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