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Question: prove $\Gamma(a)\Gamma(b) = \Gamma(a+b)B(a,b)$ using polar transformation:

my attempt:

L.H.S = $$\Gamma(a)\Gamma(b) = \int_0^{\infty}\int_0^{\infty}e^{-(x+y)}x^{a-1}y^{b-1}dxdy$$

Let $x = r\cos(\theta),y = r\sin(\theta)$

then $$\Gamma(a)\Gamma(b) = \int_0^{\pi/2}\int_0^{\infty}re^{-r(\cos(\theta)+\sin(\theta))}r^{a+b-2}\cos^{a-1}(\theta)\sin^{b-1}(\theta)drd\theta$$

$$= \int_0^{\pi/2}\cos^{a-1}(\theta)\sin^{b-1}(\theta)d\theta\int_0^{\infty}e^{-r(\cos(\theta)+\sin(\theta))}r^{a+b-1}dr $$

the first integral is almost $ = B(a,b)$ and the second integral is almost $ = \Gamma(a+b)$

but I am stuck there and cannot separate $\cos(\theta) + \sin(\theta)$ from $e$

any help is much appreciated.

AmirWG
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    Try $\Gamma(c)\color{gray}{=\int_0^\infty x^{c-1}e^{-x},dx}=2\int_0^\infty z^{2c-1}e^{-z^2},dz$ (for $c\in{a,b}$ and $z\in{x,y}$). – metamorphy Jun 22 '22 at 17:55
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    Are you required to make it wiht polar transformation? Because with the change $x=uv$ $y=u(1-v)$ it's easier – Guillermo García Sáez Jun 22 '22 at 17:59
  • @guillerminho77 yes, that's exactly what the question is asking for. – AmirWG Jun 22 '22 at 18:22
  • @metamorphy what was the transformation you used? – AmirWG Jun 22 '22 at 18:24
  • Their transformation was $x=z^2$ – FShrike Jun 22 '22 at 18:26
  • This answer might be of some help. – robjohn Jun 22 '22 at 18:33
  • There's a derivation here https://en.wikipedia.org/wiki/Beta_function#Relationship_to_the_gamma_function – Blitzer Jun 22 '22 at 18:33
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    IN your second interval, $\theta$ is a constant, so substitute $u=r(\cos \theta + \sin \theta )$ and you should get a constant times $\Gamma(a+b)$. – Blitzer Jun 22 '22 at 18:36
  • @Biltzer, yeah I know but that still does not completely solve it. – AmirWG Jun 22 '22 at 18:46
  • @@metamorphy , I tried your transformation but It does not work really – AmirWG Jun 22 '22 at 18:47
  • @metamorphy btw, using transformation $x = z^2$ should give you $2c - 2$ not $2c-1$ – AmirWG Jun 22 '22 at 19:56
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    @AmirWG: $\color{blue}{1.}$ Still $2c-1$ (you forgot $dx=2\color{red}{z},dz$). $\color{blue}{2.}$ What exactly doesn't work? You get $e^{-x^2-y^2}$ under the double integral, after which the polar substitution yields $e^{-r^2}$, and the integrations separate. $\color{blue}{0.}$ ... and, $\color{blue}{\text{btw}}$, what does $\Gamma(\color{red}{a,b})$ mean? (typo, I know ;) – metamorphy Jun 23 '22 at 02:46
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    First let $x=u^2$ and $y=v^2$, then let $u=r\cos(\theta)$ and $v=r\sin(\theta)$. – Gary Jun 23 '22 at 05:45
  • @metamorphy , gave it another try and your solution works just perfect, I would love to mark it as the correct answer. – AmirWG Jun 23 '22 at 16:01

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Clearly for $\theta\in[0,\pi/2]$, $\cos(\theta)+\sin(\theta)>0$. Let $t=r(\cos(\theta)+\sin(\theta))$ and then $r=\frac{t}{\cos(\theta)+\sin(\theta)}$. So \begin{eqnarray} &&\int_0^{\infty}e^{-r(\cos(\theta)+\sin(\theta))}r^{a+b-1}dr\\ &=&\int_0^{\infty}e^{-t}\frac{t^{a+b-1}}{[\cos(\theta)+\sin(\theta)]^{a+b}}dt\\ &=&\frac{1}{[\cos(\theta)+\sin(\theta)]^{a+b}}\int_0^{\infty}e^{-t}t^{a+b-1}dt\\ &=&\frac{1}{[\cos(\theta)+\sin(\theta)]^{a+b}}\Gamma(a+b). \end{eqnarray} Therefore \begin{eqnarray} \Gamma(a)\Gamma(b) &=& \int_0^{\pi/2}\int_0^{\infty}re^{-r(\cos(\theta)+\sin(\theta))}r^{a+b-2}\cos^{a-1}(\theta)\sin^{b-1}(\theta)drd\theta\\ &=& \int_0^{\pi/2}\cos^{a-1}(\theta)\sin^{b-1}(\theta)d\theta\int_0^{\infty}e^{-r(\cos(\theta)+\sin(\theta))}r^{a+b-1}dr\\ &=& \Gamma(a+b) \int_0^{\pi/2}\frac{\cos^{a-1}(\theta)\sin^{b-1}(\theta)}{[\cos(\theta)+\sin(\theta)]^{a+b}}d\theta\\ &=& \Gamma(a+b) \int_0^{\pi/2}\frac{\tan^{b-1}(\theta)}{[1+\tan(\theta)]^{a+b}}d\tan(\theta)\\ &\stackrel{\tan(\theta)\to t}{=}& \Gamma(a+b) \int_0^{\infty}\frac{t^{b-1}}{(1+t)^{a+b}}dt\\ &=& \Gamma(a+b)B(a,b). \end{eqnarray}

xpaul
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