Skip navigation.

Laplace transform

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


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 $

$$|e^{-st}|=|e^{-(\sigma +i\tau)t}|=|e^{-\sigma t}e^{-i\tau t}|=e^{-\sigma t}$$
$$|f(t)|\leq Me^{\sigma_0 t$$


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

From the last inequality we have

$$\lim_{\sigma \to +\infty}|F(s)|\leq \lim_{\sigma \to +\infty}\int_{0}^{+\infty}e^{-(\sigma -\sigma_0 )t}dt=\lim_{\sigma \to +\infty}\frac{1}{\sigma -\sigma_0 }=0$$
Average: 5 (1 vote)