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MathGroup Archive 2003

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Re: Re: Improper integral


On 20 Nov 2003, at 17:16, Jean-Claude Poujade wrote:

> 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
>
>
What I meant was, that you would end up with contradictions if you  
treated such an integral as a  ordinary Riemann or Lebesgue integral of  
a function. It is in fact an integral of a distribution, and these have  
to be manipulated in a different way.
As I already mentioned, the option PrincipalValue -> True in your  
question somehow escaped my attention. I think I must have noticed it  
at first but when thinking about my reply forgot about it, so I thought  
you were asking about the corresponding ordinary integral.

Andrzej Kozlowski


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