Re: Integration about a singular point of a real function
- From: glenn077@xxxxxxxxx
- Date: 21 Nov 2007 06:15:00 -0500
On 19 ,FMo](B, 18:45, Robert Israel <isr...@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx>
wrote:
glenn...@xxxxxxxxx writes:
For f(x)=1/Abs(x)^a, the integral from -1 to 1 diverges for a>=1 and
converges for a<1. Observe that for a<1 the derivative become infinite
at 0. Is this observation part of some more general theorem? E.g. if
at a single point f becomes infinite and its derivative is also
infinite, then its integral around the singularity is finite under
some conditions?
The derivative also has infinite limits at 0 for a >= 1. The fact that
|f(x)| -> infty as x -> 0 (and f'(x) exists for x <> 0) implies
lim sup_{x -> 0} |f'(x)| = infty. It's the antiderivative you want
to look at, not the derivative.
--
Robert Israel isr...@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
Ups, I'm very sorry. I meant (at the second line of the original msg)
that the derivative of abs(x)^a becomes infty if a<1 (but not if
a>=1).
So, for a general f the question/conjecture goes again:
int(1/f,{x,-1,1})=infty when D(f,x)@0=infty and 0
is an isolated singularity.
which is surely true for f=abs(x)^a, 0<a<1, but not for a>=1.
.
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