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