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Re: Integrate question
*To*: mathgroup at smc.vnet.net
*Subject*: [mg82322] Re: [mg82250] Integrate question
*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>
*Date*: Wed, 17 Oct 2007 04:12:34 -0400 (EDT)
*References*: <200710160728.DAA08846@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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