# 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 with one-sided limits 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

 $\displaystyle \mathcal L \left\{ {f' \left({t}\right)}\right\}$ $=$ $\displaystyle \int_0^{\mathop \to +\infty} e^{-s t} f' \left({t}\right) \rd t$ $\quad$ Definition of Laplace Transform $\quad$ $\displaystyle$ $=$ $\displaystyle \lim_{A \mathop \to +\infty} \int_0^A e^{-s t} f' \left({t}\right) \rd t$ $\quad$ Definition of Improper Integral on Closed Interval Unbounded Above $\quad$

Consider:

$\displaystyle \int_0^A e^{-s t} f' \left({t}\right) \rd t$

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

This means that integration by parts can be invoked:

$\displaystyle \int h j \,' \rd t = hj - \int h' j \rd t$

Here:

 $\displaystyle h$ $=$ $\displaystyle e^{-s t}$ $\quad$ $\quad$ $\displaystyle \leadsto \ \$ $\displaystyle h'$ $=$ $\displaystyle -s e^{-s t}$ $\quad$ $\quad$ $\displaystyle j\,'$ $=$ $\displaystyle f' \left({t}\right)$ $\quad$ $\quad$ $\displaystyle \leadsto \ \$ $\displaystyle j$ $=$ $\displaystyle f \left({t}\right)$ $\quad$ $\quad$

So:

 $\displaystyle \int_0^A e^{-s t} f' \left({t}\right) \rd t$ $=$ $\displaystyle \left.{e^{-s t} f \left({t}\right)}\right\vert_{t \mathop = 0}^{t \mathop = A} + s \int_0^A e^{-s t} f \left({t}\right) \rd t$ $\quad$ $\quad$

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

 $\displaystyle \mathcal L \left\{ {f' \left({t}\right)}\right\}$ $=$ $\displaystyle \lim_{A \mathop \to +\infty} e^{-s A} f \left({A}\right) - f \left({0}\right) + s \mathcal L \left\{ {f \left({t}\right)}\right\}$ $\quad$ $\quad$

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

 $\displaystyle \left\vert {f \left({A}\right)} \right \vert$ $<$ $\displaystyle Ke^{at}$ $\quad$ $\quad$ $\displaystyle \leadsto \ \$ $\displaystyle \left\vert {f \left({A}\right)} \right \vert \left \vert {e^{-s A} } \right \vert$ $<$ $\displaystyle Ke^{a t} \left\vert {e^{-sA} } \right \vert$ $\quad$ $\quad$ $\displaystyle \leadsto \ \$ $\displaystyle \left\vert {e^{-s A} f \left({A}\right)} \right \vert$ $<$ $\displaystyle \left \vert {Ke^{a t} e^{-s A} } \right \vert$ $\quad$ Modulus of Product, Exponential Tends to Zero and Infinity $\quad$ $\displaystyle$ $=$ $\displaystyle \left \vert {K e^{\left({a - s}\right)t} }\right \vert$ $\quad$ Exponent Combination Laws, definition of $A$ $\quad$ $\displaystyle$ $=$ $\displaystyle \left \vert {K e^{\left({a- \operatorname{Re} \left({s}\right) - i \operatorname{Im} \left({s}\right)}\right)t} }\right \vert$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle \left \vert {K e^{\left({a- \operatorname{Re} \left({s}\right)}\right) t} }\right \vert$ $\quad$ Modulus of Exponential is Modulus of Real Part $\quad$ $\displaystyle$ $=$ $\displaystyle K e^{\left({a- \operatorname{Re} \left({s}\right)}\right)t}$ $\quad$ Exponential Tends to Zero and Infinity $\quad$

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

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

which produces:

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

$\blacksquare$