Re: Improper integral
- To: mathgroup at smc.vnet.net
- Subject: [mg44637] Re: Improper integral
- From: "David W. Cantrell" <DWCantrell at sigmaxi.org>
- Date: Wed, 19 Nov 2003 04:59:05 -0500 (EST)
- References: <200311170838.DAA01254@smc.vnet.net> <bpd13t$c8a$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
Andrzej Kozlowski <akoz at mimuw.edu.pl> wrote:
> On 17 Nov 2003, at 17:38, Jean-Claude Poujade wrote:
> > I'm not a mathematician and I wonder why Mathematica doesn't return 0
> > for this doubly infinite improper integral :
> >
> > In[1]:=$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.
- References:
- Improper integral
- From: poujadej@yahoo.fr (Jean-Claude Poujade)
- Improper integral