Proving $\sum_{n=1}^{\infty }\frac{\cos(n)}{n^4}=\frac{\pi ^4}{90}-\frac{\pi ^2}{12}+\frac{\pi }{12}-\frac{1}{48}$ Proving $$\sum_{n=1}^{\infty }\frac{\cos(n)}{n^4}=\frac{\pi ^4}{90}-\frac{\pi ^2}{12}+\frac{\pi }{12}-\frac{1}{48}$$
I tried with Wolfram but it couldn't give me any clear value as shown below
 
The numerical value of Wolfram not different of my closed-form.
Can anyone explain how the $.5(Li_4(e^{-i}+Li_4(e^{i}))$ equal the above closed-form

By using the series
\begin{align}
\sum_{n=1}^{\infty} \frac{ \cos(nx) }{ n^{2} } = \zeta(2) - \frac{ \pi \, x}{2} + \frac{x^{2}}{4}
\end{align}
integrate with respect to $x$ from zero to $t$ to obtain
\begin{align}
\sum_{n=1}^{\infty} \frac{\sin(n t)}{n^{3}} = \zeta(2) \, t - \frac{\pi \, t^{2}}{4} + \frac{t^{3}}{12}.
\end{align}
Integrate once again in a similar manor to obtain
\begin{align}
\sum_{n=1}^{\infty} \frac{\cos(nx)}{n^{4}} = \zeta(4) - \frac{\zeta(2) \, x^{2}}{2} + \frac{\pi \, t^{3}}{12} - \frac{x^{4}}{48}.
\end{align}
Upon letting $x=1$ the above results yield
\begin{align}
\sum_{n=1}^{\infty} \frac{\cos(n)}{n^{2}} &= \zeta(2) + \frac{1 - 2 \pi}{4} \\
\sum_{n=1}^{\infty} \frac{\sin(n)}{n^{3}} &= \zeta(2) + \frac{1 - 3 \pi}{12} \\
\sum_{n=1}^{\infty} \frac{\cos(n)}{n^{4}} &= \zeta(4) - \frac{\zeta(2)}{2} + \frac{\pi}{12} - \frac{1}{48}.
\end{align}
