arXiv Analytics

Sign in

arXiv:1003.2170 [math.CA]AbstractReferencesReviewsResources

New definite integrals and a two-term dilogarithm identity

F. M. S. Lima

Published 2010-03-10, updated 2011-08-29Version 3

Among the several proofs known for $\sum_{n=1}^\infty{1/n^2} = {\pi^2/6}$, the one by Beukers, Calabi, and Kolk involves the evaluation of $\,\int_0^1 {\int_0^1{1/(1-x^2 y^2) \, dx} \, dy}$. It starts by showing that this double integral is equivalent to $\frac34 \sum_{n=1}^\infty{1/n^2}$, and then a non-trivial \emph{trigonometric} change of variables is applied which transforms that integral into $\,{\int \int}_T \: 1 \; du \, dv$, where $T$ is a triangular domain whose area is simply ${\pi^2/8}$. Here in this note, I introduce a hyperbolic version of this change of variables and, by applying it to the above integral, I find exact closed-form expressions for $\int_0^\infty{[\sinh^{-1}{(\cosh{u})}-u] d u}$, $\,\int_{\alpha}^\infty{[u-\cosh^{-1}{(\sinh{u})}] d u}$, and $\,\int_{\,\alpha/2}^\infty{\ln{(\tanh{u})} \: d u}$, where $\alpha = \sinh^{-1}(1)$. From the latter integral, I also derive a two-term dilogarithm identity.

Comments: 10 pages, 1 figure. Accepted for publication: Indagat. Mathematicae (08/29/2011)
Journal: Indagationes Mathematicae 23 (2012) 1-9
Categories: math.CA, math.NT
Subjects: 11M06, 40C10, 33B30
Related articles: Most relevant | Search more
arXiv:1911.01423 [math.CA] (Published 2019-11-04)
Two Definite Integrals That Are Definitely (and Surprisingly!) Equal
arXiv:1603.03547 [math.CA] (Published 2016-03-11)
Two Definite Integrals Involving Products of Four Legendre Functions
arXiv:1210.0041 [math.CA] (Published 2012-09-28, updated 2012-10-19)
Definite Integrals using Orthogonality and Integral Transforms