Double integral $\int_0^1\int_0^1 \frac{\sin^{-1}(xy)}{xy} dx dy$ 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.
 A: 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.
A: 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
$$1.0233110122363703231$$
A: 

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


(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


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


A: 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*}
