Derivative of Laplace Transform

From ProofWiki
Jump to: navigation, search

Theorem

Let $f: \R \to \R$ or $\R \to \C$ be a continuous function, differentiable on any closed interval $\left[{0 \,.\,.\, A}\right]$.

Let $\mathcal L \left\{{f \left({t}\right)}\right\}$ be the Laplace transform of $f$.


Then, everywhere that $\dfrac {\d \mathcal L} {\d s}$ exists:

$\dfrac \d {\d s} \mathcal L \left\{{f \left({t}\right)}\right\} = -\mathcal L \left\{{t \, f \left({t}\right)}\right\}$


Proof

\(\displaystyle \frac \d {\d s} \mathcal L \left\{ {f \left({t}\right)} \right\}\) \(=\) \(\displaystyle \frac \d {\d s} \int_0^{\to +\infty} f \left({t}\right) \, e^{-st} \rd t\) $\quad$ Definition of Laplace Transform $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \int_0^{\to +\infty} \frac {\partial} {\partial s} \left({f \left({t}\right) \, e^{-st} }\right) \rd t\) $\quad$ Definite Integral of Partial Derivative $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \int_0^{\to +\infty} f\left({t}\right) \frac {\partial}{\partial s} \left({e^{-st} }\right) \rd t\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle - \int_0^{\to +\infty} t \, f \left({t}\right) \, e^{-st} \rd t\) $\quad$ Derivative of Exponential Function $\quad$
\(\displaystyle \) \(=\) \(\displaystyle - \mathcal L \left\{ {t \, f \left({t}\right)}\right\}\) $\quad$ Definition of Laplace Transform $\quad$

$\blacksquare$


Also see


Sources