Mathematica 9 is now available
Services & Resources / Wolfram Forums / MathGroup Archive
-----

MathGroup Archive 2010

[Date Index] [Thread Index] [Author Index]

Search the Archive

Re: Help to solve an integral by using Mathematica Integrate[Sqrt[t

  • To: mathgroup at smc.vnet.net
  • Subject: [mg112852] Re: Help to solve an integral by using Mathematica Integrate[Sqrt[t
  • From: Valeri Astanoff <astanoff at gmail.com>
  • Date: Sun, 3 Oct 2010 05:41:08 -0400 (EDT)
  • References: <i84adn$gvn$1@smc.vnet.net> <i84bkk$hqk$1@smc.vnet.net> <i86v14$hqv$1@smc.vnet.net>

On 2 oct, 11:46, Hugo <hpe650... at gmail.com> wrote:
> Valeri,
>
> Suggested solution is very close to solution found in book. How did
> you get it? Please, let me know how did you implement this integral.
>

Good day,

Leonid's solution :
Re[-(2/15)
 I (2 (1 + (-1 + z) z) EllipticE[1 - z] -
 2 (1 + (-1 + z) z) EllipticE[I ArcCoth[Sqrt[z]], 1 - z] +
 z (1 + z) (EllipticF[I ArcCoth[Sqrt[z]], 1 - z] -
 EllipticK[1 - z]))];

and my own :

(1/15)*(4*(1 + (z-1)*z)*EllipticE[z] - 2*(2 + (z-3)*z)*EllipticK[z])

are numerically identical when z lies between 0 and 1.

I observed that for most rational values of z,
the solution was a linear combination of EllipticE
and EllipticK. Then I performed a polynomial fitting
of the coefficients.
Well, I confess I sort of guessed it...
--
Valeri







  • Prev by Date: Re: How to apply a list of functions
  • Next by Date: Re: How to unflatten an array ?
  • Previous by thread: Re: Help to solve an integral by using Mathematica Integrate[Sqrt[t
  • Next by thread: Finding out which fonts are available