Laplace Transform of Dirac Delta Function by Function

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Theorem

Let $\map f t: \R \to \R$ or $\R \to \C$ be a function.

Let $\map \delta t$ denote the Dirac delta function.

Let $c$ be a positive constant real number.

Let $\laptrans {\map f t} = \map F s$ denote the Laplace transform of $f$.


Then:

$\laptrans {\map \delta {t - c} \, \map f t} = e^{- s c} \, \map f c$


Proof

\(\displaystyle \laptrans {\map \delta {t - c} \, \map f t}\) \(=\) \(\displaystyle \int^{\to+\infty}_0 e^{-s t} \map \delta {t - c} \, \map f t \rd t\) Definition of Laplace Transform
\(\displaystyle \) \(=\) \(\displaystyle \int^{c^+}_{c^-} e^{-s t} \map \delta {t - c} \, \map f t \rd t\) Integrand elsewhere zero by Definition of Dirac Delta Function
\(\displaystyle \) \(=\) \(\displaystyle \int^{c^+}_{c^-} e^{-s c} \map \delta {t - c} \, \map f c \rd t\) $e^{-s t}$ and $\map f t$ are constant in interval $\closedint {c^-} {c^+}$
\(\displaystyle \) \(=\) \(\displaystyle e^{-s c} \, \map f c \int^{c^+}_{c^-} \map \delta {t - c} \rd t\) Primitive of Constant Multiple of Function
\(\displaystyle \) \(=\) \(\displaystyle e^{-s c} \, \map f c \int^{0^+}_{0^-} \map \delta {t - c} \rd \paren {t - c}\) Integration by Substitution
\(\displaystyle \) \(=\) \(\displaystyle e^{-s c} \, \map f c\) Definition of Dirac Delta Function

$\blacksquare$