TITLE: Evaluating $\sum_{n=1}^\infty \frac{n^3}{e^{2\pi n}-1}$ using inverse Mellin transform
QUESTION [14 upvotes]: inspiration on the post Evaluating $\sum_{n=1}^{\infty} \frac{n}{e^{2 \pi n}-1}$ using the inverse Mellin transform
it is possible to calculate in close form 
$$\sum _{k=1}^{\infty } -\frac{k^3}{e^{2 \pi  k}-1}=\frac{3840 \pi ^4 \psi _{e^{2 \pi }}^{(1)}(1)+480 \pi ^2 \psi _{e^{2 \pi }}^{(3)}(1)-704 \pi ^6-5760 \pi ^5+3 \Gamma \left(\frac{1}{4}\right)^8}{23040 \pi ^6}$$ using euler sum I appreciatte some comment.
I like to give another series it will interesting using elliptic theta function theory to prove it
$$\sum _{k=1}^{\infty } \frac{\left(k \left(-\log \left(\frac{\pi }{2}\right)\right)\right)^3}{e^{2 \pi  \left(k \log \left(\frac{\pi }{2}\right)\right)}+1}$$
$$\frac{\log ^4\left(\frac{\pi }{2}\right) \psi _{e^{-\frac{\pi }{\log \left(\frac{\pi }{2}\right)}}}^{(3)}(1)-\log ^4\left(\frac{\pi }{2}\right) \psi _{e^{-\frac{\pi }{\log \left(\frac{\pi }{2}\right)}}}^{(3)}\left(-\frac{\left(i \pi -\frac{\pi }{\log \left(\frac{\pi }{2}\right)}\right) \log \left(\frac{\pi }{2}\right)}{\pi }\right)}{16 \pi ^4 \log \left(\frac{\pi }{2}\right)}-\frac{1}{240} \log ^3\left(\frac{\pi }{2}\right)-\frac{7}{1920 \log \left(\frac{\pi }{2}\right)}$$
sorry for the latex type i do not to improve

REPLY [6 votes]: Suppose we seek a functional equation for the sum term
$$S(x) = \sum_{k\ge 1} 
\frac{k^5}{\exp(kx)-1}$$
which  is  harmonic and  may  be  evaluated  by inverting  its  Mellin
transform.   We  are  interested  in  possible  fixed  points  of  the
functional equation especially $S(2\pi)$.
Recall the harmonic sum identity
$$\mathfrak{M}\left(\sum_{k\ge 1} \lambda_k g(\mu_k x);s\right) =
\left(\sum_{k\ge 1} \frac{\lambda_k}{\mu_k^s} \right) g^*(s)$$
where $g^*(s)   =  \mathfrak{M}\left(g(x);s\right)$ 
is the Mellin transform of $g(x).$
In the present case we have
$$\lambda_k = k^5,
\quad \mu_k = k \quad \text{and} \quad
g(x) = \frac{1}{\exp(x)-1}.$$
We need the  Mellin transform $g^*(s)$ of $g(x)$  which is computed as
follows:
$$g^*(s) = \int_0^\infty \frac{1}{\exp(x)-1} x^{s-1} dx
= \int_0^\infty \frac{\exp(-x)}{1-\exp(-x)} x^{s-1} dx
\\ = \int_0^\infty \sum_{q\ge 1} \exp(-qx)  x^{s-1} dx
= \sum_{q\ge 1} \frac{1}{q^s} \Gamma(s)
= \Gamma(s) \zeta(s).$$

Hence the Mellin transform $Q(s)$ of $S(x)$ is given by
$$ Q(s) = \Gamma(s) \zeta(s) \zeta(s-5)
\quad\text{because}\quad
\sum_{k\ge 1} \frac{\lambda_k}{\mu_k^s} = 
\sum_{k\ge 1} \frac{k^5}{k^s} = \zeta(s-5)$$
where $\Re(s) > 6$.

