Re: elliptic integral (reloaded!)
- To: mathgroup at smc.vnet.net
- Subject: [mg75933] Re: elliptic integral (reloaded!)
- From: dimitris <dimmechan at yahoo.com>
- Date: Sun, 13 May 2007 05:57:41 -0400 (EDT)
- References: <f23ome$n1h$1@smc.vnet.net>
If I was not clear let be a little more specific!
I mean
o//ReleaseHold
produce a result in version 4 which having been checked numerically
agrees with the relevant numerical estimation (by NIntegrate).
On the other hand
o//ReleaseHold
returns the integral unevaluated in version 5.2.
David Cantrell proposed the following workaround in version 5.2.
In[2635]:=
Integrate[(x - 1)/Sqrt[(1 - x)*(x - 2)*(x^2 - 2*x + 3)], {x, 1, 3/2}]
+ Integrate[Sqrt[(1 - x)/((x - 2)*(x^2 - 2*x + 3))],
{x, 3/2, 2}]//InputForm
Out[2635]//InputForm=
(8*Sqrt[4 - I*Sqrt[2]]*EllipticF[ArcSin[Sqrt[5*I + Sqrt[2]]/2^(3/4)],
(2*Sqrt[2])/(2*I + Sqrt[2])] -
(7*I)*Sqrt[8 - (2*I)*Sqrt[2]]*EllipticF[ArcSin[Sqrt[5*I + Sqrt[2]]/
2^(3/4)], (2*Sqrt[2])/(2*I + Sqrt[2])] +
(12*I)*Sqrt[4 + I*Sqrt[2]]*EllipticF[I*(Log[2] - Log[Sqrt[2 -
(2*I)*Sqrt[2]] + I*2^(1/4)*Sqrt[2*I + Sqrt[2]]]),
(2*Sqrt[2])/(2*I + Sqrt[2])] - 3*Sqrt[8 + (2*I)*Sqrt[2]]*
EllipticF[I*(Log[2] - Log[Sqrt[2 - (2*I)*Sqrt[2]] +
I*2^(1/4)*Sqrt[2*I + Sqrt[2]]]), (2*Sqrt[2])/(2*I + Sqrt[2])] +
9*Sqrt[2*(-8 + (7*I)*Sqrt[2])]*EllipticPi[(2*Sqrt[2])/(-I +
Sqrt[2]), ArcSin[Sqrt[5*I + Sqrt[2]]/2^(3/4)],
(2*Sqrt[2])/(2*I + Sqrt[2])] - 9*Sqrt[2*(-8 +
(7*I)*Sqrt[2])]*EllipticPi[(2*Sqrt[2])/(-I + Sqrt[2]),
I*(Log[2] - Log[Sqrt[2 - (2*I)*Sqrt[2]] + I*2^(1/4)*Sqrt[2*I +
Sqrt[2]]]), (2*Sqrt[2])/(2*I + Sqrt[2])])/
(3*Sqrt[-1 + I/Sqrt[2]]*Sqrt[-8 + (7*I)*Sqrt[2]]*(-I + Sqrt[2])) +
EllipticPi[1 - I/Sqrt[2], ArcSin[Root[2 + 4*#1^2 + 3*#1^4 & , 3, 0]],
(1 - (2*I)*Sqrt[2])/3]*Root[8 + 8*#1^2 + 3*#1^4 & , 4, 0]
In[2636]:=
Chop[N[%,30]]
Out[2636]=
0=2E975261536923865518804845371749
which it looks quite mysterious to me! (3/2???)
Dimitris
=CF/=C7 dimitris =DD=E3=F1=E1=F8=E5:
> 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