Re: Re: Integrate question

*To*: mathgroup at smc.vnet.net*Subject*: [mg82362] Re: [mg82295] Re: Integrate question*From*: DrMajorBob <drmajorbob at bigfoot.com>*Date*: Thu, 18 Oct 2007 04:58:43 -0400 (EDT)*References*: <ff1pru$924$1@smc.vnet.net> <23590028.1192651562452.JavaMail.root@m35>*Reply-to*: drmajorbob at bigfoot.com

> The answer 1/16 is correct in the sense of Cauchy principal value. No, I don't think so. Integrate[x/(3 x^2 - 1)^3, {x, 0, 1}, PrincipalValue -> True] Integrate::idiv: Integral of x/(<<1>>)^3 does not converge on {0,1}. \ >> Integrate[x/(-1 + 3 x^2)^3, {x, 0, 1}, PrincipalValue -> True] Testing it numerically, I get: cauchy[epsilon_?Positive] := Integrate[x/(3 x^2 - 1)^3, {x, 0, Sqrt[1/3] - epsilon}] + Integrate[x/(3 x^2 - 1)^3, {x, Sqrt[1/3] + epsilon, 1}] cauchy /@ (3.^-Range[20]) {-0.0233868, -0.158072, -0.588358, -1.8865, -5.78329, -17.4737, \ -52.526, -157.163, -457.048, -978.052, 7680.08, 309458., 8.64827*10^6, 2.33306*10^8, 6.21267*10^9, 1.60783*10^11, 3.83616*10^12, 7.35457*10^13, 8.58974*10^14, 4.49786*10^15} Bobby On Wed, 17 Oct 2007 02:58:36 -0500, 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. 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 > > However, it would have been reasonable to expect that setting the option > PrincipalValue to True would even more easily give the desired value. But > that would be "reasonable to expect" only if one had not carefully read > the > documentation for PrincipalValue: > > "Setting PrincipalValue->True gives finite answers for integrals that had > simple pole divergences with PrincipalValue->False" > > Note the crucial word "simple". The pole at x = 1/Sqrt[3] in not simple, > and thus, after having read the documentation, it should not be a > surprise > that > > Integrate[x/(3 x^2 - 1)^3, {x, 0, 1}, PrincipalValue -> True] > > also causes version 5.2 to complain that the integral diverges. > > I consider the restriction of PrincipalValue to simple poles to be > unfortunate. Is it still that way in version 6? If so, I hope that the > utility of PrincipalValue will be broadened in some future version. > > David W. Cantrell > > -- DrMajorBob at bigfoot.com