Re: Integrate question

*To*: mathgroup at smc.vnet.net*Subject*: [mg82470] Re: Integrate question*From*: "David W.Cantrell" <DWCantrell at sigmaxi.net>*Date*: Sat, 20 Oct 2007 22:19:05 -0400 (EDT)*References*: <200710160728.DAA08846@smc.vnet.net> <ff4hpp$k0e$1@smc.vnet.net> <ff77b9$ncr$1@smc.vnet.net> <200710190909.FAA05818@smc.vnet.net> <ffcj1d$s2h$1@smc.vnet.net>

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

**References**:**Integrate question***From:*"Oskar Itzinger" <oskar@opec.org>

**Re: Integrate question***From:*"Oskar Itzinger" <oskar@opec.org>

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