Let $f$ a function $f:\mathbb{C}\to\mathbb{C},$ $$f(z)=\frac{1}{(e^{\frac{1}{z}}-1)(e^z-1)}.$$ Trying to integrate this function in a closed contour around $0$ has been impossible to me. Let $\epsilon>0\in\mathbb{R}$. $$ \int_{|z|=\epsilon}\frac{dz}{(e^{\frac{1}{z}}-1)(e^z-1)}=? $$ The function $f$ turns out to be the same after applying a otherwise useful contour integration around an essential singularity technique.
Is this "integrand" $f(z)=\frac{1}{(e^{\frac{1}{z}}-1)(e^z-1)}$, evaluated at some loop around its essential singularity at $z=0$, an integrable function? In case yes, How to find it?
1. When $\frac{1}{2\pi} < \epsilon < 2\pi$, we may use the Taylor series
$$ \frac{z}{e^z - 1} = -\frac{z}{2} + \sum_{n=0}^{\infty} \frac{B_{2n}}{(2n)!} z^{2n}, \qquad |z| < 2\pi, $$
where $B_k$'s are the Bernoulli numbers, to give the following Laurent expansion:
$$ \frac{1}{(e^{1/z}-1)(e^z - 1)} = \biggl( -\frac{z}{2} + \sum_{n=0}^{\infty} \frac{B_{2n}}{(2n)!} z^{2n} \biggr)\biggl( -\frac{1}{2z} + \sum_{n=0}^{\infty} \frac{B_{2n}}{(2n)!z^{2n}} \biggr) $$
for $\frac{1}{2\pi} < |z| < 2\pi$. So by the Residue Theorem and using $B_0 = 1$ and $B_2=\frac{1}{6}$, we get
$$ \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{\mathrm{d}z}{(e^{1/z}-1)(e^z - 1)} = -\frac{1}{2} \biggl( B_0 + \frac{B_2}{2!} \biggr) = -\frac{13}{24}. $$
This indeed confirms the conjecture in the comment.
2. As $\epsilon > 1$ increases, it will cross the poles at $\pm 2\pi i k$. So, if $2n\pi < \epsilon < 2(n+1)\pi$, then
\begin{align*} &\frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{\mathrm{d}z}{(e^{1/z}-1)(e^z - 1)} \\ &=-\frac{13}{24}+\sum_{k=1}^{n} \biggl( \underset{z=2\pi ki}{\mathrm{Res}}\,\frac{1}{(e^{1/z}-1)(e^z - 1)} + \underset{z=-2\pi ki}{\mathrm{Res}}\,\frac{1}{(e^{1/z}-1)(e^z - 1)} \biggr) \\ &=-\frac{13}{24}+\sum_{k=1}^{n} \underbrace{\biggl( \frac{1}{e^{-i/(2\pi k)} - 1} + \frac{1}{e^{i/(2\pi k)} - 1} \biggr)}_{=-1} = \boxed{-\frac{13}{24} - n}. \end{align*}