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Re: Improper integral


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


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