# Homework Help: Euler Integrals

1. Jan 31, 2007

### cepheid

Staff Emeritus
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.

Last edited: Jan 31, 2007
2. Jan 31, 2007

### 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.

Last edited: Jan 31, 2007
3. Jan 31, 2007

### 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.