Linearity of the Laplace transform

Remarks

By linearity of the integral, $\int_0^T (\alpha f + \beta g) e^{-st}\, dt = \alpha \int_0^T f e^{-st}\, dt + \beta \int_0^T g e^{-st}\, dt$ for each $T$, and the limit of a sum is the sum of the limits. So the identity holds on the half-plane $\operatorname{Re} s > \max(\sigma_c(f), \sigma_c(g))$, where $\sigma_c$ is the abscissa of convergence. Example: $\cosh t = \frac{1}{2}(e^t + e^{-t})$ has transform $\frac{1}{2}\left(\frac{1}{s - 1} + \frac{1}{s + 1}\right) = \frac{s}{s^2 - 1}$ for $\operatorname{Re} s > 1$. The transform of $\alpha f + \beta g$ may converge on a larger half-plane than those of $f$ and $g$: for $f(t) = 1 + e^t$ and $g(t) = -e^t$, both have abscissa $1$, while $f + g = 1$ has abscissa $0$.

Going deeper

For a student

Linearity is the reason the Laplace transform fits linear differential equations with constant coefficients: together with the rule for derivatives, it turns the equation into an algebraic equation for $F$. In operator language, $\mathcal{L}$ is a linear map from a space of functions (for instance those of exponential order) to a space of holomorphic functions on half-planes. Combined with Lerch's uniqueness theorem, linearity also makes the inverse transform linear.