caa a22 4500 463177843 CHVBK 20180406164830.0 cr unu---uuuuu 170326e20070801xx s 000 0 eng 10.1007/s11139-006-9003-7 doi (NATIONALLICENCE)springer-10.1007/s11139-006-9003-7 Allouche J.-P CNRS, LRI, Bâtiment 490, F-91405, Orsay Cedex, France aut Note on an integral of Ramanujan [Elektronische Daten] [J.-P. Allouche] We answer a question of Berndt and Bowman, asking whether it is possible to deduce the value of the Ramanujan integral I from the value of the Ramanujan integral J, where $$I := \int_0^1 \bigg(\frac{x^{p-1}}{1-x} - \frac{rx^{q-1}}{1-x^r} \bigg) dx \ \ \ (= \psi(q/r) - \psi(p) + \log r)$$ and $$J := \int_0^{\infty} \frac{(1+ax)^{-p} - (1+bx)^{-q}}{x} dx \ \ \ \bigg(\!\!= \psi(q) - \psi(p) + \log \frac{b}{a}\bigg).$$ We also show that the second integral can be deduced from a classical expression of the ψ function due to Dirichlet and from the classical equality $$\int_0^{\infty} (e^{-ax} - e^{-bx}) \frac{dx}{x} = \log \frac{b}{a},$$ which is a simple consequence of Frullani-Cauchy's theorem. Springer Science + Business Media, LLC, 2006 Ramanujan integral nationallicence Frullani-Cauchy theorem nationallicence The Ramanujan Journal Kluwer Academic Publishers-Plenum Publishers; http://www.springer-ny.com 14/1(2007-08-01), 39-42 1382-4090 14:1<39 2007 14 11139 https://doi.org/10.1007/s11139-006-9003-7 text/html Onlinezugriff via DOI 1 research-article jats NATIONALLICENCE 856 40 https://doi.org/10.1007/s11139-006-9003-7 text/html Onlinezugriff via DOI NATIONALLICENCE 100 1- Allouche J.-P CNRS, LRI, Bâtiment 490, F-91405, Orsay Cedex, France aut NATIONALLICENCE 773 0- The Ramanujan Journal Kluwer Academic Publishers-Plenum Publishers; http://www.springer-ny.com 14/1(2007-08-01), 39-42 1382-4090 14:1<39 2007 14 11139 Metadata rights reserved Springer special CC-BY-NC licence nationallicence BK010053 XK010053 XK010000 NATIONALLICENCE NATIONALLICENCE NL-springer