A Note on the Inverse Laplace Transformation of $f(t)$

TL;DR

This paper investigates the inverse Laplace transform, demonstrating that the condition (s) o 0 as sor does not guarantee that F(s) is the Laplace transform of a piecewise continuous exponential order function.

Contribution

It proves that the common converse condition for Laplace transforms is not sufficient to identify the original function as piecewise continuous and of exponential order.

Findings

The limit of F(s) as s approaches infinity being zero does not imply F(s) is a Laplace transform of a piecewise continuous exponential order function.

The paper clarifies a misconception about the inverse Laplace transform conditions.

It provides a counterexample to the converse of a well-known Laplace transform property.

Abstract

Let $\mathcal{L}\{f(t)\} = \int_{0}^{\infty}e^{-st}f(t)dt$ denote the Laplace transform of $f$. It is well-known that if $f(t)$ is a piecewise continuous function on the interval $t:[0,\infty)$ and of exponential order for $t > N$; then $\lim_{s\to\infty}F(s) = 0$, where $F(s) = \mathcal{L}\{f(t)\}$. In this paper we prove that the lesser known converse does not hold true; namely, if $F(s)$ is a continuous function in terms of $s$ for which $\lim_{s\to\infty}F(s) = 0$, then it does not follow that $F(s)$ is the Laplace transform of a piecewise continuous function of exponential order.