TITLE: Volume of an ideal simplex in $\mathbb{H}^3$, idea/intuition behind result
QUESTION [11 upvotes]: There is the following result on page 160 of Thurston's book "The Geometry and Topology of Three-Manifolds", as follows.

The volume of an ideal simplex in $\mathbb{H}^3$ with dihedral angles $\alpha$, $\beta$, $\gamma$ is equal to$$\Lambda(\alpha) + \Lambda(\beta) + \Lambda(\gamma),$$where $\Lambda: \mathbb{R} \to \mathbb{R}$ is the Lobachevsky function$$\Lambda(\theta) = -\int_0^\theta \log|2\sin t|\,dt.$$

What is the idea/intuition behind the proof of this result, i.e. why does the volume of an orthoscheme have a formula in terms of the Lobachevsky function? I've read the proof in Thurston's book, but I'd still be interested in hearing what others have to say on this issue.

REPLY [6 votes]: A variation of the other answer: first decompose an ideal tetrahedron into orthoschemes with two ideal vertices and (non-right) angles of the form $\alpha,\pi/2-\alpha, \alpha$. 

These orthoschemes have volume a function of $\alpha$. To compute the variation, truncate the ideal vertices by horospheres meeting the other two vertices. Then as we vary $\alpha$, only one edge contributes to Schlafli's formula, and a simple geometric computation shows that the edge length is $\log(2\sin\alpha)$. Integrating gives the Lobachevsky function.
Note: The relevent version of Schlafli's formula was proved by Hodgson in his thesis. See the Remark on p. 125 of his thesis.