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Regarding Inverse Laplace Transforms

  1. Jan 16, 2010 #1
    I was looking through some tables of Laplace Transforms on f(t) the other day, and I noticed that in all cases, as [tex]s \to \infty[/tex], [tex]F(s) \to 0[/tex]. A question that I have been trying to prove is that if [tex]\lim_{s\to\infty}F(s) = 0[/tex], then does that necessitate whether [tex]F(s)[/tex] 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 [tex]F^{(k)}\left(\frac{k}{t}\right)[/tex], but so far this has been futile)?

    Thanks (and if my question needs any clarification please let me know),
  2. jcsd
  3. Jan 18, 2010 #2
    Managed to prove it.
  4. Jan 22, 2010 #3
    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 [itex]F(s)=\frac{1}{\sqrt{s}}[/itex]. But of course we can find the inverse from the Laplace transform table.

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