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Euler Integrals

  1. Jan 31, 2007 #1

    cepheid

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

    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].


    2. Relevant equations

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

    3. 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: Jan 31, 2007
  2. jcsd
  3. Jan 31, 2007 #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: Jan 31, 2007
  4. Jan 31, 2007 #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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