Evaluate $\int_0^{\pi/2} (\operatorname{Li}_2(\sin^2(x))^2\textrm{d}x=\int_0^{\pi/2} (\operatorname{Li}_2(\cos^2(x))^2\textrm{d}x$

Question

The following problem is proposed by Cornel Ioan Valean, that is, to prove that

\begin{align} & \int_0^{\pi/2} \operatorname{Li}_{2}^{2} \left(\sin^{2}\left(x\right)\right){\rm d}x = \int_0^{\pi/2} \operatorname{Li}_{2}^{2} \left(\cos^{2}\left(x\right)\right){\rm d}x \\[5mm] = & \ 5 \pi \log(2)\zeta(3)-\frac{1}{3}\log ^2(2)\pi^3+\frac{8}{3}\log^4(2)\pi -\frac{41 }{360}\pi ^5 \\[1mm] + & \ 16 \pi \operatorname{Li}_4\left(\frac{1}{2}\right) \end{align} where $\operatorname{Li}_{s}$ is the Polylogarithm.

One of his solutions is based on exploiting the Fourier series, $$\small \operatorname{Li}_2\left(\sin^2(x)\right)=\frac{\pi^2}{6}-2\log^2(2)-2\sum_{n=1}^{\infty}\biggr(2\log(2)\frac{1}{n}+(-1)^{n-1}\frac{1}{n^2}-2\frac{\overline{H}_n}{n}\biggr)\cos(2n x), \ x\in \mathbb{R},$$ which is given in More (Almost) Impossible Integrals, Sums, and Series: A New Collection of Fiendish Problems and Surprising Solutions (2023), the sequel of (Almost) Impossible Integrals, Sums, and Series (2019), together with some simple series, which can be easily extracted by using results from these books, and the skew-harmonic number version of the classical Au-Yeung series, that is $$\sum_{n=1}^{\infty} \left(\frac{\overline{H}_n}{n}\right)^2=\frac{1}{12}\log^4(2)+\frac{5}{2}\log^2(2)\zeta(2)-\frac{13}{8}\zeta(4)+2 \operatorname{Li}_4\left(\frac{1}{2}\right),\tag1$$
where $\overline{H}_n^{(m)}=1-\frac{1}{2^m}+\cdots+(-1)^{n-1}\frac{1}{n^m}, \ m\ge1$, is the $n$th generalized skew-harmonic number of order $m$, $\zeta$ represents the Riemann zeta function, and $\operatorname{Li}_n$ denotes the Polylogarithm. I remind you that the Au-Yeung series is $\displaystyle \sum_{n=1}^{\infty} \left(\frac{H_n}{n}\right)^2=\frac{17}{4}\zeta(4)$, where $H_n=1+1/2+\cdots+1/n$ is the $n$th harmonic number. An elegant solution to $(1)$ was recently given in the paper Artwork with the Evaluation of the Skew-Harmonic Number Version of the Au-Yeung Series by C.I. Valean.

Question: What elegant ways can we employ for evaluating the integral, without using Fourier series? I would love to see many solutions with different perspectives of attacking the integral.

BONUS: Combining the reflection formula of Dilogarithm and the previous main result, it's not hard to arrive at

$$\int_0^{\pi/2}\operatorname{Li}_2(\sin^2(x)) \operatorname{Li}_2(\cos^2(x))\textrm{d}x$$ $$=\frac{23}{144}\pi ^5+\frac{4}{3}\log^4(2)\pi-9 \pi \log(2)\zeta(3) -16 \pi \operatorname{Li}_4\left(\frac{1}{2}\right),$$ which has an exotic, appealing appearance (also a simpler closed form, with one term less).

Answer 1

Here's my work;

$$I=\int_0^{\pi/2} (\operatorname{Li}_2(\sin^2(x))^2\textrm{d}x=\int_0^{\pi/2} (\operatorname{Li}_2(\cos^2(x))^2\textrm{d}x$$

$$2I=\int_0^{\pi/2} (\operatorname{Li}_2(\sin^2(x))^2+(\operatorname{Li}_2(1-\sin^2(x))^2\textrm{d}x$$

$$\color{blue}{\operatorname{Li}_2(z)+\operatorname{Li}_2(1-z)=\frac{\pi^2}{6}-\ln(z)\ln(1-z)}$$

Utilizing this above identity (taken from wikipedia)

We square that identity and get the below expression,

$$(\operatorname{Li}_2(z))^2+(\operatorname{Li}_2(1-z))^2=\frac{\pi^4}{36}+\ln^2(z)\ln^2(1-z)-\frac{\pi^2}{3}\ln(z)\ln(1-z)-2\operatorname{Li}_2(z)\operatorname{Li}_2(1-z)$$

Set $z=\sin^2 x$;

