beta - 1}\theta drd\theta\\ &= \biggl[2\int_{0}^{\pi/2}\cos^{2\alpha - 1}\theta\sin^{2\beta - 1}\theta d\theta \biggr]\biggl[2\int_{0}^{\infty}r^{2(\alpha + \beta) - 1}e^{-r^2}dr \biggr]\\ &=B(\alpha,\beta)\Gamma(\alpha + \beta) \end{split} $$

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