Re: Improper integral

*To*: mathgroup at smc.vnet.net*Subject*: [mg44657] Re: Improper integral*From*: poujadej at yahoo.fr (Jean-Claude Poujade)*Date*: Thu, 20 Nov 2003 03:16:24 -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. Presumably the reason why you > think the answer should be zero is: > > In[21]:= > Integrate[x/(1 + x^2), {x, -a, a}] > > Out[21]= > 0 > > But Integrate[x/(1 + x^2), {x, -Infinity, Infinity}] is not just the > limit of the above as a->Infinity. What has to be true is that the > limits of Integrate[x/(1 + x^2), {x, a, b}] must exist as a -> > -Infinity and b->Infinity independently of one another. This is of > course not true. If you defined the infinite integral in a different > way you could end up with all sorts of contradictions. For example, > observe that: > Limit[Integrate[x/(1 + x^2), {x, -a, 2*a}], a -> Infinity] > Log[2] [...] Ok, but such a definition would contradict the principle of symmetrical computation which Cauchy principal value integral is based on. If the definition is Limit[Integrate[[x/(1 + x^2), {x, -a, a}], a -> Infinity] could you please tell me the kind of contradiction that could occur? So, if I specify "PrincipalValue->True" I expect Mathematica to use that symmetrical definition, which is generally agreed on (see for instance the forum "Ask Dr.Math"). --- jcp

**Follow-Ups**:**Re: Re: Improper integral***From:*Andrzej Kozlowski <akoz@mimuw.edu.pl>

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