Inspired by this answer, I'm trying to show that $$\sum_{n=1}^{\infty} \frac{n}{e^{2 \pi n}-1} = \frac{1}{24} – \frac{1}{8 \pi}$$ using the inverse Mellin transform.
But the answer I get is twice as much as it should be, and I don't understand why.
EDIT:
With Marko Riedel's help, I corrected the error in my evaluation.
Since
$$ \left\{ \mathcal{M} \ \frac{x}{e^{2\pi x}-1} \right\}(s) = \int_{0}^{\infty} \frac{x^{s}}{e^{2 \pi x}-1} \, dx = (2\pi)^{-(s+1)}\Gamma(s+1)\zeta(s+1) $$ for $\operatorname{Re}(s) >1$,
we have
$$ \frac{x}{e^{2\pi x}-1}=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}(2\pi)^{-(s+1)}\Gamma(s+1)\zeta(s+1) x^{-s} \, ds , $$ where $c >1$.
Replacing $x$ with $n$ and summing both sides, we get $$ \begin{align} \sum_{n=1}^{\infty}\frac{n}{e^{2\pi n}-1} &= \frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}(2\pi)^{-(s+1)}\Gamma(s+1)\zeta(s+1)\zeta(s)\, ds \\&= \frac{1}{2 \pi i} \int_{c-i\infty}^{c+i\infty} f(s) \, ds. \end{align} $$
The integrand has simple poles at $s=-1, 0$, and $1$.
The fact that $\left|\Gamma(s)\right|$ decays exponentially fast to $0$ as $\text{Im} (s) \to \pm \infty$ allows us to shift the contour to the left.
I originally shifted the contour all the way to negative infinity.
But as Marko Riedel explains below, we want to shift the contour to the imaginary axis since the integrand is odd there.
Indeed, using the functional equation of the Riemann zeta function, we get $$ f(it) = \frac{it}{2 \pi} \sinh \left(\frac{\pi t }{2} \right) \operatorname{csch}(\pi t) \left|\zeta(1+it)\right|^{2}, \quad t \in \mathbb{R}.$$
Therefore,
$$ \int_{c-i \infty}^{c+i \infty} f(s) \, ds = 2 \pi i \ \text{Res}[f,1] + \pi i \ \text{Res}[f,0] ,$$
where
$$ \begin{align} \text{Res}[f,0] &= \lim_{s \to 0} s (2 \pi)^{-(s+1)} \Gamma(s+1) \zeta(s+1) \zeta(s) \\ &= \lim_{s\to 0} s\zeta(s+1) (2\pi)^{-(s+1)}\Gamma(s+1)\zeta(s) \\ &= 1\left(\frac{1}{2 \pi} \right)(1)\left(- \frac{1}{2} \right) \\ &=-\frac{1}{4 \pi} \end{align} $$
and
$$ \begin{align} \text{Res}[f,1] &= \lim_{s \to 1} (s-1) (2 \pi)^{-(s+1)} \Gamma(s+1) \zeta(s+1) \zeta(s) \\ &= \lim_{s\to 1}(s-1)\zeta(s) (2\pi)^{-(s+1)}\Gamma(s+1)\zeta(s+1) \\&= 1\left(\frac{1}{4 \pi^{2}}\right)(1)\left(\frac{\pi^{2}}{6}\right) \\ &=\frac{1}{24} . \end{align} $$
The result then follows.
Best Answer
In response to your query on the other thread -- I don't have the time to typeset this properly, here are some comments. The effort looks good and you should keep working on it.
Don't extract any terms out from the transform function in front of the integral. As you have noticed yourself these reappear later on in the computation, reducing readability.
Now this is the important part -- when evaluating harmonic sums you get an asymptotic expansion about zero when shifting to the left and about infinity when shifting to the right. So it is no surprise that you do not get the right value -- the expansion does not converge there (at $x=1$).
With $g(s)$ being the transform of your sum we have $$ g(s) = \left( 2\,\pi \right) ^{-s-1}\Gamma \left( s+1 \right) \zeta \left( s+1 \right) \zeta \left( s \right).$$ The residues are $$\operatorname{Res}(g(s)/x^s; s=1) = \frac{1}{24x},$$ $$\operatorname{Res}(g(s)/x^s; s=0) = -\frac{1}{4\pi},$$ $$\operatorname{Res}(g(s)/x^s; s=-1) = \frac{x}{24}.$$ Now I suggest you shift from $\Re(s)=3/2$ to $\Re(s)=0$, indenting around the pole at zero by making a half-circle of radius $\epsilon$ around said pole at $s=0.$ This only picks up half the residue, so that your result is $$\frac{1}{24\times 1} + \frac{1}{2}\left(-\frac{1}{4\pi} \right) = \frac{1}{24} - \frac{1}{8\pi}.$$ You still need to verify that $g(s)$ is odd on the imaginary axis when $x=1$ by simplifying with the functional equation of the Riemann Zeta function.
That is all for now. I hope there are no mistakes.