# Laplace Transform Convergence

1. Sep 2, 2009

### John Creighto

I've been trying to understand what the Laplace Transform means via It's definition. Even if you are trying to evaluate a simple exponential function:

$$F(s) = \mathcal{L} \left\{f(t)\right\}=\int_0^{\infty} e^{-st} f(t) \,dt.$$

it will not converge for all S in a Riemann sense. I've been trying to read a bit about measure theory and Lebesgue integration and what I found on the web hasn't made me feel like I understand much. However, I did read two things that seem to show how I might be able to make the Laplace transform converge for all S.

1) The Unit Step functions form a Lebesgue measure for the Laplace Transform.
2) Lebesgue integration can be thought of in terms of partitioning in a vertical sense instead of a horizontal sense.

Okay, so what is the Laplace Transform for a unit step function? It is:

$$u(t-\tau) \ ={ e^{-\tau s} \over s }$$

Now, we can partition an exponential function in to horizontal rectangles as follows

$$exp(\lambda t)=u(0)+\Delta y\sum_{n=1}^{\infty}u\left(t-\frac{1}{\lambda}ln\left(\ n\Delta y\right)\right)$$

Using the linearity property of the Laplace transform:

$$\mathcal{L}(exp(\lambda t))=\frac{1}{s}+\Delta y\sum_{n=1}^{\infty}{ e^{-ln(\ n\Delta y)\frac{s}{\lambda}} \over s }$$

$$\mathcal{L}(exp(\lambda t))=\frac{1}{s}+\Delta y\left(\frac{1}{\Delta y}\right)^{s/\lambda}{\sum_{n=1}^{\infty}\left(\frac{1}{n} \right)^{s/\lambda} \over s }=\frac{1}{s}+\left(\frac{1}{\Delta y}\right)^{(s/\lambda)-1}\frac{\zeta(s/\lambda)}{s}$$

(Which doesn't converge so I need to do something different :().

Last edited: Sep 2, 2009
2. Sep 2, 2009

### John Creighto

I don't think vertical partitioning helped.

Maybe if the function is $$exp(\lambda t)$$

Then I can choose as basis (measure?) Functions of the form $$exp(\beta(t-\tau))$$

That is delayed exponential (probably a negative exponential if lambda is positive).

Let:

$$b_0=f(0)exp(\beta t)$$

$$b_n=\left<f(x)-\sum_{k=o}^{n-1}b_kexp(\beta t-n \Delta t) \ , \ f(exp(\beta(t-n \Delta t))\right>$$

The inner product is taken from zero to infinity. Now if I express this sequence in the Laplace domain, can I find for any S a value of $$\beta$$ for which the sequence will converge. Also if I do this and I only approximate a few terms then what is the rate of convergence for all points in the Laplace domain.

3. Sep 3, 2009

### John Creighto

I'm not sure why the following premises in quotes is wrong but in my example I approximate f(x) arbitrarily closely and the limit in the Laplace domain doesn't converge to the correct value (or at least the limit is not well defined.

Now, given the above premise, one only has to show that the Laplace transform exists for the exponential curve that bounds the function. Without loss of generality consider the function:

$$f(x)=e^{at}$$

Let this function be bounded by:

$$f(x)=e^{bt}$$

Now I can show that (I'll prove this in another post):

$$f(x) \approx e^{bt}+\sum_1^{\infty}(e^{na\Delta t}-e^{(b+(n-1)a)\Delta t})e^{-b(t-n \Delta t)u(t-\Delta t)$$

Now, because the Laplace transform is linear we have:

$$\mathcal{L}(f(x)) \approx \frac{1}{s-b}+\sum_1^{\infty}({e^{na\Delta t}-e^{b+(n-1)a\Delta t})\frac{e^{-bn \Delta t}}{s-b}$$

or equivalently:

$$\mathcal{L}(f(x)) \approx \frac{1}{s-b}+\sum_1^{\infty}({e^{n(a-b)\Delta t}-e^{(n-1)(a-b)\Delta t})\frac{1}{s-b}$$

Writing as a geometric series:

$$\mathcal{L}(f(x)) \approx \frac{1}{s-b}+\frac{1}{s-b}\sum_1^{\infty}\left(\left({e^{(a-b)\Delta t}\right)^n-\left(e^{(a-b)\Delta t}\right)^{(n-1)}\right)$$

$$\mathcal{L}(f(x)) \approx \frac{1}{s-b}+\frac{1}{s-b}\mathop {\lim }\limits_{N \to \infty }\left(\sum_1^{\infty}\left({e^{(a-b)\Delta t}\right)^n-\sum_1^{\infty}\left(e^{(a-b)\Delta t}\right)^{(n-1)}\right)\right$$

$$\mathcal{L}(f(x)) \approx \frac{1}{s-b}+\frac{1}{s-b}\mathop {\lim }\limits_{N \to \infty }\left(\frac{(e^{(a-b)\Delta t})^{N}}{1-e^{(a-b)\Delta t}}-\frac{(e^{(a-b)\Delta t})^{N-1}}{1-(e^{(a-b)\Delta t}}\right)$$

$$\mathcal{L}(f(x)) \approx \frac{1}{s-b}+\frac{1}{s-b}\mathop {\lim }\limits_{N \to \infty }(e^{(a-b)\Delta t})^{N}\left(\frac{e^{(a-b)\Delta t}-1}{1-e^{(a-b)}}\right)$$

$$\mathcal{L}(f(x)) \approx \frac{1}{s-b}-\frac{1}{s-b}\mathop {\lim }\limits_{N \to \infty }(e^{(a-b)\Delta t})^{N}$$

Which doesn't seem like a terribly useful limit because it depends how quickly $$\Delta t$$ approach zero as N approach infinity also if b can very with the limit the expression is even more ambiguous.

4. Sep 3, 2009

### John Creighto

Actually come to think of it my approach is all wrong. So my above premmis again:

now if f(x) is bounded by a function $$ae^{at}$$ and f(x) is integrable then it should be sufficient to show that the Laplace transform of $$ae^{at}$$ exists to show that the Laplace transform of f(x) exists.

Define the Laplace as follows:

$$\mathcal{L}(f(t))=\mathop {\lim }\limits_{t \to \infty} \int_0^{t} f(t')(1-u(t'-t))e^{-st'}dt'$$

Then it is easy to show that at least for an exponential function that this converges to the Laplace transform. So, I think what I want to do is develop a limit approximation for a finite length, Compute the Laplace transform and then take the limit as L approaches infinity in the Laplace domain. So one limit in the time domain and another limit in the Laplace Domain.

5. Sep 5, 2009

### John Creighto

I found some interesting stuff:

http://www-sop.inria.fr/apics/sbpi/LtransformOlivi.pdf

Here is some more stuff on the Hankel Norm: