A friend of mine and I wanted to solve the following indefinite integral but got stuck:

$$ \int \frac{x}{e^x - 1} dx. $$

My approach:

Let

$$ I = \int \frac{x}{e^x - 1} dx.\\ \implies I = x\int \frac{dx}{e^x - 1} - \int \left ( \int \frac{dx}{e^x - 1} \right ) dx. $$

Now, let $I_2 = \int \frac{dx}{e^x - 1}$. Also, let $z = e^x \implies dx = {dz \over z}$. Then,

$$ I_2 = \int \frac{dz}{z(z-1)} \\ \implies I_2 = \int {dz \over {z - 1}} - \int {dz \over z} \\ \implies I_2 = \ln (e^x - 1) - x. $$

Substituting the value of $I_2$ in $I$, we get,

$$ I = x[\ln (e^x - 1) - x] + {x^2 \over 2} - \int \ln (e^x - 1) dx. $$

I got stuck right here. Is it possible to proceed further?

  • 2
    By any chance, were you trying to evaluate $\displaystyle\int_{0}^{\infty}\dfrac{x}{e^x-1}\,dx$? This integral evaluates to $\dfrac{\pi^2}{6}$ even though the antiderivative of $\dfrac{x}{e^x-1}$ can't be expressed with elementary functions. – JimmyK4542 Jan 4 '17 at 5:59
  • Well yeah, the original integral was what you mentioned. We just wanted to see whether we could solve the indefinite integral. – Nilabro Saha Jan 4 '17 at 6:01
up vote 3 down vote accepted

The antiderivative is not an elementary function: it can be written as $$ -\frac{{x}^{2}}{2}+x\ln \left( 1-{{\rm e}^{x}} \right) +{\it dilog} \left( 1-{{\rm e}^{x}} \right) $$

  • What is "dilog"? – Nilabro Saha Jan 4 '17 at 5:54
  • dilog is short for the Dilogarithm function – Saketh Malyala Jan 4 '17 at 5:57
  • 2
    $\text{dilog}(s) =\displaystyle \int_1^s \frac{\ln(t)}{1-t}\; dt$. $\text{dilog}(s)$ is also sometimes written as $\text{Li}_2(1-s)$. – Robert Israel Jan 4 '17 at 6:01

\begin{equation} I = \int \frac{x}{\mathrm{e}^{x}-1} dx = -I_{1} = -\int \frac{x}{1-\mathrm{e}^{x}} dx \end{equation}

Integrate by parts \begin{align} I_{1} &= \int \frac{x}{1-\mathrm{e}^{x}} dx \\ \tag{1} &= x^{2} - x \ln(1-\mathrm{e}^{x}) - \int x dx + \int \ln(1-\mathrm{e}^{x}) dx \\ \tag{2} &= x^{2} - x \ln(1-\mathrm{e}^{x}) - \frac{x^{2}}{2} - \mathrm{Li}_{2}(\mathrm{e}^{x}) \end{align} and thus \begin{equation} \int \frac{x}{\mathrm{e}^{x}-1} dx = x \ln(1-\mathrm{e}^{x}) - \frac{x^{2}}{2} + \mathrm{Li}_{2}(\mathrm{e}^{x}) \end{equation}

  1. Let $u=\mathrm{e}^{x}$ \begin{align} \int \frac{1}{1-\mathrm{e}^{x}} dx &= \int \frac{1}{u(1-u)} du \\ &= \int \left( \frac{1}{u} + \frac{1}{1-u} \right) du \\ &= \ln \frac{u}{1-u} \\ &= x - \ln(1-\mathrm{e}^{x}) \end{align}

  2. Let $u=\mathrm{e}^{x}$ \begin{align} \int \ln(1-\mathrm{e}^{x}) dx &= \int \frac{\ln(1-u)}{u} du \\ &= -\mathrm{Li}_{2}(u) \\ &= -\mathrm{Li}_{2}(\mathrm{e}^{x}) \end{align} where \begin{equation} \mathrm{Li}_{2}(z) = -\int\limits_{0}^{z} \frac{\ln(1-x)}{x} dx \end{equation} is the dilogarithm function.

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