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So I have been trying to solve this, and for some reason, i'm getting close enough to match certain terms from the given closed form but not entirely, here's my work;

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

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

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

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

Set $z=\sin^2 x$;

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

$$2I=\int_0^{\pi/2} \frac{\pi^4}{36}+\ln^2(\sin^2 x)\ln^2(1-\sin^2 x)-\frac{\pi^2}{3}\ln(\sin^2 x)\ln(1-\sin^2 x)-2\operatorname{Li}_2(\sin^2 x)\operatorname{Li}_2(1-\sin^2 x)\,\textrm{d}x$$

$$\boxed{\int_0^\frac{\pi}{2}\frac{\pi^4}{36}\,dx=\frac{\pi^5}{72}}$$

$$\boxed{\int_0^{\pi/2} \ln^2(\sin^2 x)\ln^2(\cos^2 x)\,dx=16\int_0^{\pi/2} \ln^2(\sin x)\ln^2(\cos x)\,dx=8\pi \,\left (\ln ^4(2)+\frac{\pi^4}{160}-\ln(2)\zeta(3)\right)}$$

$$\boxed{\frac{\pi^2}{3}\int_0^{\pi/2} \ln^2(\sin^2 x)\ln^2(\cos^2 x)dx=\frac{16\pi^2}{3}\int_0^{\pi/2} \ln^2(\sin x)\ln^2(\cos x)dx=\frac{8\pi^3}{3} \,\left (\ln ^4(2)+\frac{\pi^4}{160}-\ln(2)\zeta(3)\right)}$$

$$\boxed{2\int_0^{\pi/2}\operatorname{Li}_2(\sin^2(x)) \operatorname{Li}_2(\cos^2(x))\textrm{d}x=\frac{23}{72}\pi ^5+\frac{8}{3}\log^4(2)\pi-18 \pi \log(2)\zeta(3) -32\pi \operatorname{Li}_4\left(\frac{1}{2}\right)}$$

$$\boxed{2I=\frac{\pi^5}{72}+8\pi \,\left (\ln ^4(2)+\frac{\pi^4}{160}-\ln(2)\zeta(3)\right)-\frac{8\pi^3}{3} \,\left (\ln ^4(2)+\frac{\pi^4}{160}-\ln(2)\zeta(3)\right)-\left(\frac{23}{72}\pi ^5+\frac{8}{3}\log^4(2)\pi-18 \pi \log(2)\zeta(3) -32\pi \operatorname{Li}_4\left(\frac{1}{2}\right)\right)}$$

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

The closed form i'm ending up on is

$$\boxed{I=5 \pi \log(2)\zeta(3)+\frac{8}{3}\log^4(2)\pi+\ 16 \pi \operatorname{Li}_4\left(\frac{1}{2}\right)-\frac{4\pi^3}{3}\ln^4(2)-\frac{\pi^7}{120}+\frac{4\pi^3}{3}\ln^4(2)\zeta(3)-\frac{46\pi^5}{360}}$$

The given closed form is: $$\boxed{5 \pi \log(2)\zeta(3)-\frac{1}{3}\log ^2(2)\pi^3+\frac{8}{3}\log^4(2)\pi -\frac{41 }{360}\pi ^5 + \ 16 \pi \operatorname{Li}_4\left(\frac{1}{2}\right)}$$

I do not carry rigorous experience with poly-logarithms so if there are any inconsistency/flaws i would like to understand them as well.

Also, is there a closed form for $\operatorname{Li}_4 \left(\frac{1}{2}\right)?$

Amrut Ayan
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    To answer the last question, no. there is no closed form for that – polychroma Jul 28 '24 at 02:17
  • @phi-rate, thank you, i did find the confirmation here in point (17) - https://mathworld.wolfram.com/Polylogarithm.html – Amrut Ayan Jul 28 '24 at 05:21
  • The mistake is in the third boxed result. Both logs should not be squared. It's easy to spot the mistake because the weight of the integral is 5 and the weight of the third boxed result is 7 which is incorrect. There is no closed form for the quadlogarithm function of 1/2. – Ali Olaikhan Jul 31 '24 at 16:51
  • @AliShadhar, is my second boxed result right? Because all I’ve done is reused the result in the third boxed result and the integral has a weight 2 multiplied already to it due from previous steps making the weight to 7 in the third boxed result – Amrut Ayan Aug 01 '24 at 02:15
  • based on the identity you used $(\operatorname{Li}_2(\sin^2 x))^2+(\operatorname{Li}_2(1-\sin^2 x))^2=\frac{\pi^4}{36}+\ln^2(\sin^2 x)\ln^2(1-\sin^2 x)-\frac{\pi^2}{3}\ln(\sin^2 x)\ln(1-\sin^2 x)-2\operatorname{Li}_2(\sin^2 x)\operatorname{Li}_2(1-\sin^2 x)$, the integrand of the third integral is $\frac{\pi^2}{3}\ln(\sin^2 x)\ln(\cos^2 x)$ not $\frac{\pi^2}{3}\ln^2(\sin^2 x)\ln^2(\cos^2 x)$. Regarding your second integral, you can verify it on wolframalfa. – Ali Olaikhan Aug 01 '24 at 05:34
  • @AliShadhar oh yes that was a silly mistake, thank you very much sir – Amrut Ayan Aug 01 '24 at 09:54

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