TITLE: Evaluating the series $\sum_{n=1}^{\infty} \frac{n}{e^{2 \pi n}-1}$ using the inverse Mellin transform
QUESTION [37 upvotes]: 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.

REPLY [6 votes]: 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.