Improper Integral $\int_0^1\frac{\arcsin^2(x^2)}{\sqrt{1-x^2}}dx$

Question

$$I=\int_0^1\frac{\arcsin^2(x^2)}{\sqrt{1-x^2}}dx\stackrel?=\frac{5}{24}\pi^3-\frac{\pi}2\log^2 2-2\pi\chi_2\left(\frac1{\sqrt 2}\right)$$ This result seems to me digitally correct? Can we prove that the equality is exact?

Answer 1

\begin{align}\int_0^{1} \frac{\arcsin^2 x^2}{\sqrt{1-x^2}}\,dx &= \frac{1}{2}\int_0^{1} \frac{\arcsin^2 x}{\sqrt{x}\sqrt{1-x}}\,dx \tag{1}\\&= \frac{1}{2}\int_0^{\pi/2} \frac{\theta^2\cos \theta}{\sqrt{\sin \theta - \sin^2 \theta}}\,d\theta \tag{2}\\&= \frac{1}{\sqrt{2}}\int_0^{\pi/2} \frac{\left(\frac{\pi}{2} - \theta\right)^2\cos \frac{\theta}{2}}{\sqrt{1-2\sin^2 \frac{\theta}{2}}}\,d\theta \tag{3}\\&= \int_0^{\pi/2} \left(\frac{\pi}{2} - 2\arcsin \left(\frac{\sin \alpha}{\sqrt{2}}\right)\right)^2\,d\alpha \tag{4}\\&= \frac{\pi^3}{8} - 2\pi \int_0^{\pi/2} \arcsin \left(\frac{\sin \alpha}{\sqrt{2}}\right)\,d\alpha + 4\int_0^{\pi/2} \left(\arcsin \left(\frac{\sin \alpha}{\sqrt{2}}\right)\right)^2\,d\alpha \tag{5}\end{align}

where, we made $x \mapsto \sqrt{x}$ in step $(1)$. In step $(2)$ we made $\theta = \arcsin x$ and finally in $(3)$ we made the change of variable $\sin \dfrac{\theta}{2} = \dfrac{\sin \alpha}{\sqrt{2}}$.

Now we recall the famous series expansion: $\displaystyle \arcsin^2 x = \frac{1}{2}\sum\limits_{n=1}^{\infty} \dfrac{(2x)^{2n}}{n^2\binom{2n}{n}}$,

Hence, \begin{align*}\int_0^{\pi/2} \left(\arcsin \left(\frac{\sin \alpha}{\sqrt{2}}\right)\right)^2\,d\alpha &= \frac{1}{2}\sum\limits_{n=1}^{\infty} \dfrac{2^n}{n^2\binom{2n}{n}}\int_0^{\pi/2} \sin^{2n} \alpha \,d\alpha\\&= \frac{\pi}{4}\sum\limits_{n=1}^{\infty} \dfrac{1}{n^22^{n}} = \frac{\pi}{4}\operatorname{Li}_2 \left(\frac{1}{2}\right) = \frac{\pi}{8}\left(\zeta(2) - \log^2 2\right)\end{align*}

also, the infinite series expansion for $\displaystyle \arcsin x = \sum\limits_{n=0}^{\infty} \dfrac{\binom{2n}{n}x^{2n+1}}{(2n+1)4^n}$ give us,

\begin{align*}\int_0^{\pi/2} \arcsin \left(\frac{\sin \alpha}{\sqrt{2}}\right)\,d\alpha &= \frac{1}{\sqrt{2}}\sum\limits_{n=0}^{\infty} \dfrac{\binom{2n}{n}}{(2n+1)8^n}\int_0^{\pi/2} \sin^{2n+1} \alpha \,d\alpha\\&= \frac{1}{\sqrt{2}}\sum\limits_{n=0}^{\infty} \dfrac{1}{(2n+1)^2 2^n} = \chi_2 \left(\frac{1}{\sqrt{2}}\right)\end{align*}

Combining the results, $$\int_0^{1} \frac{\arcsin^2 x^2}{\sqrt{1-x^2}}\,dx = \frac{5\pi^3}{24} - \frac{\pi}{2}\log^2 2 - 2\pi \chi_2 \left(\frac{1}{\sqrt{2}}\right)$$

Answer 2

We have:

$$ \arcsin^2(z^2)=\sum_{n\geq 0}\frac{2^{2n+1} n!^2}{(2n+2)!}z^{4n+4} \tag{1}$$ hence:

$$ I = \frac{\pi}{4}\sum_{n\geq 0}\frac{n!^2 (4n+3)!}{2^{2n+1}(2n+2)!^2 (2n+1)!}=\frac{3\pi}{16}\cdot\phantom{}_5 F_4\left(1,1,1,\frac{5}{4},\frac{7}{4};\frac{3}{2},\frac{3}{2},2,2;1\right).\tag{2} $$

