We can try the harmonic analysis path. Since:

$\mathrm{log}(2\mathrm{cos}x)=\sum _{n=1}^{+\mathrm{\infty}}\frac{(-1{)}^{n+1}}{n}\mathrm{cos}(2nx),$

$\begin{array}{}\text{(1)}& \mathrm{log}(2\mathrm{sin}x)=-\sum _{n=1}^{+\mathrm{\infty}}\frac{\mathrm{cos}(2nx)}{n},\end{array}$

we have, as an example:

${\int}_{0}^{\pi /2}{\mathrm{log}}^{3}(2\mathrm{sin}x)dx=-\frac{3\pi}{4}\zeta (3)$

since

$\begin{array}{}\text{(2)}& {\int}_{0}^{\pi /2}\mathrm{cos}(2{n}_{1}x)\mathrm{cos}(2{n}_{2}x)\mathrm{cos}(2{n}_{3}x)\phantom{\rule{thinmathspace}{0ex}}dx=\frac{\pi}{8}{\delta}_{2\cdot max{n}_{i}=({n}_{1}+{n}_{2}+{n}_{3})}\end{array}$

Now:

$\begin{array}{}\text{(3)}& \pi /2-x=\frac{\pi}{4}+\frac{2}{\pi}\sum _{m=0}^{+\mathrm{\infty}}\frac{\mathrm{cos}((4m+2)x)}{(2m+1{)}^{2}}\end{array}$

hence by multiplying $(1)$ and $(3)$ we can write the Fouries cosine series of $(\pi /2-x)\mathrm{log}(2\mathrm{sin}x)$ over $(0,\pi /2)$ and grab from $(2)$ and grab from $(2)$ a combinatorial equivalent for

${\int}_{0}^{\pi /2}{((\pi /2-x)\mathrm{log}(2\mathrm{sin}x))}^{3}\phantom{\rule{thinmathspace}{0ex}}dx.$

With the aid of Mathematica I got:

$(\pi /2-x)\mathrm{log}(2\mathrm{sin}x)=\sum _{n=0}^{+\mathrm{\infty}}\frac{\mathrm{cos}((2n+1)x)}{\pi (2n+1{)}^{2}}(-(2n+1)({\psi}^{\prime}\left(\frac{2n+1}{2}\right)-{\psi}^{\prime}(-\frac{2n+1}{2}))+2(2\gamma +\psi \left(\frac{2n+1}{2}\right)+\psi (-\frac{2n+1}{2}))),$

$(\pi /2-x)\mathrm{log}(2\mathrm{sin}x)=\sum _{n=0}^{+\mathrm{\infty}}\frac{\mathrm{cos}((2n+1)x)}{\pi (2n+1{)}^{2}}(2{H}_{\frac{2n+1}{2}}-(2n+1)({\psi}^{\prime}\left(\frac{2n+1}{2}\right)-{\psi}^{\prime}(-\frac{2n+1}{2}))).$

$\begin{array}{}\text{(1)}& (\pi /2-x)\mathrm{log}(2\mathrm{sin}x)=\sum _{n=0}^{+\mathrm{\infty}}\frac{\mathrm{cos}((2n+1)x)}{\pi (2n+1{)}^{2}}(-4\mathrm{log}2+\sum _{j=0}^{n}\frac{4}{2j+1}+\frac{4}{2n+1}+(2n+1)\sum _{j=0}^{n}\frac{8}{(2j+1{)}^{2}}).\end{array}$

So we have the Fourier cosine series of $(\pi /2-x)\mathrm{log}(2\mathrm{sin}x)$ 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:

$\begin{array}{}\text{(2)}& (\pi /2-x)\mathrm{log}(2\mathrm{sin}x)=-\left(\sum _{n=1}^{+\mathrm{\infty}}\frac{\mathrm{sin}(2nx)}{n}\right)\left(\sum _{n=1}^{+\mathrm{\infty}}\frac{\mathrm{cos}(2nx)}{n}\right)\end{array}$

that directly leads to:

$\begin{array}{}\text{(3)}& f(x)=(\pi /2-x)\mathrm{log}(2\mathrm{sin}x)=-\sum _{n=1}^{+\mathrm{\infty}}\frac{{H}_{n-1}}{n}\mathrm{sin}(2nx).\end{array}$

Now since ${\int}_{0}^{\pi /2}\mathrm{sin}(2mx)dx=\frac{{\mathbb{1}}_{m\equiv 1\phantom{\rule{0.444em}{0ex}}(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}2)}}{m}$ and ${\int}_{0}^{\pi /2}\mathrm{sin}(2ax)\mathrm{sin}(2bx)dx=\frac{\pi}{4}{\delta}_{a,b}$, the first two identites are easily proven. Now the three-terms integral

${\int}_{0}^{\pi /2}\mathrm{sin}(2ax)\mathrm{sin}(2bx)\mathrm{sin}(2cx)dx$

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 , $f(x{)}^{2}$ or the integral ${\int}_{0}^{\pi /2}f(x{)}^{3}\phantom{\rule{thinmathspace}{0ex}}dx$, but still not impossible. In particular, we know that the Taylor coefficients of the powers of $\mathrm{log}(1-x)$ depends on the generalized harmonic numbers. In our case,

$-\mathrm{log}(1-x)=\sum _{n=1}^{+\mathrm{\infty}}\frac{1}{n}{x}^{n},$

$\mathrm{log}(1-x{)}^{2}=\sum _{n=2}^{+\mathrm{\infty}}\frac{2{H}_{n-1}}{n}{x}^{n},$

$-\mathrm{log}(1-x{)}^{3}=\sum _{n=3}^{+\mathrm{\infty}}\frac{3{H}_{n-1}^{2}-3{H}_{n-1}^{(2)}}{n}{x}^{n},$

$\begin{array}{}\text{(4)}& \mathrm{log}(1-x{)}^{4}=\sum _{n=4}^{+\mathrm{\infty}}\frac{4{H}_{n-1}^{3}+8{H}_{n-1}^{(3)}-12{H}_{n-1}{H}_{n-1}^{(2)}}{n}{x}^{n}\end{array}$

hence we can just find a closed form for

${\int}_{0}^{\pi /2}{x}^{3}(1-2\mathrm{cos}x{)}^{n}dx$

and sum everything through the third previous identity. Ugh.

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