```On 7 Oct 2005 16:25:18 GMT, israel@math.ubc.ca (Robert Israel) wrote:

>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:

> 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

```
```lucy wrote: