Re: Integrate yields complex results
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- Subject: [mg128007] Re: Integrate yields complex results
- From: Alexei Boulbitch <Alexei.Boulbitch at iee.lu>
- Date: Sat, 8 Sep 2012 03:10:08 -0400 (EDT)
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Hi, using Mathematica 8.0 I computed this expression:
Integrate[((1 + y^2) (1 + x^2 + y^2)^(1/2))^(-1), y]
obtaining ugly complex Logarithms which cannot be simplified even with ExpToTrig. The correct real result is given by Mathematica 5.2 and is
ArcTan[xy/((x+^2+y^2+1)^(1/2))]
How is possible to obtain this result with 8.0 version.
I even tried to calculate Real and Imaginary part, but 8.0 refuses to do that. Please help me. I need to use Mathematica for research.
Hi, Matteo,
They are no equal. Check:
int=\[Integral]1/((y^2+1) Sqrt[x^2+y^2+1]) \[DifferentialD]y
(I Log[(4 I (1 + x^2 + I y))/(x (-I + y)) + (
4 I Sqrt[1 + x^2 + y^2])/(-I + y)])/(2 x) - (
I Log[-((4 I (1 + x^2 - I y))/(x (I + y))) - (
4 I Sqrt[1 + x^2 + y^2])/(I + y)])/(2 x)
int === ArcTan[x*y/((x^2 + y^2 + 1)^(1/2))]
False
Just build this plot:
x = 1;
Plot[{int, ArcTan[x*y/((x^2 + y^2 + 1)^(1/2))] - \[Pi]}, {y, -1, 1},
PlotStyle -> {Red, Blue}]
Here the red line shows your integral, and the blue one the expression ArcTan[x*y/((x^2 + y^2 + 1)^(1/2))]. They are different.
Probably there is something more that is not accounted for, some condition, may be?
Have fun, Alexei
Alexei BOULBITCH, Dr., habil.
IEE S.A.
ZAE Weiergewan,
11, rue Edmond Reuter,
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e-mail: alexei.boulbitch at iee.lu