Intersecting the  fundamental strip and  the half-plane from  the zeta
function  term we  find  that  the Mellin  inversion  integral for  an
expansion about zero is
$$\frac{1}{2\pi i} \int_{13/2-i\infty}^{13/2+i\infty} Q(s)/x^s ds$$
which we evaluate in the left half-plane $\Re(s)<13/2.$
The two  zeta function  terms cancel the  poles of the  gamma function
term and we are left with just
$$\begin{align}
\mathrm{Res}(Q(s)/x^s; s=6) & = \frac{8\pi^6}{63x^6}  \quad\text{and}\\
\mathrm{Res}(Q(s)/x^s; s=0) & = \frac{1}{504}.
\end{align}$$
This shows that
$$S(x) = \frac{8\pi^6}{15x^6} + \frac{1}{504}
+ \frac{1}{2\pi i} \int_{-1/2-i\infty}^{-1/2+i\infty} Q(s)/x^s ds.$$
To  treat the  integral recall  the duplication  formula of  the gamma
function:
$$\Gamma(s) = 
\frac{1}{\sqrt\pi} 2^{s-1} 
\Gamma\left(\frac{s}{2}\right)
\Gamma\left(\frac{s+1}{2}\right).$$
which yields for $Q(s)$
$$\frac{1}{\sqrt\pi} 2^{s-1} 
\Gamma\left(\frac{s}{2}\right)
\Gamma\left(\frac{s+1}{2}\right)
\zeta(s) \zeta(s-5)$$
Furthermore observe  the following variant of  the functional equation
of the Riemann zeta function:
$$\Gamma\left(\frac{s}{2}\right)\zeta(s)
= \pi^{s-1/2} \Gamma\left(\frac{1-s}{2}\right)
\zeta(1-s)$$
which gives for $Q(s)$
$$\frac{1}{\sqrt\pi} 2^{s-1} 
\pi^{s-1/2}
\Gamma\left(\frac{s+1}{2}\right)
\Gamma\left(\frac{1-s}{2}\right)
\zeta(1-s)\zeta(s-5)
\\ = 
\frac{1}{\sqrt\pi} 2^{s-1} 
\pi^{s-1/2}
\frac{\pi}{\sin(\pi(s+1)/2)}
\zeta(1-s)\zeta(s-5)
\\ = 
2^{s-1}
\frac{\pi^s}{\sin(\pi(s+1)/2)}
\zeta(1-s)\zeta(s-5).$$
Now put $s=6-u$ in the remainder integral to get
$$\frac{1}{x^6}
\frac{1}{2\pi i} \int_{13/2-i\infty}^{13/2+i\infty} 
2^{5-u}
\frac{\pi^{6-u}}{\sin(\pi(7-u)/2)}
\zeta(u-5)\zeta(1-u) x^u du
\\ = \frac{64 \pi^6}{x^6}
\frac{1}{2\pi i} \int_{9/2-i\infty}^{9/2+i\infty} 
2^{u-1} \frac{\pi^{u}}{\sin(\pi(7-u)/2)}
\zeta(u-5)\zeta(1-u) (x/\pi^2/2^2)^u du.$$
Now $$\sin(\pi(7-u)/2) = \sin(\pi(-u-1)/2+4\pi)
\\ = \sin(\pi(-u-1)/2) = - \sin(\pi(u+1)/2).$$
We have shown that
$$S(x) = \frac{8\pi^6}{63x^6} +\frac{1}{504}
- \frac{64\pi^6}{x^6} S(4\pi^2/x).$$
In particular we get
$$S(2\pi) = \frac{1}{63\times 8} + \frac{1}{504}
- S(2\pi)$$
or $$S(2\pi) = \frac{1}{504}.$$
Remark. Unfortunately this method does not work for
$$S(x) = \sum_{k\ge 1} 
\frac{k^3}{\exp(kx)-1}$$
We get the functional equation
$$S(x) = \frac{\pi^4}{15x^4} -\frac{1}{240}
+ \frac{16 \pi^4}{x^4} S(4\pi^2/x).$$
which yields
$$S(2\pi) = \frac{1}{15\times 16} -\frac{1}{240}
+ S(2\pi)$$
which holds without providing any data about the value itself.