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This is a direct descendant of two other recent questions, 3D and Equivalence, both of which have been answered in skillful, interesting manners. (See also the comment [actually answer] of JimB to 3D5D, applying the methodology of 3D to a higher-order problem.)

We would now like to turn the focus from the 3D constrained integration problem posed at the very outset in 3D to a 2D one, by modifying the constraint (for absolute separability) there

Boole[Subscript[λ, 1] > Subscript[λ, 2] && Subscript[λ, 2] > Subscript[λ, 3] && Subscript[λ, 3] > 1 - Subscript[λ, 1] - Subscript[λ, 2] - Subscript[λ, 3] && Subscript[λ, 1] - Subscript[λ, 3] < 2 Sqrt[Subscript[λ, 2] (1 - Subscript[λ, 1] - Subscript[λ, 2] - Subscript[λ, 3])]]

so that the inequality

Subscript[λ, 1] - Subscript[λ, 3] < 2 Sqrt[Subscript[λ, 2] (1 - Subscript[λ, 1] Subscript[λ, 2] - Subscript[λ, 3]

becomes an equality

Subscript[λ, 1] - Subscript[λ, 3] ==2 Sqrt[Subscript[λ, 2] (1 - Subscript[λ, 1] Subscript[λ, 2] - Subscript[λ, 3]

The resultant 2D integration (formulation given at end of question) should yield the Hilbert-Schmidt probability that a "two-qubit" state lies on the boundary of the absolutely separable states, rather than as in the 3D formulation within the volume of such states.

Now, relatedly, in eq. (35) of 2009paper the inverse trigonometric function based formula ($\approx 20.9648519$)

-((3840 (-5358569267936 + 33756573946095 Sqrt[2] [Pi] - 270052591568760 Sqrt[2] ArcCot[Sqrt[2]] + 11149704525960 Sqrt[2] ArcCot[2 Sqrt[2]] + 270052591568760 Sqrt[2] ArcCot[3 + Sqrt[2]]))/(-1959684729929728 + 1601255307608064 Sqrt[2] + 1529087492782080 Sqrt[2] [Pi] - 45247615492565918250 Sqrt[2] ArcCot[Sqrt[2]] + 22619730179635540245 Sqrt[2] ArcSec[3]))

was given for the Hilbert-Schmidt area-volume ratio of the two-qubit absolutely separable states.

Now, I would like to ask, first, whether this formula can be condensed/simplified, possibly along the lines of that employed by the user yarchik in the answer to Equivalence, in which the FindIntegerNullVector command was employed.

Secondly, I would like to ask if the area-to-volume ratio formula could itself be rederived by solving the 2D constrained integration problem indicated at the beginning of this question--followed by the scaling of its result by the answer of JimB

29902415923/497664 - 50274109/(512 Sqrt[2]) - (3072529845 π)/(32768 Sqrt[2]) +(1024176615 ArcCos[1/3])/(4096 Sqrt[2])

given in 3D for the absolute separability Hilbert-Schmidt probability for the two-qubit states.

We must note, though, that the original 3D constrained integration problem posed in 3D was solved there in an unconstrained form, employing a change-of-variables

change = {Subscript[λ, 1] -> x/(1 + 2 x), Subscript[λ, 2] -> y/(1 + y) (1 + x)/(1 + 2 x), Subscript[λ, 3] -> z 1/(1 + y) (1 + x)/(1 + 2 x)};

just recently provided by N. Tessore. This transformed the problem into

Integrate[integrand2, {z, 1/2, 1}, {y, z, 2 + 2 Sqrt[1 - z] - z}, {x, y, 2 Sqrt[-((-y - 2 y^2 - y^3 + y z + 2 y^2 z + y^3 z)/(-1 + y + z)^4)] + ( 4 y + z - 3 y z - z^2)/(-1 + y + z)^2}],

where

integrand2 = (9081072000 (1 + x)^8 (x - y)^2 (1 - 2 z)^2 (y - z)^2 (-1 + y + z)^2 (z + x (-1 - y + z))^2 (-1 + z + x (y + z))^2)/((1 + 2 x)^16 (1 + y)^15)  .

The question of what would be a suitable change-of-variables in the requested dimension reduction scenario seems of interest.


