Finding the closed form for $\sum_{n=1}^{\infty }\frac{\zeta (4n)}{\beta^{4n-1}}$ Finding the closed-form 

$$\sum_{n=1}^{\infty }\frac{\zeta (4n)}{\beta^{4n-1}}$$
  for $\beta\in(1,+\infty)$.

I learned from this site many many important things but I till need more, so I need to know if there is a closed-form of this series . Thanks for any body devised me to learn from any answer.
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$\ds{\sum_{n\ =\ 1}^{\infty}\ {\zeta\pars{4n} \over \beta^{4n - 1}}:\ {\large ?}.
     \qquad\beta\in\pars{1,\infty}}$.

\begin{align}&\color{#66f}{\large%
\sum_{n\ =\ 1}^{\infty}\ {\zeta\pars{4n} \over \beta^{4n - 1}}}
=\sum_{n\ =\ 4}^{\infty}\ {\zeta\pars{n} \over \beta^{n - 1}}
-\sum_{n\ =\ 1}^{\infty}\ {\zeta\pars{4n + 1} \over \beta^{4n}}
-\sum_{n\ =\ 1}^{\infty}\ {\zeta\pars{4n + 2} \over \beta^{4n + 1}}
-\sum_{n\ =\ 1}^{\infty}\ {\zeta\pars{4n + 3} \over \beta^{4n + 2}}
\\[5mm]&=\sum_{n\ =\ 1}^{\infty}\bracks{{\zeta\pars{n + 3} \over \beta^{n + 2}}
-
{\zeta\pars{4n + 1} \over \beta^{4n}} - {\zeta\pars{4n + 2} \over \beta^{4n + 1}}- {\zeta\pars{4n + 3} \over \beta^{4n + 2}}}
\\[5mm]&=\beta\sum_{k\ =\ 1}^{\infty}\
\sum_{n\ =\ 1}^{\infty}\bracks{{1 \over \pars{\beta k}^{n + 3}}
-
{1 \over \pars{\beta k}^{4n + 1}}
-{1 \over \pars{\beta k}^{4n + 2}}- {1 \over \pars{\beta k}^{4n + 3}}}
\\[5mm]&=\beta\sum_{k\ =\ 1}^{\infty}{1 \over \pars{\beta k}^{4} - 1}
=\color{#66f}{\large{1 \over 4}\bracks{2\beta - \pi\,\cot\pars{\pi \over \beta}
-\pi\coth\pars{\pi \over \beta}}}
\end{align}
A: Since:
$$\zeta(4n)=\frac{1}{(4n-1)!}\int_{0}^{+\infty}\frac{z^{4n-1}}{e^{z}-1}\,dz \tag{1}$$
and:
$$ \sum_{n\geq 1}\frac{w^{4n-1}}{(4n-1)!}=\frac{\sinh w-\sin w}{2}\tag{2}$$
it follows that, for any $\beta>1$,
$$ \sum_{n=1}^{+\infty}\frac{\zeta(4n)}{\beta^{4n-1}} = \frac{1}{2}\int_{0}^{+\infty}\frac{\sinh(w/\beta)-\sin(w/\beta)}{e^w-1}\,dw.\tag{3}$$
For special values of $\beta$ ($\beta\in\mathbb{N}$, for instance) we can compute the last integral through the residue theorem. In general, by exploiting the inverse Laplace transform, we have:
$$ \sum_{n=1}^{+\infty}\frac{\zeta(4n)}{\beta^{4n-1}} = \frac{1}{2}\left(\beta - \frac{\pi}{2}\cot\frac{\pi}{\beta}-\frac{\pi}{2}\coth\frac{\pi}{\beta}\right).\tag{4}$$
We can achieve the same by considering the Taylor series of $x\cot x$ and $x\coth x$ in a neighbourhood of $z=0$.
