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


Sources

  • 1965: Murray R. Spiegel: Theory and Problems of Laplace Transforms ... (previous) ... (next): Chapter $1$: The Laplace Transform: Laplace Transforms of Special Functions: $13$
  • 1965: Murray R. Spiegel: Theory and Problems of Laplace Transforms ... (previous) ... (next): Appendix $\text B$: Table of Special Laplace Transforms: $113.$
  • 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 32$: Table of Special Laplace Transforms: $32.137$
  • 2009: Murray R. Spiegel, Seymour Lipschutz and John Liu: Mathematical Handbook of Formulas and Tables (3rd ed.) ... (previous) ... (next): $\S 33$: Laplace Transforms: Table of Special Laplace Transforms: $33.137.$