# Laplace Transform of Shifted Dirac Delta Function

## Theorem

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

The Laplace transform of $\map \delta {t - a}$ is given by:

$\laptrans {\map \delta {t - a} } = e^{-a s}$

## Proof 1

 $\ds \laptrans {\map \delta {t - a} }$ $=$ $\ds \int_0^{\to +\infty} e^{-s t} \map \delta {t - a} \rd t$ Definition of Laplace Transform $\ds$ $=$ $\ds e^{-s \times a}$ Integral to Infinity of Shifted Dirac Delta Function by Continuous Function $\ds$ $=$ $\ds e^{-a s}$

$\blacksquare$