# 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.

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Monthly problem 11654. Keep watching the Monthly for the solution! Yes, two solvers did it by using polylogarithm identities. And someone found essentially this in a paper: Boyd, Canad. Math. Bull. 24 (1981) 453-469. – GEdgar Apr 29 '13 at 14:28
Do you have any reference where I can find the sample solution? How do you know how two solvers did it? – Alexander Apr 30 '13 at 13:07

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:

1. 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 $$\int_{-\frac{2\pi}{3}}^{\frac{2\pi}{3}}\left(\frac{1}{ i}\int_{|u|=1}\ln J(u,v)\frac{du}{u}\right)dx=2\pi\int_{-\frac{2\pi}{3}}^{\frac{2\pi}{3}}\ln \left(4\cos^{2}\frac{x}{2}\right)dx.\tag{4}$$
2. 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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Thank you very much, very interesting solution. Do you also have an idea for the second equation? – Alexander Apr 28 '13 at 13:40
I have an intermediate expression given by a sum of four similarly-looking integrals of dilogarithm functions. Two of them I have calculated and they give a rational multiple of $\zeta(3)$. The other two look similar but I haven't finished the computation. If you are interested, I may post the progress made so far in a separate answer. Another idea: I could make a follow-up question and set a bounty for evaluation of the remaining two integrals. – Start wearing purple Apr 28 '13 at 14:59
I would be interested very much. I guess the best think is to make a follow-up question for evaluation of the ramaining integrals but it depends on you. – Alexander Apr 28 '13 at 15:33
I just analyzed your proof and one thing is still not 100% clear to me. How do you derive in (3) the term $\frac{1}{ i}\int_{|u|=1}\ln J(u,v)\frac{du}{u}$ ? especially the $\frac{1}{i}$ – Alexander Apr 28 '13 at 17:24
If $u=e^{iy}$, then $du=ie^{iy}dy=iu\,dy\Rightarrow dy=\frac{du}{iu}$. – Start wearing purple Apr 28 '13 at 17:30

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 $$A=\int_0^{\pi}\left(\int_0^{\pi}\ln I(x,y)\;dy\right)dx=\left(\int_0^{\frac{2\pi}{3}}+\int_{\frac{2\pi}{3}}^{\pi}\right)\left(\int_0^{\pi}\ln I(x,y)\;dy\right)dx.$$

1. 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}
2. 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 $$A_2=i\int_0^{\pi}\left[\mathrm{Li}_2\left(-1-e^{ix}\right)-\mathrm{Li}_2\left(-1-e^{-ix}\right)\right]dx=\frac{7}{3}\zeta(3).\tag{3}$$

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Are you sure you meant $\zeta(3)=\sum_{k=0}^{\infty}\frac{1}{(2k+1)^3}+\frac{1}{8}\zeta(3)$ ? – Alexander Apr 28 '13 at 19:37
Yes. The integral is actually proportional to the sum $\sum_{k=0}^{\infty}\frac{1}{(2k+1)^3}$. The formula that you quote expresses this sum in terms of $\zeta(3)$. I've intentionally written it in that way to give a hint on how it can be proven. – Start wearing purple Apr 28 '13 at 19:59