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Math Help - Wallis' integral

  1. #1
    Member courteous's Avatar
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    Question Wallis' integral

    How would you show that \displaystyle{  \int_0^1 (x-x^2)^n dx = \frac{(n!)^2}{(2n+1)!}  } ?
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  2. #2
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    Might not be much help but,

    \displaystyle \int_0^1 (x-x^2)^n ~dx = \int_0^1 (x(1-x))^n ~dx= \int_0^1 x^n(1-x)^n ~dx = \frac{\Gamma (n)\Gamma (n)}{\Gamma (2n)}= \frac{(n-1)!(n-1)!}{(2n-1)!}

    Now have a play around with those factorials.
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  3. #3
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by pickslides View Post
    Might not be much help but,

    \displaystyle \int_0^1 (x-x^2)^n ~dx = \int_0^1 (x(1-x))^n ~dx= \int_0^1 x^n(1-x)^n ~dx = \frac{\Gamma (n)\Gamma (n)}{\Gamma (2n)}= \frac{(n-1)!(n-1)!}{(2n-1)!}

    Now have a play around with those factorials.
    The 'Beta function' is...

    \displaystyle B(h,k)= \int_{0}^{1} x^{h-1}\ (1-x)^{k-1}\ dx , h>0, k>0 (1)

    ... and its 'connection' with the 'Gamma function' is...

    \displaystyle B(h,k)= \frac{\Gamma(h)\ \Gamma(k)}{\Gamma(h+k)} (2)

    Setting n=h+1=k+1 in (1) and (2) we obtain...

    \displaystyle \int_{0}^{1} x^{n}\ (1-x)^{n}\ dx = \frac{\Gamma(n+1)\ \Gamma(n+1)}{\Gamma(2n+2)} = \frac{(n!)^{2}}{(2n+1)!} (3)

    Kind regards

    \chi \sigma
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