Sum=Integral? Find $m$ so that $\sum _{n=1}^{\infty } \frac{n^m}{e^{2 \pi n}-1}=\int_0^{\infty } \frac{x^m}{e^{2 \pi x}-1} \, dx$ This question asked for a derivation of the interesting relation
$$\sum_{n=1}^{\infty}  \frac{n^{13}}{e^{2 \pi n}-1} = \frac{1}{24}$$
In an attempt to answer this problem I played around a bit and was surprised that the related integral
$$\int_{0}^{\infty}  \frac{n^{13}}{e^{2 \pi n}-1} \,dn= \frac{1}{24}$$
has the same nice result. Notice that the sum starts at $n=1$ and the integral starts at its "natural" boundary $0$.
Defining more generally
$$S(m) = \sum_{n=1}^{\infty}  \frac{n^m}{e^{2 \pi n}-1}\tag{1a}$$
$$i(m) = \int_{0}^{\infty}  \frac{n^m}{e^{2 \pi n}-1} \,dn\tag{1b}$$
We can ask for which $m$ sum and integral lead to equal values, i.e.
$$S(m) = i(m)\tag{2}$$
It turns out that equality seems to happen if and only if $m=4k+1, k=1,2,3...$.
This picture shows a discrete plot of the quotient $\frac{S(m)}{i(m)}$ of sum and integral.

Questions:
(1) Find all natural solutions $m$ of equation $(2)$
(2) Find an analytic continuation of $S(m)$ to real $m$ and study if the quotient $S(m)/i(m)$ is unity for other values
Hints (for derivations see):
a) the integral has a closed expression which is not restricted to natural values of $m$
$$i(m)= \frac{1}{(2 \pi )^{m+1} } \zeta (m+1) \Gamma (m+1)\tag{3}$$
Here $\zeta(x)$ is the Riemann zeta function.
b) the sum has a closed expression for $m=1,2,3,...$ as well
$$S(m)=\frac{(-1)^{m+1} \psi _{e^{-2 \pi }}^{(m)}(1)}{(2 \pi )^{m+1}}\tag{4}$$
Here $\psi _{q}^{(m)}(x)$ is the $m$-th derivative of the $q$-digamma function with respect to x.
EDIT: after a comment of @Jean Marie I saw that Marko Riedel has derived as a special case of $(4)$ for $m=4k+1$ the simplified formula
$$S(4 k+1)=\frac{B_{4 k+2}}{2 (4 k+2)}, k=1,2,3,...\tag{4a}$$
Here $B_{n}$ is the Bernoulli number.
 A: The standard method is to use that $$\frac{\pi^2}{\sin^2(\pi z)}-\sum_m \frac1{(z+m)^2}=0$$ (it is a $1$-periodic entire function which is bounded on $\Re(z)\in [0,1]$ and which vanishes at $i\infty$)
to get that for $k\ge 4$ even and $\Im(z)>0$ $$G_k(z)=\sum_{(n,m)\ne (0,0)} \frac1{(zn+m)^k}= 2\zeta(k)+\frac{4i\pi}{(k-1)!}\sum_{n\ge 1}\sum_{d\ge 1}(2i\pi d)^{k-1}e^{2i\pi dnz}$$
When $k\equiv 2\bmod 4$ we get that $G_k(i)=0$.
On the other hand $$\int_0^\infty \frac{x^{k-1}}{e^{2\pi x}-1}dx=(k-1)!(2\pi)^{-k} \zeta(k),\qquad\sum_{d\ge 1}\frac{d^{k-1}}{e^{2\pi d}-1}=\sum_{d\ge 1}\sum_{n\ge 1}d^{k-1}e^{-2\pi dn}$$

The case $k\equiv 0\bmod 4$ has a solution too. From the theory of modular forms (mainly that $E_4(z)^3-E_6(z)^2$ is a non-vanishing cusp form),  with $E_k(z)=\frac{G_k(z)}{2\zeta(k)}$: $$E_k(z)=\sum_{4a+6b=k} c_{k,a,b}E_4(z)^a E_6(z)^b$$ for some rational coefficients $c_{k,a,b}$. As $E_6(i)=0$ we get $$E_k(i) = c_{k,k/4,0}E_4(i)^{k/4}$$ The known closed-form of $E_4(i)$ in term of $\Gamma(1/4)$ implies that $E_k(i)$ is irrational, so it is $\ne 2$ and hence $\int_0^\infty \frac{x^{k-1}}{e^{2\pi x}-1}dx\ne \sum_{d\ge 1}\frac{d^{k-1}}{e^{2\pi d}-1}$.
