Find Laplace transform for $\int_{0}^{x}\sin(x-t)y(t)dt$

Question

How do I go about finding the laplace transform for $$\int_{0}^{x}\sin(x-t)y(t)dt$$ In other words how do I evaluate $$\int_{0}^{\infty}e^{-sx}{\int_{0}^{x}\sin(x-t)y(t)dt}dx$$

EDIT: Unfortunately I don't have any working because I don't know how to start. I was thinking the inner integral has something to do with convolution but the outside integral makes it confusing for me, so I was hoping for some clarification/guidance.

EDIT 2: I'm not sure if I can use the fact that the $L\{\int_{0}^{x}\sin(x-t)y(t)dt\}=L\{\sin(x-t)\}L\{y(t)\}$ - is that even true?

Answer 1

Hint:

Consider a convolution integral of function $f$ and $g$ \begin{align} f * g = \int_{0}^{t} f(t-t^{\prime}) g(t^{\prime}) \, dt^{\prime} \end{align} The Laplace transform of $f * g$ is given by \begin{align} \mathcal{L}[f * g] = \mathcal{L}[f] \; \mathcal{L}[g] \end{align}