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 $\map f {0^-} = 0$.

Sources

• 1965: Murray R. Spiegel: Theory and Problems of Laplace Transforms ... (previous) ... (next): Chapter $1$: The Laplace Transform: Some Important Properties of Laplace Transforms: $5$. Laplace transform of derivatives: Theorem $1 \text{-} 7$