Second Derivative of Laplace Transform

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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
\(\displaystyle \) \(=\) \(\displaystyle \laptrans {t^2 \, \map f t}\)

$\blacksquare$


Also see


Sources