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Homework Statement
For fun: show that
[tex] B(a,b) = \int_0^1{x^{a-1}(1-x)^{b-1}\,dx} = \frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)} [/tex]
where [itex] a > 0 [/itex], [itex] b > 0 [/itex]. Hint: start from the product [itex]\Gamma(a)\Gamma(b)[/itex] and switch to polar coordinates. The radial integral is proportional to [itex]\Gamma(a+b)[/itex].
Homework Equations
[tex] \Gamma(a) = \int_0^{\infty}{x^{a-1}e^{-x}\,dx} [/tex]
The Attempt at a Solution
[tex] \Gamma(a)\Gamma(b) = \int_0^{\infty}{x^{a-1}e^{-x}\,dx}\int_0^{\infty}{y^{b-1}e^{-y}\,dy} = \int_0^{\infty}\!\!\int_0^{\infty}{x^{a-1}e^{-x}y^{b-1}e^{-y}\,dxdy}
[/tex]
[tex] x = r\cos\theta, \ y = r\sin\theta, \ \ \ \ dxdy = rdrd\theta [/tex]
[tex] \Gamma(a)\Gamma(b) = \int_0^{2\pi}\!\!\int_0^{\infty}{(r\cos\theta)^{a-1}e^{-r\cos\theta}(r\sin\theta)^{b-1}e^{-r\sin\theta}\,rdrd\theta}
[/tex]
[tex] = \int_0^{2\pi}\!\!\int_0^{\infty}{r^{a-1}r^{b-1}(\cos\theta)^{a-1}(\sin\theta)^{b-1}e^{-r(\cos\theta+\sin\theta)}\,rdrd\theta}
[/tex]
I'm stuck here. I don't know how to sort out the exponential term (which depends on both r and [itex]\theta[/itex]) in order to obtain a separate radial integral.
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