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Let:

$$Q(A,Y) = Y^4-2 Y^2+2 A Y \sqrt{1-Y^2}-A^2+1$$

I’m curious as to whether it’s possible to find a closed-form solution for:

$$I(A)=\int_{Y_1(A)}^{Y_2(A)} \frac{1}{\sqrt{\left(1-Y^2\right) Q(A,Y)}}\,dY\\ = \int_{\sin^{-1}(Y_1(A))}^{\sin^{-1}(Y_2(A))}\frac{2 \sqrt{2}}{\sqrt{3-8 A^2+8 A \sin (2 \theta )+4 \cos (2 \theta )+\cos (4 \theta )}}\,d\theta$$

where $0 \le A \le \frac{3 \sqrt{3}}{4}$ and $-1\le Y_1(A), Y_2(A) \le 1$ are the real zeroes of $Q(A,Y)$.

I know a re-parameterisation that makes the zeroes a bit less messy than they would be in terms of $A$. If we set:

$$A(a) = \frac{\sqrt{a^3 (a+2)^3}}{2 (a+1)}$$

with $0 \le a \le 1$ then we have:

$$Y_{1,2}(a) = \frac{a^2+a\mp (a+2) \sqrt{1-a^2}}{2 (a+1)}$$

I’m not sure if there is any way to explicitly introduce these known zeroes into the integrand, as one could do if $Q(A,Y)$ were a polynomial. I do have the following relation:

$$Q(A,Y)\,Q(-A,Y)\\ = Q(A,Y)\,Q(A,-Y)\\ =\left(Y^4-2 Y^2+2 A Y \sqrt{1-Y^2}-A^2+1\right)\left(Y^4-2 Y^2-2 A Y \sqrt{1-Y^2}-A^2+1\right)\\ = P(A,Y)\,P(A,-Y)$$ where: $$P(A,Y) = Y^4+2Y^3-2Y+A^2-1\\ P(A(a),Y)= (Y-Y_1(a))\,(Y-Y_2(a))\,\left(Y^2+(a+2)Y+\frac{a (a+2)^2+2}{2 (a+1)}\right)$$

$P(A,Y)$ has exactly the same real zeroes as $Q(A,Y)$, but has the opposite sign.

Mathematica is unable to explicitly evaluate the definite integral in either form, but it returns an indefinite integral for the trigonometric form:

$$I(A,\theta) =\\ \frac{4 \sqrt{2} \sqrt{\frac{\left(r_1-r_2\right) \left(r_3-\tan (\theta )\right)}{\left(r_1-r_3\right) \left(r_2-\tan (\theta )\right)}} \left(r_1 \cos (\theta )-\sin (\theta )\right) \left(r_4 \cos (\theta )-\sin (\theta )\right) \times \\F\left(\sin ^{-1}\left(\sqrt{\frac{\left(r_2-r_4\right) \left(r_1-\tan (\theta )\right)}{\left(r_1-r_4\right) \left(r_2-\tan (\theta )\right)}}\right)|\frac{\left(r_2-r_3\right) \left(r_1-r_4\right)}{\left(r_1-r_3\right) \left(r_2-r_4\right)}\right)}{\left(r_1-r_4\right) \sqrt{\frac{\left(r_1-r_2\right) \left(r_2-r_4\right) \left(r_1-\tan (\theta )\right) \left(r_4-\tan (\theta )\right)}{\left(r_1-r_4\right){}^2 \left(r_2-\tan (\theta )\right){}^2}} \sqrt{3-8 A^2+8 A \sin (2 \theta )+4 \cos (2 \theta )+\cos (4 \theta )}}$$

where the $r_i$ are the roots of:

$$R(A,Y) = A^2 Y^4 - 2 A Y^3 + 2 A^2 Y^2 - 2 A Y + A^2 -1$$

which for $A=A(a)$ are:

$$ \begin{array}{rcl} r_1 &=& \frac{1-(a+1)^{3/2}\sqrt{1-a}}{a^{3/2} \sqrt{a+2}}\\ &=& \tan(\sin^{-1}(Y_1(a)))\\ r_2 &=& \frac{1+(a+1)^{3/2}\sqrt{1-a}}{a^{3/2} \sqrt{a+2}}\\ & =& \tan(\sin^{-1}(Y_2(a)))\\ r_3 &=& \frac{1-i (a+1) \sqrt{a^2+4 a+3}}{\sqrt{a} (a+2)^{3/2}}\\ r_4 &=& \frac{1+i (a+1) \sqrt{a^2+4 a+3}}{\sqrt{a} (a+2)^{3/2}} \end{array} $$

This formula is not very helpful in itself without a deeper understanding of how it was obtained; a naive attempt to get the definite integral from it gives imaginary results, probably because of some issue with branch cuts.

