inverse laplace of sine function

Question

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?

Answer 1

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.

Answer 2

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.