Re: Integrate question

*To*: mathgroup at smc.vnet.net*Subject*: [mg82322] Re: [mg82250] Integrate question*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>*Date*: Wed, 17 Oct 2007 04:12:34 -0400 (EDT)*References*: <200710160728.DAA08846@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>