# Laplace transform of derivative

If the real function$t\mapsto f(t)$  and its derivative are Laplace-transformable and $f$ is continuous for  $t>0$,  then

 $\displaystyle\mathcal{L}\{f^{\prime}(t)\}\;=\;s\,F(s)-\lim_{t\to 0+}\!f(t).$ (1)

Proof.  By the definition of Laplace transform and using integration by parts, the left hand side of (1) may be written

 $\int_{0}^{\infty}\!e^{-st}f^{\prime}(t)\,dt\;=\;\operatornamewithlimits{\Big{/% }}_{\!\!\!t=0}^{\,\quad\infty}\!e^{-st}f(t)+s\!\int_{0}^{\infty}\!e^{-st}f(t)% \,dt\;=\;\lim_{t\to\infty}e^{-st}f(t)-\lim_{t\to 0}e^{-st}f(t)+s\,F(s).$

The Laplace-transformability of $f$ implies that $e^{-st}f(t)$ tends to zero as $t$ increases boundlessly.  Thus the last expression leads to the right hand side of (1).

Title Laplace transform of derivative LaplaceTransformOfDerivative 2013-03-22 18:24:54 2013-03-22 18:24:54 pahio (2872) pahio (2872) 5 pahio (2872) Theorem msc 44A10 SubstitutionNotation