Laplace Transform of Derivative/Discontinuity at t = a

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 $f$ have a jump discontinuity at $t = a$.

Then:

$\laptrans {\map {f'} t} = s \laptrans {\map f t} - \map f 0 - e^{-a s} \paren {\map f {a^+} - \map f {a^-} }$

Proof

See Laplace Transform of Derivative with Finite Discontinuities and use $n = 1$ and $a_1 = a$.

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 $1 \text{-} 8$