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

\(\ds \)

\(=\)

\(\ds e^{-a s}\)

$\blacksquare$

Also see

• Laplace Transform of Dirac Delta Function

Sources

• 1965: Murray R. Spiegel: Theory and Problems of Laplace Transforms ... (previous) ... (next): Chapter $1$: The Laplace Transform: Laplace Transforms of Special Functions: $13$