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Re: Integrate question

  • To: mathgroup at smc.vnet.net
  • Subject: [mg82308] Re: Integrate question
  • From: "Jean-Marc Gulliet" <jeanmarc.gulliet at gmail.com>
  • Date: Wed, 17 Oct 2007 04:05:15 -0400 (EDT)
  • References: <ff1pru$924$1@smc.vnet.net> <4714D9C2.5090506@gmail.com>

Jean-Marc Gulliet wrote:

> 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?
>
> FWIW, Mathematica for Windows 5.2 as well as 6.0.1 cannot do it either,
> although it does not seem that hard to get the correct answer (the
> indefinite integral and the limits are evaluated in a breeze).
>
> In[1]:= int = Integrate[x/(3 x^2 - 1)^3, x]
>
> Out[1]= -(1/(12 (-1 + 3 x^2)^2))
>
> In[2]:= Limit[int, x -> 1] - int /. x -> 0
>
> Out[2]= 1/16

Sorry to have posted this nonsense (I was too quick to answer without
thinking). Indeed, Mathematica 2.1 is wrong (as I was in my previous
reply) in claiming that the definite integral is equal to 1/16, for
the integrand has a non removable discontinuity at 1/Sqrt[3] (about
0.55) and each integral are divergent on the interval {0,1/Sqrt[3]}
(-Infinity) and {1/Sqrt[3],1} (Infinity).

In[1]:= Plot[x/(3 x^2 - 1)^3, {x, 0, 1}]
Solve[12 (-1 + 3 x^2)^2 == 0, x]
% // N
Integrate[x/(3 x^2 - 1)^3, {x, 0, 1/Sqrt[3]}]
Integrate[x/(3 x^2 - 1)^3, {x, 1/Sqrt[3], 1}]

Out[1]= [ ... graphic deleted ... ]

Out[2]= {{x -> -(1/Sqrt[3])}, {x -> -(1/Sqrt[3])}, {x -> 1/Sqrt[
   3]}, {x -> 1/Sqrt[3]}}

Out[3]= {{x -> -0.57735}, {x -> -0.57735}, {x -> 0.57735}, {x ->
   0.57735}}

During evaluation of In[1]:= Integrate::idiv: Integral of x/(-1+3 \
x^2)^3 does not converge on {0,1/Sqrt[3]}. >>

Out[4]= Integrate[x/(-1 + 3*x^2)^3, {x, 0, 1/Sqrt[3]}]

During evaluation of In[1]:= Integrate::idiv: Integral of x/(-1+3 \
x^2)^3 does not converge on {1/Sqrt[3],1}. >>

Out[5]= Integrate[x/(-1 + 3*x^2)^3, {x, 1/Sqrt[3], 1}]

Sorry for the confusion,
-- 
Jean-Marc


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