But it does suggest that a human who understood the technique that produced it could perform a similar process to obtain an expression for the definite integral in terms of an elliptic integral.

The motivation here is to find a closed-form expression for the probability density function for the area of a triangle whose vertices are chosen uniformly at random from the unit circle, as discussed in this question:

Moments of area of random triangle inscribed in a circle

I believe $Prob(A) = \frac{2}{\pi^2} I(A)$, where $A$ is the area of such a triangle. Numeric integrals for this quantity give the plot below; this goes to infinity at $A=0$, and is finite and non-zero for the maximum value, $A=\frac{3 \sqrt{3}}{4}$.

Probability distribution for triangle area

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If we define:

$$\begin{array}{rcl} g(a)&=&a^2(a+2)^2-3\\ k(a)&=&(a+1)^3\sqrt{(1-a)(a+3)}\\ \end{array}$$

then, by taking limits of the antiderivative provided by Mathematica at the endpoints of the range of integration, we obtain:

$$I_0(a) = \frac{2\sqrt{2}\,(a+1)\,K\left(\frac{2 k(a) i}{g(a) + k(a) i}\right)}{\sqrt{a(a+2)}\,\sqrt{g(a) + k(a) i}}$$

Here $K$ is a complete elliptic integral of the first kind, and the convention used is that followed by Mathematica, where the argument of $K$ appears unsquared in the defining integral. The parameter $a$ is related to the original parameter $A$ by the equation stated in the question.

When $g(a)\ge 0$, which holds for $a \gt \sqrt{1+\sqrt{3}}-1 \approx 0.652892$, $I_0(a)$ is real-valued and agrees with numerical evaluations of the integral $I(a)$.

However, when $g(a)$ crosses zero, the function $K$ has a branch cut, and $I_0(a)$ jumps discontinuously to an imaginary-valued function.

This problem can be remedied across the full range for the parameter $a$ by using the analytic continuation of $K$ across the branch cut, written as $K'$, which is discussed in detail in the answer to this question on Math StackExchange:

https://math.stackexchange.com/questions/2008090/analytical-continuation-of-complete-elliptic-integral-of-the-first-kind

$$K'(m) = \frac{1}{\sqrt{m}}\left(K\left(\frac{1}{m}\right)+i K\left(1-\frac{1}{m}\right)\right)$$

$K'$ has its own, different branch cut located in a different part of the complex plane, and the argument in this application does not cross it. So if we define:

$$I_1(a) = \frac{2\sqrt{2}\,(a+1)\,K'\left(\frac{2 k(a) i}{g(a) + k(a) i}\right)}{\sqrt{a(a+2)}\,\sqrt{g(a) + k(a) i}}$$

then $I_1(a)$ is a real-valued function for $0\lt a \le 1$, and it agrees precisely with numerical evaluations of the original integral $I(a)$.

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If the purpose of this calculation is to test the geometry conjecture, comparing expansions in powers of $A$ or $a$ should be effective. The indefinite integral can readily be evaluated to any order in $A$, but for the definite integral I run into a difficulty. Take the zeroth order term $A=0=a$, when $Y_{1,2}=\pm 1$, $Q=Y^4-2Y^2+1$ and the integral over $Y$ is $$I(0)=\int_{-1}^1\frac{1}{\sqrt{(1-Y^2)(Y^4-2Y^2+1)}}\,dY,$$ which has a nonintegrable singularity at the end points (the integrand diverges as $(1\pm y)^{-3/2}$).

I have checked that the small-$a$ asymptotic of $I(a)$ is indeed a $1/\sqrt{a}$ divergence.

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  • $\begingroup$ I do expect $I(0)$ to be infinite, on geometric grounds, and numeric integrals also tend to infinity as $A$ or $a$ go to 0. There is a chance that I can make sense of Mathematica's indefinite integral and adapt it for the definite case, but it’s tricky taking limits of complicated expressions and also making sure of the right branch when taking things like square roots of complex values. $\endgroup$ – Greg Egan Jul 16 '18 at 14:09

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