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Let us order the four points $\lambda_1 \geq \lambda_2 \geq \lambda_3 \geq \lambda_4 \geq 0$ of a 3-simplex, $\lambda_1+\lambda_2+\lambda_3+\lambda_4=1$, giving us a subsection $L$. Integration over $L$ of the (Hilbert-Schmidt) measure ($N=4$) \begin{equation} \Pi_{j<k}^N(\lambda_j-\lambda_k)^2, \end{equation} gives us \begin{equation} \frac{1}{9081072000}=(2^7 \cdot 3^4 \cdot 5^3 \cdot 7^2 \cdot 11 \cdot 13)^{-1}. \end{equation} Now, let us further impose the Verstraete-Audenaert-De Moor ("absolute separability") constraint (ensuring no entanglement possible through unitary transformations), \begin{equation} \lambda_1-\lambda_3- 2 \sqrt{\lambda_2 \lambda_4} \lt 0. \end{equation} In the decade-old paper https://arxiv.org/abs/0805.0267, I reported, "making copious use of trigonometric identities involving the tetrahedral dihedral angle $\phi=\cos ^{-1}\left(\frac{1}{3}\right)$'', assisted by V. Jovovic, that the ratio of this further constrained integration of the Hilbert-Schmidt measure over $L$ to $\frac{1}{9081072000}$ gave ((34) in the cited article) later simplified to, \begin{equation} \label{HSabs} \frac{29902415923}{497664}+\frac{-3217542976+5120883075 \pi -16386825840 \tan ^{-1}\left(\sqrt{2}\right)}{32768 \sqrt{2}}\approx 0.00365826. \end{equation} I would like to be able to reverifiy this result--as the structure of its solution may aid in the further ones below.

For instance, there are many (clearly still more challenging) counterpart problems of interest, in which the Hilbert-Schmidt measure is replaced by ones based on operator-monotone functions. For example, the Bures (minimal monotone) measure (eq. (15.48) in https://pdfs.semanticscholar.org/3f28/893b7e8c5c96525493db8e3d6b09ab47f426.pdf) is proportional to \begin{equation} \frac{1}{(\lambda_1 \ldots \lambda_4)^{1/2}} \Pi_{j<k}^N\frac{(\lambda_j-\lambda_k)^2}{\lambda_j+\lambda_k}, \end{equation} while the Wigner-Yanse operator-monotone measure is proportional to \begin{equation} \frac{1}{(\lambda_1 \ldots \lambda_4)^{1/2}} \Pi_{j<k}^N\frac{(\lambda_j-\lambda_k)^2}{(\sqrt{\lambda_j}+\sqrt{\lambda_k})^{2}}. \end{equation}

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In principle, here is how the 2008 assertion that the two-qubit Hilbert-Schmidt absolute separability probability is \begin{equation} \frac{29902415923}{497664}+\frac{-3217542976+5120883075 \pi -16386825840 \tan ^{-1}\left(\sqrt{2}\right)}{32768 \sqrt{2}}\approx 0.00365826. \end{equation} can be reappraised--which was the objective of the problem posed.

Issue the five-command Mathematica sequence for a two-dimensional integration (for simplicity of presentation, we take $\lambda_1=l1$, $\lambda_2=l2$, $\lambda_3=l3$, $\lambda_4=l4$),

l4 = 1 - l1 - l2 - l3; g={l1,l2,l3,l4};HS = Product[(g[[i]]-g[[j]])^2,{i,1,3},{j,i+1,4}];t = GenericCylindricalDecomposition[ l1 >= l2 && l2 >= l3 && l3 >= l4 && l4 >= 0 && l1 - l3 <= 2 Sqrt[l2 l4], {l3, l2, l1}][[1]]; s = 9081072000 Integrate[HS Boole[t], {l2, 0, 1}, {l1, 0, 1}];

This yields a large (LeafCount=5354) output, now requiring three one-dimensional integrations of functions of l3 over the intervals $[\frac{1}{8} \left(2-\sqrt{2}\right),\frac{1}{6}]$, $[\frac{1}{6},\frac{1}{4}]$ and $[\frac{1}{4},\frac{1}{3}]$. A number of terms involving square roots of square roots in the Mathematica output resist initial integration, but can be simplified to terms involving only a single square root (https://math.stackexchange.com/questions/3375075/transform-certain-integrands-so-that-they-involve-a-single-square-root-rather-th), and then can be integrated out.

However, the resultant output is to begin still very large, involving close to two hundred terms, many with arcsines of various arguments, and will clearly require a very substantial further amount of simplifications and transformations. (As indicated in the statement of the question, in the cited 2008 paper, ``[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.) Further specifics will be provided when and if possible. But is there a more direct, less "tedious" route--and, how if possible, could the clearly still more challenging Bures and Wigner-Yanase counterparts be addressed? Probably not doable!

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