In his comment to the originally-posed question, JimB simplified (LeafCount 55 vs. 96) the area-volume ratio given above to

-((15 (-1339642316984 + 1393713065745 Sqrt[2] \[Pi] - 
2787426131490 Sqrt[2] ArcCos[1/3]))/(2 (-956877309536 + 781862943168 Sqrt[2] + 746624752335 Sqrt[2] \[Pi] - 1990999339560 Sqrt[2] ArcCos[1/3])))

As noted in my comment in response to that of JimB, I observed that FullSimplify applied to the product of this result and the earlier one of his (given above)

`29902415923/497664 - 50274109/(512 Sqrt[2]) - (3072529845 π)/(32768 Sqrt[2]) +(1024176615 ArcCos[1/3])/(4096 Sqrt[2])`

gives us simply

 (5 (-669821158492 + 1393713065745 Sqrt[2] ArcCot[2 Sqrt[2]]))/5308416 ≈0.0766949

which I now conjecture should be the result of the requested 2D integration. This integration problem is expressible as ({Subscript[[Lambda], 1] -> x, Subscript[[Lambda], 2] -> y, Subscript[[Lambda], 3] -> z}) in the form

Integrate[9081072000 (x - y)^2 (1 - 3 x + 3 y - 4 Sqrt[y - 2 x y])^2 (1 - 3 x +
 y - 2 Sqrt[y - 2 x y])^2 (2 y - 2 Sqrt[y - 2 x y])^2 (-1 + 2 x + 
2 Sqrt[y - 2 x y])^2 (x - 3 y + 2 Sqrt[y - 2 x y])^2 Boole[x > y && 3 y > x + 2 Sqrt[y - 2 x y] && 
3 x + 4 Sqrt[y - 2 x y] > 1 + 3 y], {y, 0, 1}, {x, 0, 1}].

The transformation {z -> x - 2 y + 2 Sqrt[y - 2 x y]} was used to reduce the original 3D problem to the 2D one.



In a series of three consecutive comments to the answer (956877309536 + 243 Sqrt[2] (-3217542976 + 1024176615 ArcCos[5983/6561])))/2654208 given by yarchik, I indicated that the alternative (seemingly superior) use of the transformation {y -> 1/2 (1 - x - z + Sqrt[1 - 2 x - 2 z + 4 x z])} yields a result

(5 (-1339642316984 + 1393713065745 Sqrt[2] ArcTan[(4 Sqrt[2])/7]))/5308416 

apparently exactly twice the conjecture (5 (-669821158492 + 1393713065745 Sqrt[2] ArcCot[2 Sqrt[2]]))/5308416 ≈0.0766949 stated in the question, based on eq. (35) in the cited 2009 paper and an earlier result of JimB in this context.

To re-emphasize, the motivation behind the use of the transformations was to convert the absolute separability probability inequality into an equality.

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    $\begingroup$ Shamelessly copying @yarchik 's answer (mathematica.stackexchange.com/questions/231201/…), the 3 terms {{ArcCot[2 Sqrt[2]], ArcCot[3 + Sqrt[2]], ArcCot[Sqrt[2]]} can be written as linear combinations of $\pi$ and $\phi=\cos ^{-1}\left(\frac{1}{3}\right)$ resulting in $-\frac{15 \left(-2787426131490 \sqrt{2} \phi +1393713065745 \pi \sqrt{2}-1339642316984\right)}{2 \left(-1990999339560 \sqrt{2} \phi +746624752335 \pi \sqrt{2}+781862943168 \sqrt{2}-956877309536\right)}$. $\endgroup$ – JimB Oct 4 at 22:22
  • $\begingroup$ Quite impressive simplification, JimB! LeafCount of 55 vs. original 96. REMARKABLY, FullSimplify applied to the product of your answer here and (as in the question) 29902415923/497664 - 50274109/(512 Sqrt[2]) - (3072529845 π)/(32768 Sqrt[2]) +(1024176615 ArcCos[1/3])/(4096 Sqrt[2]) gives us simply (5 (-669821158492 + 1393713065745 Sqrt[2] ArcCot[2 Sqrt[2]]))/5308416 $\approx 0.0766949$, which I suspect (more thinking needed to be sure) should be the result of the 2D integration I request in the question. $\endgroup$ – Paul B. Slater Oct 5 at 0:33
  • $\begingroup$ Maybe it's just me but I don't see a complete and compact description of the 2D problem described: the integrand is only found in a linked post rather than this post (unless I've missed something). $\endgroup$ – JimB Oct 5 at 15:58
  • $\begingroup$ I've got $\frac{149512079615}{82944}-\frac{754111635}{256 \sqrt{2}}-\frac{46087947675 \pi }{16384 \sqrt{2}}+\frac{15362649225 \cos ^{-1}\left(\frac{1}{3}\right)}{2048 \sqrt{2}}$ for the 2D result. I'll write up the steps tonight or tomorrow morning. $\endgroup$ – JimB Oct 6 at 0:14
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The answer turns out to be quite simple at the end:

