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I have the following double integral $$\int_0^{2\pi}\int_0^\infty r^{p+q-1}\cos^{p-1}({\theta})\sin^{q-1}({\theta})e^{-r(\sin{\theta}+{\cos{\theta}})}drd\theta $$ However, I can't seem to be able to isolate the required terms to prove the identity $$\Gamma(p)\Gamma(q)=B(p,q)\Gamma(p+q)$$ where $$ \Gamma(p) \equiv \int_0^\infty x^{p-1}e^{-x}dx $$ and $$ B(p,q) \equiv \int_0^1 x^{p-1}(1-x)^{q-1}dx = 2\int_0^\frac{\pi}{2}\sin^{2p-1}({2\theta})\cos^{2q-1}({2\theta})d\theta$$

Many thanks.

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  • $\begingroup$ Are you sure about the integral? I think it should be $2\int_0^{\pi/2}\int_0^\infty r^{p+q-1}\cos^{p-1}({\theta})\sin^{q-1}({\theta})e^{-r}drd\theta $. $\endgroup$
    – alexjo
    Dec 22, 2014 at 21:50

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Write the gamma function $\Gamma(p)$ as $$ \Gamma(p)=\int_0^{\infty}t^{p-1}\operatorname{e}^{-t}\operatorname{d}\!t=\int_0^{\infty}x^{2p-2}\operatorname{e}^{-x^2}2x\operatorname{d}\!x=2\int_0^{\infty}x^{2p-1}\operatorname{e}^{-x^2}\operatorname{d}\!x $$ changing the integration variable from $t$ to $x^2$. Write the product of $\Gamma(p)\Gamma(q)$ as $$ \Gamma(p)\Gamma(q)=4\int_0^{\infty}\int_0^{\infty}x^{2p-1}y^{2q-1}\operatorname{e}^{-x^2-y^2}\operatorname{d}\!x\operatorname{d}\!y.\tag 1 $$ We now rewrite the (1) in polar coordinates $(r,\theta)$ with $x=r\cos\theta,\,y=r\sin\theta$, $x^2+y^2=r^2$ and Jacobian $J=r$; in the polar coordnates the integration is over the entire first quadrant, i.e. $0\le r <\infty,\,0\le\theta\le \pi/2$. We get $$\begin{align} \Gamma(p)\Gamma(q)&=4\int_0^{\pi/2}\int_0^{\infty}(r\cos\theta)^{2p-1}(r\sin\theta)^{2q-1}\operatorname{e}^{-r^2}r\operatorname{d}\!r\operatorname{d}\!\theta\\ &=4\int_0^{\pi/2}\int_0^{\infty}(\cos\theta)^{2p-1}(\sin\theta)^{2q-1}\operatorname{e}^{-r^2}r^{2(p+q)-1}\operatorname{d}\!r\operatorname{d}\!\theta\\ &=4\int_0^{\infty}\operatorname{e}^{-r^2}r^{2(p+q)-1}\operatorname{d}\!r\int_0^{\pi/2}(\cos\theta)^{2p-1}(\sin\theta)^{2q-1}\operatorname{d}\!\theta. \end{align} $$ Making the substitution $r^2=t$ in the $r$ integration, we find $$ \int_0^{\infty}\operatorname{e}^{-r^2}r^{2(p+q)-1}\operatorname{d}\!r=\frac{1}{2}\int_0^{\infty}\operatorname{e}^{-t}t^{(p+q)-1}\operatorname{d}\!t=\frac{1}{2}\Gamma(p+q) $$ so we have $$ \Gamma(p)\Gamma(q)=\frac{1}{2}\Gamma(p+q)4\int_0^{\pi/2}(\cos\theta)^{2p-1}(\sin\theta)^{2q-1}\operatorname{d}\!\theta $$ that is $$ \frac{\Gamma(p)\Gamma(q)}{\Gamma(p+q)}=B(p,q)=2\int_0^{\pi/2}(\cos\theta)^{2p-1}(\sin\theta)^{2q-1}\operatorname{d}\!\theta. $$

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  • $\begingroup$ Many thanks for your help kind sir! Much appreciated, and my mistake with the initial integral, it is now corrected. $\endgroup$
    – user211337
    Dec 23, 2014 at 11:13

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