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Re: Improper integral
*To*: mathgroup at smc.vnet.net
*Subject*: [mg44652] Re: Improper integral
*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>
*Date*: Wed, 19 Nov 2003 04:59:17 -0500 (EST)
*References*: <6C03C616-19CE-11D8-864F-00039311C1CC@mimuw.edu.pl>
*Sender*: owner-wri-mathgroup at wolfram.com
More on this theme: can somebody explain what the concept of a
"principal value" of an integral is good for? I have been a
professional mathematician for years and have been involved in several
different areas of research, and yet never came across any use for it.
I have more then a dozen texts on analysis yet none of them mentions
it. The only books where I can find it mentioned are books for
physicists and engineers (one is the well known text by Riley, Hobson
and Bence, the other a book in Polish) and they both give one line
definitions without any examples of use (and do not mention poles at
infinity). At first sight it seems a pretty trivial and useless
concept, so I would like to know if it really has any serious
applications.
Andrzej Kozlowski
On 18 Nov 2003, at 22:52, Andrzej Kozlowski wrote:
> 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 mimuw.edu.pl> 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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