TITLE: Double integral $\int_0^1\int_0^1 \frac{\sin^{-1}(xy)}{xy} dx dy$
QUESTION [5 upvotes]: 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.

REPLY [2 votes]: 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.