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Math Help - improper integral can anyone calculate this?

  1. #1
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    improper integral can anyone calculate this?

    \int_{0}^{\infty}\frac{x^2}{e^x-1}dx
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  2. #2
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    We normally like to see some work done. Anyway, you are going to need to rewrite the integral like so
    \lim_{t\to \infty} \int_0^t\! x^2 (e^x -1)^{-1}\, dx
    and then integrate by parts. Let u = x^2 and then dv = (e^x-1)^{-1}\, dx.
    See what you can do with that.
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  3. #3
    MHF Contributor chisigma's Avatar
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    With the successive substitutions x= \ln \xi and \xi = \frac{1}{t} the integral becomes...

    \displaystyle \int_{0}^{\infty} \frac{x^{2}}{e^{x}-1}\ dx = \int_{0}^{1} \frac{\ln^{2} t}{1-t}\ dt (1)

    Now with a little of patience You can verify that is...

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

    ... and because for |t|<1 is...

    \displaystyle \frac{1}{1-t} = \sum_{m=0}^{\infty} t^{m} (3)

    ... combining (1), (2) and (3) we arrive to write...

    \displaystyle \int_{0}^{1} \frac{\ln^{2} t}{1-t}\ dt = \sum_{m=0}^{\infty} \int_{0}^{1} t^{m}\ \ln^{2} t\ dt = 2\ \sum_{m=0}^{\infty} \frac{1}{(m+1)^{3}} = 2\ \zeta (3) (4)

    ... where \zeta(*) is the 'Riemann Zeta Function'...

    Kind regards

    \chi \sigma
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