inverse laplace of sine function I use the transformation rule
$L(f(t)*t^n) = F^{(n)}(s)(-1)^n$
to find out the inverse Laplace of $\sin(s)$.
$F(s) = \sin(s)$
$F''(s)=-F(s)$
$L(f(t)*t^2) = F''(s) = -F(s) = -L(f(t))$
$L(f(t)(1+t^2))= 0$
Here I can only see the trivial solution $f(t) = 0$. Is there any function whose Laplace is zero, other than zero? I mean I wonder where is my error?

If $\mathcal L\{f\} = F(s)$ vanishes on an infinite sequence of points that are
located at equal intervals along a line parallel to the real axis
$$
 F(s_0+n\sigma)=0\qquad (\sigma >0, n=1,\,2,\,\ldots)
$$
$s_0$ being a point of convergence of $\mathcal L\{f\}$; then it follows that $f(t)$ is a nullfunction.
So it follows that a Laplace transform $F(s)\neq 0$ cannot be periodic.

Thus $\sin(s)$ cannot be the Laplace transform of a function.

