I'm reading through Sidney Redner's lectures on first passage processes (lectures at this link) and I've been stuck on the convolution steps he takes to go from equation (3.1) to (3.2). He starts by relating the first passage probability $F(\textbf{r}, t)$ to the occupation probability $P(\textbf{r}, t)$ by the following equation: $$ P(\textbf{r}, t) = \int_{0}^{t}{F(\textbf{r}, t')P(\textbf{0}, t-t')dt'} + \delta_{\textbf{r}, \textbf{o}}\delta_{t, 0}\text{.} $$
I Laplace transform $P(\textbf{r}, t)$ as he suggests, but I end up with $$ P(\textbf{r}, s) = \int_{0}^{s}{F(\textbf{r}, u)P(\textbf{0}, s-u)du} + \delta_{\textbf{r}, \textbf{0}} $$
instead of $P(\textbf{r}, s) = F(\textbf{r}, s)P(\textbf{0}, s) + \delta_{\textbf{r}, \textbf{0}}$. I got to this point by substituting $P(\textbf{r}, t)$ into the Laplace transform, splitting the integral into two terms, then having the term without the delta functions look like $$ P(\textbf{r}, s) = \int_{t=0}^{t=\infty}{\int_{t'=0}^{t'=t}{F(\textbf{r}, t)P(\textbf{0}, t-t')e^{-st}{dt}'}{dt}}\text{,} $$
where the $t'$ integral is a convolution, so the outer integral is the Laplace transform of the convolution of $F(\textbf{r}, t)$ and $P(\textbf{0}, t)$: $$ P(\textbf{r}, t) = \int_{t=0}^{t=\infty}{e^{-st}(F(\textbf{r}, t)*P(\textbf{0}, t))dt}=F(\textbf{r}, s) * P(\textbf{0}, s)\text{.} $$
I'm not sure if I've made a mistake somewhere or if there's a connection between my final expression and Redner's - any help would be greatly appreciated!