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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:

$$ \begin{align} J_1 = & \int_0^1 \int_0^1 {\frac {\arcsin \left( \sqrt {1-s}\sqrt {y} \right) }{\sqrt {1-y} \sqrt {sy-y+1}}}\,ds\,dy = 2\pi -2\pi \ln 2. \\ J_2 = & \int_0^1 \int_0^1 {\frac {\arcsin \left( \sqrt {1-s}\sqrt {y} \right) }{\sqrt {1-s} \sqrt {y}\sqrt {sy-y+1}}}\,ds\,dy = -\frac{7}{4}\zeta\left( 3 \right)+\frac{1}{2}{\pi }^{2}\ln 2. \end{align}$$

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  • $\begingroup$ Can you not reset the equation with $$sy-y+1=1-\left(\sqrt{y}\sqrt{1-s}\right)^2$$ in the denominator? $\endgroup$ – Chinny84 Sep 19 '14 at 7:15
  • $\begingroup$ The inner term has a closed form solution but I gave up with the outer integral. $\endgroup$ – Claude Leibovici Sep 19 '14 at 7:36
  • $\begingroup$ @ClaudeLeibovici Same here. But maybe there is another approach. $\endgroup$ – user153012 Sep 19 '14 at 7:39
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    $\begingroup$ Could you please give me links or papers for solving $J_1$ & $J_2$? I'm curious. Thanks... $\endgroup$ – Anastasiya-Romanova 秀 Sep 19 '14 at 8:13
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    $\begingroup$ @user153012 Here resolved by Seraphim mathematica.gr/forum/viewtopic.php?f=9&t=59475 $\endgroup$ – whitexlotus Aug 20 '17 at 18:11
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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. =)

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  • $\begingroup$ quite Interesting! +1 $\endgroup$ – Math-fun Aug 1 '15 at 13:52
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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!

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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)

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  • $\begingroup$ Thank you for the answer, but this is a partial answer even for the inner integral. I could get it using Mathematica or Maple. Wolfram Integrator also could solve it. This is far-far the easier part of the problem. The complications come just from here, and maybe this way is also not the best, because maybe there are other starts for the problem, thats make the inner part easier. By the way I solved the integral, but in the solution I have to use generalized hypergeometric function or Meijer G, or $\operatorname{Li}_2$ and this is what I want to avoid. $\endgroup$ – user153012 Sep 28 '14 at 9:13

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