Re: Re: Integrate question

• To: mathgroup at smc.vnet.net
• Subject: [mg82428] Re: [mg82413] Re: Integrate question
• From: DrMajorBob <drmajorbob at bigfoot.com>
• Date: Sat, 20 Oct 2007 05:47:40 -0400 (EDT)
• References: <200710160728.DAA08846@smc.vnet.net> <ff4hpp\$k0e\$1@smc.vnet.net> <ff77b9\$ncr\$1@smc.vnet.net> <1870936.1192798385701.JavaMail.root@m35>
• Reply-to: drmajorbob at bigfoot.com

```Acceptable in what sense? The result is exactly the same.

Bobby

On Fri, 19 Oct 2007 04:09:09 -0500, Oskar Itzinger <oskar at opec.org> 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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