Linearity

The Laplace transform is a linear operator:

$$\zbox{\alpha x(t) + \beta y(t) \longleftrightarrow \alpha X(s) + \beta Y(s)}$$

Proof: Let

$$w(t) = \alpha x(t) + \beta y(t),$$

where $\alpha$ and $\beta$ are real or complex constants. Then

$$\begin{eqnarray*} W(s) &\mathrel{\stackrel{\mathrm{\Delta}}{=}}& {\cal L}_s\{w\} \mathrel{\stackrel{\mathrm{\Delta}}{=}}{\cal L}_s\{\alpha x(t) + \beta y(t)\}\\ [10pt] &\mathrel{\stackrel{\mathrm{\Delta}}{=}}& \int_0^\infty \left[\alpha x(t) + \beta y(t)\right] e^{-st} dt\\ [10pt] &=& \alpha \int_0^\infty x(t) e^{-st} dt + \beta \int_0^\infty y(t) e^{-st} dt\\ [5pt] &\mathrel{\stackrel{\mathrm{\Delta}}{=}}& \alpha X(s) + \beta Y(s). \end{eqnarray*}$$

Thus, linearity of the Laplace transform follows immediately from linearity of integration