Re: Re: Integrate question
- To: mathgroup at smc.vnet.net
- Subject: [mg82354] Re: [mg82289] Re: [mg82250] Integrate question
- From: DrMajorBob <drmajorbob at bigfoot.com>
- Date: Thu, 18 Oct 2007 04:54:38 -0400 (EDT)
- References: <200710160728.DAA08846@smc.vnet.net> <31324488.1192621144501.JavaMail.root@m35>
- Reply-to: drmajorbob at bigfoot.com
I wonder if this is what you mean by "regularized integral":
indef[x_] = Integrate[x/(3 x^2 - 1)^3, x]
indef[1] - indef[0]
-(1/(12 (-1 + 3 x^2)^2))
1/16
But that's meaningless, isn't it, since "indef" isn't continuous on the
range?
Plot[indef[x], {x, 0, 1}]
Bobby
On Wed, 17 Oct 2007 02:55:32 -0500, Daniel Lichtblau <danl at wolfram.com>
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?
>>
>> Thanks.
>
> There is a nonintegrable singularity at 1/Sqrt[3]. You can check this
> via Series.
>
> In[6]:= Series[x/(3*x^2 - 1)^3, {x,1/Sqrt[3],2}]
>
> 1 1 1
> Out[6]= -------------------- - ---------------------------- + ----------
> -
> 1 3 1 2 32 Sqrt[3]
> 72 (-(-------) + x) 48 Sqrt[3] (-(-------) + x)
> Sqrt[3] Sqrt[3]
>
> 1 1 2
> 5 (-(-------) + x) 9 Sqrt[3] (-(-------) + x)
> Sqrt[3] Sqrt[3]
> 1 3
> > ------------------ + --------------------------- + O[-(-------)
> + x]
> 128 256 Sqrt[3]
>
> To get a regularized integral you can use GenerateConditions->False.
>
> In[7]:= Integrate[x/(3*x^2 - 1)^3, {x,0,1}, GenerateConditions->False]
>
> 1
> Out[7]= --
> 16
>
> I realize this is an overloaded option and expect that in some future
> version there will be a cleaner way to request regularized results of
> definite integration.
>
>
> Daniel Lichtblau
> Wolfram Research
>
>
--
DrMajorBob at bigfoot.com
- References:
- Integrate question
- From: "Oskar Itzinger" <oskar@opec.org>
- Integrate question