I recently gave a try to show that
$$\sum_{n=1}^{+\infty} \frac{n}{e^{2\pi n}-1}=\frac{1}{24}-\frac{1}{8\pi} $$
without using the Theta function or Mellin transform, but I ended up with twice the result and I can't figure out where my mistake was. For my approach, I need to introduce some Lemmas to help the proof:
(1): $\forall z \in \mathbb{R}^*, \frac{\pi z}{\sinh(\pi z)}=\int_{0}^{\infty} \frac{\cos(2zx)}{\cosh^2(x)}dx$
(2): $\forall z \in \mathbb{R} , \sum_{n=1}^{+\infty} \cos(2nz)e^{-n\pi} = \frac{\cos(2z)-e^{-\pi}}{2(\cosh(\pi)-\cos(2z))}$
(3): $\forall z \in \mathbb{R} , \frac{\sinh(\pi)}{4(\cosh(\pi)-\cos(2z))}=\frac{\pi}{2(\pi^2+4z^2)}+\int_{0}^{\infty} \frac{\sin(\pi t)\cosh(2zt)}{e^{2\pi t}-1} dt $
(4): $ \forall z \in \mathbb{R} , \frac{\pi}{\pi^2+4z^2}=\int_{0}^{\infty} \cos(2zu)e^{-\pi u} du $
(5): $ \frac{\pi^2}{6}=\int_{0}^{\infty} \frac{w}{e^w-1}dw$
Here is my approach:
$$\begin{align} I&=\sum_{n=1}^{+\infty} \frac{n}{e^{2\pi n}-1}\\ &=\frac{1}{2}\sum_{n=1}^{+\infty} \frac{n}{\sinh(\pi n)}e^{-\pi n}\\ &=\frac{1}{2\pi}\sum_{n=1}^{+\infty} \frac{\pi n}{\sinh(\pi n)}e^{-\pi n} \end{align} $$
Then using Lemmas (1) and (2) we get:
$$ \begin{align} I&=\frac{1}{2\pi}\sum_{n=1}^{+\infty}e^{-\pi n}\int_{0}^{\infty}\frac{\cos(2nx)}{\cosh^2(x)}dx\\ &=\frac{1}{2\pi}\int_{0}^{\infty}\sum_{n=1}^{+\infty}\frac{\cos(2nx)}{\cosh^2(x)}e^{-\pi n}dx\\ &=\frac{1}{4\pi}\int_{0}^{\infty}\frac{\cos(2x)-e^{-\pi}}{\cosh(\pi)-\cos(2x)} \frac{dx}{\cosh^2(x)}\\ \end{align} $$
From here, we can remove the cosine function in the numerator:
$I=\frac{\cosh(\pi)-e^{-\pi}}{4\pi}\int_{0}^{\infty}\frac{1}{\cosh(\pi)-\cos(2x)} \frac{dx}{\cosh^2(x)}-\frac{1}{4\pi}\int_{0}^{\infty}\frac{dx}{\cosh^2(x)}=\frac{\sinh(\pi)}{4\pi}\int_{0}^{\infty}\frac{1}{\cosh(\pi)-\cos(2x)} \frac{dx}{\cosh^2(x)}-\frac{1}{4\pi}$
Using Lemma (3), then (1) on $\cosh(2xt)=\cos(2ixt)$:
$I=\frac{1}{\pi}\int_{0}^{\infty}\int_{0}^{\infty}\frac{\sin(\pi t)}{e^{2\pi t}-1}\frac{\cosh(2xt)}{\cosh^2(x)} dxdt+\frac{1}{2}\int_{0}^{\infty}\frac{1}{\pi^2+4x^2}\frac{dx}{\cosh^2(x)}-\frac{1}{4\pi}=\int_{0}^{\infty}\frac{t}{e^{2\pi t}-1}dt+\frac{1}{2}\int_{0}^{\infty}\frac{1}{\pi^2+4x^2}\frac{dx}{\cosh^2(x)}-\frac{1}{4\pi}$
$=\frac{1}{4\pi^2}\int_{0}^{\infty}\frac{t}{e^t-1}dt+\frac{1}{2}\int_{0}^{\infty}\frac{1}{\pi^2+4x^2}\frac{dx}{\cosh^2(x)}-\frac{1}{4\pi}=\frac{1}{24}-\frac{1}{4\pi}+ \underbrace{\frac{1}{2}\int_{0}^{\infty}\frac{1}{\pi^2+4x^2}\frac{dx}{\cosh^2(x)}}_{:=J}$
Let's evaluate the last integral $J$, with Lemmas (4) and (1) :
$$\begin{align} J&=\frac{1}{2}\int_{0}^{\infty}\frac{1}{\pi^2+4x^2}\frac{dx}{\cosh^2(x)}\\ &=\frac{1}{2\pi}\int_{0}^{\infty}\int_{0}^{\infty}\frac{\cos(2xu)}{\cosh^2(x)} e^{-\pi u} dxdu\\ &=\frac{1}{2}\int_{0}^{\infty}\frac{u}{\sinh(\pi u)} e^{-\pi u}du\\ &=\frac{1}{4\pi^2}\int_{0}^{\infty}\frac{u}{e^u-1}du=\frac{1}{24} \end{align}$$
Finally, $I=\frac{1}{12}-\frac{1}{4\pi}$, which is twice the correct result.
The application of the Lamma (3) is questionable, resulting in the undefined double integral below
$$\int_{0}^{\infty}\int_{0}^{\infty}\frac{\sin \pi t}{e^{2\pi t}-1}\frac{\cosh2xt}{\cosh^2x} dx\ dt $$