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

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

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

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

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

Out=
True

Even application of the Newton-Leibniz formula fails.

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

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

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

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

Dimitris

```

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