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MathGroup Archive 2003

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Re: Improper integral

  • To: mathgroup at
  • Subject: [mg44614] Re: [mg44608] Improper integral
  • From: Andrzej Kozlowski <akoz at>
  • Date: Tue, 18 Nov 2003 06:41:43 -0500 (EST)
  • References: <>
  • Sender: owner-wri-mathgroup at

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:

Integrate[x/(1 + x^2), {x, -a, a}]


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]


and so on.

Andrzej Kozlowski
Chiba, Japan

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