Laplace Transform of Higher Order Derivatives

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 $f$ be $n$ times differentiable on said intervals.

Let $f,f',\ldots,f^{\left({n-1}\right)}$ be continuous and $f^{\left({n}\right)}$ piecewise continuous on said intervals.

Let $f, f',\ldots, f^{\left({n-1}\right)}$ be of exponential order $a$.

Let $\mathcal Lf = F$ be the Laplace transform.

Then $\mathcal L \left\{{f^{\left({n}\right)}}\right\}$ exists for $\operatorname{Re}\left({s}\right) > a$, and:


 * $\displaystyle \mathcal L \left\{{f^{\left({n}\right)} \left({t}\right)}\right\}= s^n \mathcal L \left\{{f\left({t}\right)}\right\} - \sum_{j \mathop = 1}^n s^{j-1} f^{\left({n-j}\right)}\left({0}\right)$


 * $= s^n F\left({s}\right) - s^{n-1}f\left({0}\right) - s^{n-2}f'\left({0}\right) - s^{n-3}f''\left({0}\right) - \ldots - sf^{\left({n-2}\right)}\left({0}\right) - f^{\left({n-1}\right)}\left({0}\right)$

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

Basis for the Induction
By hypothesis $f$ is of exponential order $a$.

Thus Laplace Transform of Derivative can be invoked:


 * $\mathcal L \left\{{f'\left({t}\right)}\right\} = s \mathcal L \left\{{f\left({t}\right)}\right\} - f\left({0}\right)$

This is the basis for the induction.

Induction Hypothesis
Fix $n - 1 \in \N$ with $n \ge 2$.

Assume:


 * $\displaystyle \mathcal L \left\{{f^{\left({n-1}\right)} \left({t}\right)}\right\} = s^{n-1} \mathcal L \left\{{f\left({t}\right)}\right\} - \sum_{j \mathop = 1}^{n-1} s^{j-1} f^{\left({n-1-j}\right)}\left({0}\right)$

This is our induction hypothesis.

Induction Step
This is our induction step:

By hypothesis $f^\left({n-1}\right)$ is of exponential order $a$

Thus Laplace Transform of Derivative can be invoked:

The result follows by the Principle of Mathematical Induction.

Also see

 * Higher Order Derivatives of Laplace Transform