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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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