Knowledge of $\frac{x}{e^{x}-1}$ I wonder if anyone has seen the function $\frac{x}{exp(x)-1}$ in some application or otherwise has some information about it.
A visual inspection e.g. https://www.wolframalpha.com/input/?i=x%2F%28exp%28x%29-1%29 demonstrates it is convex.
Erling
 A: The coefficents of the Taylor series of $\frac x{\exp(x)-1}$ at $0$ are the Bernoulli numbers. More precisely:$$\frac x{\exp(x)-1}=\sum_{n=0}^\infty\frac{B_nx^n}{n!}.$$And they have lots of applications.
A: The way to see that it is convex is to show that it is always decreasing and concave up.
To show decreasing, you differentiate it, and you get $$\frac{(e^x-1)-xe^x}{(e^x-1)^2}$$
Assuming $x\neq 0$, the bottom is positive, and does not affect the sign, so we need to show that the top is always negative.  In other words,
$$e^x-1-xe^x<0$$
$$e^x(1-x)<1$$
$$e^x<1/(1-x)$$
Expanding a Taylor expansion, we see that in fact,
$$1+x+x^2/2+x^3/6+\ldots<1+x+x^2+x^3+\ldots{\rm\ \ for\ all\ }x\neq 0$$
Then, we can do a similar analysis on the second derivative.  This derivative is:$$
\left(\frac{e^x(1-x)-1}{(e^x-1)^2}\right)'=\frac{(e^x+e^x(1-x))(e^x-1)^2-(e^x(1-x)-1)(2e^x(e^x-1))}{(e^x-1)^4}=\frac{e^x(2-x)(e^x-1)-2e^{2x}(1-x)+2e^x}{(e^x-1)^3}=\frac{e^x(2e^x-xe^x-2+x-2e^x+2xe^x+2)}{(e^x-1)^3}=\frac{e^x(xe^x+x)}{(e^x-1)^3}=\frac{xe^x(e^x+1)}{(e^x-1)^3}
$$
To show that this is always positive (for $x\neq 0$), we note that $e^x$ and $e^x+1$ are always positive, so they don't affect the sign.  Also, it's clear that $x$ and $(e^x-1)$ are both negative for $x<0$, and positive for $x>0$.  Therefore, this quotient is always positive, proving what you noticed in your post.
A: The integral $$\int_0^{\infty}\frac{x}{e^x-1}dx$$ is $\zeta(2)$, which is famously equal to $\frac{\pi^2}{6}$ (Basel Problem).
This is actually a fairly important integral. For example, in physics you have Planck's Law, which is to do with blackbody radiation (which led to the development of quantum mechanics). Using Planck's Law, you can derive the Stefan-Boltzmann Law. This involves evaluating a multiple of the integral
$$\int_0^{\infty}\frac{x^3}{e^x-1}dx$$
which is equal to $$6\zeta(4)=\frac{\pi^4}{15}$$
There are various ways of showing this, but one method involves computing the Fourier series for $x^4$ in $[-\pi,\pi]$, which is
$$x^4=\frac{\pi^4}{5}+\sum_{n=1}^{\infty}\left(\frac{8n^2\pi^2-48}{n^4}\right)\cos 
n\pi\cos nx$$
Leading to
$$\pi^4=\frac{\pi^4}{5}+8\pi^2\zeta(2)-48\zeta(4)$$
(Using the series definition of $\zeta$)
And as I said earlier,
$$\zeta(2)=\int_0^{\infty}\frac{x}{e^x-1}dx$$
Again, we can consider the Fourier Series of $x^2$ in $[-\pi,\pi]$:
$$x^2=\frac{\pi^2}{3}+\sum_{n=1}^{\infty}\frac{4\cos n\pi\cos nx}{n^2}$$
Meaning
$$\pi^2=\frac{\pi^2}{3}+4\zeta(2)\Rightarrow\zeta(2)=\frac{\pi^2}{6}$$
