<s>[INST] Clausen and Riemann zeta function This is an exercise from the American Monthly Problems from last year.
I would like prove two formulas:
(1) $\int_0^{2\pi}\int_0^{2\pi}\log(3+2\cos(x)+2\cos(y)+2\cos(x-y)) dxdy=8\pi Cl(\frac{\pi}{3})$
(2) $\int_0^{\pi}\int_0^{\pi}\log(3+2\cos(x)+2\cos(y)+2\cos(x-y)) dxdy=\frac{28}{3}\zeta(3)$
$Cl(\phi)=\sum_{n=1}^{\infty}\frac{\sin(n\phi)}{n^2}$
The first I did was to reqwirte $Cl(\phi)$. I took the derivative and received $Cl(\phi)=-\int_0^{\phi}\log(2\sin(\frac{t}{2}))dt$ for $0\le\phi\le\pi$
For (2) I have no idea. It seems to me these integrals deliver a nice approximation mathod for $Cl(\frac{\pi}{3})$ and $\zeta(3)$ but I could not find the formulas somewhere else. 
 [/INST] First identity. Let us rewrite the integrand in several ways:
\begin{align}I(x,y)=3+2\cos x+2\cos y+2\cos(x-y)=\\
\tag{1} =1+4\cos\frac{x+y}{2}\cos\frac{x-y}{2}+4\cos^2\frac{x-y}{2}=\\
\tag{2}=\left(u+v+1\right)\left(u+\frac{v}{v+1}\right)\frac{v+1}{uv}=J(u,v),
\end{align}
where $u=e^{iy}$, $v=e^{ix}$. Equality (1) shows that $I(x,y)\geq0$ for $x,y\in[0,2\pi]$. Moreover, $I(x,y)=0$ only if (a) $\cos\frac{x+y}{2}=1$, $\cos\frac{x-y}{2}=-\frac12$ or (b) $\cos\frac{x+y}{2}=-1$, $\cos\frac{x-y}{2}=\frac12$. It is easy to check that the solutions of (b) are $(x,y)=\left(\frac{2\pi}{3},\frac{4\pi}{3}\right)$ and $(x,y)=\left(\frac{4\pi}{3},\frac{2\pi}{3}\right)$, whereas (a) has no solutions.
Since $I(x,y)$ is $2\pi$-periodic in $x$, we can replace $\int_{0}^{2\pi}dx$ by $\int_{-\frac{2\pi}{3}}^{\frac{2\pi}{3}}dx+\int_{\frac{2\pi}{3}}^{\frac{4\pi}{3}}dx$ and then write
\begin{align}
\int_0^{2\pi}\left(\int_0^{2\pi}\ln I(x,y)\;dy\right)dx=\left(\int_{-\frac{2\pi}{3}}^{\frac{2\pi}{3}}+\int_{\frac{2\pi}{3}}^{\frac{4\pi}{3}}\right)\left(\frac{1}{ i}\int_{|u|=1}\ln J(u,v)\frac{du}{u}\right)dx\tag{3}
\end{align}
The indefinite integral $\int\ln J(u,v)\frac{du}{u}$ can be expressed in terms of dilogarithms. Further, since we integrate over the unit circle, only differences of such dilogarithms on their different branches will appear in the answer, and such differences are expressed in terms of elementary functions (logarithms). 
Let us now work this out explicitly in a more elementary way: 


*

*First assume that $x\in\left(-\frac{2\pi}{3},\frac{2\pi}{3}\right)$, then $|1+v|>1$ and $|v/(1+v)|<1$. Make the change of variables $u=\frac{v}{v+1}w$, then
$$ \int_{|u|=1}\ln J(u,v)\frac{du}{u}=\int_{C_1}\ln\Bigl[(1+w^{-1})(\overbrace{4\cos^2\frac{x}{2}}^{>1}+w)\Bigr]\frac{dw}{w},$$
where $C_1$ is a circle centered at $0$ of radius $R_1$ such that $1<R_1<4\cos^2\frac{x}{2}$. The branch of the logarithm is fixed so that for $w=R_1$ the logarithm is real. Now by residues we have
\begin{align}
\int_{C_1}\frac{\ln(1+w^{-1})}{w}dw=0,\qquad\\
\int_{C_1}\frac{\ln(4\cos^2\frac{x}{2}+w)}{w}dw=2\pi i\ln 4\cos^{2}\frac{x}{2}.
\end{align}
(In the first integral we shrink the contour around $\infty$, in the second around $0$). Therefore we get
$$ \int_{-\frac{2\pi}{3}}^{\frac{2\pi}{3}}\left(\frac{1}{ i}\int_{|u|=1}\ln J(u,v)\frac{du}{u}\right)dx=2\pi\int_{-\frac{2\pi}{3}}^{\frac{2\pi}{3}}\ln \left(4\cos^{2}\frac{x}{2}\right)dx.\tag{4}$$

*Similarly, for $x\in\left(\frac{2\pi}{3},\frac{4\pi}{3}\right)$ we have $|1+v|<1$ and $|v/(1+v)|>1$. Now we make in the complex integral over $u$ in (3) the change of variables $u=(v+1)w$ so that
$$ \int_{|u|=1}\ln J(u,v)\frac{du}{u}=\int_{C_2}\ln\Bigl[(1+w^{-1})(1+\overbrace{4\cos^2\frac{x}{2}}^{<1}\;w)\Bigr]\frac{dw}{w},$$
where $C_2$ is a circle centered at $0$ of radius $R_2$ such that $1<R_2<(4\cos^2\frac{x}{2})^{-1}$. But now, similarly to the above, by residues:
$$\int_{C_2}\frac{\ln(1+w^{-1})}{w}dw=
\int_{C_2}\frac{\ln(1+4\cos^2\frac{x}{2}w)}{w}dw=0,$$
so that
$$ \int_{\frac{2\pi}{3}}^{\frac{4\pi}{3}}\left(\frac{1}{ i}\int_{|u|=1}\ln J(u,v)\frac{du}{u}\right)dx=0.\tag{5}$$


Substituting (4) and (5) into (3), we obtain
\begin{align}
\int_0^{2\pi}\left(\int_0^{2\pi}\ln I(x,y)\;dy\right)dx=2\pi\int_{-\frac{2\pi}{3}}^{\frac{2\pi}{3}}\ln \left(4\cos^{2}\frac{x}{2}\right)dx=8\pi\int_{0}^{\frac{2\pi}{3}}\ln \left|2\cos\frac{x}{2}\right|dx=\\=8\pi\int_{\pi/3}^{\pi}\ln \left|2\sin\frac{x}{2}\right|dx=8\pi\left[\mathrm{Cl}_2\left(\frac{\pi}{3}\right)-\underbrace{\mathrm{Cl}_2\left(\pi\right)}_{=0}\right]=8\pi\,\mathrm{Cl}_2\left(\frac{\pi}{3}\right),
\end{align}
where at the last step we have used the standard integral representation of the Clausen function. $\blacksquare$
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