Closed form expression of odd series 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. 
 A: 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 $z\neq 1$.
A: 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.
