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Re: Integrate question
*To*: mathgroup at smc.vnet.net
*Subject*: [mg82349] Re: Integrate question
*From*: "Oskar Itzinger" <oskar at opec.org>
*Date*: Thu, 18 Oct 2007 04:52:04 -0400 (EDT)
*References*: <200710160728.DAA08846@smc.vnet.net> <ff4hpp$k0e$1@smc.vnet.net>
Hmm, from Mathematica 2.1 Help:
Integrate can evaluate definite integrals whenever the correct result can be
found by taking limits
of the indefinite form at the endpoints.
?
/oskar
"Andrzej Kozlowski" <akoz at mimuw.edu.pl> wrote in message
news:ff4hpp$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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