Asymptotic expansion of a series

Question

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 $o(1)$)

Answer 1

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.

Answer 2

Note that $$\sum_{m=1}^\infty \frac{m}{q^{m+1}} = \frac{1}{(q-1)^2}, \qquad |q|>1.$$

With $q= e^{nx}$, we have $$f(x)= \sum_{n=1}^\infty \sum_{m=1}^\infty \frac{m e^{-n (m+1) x}}{n} =-\sum_{m=1}^\infty m \log\left(1-e^{-(1+m)x}\right). $$

Next, we use the Euler-Maclaurin formula: $$f(x) \sim -\int_1^\infty \!dz\,z \log\left(1-e^{-(1+z)x}\right) - \overbrace{\frac{\log(1-e^{-2x})}{2}}^{\log x+ \mathcal{O}(1)} + \frac{1}{12} \overbrace{\left[ \frac{x}{e^{2x}-1}+\log \left(1-e^{-2 x}\right) \right]}^{\log x+\mathcal{O}(1)} + R$$

With $$|R| \leq C \int_1^{\infty} \left| g''(z) \right|dz$$ where $g(z) = z \log\left(1-e^{-(1+z)x}\right) $. Numerical calculation shows $R = \mathcal{O}(1)$ though I have no proof yet.

Thus we have $$f(x) \sim - \int_1^\infty \!dz\,z \log\left(1-e^{-(1+z)x}\right) = \frac{1}{x^2} \int_{2x}^\infty\!dx\,(x-y) \log(1-e^{-y}).$$

We need to evaluate $$\begin{align}\int_{2x}^\infty\!dy\,(x-y) \log(1-e^{-y}) &= x \underbrace{\int_{0}^\infty\!dy\, \log(1-e^{-y})}_{-\pi^2/6} -\underbrace{\int_{0}^\infty\!dy\, y\log(1-e^{-y})}_{\zeta(3)}\\ &\quad+ \underbrace{\int_0^{2x} \!dy\,(y-x)\log(1-e^{-y})}_{\mathcal{O}(x^2)}\end{align} .$$

In conclusion, we have $$f(x) \sim \frac{\zeta(3)}{x^2}- \frac{\pi^2}{6x} - \frac{5}{12}\log x+ \mathcal{O}(1).$$

Answer 3

The asymptotic series has an infinite coefficient ($\frac{5}{12} \zeta(1)$) for the $O(1)$ term. The behavior of the series in the limit as $x \rightarrow 0$ is more complicated, having a term with a factor of $\log{x}$, I suspect.