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Re: Re: Improper integral
*To*: mathgroup at smc.vnet.net
*Subject*: [mg44768] Re: [mg44652] Re: Improper integral
*From*: Garry Helzer <gah at math.umd.edu>
*Date*: Tue, 25 Nov 2003 00:45:37 -0500 (EST)
*Sender*: owner-wri-mathgroup at wolfram.com
When computing inverse Fourier transforms the principal value may give
the answer even when the integral does not converge. On p. 386 of
Churchill's Operational Mathematics there is an example which turns out
to be the very integral in question. Have we come full circle?
On Wednesday, November 19, 2003, at 04:59 AM, Andrzej Kozlowski wrote:
> 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.
>>>
>>
>
>
Garry Helzer
Department of Mathematics
University of Maryland
1303 Math Bldg
College Park, MD 20742-4015
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