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:
$$
\begin{align}
J_1 = & \int_0^1 \int_0^1 {\frac {\arcsin \left( \sqrt {1-s}\sqrt {y} \right) }{\sqrt {1-y} \sqrt {sy-y+1}}}\,ds\,dy = 2\pi -2\pi \ln 2. \\
J_2 = & \int_0^1 \int_0^1 {\frac {\arcsin \left( \sqrt {1-s}\sqrt {y} \right) }{\sqrt {1-s} \sqrt {y}\sqrt {sy-y+1}}}\,ds\,dy = -\frac{7}{4}\zeta\left( 3 \right)+\frac{1}{2}{\pi }^{2}\ln 2.
\end{align}$$
 A: This is not an answer, but I would like to share what I did on this.
Let $s=1-w^2$ and $y=z^2$ to have
$$I=4\int_0^1\int_0^1\frac{wz\arcsin wz}{\sqrt{1-z^2}(1-w^2z^2)}dwdz$$ 
Now using $wz=u$ and $z=v$ we obtain
$$I=4\int_0^1\int_0^v\frac{u\arcsin u}{v\sqrt{1-v^2}(1-u^2)}dudv$$
Approach $1$: write both $\arcsin u$ and $(1-u^2)^{-1}$ in terms of Taylor's expansions and integrate. This results in an ugly double sum. 
Approach $2$: change the order of integration to obtain
$$\begin{align}I&=4\int_0^1\int_{v=u}^1\frac{u\arcsin u}{v\sqrt{1-v^2}(1-u^2)}dvdu\\
&=4\int_0^1\frac{u\arcsin u}{(1-u^2)}\ln{\frac{1+\sqrt{1-u^2}}{u}} du\\
&=16\int_0^{\pi/4}x \tan{2x}\ln{\cot{x}} du
\end{align}$$
where for the last equality I used $u=\sin x$ and made some trigonometric manipulations. 
I tried different Taylor's expansions for this second approach but did not manage to "see" a neatly summable result. Out of what I tried for this second what I liked most was an expansion of $\ln \cot x$ around $\frac{\pi}{4}$, but no success so far.
I thought to do some integrations by part and that might work ... maybe somebody else would do it faster though!
A: I think Math-fun's second approach based on changing the order of integration is a good strategy. Appropriate use of substitutions and trig identities along the way clean up a lot of the resulting "mess": 
$$\begin{align}
\mathcal{I}
&=\int_{0}^{1}\mathrm{d}y\int_{0}^{1}\mathrm{d}s\,\frac{\arcsin{\left(\sqrt{1-s}\sqrt{y}\right)}}{\left(sy-y+1\right)\sqrt{1-y}}\\
&=\int_{0}^{1}\mathrm{d}y\int_{0}^{1}\mathrm{d}t\,\frac{\arcsin{\left(\sqrt{ty}\right)}}{\left(1-ty\right)\sqrt{1-y}};~~~\small{\left[1-s=t\right]}\\
&=\int_{0}^{1}\mathrm{d}y\int_{0}^{y}\mathrm{d}u\,\frac{\arcsin{\left(\sqrt{u}\right)}}{\left(1-u\right)y\sqrt{1-y}};~~~\small{\left[yt=u\right]}\\
&=\int_{0}^{1}\mathrm{d}u\int_{u}^{1}\mathrm{d}y\,\frac{\arcsin{\left(\sqrt{u}\right)}}{\left(1-u\right)y\sqrt{1-y}}\\
&=\int_{0}^{1}\mathrm{d}u\,\frac{\arcsin{\left(\sqrt{u}\right)}}{1-u}\int_{0}^{\sqrt{1-u}}\frac{2\,\mathrm{d}x}{1-x^2};~~~\small{\left[\sqrt{1-y}=x\right]}\\
&=\int_{0}^{1}\mathrm{d}u\,\frac{2\arcsin{\left(\sqrt{u}\right)}}{1-u}\cdot\operatorname{arctanh}{\left(\sqrt{1-u}\right)}\\
&=\int_{0}^{1}\frac{2\arcsin{\left(\sqrt{1-v}\right)}\operatorname{arctanh}{\left(\sqrt{v}\right)}}{v}\,\mathrm{d}v;~~~\small{\left[1-u=v\right]}\\
&=\int_{0}^{1}\frac{4\arcsin{\left(\sqrt{1-w^2}\right)}\operatorname{arctanh}{\left(w\right)}}{w}\,\mathrm{d}w;~~~\small{\left[\sqrt{v}=w\right]}\\
&=4\int_{0}^{1}\frac{\arccos{\left(w\right)}\operatorname{arctanh}{\left(w\right)}}{w}\,\mathrm{d}w\\
&=4\int_{0}^{1}\frac{\operatorname{Li}{\left(w\right)}-\operatorname{Li}{\left(-w\right)}}{2\sqrt{1-w^2}}\,\mathrm{d}w\\
&=4\,{_4F_3}{\left(\frac12,\frac12,1,1;\frac32,\frac32,\frac32;1\right)}.\\
\end{align}$$
And so we see that the above integral is intimately connected to this fun problem, which has generated so much discussion and so many spin-off questions that it wouldn't make sense for me to try to rehash everything here. And given the participation of this question's author in said discussions, I can't help but wonder if he suspected this integral's closed form value all along. =)
A: 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)
