I have seen a derivation of the Laplace transform of $\sin(\omega t)$ that was done as follows. Since the sine is the solution of the ODE $y''+\omega^2 y=0$ with the initial conditions $y(0)=0$ and $y'(0)=\omega$, we let $\mathcal{L}[y](s) = F(s)$ and take the Laplace transform of the ODE to get $$ s^2 F(s) - \omega + \omega^2 F(s) = 0 \iff F(s) = \frac{\omega}{s^2 + \omega^2} $$
However, Wikipedia lists $$ \mathcal{L}\left[ \sin(\omega t) \mathcal{U}(t)\right] = \frac{\omega}{s^2 + \omega^2}, $$ where $\mathcal{U}(\cdot)$ represents the Heaviside function. Where does this function come from in the transform table?
My guess: The typical Laplace Transform is given by $F(s)=\int_0^\infty e^{-ts}f(t)\,dt$, but the main assumption is that since the variable is time, $f(t)$ is assumed to just be zero for $t<0$. Technically, the definition of Laplace transform is $$ F(s)=\int_{-\infty}^\infty e^{-ts}f(t)\,dt, $$ where if $s=2\pi \xi i$, this is obviously the Fourier Transform. In your link, throwing in $u(t)$ next to every function just ensures that it doesn't matter which definition of Laplace Transform you use.