The result of the three-dimensional integration
Integrate[9081072000 (Subscript[λ, 1] - Subscript[λ,
2])^2 (Subscript[λ, 1] - Subscript[λ,
3])^2 (Subscript[λ, 2] - Subscript[λ, 3])^2 (-1 +
2 Subscript[λ, 1] + Subscript[λ, 2] +
Subscript[λ, 3])^2 (-1 + Subscript[λ, 1] +
2 Subscript[λ, 2] + Subscript[λ, 3])^2 (-1 +
Subscript[λ, 1] + Subscript[λ, 2] +
2 Subscript[λ, 3])^2 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])]], {Subscript[λ, 3], 0, 1}, {Subscript[λ, 2], 0, 1}, {Subscript[λ, 1], 0, 1}],
that is,
for the two-qubit Hilbert-Schmidt absolute separability probability apparently can be expressed as
\begin{equation} \label{HSabs} \frac{29902415923}{497664}+\frac{-3217542976+5120883075 \pi -16386825840 \tan ^{-1}\left(\sqrt{2}\right)}{32768 \sqrt{2}} = \end{equation} \begin{equation} \frac{32(29902415923 - 24433216974 \sqrt{2})+248874917445 \sqrt{2}(5 \pi - 16 \tan ^{-1}\left(\sqrt{2}\right))}{2^{16} \cdot 3^5} \approx 0.00365826 \end{equation}
QuantumComputingStackExchangeQuestion
Can this be explicitly confirmed using Mathematica?
Through use of the transformation,
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)};
Nicolas Tessore has now reported to me that he was able to convert the 3D integral into an unconstrained one of the form,
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) .
Let me indicate here that the indicated result was obtained in the 2009paper
I obtained this result (eq. (34) there), but the now-requested step-by-step process was not detailed. Comments of present interest there were that `[C]opious use was made of trigonometric identities involving the tetrahedral dihedral angle $\phi=\cos ^{-1}\left(\frac{1}{3}\right)$, assisted by V. Jovovic" and that use was made of the Sloane website sequence A025172--"Let phi = arccos(1/3), the dihedral angle of the regular tetrahedron. Then cos(n*phi) = a(n)/3^n". (This sequence is "[u]sed when showing that the regular simplex is not "scisssors-dissectable" to a cube, thus answering Hilbert's third problem.".)
These comments led me to consult my email archives.
On April 21, 2008 I wrote to Vladeta Jovovic (and also Wouter Meeussen and Neil Sloane) the following:
"Dear Drs. Jovocic/Meeussen/Sloane:
I been doing some analyses in which I've been able to simplify several trigonometric terms using the relation
cos(n*phi) =a(n)/3^n
where phi=ArcCos(1/3)
reported in Sloane's Superseeker sequence
A025172.
I have two further terms ArcTan[(1/329 )(729 - 460 Sqrt[2])]
and ArcTan[(1/7) (9 + 4 Sqrt[2])]
, which also clearly pertain, since 329/729 =a(6)/3^6
and -7/9 =a(2)/3^2
.
But I don't see how to manipulate them to reexpress/simplify them in terms of phi, which I presume is possible/natural.
Perhaps you have some insights in this matter?
Sincerely,
Paul B. Slater
P. S. I also have the term
ArcTan[(1/7) (-3 + Sqrt[2])]
which perhaps also has some simpler form."
I received replies:
"for n from 0 to 10 do q:=tan(-n*phi):print(expand(q));od:
0
1/2
-2 2
1/2
4 2
------
7
1/2
10 2
- -------
23
1/2
56 2
-------
17
1/2
22 2
-------
241
1/2
460 2
- --------
329
1/2
1118 2
---------
1511
1/2
1904 2
- ---------
5983
1/2
13870 2
----------
1633
1/2
10604 2
----------
57113
V.
and
phi=ArcCos(1/3)
ArcTan[(1/329 )(729 - 460 Sqrt[2])]
5Pi/4 - 3phi
ArcTan[(1/7) (9 + 4 Sqrt[2])],
3*Pi/4 - phi.
Best regards, Vladeta"
Within the next week, V. Jovovic also wrote:
ArcTan[(1/7) (-9 + 4 Sqrt[2])]
Pi/4-phi
ArcTan[(1/7) (-3 + Sqrt[2])]
Pi/8-phi/2
ArcSin[(1/6) (4 + Sqrt[2])]
= 3*Pi/4 - phi
and
ArcCsc[3/17 Sqrt[52 + 14 Sqrt[2]]]
5*Pi/8-phi
ArcTan[7/(3 + Sqrt[2])]
Pi/8+phi/2
ArcTan[1/(3 + Sqrt[2])]
- Pi/8+phi/2
ArcCsc[Sqrt[6 (2 + Sqrt[2])]]
5*Pi/8-phi
Although this 2008 email correspondence was clearly central to the obtaining of the indicated formula (for which a Mathematica demonstration is requested), it is presently not clear to me in what manner the results discussed there were obtained and further employed. (Also, apparently this 2008 correspondence was carried on after(!) I had been able to perform the desired 3D integration, and had a result for which some simplification--using the Jovovic transformations--was possible.)
Integrate
byNIntegrate
you'll get the numerical verification...=0.00365826
$\endgroup$Integrate
. $\endgroup$Integrate
but it doesn't evaluate. Additional I checked the integration range (defined by Block) which is composed by three planes. Perhaps splittingIntegrate
helps solving symbolically. $\endgroup$LeafCount[Integrate[the integrand under consideration,{Subscript[\[Lambda], 1], 0, 1}, Assumptions -> Subscript[\[Lambda], 2] >= 0 && Subscript[\[Lambda], 2] <= 1 && Subscript[\[Lambda], 3] >= 0 && Subscript[\[Lambda], 3] <= 1]]
performs 5177. To much for a next integration. $\endgroup$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
in the updatedIntegrate[integrand2...]
statement can be simplified to(4 y + z - 3 y z - z^2 + 2 (1 + y) Sqrt[y - y z])/(-1 + y + z)^2
. $\endgroup$