# We already know that int_(0)^(pi/2) x ln(2 cos x)dx=(7)/(16)zeta(3), int_(0)^(pi/2)x^2 ln^2(2cos x)dx=(11pi)/(16)zeta(4). Does the following integral admit a closed form? int_(0)^(pi/2)x^3 ln^3(2cos x)dx

A closed form for ${\int }_{0}^{\pi /2}{x}^{3}{\mathrm{ln}}^{3}\left(2\mathrm{cos}x\right)\phantom{\rule{mediummathspace}{0ex}}\mathrm{d}x$
$\begin{array}{rl}& {\int }_{0}^{\pi /2}x\mathrm{ln}\left(2\mathrm{cos}x\right)\phantom{\rule{mediummathspace}{0ex}}\mathrm{d}x=-\frac{7}{16}\zeta \left(3\right),\\ \\ & {\int }_{0}^{\pi /2}{x}^{2}{\mathrm{ln}}^{2}\left(2\mathrm{cos}x\right)\phantom{\rule{mediummathspace}{0ex}}\mathrm{d}x=\frac{11\pi }{16}\zeta \left(4\right).\end{array}$
Does the following integral admit a closed form?
$\begin{array}{rl}& {\int }_{0}^{\pi /2}{x}^{3}{\mathrm{ln}}^{3}\left(2\mathrm{cos}x\right)\phantom{\rule{mediummathspace}{0ex}}\mathrm{d}x\end{array}$
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Ashly Sanford
Proposition.
${\int }_{0}^{\pi /2}{x}^{3}{\mathrm{ln}}^{3}\left(2\mathrm{cos}x\right)\phantom{\rule{mediummathspace}{0ex}}\mathrm{d}x=\frac{45}{512}\zeta \left(7\right)-\frac{3{\pi }^{2}}{16}\zeta \left(5\right)-\frac{5{\pi }^{4}}{64}\zeta \left(3\right)-\frac{9}{4}\zeta \left(\overline{5},1,1\right)$
where $\zeta \left(\overline{p},1,1\right)$ is the colored MZV (Multi Zeta Values) function of depth 3 and weight $p+2$ given by
$\zeta \left(\overline{p},1,1\right):=\sum _{n=1}^{\mathrm{\infty }}\frac{\left(-1{\right)}^{n}}{{n}^{p}}\sum _{k=1}^{n-1}\frac{{H}_{k-1}}{k}$
belonging to a family of functions introduced by L. Euler and also called Euler(-Zagier) sums.
We have a general result.
Theorem. Let $\ell$ be any positive integer. Then
${\int }_{0}^{\pi /2}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}{x}^{2\ell +1}\phantom{\rule{negativethinmathspace}{0ex}}{\mathrm{ln}}^{2\ell +1}\left(2\mathrm{cos}x\right)\mathrm{d}x\in \mathbb{Q}\phantom{\rule{negativethinmathspace}{0ex}}\left(\zeta \left(4\ell +3\right),\zeta \left(2\right)\zeta \left(4\ell +1\right),...,\zeta \left(4\ell \right)\zeta \left(3\right),\zeta \left(\overline{2\ell +3},\left\{1{\right\}}_{2\ell }\right)\right)$
It is remarkable that there is only one constant
$\zeta \left(\overline{2\ell +3},\left\{1{\right\}}_{2\ell }\right)=\sum _{{n}_{1}>...>{n}_{2\ell +1}>0}\frac{\left(-1{\right)}^{{n}_{1}}}{{n}_{1}^{2\ell +3}{n}_{2}\cdots {n}_{2\ell +1}}$
for each integral of the considered form. The question of whether one can reduce this constant to colored MZVs/MZVs of lower depths is still subject to a conjecture (Zagier).
###### Did you like this example?
easternerjx
We can try the harmonic analysis path. Since:
$\mathrm{log}\left(2\mathrm{cos}x\right)=\sum _{n=1}^{+\mathrm{\infty }}\frac{\left(-1{\right)}^{n+1}}{n}\mathrm{cos}\left(2nx\right),$
$\begin{array}{}\text{(1)}& \mathrm{log}\left(2\mathrm{sin}x\right)=-\sum _{n=1}^{+\mathrm{\infty }}\frac{\mathrm{cos}\left(2nx\right)}{n},\end{array}$
we have, as an example:
${\int }_{0}^{\pi /2}{\mathrm{log}}^{3}\left(2\mathrm{sin}x\right)dx=-\frac{3\pi }{4}\zeta \left(3\right)$
since
$\begin{array}{}\text{(2)}& {\int }_{0}^{\pi /2}\mathrm{cos}\left(2{n}_{1}x\right)\mathrm{cos}\left(2{n}_{2}x\right)\mathrm{cos}\left(2{n}_{3}x\right)\phantom{\rule{thinmathspace}{0ex}}dx=\frac{\pi }{8}{\delta }_{2\cdot max{n}_{i}=\left({n}_{1}+{n}_{2}+{n}_{3}\right)}\end{array}$
Now:
$\begin{array}{}\text{(3)}& \pi /2-x=\frac{\pi }{4}+\frac{2}{\pi }\sum _{m=0}^{+\mathrm{\infty }}\frac{\mathrm{cos}\left(\left(4m+2\right)x\right)}{\left(2m+1{\right)}^{2}}\end{array}$
hence by multiplying $\left(1\right)$ and $\left(3\right)$ we can write the Fouries cosine series of $\left(\pi /2-x\right)\mathrm{log}\left(2\mathrm{sin}x\right)$ over $\left(0,\pi /2\right)$ and grab from $\left(2\right)$ and grab from $\left(2\right)$ a combinatorial equivalent for
