Closed form expression of odd series

Question

Does there exist a closed-form expression for the series below?

$$\sum_{r=1}^{\infty} \frac{e^{-rx}}{\ln(r+1)}$$

I’m pretty sure this series converges for $x \geq 0$, though I haven’t checked myself. Even so, I’m not sure that there’s even an expression for this. Just wondering if there was.

Answer 1

Consider $\;z:=e^{-x}\,$ for $\,x>0\,$ (i.e. $\;z<1$) then you want : $$S_z:=\sum_{k=1}^{\infty} \frac{z^k}{\ln(k+1)}$$ Since $\;\displaystyle \frac 1{\ln(k+1)}=\int_0^\infty (k+1)^{-s} ds\ \;$ for $k>0\;$ we may rewrite $S_z$ as : $$S_z:=\sum_{k=1}^{\infty} \int_0^\infty \frac{z^k}{(k+1)^s}\,ds$$ I'll let you prove that $\sum$ and $\int$ may be exchanged to get : \begin{align} S_z&=\frac 1z\int_0^\infty \sum_{k=1}^{\infty} \frac{z^{k+1}}{(k+1)^s}\,ds\\ &=\frac 1{z}\int_0^\infty\operatorname{Li_s}(z)-z\;ds\\ \end{align} with $\operatorname{Li_s}$ the polylogarithm function (evaluation using alpha).

Of course this doesn't help much for a closed form. Integrating polylogarithms relatively to their first parameter (denominator's power) is not easier than integrating zeta functions but it could define a new (possibly interesting) special function.
Analytic continuation may be used too to evaluate your series for any value of $z\neq 1$.

Answer 2

It's rather unlikely to have a closed form. For $x > 0$, it does converge: just use the comparison test with a geometric series. For $x=0$, it does not converge.