On Tue, Dec 12, 2006 at 08:16:48AM -0800, Daniel Lakeland wrote:
> On Tue, Dec 12, 2006 at 03:33:30PM +0000, miguel lopez wrote:
> > Daniel Lakeland <dlakelan <at> street-artists.org> writes:
> >
> > >
> > > On Tue, Dec 12, 2006 at 02:09:27PM +0000, miguel lopez wrote:
> > >
> > > > I think that there is a long life for mathematician,
> > >just not to be replaced by a computer program :).
> > > >
> > > > Best wishes to the list.
> > >
> > > Indeed. I think it should be possible to prove that:
> > if
> > > integral(f(x),x,0,inf) diverges then
> > integral(periodic_function(x) * f(x)) should also diverge.
> >
> > but I haven't tried yet.
> > >
> > > In any case, I'll file this as a bug ...
> > >
> > You can not prove it: it is false:
> >
> >
> > f(x) = 1/x diverge, f(x) * sin(x) = sin(x)/x converge.
> > (30 seconds time in ultramaxima code, to discover this :)
>
> Whoops. :-) That was exactly the first example I posted...
>
> But I guess what I meant was if f(x) is not absolutely integrable on
> [0,inf) then periodic_function(x) * f(x) should not be absolutely
> integrable. (this is where my initial questions for maxima were coming
> from)
>
> The idea being that although oscillations around 0 may cause an
> integral to converge, absolute integrability is dependant on the speed
> at which f(x), the nonperiodic function, goes to zero at infinity. A
> periodic function won't affect that rate.
>
> It's intuitive but not rigorous, perhaps you have a counterexample in
> your pocket :-)
Ok, thanks to a different off-list example from stavros I see that for
my intuitive idea to hold requires at least that the periodic function
have non-zero integral over one period. otherwise a periodic function
like one that is 1 at every integer and 0 everywhere else will
obviously hammer the integral to zero.
--
Daniel Lakeland
dlakelan at street-artists.org
http://www.street-artists.org/~dlakelan