# Formally confirm a formula for a certain three-dimensional constrained integral over the unit cube

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,

MathematicaFormulation

this seems quite daunting (very large intermediate outputs). Some simple attempts at transformation of variables did not seem initially productive.

Some ten years ago,

2009paper

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.".)

A025172sequence

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."

"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.

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 $$\begin{equation} (2 x+1)^{16} (y+1)^{15}. \end{equation}$$

The answer given by JimB $$\begin{equation} \frac{29902415923}{497664}-\frac{50274109}{512 \sqrt{2}}-\frac{3072529845 \pi }{32768 \sqrt{2}}+\frac{1024176615 \cos ^{-1}\left(\frac{1}{3}\right)}{4096 \sqrt{2}} \end{equation}$$ or in Mathematica notation,
29902415923/497664 - 50274109/(512 Sqrt) - (3072529845 π)/(32768 Sqrt) +(1024176615 ArcCos[1/3])/(4096 Sqrt)
`