$(\operatorname{Li}_2(\sin^2 x))^2+(\operatorname{Li}_2(\cos^2 x))^2=\frac{\pi^4}{36}+\ln^2(\sin^2 x)\ln^2(\cos^2 x)-\frac{\pi^2}{3}\ln(\sin^2 x)\ln(\cos^2 x)-2\operatorname{Li}_2(\sin^2 x)\operatorname{Li}_2(\cos^2 x)$

Substitute above mess into the integral, we are going to split this down to simpler integrals,

$2I=\int_0^{\pi/2} \frac{\pi^4}{36}+\ln^2(\sin^2 x)\ln^2(\cos^2 x)-\frac{\pi^2}{3}\ln(\sin^2 x)\ln(\cos^2 x)-2\operatorname{Li}_2(\sin^2 x)\operatorname{Li}_2(\cos^2 x)\,\textrm{d}x$

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Distributing the integral over all terms,

$2I=\int_0^{\pi/2} \frac{\pi^4}{36}\,dx+\int_0^{\pi/2} \ln^2(\sin^2 x)\ln^2(\cos^2 x)\,dx-\frac{\pi^2}{3}\int_0^{\pi/2} \ln(\sin^2 x)\ln(\cos^2 x)\,dx-2\int_0^{\pi/2} \operatorname{Li}_2(\sin^2 x)\operatorname{Li}_2(\cos^2 x)\,\textrm{d}x$

$I=\int_0^{\pi/2} \frac{\pi^4}{72}\,dx+\frac12\int_0^{\pi/2} \ln^2(\sin^2 x)\ln^2(\cos^2 x)\,dx-\frac{\pi^2}{6}\int_0^{\pi/2} \ln(\sin^2 x)\ln(\cos^2 x)\,dx-\int_0^{\pi/2} \operatorname{Li}_2(\sin^2 x)\operatorname{Li}_2(\cos^2 x)\,\textrm{d}x$

$I=\int_0^{\pi/2} \frac{\pi^4}{72}\,dx+8\int_0^{\pi/2} \ln^2(\sin x)\ln^2(\cos x)\,dx-\frac{2\pi^2}{3}\int_0^{\pi/2} \ln(\sin x)\ln(\cos x)\,dx-\int_0^{\pi/2} \operatorname{Li}_2(\sin^2 x)\operatorname{Li}_2(\cos^2 x)\,\textrm{d}x$

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Now we start to evaluate one by one,

$\boxed{\int_0^\frac{\pi}{2}\frac{\pi^4}{72}\,dx=\frac{\pi^5}{144}}\tag1$

$\boxed{8\int_0^{\pi/2} \ln^2(\sin x)\ln^2(\cos x)\,dx=4\pi\ln^4(2)+\frac{4\pi^5}{160}-4\pi\ln(2)\zeta(3)}\tag2$

$\boxed{\frac{2\pi^2}{3}\int_0^{\pi/2} \ln(\sin x)\ln(\cos x)dx=\frac{\pi^3}{3}\ln^2(2)-\frac{\pi^5}{72}}\tag3$

$\boxed{\int_0^{\pi/2}\operatorname{Li}_2(\sin^2(x)) \operatorname{Li}_2(\cos^2(x))\textrm{d}x=\frac{23}{144}\pi ^5+\frac{4}{3}\log^4(2)\pi-9 \pi \log(2)\zeta(3) -16 \pi \operatorname{Li}_4\left(\frac{1}{2}\right)}\tag4$

The above result $(4)$ was taken from OP's BONUS section

Substituting all our above results formed in $(1),(2),(3),(4)$;

$I=\frac{\pi^5}{144}+4\pi\ln^4(2)+\frac{4\pi^5}{160}-4\pi\ln(2)\zeta(3)-\left(\frac{\pi^3}{3}\ln^2(2)-\frac{\pi^5}{72}\right)-\left(\frac{23\pi^5}{144}+\frac{4}{3}\log^4(2)\pi-9 \pi \log(2)\zeta(3) -16 \pi \operatorname{Li}_4\left(\frac{1}{2}\right)\right)$

$I=\left(\frac{\pi^5}{144}+\frac{4\pi^5}{160}+\frac{\pi^5}{72}-\frac{23\pi^5}{144}\right)+\left(4\pi\ln^4(2)-\frac{4}{3}\log^4(2)\pi \right)-4\pi\ln(2)\zeta(3)+9 \pi \log(2)\zeta(3) -\frac{\pi^3}{3}\ln^2(2)+16 \pi \operatorname{Li}_4\left(\frac{1}{2}\right)$

Put in the mess from above and we end up on this;

$$\boxed{\color{red}{I=5 \pi \log(2)\zeta(3)-\frac{1}{3}\log ^2(2)\pi^3+\frac{8}{3}\log^4(2)\pi -\frac{41 }{360}\pi ^5 + \ 16 \pi \operatorname{Li}_4\left(\frac{1}{2}\right)}}$$

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Edit:

I found the two results as follows;

The first result is from here (2). [Basically exploiting the derivative of the beta function]

I used the technique for evaluating $(2)$ and used it to evaluate $(3)$.