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

I am trying to calculate $$I=\int_0^\frac{\pi}{2} \arcsin(\sqrt{\sin x}) dx$$ So far I have done the following. First I tried to let $\sin x= t^2$ then: $$I=2\int_0^1 \frac{x\arcsin x}{\sqrt{1-x^4}}dx =\int_0^1 (\arcsin^2 x)'\frac{x}{\sqrt{1+x^2}}dx$$ $$=\frac{\pi^2}{8}-\int_0^1 \frac{\arcsin^2 x}{(1+x^2)^{3/2}}dx$$ We can expand into power series the integral, we have: $\arcsin^2z=\sum\limits_{n\geq1}\frac {2^{2n-1}z^{2n}}{n^2\binom {2n}n}$ and using the binomial series for $(1+x^2)^{-3/2}$ will result in: $$\sum_{n\geq1}\frac{2^{2n-1}x^{2n}}{n^2\binom {2n}n}\sum_{k\ge 0}\binom{-3/2}{k}x^{2k}$$ But I dont know how to simplify this. I tried one more thing, letting $\sin x= \sin^2 t$ gives: $$I=2\int_0^\frac{\pi}{2}\frac{x\sin x}{\sqrt{1+\sin^2 x}}dx$$ Since $\int \frac{\sin x}{\sqrt{1+\sin^2x}}dx=-\arcsin\left(\frac{\cos x}{\sqrt 2} \right)+C$ we can integrate by parts to obtain: $$I=2\int_0^\frac{\pi}{2}\arcsin\left(\frac{\cos x}{\sqrt 2}\right)dx=2\int_0^\frac{\pi}{2}\arcsin\left(\frac{\sin x}{\sqrt 2}\right)dx$$ But I am stuck, so I would appreciate some help.

Edit: By letting $\frac{\sin x}{\sqrt 2} =t$ We get: $$I=2\int_0^\frac1{\sqrt{2}} \frac{\arcsin x}{\sqrt{\frac12-x^2}}dx=2\text{Li}_2\left(\frac1{\sqrt 2}\right)-\frac{\pi^2}{24}+\frac{\ln^2 2}{4}$$ Where the latter integral was evaluated with wolfram. I would love to see a proof for that.

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}

• Good ol' Feynman's trick always wins $\color{green}{\checkmark}$ – Jack D'Aurizio Aug 9 '18 at 1:03

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 $g(x)\in\{1,\log(x),\log(2-x)\}$.

Mathematica gives:

$$\frac{1}{24} \left(-6 \text{Li}_2\left(\frac{1}{2}-\frac{1}{\sqrt{2}}\right)+6 \text{Li}_2\left(3-2 \sqrt{2}\right)+4 \pi ^2-3 \log ^2(2)+3 \log ^2\left(\sqrt{2}-1\right)+3 \log ^2\left(3+2 \sqrt{2}\right)+ \log (64) \log \left(\sqrt{2}-1\right)+6 \sinh ^{-1}(1)^2-12 \log \left(2 \left(1+\sqrt{2}\right)\right) \sinh ^{-1}(1)-6 i \pi \left(2 \sinh ^{-1}(1)-\log \left(3+2 \sqrt{2}\right)\right)\right)$$

which strongly suggests that hand calculation will be extremely difficult and error prone. Incidentally, the numerical value of that expression is $1.5122$, and the value of the numerical evaluation of the original is also $1.5122$, which strongly suggests the answer is correct.

Here is a graph of the function, and shaded value of the integral, which suggests there isn't a major error:

• Mathematica is renown for not dealing optimally with polylogarithms or high-order hypergeometric functions, so the fact that Mathematica is not able to simply the shown output does not really imply anything significative. – Jack D'Aurizio Aug 9 '18 at 0:21