Laplace Transform of Derivative/Discontinuity at t = 0

Theorem
Let $f: \R \to \R$ or $\R \to \C$ be a continuous function, differentiable on any interval of the form $0 < t \le A$.

Let $f$ be of exponential order $a$.

Let $f'$ be piecewise continuous with one-sided limits on said intervals.

Let $\laptrans f$ denote the Laplace transform of $f$.

Let $f$ fail to be continuous at $t = 0$, but let:
 * $\ds \lim_{t \mathop \to 0} \map f t = \map f {0^+}$

exist.

Then $\laptrans f$ exists for $\map \Re s > a$, and:


 * $\laptrans {\map {f'} t} = s \laptrans {\map f t} - \map f {0^+}$

Proof
See Laplace Transform of Derivative/Discontinuity at t = a and use $a = 0$ and $\map f {0^-} = 0$.