Evaluating the series $\sum_{n=1}^{\infty} \frac{n}{e^{2 \pi n}-1}$ using the inverse Mellin transform

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.

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.

1. 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.

2. 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 $x=1$).

3. 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.

There is another missing piece (in response to the comments) that must be added here, and that is the proof of the claim that $$\frac{1}{\pi^s}\Gamma(s)(s-1/4)\zeta(2s)$$ is odd and purely imaginary on the line $\Re(s) = 1/4.$ This is due to a deep connection to the functional equation of the Riemman zeta function. We will show that $$\frac{1}{\pi^s}\Gamma(s)\zeta(2s)$$ is even and real when $\Re(s) = 1/4.$ The initial claim then follows.

We have from the proof of the functional equation of the Riemann zeta function that $$ \frac{1}{\pi^s}\Gamma(s)\zeta(2s) = -\frac{1}{2s(1-2s)} + \int_1^\infty (x^{s-1} + x^{-s-1/2}) \left(\sum_{q\ge 1} e^{-\pi q^2 x}\right)dx.$$ (The first term in the integral here is a straightforward Mellin transform while the second one requires more work.)

Putting $s=1/4+it$, we have that $$ \frac{1}{\pi^s}\Gamma(s)\zeta(2s) = -\frac{4}{1+16t^2} + \int_1^\infty (x^{-3/4+it} + x^{-3/4-it}) \left(\sum_{q\ge 1} e^{-\pi q^2 x}\right)dx.$$ But clearly $$x^{it} + x^{-it} = \exp(it\log x) + \exp(-it\log x) = 2\cos(t\log x)$$ is a real number and even in $t$ as is the fractional term in front and the result follows.