Euler Integrals

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cepheid
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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.
 
Last edited:

Answers and Replies

  • #2
dextercioby
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There's a trick here:

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

Make the substitution [itex] x=u^{2} [/itex] and get

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

Do the same for

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

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

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

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

[tex] \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 [/tex]

and finally make the substitution [itex] \cos^{2}\varphi=z [/itex]

You'll get the identity easily, just watch the "-" signs and reversing the order of integration.
 
Last edited:
  • #3
HallsofIvy
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The [itex]\theta[/itex] integral is NOT from 0 to [itex]2\pi[/itex], it is from o to [itex]\pi /2[/itex]. 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.
 

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