# Laplace Transform of Second Derivative

## 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 twice differentiable.

Let $f'$ be continuous and $f''$ piecewise continuous with one-sided limits on said intervals.

Let $f, f'$ be of exponential order $a$.

Let $\mathcal L$ be the Laplace transform.

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

$\mathcal L \left\{{f'' \left({t}\right)}\right\} = s^2 \mathcal L \left\{{f \left({t}\right)}\right\} - s f \left({0}\right) - f' \left({0}\right)$

## Proof

 $\displaystyle \mathcal L \left\{ {f'' \left({t}\right)}\right\}$ $=$ $\displaystyle s \mathcal L \left\{ {f' \left({t}\right)}\right\} - f' \left({0}\right)$ $\quad$ Laplace Transform of Derivative $\quad$ $\displaystyle$ $=$ $\displaystyle s \left({s \mathcal L \left\{ {f \left({t}\right)}\right\} - f \left({0}\right)}\right) - f' \left({0}\right)$ $\quad$ Laplace Transform of Derivative $\quad$ $\displaystyle$ $=$ $\displaystyle s^2 \mathcal L \left\{ {f \left({t}\right)}\right\} - s f \left({0}\right) - f' \left({0}\right)$ $\quad$ $\quad$

$\blacksquare$