Laplace transform of $u(x,t)\sin{x}$

Question

I am trying to work out what the Laplace transform of $u(x,t)\sin{x}$ is.

I have worked it out as $L(u(x,t)\sin{x})=U(x,s)\sin{x}$, where $L(u(x,t))=U(x,s)$, using definition:

$$\int_0^{\infty} e^{-st}f(x)dt$$

In one of the comments on Solving $u_{t}+u_{xxx} = u\sin{x}$ using Laplace transforms , in a comment it is indicated that the Laplace transform of $u(x)\sin{x}$ is not $U(s)\sin{x}$.