(5 (956877309536 + 243 Sqrt[2] (-3217542976 + 1024176615 ArcCos[5983/6561])))/2654208
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  • $\begingroup$ If LeafCount is used as an indicator of simplicity, that's the smallest value yet. That also demonstrates the trigonometric identity of ArcCos[5983/6561] == 8 ArcCos[1/3] - 3 \[Pi]. $\endgroup$ – JimB Oct 8 at 16:17
  • $\begingroup$ The last sentence of the question is :"The transformation {z -> x - 2 y + 2 Sqrt[y - 2 x y]} was used to reduce the original 3D problem to the 2D one." It turns out that if instead one employs {y -> 1/2 (1 - x - z + Sqrt[1 - 2 x - 2 z + 4 x z])}, the resultant integration is speedily carried out, yielding now rather than (5 (956877309536 + 243 Sqrt[2] (-3217542976 + 1024176615 ArcCos[5983/6561])))/2654208 $\approx 0.109748$, the answer (5 (-1339642316984 + 1393713065745 Sqrt[2] ArcTan[(4 Sqrt[2])/7]))/5308416 $\approx 0.1533898$. This is exactly equal to twice the conjecture (continued) $\endgroup$ – Paul B. Slater Oct 10 at 15:20
  • $\begingroup$ (5 (-669821158492 + 1393713065745 Sqrt[2] ArcCot[2 Sqrt[2]]))/5308416 ≈0.0766949 also put forth in the question, based as indicated on a previous answer of JimB and the formula (35) for the area-volume ratio given in the 2009 paper. This identity can--it would seem--be confirmable through, as seems to have become standard, use of the FindIntegerNullVector command, first employed in this context by yarchik. $\endgroup$ – Paul B. Slater Oct 10 at 15:28
  • $\begingroup$ Also, the second transformation seems less problematical in the (probability) sense that the indicated 2D integration, omitting the constraint, yields the real-valued outcome $\frac{35587695}{4}$, rather than the complex-valued one $-\frac{2391273386563840}{96577}-\frac{516339750748160 i}{96577}$, yielded by the first transformation. $\endgroup$ – Paul B. Slater Oct 10 at 15:39
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To reiterate (this was not explicit in the original of the question posed), the 2D constrained integration problem I am seeking to solve takes the form

Integrate[9081072000 (x - y)^2 (1 - 3 x + 3 y - 4 Sqrt[y - 2 x y])^2 (1 - 3 x +y - 2 Sqrt[y - 2 x y])^2 (2 y - 2 Sqrt[y - 2 x y])^2 (-1 + 2 x + 2 Sqrt[y - 2 x y])^2 (x - 3 y + 2 Sqrt[y - 2 x y])^2 Boole[x > y && 3 y > x + 2 Sqrt[y - 2 x y] && 3 x + 4 Sqrt[y - 2 x y] > 1 + 3 y], {y, 0, 1}, {x, 0, 1}]

In the previous answer, the outer integration was first performed, and the denestSqrt procedure of Carl Woll (in his answer to denestSqrt) applied to the univariate result, followed by integration over $y$, yielding the result (LeafCount of 193)

