Euler Integrals

[cepheid]

1. The problem statement, all variables and given/known data

For fun: show that

B(a,b) = \int_0^1{x^{a-1}(1-x)^{b-1}\,dx} = \frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}

where a > 0 , b > 0 . Hint: start from the product \Gamma(a)\Gamma(b) and switch to polar coordinates. The radial integral is proportional to \Gamma(a+b).


2. Relevant equations

\Gamma(a) = \int_0^{\infty}{x^{a-1}e^{-x}\,dx}

3. The attempt at a solution

\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}


x = r\cos\theta, \ y = r\sin\theta, \ \ \ \ dxdy = rdrd\theta

\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}


= \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}


I'm stuck here. I don't know how to sort out the exponential term (which depends on both r and \theta) in order to obtain a separate radial integral.

[dextercioby]

There's a trick here:

\Gamma (a)=\int_{0}^{\infty} x^{a-1} e^{-x} {} dx

Make the substitution x=u^{2} and get

\Gamma (a)=2\int_{0}^{\infty} u^{2a-1} e^{-u^{2}} {} du

Do the same for

\Gamma (b)=2\int_{0}^{\infty} v^{2b-1} e^{-v^{2}} {} dv

Consider the product of the 2 integrals+ Fubini's theorem to get

\Gamma (a)\Gamma (b)=4\int_{0}^{\infty} \int_{0}^{\infty} u^{2a-1} v^{2v-1} e^{-(u^{2}+v^{2})} {} du {} dv

Switch to polar coordinates (r,\varphi) and perform the "r" integration to get

\Gamma (a)\Gamma (b)=2\Gamma (a+b)\int_{0}^{\pi/2} \left(\cos^{2}\varphi\right)^{\frac{2a-1}{2}} \left(\sin^{2}\varphi\right)^{\frac{2b-1}{2}} \ d\varphi

and finally make the substitution \cos^{2}\varphi=z

You'll get the identity easily, just watch the "-" signs and reversing the order of integration.

[HallsofIvy]

The \theta integral is NOT from 0 to 2\pi, it is from o to \pi /2. I'm not sure why you are doing that however. The problem does not ask you to evaluate the integrals, just show that the two sides are the same.