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Math Help - Hard integration problem

  1. #1
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    Hard integration problem

    The question:
    Show that \int_0^1 {x^m(1 - x)^n \ dx = \frac{m!n!}{(m + n + 1)!} for all nonnegative integers m and n.

    My attempt:
    Using integration by parts:
    u = x^m
    u' = mx^{m - 1}
    v' = (1 - x)^n
    v = -\frac{(1 - x)^{n + 1}}{n + 1}
    I_{m,n} = _0^1[-x^m\frac{(1 - x)^{n + 1}}{n + 1}] + m\int_0^1 {x^{m - 1}\frac{(1 - x)^{n + 1}}{n + 1}}
    = _0^1[-x^m\frac{(1 - x)^{n + 1}}{n + 1}] + \frac{m}{n + 1}\int_0^1 {x^{m - 1}(1 - x)^{n + 1}}
    = \frac{m}{n + 1}\int_0^1 {x^{m - 1}(1 - x)^{n + 1}}
    = \frac{m}{n + 1} I_{m-1, n+1}

    I think that's right to far. I'm not sure what to do from here. Any assistance would be most appreciated!
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  2. #2
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    I don't think that's going to work.

    What you have is a special case of the Beta function. Try looking at the Wiki page on that; it shows how the beta function relates to the gamma function.
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  3. #3
    Behold, the power of SARDINES!
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    I have a different method, but I don't know if it will help you.

    Here we go

    The convolution of two functions f(t),g(t) is defined to be

    \displaystyle f*g=\int_{0}^{t}f(t-\tau)g(\tau)d\tau

    using this

    \displaystyle t^n*t^m=\int_{0}^{t}(t-\tau)^n\tau^m d\tau=t^n\int_{0}^{t}(1-\frac{\tau}{t})^n\tau^md\tau

    now sub \displaystyle u=\frac{\tau}{t} \implies tdu=d\tau

    \displaystyle t^n\int_{0}^{1}(1-u)^n(tu)^mtdu= t^{n+m+1}\int_{0}^{1}(1-u)^nu^mdu

    Now taking the Laplace transform of both sides gives

    \displaystyle \mathcal{L}(t^n*t^m)=\mathcal{L}\left((t^{n+m+1}\i  nt_{0}^{1}(1-u)^nu^mdu\right)=\left( \int_{0}^{1}(1-u)^nu^mdu\right)\mathcal{L}(t^{n+m+1})

    \displaystyle \frac{n!}{s^{n+1}} \cdot \frac{m!}{s^{m+1}}=\left( \int_{0}^{1}(1-u)^nu^mdu\right)\frac{(n+m+1)!}{s^{n+m+2}}

    Now solving for the integral gives

    \displaystyle \int_{0}^{1}(1-u)^nu^mdu=\frac{n!m!}{(n+m+1)!}
    Last edited by TheEmptySet; December 6th 2010 at 09:40 PM. Reason: Latex error
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  4. #4
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    Thanks. I'm not familiar with those techniques, I think your method is a bit beyond my level. I'm sure there's a way to do this with more basic methodology.
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  5. #5
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    How about this:

    \begin{aligned}<br />
I_{m,n} & =\frac{m}{n+1}I_{m-1,n+1}\\<br />
 & =\frac{m(m-1)(m-2)...1}{(n+1)(n+2)....(n+m)}I_{0,n+m}\\<br />
 & =\frac{m!n!}{(n+m)!}I_{0,n+m}\end{aligned}


    Then you just have to work out $I_{0,n+m}$ and you have the answer...
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  6. #6
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by Glitch View Post
    The question:
    Show that \int_0^1 {x^m(1 - x)^n \ dx = \frac{m!n!}{(m + n + 1)!} for all nonnegative integers m and n...
    In...

    Beta Function -- from Wolfram MathWorld

    ... You can verify that...

    \displaystyle \int_{0}^{1} u^{m}\ (1-u)^{n}\ du = B(m+1,n+1) = \frac{m!\ n!}{(m+n+1)!} (1)



    Merry Christmas from Italy

    \chi \sigma
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