Re: Improper integral

*To*: mathgroup at smc.vnet.net*Subject*: [mg44652] Re: Improper integral*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>*Date*: Wed, 19 Nov 2003 04:59:17 -0500 (EST)*References*: <6C03C616-19CE-11D8-864F-00039311C1CC@mimuw.edu.pl>*Sender*: owner-wri-mathgroup at wolfram.com

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. >> >