Re: Integrate question
- To: mathgroup at smc.vnet.net
- Subject: [mg82413] Re: Integrate question
- From: "Oskar Itzinger" <oskar at opec.org>
- Date: Fri, 19 Oct 2007 05:09:09 -0400 (EDT)
- Organization: Tele2UTA Telecommunications GmbH
- References: <200710160728.DAA08846@smc.vnet.net> <ff4hpp$k0e$1@smc.vnet.net> <ff77b9$ncr$1@smc.vnet.net>
Would setting u=3 x^2 - 1 and integrating (1/6)/u^3 on [-1,2] be acceptable? /oskar "Oskar Itzinger" <oskar at opec.org> wrote in message news:ff77b9$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 > > > > > > > > > > >
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- From: "Oskar Itzinger" <oskar@opec.org>
- Integrate question