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Thread: Raabe's Integral

  1. #1
    MHF Contributor chiph588@'s Avatar
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    Raabe's Integral

    Here's a fun challenge!

    Evaluate Raabe's integral:

    $\displaystyle \int_{0}^{1} \log\Gamma(x)dx $


    (This used to be my signature. )
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  2. #2
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    it's quite easy, first note that $\displaystyle \int_{0}^{1}{\ln \left( \Gamma (x) \right)\,dx}=\frac{1}{2}\left( \ln \pi -\int_{0}^{1}{\ln (\sin \pi x)\,dx} \right),$ for the last integral put $\displaystyle t=\pi x$ so the last challenge is to compute $\displaystyle \int_{0}^{\pi }{\ln (\sin x)\,dx},$ which comes from the fact that $\displaystyle \int_{0}^{\frac{\pi }{2}}{\ln (\sin x)\,dx}=-\frac{\pi }{2}\ln 2.$

    eventually, your integral equals $\displaystyle \frac12\ln(2\pi).$
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  3. #3
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by Krizalid View Post
    it's quite easy, first note that $\displaystyle \int_{0}^{1}{\ln \left( \Gamma (x) \right)\,dx}=\frac{1}{2}\left( \ln \pi -\int_{0}^{1}{\ln (\sin \pi x)\,dx} \right),$ for the last integral put $\displaystyle t=\pi x$ so the last challenge is to compute $\displaystyle \int_{0}^{\pi }{\ln (\sin x)\,dx},$ which comes from the fact that $\displaystyle \int_{0}^{\frac{\pi }{2}}{\ln (\sin x)\,dx}=-\frac{\pi }{2}\ln 2.$

    eventually, your integral equals $\displaystyle \frac12\ln(2\pi).$
    Correct! The hard part of the problem though is showing $\displaystyle \int_{0}^{\frac{\pi }{2}}{\ln (\sin x)\,dx}=-\frac{\pi }{2}\ln 2$.
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  4. #4
    Super Member Random Variable's Avatar
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    $\displaystyle \int_{0}^{1}{\ln \left( \Gamma (x) \right)\,dx}=\frac{1}{2}\left( \ln \pi -\int_{0}^{1}{\ln (\sin \pi x)\,dx} \right) $

    What exactly was done here? Integration by parts?
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  5. #5
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by Random Variable View Post
    $\displaystyle \int_{0}^{1}{\ln \left( \Gamma (x) \right)\,dx}=\frac{1}{2}\left( \ln \pi -\int_{0}^{1}{\ln (\sin \pi x)\,dx} \right) $

    What exactly was done here? Integration by parts?
    He used the Gamma function's reflection formula.

    Gamma function - Wikipedia, the free encyclopedia
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  6. #6
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    Quote Originally Posted by chiph588@ View Post
    Correct! The hard part of the problem though is showing $\displaystyle \int_{0}^{\frac{\pi }{2}}{\ln (\sin x)\,dx}=-\frac{\pi }{2}\ln 2$.
    i've solved that one before, it should be somewhere.

    no time now to search for, but i'd take a look tomorrow.

    Quote Originally Posted by chiph588@ View Post
    He used the Gamma function's reflection formula.

    Gamma function - Wikipedia, the free encyclopedia
    yeah, that's why i said it's an easy problem, because it turns to be to another known problem.
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  7. #7
    Super Member Random Variable's Avatar
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    Quote Originally Posted by chiph588@ View Post
    He used the Gamma function's reflection formula.

    Gamma function - Wikipedia, the free encyclopedia
    $\displaystyle \Gamma(x) \Gamma(1-x) = \frac{\pi}{\sin \pi x} $

    $\displaystyle \ln \Gamma(x) + \ln \Gamma(1-x) = \ln \pi - \ln \sin \pi x $

    What do you do with the second term on the left?
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  8. #8
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by Random Variable View Post
    $\displaystyle \Gamma(x) \Gamma(1-x) = \frac{\pi}{\sin \pi x} $

    $\displaystyle \ln \Gamma(x) + \ln \Gamma(1-x) = \ln \pi - \ln \sin \pi x $

    What do you do with the second term on the left?
    Perform $\displaystyle u $ substitution by letting $\displaystyle u=1-x $ to get
    $\displaystyle \int_{0}^{1} \log\Gamma(1-x)dx = \int_{0}^{1} \log\Gamma(u)du $.

    You could also look at it this way: $\displaystyle \{\Gamma(x)|x\in[0,1]\} = \{\Gamma(1-x)|x\in[0,1]\} $, so when we integrate we can swap the two functions.
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  9. #9
    Super Member Random Variable's Avatar
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    Then shouldn't it be the following:

    $\displaystyle \int^{1/2}_{0} \ln \Gamma(x) \ dx + \int^{1/2}_{0} \ln \Gamma(1-x) \ dx = \int_{0}^{1/2} (\ln \pi - \ln \sin \pi x) \ dx $

    $\displaystyle \int^{1/2}_{0} \ln \Gamma(x) \ dx + \int^{1}_{1/2} \ln \Gamma(t) \ dt = \int_{0}^{1/2} (\ln \pi - \ln \sin \pi x) \ dx $

    EDIT: $\displaystyle \int^{1}_{0} \ln \Gamma(x) \ dx = \frac{1}{2} \ln \pi - \int^{1/2}_{0} \ln \sin \pi x \ dx $

    Actually, I think that might be equivalent.
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  10. #10
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by Random Variable View Post
    Then shouldn't it be the following:

    $\displaystyle \int^{1/2}_{0} \ln \Gamma(x) \ dx + \int^{1/2}_{0} \ln \Gamma(1-x) \ dx = \int_{0}^{1/2} (\ln \pi - \ln \sin \pi x) \ dx $

    $\displaystyle \int^{1/2}_{0} \ln \Gamma(x) \ dx + \int^{1}_{1/2} \ln \Gamma(t) \ dt = \int_{0}^{1/2} (\ln \pi - \ln \sin \pi x) \ dx $

    EDIT: $\displaystyle \int^{1}_{0} \ln \Gamma(x) \ dx = \frac{1}{2} \ln \pi - \int^{1/2}_{0} \ln \sin \pi x \ dx $

    Actually, I think that might be equivalent.
    Your method works too, thus showing $\displaystyle \int^{\frac{1}{2}}_{0} \log (\sin (\pi x)) dx = \frac{1}{2}\int^{1}_{0} \log (\sin (\pi x)) dx $.
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  11. #11
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    Take a look here , the integral can be generalized to that has a new lower and upper limit :


    http://www.mathhelpforum.com/math-he...egral-bee.html
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  12. #12
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