${\int }_{0}^{\pi /2}{\left(\left(\pi /2-x\right)\mathrm{log}\left(2\mathrm{sin}x\right)\right)}^{3}\phantom{\rule{thinmathspace}{0ex}}dx.$
With the aid of Mathematica I got:
$\left(\pi /2-x\right)\mathrm{log}\left(2\mathrm{sin}x\right)=\sum _{n=0}^{+\mathrm{\infty }}\frac{\mathrm{cos}\left(\left(2n+1\right)x\right)}{\pi \left(2n+1{\right)}^{2}}\left(-\left(2n+1\right)\left({\psi }^{\prime }\left(\frac{2n+1}{2}\right)-{\psi }^{\prime }\left(-\frac{2n+1}{2}\right)\right)+2\left(2\gamma +\psi \left(\frac{2n+1}{2}\right)+\psi \left(-\frac{2n+1}{2}\right)\right)\right),$
$\left(\pi /2-x\right)\mathrm{log}\left(2\mathrm{sin}x\right)=\sum _{n=0}^{+\mathrm{\infty }}\frac{\mathrm{cos}\left(\left(2n+1\right)x\right)}{\pi \left(2n+1{\right)}^{2}}\left(2{H}_{\frac{2n+1}{2}}-\left(2n+1\right)\left({\psi }^{\prime }\left(\frac{2n+1}{2}\right)-{\psi }^{\prime }\left(-\frac{2n+1}{2}\right)\right)\right).$
$\begin{array}{}\text{(1)}& \left(\pi /2-x\right)\mathrm{log}\left(2\mathrm{sin}x\right)=\sum _{n=0}^{+\mathrm{\infty }}\frac{\mathrm{cos}\left(\left(2n+1\right)x\right)}{\pi \left(2n+1{\right)}^{2}}\left(-4\mathrm{log}2+\sum _{j=0}^{n}\frac{4}{2j+1}+\frac{4}{2n+1}+\left(2n+1\right)\sum _{j=0}^{n}\frac{8}{\left(2j+1{\right)}^{2}}\right).\end{array}$
So we have the Fourier cosine series of $\left(\pi /2-x\right)\mathrm{log}\left(2\mathrm{sin}x\right)$ but the path does not look promising from here. However, if we replace $\pi /2-x$ with a periodic continuation we get the way nicer identity:
$\begin{array}{}\text{(2)}& \left(\pi /2-x\right)\mathrm{log}\left(2\mathrm{sin}x\right)=-\left(\sum _{n=1}^{+\mathrm{\infty }}\frac{\mathrm{sin}\left(2nx\right)}{n}\right)\left(\sum _{n=1}^{+\mathrm{\infty }}\frac{\mathrm{cos}\left(2nx\right)}{n}\right)\end{array}$
$\begin{array}{}\text{(3)}& f\left(x\right)=\left(\pi /2-x\right)\mathrm{log}\left(2\mathrm{sin}x\right)=-\sum _{n=1}^{+\mathrm{\infty }}\frac{{H}_{n-1}}{n}\mathrm{sin}\left(2nx\right).\end{array}$
Now since ${\int }_{0}^{\pi /2}\mathrm{sin}\left(2mx\right)dx=\frac{{\mathbb{1}}_{m\equiv 1\phantom{\rule{0.444em}{0ex}}\left(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}2\right)}}{m}$ and ${\int }_{0}^{\pi /2}\mathrm{sin}\left(2ax\right)\mathrm{sin}\left(2bx\right)dx=\frac{\pi }{4}{\delta }_{a,b}$, the first two identites are easily proven. Now the three-terms integral
${\int }_{0}^{\pi /2}\mathrm{sin}\left(2ax\right)\mathrm{sin}\left(2bx\right)\mathrm{sin}\left(2cx\right)dx$
is a linear combination of $\frac{1}{a+b+c},\frac{1}{-a+b+c},\frac{1}{a-b+c},\frac{1}{a+b-c}$ depending on the parity of $a,b,c$, so it is quite difficult to find, explicitly, the Fourier cosine series of , $f\left(x{\right)}^{2}$ or the integral ${\int }_{0}^{\pi /2}f\left(x{\right)}^{3}\phantom{\rule{thinmathspace}{0ex}}dx$, but still not impossible. In particular, we know that the Taylor coefficients of the powers of $\mathrm{log}\left(1-x\right)$ depends on the generalized harmonic numbers. In our case,
$-\mathrm{log}\left(1-x\right)=\sum _{n=1}^{+\mathrm{\infty }}\frac{1}{n}{x}^{n},$
$\mathrm{log}\left(1-x{\right)}^{2}=\sum _{n=2}^{+\mathrm{\infty }}\frac{2{H}_{n-1}}{n}{x}^{n},$
$-\mathrm{log}\left(1-x{\right)}^{3}=\sum _{n=3}^{+\mathrm{\infty }}\frac{3{H}_{n-1}^{2}-3{H}_{n-1}^{\left(2\right)}}{n}{x}^{n},$
$\begin{array}{}\text{(4)}& \mathrm{log}\left(1-x{\right)}^{4}=\sum _{n=4}^{+\mathrm{\infty }}\frac{4{H}_{n-1}^{3}+8{H}_{n-1}^{\left(3\right)}-12{H}_{n-1}{H}_{n-1}^{\left(2\right)}}{n}{x}^{n}\end{array}$
hence we can just find a closed form for
${\int }_{0}^{\pi /2}{x}^{3}\left(1-2\mathrm{cos}x{\right)}^{n}dx$
and sum everything through the third previous identity. Ugh.