Answer 3

$$\displaystyle{\int\limits_{0}^{1}{\frac{\left( \arcsin x^{2} \right)^{2}}{\sqrt{1-x^{2}}}dx}=\frac{1}{2}\int\limits_{0}^{1}{\frac{\left( \arcsin x \right)^{2}}{\sqrt{x}\sqrt{1-x}}dx}=\frac{1}{2}\int\limits_{0}^{\frac{\pi }{2}}{\frac{x^{2}\cos x}{\sqrt{\sin x}\sqrt{1-\sin x}}dx}=\frac{\sqrt{2}}{2}\int\limits_{0}^{\frac{\pi }{2}}{\frac{\left( \frac{\pi }{2}-x \right)^{2}\cos \frac{x}{2}}{\sqrt{1-2\sin ^{2}\frac{x}{2}}}dx}}$$

$$\displaystyle{=\int\limits_{0}^{\frac{\pi }{2}}{\left( \frac{\pi }{2}-2\arcsin \left( \frac{\sin x}{\sqrt{2}} \right) \right)^{2}dx}=\left( \frac{\pi }{2} \right)^{2}\int\limits_{0}^{\frac{\pi }{2}}{dx}-2\pi \int\limits_{0}^{\frac{\pi }{2}}{\arcsin \left( \frac{\sin x}{\sqrt{2}} \right)dx}+4\int\limits_{0}^{\frac{\pi }{2}}{\arcsin ^{2}\left( \frac{\sin x}{\sqrt{2}} \right)dx}}$$

$$\displaystyle{=\frac{\pi ^{3}}{8}-2\pi \underbrace{\int\limits_{0}^{\frac{\pi }{2}}{\arcsin \left( \frac{\sin x}{\sqrt{2}} \right)dx}}_{I_{1}}+4\underbrace{\int\limits_{0}^{\frac{\pi }{2}}{\arcsin ^{2}\left( \frac{\sin x}{\sqrt{2}} \right)dx}}_{I_{2}}}$$

$$\displaystyle{I_{1}=\int\limits_{0}^{\frac{\pi }{2}}{\arcsin \left( \frac{\sin x}{\sqrt{2}} \right)dx}=\frac{\sqrt{2}}{2}\sum\limits_{j=0}^{+\infty }{\frac{\left( \begin{matrix} 2j \\ j \\ \end{matrix} \right)}{\left( 2j+1 \right)2^{3j}}\int\limits_{0}^{\frac{\pi }{2}}{sin^{2j+1}xdx}}=\frac{\sqrt{2}}{2}\sum\limits_{j=0}^{+\infty }{\frac{\left( \begin{matrix} 2j \\ j \\ \end{matrix} \right)}{\left( 2j+1 \right)2^{3j}}\cdot \frac{\sqrt{\pi }\Gamma \left( j+1 \right)}{2\Gamma \left( j+\frac{3}{2} \right)}}}$$

$$\displaystyle{=\frac{\sqrt{2}}{2}\sum\limits_{j=0}^{+\infty }{\frac{\left( \begin{matrix} 2j \\ j \\ \end{matrix} \right)}{\left( 2j+1 \right)2^{3j}}\cdot \frac{\sqrt{\pi }j!}{2\frac{\left( 2\left( j+1 \right) \right)!}{2^{2\left( j+1 \right)}\left( j+1 \right)!}\sqrt{\pi }}}=\sqrt{2}\sum\limits_{j=0}^{+\infty }{\frac{\left( 2j \right)!}{\left( j! \right)^{2}\left( 2j+1 \right)2^{j}}\cdot \frac{j!\left( j+1 \right)!}{\left( 2j+1+1 \right)!}}}$$

$$\displaystyle{=\sqrt{2}\sum\limits_{j=0}^{+\infty }{\frac{\left( 2j \right)!}{\left( j! \right)^{2}\left( 2j+1 \right)2^{j}}\cdot \frac{j!\left( j+1 \right)!}{\left( 2j+2 \right)\left( 2j+1 \right)\left( 2j \right)!}}=\frac{\sqrt{2}}{2}\sum\limits_{j=0}^{+\infty }{\frac{1}{\left( 2j+1 \right)^{2}2^{j}}}}$$

$$\displaystyle{=\frac{\sqrt{2}}{2}\sum\limits_{j=0}^{+\infty }{\frac{\left( \frac{1}{2} \right)^{j}}{\left( 2j+1 \right)^{2}}}=\frac{\sqrt{2}}{2}\left( 1+\frac{\frac{1}{2}}{3^{2}}+\frac{\left( \frac{1}{2} \right)^{2}}{5^{2}}+\frac{\left( \frac{1}{2} \right)^{3}}{7^{2}}+\frac{\left( \frac{1}{2} \right)^{4}}{9^{2}}+\frac{\left( \frac{1}{2} \right)^{5}}{11^{2}}+... \right)}$$

