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
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.
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 $\pi$).
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{{\displaystyle #1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\sr}[2]{\,\,\,\stackrel{{#1}}{{#2}}\,\,\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}
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}
Let $k$ be a nonnegative integer.
Contour integration shows that $$ \sum_{n=1}^{\infty}\frac{\sin (n)}{n^{2k-1}} = - \frac{1}{2} \, \Im\, \operatorname{Res} \left[\frac{\pi \left(\cot (\pi z)-\color{red}{i} \right)e^{iz}}{z^{2k-1}},0 \right]$$ and $$\sum_{n=1}^{\infty} \frac{\cos (n)}{n^{2k}} = - \frac{1}{2} \, \Re \, \operatorname{Res} \left[\frac{\pi \left(\cot (\pi z)-\color{red}{i} \right)e^{iz}}{z^{2k}},0 \right].$$
See my previous answer here for an explanation of why we use $\cot(\pi z)-i$ and not just $\cot(\pi z)$.
Basically it has to do with the fact that the magnitude of $e^{iz}$ is unbounded in the lower half-plane.
At the origin, we have the Laurent series expansion $$ \begin{align} \pi \left(\cot(\pi z)-i \right)e^{iz} &=\small \left(\frac{1}{z}- i \pi - 2\zeta(2)z -2 \zeta(4) x^{3} + O(z^{5}) \right) \left( 1+iz- \frac{z^{2}}{2!}- \frac{iz^{3}}{3!} + \frac{z^{4}}{4!} + O(z^{5}) \right) \\ & = \small \frac{1}{z} + i \left(1-\pi \right) + \left(- \frac{1}{2!} +\pi - 2\zeta(2) \right)z + i \left(- \frac{1}{3!} + \frac{\pi}{2!} -2 \zeta(2) \right)z^{2} \\ &+ \small \left(\frac{1}{4!} -\frac{\pi}{3!} + \frac{2 \zeta(2)}{2!} - 2 \zeta(4) \right)z^{3} + O(z^{4}) . \end{align}$$
Therefore, $$\sum_{n=1}^{\infty} \frac{\sin (n)}{n} = \frac{\pi -1}{2}, $$
$$\sum_{n=1}^{\infty} \frac{\cos (n)}{n^{2}} = \frac{1}{4} - \frac{\pi}{2} + \zeta(2), $$
$$\sum_{n=1}^{\infty} \frac{\sin (n)}{n^{3}} = \frac{1}{12} - \frac{\pi}{4} + \zeta(2), $$
$$\sum_{n=1}^{\infty} \frac{\cos (n)}{n^{4}} = -\frac{1}{48} + \frac{\pi}{12} - \frac{\zeta(2)}{2} + \zeta(4), $$ and so on.
We will evaluate the general Fourier series
\begin{align*} \sum_{n=1}^\infty \dfrac{\cos (n)}{n^{2k}} \end{align*}
for $k=1,2,\dots$. We have:
\begin{align*} & \sum_{n=0}^\infty \dfrac{\cos (n)}{n^{2k}} \nonumber \\ & = \frac{1}{(2k-1)!} \sum_{n=1}^\infty \dfrac{\cos (n)}{n^{2k}} \int_0^\infty u^{2k-1} e^{-u} du \nonumber \\ & = \frac{1}{(2k-1)!} \sum_{n=1}^\infty \cos (n) \int_0^\infty u^{2k-1} e^{-nu} du \nonumber \\ & = \frac{1}{(2k-1)!} \frac{1}{2} \sum_{n=1}^\infty \int_0^\infty (e^{-nu + i n} + e^{-nu - i n}) u^{2k-1} du \nonumber \\ & = \frac{1}{2 (2k-1)!} \int_0^\infty \sum_{n=1}^\infty (e^{-nu + i n} + e^{-nu - i n}) u^{2k-1} du \nonumber \\ & = \frac{1}{2 (2k-1)!} \int_0^\infty \left( \dfrac{1}{e^{u - i} - 1} + \dfrac{1}{e^{u + i} - 1} \right) u^{2k-1} du \end{align*}
To justify interchanging summation and integration in the above, we can use Fubini (see details at end)
From the above
\begin{align*} \sum_{n=0}^\infty \dfrac{\cos (n)}{n^{2k}} & = \frac{1}{2 (2k-1)!} \int_0^\infty \left( \dfrac{1}{e^{u - i} - 1} + \dfrac{1}{e^{u + i} - 1} \right) u^{2k-1} du \nonumber \\ & = \frac{1}{2 (2k-1)!} \int_0^\infty \left( \dfrac{e^{i} u^{2k-1}}{e^u - e^{i}} + \dfrac{e^{-i} u^{2k-1}}{e^u - e^{-i}} \right) du \nonumber \\ & = \frac{1}{2 (2k)!} \int_0^\infty \left( \dfrac{e^{i} e^u u^{2k}}{(e^u - e^{i})^2} + \dfrac{e^{-i} e^u u^{2k}}{(e^u - e^{-i})^2} \right) du \nonumber \\ & = \frac{1}{4 (2k)!} \int_{-\infty}^\infty \left( \dfrac{e^{i} e^u u^{2k}}{(e^u - e^{i})^2} + \dfrac{e^{-i} e^u u^{2k}}{(e^u - e^{-i})^2} \right) du \nonumber \\ & = \frac{1}{2 (2k)!} Re \int_{-\infty}^\infty \dfrac{e^{i} e^u u^{2k}}{(e^u - e^{i})^2} du \end{align*}
where I have done an integration by parts. Extending the range of integration is valid if you use both terms.
