Closed-form of integral $\int_0^1 \int_0^1 \frac{\arcsin\left(\sqrt{1-s}\sqrt{y}\right)}{\sqrt{1-y} \cdot (sy-y+1)}\,ds\,dy $ I'm looking for a closed form of this definite iterated integral.
$$I = \int_0^1 \int_0^1 \frac{\arcsin\left(\sqrt{1-s}\sqrt{y}\right)}{\sqrt{1-y} \cdot (sy-y+1)}\,ds\,dy. $$
From Vladimir Reshetnikov we already know it, that the numerical value of it is
$$I\approx4.49076009892257799033708885767243640685411695804791115741588093621176851...$$
There are similar integrals having closed forms:
$$
\begin{align}
J_1 = & \int_0^1 \int_0^1 {\frac {\arcsin \left( \sqrt {1-s}\sqrt {y} \right) }{\sqrt {1-y} \sqrt {sy-y+1}}}\,ds\,dy = 2\pi -2\pi \ln 2. \\
J_2 = & \int_0^1 \int_0^1 {\frac {\arcsin \left( \sqrt {1-s}\sqrt {y} \right) }{\sqrt {1-s} \sqrt {y}\sqrt {sy-y+1}}}\,ds\,dy = -\frac{7}{4}\zeta\left( 3 \right)+\frac{1}{2}{\pi }^{2}\ln 2.
\end{align}$$

I think in case of iterated integration you have to bear with Meijer G or Li2 or Li3 etc. Here (in case of iterated integration) we integrate the inner part with respect to 's' treating 'y' as a constant. 
Here let $$\sin \:^{-1}\left(\sqrt{1-s}\sqrt{y}\right)=t$$
$$-\frac{\sqrt{y}}{2\sqrt{1-s}\sqrt{y\left(s-1\right)+1}}ds=dt$$
So the intitial integral (the inner integral) changes into:
$$\int _0^{\frac{\pi }{2}}\left(\frac{-2\sin \left(t\right)t}{\sqrt{1-y}\sqrt{sy-y+1}}\right)dt\:=\int _0^{\frac{\pi }{2}}\:\left(\frac{-2\left(t\right)\tan \left(t\right)}{\sqrt{1-y}}\right)dt$$ ------->(EQ.1)
As:
$$\sin \left(t\right)=\sqrt{y\left(1-s\right)}$$
We have $$\sqrt{sy-y+1}=\sqrt{1-\left(y+ys\right)}\sqrt{1-\sin ^2\left(t\right)}=\cos \left(t\right)$$
And hecne we have EQ.1.
Integrating EQ.1 we have:
$$-\frac{i\left(\text{Li}_2\left(-e^{2ix}\right)+x^2+2ix\ln \left(1+e^{2ix}\right)\right)}{\sqrt{1-a}}+C$$
You further apply the limits. From here you take it home. (Hope this helps a bit)
