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For a given a Laplace transform $F(s)$ of a function $f(t)$, it can be shown that $$\mathfrak{L}\{t^n f(t)\}, n \in \Bbb{N} = (-1)^n F^{(n)}(s)$$ It then follows that for any function $F(s)$ where $$\lim_{n\to\infty}D^{n}F(s) = 0$$ Where D is the differential operator $\frac{d}{ds}$, has an inverse Laplace transform $$\mathfrak{L^{-1}}\{F(s)\}= 0$$ So for a given function $F(s)$ to have a non-zero inverse Laplace transform, the second equation must be false. My question is if there is a way to succinctly express this condition. For example, my initial thought is that $F(s)$ must be an eigenfunction of the differential operator, however I'm not sure if this is correct.

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  • $\begingroup$ A counter example to my original though is F(s) = sin(s), where F(s) is not an eigenfunction of the differential operator, however D^nF(s) does not tend towards zero as n tends towards infinity. $\endgroup$ – Corsair64 Apr 5 at 1:12

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