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Re: Improper integral

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  • Subject: [mg44638] Re: Improper integral
  • From: Andrzej Kozlowski <akoz at>
  • Date: Wed, 19 Nov 2003 04:59:05 -0500 (EST)
  • Sender: owner-wri-mathgroup at

In fact,  I somehow did not notice the "PrincipalValue->True" in the  
Jean-Clause's question, but Mathematica's documentation says:

  Setting PrincipalValue->True gives finite answers for integrals that  
had single pole divergences with PrincipalValue->False.

It is not clear form this if poles at infinity are included  but this  
example suggests that probably not. It is also not obvious to me that  
they ought to be included; although I myself have already long since  
forgotten this stuff, a couple of well-known text books I looked at do  
not mention them when defining "principal value".

Andrzej Kozlowski

On 18 Nov 2003, at 22:08, David W. Cantrell wrote:

> Andrzej Kozlowski <akoz at> wrote:
>> On 17 Nov 2003, at 17:38, Jean-Claude Poujade wrote:
>>> 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 ?
>> No it is the same, and it is correct.
> Correct??
> Well, at least it's not blatantly wrong. Mathematica makes the  
> statement
> "does not converge" about Integrate[x/(1+x^2),{x,-Infinity,Infinity}],
> rather than about its Cauchy principal value. But Mathematica never  
> gets
> the answer to the question that was asked. If a student gave me the  
> same
> response to that question on a test, they would get little (if any)  
> partial
> credit.
>> 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.
> Right. But Jean-Claude's question was specifically about that  
> integral's
> _Cauchy principal value_, which is precisely the limit you mentioned.  
> Thus,
> In[1]:= Limit[Integrate[x/(1+x^2),{x,-a,a}],a->Infinity]
> Out[1]= 0
> is a way for us to assist Mathematica so that it can give the correct
> answer for the Cauchy principal value.
> But of course, we shouldn't have to assist Mathematica in this way!
> David Cantrell
>> 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]
>> and so on.

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