Differentiation

The differentiation theorem for Laplace transforms states that

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

where ${\dot x}(t) \isdef \frac{d}{dt}x(t)$, and $x(t)$ is any differentiable function that approaches zeros as $t$ goes to infinity. In operator notation,

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

Proof: This follows immediately from integration by parts:

$$\begin{eqnarray*} {\cal L}_{s}\{{\dot x}\} &\isdef & \int_{0}^\infty {\dot x}(t)... ...y} - \int_{0}^\infty x(t) (-s)e^{-s t} dt\\ &=& s X(s) - x(0) \end{eqnarray*}$$

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

Corollary: Integration Theorem

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

Thus, successive time derivatives correspond to successively higher powers of $s$, and successive integrals with respect to time correspond to successively higher powers of $1/s$.