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

Tim Wescott wrote:
> Robert Israel wrote:

> > You might also try the case where f is not a function, but a
> > distribution.  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.

> What's the difference between a function and a distribution?  Judging
> from the Dirac delta distribution I gather that a distribution is
> something that you can sometimes treat as a function if you hold your
> mouth right, but that's hardly a rigorous definition.

A distribution is a continuous linear functional on the space of
C^infinity functions with compact support.  For the purposes of
Fourier and Laplace transforms, one generally restricts attention
to tempered distributions.  See e.g.
<http://en.wikipedia.org/wiki/Tempered_distribution>

 Robert Israel                                israel@math.ubc.ca
 Department of Mathematics        http://www.math.ubc.ca/~israel
 University of British Columbia            Vancouver, BC, Canada

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).
> 
> You might also try the case where f is not a function, but a 
> distribution.  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
> 
What's the difference between a function and a distribution?  Judging 
from the Dirac delta distribution I gather that a distribution is 
something that you can sometimes treat as a function if you hold your 
mouth right, but that's hardly a rigorous definition.

-- 

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

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).

You might also try the case where f is not a function, but a 
distribution.  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

Thanks a lot Robert!
Will it possibly be some Laplace Transform of an non-exponentially
bounded function?

>>>>> "dave" == dave  <david_lawrence_petry@yahoo.com> writes:

    dave> Raymond Toy wrote:
    >> >>>>> "dave" == dave  <david_lawrence_petry@yahoo.com> writes:
    >> 
    dave> lucy wrote:
    >> >> sin(s)/(s^2+4),
    >> >>
    >> >> can it be a proper Laplace transform?
    >> 
    dave> I don't think so. The Laplace Inversion formula doesn't
    dave> work for it.
    >> 
    >> What fails?
    >> 
    >> I took a quick stab at it by replacing sin(s) with its power series.
    >> Then we have terms of the form s^(2*k+1)/(s^2+4), for which we know
    >> the inverse Laplace transform.

    dave> Really? What is the inverse Laplace transform of  s^5/(s^2+4) ?

I was using the relationship f'(t) -> sF(s) - f(+0).  But there are
probably conditions that need to be satisfied for this to be true.  I
dont' have any transform theory book handy, other than a short table
of transforms.

Ray

In article <1128377002.944857.263010@z14g2000cwz.googlegroups.com>,
lucy <losemind@yahoo.com> wrote:
>sin(s)/(s^2+4),

>can it be a proper Laplace transform?

>why?

The Laplace transform F(s) of an exponentially bounded function
f(t) (say with |f(t)| <= K exp(Bt)) converges for Re(s) > B with
|F(s)| <= K/(Re s - B).  But |sin(s)| ~ exp(|Im(s)|)/2 as 
|Im(s)| -> infty.  So sin(s)/(s^2+4) is not the Laplace transform
of an exponentially bounded function.  

Robert Israel                                israel@math.ubc.ca
Department of Mathematics        http://www.math.ubc.ca/~israel 
University of British Columbia            Vancouver, BC, Canada

Raymond Toy wrote:
> >>>>> "dave" == dave  <david_lawrence_petry@yahoo.com> writes:
>
>     dave> lucy wrote:
>     >> sin(s)/(s^2+4),
>     >>
>     >> can it be a proper Laplace transform?
>
>     dave> I don't think so. The Laplace Inversion formula doesn't
>     dave> work for it.
>
> What fails?
>
> I took a quick stab at it by replacing sin(s) with its power series.
> Then we have terms of the form s^(2*k+1)/(s^2+4), for which we know
> the inverse Laplace transform.

Really? What is the inverse Laplace transform of  s^5/(s^2+4) ?

"Proper" (I'm not sure what that means) Laplace transforms usually
tend to 0 as s goes to infinity.


> I think the resulting infinite series
> converges.  But I did leave out all the terms having to do with the
> initial values, so perhaps it doesn't really converge.

I suspect you left out the key parts.

>>>>> "dave" == dave  <david_lawrence_petry@yahoo.com> writes:

    dave> lucy wrote:
    >> sin(s)/(s^2+4),
    >> 
    >> can it be a proper Laplace transform?

    dave> I don't think so. The Laplace Inversion formula doesn't
    dave> work for it. 

What fails?

I took a quick stab at it by replacing sin(s) with its power series.
Then we have terms of the form s^(2*k+1)/(s^2+4), for which we know
the inverse Laplace transform.  I think the resulting infinite series
converges.  But I did leave out all the terms having to do with the
initial values, so perhaps it doesn't really converge.

Ray