-((5 (-1197989196206888748236188141460 + 
461683252327647484985967104198 Sqrt[7] + 
733143027137266392522295640220 Sqrt[2] ArcCsc[3] - 
1759553227963208925524554251975 Sqrt[2]
  ArcSin[1/9 (4 - Sqrt[7])]))/1954710932452630508961792) - (5 (493289889307851856183541978996 + 575808694228213966079710592832 Sqrt[2] -461683252327647484985967104198 Sqrt[7] - 1466024211975381520156897153311360 Sqrt[2] ArcCsc[3/Sqrt[2]] + 1466024211975381520156897153311360 Sqrt[2]ArcCsc[Sqrt[12 - 6 Sqrt[2]]] - 1467783765203344729082421707563335 Sqrt[2]ArcCsc[6 Sqrt[3/(34 + 5 Sqrt[2] - 4 Sqrt[7] - 8 Sqrt[14])]]+578188180294616824015749120 Sqrt[2] ArcSin[Root[1 - 24 #^2 + 72 #^4& , 2, 0]] + 293845361868970757303978720655 Sqrt[2] ArcSin[Root[1 - 24 #^2 + 72 #^4& , 3, 0]]))/1954710932452630508961792

Now, we have found that if we first perform the transformation {y -> -(Z^2/(-1 + 2 x))}, the problem is transformed into

Integrate[1/y^11 141891750 (y - Z)^2 Z^3 (-2 y + Z)^2 (y - 2 y^2 + 4 y Z - 3 Z^2)^2 (y - 6 y^2 + 8 y Z - 3 Z^2)^2 (-y + 2 y^2 + Z^2)^2 (6 y^2 + Z^2 - y (1 + 4 Z))^2 Boole[y (2 y^2 + Z^2) < y^2 && y (6 y^2 + Z^2 - y (1 + 4 Z)) > 0 && y (y - 6 y^2 + 8 y Z - 3 Z^2) > 0], {y, 0, 1}, {Z, 0, 1/3}]

Mathematica now performs the double integration (not now requiring the denestSqrt procedure), yielding (LeafCount of 146)

-(1/61046784) 5 (-22008178119328 + 17982847692864 Sqrt[2] + 22896492404940 Sqrt[2] ArcCsc[3] - 45784807319278720 Sqrt[2] ArcCsc[3/Sqrt[2]] +45784807319278720 Sqrt[2] ArcCsc[Sqrt[12 - 6 Sqrt[2]]] - 45839759212195795 Sqrt[2]ArcCsc[6 Sqrt[3/(34 + 5 Sqrt[2] - 4 Sqrt[7] - 8 Sqrt[14])]] - 54951892917075 Sqrt[2] ArcSin[1/9 (4 - Sqrt[7])] +  9158908107195 Sqrt[2] ArcSin[Root[1 - 24 #^2 + 72 #^4& , 3, 0]])

However, both results contain the Root command, perhaps impeding possible simplification along the lines of user yarchik--in applying the FindIntegerNullVector command--in his answer to Establish .

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  • $\begingroup$ Root[1 - 24 #^2 + 72 #^4 &, 3, 0] // ToRadicals results in $\frac{1}{2} \sqrt{\frac{1}{3} \left(2-\sqrt{2}\right)}$. $\endgroup$ – JimB Oct 6 at 0:23
  • $\begingroup$ And all of the ArcSin and ArcCsc terms simplify to linear combinations of $\pi$ and ArcCos[1/3] using the FindIntegerNullVector function. $\endgroup$ – JimB Oct 6 at 0:25
  • $\begingroup$ Terrific--JimB! Curious as to how a final, simplified form will compare/pertain to my conjecture of (5 (-669821158492 + 1393713065745 Sqrt[2] ArcCot[2 Sqrt[2]]))/5308416 ≈0.0766949. $\endgroup$ – Paul B. Slater Oct 6 at 1:15
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    $\begingroup$ Using the rule ArcCot[2 Sqrt[2]] -> (\[Pi] - 2 ArcCos[1/3])/2, then your equation transforms to $-\frac{837276448115}{1327104}+\frac{86031670725 \pi }{65536 \sqrt{2}}-\frac{86031670725 \cos ^{-1}\left(\frac{1}{3}\right)}{32768 \sqrt{2}}$. $\endgroup$ – JimB Oct 6 at 3:17
  • $\begingroup$ I'm sure I'm being stupid here, JimB--but while I find ArcCot[2 Sqrt[2]] -> ([Pi] - 2 ArcCos[1/3])/2 in the immediately preceding comment directly copyable into a notebook, the second expression I do not, and I have to proceed piecemeal. If I click on it, a surrounding box appears, which I do not know how to effectively deal with. $\endgroup$ – Paul B. Slater Oct 7 at 15:54

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