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Thread: a couple more definite integrals

  1. #1
    Super Member Random Variable's Avatar
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    a couple more definite integrals

    $\displaystyle \displaystyle \int_{0}^{1} \ln(1+x) \ln(1-x) \ dx $

    If the above integral is too easy, also try to evaluate$\displaystyle \displaystyle \int_{0}^{1} \ln(1+x) \ln (x) \ln(1-x) \ dx $ (though I personally wouldn't recommend it ).
    Last edited by Random Variable; Mar 29th 2011 at 07:25 PM.
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  2. #2
    MHF Contributor chisigma's Avatar
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    With the substitution $\displaystyle t=1-x$ the integral becomes...

    $\displaystyle \displaystyle I= \int_{0}^{1} \ln t\ \ln (2-t)\ dt$ (1)

    Now is...

    $\displaystyle \displaystyle \ln (2-t)= \ln 2 - \sum_{n=1}^{\infty} \frac{t^{n}}{n\ 2^{n}}$ (2)

    ... and remembering the general formula...

    $\displaystyle \displaystyle \int_{0}^{1} t^{n}\ \ln t\ dt = -\frac{1}{(n+1)^{2}}$ (3)

    ... You obtain...

    $\displaystyle \displaystyle I= -\ln 2 + \sum_{n=1}^{\infty} \frac{1}{n\ (n+1)^{2}\ 2^{n}}$ (4)

    To extablish if the sum of the series in (4) can be expressed as function of known constants it is necessary some more work...

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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  3. #3
    Super Member Random Variable's Avatar
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    If you make a slightly different substitution, you'll end up having to evaluate a much simpler series.


    $\displaystyle \displaystyle \int^{1}_{0} \ln (1+x) \ln(1-x) \ dx = \frac{1}{2} \int^{1}_{-1} \ln (1+x) \ln(1-x) \ dx $


    Now let $\displaystyle \displaystyle 2t = 1+x $


    $\displaystyle \displaystyle = \int^{1}_{0} \ln (2t) \ln(2-2t) \ dt = \int^{1}_{0} \big(\ln 2 + \ln t\big) \big(\ln 2 + \ln(1-t) \big) \ dt$

    $\displaystyle \displaystyle = \ln^{2}(2) \int ^{1}_{0} \ dt + \ln 2 \int^{1}_{0} \ln(1-t) \ dt + \ln 2 \int^{1}_{0} \ln t \ dt + \int^{1}_{0} \ln t \ln(1-t) \ dt $

    The first three integrals are straightforward, and the last one is equivalent to evaluating $\displaystyle \displaystyle \sum^{\infty}_{n =1} \frac{1}{n(n+1)^2} $ which is much easier to evaluate than $\displaystyle \displaystyle \sum^{\infty}_{n =1} \frac{1}{n(n+1)^{2}2^{n}} $ .


    The final answer is $\displaystyle \displaystyle \ln^{2}(2) - 2 \ln 2 + 2 - \frac{\pi^{2}}{6} $
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