A closed form for $\int_{0}^{\pi/2} x^3 \ln^3(2 \cos x)\:\mathrm{d}x$

Proposition. $$ \int_{0}^{\pi/2} x^3 \ln^3(2 \cos x)\:{\mathrm{d}}x = \frac{45}{512} \zeta(7)-\frac{3\pi^2}{16} \zeta(5)-\frac{5\pi^4}{64} \zeta(3)-\frac{9}{4}\zeta(\bar{5},1,1) $$

where $\zeta(\bar{p},1,1)$ is the colored MZV (Multi Zeta Values) function of depth 3 and weight $p+2$ given by $$ \zeta(\bar{p},1,1) : = \sum_{n=1}^{\infty} \frac{(-1)^n}{n^{p}}\sum_{k=1}^{n-1}\frac{H_{k-1}}{k} $$ 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 $$ \int_{0}^{\pi/2}\! \! x^{2\ell+1}\! \ln^{2\ell+1}(2 \cos x){\mathrm{d}}x \in \mathbb{Q} \!\left( \zeta(4\ell+3),\zeta(2)\zeta(4\ell+1), ... ,\zeta(4\ell)\zeta(3), \zeta(\overline{2\ell+3},\{1\}_{2\ell})\right) $$

It is remarkable that there is only one constant $$ \zeta(\overline{2\ell+3},\{1\}_{2\ell})=\sum_{n_{1}> ...>n_{2\ell+1}>0} \frac{(-1)^{n_1}}{n_1^{2\ell+3} n_2\cdots n_{2\ell+1}} $$ 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).

This paper may be useful, another one and of course Hoffman's site which has a list of many related references.

We can try the harmonic analysis path. Since: $$\log(2\cos x)=\sum_{n=1}^{+\infty}\frac{(-1)^{n+1}}{n}\cos(2nx),$$ $$\log(2\sin x)=-\sum_{n=1}^{+\infty}\frac{\cos(2nx)}{n},\tag{1}$$ we have, as an example: $$\int_{0}^{\pi/2}\log^3(2\sin x)dx=-\frac{3\pi}{4}\zeta(3)$$ since $$\int_{0}^{\pi/2}\cos(2n_1 x)\cos(2n_2 x)\cos(2n_3 x)\,dx = \frac{\pi}{8}\delta_{2\cdot\max n_i=(n_1+n_2+n_3)}\tag{2}$$ Now: $$\pi/2-x = \frac{\pi}{4}+\frac{2}{\pi}\sum_{m=0}^{+\infty}\frac{\cos((4m+2)x)}{(2m+1)^2}\tag{3}$$ hence by multiplying $(1)$ and $(3)$ we can write the Fouries cosine series of $(\pi/2-x)\log(2\sin x)$ over $(0,\pi/2)$ and grab from $(2)$ a combinatorial equivalent for $$\int_{0}^{\pi/2}\left((\pi/2-x)\log(2\sin x)\right)^3\,dx.$$ With the aid of Mathematica I got: $$(\pi/2-x)\log(2\sin x)=\sum_{n=0}^{+\infty}\frac{\cos((2n+1)x)}{\pi(2n+1)^2}\left(-(2n+1)\left(\psi'\left(\frac{2n+1}{2}\right)-\psi'\left(-\frac{2n+1}{2}\right)\right)+2\left(2\gamma+\psi\left(\frac{2n+1}{2}\right)+\psi\left(-\frac{2n+1}{2}\right)\right)\right),$$

$$(\pi/2-x)\log(2\sin x)=\sum_{n=0}^{+\infty}\frac{\cos((2n+1)x)}{\pi(2n+1)^2}\left(2 H_{\frac{2n+1}{2}}-(2n+1)\left(\psi'\left(\frac{2n+1}{2}\right)-\psi'\left(-\frac{2n+1}{2}\right)\right)\right).$$

$$(\pi/2-x)\log(2\sin x)=\sum_{n=0}^{+\infty}\frac{\cos((2n+1)x)}{\pi(2n+1)^2}\left(-4\log 2+\sum_{j=0}^n\frac{4}{2j+1}+\frac{4}{2n+1}+(2n+1)\sum_{j=0}^{n}\frac{8}{(2j+1)^2}\right).\tag{1}$$

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

$$(\pi/2-x)\log(2\sin x)=-\left(\sum_{n=1}^{+\infty}\frac{\sin(2nx)}{n}\right)\left(\sum_{n=1}^{+\infty}\frac{\cos(2nx)}{n}\right)\tag{2}$$

that directly leads to:

$$f(x)=(\pi/2-x)\log(2\sin x)=-\sum_{n=1}^{+\infty}\frac{H_{n-1}}{n}\sin(2nx).\tag{3}$$

Now since $\int_{0}^{\pi/2}\sin(2mx)dx=\frac{\mathbb{1}_{m\equiv 1\pmod{2}}}{m}$ and $\int_{0}^{\pi/2}\sin(2ax)\sin(2bx)dx=\frac{\pi}{4}\delta_{a,b}$, the first two identites are easily proven. Now the three-terms integral $$\int_{0}^{\pi/2}\sin(2ax)\sin(2bx)\sin(2cx)dx$$ 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 $\int_{0}^{\pi/2}f(x)^3\,dx$, but still not impossible. In particular, by this answer we know that the Taylor coefficients of the powers of $\log(1-x)$ depends on the generalized harmonic numbers. In our case, $$-\log(1-x)=\sum_{n=1}^{+\infty}\frac{1}{n}x^n,$$ $$\log(1-x)^2 = \sum_{n=2}^{+\infty}\frac{2H_{n-1}}{n}x^n,$$ $$-\log(1-x)^3 = \sum_{n=3}^{+\infty}\frac{3H_{n-1}^2-3H_{n-1}^{(2)}}{n}x^n,$$ $$\log(1-x)^4 = \sum_{n=4}^{+\infty}\frac{4H_{n-1}^3+8H_{n-1}^{(3)}-12 H_{n-1}H_{n-1}^{(2)}}{n}x^n\tag{4}$$ hence we can just find a closed form for $$\int_{0}^{\pi/2}x^3 (1-2\cos x)^n dx$$ and sum everything through the third previous identity. Ugh.