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Re: Integrate yields complex results

  • To: mathgroup at smc.vnet.net
  • 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,
L-5326 Contern, LUXEMBOURG

Office phone :  +352-2454-2566
Office fax:       +352-2454-3566
mobile phone:  +49 151 52 40 66 44

e-mail: alexei.boulbitch at iee.lu







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