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Math Help - Tricky Integral

  1. #1
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    Tricky Integral

    Hi I'm doing a question to find the pressure of a gas made of massless bosons. I've done everything else and now the question has resulted in me needing to evaluate this integral
    \int_{0}^{\infty}\frac{z^2}{e^{z}+1}

    This I'm pretty stuck on I've tried so far to multiply the integrand by

    \frac{e^z-1}{e^z-1}

    which gives me a second term which is easy to evaluate in terms of the zeta function but again a first term I can't seem to be able to do much with.

    Any hints in the right direction would be much appreciated!

    Thank you =)
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  2. #2
    MHF Contributor chisigma's Avatar
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    With the substitution e^{-x}= 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)

    With a little of patience You can arrive to the 'general formula'...

    \displaystyle \int t^{m}\ \ln^{n} t \ dt = t^{m+1}\ \sum_{k=0}^{\infty} \frac{(-1)^{k}}{(m+1)^{k+1}}\ \frac{n!}{(n-k)!}\ \ln^{n-k} t + c (2)

    ... and from (2) to the 'particular formula'...

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

    Very well!... now using (3) You can 'meet the goal'...

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

    \displaystyle = 2\ \{\sum_{k=1}^{\infty} \frac{1}{k^{3}} - 2\ \sum_{k=1}^{\infty} \frac{1}{(2k)^{3}} \} = 2\ (1-\frac{1}{4}) \ \zeta(3) = \frac{3}{2}\ \zeta (3) (4)

    ... where \zeta(s) is the so called 'Riemann Zeta Function'...

    Kind regards

    \chi \sigma
    Thanks from thelostchild
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  3. #3
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    Thank you very much - nice to see how you would deal with a problem like this in general as well
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