Define
\begin{align*} I (\alpha) = \int_{-\infty}^\infty \dfrac{e^{i} e^{\alpha u} e^u}{(e^u - e^{i})^2} du \end{align*}
for $-\frac{1}{2} \leq \alpha \leq \frac{1}{2}$. Then
\begin{align*} \sum_{n=1}^\infty \dfrac{\cos (n)}{n^{2k}} = \frac{1}{2 (2k)!} \left. \dfrac{\partial^{2k}}{\partial \alpha^{2k}} Re I (\alpha) \right|_{\alpha =0} \qquad (*) . \end{align*}
In order to evaluate $I(\alpha)$, consider the contour in the figure.
and the integral
\begin{align*} \oint_C \dfrac{e^{i} e^{\alpha z} e^z}{(e^z - e^{i})^2} \end{align*}
whose integrand has pole at $i$. The integral along the vertical edges vanishes as:
\begin{align*} f(z) = \dfrac{e^{i} e^{(\alpha + 1) (u+iv)}}{(e^{u+iv} + e^{i})^2} = \begin{cases} e^{i} e^{(\alpha - 1) (u+iv)} & u \rightarrow \infty \\ \dfrac{e^{(\alpha + 1) (u+iv)}}{e^{i}} & u \rightarrow - \infty \\ \end{cases} \end{align*}
So that
\begin{align*} \oint_C \dfrac{e^{i} e^{\alpha z} e^z}{(e^z - e^{i})^2} & = \int_{-\infty}^\infty \dfrac{e^{i} e^{\alpha u}}{(e^u - e^{i})^2} du - e^{\alpha 2 \pi i} \int_{-\infty+2 \pi i}^{\infty+ 2 \pi i} \dfrac{e^{i} e^{\alpha u}}{(e^u - e^{2 \pi x i})^2} du \nonumber \\ & = (1-e^{\alpha 2 \pi i}) \int_{-\infty}^\infty \dfrac{e^{i} e^{\alpha u}}{(e^u - e^{i})^2} du \end{align*}
Which rearranged is
\begin{align*} \int_{-\infty}^\infty \dfrac{e^{i} e^{\alpha u} e^u}{(e^u - e^{i})^2} du & = \frac{2 \pi i }{1-e^{\alpha 2 \pi i}} \frac{1}{2 \pi i} \oint_C \dfrac{e^{i} e^{\alpha z} e^z}{(e^z - e^{i})^2} dz \nonumber \\ & = \frac{2 \pi i}{1-e^{\alpha 2 \pi i}} Res [f(z)] \end{align*}
By expand the integrand in powers of $z-z_0$ about pole $z_0= i$ we find $Res [f(z)] = \alpha e^{\alpha i}$. Plugging this in the above
\begin{align*} I(\alpha) = \int_{-\infty}^\infty \dfrac{e^{i} e^{\alpha u}}{(e^u - e^{i})^2} du = - e^{\alpha i - \alpha \pi i} \dfrac{\alpha \pi}{\sin \alpha \pi} \end{align*}
We can take the case $k=2$ and Taylor expand in $\alpha$ to obtain the coefficient of $\alpha^4$. Doing so and using $(*)$ we obtain
\begin{align*} \sum_{n=1}^\infty \dfrac{\cos (n)}{n^4} = \frac{\pi^4}{90} - \frac{\pi^2}{12} + \frac{\pi}{12} - \frac{1}{48} . \end{align*}
NOTE: This expression
\begin{align*} - e^{\alpha 2 \pi x i - \alpha \pi i} \dfrac{\alpha \pi}{\sin \alpha \pi} \end{align*}
is related to the generating function for Bernoulli polynomials, as we establish in a moment. We will use this to prove $\sum_{n=1}^\infty \dfrac{\cos (n)}{n^{2k}} = \frac{(-1)^{k+1} (2 \pi)^{2k}}{2 (2k)!} B_{2k} (\frac{1}{2 \pi})$ where $B_{2k} (x)$ is the $2k-$th Bernoulli polynomial.
