TITLE: Asymptotic expansion of a series
QUESTION [11 upvotes]: I am interested in the asymptotics, as $x$ tends to $0$, of
$$f(x) = \sum_{n=1}^\infty \frac{1}{n}\frac{1}{(e^{nx}-1)^2}$$
This function is well defined for every $x > 0$ (for example, use $e^{nx}-1 \geq nx$).
Furthermore, Lebesgue's dominated convergence theorem shows that, 
$$
f(x) \sim \frac{\zeta(3)}{x^2}
$$
as $x$ tends to $0$, where  $\displaystyle\zeta(3) = \sum_{n=1}^\infty \frac{1}{n^3}$
is Apéry's constant.
Could you help me get a more precise asymptotic expansion of $f(x)$ as $x$ tends to $0$ ?
(best would be with $o(1)$)

REPLY [4 votes]: This series may also be evaluated by calculating its Mellin transform and inverting that to get the asymptotic expansion.
We have $$ f(x) = \sum_{n\ge 1} \frac{1}{n} \frac{1}{(e^{nx}-1)^2} =
\sum_{n\ge 1} \frac{1}{n} \frac{1}{e^{2nx}} \frac{1}{(1-e^{-nx})^2}$$
which gives
$$ f(x) = \sum_{n\ge 1} \frac{1}{n} \frac{1}{e^{2nx}} \sum_{k\ge 0} (k+1) e^{-knx} =
\sum_{n\ge 1} \frac{1}{n} \sum_{k\ge 0} (k+1) e^{-(k+2)nx}.$$
Now recall that $$\mathfrak{M}(e^{-qx};s) = \Gamma(s) \frac{1}{q^s}$$
so that $$f^*(s) = \mathfrak{M}(f(x);s) = \Gamma(s) 
\sum_{n\ge 1} \frac{1}{n} \sum_{k\ge 0} (k+1) \frac{1}{(k+2)^s n^s} =
\Gamma(s) \zeta(s+1)
\left( - \sum_{k\ge 0}  \frac{1}{(k+2)^s} + \sum_{k\ge 0}  \frac{k+2}{(k+2)^s} \right)$$
and finally
$$\Gamma(s) \zeta(s+1) \left( 1 - \zeta(s) -1 + \zeta(s-1) \right)
= \Gamma(s) \zeta(s+1) \left( \zeta(s-1) - \zeta(s) \right).$$
The Mellin inversion integral is
$$ f(x) = \int_{4-i\infty}^{4+i\infty}  \frac{f^*(s)}{x^s} ds.$$ 
Introduce $$ L(s) =  \frac{f^*(s)}{x^s} .$$ The asymptotic expansion may now be read off from the residues at the poles:
$$\operatorname{Res}_{s=2} L(s) = \frac{\zeta(3)}{x^2},$$
$$\operatorname{Res}_{s=1} L(s) = -\frac{\pi^2}{6x},$$
$$\operatorname{Res}_{s=0} L(s) = 
\zeta'(-1) + \frac{1}{2}\log(2\pi) - \frac{5}{12}\log x.$$
Sum these to get the asymptotic expansion. There are some additional terms that are generated by the poles of the Gamma function.