Laplace Transform

Question

How can one show that 1/$e^s$ is not the laplace transform of any function?

Note that function here does not include distributions like dirac delta function.

Answer 1

Well, since $\hat{f}(s) = e^{-s} \implies f(t) = \delta(t-1)$, then, yes, there is no non-distribution function that is the ILT of $\hat{f}(s) = e^{-s}$.

Answer 2

Suppose that $e^{-s}$ is the Laplace transform of some function $f(t)$. Then, $$f(t) = \frac{1}{2\pi i} \lim_{T \to \infty} \int_{\gamma -iT}^{\gamma+iT} e^{st}e^{-s}\ ds = \delta(t-1).$$

By uniqueness of the inverse Laplace transform, this is the only such "function."