Integral $\int_0^\frac{\pi}{2} \arcsin(\sqrt{\sin x}) dx$

Solution 1:

Write $$ I(t)=\int_0^{\frac{1}{\sqrt{2}}} \frac{2\arcsin(tx)}{\sqrt{\frac{1}{2}-x^2}} \, {\rm d}x $$ and calculate \begin{align} I'(t) &= \int_0^{\frac{1}{\sqrt{2}}} \frac{2x}{\sqrt{\left(\frac{1}{2}-x^2\right)\left(1-(tx)^2\right)}} \, {\rm d}x \\ &= \frac{\log\left(\sqrt{2}+t\right)-\log\left(\sqrt{2}-t\right)}{t} \\ &= \frac{{\rm Li}_1 \left(\frac{t}{\sqrt{2}}\right) - {\rm Li}_1 \left(-\frac{t}{\sqrt{2}}\right)}{t}\, . \end{align}

Then \begin{align} I(1) &= \int_0^1 I'(t) \, {\rm d}t \\ &={\rm Li}_2 \left(\frac{1}{\sqrt{2}}\right) - {\rm Li}_2 \left(-\frac{1}{\sqrt{2}}\right) \, . \end{align}

Solution 2:

Work in progress.

Since $\arcsin(x)=\sum_{n\geq 0}\frac{\binom{2n}{n}}{(2n+1)4^n}x^{2n+1}$ for any $x\in[-1,1]$ and $$ \int_{0}^{\pi/2}\left(\sin x\right)^{n+1/2}\,dx=\frac{\sqrt{\pi}}{2}\cdot\frac{\Gamma\left(\frac{n}{2}+\frac{3}{4}\right)}{\Gamma\left(\frac{n}{2}+\frac{5}{4}\right)} $$ we "just" need an explicit value for the series $$ \sqrt{\frac{2}{\pi}}\sum_{n\geq 0}\frac{2^n \Gamma\left(\frac{n}{2}+\frac{3}{4}\right)^2}{(2n+1)^2 \Gamma(n+1)} $$ which is given by a linear combination of two $\phantom{}_4 F_3(\ldots;1)$ functions with quarter-integer parameters, namely $\phantom{}_4 F_3\left(\frac{1}{4},\frac{1}{4},\frac{3}{4},\frac{3}{4};\frac{1}{2},\frac{5}{4},\frac{5}{4}; 1\right)$ and $\phantom{}_4 F_3\left(\frac{3}{4},\frac{3}{4},\frac{5}{4},\frac{5}{4}; \frac{3}{2},\frac{7}{4},\frac{7}{4};1\right)$.

Fourier-Legendre expansions revealed to be extremely effective in dealing with such objects: for instance all the functions $\frac{\arcsin\sqrt{x}}{\sqrt{x}},\frac{1}{\sqrt{1-x^2}},\frac{1}{\sqrt{1-x^4}},K(x)$ have reasonably simple FL-expansions, opposed to the moderate complexity of their Maclaurin series. This observation allowed Campbell, Cantarini, Di Trani, Sondow and I to exhibit many surprising identities about $\phantom{}_3 F_2(\ldots;1)$ and $\phantom{}_4 F_3(\ldots;1)$ in terms of polylogarithms. My bet is that the same occurs here. With a step of integration by parts we have

$$ I = \int_{0}^{1}\frac{2x\arcsin x}{\sqrt{1-x^4}}\,dx = \frac{\pi^2}{4}-\int_{0}^{1}\frac{\arcsin(x^2)}{\sqrt{1-x^2}}\,dx$$ which is extremely good in simplifying the hypergeometric structure:

$$ I = \frac{\pi^2}{4}-\sum_{n\geq 0}\frac{\binom{2n}{n}}{(2n+1)4^n}\int_{0}^{\pi/2}\left(\sin x\right)^{4n+2}\,dx $$ leads to $$ I = \frac{\pi^2}{4}-\frac{\pi}{2}\sum_{n\geq 0}\frac{\binom{2n}{n}\binom{4n+2}{2n+1}}{(2n+1)4^{3n+1}}=\frac{\pi^2}{4}-\frac{\pi}{4}\sum_{n\geq 0}\frac{\binom{2n}{n}\binom{4n}{2n}}{4^{3n}}\cdot\frac{4n+1}{(2n+1)^2} $$ where the last series is blatantly related to Legendre function $P_{-1/4}$. Indeed, according to Mathematica's notation for the complete elliptic integrals (i.e. the argument is the elliptic modulus) $$ \sum_{n\geq 0}\frac{\binom{2n}{n}\binom{4n}{2n}}{4^{3n}}z^{2n} = \frac{2}{\pi\sqrt{1+z}}\,K\left(\frac{2z}{1+z}\right) $$ and the given problem boils down to computing $$ \int_{0}^{1}\frac{1}{\sqrt{1+z}}\,K\left(\frac{2z}{1+z}\right)\,dz\quad\text{and}\quad \int_{0}^{1}\frac{\log z}{\sqrt{1+z}}\,K\left(\frac{2z}{1+z}\right)\,dz.$$ The substitution $z\mapsto\frac{x}{2-x}$ leads to three integrals which are simple to tackle through the FL machinery, namely $\int_{0}^{1}\frac{K(x)}{(2-x)^{3/2}}g(x)\,dx$ where $g(x)\in\{1,\log(x),\log(2-x)\}$.