Double integral $\int_0^1\int_0^1 \frac{\sin^{-1}(xy)}{xy} dx dy$

Question

I'm trying to find the value of the integral $$I=\int_0^1\int_0^1 \frac{\sin^{-1}(xy)}{xy} \,\mathrm dx\,\mathrm dy.$$ I tried to use that fact that for $x\in (-1,1)$, $\arcsin(x)=\sum_{n\geq 0}\frac{\binom{2n}{n}}{4^n(2n+1)}x^{2n+1}$, but it just makes things more complicated.

Answer 1

Notice for any $x,y \in [0,1]$, $xy$ takes values in $[0,1]$. Furthermore, for any $t \in [0,1]$, we have $$A(t) \stackrel{def}{=} \verb/Area/\big\{ (x,y) \in [0,1]^2 : xy \le t \big\} = \int_0^1 \min\big\{ 1, \frac{t}{x} \big\} dx = t - t\log t$$ Using this, we can rewrite the integral $I$ as $$I = \int_0^1 \frac{\sin^{-1}(t)}{t} A'(t) dt = - \int_0^1 \frac{\sin^{-1}(t)}{t}\log t dt = -\frac12 \int_0^1 \sin^{-1}(t) (\log^2(t))' dt$$ Integrate by part, we get

$$\begin{align}I = \frac12 \int_0^1 \frac{\log^2(t)}{\sqrt{1-t^2}} dt &= \frac18 \frac{\partial^2}{\partial\alpha^2}\left[ \int_0^1 \frac{t^{2\alpha}}{\sqrt{1-t^2}}dt \right]_{\alpha=0}\\ \color{blue}{u = t^2 \rightarrow}\quad &= \frac{1}{16}\frac{\partial^2}{\partial\alpha^2} \left[ \int_0^1 u^{\alpha-\frac12}(1-u)^{-\frac12}du \right]_{\alpha=0}\\ \color{blue}{\text{beta integral}\rightarrow}\quad &= \frac{1}{16}\frac{\partial^2}{\partial\alpha^2} \left[ \frac{\Gamma(\alpha+\frac12)\Gamma(\frac12)}{\Gamma(\alpha+1)} \right]_{\alpha=0}\\ &= \frac{1}{16}\frac{\Gamma(\frac12)^2}{\Gamma(1)}\left[ \psi'\left(\frac12\right)-\psi'(1)+ \left(\psi\left(\frac12\right)-\psi(1)\right)^2 \right]\\ &= \frac{\pi}{16}\left[ \frac{\pi^2}{2} - \frac{\pi^2}{6} + \left((-2\log(2) - \gamma) - (-\gamma)\right)^2 \right]\\ &= \frac{\pi}{16}\left[ \frac{\pi^2}{3} + \log^2(4)\right] \end{align} $$ where $\Gamma(x)$ is the gamma function and $\psi(x) = \frac{\Gamma'(x)}{\Gamma(x)}$ is the digamma function.

Answer 2

Using your series and integrating term-by-term, I get $$\sum_{n=0}^\infty \frac{2n \choose n}{4^n (2n+1)^3}$$ which according to Maple is $${\mbox{$_4$F$_3$}(1/2,1/2,1/2,1/2;\,3/2,3/2,3/2;\,1)}$$ I don't know if there's a simpler closed form. It's approximately $$1.0233110122363703231$$

Answer 3

$$I=\int_0^1\int_0^1 \frac{\sin^{-1}(xy)}{xy}$$

(1)Do term by term Integration (with your series Expansion of $arcsin$) and

(2)Afterwards use $\int_0^1x^{2 n}\log^2(x)=2(2n+1)^{-3}$

(3) Exploit the Taylorexpansion of the square root,

then you will get that

$$ I=\frac{1}{2}\int_0^1 \frac{\log^2(x)}{\sqrt{1-x^2}} $$

no you can use the ideas of @tired here:

Finding $\int^{1}_{0}\frac{\ln^2(x)}{\sqrt{4-x^2}}dx$

you will find

$$ I=\frac{\pi^3}{48}+\frac{3 \pi \log(4)}{48} $$

Answer 4

Ooops I did not realize there were already two good answers. But since I already typed it, here is yet another solution. Using the change of variables $t=xy$, $dt=y\,dx$ you get \begin{align*} & \int_{0}^{1}\int_{0}^{1}\frac{\arcsin(xy)}{xy}\,dxdy\\ & =\int_{0}^{1}\frac{1}{y}\int_{0}^{y}\frac{\arcsin t}{t}\,dtdy \end{align*} and using $$ \frac{\arcsin t}{t}=\sum_{k=0}^{\infty}\frac{(2k-1)!!}{(2k)!!}\frac{t^{2k}% }{2k+1}% $$ you find \begin{align*} & \sum_{k=0}^{\infty}\frac{(2k-1)!!}{(2k)!!}\frac{1}{2k+1}\int_{0}^{1}\frac {1}{y}\int_{0}^{y}t^{2k}\,dtdy\\ & =\sum_{k=0}^{\infty}\frac{(2k-1)!!}{(2k)!!}\frac{1}{2k+1}\int_{0}^{1}% \frac{1}{y}\frac{y^{2k+1}}{2k+1}\,dy\\ & =\sum_{k=0}^{\infty}\frac{(2k-1)!!}{(2k)!!}\frac{1}{(2k+1)^{2}}\int_{0}% ^{1}y^{2k}\,dy\\ & =\sum_{k=0}^{\infty}\frac{(2k-1)!!}{(2k)!!}\frac{1}{(2k+1)^{3}}% \end{align*}