inverse Laplace transform of a piecewise defined function

I understand the conditions for the existence of the inverse Laplace transforms are $\lim_{s\to\infty}F(s) = 0$ and $\lim_{s\to\infty}(sF(s))<\infty$.

I am interested in finding the inverse Laplace transform of a piecewise defined function defined, such as

$$F(s) =\begin{cases} 1-s &\text{ if }0\le s\le1 \text{ and}\\ 0&\text{ if } s>1 \end{cases}$$ Clearly the limits above do satisfy the existence of the inverse condition, but I'm not sure how to determine the inverse.

I'm not sure whether the Bromwich integral method can be applied, since it would appear that if I choose gamma between 0 and 1 the function to integrate is (1-s), whereas if I choose gamma > 1 then the Bromwich integral is obviously 0. I'm also not sure whether Post's inversion formula can be used since I'm not sure I understand how to evaluate high-order derivatives of a function which is not differentiable at $s = 1$. Clearly for a finite $k$, the $k$th order derivative of $F$ exists for all $s$ except $1$, but how about as $k\to\infty$?

Finally, just wondering if the two conditions I listed initially (the two limits) are sufficient for the inverse Laplace transform of $F(s)$ to exist.

• You can maybe just use your definition of inverse Laplace transform.
– N74
Feb 15 '17 at 21:00
• $F(s)$ has no singularities.
– Paul
Feb 15 '17 at 22:06
• N74, not sure I understand your answer. Feb 15 '17 at 23:37
• Paul, thanks for your comment. I've updated the question above. Feb 15 '17 at 23:39

I'm still not sure about how to arrive analytically at the inverse Fourier transform, but I have played around trying to fit a sum of exponentials to F(s), using F(s) values for s from 0 to 10 with a step size of 1e-3 and using 150 bins for the t from 0.005 to 30. The f(t) determined by least-squares fitting looks something like this (on a log10 scale for the x-axis):  