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

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 $$\displaystyle \frac{2}{\pi}\int_0^{\frac{\pi}{2}}\operatorname{Li}_2\left(\pm256w\sin^8x\right)dx , \;\; |w|\leq 1/256$$

## 1 Answer

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

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.

• Thank you for putting down you efforts to provide an alternative way for the results. Your final results still seems mysterious to me (though by WA check they are found to be correct). Regarding the hypergeometric one which i had too mentioned in the cited paper however, the second one seems much curious to me. – Naren Apr 13 at 20:35
• @Naren Thank you for the comment! Typo fixed, links added. – Yuri Negometyanov Apr 13 at 22:48
• @Naren Thank you for the bounty. Is something unclear? – Yuri Negometyanov Apr 15 at 21:07