Re: Re: Improper integral

*To*: mathgroup at smc.vnet.net*Subject*: [mg44690] Re: [mg44657] Re: Improper integral*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>*Date*: Fri, 21 Nov 2003 05:13:13 -0500 (EST)*References*: <200311170838.DAA01254@smc.vnet.net> <bpd13t$c8a$1@smc.vnet.net> <200311200816.DAA01517@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

On 20 Nov 2003, at 17:16, Jean-Claude Poujade wrote: > Andrzej Kozlowski <akoz at mimuw.edu.pl> wrote in message > news:<bpd13t$c8a$1 at smc.vnet.net>... >> On 17 Nov 2003, at 17:38, Jean-Claude Poujade wrote: >> >>> Bonjour le groupe, >>> >>> I'm not a mathematician and I wonder why Mathematica doesn't return 0 >>> for this doubly infinite improper integral : >>> >>> In[1]:=$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 ? >>> --- >>> jcp >>> >> No it is the same, and it is correct. 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. 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] > [...] > Ok, but such a definition would contradict the principle of symmetrical > computation which Cauchy principal value integral is based on. > If the definition is > Limit[Integrate[[x/(1 + x^2), {x, -a, a}], a -> Infinity] > could you please tell me the kind of contradiction that could occur? > So, if I specify "PrincipalValue->True" I expect Mathematica > to use that symmetrical definition, which is generally agreed on > (see for instance the forum "Ask Dr.Math"). > --- > jcp > > What I meant was, that you would end up with contradictions if you treated such an integral as a ordinary Riemann or Lebesgue integral of a function. It is in fact an integral of a distribution, and these have to be manipulated in a different way. As I already mentioned, the option PrincipalValue -> True in your question somehow escaped my attention. I think I must have noticed it at first but when thinking about my reply forgot about it, so I thought you were asking about the corresponding ordinary integral. Andrzej Kozlowski

**References**:**Improper integral***From:*poujadej@yahoo.fr (Jean-Claude Poujade)

**Re: Improper integral***From:*poujadej@yahoo.fr (Jean-Claude Poujade)