generalization of (Rogers) dilogarithm

Question

Let $C$ and $S$ be abbreviations for $\cosh$ and $\sinh$, and consider the following function:

$$f(x,y) = \int_{-y\le r+l \le y} \frac{ (C(x)S(l)C(r) - C(l)S(r))(C(x)C(l)S(r)-S(l)C(r)) } {(C(x)C(l)C(r) - S(l)S(r))^2-1} dl dr$$

If $y=\infty$, this specializes (I think!) to $4\mathcal{L}(1/C^2(x/2))$ where $\mathcal{L}$ is the Rogers dilogarithm (maybe some constants and factors are missing). The question is whether the function $f$ is studied anywhere. References would be appreciated.

Note: This function arises as the volume of a certain region in the unit tangent bundle of a hyperbolic surface; therefore I am not looking for an answer which just translates it back into its geometric origin.

Answer 1

There are very similar integrals (coming from the same source, shockingly) in arXiv:1002.1905 (Bridgeman/Kahn), where they seem to be evaluated in closed form.

Answer 2

This integral is expressible in terms of elementary functions. At least the corresponding indefinite integral. I won't say anything about convergence.

1. Use hyperbolic double angle formulas to express all hyperbolic functions of $r$ and $l$ in terms of $\cosh(r\pm l)$ and $\sinh(r\pm l)$.
2. Use the new integration variables $Z_\pm$, where $\sinh(r\pm l)=(Z_\pm-Z_\pm^{-1})/2$ and $\cosh(r\pm l)=(Z_\pm+Z_\pm^{-1})/2$.
3. The resulting integral is now rational in $Z_\pm$ with bounds $0\le Z_-\le\infty$ and $Z_+(-y)\le Z_+\le Z_+(y)$.
4. The pole structure becomes especially nice. Decompose into partial fractions wrt $Z_+$ and integrate. Decompose into partial fractions wrt $Z_-$ and integrate again.

Doing this in a Maxima session gives (up to factors of 2): \begin{gather} -{{(C^2(x)+2 C(x)+1) \log { Z_+} \log { Z_-}}\over{4}} \cr -{{(4 C^2(x)-8 C(x)+4) { Z_+}^2 \log { Z_+}+(-C^2 (x)+2 C(x)-1) { Z_+}^4+C^2(x )-2 C(x)+1}\over{32 { Z_+}^2 ( { Z_-}+1)}} \cr +{{(4 C^2(x)-8 C(x )+4) { Z_+}^2 \log { Z_+}+(-C^2(x )+2 C(x)-1) { Z_+}^4+C^2(x)- 2 C(x)+1}\over{32 { Z_+}^2 ({ Z_-}-1 )}} \end{gather}

Hopefully, I hadn't made any typos on the way to the above answer. In any case, the procedure to evaluate this integral correctly should be about the same.