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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 $o(1)$)

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Unclear what you mean by "better". Better approximation over an interval close to zero? Better behavior away from zero? – Ron Gordon Jan 3 '13 at 16:41
I edited my post according to your remark. – Siméon Jan 3 '13 at 16:43
up vote 4 down vote accepted

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.

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Working with Maple the numeric evidence for the log term is very strong. Subtract out everything else and what you get is precisely the log term (numerically). Back in two hours. – Marko Riedel Jan 3 '13 at 20:01
Looks great too, but I don't know this technique of Mellin transform yet. Where can I learn it ? – Siméon Jan 3 '13 at 22:09
This is from page 48 of the seminal paper Average Case Analysis of Algorithms: Mellin Transform Asymptotics by Flajolet and Sedgewick. – Marko Riedel Jan 3 '13 at 22:15

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

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Excellent analysis! – Ron Gordon Jan 3 '13 at 19:11
@rlgordonma: I found a mistake. I will fix it... – Fabian Jan 3 '13 at 19:31
This looks great, thanks ! I will check up the details tomorrow. How do you know a priori that the Euler-Maclaurin formula will work with $z\log(1-e^{-(1+z)x})$ but not with $\dfrac{1}{z}\dfrac{1}{(e^{zx}-1)^2}$ ? – Siméon Jan 3 '13 at 22:07
@Ju'x: I didn't. I first tried with $\dfrac{1}{z}\dfrac{1}{(e^{zx}-1)^2}$ and it didn't work. So I transformed it to something which is less divergent for small $z$. – Fabian Jan 3 '13 at 23:03
Or was your question why Maclaurin does not work for the initial sum. The reason is that the error term in that case cannot be bounded for small $x$... – Fabian Jan 3 '13 at 23:16

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.

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