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: [mg112834] Re: Help to solve an integral by using Mathematica Integrate[Sqrt[t
  • From: Hugo <hpe650820 at gmail.com>
  • Date: Sat, 2 Oct 2010 05:46:07 -0400 (EDT)
  • References: <i84adn$gvn$1@smc.vnet.net> <i84bkk$hqk$1@smc.vnet.net>

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.

Thanks,

Hugo


On Oct 1, 5:03 am, Valeri Astanoff <astan... at gmail.com> wrote:
> On 1 oct, 11:42, Leonid Shifrin <lsh... at gmail.com> wrote:
>
>
>
>
>
> > To my surprise, Mathematica has been very reluctant to help here, despite
> > the
> > seemingly simple form of the integral. Here is what I did anyway:
>
> > 1. Find an indefinite integral:
>
> > In[94]:= expr = Integrate[Sqrt[t (1 - t) (z - t)], t]
>
> > Out[94]= (Sqrt[(-1 + t) t (t - z)] (2 t (-1 + 3 t - z) + (
> >    2 (-1 + t) (-((2 t (t - z) (1 - z + z^2))/(-1 + t)^2) - (
> >       2 I Sqrt[t/(-1 + t)] Sqrt[(
> >        t - z)/(-1 + t)] (1 - z + z^2) EllipticE[
> >         I ArcSinh[1/Sqrt[-1 + t]], 1 - z])/Sqrt[-1 + t] + (
> >       I Sqrt[t/(-1 + t)] Sqrt[(t - z)/(-1 + t)]
> >         z (1 + z) EllipticF[I ArcSinh[1/Sqrt[-1 + t]], 1 - z])/
> >       Sqrt[-1 + t]))/(t - z)))/(15 t)
>
> > 2. Find a limit on the lower and (at zero):
>
> > In[95]:= expr0 = Limit[expr , t -> 0]
>
> > Out[95]= -(2/  15) I (2 (1 - z + z^2) EllipticE[1 - z] -
> >    z (1 + z) EllipticK[1 - z])
>
> > 3. Make a substitution z->t-q in the indefinite integral result:
>
> > In[96]:= expr1 = expr /. z -> t - q
>
> > Out[96]= (Sqrt[
> >  q (-1 + t) t] (2 t (-1 + q + 2 t) + (
> >    2 (-1 + t) (-((2 q t (1 + q - t + (-q + t)^2))/(-1 + t)^2) - (
> >       2 I Sqrt[q/(-1 + t)] Sqrt[
> >        t/(-1 + t)] (1 + q - t + (-q + t)^2) EllipticE[
> >         I ArcSinh[1/Sqrt[-1 + t]], 1 + q - t])/Sqrt[-1 + t] + (
> >       I Sqrt[q/(-1 + t)] Sqrt[
> >        t/(-1 + t)] (-q + t) (1 - q + t) EllipticF[
> >         I ArcSinh[1/Sqrt[-1 + t]], 1 + q - t])/Sqrt[-1 + t]))/
> >    q))/(15 t)
>
> > 4. Simplify it:
> > In[97]:= expr2 = FullSimplify[expr1, t > 0 && t < 1 && q > 0 && q <=
 1=
> ]
>
> > Out[97]= (Sqrt[
> >  q (-1 + t) t] (-((2 t (1 + q (3 + 2 q - 5 t) + t))/(-1 + t)) - (
> >    2 I t (2 (1 + q + q^2 - 2 q t + (-1 + t) t) EllipticE[
> >         I ArcCoth[Sqrt[t]],
> >         1 + q - t] - (-1 + q - t) (q - t) EllipticF[
> >         I ArcCoth[Sqrt[t]], 1 + q - t]))/Sqrt[q (-1 + t) t]))/(15 t)
>
> > 5. Get the result for the upper end by substituting q->0 and then t->z:
>
> > In[98]:= expr3 = (expr2 // Apart) /. q -> 0 /. t -> z
>
> > Out[98]= -(4/15) I EllipticE[I ArcCoth[Sqrt[z]], 1 - z] -
> >  2/15 I z^2 (2 EllipticE[I ArcCoth[Sqrt[z]], 1 - z] -
> >     EllipticF[I ArcCoth[Sqrt[z]], 1 - z]) +
> >  2/15 I z (2 EllipticE[I ArcCoth[Sqrt[z]], 1 - z] +
> >     EllipticF[I ArcCoth[Sqrt[z]], 1 - z])
>
> > 6. Get the final result:
>
> > In[99]:= expr4 = FullSimplify[expr3 - expr0]
>
> > Out[99]=
> >  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]))
>
> > Now, it turns out that the sign is wrong. All my attempts to verify the
> > correctness of
> > this analytically failed (I did not try too hard though). Neither was I able
> > to reduce it
> > to the manifestly real form.
>
> > The final form of the result is then (correcting the sign and taking the
> > real part):
>
> > exprint[z_] :=
> >   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]))];
>
> > I did compare it to the result of numerical integration:
>
> > exprintN[z_?NumericQ] :=
> >  NIntegrate[Sqrt[t (1 - t) (z - t)], {t, 0, z}]
>
> > Plot[{exprint[z], exprintN[z]}, {z, 0, 1}]
>
> > Plot[{exprint[z] - exprintN[z]}, {z, 0, 1}]
>
> > And they seem to agree, but that's about all I could squeeze out of it.
>
> > Hope this helps.
>
> > Regards,
> > Leonid
>
> > On Thu, Sep 30, 2010 at 12:51 PM, Hugo <hpe650... at gmail.com> wrote:
> > > Could anybody help me to implement this integral in Mathematica?
> > > Integrate[Sqrt[t (1-t) (z-t)],{t,0,z}] where z and t are real with
> > > intervals 0<t<1 and 0<z<1. I'd really appreciate any help for this
> > > problem.- Masquer le texte des messages pr=E9c=E9dents -
>
> > - Afficher le texte des messages pr=E9c=E9dents -
>
> I suggest this form :
>
> (1/15)*(4*(1 + (z-1)*z)*EllipticE[z] - 2*(2 + (z-3)*z)*EllipticK[z])
>
> v.a.



  • Prev by Date: Re: How to unflatten an array ?
  • Next by Date: Re: Question on Solve
  • Previous by thread: Re: Help to solve an integral by using Mathematica Integrate[Sqrt[t
  • Next by thread: Re: Help to solve an integral by using Mathematica Integrate[Sqrt[t