Show that the following function is Lebesgue integrable.

Question

\begin{equation} \int_0^{\infty} \frac{x}{e^x-1} dx \end{equation}

I know that this function has been tackled from other perspectives, but I haven't been able to find anything on its Lebesgue integrability.

It seems to me that it is possible to factor the function in such a way as to show that the function is the product of another function and the sum of a geometric series, and so, it is Lebesgue integrable, by the monotone convergence theorem, but I am having trouble proving this to be true, and evaluating the resulting integral.

Answer 1

Near $x=0$, the function may be extended by continuity, since $$ \frac{x}{e^x-1} \sim 1 $$ On the other hand we have, as $x \to +\infty$, $$ \frac{x^3}{e^x-1} \to 0 $$ then there exists an $M>0$ such that $$ \left|\frac{x^3}{e^x-1} \right|<1, \quad x>M, $$ or equivalently $$ \left|\frac{x}{e^x-1} \right|<\frac1{x^2}, \quad x>M, $$ giving the convergence near $+\infty$ and your initial integral is convergent.

Answer 2

At $x=0$ it has a removable singularity: the limit is $1$. So, it is integrable on $[0,1]$.

On $[1,\infty)$ we have that $\frac{x}{e^x-1}\leq \frac{x}{(1+x/3)^3-1}=\frac{1}{1+x/3+x^2/9}\leq \frac{9}{x^2}.$