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I want to find the value (desirably to high-precision) of the parameter $i$ for which the constrained 3d integration

972 NIntegrate[Boole[jj0 && (P1 > n1/3^(i) || P2 > 16/9)], {Q1, 0, o}, {Q2, 0, 
o}, {Q3, 0, o}]/13

equals 1, where the integration limit $o=\infty$. (The sought value--my investigation so far shows--lies between 6.73746 and 6.73748, yielding 0.374657 and 1.02754, respectively. It seems as a function of $i$, the integral is sharply peaked in this neighborhood. Plots might be of interest.)

Now, the constraint jj0 is

Q1 > 0 && Q2 > 0 && Q3 > 0 && Q1 + 3 Q2 + 2 Q3 < 1

and P1 is

(65536 (Q1 - Q3)^12 (1 - 9 Q2 - 6 Q3 +  3 (Q1^2 + 9 Q2^2 + 6 Q2 Q3 + 4 Q3^2 + 
  Q1 (-1 + 3 Q2 + 4 Q3)))^2)/43046721

and P2 is

(4 Abs[Q1 - Q3] + 4/3 Sqrt[Abs[
1 - 9 Q2 - 6 Q3 + 
 3 (Q1^2 + 9 Q2^2 + 6 Q2 Q3 + 4 Q3^2 + Q1 (-1 + 3 Q2 + 4 Q3))]])^2

Further, $n1=\frac{268435456}{29195279392693578129}=\frac{2^{28}}{3^{16} \cdot 7^{14}}$.

My conjecture is that $\frac{n1}{3^i}$ should assume a "simple" structure, possibly a rational number somewhat like $n1$ itself. (Of course, $\frac{n1}{3^i}$ can be reparameterized--I have no strong case for the use of this form--although $i \approx 6.73746$ is of a "pleasant magnitude".)

I don't think that FindRoot can be used here--since doing so seems to yield singular jacobians (FindRoot::jsing).

Perhaps some "binary search" procedure might be in order.

Of course, the WorkingPrecision option (and other options) can be employed in the numerical integration.

This is a quantum-information-theoretic problem pertaining to the "entanglement probability" of $\frac{13}{27}=\frac{36}{\frac{972}{13}}$ for the "two-qutrit Hiesmayr-L{\"o}ffler magic simplices of Bell states"

[https://arxiv.org/abs/1905.09228][1]

P1 and P2 pertain to the pair of entanglement constraints in

[https://arxiv.org/abs/1708.05336][2]

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  • $\begingroup$ You could start by limiting the integration domain: instead of using jj0 you can NIntegrate[..., {Q1, 0, 1}, {Q2, 0, (1-Q1)/3}, {Q3, 0, (1-Q1-3*Q2)/2}]. With this I get different results from yours, and a result of $i\approx5.1989$. $\endgroup$
    – Roman
    Mar 31 '20 at 20:16
  • $\begingroup$ Yes, Roman I see you can get those limits using GenericCylindricalDecomposition[jj0, {Q1, Q2, Q3}]. But they would seem to be implicit in the jj0 constraint, so what explanation for the difference in results? Also, I get "has evaluated to non-numerical values for all sampling points..." when I attempt the NIntegrate. $\endgroup$ Apr 1 '20 at 0:51
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Well, I am certainly having difficulty achieving the desired strong numerical precision here, but to the extent I can, coupled with my conjecture "that $\frac{n1}{3^i}$ should assume a "simple" structure, possibly a rational number somewhat like $n1$ itself", leads me to believe that \begin{equation} \frac{n1}{3^i}= \frac{134217728}{23910933822616040487651}=\frac{2^{27}}{3^{18} \cdot 7^{15} \cdot13}. \end{equation} So, the exponent of 3 would be \begin{equation} i=\frac{\log (1638)}{\log (3)} \approx 6.73689$, \end{equation} where $1638 = 2 \cdot 3^2 \cdot 7 \cdot 13$. (The remark in the question that "the sought value--my investigation so far shows--lies between 6.73746 and 6.73748", thus now seems slightly amiss.)

Let us also recall that \begin{equation} n1=\frac{134217728}{23910933822616040487651}=\frac{2^{28}}{3^{16} \cdot 7^{14}}. \end{equation}

So, in this setup, $3^i= 1648$.

Of course, it would be nice to have greater numerical support for this line of reasoning than I now possess.

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