Need help with this difficult Integral involving exponentials

[DiogenesTorch]

Homework Statement

This is the integral
<br /> \int \frac{x^2}{e^x-1}dx<br />

Homework Equations

Can this even be solved in closed form?

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:

<br /> \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 <br />

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

Any ideas or suggestions much appreciated :)

 

[DiogenesTorch]

Is this acceptable?

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

<br /> \begin{align*}<br /> \frac{1}{(e^x-1)} &amp;= \frac{e^{-x}}{1-e^{-x}} \\<br /> &amp;= e^{-x}\frac{1}{1-e^{-x}} \\<br /> &amp;=e^{-x}\sum_{n=0}^{\infty}e^{-nx} \\<br /> &amp;=\sum_{n=0}^{\infty}e^{-(n+1)x} \\<br /> &amp;=\sum_{n=1}^{\infty}e^{-nx}<br /> \end{align*}<br />

Then subbing this series into the original integral

<br /> \begin{align*}<br /> \int \frac{x^2}{e^x-1}dx &amp;= \int x^2 \Biggr( \sum_{n=0}^{\infty}e^{-(n+1)x} \Biggr) dx \\<br /> &amp;= \sum_{n=0}^{\infty} \int x^2 \Biggr( e^{-(n+1)x} \Biggr) dx \\<br /> &amp;= \sum_{n=1}^{\infty} \int x^2 e^{-nx} dx<br /> \end{align*}<br />

 

[micromass]

You cannot solve it in closed form, you need the polylogarithm function. You can find series solutions though.

 

[DiogenesTorch]

Thanks micromass,

So is the series solution above okay are am I totally off on it?

 

[Saitama]

Thanks micromass,

So is the series solution above okay are am I totally off on it?

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?

 

[DiogenesTorch]

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?

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 $0<x<\infty$.

 

[dirk_mec1]

This integral has a solution if the boundary is from 0 to infinity but I do not know if this is relevant to you.