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Re: Integrate question

  • To: mathgroup at smc.vnet.net
  • Subject: [mg82348] Re: Integrate question
  • From: "David W.Cantrell" <DWCantrell at sigmaxi.net>
  • Date: Thu, 18 Oct 2007 04:51:34 -0400 (EDT)
  • References: <ff1pru$924$1@smc.vnet.net> <ff4g60$f0f$1@smc.vnet.net>

"David W.Cantrell" <DWCantrell at sigmaxi.net> wrote:
> "Oskar Itzinger" <oskar at opec.org> wrote:
> > Mathematica 5.2 under IRIX complains that
> >
> > Integrate[x/(3 x^2 - 1)^3,{x,0,1}]
> >
> > doesn't converge on [0,1].
>
> Actually, that's good! The integral doesn't converge. The integrand has a
> pole at x = 1/Sqrt[3].
>
> > However, Mathematica 2.1 under Windows gives the corrrect answer,
> > (1/16).
>
> The answer 1/16 is correct in the sense of Cauchy principal value.

Sorry!! That's false, as at least two respondents have already pointed out.
Although regularizing the integral gives 1/16, as mentioned by Daniel
Lichtblau, that is not a Cauchy principal value. Consequently, the next two
times I mention "Cauchy principal value", it should actually have been
"regularized value" instead.

> But I
> think that the behavior of version 5.2 is better, that is, if an integral
> doesn't converge in the usual sense, complain that it doesn't converge,
> rather than giving a Cauchy principal value which was not specifically
> requested.
>
> > When did Mathematica lose the ability to do said integral?
>
> Of course you can get the Cauchy principal value easily in version 5.2.
> The quick and dirty method is to get the antiderivative and then just use
> Newton-Leibniz:
>
> In[3]:= Integrate[x/(3 x^2 - 1)^3, x]
>
> Out[3]= -(1/(12*(-1 + 3*x^2)^2))
>
> In[4]:= (% /. x - > 1) - (% /. x -> 0)
>
> Out[4]= 1/16

Unfortunately, I had overlooked the even easier way, mentioned by Daniel,
for getting the regularized value: using GenerateConditions -> False as an
option in the definite integral.

David W. Cantrell


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