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

Solution 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.