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MathGroup Archive 2007

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elliptic integral

  • To: mathgroup at smc.vnet.net
  • Subject: [mg75733] elliptic integral
  • From: dimitris <dimmechan at yahoo.com>
  • Date: Wed, 9 May 2007 04:38:02 -0400 (EDT)

Consider the function

In[17]:=
f[x_] := Sqrt[(1 - x)/((x - 2)*(x^2 - 2*x + 3))]

In[20]:=
Plot[f[x], {x, 1, 2}]

The following definite integral can be evaluated in version 4.0.
However, version 5.2 fails to obtain a closed form solution.

I really appreciate if someone will try it to version 6.

(*version 5.2*)
In[28]:=
Integrate[f[x], {x, 1, 2}]

Out[28]=
Integrate[Sqrt[(1 - x)/((-2 + x)*(3 - 2*x + x^2))], {x, 1, 2}]

The indefinite integral can be done. But the returned result
is much more complicated than the respective antiderivative
by version 4.0.

(*version 5.2*)
In[31]:=
F[x_] = Integrate[Sqrt[(1 - x)/((x - 2)*(x^2 - 2*x + 3))], x]

Out[31]=
(2^(3/4)*(-2 + x)*Sqrt[(I + Sqrt[2] - I*x)/((-I + Sqrt[2])*(-1 +
x))]*Sqrt[(-I + Sqrt[2] + I*x)/((I + Sqrt[2])*(-1 + x))]*
   Sqrt[(1 - x)/(-6 + 7*x - 4*x^2 + x^3)]*EllipticPi[1 - I/Sqrt[2],
ArcSin[2^(1/4)*Sqrt[(-2 + x)/((-I + Sqrt[2])*(-1 + x))]],
    (-I + Sqrt[2])/(I + Sqrt[2])])/Sqrt[(-2 + x)/((-I + Sqrt[2])*(-1 +
x))]

In[37]:=
Simplify[D[F[x], x] == f[x]]

Out[37]=
True

Even application of the Newton-Leibniz formula fails.

In[40]:=
Plot[F[x], {x, 1, 2}, Frame -> {True, True, False, False}]

In[47]:=
Limit[F[x], x -> 2,Direction->1] - Limit[F[x], x -> 1,Direction->-1]

Out[47]=
-Limit[(2^(3/4)*(-2 + x)*Sqrt[(I + Sqrt[2] - I*x)/((-I + Sqrt[2])*(-1
+ x))]*Sqrt[(-I + Sqrt[2] + I*x)/((I + Sqrt[2])*(-1 + x))]*Sqrt[(1 -
x)/(-6 + 7*x - 4*x^2 + x^3)]*EllipticPi[1 - I/Sqrt[2],
ArcSin[2^(1/4)*Sqrt[(-2 + x)/((-I + Sqrt[2])*(-1 + x))]],
      (-I + Sqrt[2])/(I + Sqrt[2])])/Sqrt[(-2 + x)/((-I + Sqrt[2])*(-1
+ x))], x -> 1,Direction->-1]

(unfortunately I don't have installed version 4.0 in this computer).

Dimitris



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