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Math Help - Integration of the product of two logarithms

  1. #1
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    Integration of the product of two logarithms

    Find \int\limits_{0}^{1}ln(x)ln(1-x)\, dx

    According to Wolfram alpha, the answer is 2 - \frac{\pi^2}{6}

    Is there a way of arriving at this result without using dilogarithms? I've tried using integration by parts and logarithmic series, but to no avail.
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  2. #2
    MHF Contributor chisigma's Avatar
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    As in...

    http://www.mathhelpforum.com/math-he...te-159304.html

    ...You start with the general formula...

    \displaystyle \int_{0}^{1} x^{m}\ \ln^{n} x\ dx = (-1)^{n}\ \frac{n!}{(m+1)^{n+1}} (1)

    ... and because for |x|<1...

    \displaystyle \ln (1-x)= - \sum_{m=1}^{\infty} \frac{x^{m}}{m} (2)

    ... combining (1) and (2) You have...

    \displaystyle \int_{0}^{1} \ln x\ \ln (1-x)\ dx = \sum_{m=1}^{\infty} -\frac{1}{m}\ \int_{0}^{1} x^{m}\ \ln x\ dx=

    \displaystyle = \sum_{m=1}^{\infty} \frac{1}{m\ (m+1)^{2}} = \sum_{m=1}^{\infty} \frac{1}{m\ (m+1)} - \sum_{m=1}^{\infty} \frac{1}{(m+1)^{2}} = 1+1-\frac{\pi^{2}}{6} (3)

    Kind regards

    \chi \sigma
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