From: JeffLeader@MindSpring.com (Jeffery J. Leader)
Subject: Re: Cauchy principal value
Date: Sat, 06 Feb 1999 08:19:20 GMT
Newsgroups: sci.math

John Saccente <brennan@lcc.net> wrote:
>What is
>the significance of the CPV?  How does it differ from an ordinary
>improper integral?

The integral of f(x) from -inf. to inf. exists only if the integral of
f(x) from -inf. to 0 and the integral of f(x) from 0 to inf. both
exist.  Hence, the doubly improper integral of f(x)=x fails to exist
(the left half gives -inf., the right half gives +inf.).  The CPV is
the limit (as B-->inf.) of the integral from -B to B of f(x).  Hence,
the CPV of the doubly improper integral of f(x)=x exists because the
integral from -B to B of f(x)=x is always zero (the neg. half cancels
out the pos. half).

So, we use the CPV when--say, on physical grounds--we want integrals
like that to be zero (or whatever), not undefined.  Examples include
the Cauchy dist. in statistics, which has a mean in the CPV sense but
not in the classical sense, and examples in PDEs (say, via complex
maps) where this is a physically meaningful interpretation.  The usual
sense of the doubly improper integral is a more restrictive
notion--notice that the CPV can give a finite value to the integral of
a function which is unbounded as x-->+/- infinity, which may seem a
bit strange.
==============================================================================

From: rusin@vesuvius.math.niu.edu (Dave Rusin)
Subject: Re: Cauchy principal value
Date: 6 Feb 1999 19:41:18 GMT
Newsgroups: sci.math

John Saccente <brennan@lcc.net> wrote:
>What is
>the significance of the CPV?  How does it differ from an ordinary
>improper integral?

Jeffery J. Leader <JeffLeader@MindSpring.com> wrote:

[ valid answer deleted; it carried the reminder that ]
>The CPV is the limit (as B-->inf.) of the integral from -B to B of f(x). 

One can also define the CPV for integrals around a singularity, e.g.
		integral_[-1,1]  (1/x) dx

Leader continued,
>Notice that the CPV can give a finite value to the integral of
>a function which is unbounded as x-->+/- infinity, which may seem a
>bit strange.

Yes, and what's worse, the CPV is _not_ a "stable" value for the integral;
if you evaluate the integral using change-of-variables you might get a
different answer. For example, it seems absolutely reasonable to allow
the substitution  x = u+1  in the integral  integral_(-oo,oo)  x dx, but
upon doing so we obtain a CPV not of zero but of
	lim_{B -> oo}  [ (B+1)^2/2 - (B-1)^2/2 ] = +oo

dave