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. 
 A: 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. $\blacksquare$
A: Following the comments, I post here some progress on the 2nd identity. The main results are (1) and (2), and it remains to prove (3). Probably I will finish the calculation later or set a bounty if nobody comes up with a simpler idea.
Second identity: Using the same notation as above, we can rewrite the integral as
$$ A=\int_0^{\pi}\left(\int_0^{\pi}\ln I(x,y)\;dy\right)dx=\left(\int_0^{\frac{2\pi}{3}}+\int_{\frac{2\pi}{3}}^{\pi}\right)\left(\int_0^{\pi}\ln I(x,y)\;dy\right)dx.$$


*

*
For $x\in(0,2\pi/3)$ we have $|2\cos\frac{x}{2}|>1$ and can write (using (2) from the above)
\begin{align}
F_+(x)=\int_0^{\pi}\ln I(x,y)\;dy=\\=\int_0^{\pi}\left[\ln\left(1+\frac{1}{2\cos\frac{x}{2}}e^{-i(y-x/2)}\right)+\ln\left(1+\frac{1}{2\cos\frac{x}{2}}e^{i(y-x/2)}\right)+\ln 4\cos^2\frac{x}{2}\right]dy=\\
=i\left[\mathrm{Li}_2\left(\frac{e^{-ix/2}}{2\cos\frac{x}{2}}\right)-\mathrm{Li}_2\left(-\frac{e^{-ix/2}}{2\cos\frac{x}{2}}\right)+\mathrm{Li}_2\left(-\frac{e^{ix/2}}{2\cos\frac{x}{2}}\right)-\mathrm{Li}_2\left(\frac{e^{ix/2}}{2\cos\frac{x}{2}}\right)\right]+\\+\pi \ln 4\cos^2\frac{x}{2}.
\end{align}
At the last step we have used the standard evaluations of $\mathrm{Li}_2(z)$:
\begin{align}
\int_0^{\pi}\ln\left(1+re^{\pm iy}\right)dy=\pm i\Bigl[\mathrm{Li}_2(r)-\mathrm{Li}_2(-r)\Bigr],\qquad |r|<1.
\end{align}

*Similarly, for $x\in(2\pi/3,\pi)$ we have $|2\cos\frac{x}{2}|<1$, so that
\begin{align}
F_-(x)=\int_0^{\pi}\ln I(x,y)\;dy=\qquad\\=\int_0^{\pi}\left[\ln\left(1+2\cos\frac{x}{2}e^{-i(y-x/2)}\right)+\ln\left(1+2\cos\frac{x}{2}e^{i(y-x/2)}\right)\right]dy=\qquad\\
=i\left[\mathrm{Li}_2\left(2\cos\frac{x}{2}e^{-ix/2}\right)-\mathrm{Li}_2\left(-2\cos\frac{x}{2}e^{-ix/2}\right)+\mathrm{Li}_2\left(-2\cos\frac{x}{2}e^{ix/2}\right)-\mathrm{Li}_2\left(2\cos\frac{x}{2}e^{ix/2}\right)\right].
\end{align}
$$=i\left[\mathrm{Li}_2\left(1+e^{-ix}\right)-\mathrm{Li}_2\left(1+e^{ix}\right)-\mathrm{Li}_2\left(-1-e^{-ix}\right)+\mathrm{Li}_2\left(-1-e^{ix}\right)\right].$$


The functions $F_+(x)$ and $F_-(x)$ can be analytically continued outside their respective domains of definition. It turns out that these continuations coincide in an open neibourhood of the real interval $x\in(0,\pi)$. This is after all not very surprising since both $F_{\pm}(x)$ represent the same integral $\int_0^{\pi}\ln I(x,y)\;dy$. In order to show that, indeed, $F_+(x)=F_-(x)$ for $x\in(0,\pi)$, one should use the identities of type $\mathrm{Li}_2(z)+\mathrm{Li}_2(z^{-1})=-\frac{\pi^2}{6}-\frac12 \ln^2(-z)$. [Or simply verify that the derivatives (given by elementary functions) of $F_{\pm}(x)$ coincide and compute both functions at some particular point].
Now, since $F_+(x)=F_-(x)$ on $x\in(0,\pi)$, we can write
\begin{align}
A=\int_0^{\frac{2\pi}{3}}F_+(x)dx+\int_{\frac{2\pi}{3}}^{\pi}F_-(x)dx=\int_0^{\pi}F_-(x)dx=\\=
i\int_0^{\pi}\left[\mathrm{Li}_2\left(1+e^{-ix}\right)-\mathrm{Li}_2\left(1+e^{ix}\right)-\mathrm{Li}_2\left(-1-e^{-ix}\right)+\mathrm{Li}_2\left(-1-e^{ix}\right)\right]\tag{1}dx.
\end{align}
N.B.: Dilogarithms $\mathrm{Li}_2(z)$ are defined on their main sheet: the cut plane $z\in\mathbb{C}\backslash[1,\infty)$.
Let us show how to compute the first two integrals in (1). This can be done as follows. First, use the
identity $\mathrm{Li}_2(z)+\mathrm{Li}_2(1-z)=-\ln z\ln(1-z)+\frac{\pi^2}{6}$ to rewrite these integrals as
\begin{align}
A_1=i\int_0^{\pi}\left[\mathrm{Li}_2\left(1+e^{-ix}\right)-\mathrm{Li}_2\left(1+e^{ix}\right)\right]dx=\\=
i\int_0^{\pi}\left[\mathrm{Li}_2\left(-e^{ix}\right)-\mathrm{Li}_2\left(-e^{-ix}\right)-i(\pi-x)\ln4\cos^2\frac{x}{2}\right]dx.
\end{align}
For the first two terms, introduce $z=e^{ix}$ and deform the demi-circles of integration to diameters. This yields
$$i\int_0^{\pi}\left[\mathrm{Li}_2\left(-e^{ix}\right)-\mathrm{Li}_2\left(-e^{-ix}\right)\right]dx=2\int_{-1}^{1}\frac{\mathrm{Li}_2(z)}{z}dz=
\frac{7}{2}\zeta(3),$$
where the last equality is obtained by expanding $\mathrm{Li}_2(z)$ into series, computing the integrals and using that
$\zeta(3)=\sum_{k=0}^{\infty}\frac{1}{(2k+1)^3}+\frac{1}{8}\zeta(3)$. Next, in $\int_0^{\pi}(\pi-x)\ln4\cos^2\frac{x}{2}\,dx$,
integrating once by parts to kill the linear term, we obtain one half of the initial integral $A_1$. Therefore,
$$A_1=i\int_0^{\pi}\left[\mathrm{Li}_2\left(1+e^{-ix}\right)-\mathrm{Li}_2\left(1+e^{ix}\right)\right]dx=7\zeta(3)\tag{2}$$
and now it remains to show that
$$A_2=i\int_0^{\pi}\left[\mathrm{Li}_2\left(-1-e^{ix}\right)-\mathrm{Li}_2\left(-1-e^{-ix}\right)\right]dx=\frac{7}{3}\zeta(3).\tag{3}
$$
