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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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