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Math Help - Computing an Integral

  1. #1
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    Computing an Integral

    I cant figure out this integral............ please help

    \int_0^z\ln\Gamma(x+1)dx

    Thanks in advance..
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Arka View Post
    I cant figure out this integral............ please help

    \int_0^z\ln\Gamma(x+1)dx

    Thanks in advance..
    That's easy:

    \displaystyle \int_0^z\ln\Gamma(x+1)dx=\frac{z}{2}\ln(2\pi)-\frac{z(z+1)}{2}+z\ln\Gamma(z+1) \displaystyle -\ln\left\{(2\pi)^{\frac{z}{2}}\exp\left(\frac{-z(z+1)}{2}-\frac{\gamma z^2}{2}\right)\prod_{k=1}^{\infty}\left\{\left(1+\  frac{z}{k}\right)^k\exp\left(-z+\frac{z^2}{2k}\right)\right\}\right\}
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  3. #3
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by Arka View Post
    I cant figure out this integral............ please help

    \int_0^z\ln\Gamma(x+1)dx

    Thanks in advance..
    In a previous post [I don't remember exactly ... but I think is in the 'math challenge forum'...] there is a [non elementar..] expansions I found some years ago...

    \displaystyle \ln \Gamma (1+x)= - \gamma\ x + \sum_{n=2}^{\infty} (-1)^{n} \zeta(n)\ \frac{x^{n}}{n} (1)

    ... where \gamma is the 'Euler's constant' and \zeta (*) the 'Riemann zeta function'... integrating (1) 'term by term' You obtain...

    \displaystyle \int_{0}^{z} \ln \Gamma (1+x)\ dx= - \gamma\ \frac{z^{2}}{2} + \sum_{n=2}^{\infty} (-1)^{n} \zeta(n)\ \frac{z^{n+1}}{n\ (n+1)} (2)

    Kind regards

    \chi \sigma
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  4. #4
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    Quote Originally Posted by Drexel28 View Post
    That's easy:

    \displaystyle \int_0^z\ln\Gamma(x+1)dx=\frac{z}{2}\ln(2\pi)-\frac{z(z+1)}{2}+z\ln\Gamma(z+1) \displaystyle -\ln\left\{(2\pi)^{\frac{z}{2}}\exp\left(\frac{-z(z+1)}{2}-\frac{\gamma z^2}{2}\right)\prod_{k=1}^{\infty}\left\{\left(1+\  frac{z}{k}\right)^k\exp\left(-z+\frac{z^2}{2k}\right)\right\}\right\}
    Would you please show me the steps?........because I couldn't really get you.
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