Re: cauchy principal value double integral

*To*: mathgroup at smc.vnet.net*Subject*: [mg128966] Re: cauchy principal value double integral*From*: Roland Franzius <roland.franzius at uos.de>*Date*: Thu, 6 Dec 2012 04:56:53 -0500 (EST)*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com*Delivered-to*: l-mathgroup@wolfram.com*Delivered-to*: mathgroup-newout@smc.vnet.net*Delivered-to*: mathgroup-newsend@smc.vnet.net*References*: <k9hn9d$jg$1@smc.vnet.net> <k9kei0$66i$1@smc.vnet.net>

Am 04.12.2012 10:08, schrieb daniel.lichtblau0 at gmail.com: > On Monday, December 3, 2012 2:18:53 AM UTC-6, Alex Krasnov wrote: >> I am unclear how to interpret the following result (Mathematica 8.0.4): >> >> >> >> In: Integrate[1/(x^2+y^2), {x, -1, 1}, {y, -1, 1}, PrincipalValue -> True] >> >> Out: -4*Catalan >> >> >> >> How is Cauchy principal value defined in this case? Since the integrand is >> >> circularly symmetric around (0,0), excluding a shrinking neighborhood >> >> around (0,0) is not useful. Perhaps this result is merely an issue with >> >> option handling for multiple integrals? >> >> >> >> Alex > > That does indeed look like a bug. > The integral doesnt exist and what a Cauchy pincipal value option in two dimensions may mean to the Integrate subsystem remains obscure. In polar coordinates the integrand is dx /\ dy /(x^2+y^2) = dr /\ dphi /r Any integral containing the origin diverges logarithmically. -- Roland Franzius