Laplace Transform of Derivative

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

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

Let $f'$ be piecewise continuous on said intervals.

Let $\mathcal L$ be the Laplace transform.

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


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

Proof
Consider:


 * $\displaystyle \int_0^A e^{-st}f'\left({t}\right) \, \mathrm dt$

By hypothesis, $f'$ is piecewise continuous.

So by Piecewise Continuous Function is Riemann Integrable, this integral exists.

This means that integration by parts can be invoked:


 * $\displaystyle \int hj\,' \, \mathrm dt = hj - \int h'j \, \mathrm dt$

Here:

So:

Now, take the limit as $t = A \to +\infty$:

Recall that $f$ is of exponential order $a$:

This implies, from Complex Exponential Tends to Zero and the Squeeze Theorem:


 * $\displaystyle \lim_{A \mathop \to +\infty} e^{-sA}f\left({A}\right) = 0$

which produces:


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

Also see

 * Laplace Transform of Second Derivative
 * Laplace Transform of Higher Order Derivatives