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}