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Really tricky integral

  1. Mar 23, 2013 #1
    1. The problem statement, all variables and given/known data
    I=\int_{0}^{\infty} \frac{x^3}{e^x-1}\ln(e^x-1)\,dx
    2. Relevant equations

    Any identity involving [itex]\zeta(s)[/itex].

    3. The attempt at a solution
    I noted that:
    \ln(e^x - 1) = \ln(x) + \sum_{n=0}^{\infty} \frac{B_{n+1}(1)}{(n+1)^2n!}x^{n+1}
    Therefore, the integral can be split into two parts:
    I = \int_{0}^{\infty} \frac{x^3}{e^x-1}\ln(x)\,dx + \int_{0}^{\infty} \frac{x^3}{e^x-1}\sum_{n=0}^{\infty} \frac{B_{n+1}(1)}{(n+1)^2n!}x^{n+1}\,dx = I_{1} +I_{2}

    Ignoring the first one for now,
    I_{2} = \sum_{n=0}^{\infty} \frac{B_{n+1}(1)}{(n+1)^2n!} \int_{0}^{\infty} \frac{x^{n+4}}{e^x-1}\,dx = \sum_{n=0}^{\infty} \frac{B_{n+1}(1)}{(n+1)^2n!} \Gamma(n+5)\zeta(n+5) = \sum_{n=0}^{\infty} B_{n+1}(1) \frac{(n+4)(n+3)(n+2)}{n+1}\zeta(n+5) = 12\zeta(5) + \sum_{m=0}^{\infty} B_{2m}\frac{(2m+5)(2m+4)(2m+3)}{2m+2}\zeta(2m+6)
    Now I suppose that you can use the relationship between [itex]B_{2m}[/itex] and [itex]\zeta(2m)[/itex] but I don't see how that would help. I want to believe that [itex]I_{1}[/itex] can be done via contour integration, but I can't come up with anything. Thanks for reading this mess.
  2. jcsd
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