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[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. >