Regarding Inverse Laplace Transforms

[flouran]

I was looking through some tables of Laplace Transforms on f(t) the other day, and I noticed that in all cases, as $$s \to \infty$$, $$F(s) \to 0$$. A question that I have been trying to prove is that if $$\lim_{s\to\infty}F(s) = 0$$, then does that necessitate whether $$F(s)$$ can undergo an inverse Laplace transform (i.e. by the Bromwich integral)?

I suspect that the answer is "no", but if anyone has some attempt at a proof I would appreciate it (my idea would be to use Post's inversion formula and utilizing the Grunwald-Letnikov differintegral for evaluating $$F^{(k)}\left(\frac{k}{t}\right)$$, but so far this has been futile)?

Thanks (and if my question needs any clarification please let me know),
flouran

 

[flouran]

I was looking through some tables of Laplace Transforms on f(t) the other day, and I noticed that in all cases, as $$s \to \infty$$, $$F(s) \to 0$$. A question that I have been trying to prove is that if $$\lim_{s\to\infty}F(s) = 0$$, then does that necessitate whether $$F(s)$$ can undergo an inverse Laplace transform (i.e. by the Bromwich integral)?

No.

I suspect that the answer is "no", but if anyone has some attempt at a proof I would appreciate it (my idea would be to use Post's inversion formula and utilizing the Grunwald-Letnikov differintegral for evaluating $$F^{(k)}\left(\frac{k}{t}\right)$$, but so far this has been futile)?

Managed to prove it.

 

[matematikawan]

I was looking through some tables of Laplace Transforms on f(t) the other day, and I noticed that in all cases, as $$s \to \infty$$, $$F(s) \to 0$$. A question that I have been trying to prove is that if $$\lim_{s\to\infty}F(s) = 0$$, then does that necessitate whether $$F(s)$$ can undergo an inverse Laplace transform (i.e. by the Bromwich integral)?

Interesting. One book that I read gives |sF(s)| < M as a sufficient condition for the Bromwich integral to exist. With this condition it is not possible to invert $F(s)=\frac{1}{\sqrt{s}}$. But of course we can find the inverse from the Laplace transform table.

Is there a more less restricted sufficient condition? As you stated $\lim_{s\to\infty}F(s) = 0$ doesn't works.