$$\displaystyle{=\frac{1}{\sqrt{2}}+\frac{\left( \frac{1}{\sqrt{2}} \right)^{3}}{3^{2}}+\frac{\left( \frac{1}{\sqrt{2}} \right)^{5}}{5^{2}}+\frac{\left( \frac{1}{\sqrt{2}} \right)^{7}}{7^{2}}+\frac{\left( \frac{1}{2} \right)^{9}}{9^{2}}+\frac{\left( \frac{1}{2} \right)^{11}}{11^{2}}+...}$$

$$\displaystyle{=\left( \frac{1}{\sqrt{2}}+\frac{\left( \frac{1}{\sqrt{2}} \right)^{2}}{2^{2}}+\frac{\left( \frac{1}{\sqrt{2}} \right)^{3}}{3^{2}}+\frac{\left( \frac{1}{\sqrt{2}} \right)^{4}}{4^{2}}+\frac{\left( \frac{1}{\sqrt{2}} \right)^{5}}{5^{2}}+... \right)-\frac{1}{2^{2}}\left( \frac{\frac{1}{2}}{1}+\frac{\left( \frac{1}{2} \right)^{2}}{2^{2}}+\frac{\left( \frac{1}{2} \right)^{3}}{3^{2}}+\frac{\left( \frac{1}{2} \right)^{4}}{4^{2}}+\frac{\left( \frac{1}{2} \right)^{5}}{5^{2}}+... \right)}$$

$$\displaystyle{=Li_{2}\left( \frac{1}{\sqrt{2}} \right)-\frac{1}{2^{2}}\left( \frac{1}{2}+\frac{\left( \frac{1}{2} \right)^{2}}{2^{2}}+\frac{\left( \frac{1}{2} \right)^{3}}{3^{2}}+\frac{\left( \frac{1}{2} \right)^{4}}{4^{2}}+\frac{\left( \frac{1}{2} \right)^{5}}{5^{2}}+... \right)}$$

$$\displaystyle{=Li_{2}\left( \frac{1}{\sqrt{2}} \right)-\frac{1}{2^{2}}Li_{2}\left( \frac{1}{2} \right)=Li_{2}\left( \frac{1}{\sqrt{2}} \right)-\frac{1}{4}\cdot \frac{\pi ^{2}-6\log ^{2}2}{12}}$$

$$\displaystyle{I_{2}=\int\limits_{0}^{\frac{\pi }{2}}{\arcsin ^{2}\left( \frac{\sin x}{\sqrt{2}} \right)dx}=\sum\limits_{j=1}^{+\infty }{\frac{2^{j-1}}{j^{2}\left( \begin{matrix} 2j \\ j \\ \end{matrix} \right)}\int\limits_{0}^{\frac{\pi }{2}}{sin^{2j}xdx}}=\sum\limits_{j=1}^{+\infty }{\frac{2^{j-1}}{j^{2}\left( \begin{matrix} 2j \\ j \\ \end{matrix} \right)}\cdot \frac{\sqrt{\pi }\Gamma \left( j+\frac{1}{2} \right)}{2\Gamma \left( j+1 \right)}}}$$

$$\displaystyle{=\frac{\pi }{2}\sum\limits_{j=1}^{+\infty }{\frac{2^{j-1}}{j^{2}\left( \begin{matrix} 2j \\ j \\ \end{matrix} \right)}\cdot \frac{\left( 2j \right)!}{2^{2j}\left( j! \right)^{2}}}=\frac{\pi }{2}\sum\limits_{j=1}^{+\infty }{\frac{2^{j-1}}{j^{2}}\cdot \frac{1}{2^{2j}}}=\frac{\pi }{4}\sum\limits_{j=1}^{+\infty }{\frac{1}{j^{2}2^{j}}}=\frac{\pi }{4}\left( -\frac{1}{2}\log ^{2}2+\frac{\zeta \left( 2 \right)}{2} \right)}$$

$$\displaystyle{=\frac{\pi }{8}\left( -\log ^{2}2+\frac{\pi ^{2}}{6} \right)}$$

$$\displaystyle{\int\limits_{0}^{1}{\frac{\left( \arcsin x^{2} \right)^{2}}{\sqrt{1-x^{2}}}dx}=\frac{\pi ^{3}}{8}-2\pi \left( Li_{2}\left( \frac{1}{\sqrt{2}} \right)-\frac{1}{4}\cdot \frac{\pi ^{2}-6\log ^{2}2}{12} \right)+4\cdot \frac{\pi }{8}\left( -\log ^{2}2+\frac{\pi ^{2}}{6} \right)}$$

$$\displaystyle{=\frac{\pi ^{3}}{4}-\frac{3\pi }{4}\cdot \log ^{2}2-2\pi Li_{2}\left( \frac{1}{\sqrt{2}} \right)}$$