Re: Integrate question

*To*: mathgroup at smc.vnet.net*Subject*: [mg82348] Re: Integrate question*From*: "David W.Cantrell" <DWCantrell at sigmaxi.net>*Date*: Thu, 18 Oct 2007 04:51:34 -0400 (EDT)*References*: <ff1pru$924$1@smc.vnet.net> <ff4g60$f0f$1@smc.vnet.net>

"David W.Cantrell" <DWCantrell at sigmaxi.net> wrote: > "Oskar Itzinger" <oskar at opec.org> wrote: > > Mathematica 5.2 under IRIX complains that > > > > Integrate[x/(3 x^2 - 1)^3,{x,0,1}] > > > > doesn't converge on [0,1]. > > Actually, that's good! The integral doesn't converge. The integrand has a > pole at x = 1/Sqrt[3]. > > > However, Mathematica 2.1 under Windows gives the corrrect answer, > > (1/16). > > The answer 1/16 is correct in the sense of Cauchy principal value. Sorry!! That's false, as at least two respondents have already pointed out. Although regularizing the integral gives 1/16, as mentioned by Daniel Lichtblau, that is not a Cauchy principal value. Consequently, the next two times I mention "Cauchy principal value", it should actually have been "regularized value" instead. > But I > think that the behavior of version 5.2 is better, that is, if an integral > doesn't converge in the usual sense, complain that it doesn't converge, > rather than giving a Cauchy principal value which was not specifically > requested. > > > When did Mathematica lose the ability to do said integral? > > Of course you can get the Cauchy principal value easily in version 5.2. > The quick and dirty method is to get the antiderivative and then just use > Newton-Leibniz: > > In[3]:= Integrate[x/(3 x^2 - 1)^3, x] > > Out[3]= -(1/(12*(-1 + 3*x^2)^2)) > > In[4]:= (% /. x - > 1) - (% /. x -> 0) > > Out[4]= 1/16 Unfortunately, I had overlooked the even easier way, mentioned by Daniel, for getting the regularized value: using GenerateConditions -> False as an option in the definite integral. David W. Cantrell