Volume of the regular ideal tetrahedron

Apply a Euclidean similarity of the boundary plane, extended by the same scaling of height, sending the circumcircle of the three finite ideal vertices to the unit circle centered at $0$; this upper half-space isometry fixes $\mathrm{\infty }$. The three vertices become points $p,q,r$ on that circle and remain equilateral. Hence $(0,0)$ lies in the interior of their triangle, so the vertical ray from $(0,0,1)$ to $\mathrm{\infty }$ lies inside the tetrahedron.

On the unit-hemisphere face, draw the perpendicular arcs from $(0,0,1)$ to its three edges and the arcs from $(0,0,1)$ to $p,q,r$. Cone the six resulting sectors to $\mathrm{\infty }$. Because the vertical ray lies in the tetrahedron, the six simplices have disjoint interiors and cover the tetrahedron; their common faces have three-dimensional volume zero. Each simplex has two finite vertices and two ideal vertices. Its three right dihedral angles come from the perpendicular cuts; among its other angles, two equal the corresponding original tetrahedron edge angle $\zeta$ and the third is $\pi \mathrm{/}2-\zeta$. The tetrahedron is regular, so [F2] gives $\zeta =\pi \mathrm{/}3$ for each of the six simplices.

Normalize one simplex by a hyperbolic isometry so its ideal vertices are $\mathrm{\infty }$ and $(\cos \zeta ,\sin \zeta )$ and one finite vertex is $(0,0,1)$. The three right angles place the projected right-angle vertex at $(\cos \zeta ,0)$: the two legs are horizontal and vertical. The angle at the origin is $\zeta$, so the other ideal point has coordinates $(\cos \zeta ,\sin \zeta )$. The projected triangle is the intersection of the half-planes $x\ge 0$, $y\ge 0$, $x\le \cos \zeta$ and $y\le x\tan \zeta$. The unit hemisphere through $(0,0,1)$ has equation $x^2+y^2+t^2=1$, and the tetrahedral side is above it. Thus the simplex is exactly ${\mathcal{O}}_{\zeta }$ from [F3] with $\zeta =\pi \mathrm{/}3$.

By [F3], each of the six simplices has volume $\frac{1}{2}\mathrm{\Lambda }(\pi \mathrm{/}3)$. Their interiors are disjoint and their union is the tetrahedron up to faces of volume zero, so additivity gives $\mathrm{Vol}=6\cdot \frac{1}{2}\mathrm{\Lambda }(\pi \mathrm{/}3)=3\mathrm{\Lambda }(\pi \mathrm{/}3)$. Substitution of the absolutely convergent series in [F4] gives the second displayed expression; absolute convergence follows by comparison with ${\sum }_{n\ge 1}{n}^{-2}$.