# Need help with this difficult Integral involving exponentials

1. Jun 12, 2014

### DiogenesTorch

1. The problem statement, all variables and given/known data

This is the integral
$$\int \frac{x^2}{e^x-1}dx$$

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:

$$\int \frac{x^2}{e^x-1}dx = x^2 \ln{(e^x-1)} -\frac{5x^3}{3}-2 \int x \ln{(e^x-1)} dx$$

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 :)

2. Jun 12, 2014

### DiogenesTorch

Is this acceptable?

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

\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*}

Then subbing this series into the original integral

\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*}

3. Jun 12, 2014

### micromass

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

4. Jun 12, 2014

### DiogenesTorch

Thanks micromass,

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

5. Jun 12, 2014

### Saitama

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?

6. Jun 12, 2014

### DiogenesTorch

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 $0<x<\infty$.

7. Jun 13, 2014

### 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.