The result of the three-dimensional constrained integration (for the Hilbert-Schmidt two-qubit absolute separability probability) over the unit cube $[0,1]^3$ \begin{equation} \label{one} \int_0^1 \int_0^1 \int_0^1 9081072000 \left(\lambda _1-\lambda _2\right){}^2 \left(\lambda _1-\lambda _3\right){}^2 \left(\lambda _2-\lambda _3\right){}^2 \left(2 \lambda _1+\lambda _2+\lambda _3-1\right){}^2 \left(\lambda _1+2 \lambda _2+\lambda _3-1\right){}^2 \left(\lambda _1+\lambda _2+2 \lambda _3-1\right){}^2 \text{Boole}\left[\lambda _1>\lambda _2\land \lambda _2>\lambda _3\land \lambda _3>-\lambda _1-\lambda _2-\lambda _3+1\land \lambda _1-\lambda _3<2 \sqrt{\lambda _2 \left(-\lambda _1-\lambda _2-\lambda _3+1\right)}\right] \mbox{d} \lambda_3 \mbox{d} \lambda_2 \mbox{d} \lambda_1 \end{equation} can be expressed (QuantumComputing) 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}
Can a formal derivation of this result be given (yielding possibly a re-expressed formula of interest)?
As a Mathematica question,
this seems quite daunting (very large intermediate outputs). Some simple attempts at transformation of variables did not seem initially productive.
Some ten years ago,
I obtained this result (eq. (34) there), but the now-requested formalized 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 Sqrt2)]
and
ArcTan[(1/7) (9 + 4 Sqrt2)],
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 + Sqrt2)]
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 Sqrt2)]
= 5Pi/4 - 3phi
ArcTan[(1/7) (9 + 4 Sqrt2)],
= 3*Pi/4 - phi.
Best regards, Vladeta"
Within the next week, V. Jovovic also wrote:
" ArcTan[(1/7) (-9 + 4 Sqrt2)]
= Pi/4-phi
ArcTan[(1/7) (-3 + Sqrt2)]
= Pi/8-phi/2 "
" ArcSin[(1/6) (4 + Sqrt2)] = 3*Pi/4 - phi "
and
" ArcCsc[3/17 Sqrt[52 + 14 Sqrt2]]
= 5*Pi/8-phi
ArcTan[7/(3 + Sqrt2)]
= Pi/8+phi/2
ArcTan[1/(3 + Sqrt2)]
= - Pi/8+phi/2
ArcCsc[Sqrt[6 (2 + Sqrt2)]]
= 5*Pi/8-phi
"
Although this email correspondence was clearly central to the obtaining of the indicated formula (for which a formalized proof is requested) it is presently not clear to me in what manner they were employed.
Nicolas Tessore has now shown that the 3D-constrained integral that is the subject of this question can be converted, using the transformation, \begin{equation} \left\{\lambda _1\to \frac{x}{2 x+1},\lambda _2\to \frac{(x+1) y}{(2 x+1) (y+1)},\lambda _3\to \frac{(x+1) z}{(2 x+1) (y+1)}\right\} \end{equation} to an unconstrained form, \begin{equation} \int_{\frac{1}{2}}^1 \int_z^{-z+2 \sqrt{1-z}+2} \int_y^{2 \sqrt{-\frac{y^3 z+2 y^2 z-y^3-2 y^2+y z-y}{(y+z-1)^4}}+\frac{-3 y z+4 y-z^2+z}{(y+z-1)^2}} S \hspace{.1in} \mbox{d} z \mbox{d} y \mbox{d} x, \end{equation} where \begin{equation} S = \end{equation} \begin{equation} 9081072000 (x+1)^8 (1-2 z)^2 (x-y)^2 (y-z)^2 (y+z-1)^2 (x (-y+z-1)+z)^2 (x (y+z)+z-1)^2 \end{equation} divided by \begin{equation} (2 x+1)^{16} (y+1)^{15}. \end{equation}