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Re: Re: Integrate question

  • To: mathgroup at smc.vnet.net
  • Subject: [mg82362] Re: [mg82295] Re: Integrate question
  • From: DrMajorBob <drmajorbob at bigfoot.com>
  • Date: Thu, 18 Oct 2007 04:58:43 -0400 (EDT)
  • References: <ff1pru$924$1@smc.vnet.net> <23590028.1192651562452.JavaMail.root@m35>
  • Reply-to: drmajorbob at bigfoot.com

> The answer 1/16 is correct in the sense of Cauchy principal value.

No, I don't think so.

Integrate[x/(3 x^2 - 1)^3, {x, 0, 1}, PrincipalValue -> True]

Integrate::idiv: Integral of x/(<<1>>)^3 does not converge on {0,1}. \
>>

Integrate[x/(-1 + 3 x^2)^3, {x, 0, 1}, PrincipalValue -> True]

Testing it numerically, I get:

cauchy[epsilon_?Positive] :=
  Integrate[x/(3 x^2 - 1)^3, {x, 0, Sqrt[1/3] - epsilon}] +
   Integrate[x/(3 x^2 - 1)^3, {x, Sqrt[1/3] + epsilon, 1}]
cauchy /@ (3.^-Range[20])

{-0.0233868, -0.158072, -0.588358, -1.8865, -5.78329, -17.4737, \
-52.526, -157.163, -457.048, -978.052, 7680.08, 309458., 8.64827*10^6,
   2.33306*10^8, 6.21267*10^9, 1.60783*10^11, 3.83616*10^12,
  7.35457*10^13, 8.58974*10^14, 4.49786*10^15}

Bobby

On Wed, 17 Oct 2007 02:58:36 -0500, 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. 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
>
> However, it would have been reasonable to expect that setting the option
> PrincipalValue to True would even more easily give the desired value. But
> that would be "reasonable to expect" only if one had not carefully read  
> the
> documentation for PrincipalValue:
>
> "Setting PrincipalValue->True gives finite answers for integrals that had
> simple pole divergences with PrincipalValue->False"
>
> Note the crucial word "simple". The pole at x = 1/Sqrt[3] in not simple,
> and thus, after having read the documentation, it should not be a  
> surprise
> that
>
> Integrate[x/(3 x^2 - 1)^3, {x, 0, 1}, PrincipalValue -> True]
>
> also causes version 5.2 to complain that the integral diverges.
>
> I consider the restriction of PrincipalValue to simple poles to be
> unfortunate. Is it still that way in version 6? If so, I hope that the
> utility of PrincipalValue will be broadened in some future version.
>
> David W. Cantrell
>
>



-- 

DrMajorBob at bigfoot.com


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