generalization of (Rogers) dilogarithm - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T03:34:20Z http://mathoverflow.net/feeds/question/51343 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/51343/generalization-of-rogers-dilogarithm generalization of (Rogers) dilogarithm Danny Calegari 2011-01-06T21:25:59Z 2011-05-13T12:22:19Z <p>Let $C$ and $S$ be abbreviations for $\cosh$ and $\sinh$, and consider the following function:</p> <p>$$f(x,y) = \int_{-y\le r+l \le y} \frac{ (C(x)S(l)C(r) - C(l)S(r))(C(x)C(l)S(r)-S(l)C(r)) } {(C(x)C(l)C(r) - S(l)S(r))^2-1} dl dr$$</p> <p>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.</p> <p>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.</p> http://mathoverflow.net/questions/51343/generalization-of-rogers-dilogarithm/51355#51355 Answer by Igor Rivin for generalization of (Rogers) dilogarithm Igor Rivin 2011-01-06T23:26:53Z 2011-01-06T23:47:23Z <p>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.</p> http://mathoverflow.net/questions/51343/generalization-of-rogers-dilogarithm/57332#57332 Answer by Igor Khavkine for generalization of (Rogers) dilogarithm Igor Khavkine 2011-03-04T09:50:09Z 2011-03-04T09:50:09Z <p>This integral is expressible in terms of elementary functions. At least the corresponding indefinite integral. I won't say anything about convergence.</p> <ol> <li>Use hyperbolic double angle formulas to express all hyperbolic functions of $r$ and $l$ in terms of $\cosh(r\pm l)$ and $\sinh(r\pm l)$.</li> <li>Use the new integration variables $Z_\pm$, where $\sinh(r\pm l)=(Z_\pm-Z_\pm^{-1})/2$ and $\cosh(r\pm l)=(Z_\pm+Z_\pm^{-1})/2$.</li> <li>The resulting integral is now rational in $Z_\pm$ with bounds $0\le Z_-\le\infty$ and $Z_+(-y)\le Z_+\le Z_+(y)$.</li> <li>The pole structure becomes especially nice. Decompose into partial fractions wrt $Z_+$ and integrate. Decompose into partial fractions wrt $Z_-$ and integrate again.</li> </ol> <p>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}</p> <p>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.</p>