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.