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Math Help - Challenging Integral?.

  1. #1
    Eater of Worlds
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    Challenging Integral?.

    This may not be that bad, but give it a go for those who wish.

    Let's see what interesting methods are come up with.

    \int_{0}^{1}ln(x)ln(1-x)dx
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  2. #2
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by galactus View Post
    This may not be that bad, but give it a go for those who wish.

    Let's see what interesting methods are come up with.

    \int_{0}^{1}ln(x)ln(1-x)dx
    We know that -\sum_{n=0}^{\infty}\frac{x^{n+1}}{n+1}=\ln(1-x)

    so then I think what you can do is this

    -\int{\ln(x)}\sum_{n=0}^{\infty}\frac{x^{n+1}}{n+1}  dx=-\sum_{n=0}^{\infty}\frac{1}{n+1}\int\ln(x)x^{n+1}

    \int_0^{1}\ln(x)\cdot{x^{n+1}}=\bigg[x^{n}\bigg[\frac{x^2\ln(x)}{n+2}-\frac{x^2}{(n+2)^2}\bigg]\bigg]\bigg|_0^{1}=\frac{-1}{(n+2)^2}

    so we have

    \sum_{n=0}^{\infty}\frac{1}{(n+1)(n+2)^2}=2-\frac{\pi^2}{6}

    by PFD and use of a riemann zeta table =D

    NOTE: what I mean by PFD was that after I decomposed it it gave a telescoping series and a riemann zeta function, which came out to be 1-\bigg(\frac{\pi^2}{6}-1\bigg), the latter riemann zeta was gotten using the CRC table of values
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  4. #4
    Eater of Worlds
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    Very good. I knew you fellas would chew it up and spit it out.

    Here is what I done. Which is similar to you fellas line of thought.

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

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

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

    \sum_{k=1}^{\infty}\frac{1}{k(k+1)^{2}}

    \sum_{k=1}^{\infty}\frac{1}{k(k+1)}-\sum_{k=1}^{\infty}\frac{1}{(k+1)^{2}}

    =1-\left(\frac{{\pi}^{2}}{6}-1\right)

    =\boxed{2-\frac{{\pi}^{2}}{6}}
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