Help to solve Integrate[Sqrt[t (1 - t) (z - t)], t]
- To: mathgroup at smc.vnet.net
- Subject: [mg114498] Help to solve Integrate[Sqrt[t (1 - t) (z - t)], t]
- From: Bert RAM Aerts <bert.ram.aerts at gmail.com>
- Date: Sun, 5 Dec 2010 21:56:54 -0500 (EST)
With Mathematica 8 for Linux x86_64 step 4 from the solution below
does not give a result after half an hour.
Are there any options needed?
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Leonid Shifrin wrote on Oct 1st:
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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