# Laplace Transform of Dirac Delta Function by Function

## 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

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

$\blacksquare$