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
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
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\begin{align}
& \color{#44f}{\sum_{n = 1}^{\infty}
{\cos\pars{n} \over n^{4}}} =
\Re\sum_{n = 1}^{\infty}
{\pars{\expo{\ic}}^{n} \over n^{4}} = \Re\on{Li}_{4}\pars{\expo{\ic}}
\\[5mm] = & \
{1 \over 2}\left[%
\on{Li}_{4}\pars{\exp\pars{2\pi\ic{1 \over 2\pi}}}\right.
\\[2mm] & \left. \phantom{AA}+
\pars{-1}^{4}\,
\on{Li}_{4}\pars{%
\exp\pars{-2\pi\ic{1 \over 2\pi}}}
\right]
\\[5mm] = & \
{1 \over 2}\bracks{-\,{\pars{2\pi\ic}^{4} \over 4!}\on{B}_{4}\pars{1 \over 2\pi}}\tag{1}\label{1}
\end{align}
$\ds{\on{Li}_{s}\ \mbox{and}\ \on{B}_{n}}$ are the Polylogarithm Function and a Bernoulli Polynomial, respectively.
The bracket enclosed identity in
(\ref{1}) is the $\ds{Jonqui\grave{e}re\ Inversion\ Formula}$.
$$
\mbox{Note that}\quad
\on{B}_{4}\pars{x} = x^{4} - 2x^{3} + x^{2} - {1 \over 30}
$$
Therefore, with (\ref{1}),
\begin{align}
& \color{#44f}{\sum_{n = 1}^{\infty}
{\cos\pars{n} \over n^{4}}}
\\[5mm] = & \
{1 \over 2}\braces{-\,{16\pi^{4} \over 24}
\bracks{\pars{1 \over 2\pi}^{4} - 2\pars{1 \over 2\pi}^{3} + \pars{1 \over 2\pi}^{2} - {1 \over 30}}}
\\[5mm] = & \
\bbx{\color{#44f}{{\pi^{4} \over 90}- {\pi^{2} \over 12} + {\pi \over 12} - {1 \over 48}}} \approx 0.5008
\\ &
\end{align}
A: 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}
A: Start with :
$$\tag{1}f(x):=\sum_{n=1}^{\infty }\frac{\cos(n\,x)}{n^4}$$
and observe that (from the "Sawtooth Wave" Fourier series with $L=2\pi$ and $x=2X$) :
$$\tag{2}f^{(3)}(x)=\sum_{n=1}^{\infty }\frac{\sin(n\,x)}{n}=\frac{\pi-x}2,\quad \text{for}\;x\in(0,2\pi)$$
At this point it remains only to integrate $(2)$ three times with the appropriate constant of integration :
$$\tag{3}f^{(2)}(x)=\sum_{n=1}^{\infty }\frac{-\cos(n\,x)}{n^2}=C+\frac{\pi}2x-\frac 14x^2$$
with $\;\displaystyle C=f^{(2)}(0)=-\zeta(2)=-\frac{\pi^2}6$.
$$\tag{4}f'(x)=\sum_{n=1}^{\infty }\frac{-\sin(n\,x)}{n^3}=-\frac{\pi^2}6x+\frac{\pi}4x^2-\frac 1{12}x^3$$
(since $f'(0)=0$)
and the final solution :
$$\tag{5}f(x)=\sum_{n=1}^{\infty }\frac{\cos(n\,x)}{n^4}=\zeta(4)-\frac{\pi^2}{12}x^2+\frac{\pi}{12}x^3-\frac 1{48}x^4,\quad \text{for}\;x\in(0,2\pi)$$
Of course $x=1$ and $\zeta(4)=\dfrac{\pi^4}{90}$ will return the wished conclusion.
Generalization
It is clear that $\;\displaystyle\sum_{n=1}^{\infty }\frac{\cos(n\,x)}{n^{2m}}\;$ and $\;\displaystyle\sum_{n=1}^{\infty }\frac{\sin(n\,x)}{n^{2m-1}}\;$ may be evaluated by this method for $x\in(0;2\pi)$ and $m$ any positive integer. The polynomials obtained are well known since they correspond to the Bernoulli polynomials as given in Abramowitz and Stegun $(23.1.18)$ and $(23.1.17)$.
The remaining cases $\;\displaystyle\sum_{n=1}^{\infty }\frac{\cos(n\,x)}{n^{2m-1}}\;$ and $\;\displaystyle\sum_{n=1}^{\infty }\frac{\sin(n\,x)}{n^{2m}}\;$ are not so easy to evaluate and got the name of Clausen functions $\;\operatorname{Cl}_{2m-1}(x)\;$ and $\;\operatorname{Cl}_{2m}(x)$.
The origin of the difficulty is that $\;\displaystyle\operatorname{Cl}_1(x):=\sum_{n=1}^{\infty }\frac{\cos(n\,x)}{n}=-\ln(2\,\sin(x/2))\;$ (see for example here) implying that the integrals will be harder to evaluate (except for specific fractions of $\pi$).
