Differentiation

The differentiation theorem for Laplace transforms:

$$\zbox{{\dot x}(t) \leftrightarrow s X(s) - x(0)}$$

where ${\dot x}(t) \mathrel{\stackrel{\mathrm{\Delta}}{=}}\frac{d}{dt}x(t)$ , and $x(t)$ is any differentiable function that approaches zero as $t$ goes to infinity.

Operator notation:

$$\zbox{{\cal L}_{s}\{{\dot x}\} = s X(s) - x(0).}$$

Proof: Immediate from integration by parts:

$$\begin{eqnarray*} {\cal L}_{s}\{{\dot x}\} &\mathrel{\stackrel{\mathrm{\Delta}}{=}}& \int_{0}^\infty {\dot x}(t) e^{-s t} dt\\ [10pt] &=& \left. x(t)e^{-s t}\right\vert _{0}^{\infty} - \int_{0}^\infty x(t) (-s)e^{-s t} dt\\ [10pt] &=& s X(s) - x(0) \end{eqnarray*}$$

since $x(\infty)=0$ by assumption

Corollary: Integration Theorem:

$$\zbox{{\cal L}_{s}\left\{\int_0^t x(\tau)d\tau\right\} = \frac{X(s)}{s}}$$