Evaluate $\sum_{\ell=1}^\infty \frac{1}{\ell(e^{2\pi \ell}-1)}$

Question

I am trying to find if there is a simple expression for the following series $$ S=\frac{1}{2}\sum_{\ell=1}^\infty\frac{\coth(\pi\ell)-1}{\ell}=\sum_{\ell=1}^\infty \frac{1}{\ell(e^{2\pi \ell}-1)}\,. $$ In principle I think it's a good idea to get rid of the $\ell$ in the denominator, for instance consindering $$ S(a)=\sum_{\ell=1}^\infty \frac{1}{\ell(e^{2\pi a \ell}-1)}\,,\qquad S'(a)=-\frac{\pi}{2} \sum_{\ell=1}^\infty \frac{1}{\sinh^2(\pi a\ell)}\,, $$ but this doesn't seem to help much. I also tried comparing with an integral, writing $$ S = \sum_{\ell=0}^\infty \frac{1}{(\ell+1)(e^{2\pi(\ell+1)}-1)} = \int_0^\infty \frac{d\ell}{(\ell+1)(e^{2\pi(\ell+1)}-1)} + \frac{1}{2(e^{2\pi}-1)} + C $$ where $C$ can be in principle calculated with the Euler-MacLaurin or Abel-Plana formulas. Unfortunately, also these integrals seem pretty hard... Any suggestion?

Answer 1

Theory of modular forms and elliptic curves. For $\Im(z)>0$

$$f(z)=\sum_{l=1}^\infty \frac1{l (e^{-2i\pi lz}-1)}=\sum_{n=1}^\infty \sigma_{-1}(n) e^{2i\pi nz}= -\sum_{m\ge 1} \log(1-e^{2i\pi mz})= -\frac1{24}\log \left(\frac{\Delta(z)}{e^{2i\pi z}}\right)$$

From that $\Delta(z)$ is a weight $12$ cusp form we get the relation with Eisenstein series

$$\Delta(z)=\frac{E_4(z)^3-E_6(z)^2}{1728}$$

$E_6(i)=0$ and $E_4(i)$ has a closed-form in term of $\Gamma(1/4)$.

This closed-form follows from $1=\int_0^1 dz=\int_0^1 \frac{d\wp_i(z)}{\wp_i'(z)}=\int_C \frac{dx}{\sqrt{4x^3-g_2(i) x}}$ which reduces to $\frac{A}{E_4(i)}\int_0^1 \frac{dx}{x^3-x}=\frac{A}{E_4(i)}\ B(1/2,1/4)$ (the beta function)