Re: Re: Integrate question

*To*: mathgroup at smc.vnet.net*Subject*: [mg82428] Re: [mg82413] Re: Integrate question*From*: DrMajorBob <drmajorbob at bigfoot.com>*Date*: Sat, 20 Oct 2007 05:47:40 -0400 (EDT)*References*: <200710160728.DAA08846@smc.vnet.net> <ff4hpp$k0e$1@smc.vnet.net> <ff77b9$ncr$1@smc.vnet.net> <1870936.1192798385701.JavaMail.root@m35>*Reply-to*: drmajorbob at bigfoot.com

Acceptable in what sense? The result is exactly the same. Bobby On Fri, 19 Oct 2007 04:09:09 -0500, Oskar Itzinger <oskar at opec.org> wrote: > Would setting > > u=3 x^2 - 1 > > and integrating > > (1/6)/u^3 on [-1,2] > > be acceptable? > > /oskar > > "Oskar Itzinger" <oskar at opec.org> wrote in message > news:ff77b9$ncr$1 at smc.vnet.net... >> Hmm, from Mathematica 2.1 Help: >> >> Integrate can evaluate definite integrals whenever the correct result >> can > be >> found by taking limits >> of the indefinite form at the endpoints. >> >> ? >> >> /oskar >> >> "Andrzej Kozlowski" <akoz at mimuw.edu.pl> wrote in message >> news:ff4hpp$k0e$1 at smc.vnet.net... >> > >> > On 16 Oct 2007, at 16:28, Oskar Itzinger wrote: >> > >> > > Mathematica 5.2 under IRIX complains that >> > > >> > > Integrate[x/(3 x^2 - 1)^3,{x,0,1}] >> > > >> > > doesn't converge on [0,1]. >> > > >> > > However, Mathematica 2.1 under Windows gives the corrrect answer, >> > > (1/16). >> > > >> > > When did Mathematica lose the ability to do said integral? >> > > >> > > Thanks. >> > > >> > > >> > > >> > >> > >> > The reason is that Mathematica 2.1 was wrong and Mathematica 5.2 is >> > much more careful and right. What Mathematica 2.1 did here was simply: >> > >> > Subtract @@ (Integrate[x/(3 x^2 - 1)^3, x] /. {{x -> 1}, {x -> 0}}) >> > 1/16 >> > >> > in other words, it applied the Newton-Leibnitz rule in a mindless >> > way. Later versions are more intelligent and see that the singularity >> at >> > =CE=B1 = Last[x /. Solve[3*x^2 - 1 == 0, x]] >> > 1/Sqrt[3] >> > >> > One can also see this graphically (of course!): >> > >> > Plot[x/(3 x^2 - 1)^3, {x, 0, 1}] >> > >> > >> > the integral still might exist in the sense of Cauchy PrincipalValue >> > but we see that it does not: >> > >> > Integrate[x/(3*x^2 - 1)^3, {x, 0, 1}, PrincipalValue -> True] >> > Integrate::idiv:Integral of x/(3 x^2-1)3 does not converge on {0,1}. >> >> >> > Integrate[x/(3*x^2 - 1)^3, {x, 0, 1}, PrincipalValue -> True] >> > >> > If you still don't beleive it, you can do it "by hand": >> > >> > int = FullSimplify[Integrate[x/(3*x^2 - 1)^3, >> > {x, 0, 1/Sqrt[3] - =CE=B5}] + >> > Integrate[x/(3*x^2 - 1)^3, >> > {x, 0, 1/Sqrt[3] + =CE=B5}], =CE=B5 > 0] >> > (9*=CE=B5^2*(3*=CE=B5^4 - 8*=CE=B5^2 + 5) - 4)/ >> > (18*(3*=CE=B5^3 - 4*=CE=B5)^2) >> > >> > Limit[int, =CE=B5 -> 0] >> > -=E2=88=9E >> > >> > >> > Andrzej Kozlowski >> > >> > >> > >> > >> >> >> > > > > -- DrMajorBob at bigfoot.com

**References**:**Integrate question***From:*"Oskar Itzinger" <oskar@opec.org>