# Second Derivative of Laplace Transform

## Theorem

Let $f: \R \to \R$ or $\R \to \C$ be a continuous function, twice differentiable on any closed interval $\closedint 0 a$.

Let $\laptrans f = F$ denote the Laplace transform of $f$.

Then, everywhere that $\dfrac {\d^2} {\d^2 s} \laptrans f$ exists:

$\dfrac {\d^2} {\d^2 s} \laptrans {\map f t} = \laptrans {t^2 \, \map f t}$

## Proof

 $\displaystyle \dfrac {\d^2} {\d^2 s} \laptrans {\map f t}$ $=$ $\displaystyle \map {\frac \d {\d s} } {\dfrac \d {\d s} \laptrans {\map f t} }$ Definition of Second Derivative $\displaystyle$ $=$ $\displaystyle \map {\frac \d {\d s} } {-\laptrans {t \, \map f t} }$ Derivative of Laplace Transform $\displaystyle$ $=$ $\displaystyle -\frac \d {\d s} \laptrans {t \, \map f t}$ $\displaystyle$ $=$ $\displaystyle -\paren {-\laptrans {t \paren {t \, \map f t} } }$ Derivative of Laplace Transform $\displaystyle$ $=$ $\displaystyle \laptrans {t^2 \, \map f t}$

$\blacksquare$