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?
 A: OK I got it. Whenever we take a derivative of something we must lose some information. Here the rule is a specific part of the general thing.
If we say
$L(tf(t))=-F'(s)$
then the most general form of F(s) is
$F(s)=L(f(t))+C$
for arbitrary C. Thus, considering the previous equation
$L(t^2f(t))=F''(s)$
$F(s)=L(f(t))+as+b$
$F''(s)=-F(s)=L(t^2f(t))=-L(f(t))-as-b$
$L(f(t)(t^2+1))=-(as+b)$
$f(t)=[L^{-1}(as+b)]/(t^2+1)$
I've ignored the minus sign due to arbitrary a and b.
$f(t)=[aL^{-1}(s) + L^{-1}(b)]/(t^2+1)$
$f(t)=a\delta'(t)/(t^2+1) + b\delta(t)/(t^2+1)$
a,b are arbitrary values. Since our differential equation has 2 degrees of ambiguity it seems convenient to have it in solution, too.
I think one of them is for sine and the other is for cosine.
A: 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.

