On 7 Oct 2005 16:25:18 GMT, israel@math.ubc.ca (Robert Israel) wrote:>In article <1128664482.687884.120780@z14g2000cwz.googlegroups.com>, >lucy <losemind@yahoo.com> wrote: >>Thanks a lot Robert! >>Will it possibly be some Laplace Transform of an non-exponentially >>bounded function? > >The Laplace Transform is defined by F(s) = int_0^infty f(t) exp(-st) dt, >when that improper integral converges. f doesn't have to be >exponentially bounded, but it helps. > >Now let's say the integral converges absolutely for s = s_0 (real), i.e. >B = int_0^infty |f(t)| exp(-s_0 t) dt < infty. Then |F(s)| <= B >whenever Re(s) >= s_0. > >In your case F(s) is unbounded for Re(s) >= s_0, so the integral can't >ever converge absolutely. I guess there's a possibility that the integral >could converge conditionally for s in some domain (it's easy to arrange >for some particular s, but hard to imagine how to do it for s in an open >set).Actually it's not so hard. Take s_0 = 0 for convenience. Suppose that f is locally (Lebesgue) integrable on [0, infinity). Let I(A) = int_0^A f(t) dt, and suppose that I(A) -> L as A -> infinity. Then I is bounded, so an integration by parts shows that (as an improper Riemann integral) F(s) converges for Re(s) > 0, in fact F(s) = s int_0^infinity I(t) exp(-st) dt. If f takes is large enough in absolute value (but wildly oscillating so that I(A) converges) then the integral defining F(s) will fail to be absolutely convergent for any s with Re(s) > 0. Of course the same argument shows that |F(s)| is bounded by c(1 + |Im(s)|), so you don't get sin(s)/(s^2+4) this way. (The bound shows that F _is_ holomorphic in the right half-plane...)>You might also try the case where f is not a function, but a >distribution.If f is a distribution it's not clear to me what conditional convergence would even mean - we'd have to define it using a limit of smooth cutoff functions, and it seems like the answer might depend on the choice of cutoff function. If f is a tempered distribution (with support in (0, infinity), say) then I _think_ it's clear that F has polynomial growth - don't quote me on that...>However, even allowing a finite number of derivatives >of functions for which the integral converges absolutely at s_0 will >make the Laplace transform bounded by a polynomial in |s| for >Re(s) > s_0, so you still can't get sin(s)/(s^2+4) that way. > >Robert Israel israel@math.ubc.ca >Department of Mathematics http://www.math.ubc.ca/~israel >University of British Columbia Vancouver, BC, Canada************************ David C. Ullrich