Closed form of $\sum_{n=1}^{\infty} \frac{1}{n(e^{2\pi n}+1)}$

Question

In this question, there's a quite simple closed form mentioned for the series $$\sum_{n=1}^{\infty} \frac{1}{n(e^{2\pi n}-1)}$$

However I'm wondering if there exists any simple closed form known for the series of type

$$\sum_{n=1}^{\infty} \frac{1}{n(e^{\pi n}+1)}, \quad \sum_{n=1}^{\infty} \frac{1}{n(e^{2\pi n}+1)},\quad \sum_{n=1}^{\infty} \frac{1}{n(e^{(2k+1)n\pi }+1)},\quad \sum_{n=1}^{\infty} \frac{1}{n(e^{(2k)n\pi}+1)}$$

I've been thinking about thee kinds of series from quite some time and so far I have tried contour integration, cotangent partial fraction, and applying Poisson summation or converting to an integral but so far no success. Any help would be highly appreciated!

Answer 1

With help of Mathematica I have:

$$\sum _{n=1}^{\infty } \frac{1}{n \left(e^{a n \pi }-1\right) }=\\\sum _{n=1}^{\infty } \left(\sum _{m=1}^{\infty } \frac{e^{-a n m \pi }}{n}\right)=\\\sum _{m=1}^{\infty } \left(\sum _{n=1}^{\infty } \frac{e^{-a n m \pi }}{n}\right)=\\\sum _{m=1}^{\infty } -\ln \left(1-e^{-a m \pi }\right)=\\-\frac{1}{24} (a \pi )-\ln \left(\eta \left(\frac{i a}{2}\right)\right)$$ and $$\sum _{n=1}^{\infty } \frac{1}{n (\exp (a \pi n)+1)}=\frac{a \pi }{8}-\ln \left(\eta \left(\frac{i a}{2}\right)\right)+2 \ln (\eta (i a))$$

where $a>0$ and $\eta \left(\frac{i a}{2}\right)$ is: Dedekind eta modular elliptic function.

MMA code:

Sum[1/(n*(Exp[a Pi n] - 1)), {n, 1, Infinity}] == -((a \[Pi])/24) - Log[DedekindEta[(I a)/2]]

Sum[1/(n*(Exp[a Pi n] + 1)), {n, 1, Infinity}] == (a \[Pi])/8 - Log[DedekindEta[(I a)/2]] + 2 Log[DedekindEta[I a]]