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 $\laptrans f$ denote the Laplace transform of $f$. Let $f'$ be piecewise continuous with one-sided limits on said intervals.

Then $\laptrans f$ exists for $\map \Re s > a$, and:


 * $\laptrans {\map {f'} t} = s \laptrans {\map f t} - \map f 0$

Proof
Consider:


 * $\ds \int_0^A e^{-s t} \map {f'} t \rd t$

By hypothesis, $f'$ is piecewise continuous with one-sided limits.

So by Piecewise Continuous Function with One-Sided Limits is Darboux Integrable, this integral exists.

This means that integration by parts can be invoked:


 * $\ds \int h j\,' \rd t = h j - \int h' j \rd t$

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 for Functions:


 * $\ds \lim_{A \mathop \to +\infty} e^{-s A} \map f A = 0$

which produces:


 * $\laptrans {\map {f'} t} = s \laptrans {\map f t} - \map f 0$

Also see

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