Evaluate alternatively $\int_0^{\frac{\pi}{2}}\operatorname{Li}_2\left(\pm\sin^4x\right)dx$

Question

In my recent article here at Researchgate on page no 9, I was able to deduce the following two identities from the general result (which I treat it as Theorem 3.2). Those two identities are

$$\frac{2}{\pi}\int_0^{\frac{\pi}{2}}\operatorname{Li}_2\left(\sin^4x\right)dx =\frac{7\pi^2}{12}-\frac{25}{4}\ln^2(2)-4\operatorname{Li}_2\left(\frac{1}{\sqrt 2}\right)+2\operatorname{Li}_2\left(\frac{2-\sqrt 2}{4}\right)-\operatorname{arcsinh}^2(1)+3\ln(2)\operatorname{arcsinh}(1)\approx 0.581222\cdots$$ also

$$\frac{2}{\pi}\int_0^{\frac{\pi}{2}}\operatorname{Li}_2\left(-\sin^4x\right)dx=\operatorname{Li}_2\left(\frac{1-\sqrt 2}{2}\right)+4\operatorname{Li}_2\left(-\sqrt{\frac{1+\sqrt 2}{2}}\right)+\frac{\pi^2}{3}-\frac{19}{4}\ln^2(2)+2\operatorname{Li}_2\left(\frac{\sqrt 2-\sqrt{1+\sqrt 2}}{2\sqrt 2}\right)-\frac{\operatorname{arcsinh}^2(1)}{2}-\ln^2\left(\sqrt {2}+\sqrt{1+\sqrt{2}}\right)-3\ln(2)\ln\left(\sqrt{1+\sqrt 2}-\sqrt {2}\right)\approx -0.32379\cdots$$

These two beautiful identities are the special cases of general result $$\frac{2}{\pi}\int_0^{\frac{\pi}{2}}\operatorname{Li}_2\left(\pm16v\sin^4x\right)=\text{Result of theorem 3.2}$$for all $v\leq 1/16$. However, my interest is to seek an alternative path for both general result as well as those two special cases mentioned above other than technique used in the article. If there are other ways I would be glad to known. Using the same technique one can easily find the closed form for the integral of type $$\displaystyle \frac{2}{\pi}\int_0^{\frac{\pi}{2}}\operatorname{Li}_2\left(\pm256w\sin^8x\right)dx , \;\; |w|\leq 1/256$$

Answer 1

$\color{green}{\textbf{Version of 14.04.21.}}$

The given integrals can be expressed via the generalized hypergeometric function:

\begin{align} &I_\pm =\dfrac2\pi \int\limits_0^{\large^\pi/_2}\operatorname{Li_2}(\pm\sin^4x) =\dfrac2\pi\sum\limits_{k=1}^\infty \,\dfrac{(\pm1)^k}{k^2} \int\limits_0^{\large^\pi/_2}\sin^{4k}x\,\text dx =\dfrac1\pi\sum\limits_{k=1}^\infty \,\dfrac{(\pm1)^k}{k^2}\,\operatorname B\left(\dfrac12,2k+\dfrac12\right),\\[4pt] &I_+ = \dfrac1\pi\sum\limits_{k=1}^\infty \,\dfrac1{k^2}\,\operatorname B\left(\dfrac12,2k+\dfrac12\right) = \dfrac38\operatorname{_5 F_4}\left(1, 1, 1, \dfrac54, \dfrac74;\dfrac32, 2, 2, 2;1\right),\\[4pt] &I_+ \approx 0.5081222068073732302023528705866145091793935116040652847025683994... \\[4pt] &I_-=\dfrac1\pi\sum\limits_{k=1}^\infty \,\dfrac{(-1)^k}{k^2}\,\operatorname B\left(\dfrac12,2k+\dfrac12\right) = -\dfrac38\operatorname{_5 F_4}\left(1, 1, 1, \dfrac54, \dfrac74;\dfrac32, 2, 2, 2;-1\right),\\[4pt] &I_- \approx -0.323792143703370467535820065099172050645186814259075110252039257... \end{align} (see also WA integration 1, WA approximation 1, WA integration 2, WA approximation 2).

On the other hand, \begin{align} &I_+ =\dfrac4\pi\sum\limits_{k=1}^\infty \,\dfrac{1+(-1)^k}{k^2} \int\limits_0^{\large^\pi/_2}\sin^{2k}x\,\text dx = \dfrac2\pi\sum\limits_{k=1}^\infty \,\dfrac{1+(-1)^k}{k^2}\,\operatorname B\left(\dfrac12,k+\dfrac12\right),\\[4pt] &I_+ = \left(\dfrac{\pi^2}3 - 4\ln^2 2\right) +\left(-2\ln^2(2\sqrt2-2) + 4\operatorname{Li_2}\left(\dfrac{1-\sqrt2}2\right)\right); \\[4pt] &I_- = \dfrac4\pi \Re\sum\limits_{k=1}^\infty \,\dfrac{i^k}{k^2} \int\limits_0^{\large^\pi/_2}\sin^{2k}x\,\text dx = \dfrac4\pi\Re\sum\limits_{k=1}^\infty \,\dfrac{i^k}{k^2}\,\operatorname B\left(\dfrac12,k+\dfrac12\right),\\[4pt] &I_- =-\ln^2\dfrac{1 + \sqrt2 + \sqrt{2\sqrt2+2}}4 + 4\operatorname{arccot}^2\left(1 + \sqrt2 + \sqrt{2\sqrt2+2}\right)\\[4pt] & + 8\Re\operatorname{Li_2}\left(\dfrac{1 - \sqrt{1 - i}}2\right),\\[4pt] \end{align} with the same numerical results (see WA test1, WA test2).

The additionall links are: Int1Sum1, Int1Sum2, Int2Sum, Int2ReSum, Int2Final.

Edit

Also, $$I_- =-\ln^2\dfrac t4 + 4\operatorname{arccot}^2 t + 8\Re\operatorname{Li_2}\left(\dfrac{e^{\large i\arctan t+i\pi}}{2\sqrt t}\right),$$ where $$t = 1 + \sqrt2 + \sqrt{2\sqrt2+2}.$$

Additional links: AbsLiData, ArgLiData.