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MathGroup Archive 2000

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Poisson integral with Mathematica

  • To: mathgroup at smc.vnet.net
  • Subject: [mg23037] Poisson integral with Mathematica
  • From: "Francois GRIMAL" <grimal at marius.univ-mrs.fr>
  • Date: Thu, 13 Apr 2000 02:43:24 -0400 (EDT)
  • Organization: Universite de la Mediterranee Aix en Provence
  • Sender: owner-wri-mathgroup at wolfram.com

Mathematica 3.0 has some problems with the Poisson integrals. If you
evaluate a cell like :

 Simplify[1/(2Pi)Integrate[(1+z Exp[-I u])/(1-z Exp[-I u]),{u,0,2 Pi}]]

Mathematica give the answer -1 wich is true if |z|>1 but false if |z|<1.

Now, if you evaluate the cell :

Simplify[1/(2Pi)Integrate[(1+z Exp[-I u])/(-(1-z Exp[-I u])),{u,0,2 Pi}]]

Mathematica answers -1 again but if you evaluate :

Simplify[1/(2Pi)Integrate[-(1+z Exp[-I u])/((1-z Exp[-I u])),{u,0,2 Pi}]] it
answers 1 !

The problem persists with Mathematica 4.0

How does it work ?





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