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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:

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

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:

[tex] u(t-\tau) \ ={ e^{-\tau s} \over s } [/tex]

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

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

Using the linearity property of the Laplace transform:

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

[tex] \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}[/tex]

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

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

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:

[tex] u(t-\tau) \ ={ e^{-\tau s} \over s } [/tex]

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

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

Using the linearity property of the Laplace transform:

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

[tex] \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}[/tex]

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

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