Is there a closed-form of $\sum_{n=1}^{\infty} \frac{\sin(n)}{n^4}$ Is there a closed-form summation result for Fourier series:
$$\sum_{n=1}^{\infty}\frac{\sin(n)}{n^4}?\tag{1}$$
I tried using  available result of the following (odd) function :
$$\frac{\pi-x}{2}=\sum_{n\geq 1}\frac{\sin(nx)}{n}.\tag{2}$$
However I couldn't get the series to agree because when I integrate the formula to evaluate coefficients, the powers of $n$ become odd.. 

Consider the series
\begin{align}
f(x) = \sum_{n=1}^{\infty} \frac{\sin(n x)}{n^{4}}
\end{align}
for which upon differentiation the following is obtained.
\begin{align}
f'(x) &= \sum_{n=1}^{\infty} \frac{\cos(n x)}{n^{3}} \\
f''(x) &= - \sum_{n=1}^{\infty} \frac{\sin(n x)}{n^{2}} \\
f'''(x) &= - \sum_{n=1}^{\infty} \frac{\cos(n x)}{n} = \frac{1}{2} \left( \ln(1 - e^{i x}) + \ln(1 - e^{- i x}) \right).
\end{align}
By integration the following is obtained
\begin{align}
f''(x) &= \frac{i}{2} \left( Li_{2}(e^{i x}) - Li_{2}(e^{-i x}) \right) + c_{2} \\
f'(x) &= \frac{1}{2} \left( Li_{3}(e^{i x}) - Li_{3}(e^{- i x}) \right) + c_{2} x + c_{1} \\
f(x) &= \frac{-i}{2} \left( Li_{4}(e^{i x}) - Li_{4}(e^{-i x}) \right) + \frac{c_{2} x^{2}}{2} + c_{1} x + c_{0}
\end{align}
where $Li_{m}(x)$ is the polylogarithm function. In order to determine the constants of integration evaluation of the series is required. This is determined by setting $x = 0$. 
\begin{align}
f(0) &= 0 = c_{0} \\
f'(0) &= \zeta(3) = c_{1} \\
f''(0) &= 0 = c_{2}.
\end{align}
The result is then
\begin{align}
\sum_{n=1}^{\infty} \frac{\sin(n x)}{n^{4}} &= \frac{-i}{2} \left( Li_{4}(e^{i x}) - Li_{4}(e^{-i x}) \right) + \zeta(3) \, x 
\end{align}
