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

preprint2011arXivOpen access

Abstract

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

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