       • 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:=
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//InputForm=
(8*Sqrt[4 - I*Sqrt]*EllipticF[ArcSin[Sqrt[5*I + Sqrt]/2^(3/4)],
(2*Sqrt)/(2*I + Sqrt)] -
(7*I)*Sqrt[8 - (2*I)*Sqrt]*EllipticF[ArcSin[Sqrt[5*I + Sqrt]/
2^(3/4)], (2*Sqrt)/(2*I + Sqrt)] +
(12*I)*Sqrt[4 + I*Sqrt]*EllipticF[I*(Log - Log[Sqrt[2 -
(2*I)*Sqrt] + I*2^(1/4)*Sqrt[2*I + Sqrt]]),
(2*Sqrt)/(2*I + Sqrt)] - 3*Sqrt[8 + (2*I)*Sqrt]*
EllipticF[I*(Log - Log[Sqrt[2 - (2*I)*Sqrt] +
I*2^(1/4)*Sqrt[2*I + Sqrt]]), (2*Sqrt)/(2*I + Sqrt)] +
9*Sqrt[2*(-8 + (7*I)*Sqrt)]*EllipticPi[(2*Sqrt)/(-I +
Sqrt), ArcSin[Sqrt[5*I + Sqrt]/2^(3/4)],
(2*Sqrt)/(2*I + Sqrt)] - 9*Sqrt[2*(-8 +
(7*I)*Sqrt)]*EllipticPi[(2*Sqrt)/(-I + Sqrt),
I*(Log - Log[Sqrt[2 - (2*I)*Sqrt] + I*2^(1/4)*Sqrt[2*I +
Sqrt]]), (2*Sqrt)/(2*I + Sqrt)])/
(3*Sqrt[-1 + I/Sqrt]*Sqrt[-8 + (7*I)*Sqrt]*(-I + Sqrt)) +
EllipticPi[1 - I/Sqrt, ArcSin[Root[2 + 4*#1^2 + 3*#1^4 & , 3, 0]],
(1 - (2*I)*Sqrt)/3]*Root[8 + 8*#1^2 + 3*#1^4 & , 4, 0]

In:=
Chop[N[%,30]]

Out=
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:=
> 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

```

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