       Re: Improper integral

• To: mathgroup at smc.vnet.net
• Subject: [mg44638] Re: Improper integral
• From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
• Date: Wed, 19 Nov 2003 04:59:05 -0500 (EST)
• Sender: owner-wri-mathgroup at wolfram.com

```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:=\$Version
>>> Out=4.1 for Microsoft Windows (November 2, 2000)
>>>
>>> In:=Integrate[x/(1+x^2),{x,-Infinity,Infinity},PrincipalValue-
>>> >True]
>>> Integrate::idiv[...]does not converge[...]
>>> Out:=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:=
>> Integrate[x/(1 + x^2), {x, -a, a}]
>>
>> Out=
>> 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:= Limit[Integrate[x/(1+x^2),{x,-a,a}],a->Infinity]
>
> Out= 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
>>
>> and so on.
>

```

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