Bernoulli polynomials
We turn to the evaluation of the general sum, $\sum_{n=1}^\infty \frac{\cos (n)}{n^{2k}}$. The generating function for the Bernoulli polynomials is
\begin{align*} \dfrac{t e^{xt}}{e^t-1} = \sum_{k=0}^\infty B_k (x) \frac{t^k}{k!} . \end{align*}
We have
\begin{align*} \dfrac{t e^{xt}}{\sinh \frac{t}{2}} = \dfrac{2t e^{ t ( x+\frac{1}{2} ) }}{e^t-1} = 2 \sum_{k=0}^\infty B_k (x+\frac{1}{2}) \frac{t^k}{k!} \end{align*}
We perform $t \mapsto i 2 \pi t$ and $x \mapsto x-\frac{1}{2}$,
\begin{align*} \dfrac{\pi t}{\sin \pi t} e^{i 2 \pi t x - i \pi t} = \sum_{k=0}^\infty B_k (x) \frac{(i 2 \pi t)^k}{k!} \end{align*}
If we put $x = \frac{1}{2 \pi}$, we recognise the LHS as our expression for $-I(\alpha)$ with $\alpha$ replaced with $t$. Taking the real part and putting $x = \frac{1}{2 \pi}$:
\begin{align*} Re \dfrac{\pi t}{\sin \pi t} e^{i t - i \pi t} = \sum_{k=0}^\infty B_{2k} (\frac{1}{2 \pi}) \frac{(-1)^k (2 \pi)^{2k}}{(2k)!} t^{2k} \end{align*}
Using this in $(*)$ we obtain the general formula:
\begin{align*} \sum_{n=1}^\infty \dfrac{\cos (n)}{n^{2k}} & = \left. - \frac{1}{2 (2k)!} \dfrac{\partial^{2k}}{\partial \alpha^{2k}} Re \; e^{\alpha i - \alpha \pi i} \dfrac{\alpha \pi}{\sin \alpha \pi} \right|_{\alpha=0} \nonumber \\ & = \frac{(-1)^{k+1} (2 \pi)^{2k}}{2 (2k)!} B_{2k} (\frac{1}{2 \pi}) \end{align*}
For $k=2$,
\begin{align*} \sum_{n=1}^\infty \dfrac{\cos (n)}{n^4} = \frac{(-1) (2 \pi)^4}{2 (4)!} B_4 (\frac{1}{2 \pi}) \end{align*}
where
\begin{align*} B_4 (x) = x^4 - 2 x^3 + x^2 - \frac{1}{30} \end{align*}
which is the same result as quote earlier.
NOTE: It is not difficult to modify the above derivation to prove that:
\begin{align*} \sum_{n=1}^\infty \dfrac{\cos (2 \pi n x)}{n^{2k}} = \frac{(-1)^{k+1} (2 \pi)^{2k}}{2 (2k)!} B_{2k} (x) \end{align*}
Details: justify interchanging summation and integration
To justify interchanging summation and integration in the above, we can use Fubini:
\begin{align*} \frac{1}{(2k-1)!} \sum_{n=1}^\infty \int_0^\infty | \cos (n) u^{2k-1} e^{-nu} | du & \leq \frac{1}{(2k-1)!} \sum_{n=1}^\infty \int_0^\infty u^{2k-1} e^{-nu} du \nonumber \\ & = \sum_{n=1}^\infty \frac{1}{n^{2k}} \nonumber \\ & \leq \sum_{n=1}^\infty \frac{1}{n^2} \nonumber \\ & < 1 + \lim_{N \rightarrow \infty} \sum_{n=2}^N \frac{1}{n (n-1)} \nonumber \\ & = 1 + \lim_{N \rightarrow \infty} \sum_{n=2}^N \left( \frac{1}{n-1} - \frac{1}{n} \right) \nonumber \\ & = 1 + 1 - \lim_{N \rightarrow \infty} \frac{1}{N} = 2 < \infty \end{align*}
