I am learning how to use Laplace transforms to solve PDEs. In case of an expression like
\begin{equation} \int_{r}^{\infty} f(x,t) g(x) dx \end{equation}
the resulting Laplace transform (with respect to t) is relatively simple if the expression is first differentiated by $r$:
\begin{equation} \frac{\partial}{\partial r} \int_{r}^{\infty} f(x,t) g(x) dx = -f(r, t) g(r) \end{equation}
and
\begin{equation} \mathcal{L} ( -f(r, t) g(r) ) = -g(r) F(r,s) \end{equation}
Now I have been working with a similar expression:
\begin{equation} \int_{r}^{\infty} f(x,t) g(x-r) dx \end{equation}
However, in this case, the derivative with respect to $r$ does not simplify the expression:
\begin{equation} \frac{\partial}{\partial r} \int_{r}^{\infty} f(x,t) g(x-r) dx = -f(r, t) g(0) - \int_{r}^{\infty} f(x,t) g'(x-r) dx \end{equation}
and I could not obtain a simplified expression with the Laplace transform. Is there another way to simplify the expression above?
EDIT:
It occurred to me that the term above is a convolution of $f$ and $g$:
\begin{equation} (f \ast g)(r) = \int_{r}^{\infty} f(x,t) g(x-r) dx \end{equation}
In this case, is it valid to say that the Laplace transform (this time with respect to $r$) can be defined as:
\begin{equation} \mathcal{L} \left( \int_{r}^{\infty} f(x,t) g(x-r) dx \right) = F(s,t)G(s) ? \end{equation}