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Re: Integrate question

  • To: mathgroup at smc.vnet.net
  • Subject: [mg82514] Re: Integrate question
  • From: "David W.Cantrell" <DWCantrell at sigmaxi.net>
  • Date: Tue, 23 Oct 2007 05:29:56 -0400 (EDT)
  • References: <20071020082209.763$0x_-_@newsreader.com> <13065124.1192936090864.JavaMail.root@m35> <fff57v$f10$1@smc.vnet.net>

DrMajorBob <drmajorbob at bigfoot.com> wrote:
> > principal value sense. But this actually confirms the other point I
> > made: that one cannot exclude the possibility that there is a sense
> > in which the answer 1/16 is "acceptable". However, and this is the
> > key point, the fact that such a sense exists is only a curiosity
> > unless this "sense" fits the intended purpose of this "integration".
>
> I gotta agree with Andrzej,

The way you say that makes it seem as though you think I disagree with
Andrzej about his "key point". But of course I don't, and never have.

> and I'll repeat my earlier comment that you
> can make the integral anything you want by approaching the pole
>
> (in)appropriately from left and right.

But in dealing with Cauchy principal value, one is not at liberty to have
two epsilons which approach the pole differently. Of course, if you do what
you indicate below, you can get any value you wish. Perhaps that could be
called Bobby's arbitrary value method, but that's not Cauchy principal
value. If the latter exists, it is unique.

> If you want the result to be Pi, choose epsilon1 so that the right
> integral is Pi/2 + n and epsilon2 so that the left integral is Pi/2 - n.
> e1 and e2 are not equal, but they both tend to zero as n -> Infinity, and
> the partial integral is Pi for every n.
>
> A version of this is precisely what happened in the change of variables:
> unequal epsilons (original problem) mapped to equal epsilons (new
> problem), so the new problem has a finite Cauchy principal value when the
> original did not. I suspect there's a change of variables that
> accomplishes the same trick for ANY integrand whose antiderivative tends
> to +Infinity on one side of the pole and -Infinity on the other.
>
> Integrals that aren't invariant (even as to the question of convergence)
> under a change of variables?

Then perhaps you would prefer using the regularized integral, mentioned by
Daniel Lichtblau, since its value is invariant.

> Yikes. Have fun with that.

I can't ever remember _using_ either a Cauchy principal value or a
regularized integral myself. I was merely trying to assist the OP in
providing a sense in which the value of the integral could be said to be
1/16. That seemed to be what he wanted. Whether such a sense fits his
intended purpose is a matter which must be left to the OP; he didn't
provide enough information for us to make that assessment for him.

For better or worse, it seems that regularized integrals are used in
physics (quantum mechanics, in particular) etc.

David


> On Sat, 20 Oct 2007 21:21:08 -0500, Andrzej Kozlowski <akoz at mimuw.edu.pl>
> wrote:
>
> > You are right of course, I should have checked it "by hand" (as I did
> > in my original response) before writing it was not true in the Cauchy
> > principal value sense. But this actually confirms the other point I
> > made: that one cannot exclude the possibility that there is a sense
> > in which the answer 1/16 is "acceptable". However, and this is the
> > key point, the fact that such a sense exists is only a curiosity
> > unless this "sense" fits the intended purpose of this "integration".
> > This is why I am usually not very impressed when someone asks a
> > question to which, in the intended context, the answer is "no" and
> > then someone writes to point out that there is some other context in
> > which the answer is "yes". There is a countless number of situations
> > of this kind and it is easy to multiply examples. You can take almost
> > any theorem of elementary analysis and find that in some more
> > abstract context it is not valid. The first example that comes to my
> > mind (just as I am writing this): the familiar and even "obvious"
> > fact that a smooth function is continuous is not true in the context
> > of analysis on locally convex spaces more general than Banach spaces
> > (the Frolicher-Kriegl theory). But this is completely irrelevant to
> > anyone who never deals with anything other than finite dimensional
> > euclidean spaces. Similarly, the fact that this particular integral
> > (after the change of variable) has Cauchy Principal value 1/16 is
> > probably irrelevant - "probably" because the OP never told us what
> > sort of "integral" he had in mind.
> >
> > Andrzej Kozlowski
> >
> >
> > On 20 Oct 2007, at 21:22, David W. Cantrell wrote:
> >
> >> [Message also posted to: comp.soft-sys.math.mathematica]
> >>
> >> Andrzej Kozlowski <akoz at mimuw.edu.pl> wrote:
> >>>   It represents the same mathematica object, so you will get the sam=
> e
> >>> answer:
> >>>
> >>> Integrate[1/6/u^3, {u, -1, 2}]
> >>> Integrate::idiv:Integral of 1/u3 does not converge on {-1,2}. >>
> >>> Integrate[1/(6*u^3), {u, -1, 2}]
> >>>
> >>> What else did you expect? It's the same and therefore 1/16 it is
> >>> equally "unacceptable" or "non-acceptable" as an answer as before.
> >>
> >> Of course, you are correct in saying that 1/16 is unacceptable for
> >> Integrate[1/6/u^3, {u, -1, 2}] since nothing other than Mathematica's=
>
> >> standard integration was specified. But curiously, there is a sense
> >> (see
> >> below) in which 1/16 is now the correct answer, even though it
> >> wasn't for
> >> the original integral in terms of x.
> >>
> >>> The integral does not converge in any usual mathematical sense : as
> >>> Riemann or Lebesgue integral or even in the sense of "Principal
> >>> value".
> >>
> >> No, it _does_ converge in the sense of Cauchy principal value. Of
> >> course,
> >> as I noted before, PrincipalValue in the current version of
> >> Mathematica
> >> doesn't handle anything but _simple_ poles. But the integral in
> >> terms of
> >> u converges to 1/16 in the sense of Cauchy principal value,
> >> although the
> >> original integral in terms of x did not:
> >>
> >> In[11]:= Simplify[Assuming[0 < e < 1,
> >> Integrate[1/(6*u^3), {u, -1, -e}] + Integrate[1/(6*u^3), {u, e, 2}]]]=
>
> >>
> >> Out[11]= 1/16
> >>
> >> David
> >>
> >>
> >>> On 19 Oct 2007, at 18:09, Oskar Itzinger wrote:
> >>>
> >>>> Would setting
> >>>>
> >>>> u=3 x^2 - 1
> >>>>
> >>>> and integrating
> >>>>
> >>>> (1/6)/u^3 on [-1,2]
> >>>>
> >>>> be acceptable?
> >>>>
> >>>> /oskar
> >>>>
> >>>> "Oskar Itzinger" <oskar at opec.org> wrote in message
> >>>> news:ff77b9$ncr$1 at smc.vnet.net...
> >>>>> Hmm, from Mathematica 2.1 Help:
> >>>>>
> >>>>> Integrate can evaluate definite integrals whenever the correct
> >>>>> result can be found by taking limits
> >>>>> of the indefinite form at the endpoints.
> >>>>>
> >>>>> ?
> >>>>>
> >>>>> /oskar
> >>>>>
> >>>>> "Andrzej Kozlowski" <akoz at mimuw.edu.pl> wrote in message
> >>>>> news:ff4hpp$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
> >
> >
> >
>
> -- =
>
> DrMajorBob at bigfoot.com


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