Re: elliptic integral

Hi David!

Your solution has a lot of magic; doesn't it?

Ok I even tried by myself to write the integrand as [(x - 1)/Sqrt[(1 -
x)*(x - 2)*(x^2 - 2*x + 3)]
but 3/2???
How you came to the conclusion and you broke the integral?
Do I miss something or it was a tactic based to your experience?

Dimitris

Ο/Η David W. Cantrell έγραψε:
dimitris <dimmechan@xxxxxxxxx> wrote:
Hello to all!

I have the definite integral

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

Working with Mathematica...

Version 4. succeeds in getting a closed form result
(agreement with NIntegrate).
Version 5.2 returns the integral unevalueated.

Any ideas on how I could obtain an analytic result
(by CAS or hand)?

Huh? Didn't you _already_ get such a result using Mathematica 4?

But anyway, here's a way to get such a result using Mathematica 5.2:

In[4]:=
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}]

Out[4]=
EllipticPi[1 - I/Sqrt[2], ArcSin[Root[2 + 4*#1^2 + 3*#1^4 & , 3]],
(1/3)*(1 - 2*I*Sqrt[2])]*Root[8 + 8*#1^2 + 3*#1^4 & , 4] +
(EllipticF[ArcSin[Root[3 + 4*#1^2 + 2*#1^4 & , 3]], (1/3)*(1 -
2*I*Sqrt[2])] - EllipticPi[(1/3)*(2 - I*Sqrt[2]), ArcSin[Root[3 + 4*#1^2
+ 2*#1^4 & , 3]], (1/3)*(1 - 2*I*Sqrt[2])])*Root[8 + 8*#1^2 + 3*#1^4 & , 4]

In[5]:= Chop[N[%]]

Out[5]= 0.9752615369238598

Of course, I'm curious to know how Mathematica 6 does with this integral.
After all, writing it as I did in In[4] should not be necessary. We should
be able to ask straightforwardly for
Integrate[Sqrt[(1 - x)/((x - 2)*(x^2 - 2*x + 3))], {x, 1, 2}]
and get a result.

David

.

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