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Need help with this difficult Integral involving exponentials

  1. Jun 12, 2014 #1
    1. The problem statement, all variables and given/known data

    This is the integral
    [tex]
    \int \frac{x^2}{e^x-1}dx
    [/tex]


    2. Relevant equations
    Can this even be solved in closed form?


    3. The attempt at a solution
    Only method I can think of is integration by parts over numerous steps. I did that until I get the following:

    [tex]
    \int \frac{x^2}{e^x-1}dx = x^2 \ln{(e^x-1)} -\frac{5x^3}{3}-2 \int x \ln{(e^x-1)} dx
    [/tex]

    Now I am stuck on the last integral above: [itex]\int x \ln{(e^x-1)} dx[/itex] and can't find how to integrate it.

    Any ideas or suggestions much appreciated :)
     
  2. jcsd
  3. Jun 12, 2014 #2
    Is this acceptable?

    Can we do this? If x is taken to always be positive then:

    [tex]
    \begin{align*}
    \frac{1}{(e^x-1)} &= \frac{e^{-x}}{1-e^{-x}} \\
    &= e^{-x}\frac{1}{1-e^{-x}} \\
    &=e^{-x}\sum_{n=0}^{\infty}e^{-nx} \\
    &=\sum_{n=0}^{\infty}e^{-(n+1)x} \\
    &=\sum_{n=1}^{\infty}e^{-nx}
    \end{align*}
    [/tex]

    Then subbing this series into the original integral

    [tex]
    \begin{align*}
    \int \frac{x^2}{e^x-1}dx &= \int x^2 \Biggr( \sum_{n=0}^{\infty}e^{-(n+1)x} \Biggr) dx \\
    &= \sum_{n=0}^{\infty} \int x^2 \Biggr( e^{-(n+1)x} \Biggr) dx \\
    &= \sum_{n=1}^{\infty} \int x^2 e^{-nx} dx
    \end{align*}
    [/tex]
     
  4. Jun 12, 2014 #3

    micromass

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    You cannot solve it in closed form, you need the polylogarithm function. You can find series solutions though.
     
  5. Jun 12, 2014 #4
    Thanks micromass,

    So is the series solution above okay are am I totally off on it?
     
  6. Jun 12, 2014 #5
    The series solution looks fine, I don't think you can express the result of final integral without the help of error function. Note that the result is valid when ##0<x<\infty##.

    Are you sure you were asked for an indefinite integral instead of a definite one?
     
  7. Jun 12, 2014 #6
    Thanks Pranav.

    Looking back this has to be only for definite integrals as ##x## was a change of variable for a frequency relationship, ##x=\frac{\hbar\omega}{k_b T}##. So it has to be only for definite integrals over ##0<x<\infty##.
     
  8. Jun 13, 2014 #7
    This integral has a solution if the boundary is from 0 to infinity but I do not know if this is relevant to you.
     
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