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Re: Integrate question

  • To: mathgroup at smc.vnet.net
  • Subject: [mg82273] Re: [mg82250] Integrate question
  • From: DrMajorBob <drmajorbob at bigfoot.com>
  • Date: Wed, 17 Oct 2007 03:47:19 -0400 (EDT)
  • References: <29429378.1192522385650.JavaMail.root@m35>
  • Reply-to: drmajorbob at bigfoot.com

Plot that function:

Plot[x/(3 x^2 - 1)^3, {x, 0, 1}]

and you'll see that the integrand is unbounded on both sides of 1/Sqrt[3].  
The pole is of order 3, so no, the integral does NOT converge.

It's possible, of course, that negative areas on one side cancel positive  
on the other, so that the following limit might be finite:

epsilon =.
Integrate[x/(3 x^2 - 1)^3, {x, 0, 1/Sqrt[3] - epsilon}] +
   Integrate[x/(3 x^2 - 1)^3, {x, 1/Sqrt[3] + epsilon, 1}] // Simplify
Limit[%, epsilon -> 0]

(-32 Sqrt[3] + 144 epsilon - 216 epsilon^3 +
  81 epsilon^5)/(144 epsilon (4 - 3 epsilon^2)^2)

-\[Infinity]

But, as you see, it isn't. Here's a partial verification:

Table[
  Integrate[x/(3 x^2 - 1)^3, {x, 0, 1./Sqrt[3] - 10^-n}] +
   Integrate[x/(3 x^2 - 1)^3, {x, 1./Sqrt[3] + 10^-n, 1}], {n, 1, 10}]

{-0.181712, -2.34348, -23.9916, -238.35, -257.148, 2.12135*10^6,
  2.11647*10^9, 1.85626*10^12, 6.40027*10^14, 1.02864*10^16}

and here's another:

Table[
  Quiet@NIntegrate[x/(3 x^2 - 1)^3, {x, 0, 1./Sqrt[3] - 10^-n}] +
   Quiet@NIntegrate[x/(3 x^2 - 1)^3, {x, 1./Sqrt[3] + 10^-n, 1}], {n,
   1, 12}]

{-0.181712, -2.34349, -23.9938, -240.5, -2405.57, -24058.2, -242718., \
-4.5604*10^6, -2.39253*10^9, -2.15502*10^12, -2.15478*10^15, \
-2.15494*10^18}

Even if this limit WERE finite, that wouldn't imply convergence for the 
original integral, since there's no reason to use those particular limits  
on the left and right (equal gaps around the pole). Indeed, we can make 
the integral anything we want, by choosing limits accordingly.

If we want the integral to be Pi, and epsilon1 > 0 is chosen so that the 
right integral is equal to 10^n (any n), we can chose epsilon2 > 0 for the  
left integral to make it equal to Pi - 10^n, which makes the sum equal to  
Pi. A sequence of epsilon1, epsilon2 values chosen this way (both  
converging to zero) make the limit equal to Pi.

So... versions 5.2 and 6 are correct; version 2.1 was wrong.

Bobby

On Tue, 16 Oct 2007 02:28:10 -0500, Oskar Itzinger <oskar at opec.org> 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.
>
>
>
>



-- 

DrMajorBob at bigfoot.com


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