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 $\mathcal Lf$, the Laplace transform of $f$, exist.

Then, everywhere that $\mathcal Lf$ is $n$ times differentiable:


 * $\dfrac {\mathrm d^n} {\mathrm ds} \mathcal L \left\{ {f \left({t}\right)}\right\} = \begin{cases}

- \mathcal L \left\{ {t^n \, f \left({t}\right)}\right\} & : \text{n odd}\\ \ \ \ \mathcal L \left\{ {t^n \, f \left({t}\right)}\right\} & : \text{n even} \end{cases} = \left({-1}\right)^n \mathcal L \left\{{t^n \, f \left({t}\right)}\right\}$

Proof
That:


 * $\left({-1}\right)^n = \begin{cases}

-1 & : \text{n odd}\\ \ \ \ 1 & : \text{n even} \end{cases}$

follows from Even Power is Non-Negative.

For the left equality, the proof proceeds by induction on $n$, the order of the derivative of $\mathcal Lf$.

Basis for the Induction
The case $n = 0$ is verified as follows:

This is the basis for the induction.

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

Assume:


 * $\displaystyle \frac {\mathrm d^n} {\mathrm ds} \mathcal L \left\{ {f \left({t}\right)}\right\} = \begin{cases}

- \mathcal L \left\{ {t^n \, f \left({t}\right)}\right\} & : \text{n odd}\\ \ \ \ \mathcal L \left\{ {t^n \, f \left({t}\right)}\right\} & : \text{n even} \end{cases}$

This is our induction hypothesis.

Induction Step
This is our induction step:

Suppose $n+1$ is odd.

Then $n$ is even.

So:

Now suppose $n+1$ even and $n$ odd.

Then:

The result follows by the Principle of Mathematical Induction.

Also see

 * Laplace Transform of Higher Order Derivatives