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

  • To: mathgroup at smc.vnet.net
  • Subject: [mg112809] Re: Help to solve an integral by using Mathematica Integrate[Sqrt[t
  • From: Leonid Shifrin <lshifr at gmail.com>
  • Date: Fri, 1 Oct 2010 05:42:12 -0400 (EDT)

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 <hpe650820 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.
>
>



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