Laplace transform

Posted December 8th, 2007 by Structure

Laplace transform is a powerful tool which allows reducing relations between functions and its derivatives to some algebraic relations. Of course , former statement gives just a little idea about the power of Laplace transform.
Laplace transform is a correspondence which associate to a measurable function with a certain growth at infinite a complex olomorph function in a semi space in the complex plan.

Let $f:R\rightarrow C$ a function with properties:

(i)f is a measurable function
(ii) $\forall t<0, f(t)=0$ or $supp f\subseteq[0,+\infty)$
(iii)there is a couple or real constants $M>0,\sigma_0\in R$ such as

$|f(t)|\leq Me^{\sigma_0 t$

For such a function we take a complex $s\in C$ with $re(s)>\sigma_0$ and define

$F(s)=\mathcal{L}(f(t))(s)=\int_{0}^{+\infty}f(t)e^{-st}dt$

The function $g(t)=f(t)e^{-st}$ is in the Lebesgue space $L^1(R)$ and using theorem of differentiability of integral with parameter we have, as $\frac{\partial }{\partial \overline{s}}e^{-st}=0$

$\frac{\partial F}{\partial \overline{s}}(s)=\frac{\partial }{\partial \overline{s}}\mathcal{L}(f(t))(s)=\int_{0}^{+\infty}f(t)\frac{\partial e^{-st}}{\partial \overline{s}}dt=0$

This is valid for $s\in C, re(s)>\sigma_0$ as for such a s we have $s=\sigma +i\tau$

$|e^{-st}|=|e^{-(\sigma +i\tau)t}|=|e^{-\sigma t}e^{-i\tau t}|=e^{-\sigma t}$

$|f(t)|\leq Me^{\sigma_0 t$

and

$|f(t)e^{-st}|\leq Me^{\sigma_0 t}e^{-\sigma t}=e^{-(\sigma -\sigma_0 )t}\in L^1([0,+\infty))$

From the last inequality we have

$\lim_{\sigma \to +\infty}|F(s)|\leq \lim_{\sigma \to +\infty}\int_{0}^{+\infty}e^{-(\sigma -\sigma_0 )t}dt=\lim_{\sigma \to +\infty}\frac{1}{\sigma -\sigma_0 }=0$

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