# A pair of integrals involving square roots and inverse trigonometric functions over the unit disk

I would like to perform the integration ($$u \in [0,1]$$), $$$$\int_{a=-1}^1 \int_{b=-\sqrt{1-a^2}}^{\sqrt{1-a^2}} u^2 \sqrt{-a^2 u^2-b^2+1} \tan ^{-1}\left(\frac{\left| a\right| }{\sqrt{-a^2-b^2+1}}\right) db da .$$$$

I also have a companion problem $$$$\int_{a=-1}^1 \int_{b=-\sqrt{1-a^2}}^{\sqrt{1-a^2}} u \sqrt{-a^2-b^2+1} \tan ^{-1}\left(u \left| a\right| \sqrt{-\frac{1}{a^2 u^2+b^2-1}}\right) db da .$$$$

Solutions to these two problems would provide a resolution to the question "Compute a certain "separability probability" via a constrained 4D integration over $$[-1,1]^4$$", posed in https://mathematica.stackexchange.com/questions/193337/compute-a-certain-separability-probability-via-a-constrained-4d-integration-ov

Now, for the component functions of the first integrand, we have $$$$\int_{a=-1}^1 \int_{b=-\sqrt{1-a^2}}^{\sqrt{1-a^2}} u^2 \sqrt{-a^2 u^2-b^2+1} db da = \frac{4}{3} u \left(\sqrt{1-u^2} u+\sin ^{-1}(u)\right),$$$$ and $$$$\int_{a=-1}^1 \int_{b=-\sqrt{1-a^2}}^{\sqrt{1-a^2}} \tan ^{-1}\left(\frac{\left| a\right| }{\sqrt{-a^2-b^2+1}}\right) db da=\frac{1}{2} (\pi -2) \pi.$$$$ Can some form of integration-by-parts--or other methodology--be performed? (I realize that more may certainly be needed in addition to the two above results.)

The much-viewed question https://math.stackexchange.com/questions/1167346/integration-by-parts-for-a-double-integral seems relevant here.

• To get rid of at least one integral, first change $a, b$ to polar coordinates $r, \theta$, and change again to $x, y$ with $r = x/\sqrt{1+x^2}, \cos(\theta) = y/x$ with bounds $x \in [0, \infty)$ and $y \in [0, x]$. Switching the integration order to $y \in [0, \infty)$, $x \in [y, \infty)$ then allows you to perform the $x$-integral, and $$\int_0^\infty \! \frac{8 u^2 \sqrt{1 + (1-u^2) \, y^2} \tan^{-1}(y)}{3 (1 + y^2)^2} \, dy$$ remains. Not sure how helpful that actually is though. – student Mar 18 '19 at 19:52

Further to my comment, one can set $$y = \frac{z}{\sqrt{1-z^2}}$$ to obtain $$\frac{8}{3} \, u^2 \int_0^1 \! \sqrt{1 - u^2 z^2} \sin^{-1}(z) \, dz \;,$$ which Mathematica evaluates to $$\frac{2}{3} \, u \, \bigl(\operatorname{Li}_2(-u) - \operatorname{Li}_2(u) + \pi \sqrt{1 - u^2} \, u - (1 - u^2) \tanh^{-1}(u) - u + \pi \sin^{-1}(u)\bigr) \;,$$ where $$\operatorname{Li}_2(x)$$ is the dilogarithm.

A very similar derivation for your companion problem yields the final integral $$\frac{8}{3} \, u \int_0^1 \! \sqrt{1 - z^2} \sin^{-1}(u z) \, dz$$ which Mathematica evaluates to $$-\frac{2}{3u} \, \bigl(u^2 \, \operatorname{Li}_2(-u) - u^2 \, \operatorname{Li}_2(u) - (1 - u^2) \tanh^{-1}(u) + u\bigr) \;.$$

For completeness, we can also do the third integral (from the comments) in exactly the same way, yielding $$\frac{8}{\pi} \, u^2 \int_0^1 \! \sqrt{1 - u^2 z^2} \, dz$$ as the final integral, which evaluates to $$\frac{4}{\pi} \, u \, \bigl(\sqrt{1 - u^2} \, u + \sin^{-1}(u)\bigr) \;.$$

Bonus round: If $$I_1$$, $$I_2$$, $$I_3$$ are the three results above, and $$A = \frac{6}{\pi^2} \, (I_2 - I_1) + I_3 \;, \\ B = \frac{12 u^2 \left(u^6+9 u^4-9 u^2-12 \left(u^4+u^2\right) \log (u)-1\right)}{\left(u^2-1\right)^5} \;,$$ then we are also interested in the integral of $$AB$$ over $$u \in [0,1]$$. That seems quite impossible, but we are in luck: integrating twice by parts, differentiating each time the factor with dilogarithms, yields a simple integral in $$u$$ that can be evaluated. The result is $$\frac{256-168 \zeta (3)}{\pi ^2}-38+48 \log (2) \approx 0.747925 \;.$$

• Great! Quick numerical integrations give me agreement with both of your results to eight decimal places. I will incorporate your findings in my related question mathematica.stackexchange.com/questions/193337/… – Paul B. Slater Mar 18 '19 at 22:47
• For simplicity, I omitted certain numerical factors that appear in the subject matter context mathematica.stackexchange.com/questions/193337/…. In that context, we have to add together the two results of Student multiplying them by $6/\pi^2$ and add them to a companion third term (continued in next comment) – Paul B. Slater Mar 19 '19 at 16:50
• we had already been able to calculate ourselves—that is, the integral of $\frac{3 u^2 \sqrt{-a^2 u^2-b^2+1}}{\pi }$—giving us $\frac{2 \left(\sqrt{(1-u^2) u^2}+\sin ^{-1}\left(u\right)\right)}{\pi }$. This sum of the three terms proves—rather surprisingly—to be precisely equal to the “two-rebit Lovas-Andai separability function” (eq. (2) in arxiv.org/abs/1701.01973). – Paul B. Slater Mar 19 '19 at 16:50
• Thanks, student. What would be of major importance to the underlying "separability probability question" would be the integration over $u \in [0,1]$ of the product of each of your two functions with $\frac{12 u^2 \left(u^6+9 u^4-9 u^2-12 \left(u^4+u^2\right) \log (u)-1\right)}{\left(u^2-1\right)^5}$. The result of the first integration times $\frac{6}{\pi^2}$ appears (via numeric means) to equal $-\frac{41 C}{2}-69-\pi (-22+\pi -37 \log (2)+15 \log (3))$, where Catalan's constant is indicated. Similarly, the result of the second integration times $\frac{6}{\pi^2}$ appears – Paul B. Slater Mar 20 '19 at 3:12
• (again via numeric means) to equal $\frac{1}{6} (-191 C-395+\pi (139-15 \pi +125 \log (2)+\log (27))) \approx 0.20672$. (I was steered to these two conjectures by wolframalpha.com.) The result of the second integration minus the result of the first integration plus $\frac{919}{5}-264 \log (2)$ should equal the sought "rebit-retrit separability probability" of $-\frac{34 C}{3}+\frac{5609}{30}-264 \log (2)+\frac{1}{6} \pi (7-9 \pi -97 \log (2)+93 \log (3)) \approx 0.747925$. My attempts to use Mathematica to perform these computations has so far been unsuccessful. – Paul B. Slater Mar 20 '19 at 3:25