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MathGroup Archive 2007

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Re: elliptic integral (reloaded!)

  • To: mathgroup at smc.vnet.net
  • Subject: [mg75894] Re: [mg75851] elliptic integral (reloaded!)
  • From: DrMajorBob <drmajorbob at bigfoot.com>
  • Date: Sun, 13 May 2007 05:37:35 -0400 (EDT)
  • References: <1673653.1178955431279.JavaMail.root@m35>
  • Reply-to: drmajorbob at bigfoot.com

> In[9]:=
> o = HoldForm[Integrate[Sqrt[(1 - x)/((x - 2)*(x^2 - 2*x + 3))], {x, 1,
> 2}]]

The result of entering

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

in version 6 is

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

HoldForm explicitly says NOT to do any evaluation, so version 5 and 6  
don't do any. They don't even remove the HoldForm wrapper.

If version 4 did evaluate the integral for that statement... it shouldn't  
have.

If I enter

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

in version 6, I get

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

(no actual result) for a different reason... because Mathematica can't  
integrate it.

I suspect, when true experts weigh in, they may say the version 4 answer

is incorrect. I suggest you test that theory, if you can. Or maybe THAT 
 

result is correct, but the version 4 method responsible for getting that
  

one right sometimes got OTHER answers wrong.

Bobby

On Sat, 12 May 2007 01:59:32 -0500, dimitris <dimmechan at yahoo.com> wrote:

> I guess saying many things prevent forumists from answering?!
>
> Anyway...
>
> In[9]:=
> o = HoldForm[Integrate[Sqrt[(1 - x)/((x - 2)*(x^2 - 2*x + 3))], {x, 1,
> 2}]]
>
> Version 4. succeeds in getting a closed form result.
> Version 5.2 returns the integral unevalueated.
>
> I just want to know what version 6 does!
>
> Thanks a lot!
>
> Dimitris
>
>
>



-- 

DrMajorBob at bigfoot.com


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