Laplace transform of a derivative

Statement

Let $f : [0, \infty) \to \mathbb{C}$ be continuously differentiable, and let $M, a$ be real constants with $|f(t)| \le M e^{a t}$ for all $t \ge 0$. Let $F$ be the Laplace transform of $f$. Then for each $s \in \mathbb{C}$ with $\operatorname{Re} s > a$, the Laplace transform of $f'$ converges at $s$ and $$\mathcal{L}\{f'\}(s) = s F(s) - f(0).$$

Remarks

Proof: integration by parts gives $\int_0^T f'(t) e^{-st}\, dt = f(T) e^{-sT} - f(0) + s \int_0^T f(t) e^{-st}\, dt$. The term $f(T) e^{-sT}$ tends to $0$ for $\operatorname{Re} s > a$. Applied twice, under the same hypotheses on $f$ and $f'$: $\mathcal{L}\{f''\}(s) = s^2 F(s) - s f(0) - f'(0)$. This rule is the basis of solving initial value problems with the Laplace transform. The initial values enter the transformed equation directly.

Explanations

L1 human

Differentiation in time becomes multiplication by $s$ after the transform, with a correction by the initial value. So a differential equation becomes an algebraic equation for $F$.