Higher Order Derivatives of Laplace Transform

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

Let $\dfrac {\partial f}{\partial s}$, the partial derivative of $f$ with respect to $s$, exist and be continuous on said intervals.

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

Then, everywhere that $\laptrans f$ exists and is $n$ times differentiable:


 * $\dfrac {\d^n} {\d s^n} \laptrans {\map f t} = \paren {-1}^n \laptrans {t^n \, \map f t}$

Proof
The proof proceeds by induction on $n$, the order of the derivative of $\laptrans f$.

The case $n = 0$ is verified as follows:

Basis for the Induction
The case $n = 1$ is demonstrated in Derivative of Laplace Transform:


 * $\dfrac \d {\d s} \laptrans {\map f t} = -\laptrans {t \, \map f t}$

This is the basis for the induction.

Induction Hypothesis
Fix $n \in \N$ with $n \ge 0$.

Assume:


 * $\dfrac {\d^n} {\d s^n} \laptrans {\map f t} = \paren {-1}^n \laptrans {t^n \, \map f t}$

This is our induction hypothesis.

Induction Step
This is our induction step:

The result follows by the Principle of Mathematical Induction.

Also see

 * Laplace Transform of Higher Order Derivatives