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Math Help - gamma integral

  1. #1
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    gamma integral

    i am trying to express the following in form of a gamma integral but so far i have been unsuccessful:
    - \int_0^1 x^k ln(x)
    some of the substitutions i have tried are:  x = \frac{1}{t} ,  =t ,  e^{-t} , = log(\frac{1}{t}) ,  = e^{-\frac{1}{t}} and even  x^k = t , any pointer would be appreciated.
    i am tring to prove the following result:
     -\int_0^1 x^k ln(x) = \frac{1}{(k+1)^2}


    edit i omitted the variable of integration so:
    - \int_0^1 x^k ln(x) dx
    for the first, and
     -\int_0^1 x^k ln(x) dx = \frac{1}{(k+1)^2}
    for the last integrals respectively.
    Last edited by phycdude; November 7th 2011 at 01:36 AM. Reason: omisions
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Re: gamma integral

    Quote Originally Posted by phycdude View Post
    i am tring to prove the following result:
     -\int_0^1 x^k ln(x) = \frac{1}{(k+1)^2}
    Consider I(k)=\int_{0}^1x^{k}\;dx=\dfrac{1}{k+1} . Using parametric derivation, I'(k)=\int_{0}^1x^{k}\ln x\;dx . Conclude.
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  3. #3
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    Re: gamma integral

    i'd love to say i understand you method, but i do not, that said, im working on a topic in mathematicsl physics (gamma fuctions, beta fucntions and such), hence why i'm trying to convert this problem into the general form of gamma a function.
    Quote Originally Posted by FernandoRevilla View Post
    Consider I(k)=\int_{0}^1x^{k}\;dx=\dfrac{1}{k+1} . Using parametric derivation, I'(k)=\int_{0}^1x^{k}\ln x\;dx . Conclude.

    edit:
    see the attached scan of arfken and weber i think, sixth ed.
    Attached Thumbnails Attached Thumbnails gamma integral-gammafunc.png  
    Last edited by phycdude; November 7th 2011 at 03:04 AM. Reason: adding an attachement
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  4. #4
    MHF Contributor FernandoRevilla's Avatar
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    Re: gamma integral

    In such a case, using the substitution u=-(k+1)\log x you'll obtain:

    -\int_0^1x^k\ln x\;dx=\ldots=\frac{1}{(k+1)^2}\int_0^{+\infty}e^{-u}u\;du=\frac{\Gamma (2)}{(k+1)^2}=\frac{1}{(k+1)^2}
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  5. #5
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    Re: gamma integral

    Thats brilliant, thanks
    Quote Originally Posted by FernandoRevilla View Post
    In such a case, using the substitution u=-(k+1)\log x you'll obtain:

    -\int_0^1x^k\ln x\;dx=\ldots=\frac{1}{(k+1)^2}\int_0^{+\infty}e^{-u}u\;du=\frac{\Gamma (2)}{(k+1)^2}=\frac{1}{(k+1)^2}
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  6. #6
    MHF Contributor chisigma's Avatar
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    Re: gamma integral

    Quote Originally Posted by phycdude View Post
    i am trying to express the following in form of a gamma integral but so far i have been unsuccessful:
    - \int_0^1 x^k ln(x)
    some of the substitutions i have tried are:  x = \frac{1}{t} ,  =t ,  e^{-t} , = log(\frac{1}{t}) ,  = e^{-\frac{1}{t}} and even  x^k = t , any pointer would be appreciated.
    i am tring to prove the following result:
     -\int_0^1 x^k ln(x) = \frac{1}{(k+1)^2}
    Integration by parts permits an elementary solution...

    \int_{0}^{1}x^{k}\ \ln x\ dx = |\frac{x^{k+1}}{k+1}\ \ln x|_{0}^{1} - \frac{1}{k+1}\ \int_{0}^{1} x^{k}\ dx = - \frac{1}{(k+1)^{2}} (1)

    Kind regards

    \chi \sigma
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