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Thread: Challenge integral

  1. #1
    MHF Contributor Bruno J.'s Avatar
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    Challenge integral

    This one was given to me as a challenge. I was happy I could solve it!

    Show that
    $\displaystyle \int_0^1\frac{\log(x)\log(1-x)}{x}\: dx = \zeta(3) = \sum_{n=1}^\infty\frac{1}{n^3}$
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  2. #2
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    It's quite easy.

    $\displaystyle \begin{aligned}
    \int_{0}^{1}{x^{j-1}\ln (x)\,dx}&=-\int_{0}^{1}{\int_{x}^{1}{\frac{x^{j-1}}{t}\,dt}\,dx} \\
    & =-\int_{0}^{1}{\int_{0}^{t}{\frac{x^{j-1}}{t}\,dx}\,dt} \\
    & =-\frac{1}{j}\int_{0}^{1}{t^{j-1}\,dt} \\
    & =-\frac{1}{j^{2}}.
    \end{aligned}$

    Hence we get $\displaystyle \int_{0}^{1}{\frac{\ln (x)\ln (1-x)}{x}\,dx}=-\sum\limits_{j=1}^{\infty }{\left( \frac{1}{j}\int_{0}^{1}{x^{j-1}\ln (x)\,dx} \right)}=\sum\limits_{j=1}^{\infty }{\frac{1}{j^{3}}},$ as required. $\displaystyle \quad\blacksquare$
    Last edited by Krizalid; Aug 4th 2009 at 06:25 AM.
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  3. #3
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    First, use integration by parts with $\displaystyle u=\ln(x)\ln(1-x)$, $\displaystyle dv=\frac{dx}{x}$, so that $\displaystyle du=\frac{\ln(1-x)}{x}-\frac{\ln(x)}{1-x}$, $\displaystyle v=\ln{x}$, and
    $\displaystyle \int_0^1\frac{\ln(x)\ln(1-x)}{x}\,dx=\left[\ln^2(x)\ln(1-x)\right]_0^1-\int_0^1\left(\frac{\ln(1-x)}{x}-\frac{\ln(x)}{1-x}\right)\ln(x)\,dx$
    $\displaystyle \int_0^1\frac{\ln(x)\ln(1-x)}{x}\,dx=0-\int_0^1\frac{\ln(x)\ln(1-x)}{x}\,dx+\int_0^1\frac{\ln^2(x)}{1-x}\,dx$
    $\displaystyle 2\int_0^1\frac{\ln(x)\ln(1-x)}{x}\,dx=\int_0^1\frac{\ln^2(x)}{1-x}\,dx$
    $\displaystyle \int_0^1\frac{\ln(x)\ln(1-x)}{x}\,dx=\frac{1}{2}\int_0^1\frac{\ln^2(x)}{1-x}\,dx$
    Now, let $\displaystyle u=-\ln(x)$: x=0 becomes $\displaystyle u=\infty$, x=1 becomes $\displaystyle u=0$, and $\displaystyle x=e^{-u}$, and $\displaystyle dx=-e^{-u}\,du$, so our right-hand integral becomes
    $\displaystyle \frac{1}{2}\int_0^1\frac{\ln^2(x)}{1-x}\,dx=\frac{1}{2}\int_{\infty}^0\frac{(-u)^2}{1-e^{-u}}\cdot-e^{-u}\,du$
    $\displaystyle =\frac{1}{2}\int_0^{\infty}\frac{u^2e^{-u}}{1-e^{-u}}\,du$
    $\displaystyle =\frac{1}{2}\int_0^{\infty}\frac{u^2}{e^u-1}\,du$.
    Now,
    $\displaystyle \int_0^{\infty}\frac{x^{s-1}}{e^x-1}\,dx=\Gamma(s)\zeta(s)$ (see here).
    So
    $\displaystyle \frac{1}{2}\int_0^{\infty}\frac{u^2}{e^u-1}\,du=\frac{1}{2}\Gamma(3)\zeta(3)$
    $\displaystyle =\frac{1}{2}(2!)\zeta(3)$
    $\displaystyle =\zeta(3)$.


    -Kevin C.
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  4. #4
    Super Member Random Variable's Avatar
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    $\displaystyle \int_{0}^{1}{\frac{\ln (x)\ln (1-x)}{x}\,dx}=-\sum\limits_{j=1}^{\infty }{\left( \frac{1}{j}\int_{0}^{1}{x^{j-1}\ln (x)\,dx} \right)} $

    I don't understand what you did here.
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  5. #5
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    $\displaystyle \ln (1-x)=-\sum\limits_{j=1}^{\infty }{\frac{x^{j}}{j}},$ and then switch sum and integral.
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  6. #6
    Moo
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    Quote Originally Posted by Krizalid View Post
    $\displaystyle \ln (1-x)=-\sum\limits_{j=1}^{\infty }{\frac{x^{j}}{j}},$ and then switch sum and integral.
    This kind of stuff has to be justified precisely huh...
    (It has been brought to my attention just a while ago lol)
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  7. #7
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    you know the justification, and it's quite known and I won't do it.
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