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

**References**:**Improper integral***From:*poujadej@yahoo.fr (Jean-Claude Poujade)