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Re: Integrate question
Andrzej Kozlowski <akoz at mimuw.edu.pl> wrote:
> It represents the same mathematica object, so you will get the same
> 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
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