# 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.

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 $z\neq 1$.

• Holy I didn’t see that coming. Nice super awesome thanks for this beautiful answer. – Horus Mar 13 '18 at 12:35
• On a side note can’t this already be considered an analytic continuation already? Seeing that I could sub x as any value not equal to zero, doesn’t this already extend it to the rest of the complex values? – Horus Mar 13 '18 at 13:03
• Glad you liked this answer @Horus and yes for $|z|>1$ we may use the $\;\operatorname{Li}_s(z)/z-1\;$ integral for the analytic continuation even if the definition of $\,S_z\,$ and the series defining the polylogarithm $\;\displaystyle \sum_{k=1}^{\infty} \frac{z^k}{k^s}\;$ are divergent. Excellent continuation, – Raymond Manzoni Mar 13 '18 at 13:56
• That’s another upvote for me haha – Horus Mar 13 '18 at 15:03

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.

• Wouldn’t it also converge for $x=0$ too since using the integral test gives it a finite value? – Horus Mar 5 '18 at 20:05
• No. $\int_1^\infty \frac{dr}{\ln(r+1)}$ diverges. – Robert Israel Mar 5 '18 at 20:07
• Wait no sorry I remembered the integral wrong. Yes it doesn’t converge at zero. – Horus Mar 5 '18 at 20:07
• It does converge at 0, but not at infinity. – user Mar 5 '18 at 20:09
• Seriously? Based on integral test for convergence it should diverge at 0 but converge to zero at infinity no? @user – Horus Mar 5 '18 at 20:16