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MathGroup Archive 2003

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Re: Improper integral

  • To: mathgroup at smc.vnet.net
  • Subject: [mg44642] Re: Improper integral
  • From: poujadej at yahoo.fr (Jean-Claude Poujade)
  • Date: Wed, 19 Nov 2003 04:59:09 -0500 (EST)
  • References: <200311170838.DAA01254@smc.vnet.net> <bpd13t$c8a$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Andrzej Kozlowski <akoz at mimuw.edu.pl> wrote in message news:<bpd13t$c8a$1 at smc.vnet.net>...
> On 17 Nov 2003, at 17:38, Jean-Claude Poujade wrote:
> 
> > Bonjour le groupe,
> >
> > I'm not a mathematician and I wonder why Mathematica doesn't return 0
> > for this doubly infinite improper integral :
> >
> > In[1]:=$Version
> > Out[1]=4.1 for Microsoft Windows (November 2, 2000)
> >
> > In[2]:=Integrate[x/(1+x^2),{x,-Infinity,Infinity},PrincipalValue->True]
> > Integrate::idiv[...]does not converge[...]
> > Out[2]:=Integrate[x/(1+x^2),{x,-Infinity,Infinity},PrincipalValue- 
> > >True]
> >
> > maybe it's different with Mathematica 5.0 ?
> > ---
> > jcp
> >
> No it is the same, and it is correct. 
[...]
Let me just quote the "Ask Dr.Math" Forum :
You have just described what is known as 
the Cauchy principal value integral. 
There are essentially two types:  
the first is the doubly infinite improper integral, e.g.,

     G = Integrate[F[x], {x,-Infinity,Infinity}]

for which the Cauchy principal value is defined as

     PV(G) = Limit[Integrate[F[x], {x,-R,R}], R -> Infinity].
[...]
So what should I think ?
---
jcp


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