While learning how to compute the product of two Laplace transform and the inverse transform, I faced with this equality : $$\int_{0}^{\infty} \int_{0}^{\infty} e^{-s(x+y)}f(x)g(y) \ dx \ dy = \int_{0}^{\infty} \int_{0}^{t} e^{-st}f(t-u)g(u) \ du \ dt$$ with the change of variables $x=t-u$ , $u=y$ . The two things that I can't understand why this equality holds are:
$1-$ change in upper-bound, i.e. $\infty \to t$ and
$2-$ change in differentials i.e. $dx \ dy = du \ dt$.
I also attempt 'separating' the l.h.s to the product of two integrals but failed to reach the equality. That is $$\int_{0}^{\infty} \int_{0}^{\infty} e^{-s(x+y)}f(x)g(y) \ dx \ dy = \Big( \int_{0}^{\infty} e^{-sx}f(x) \ dx \Big) \Big( \int_{0}^{\infty} e^{-sy}g(y) \ dy \Big) = \Big( \int_{u}^{\infty} e^{-s(t-u)}f(t-u) \ dt \Big) \Big( \int_{0}^{\infty} e^{-su}g(u) \ du \Big).$$
Please help me. Thank you!
PS The book is trying convince the equality by just the following pic and no good explanation at all. The problem is that not only I can't figure that out by the pic but also I would rather have a 'calculated' solution.
The meaning of the picture presented in the book is the following. You are integrating in $x,y$ both from $0$ to $\infty$ so you are integrating basically in the first quadrant $\{x\ge0,y\ge0\}$. Then you apply tha change of variables $t=x+y$ , $u=y$ and since $x>0$ and $y>0$ you have $$t\ge u.$$ $$t\ge0,u\ge0$$ This means that you are now integrating on the first quadrant, but only where $t\ge u$, that is precisely the zone that is coloured.
Once you accept that we are integrating over the set $\{t\ge0,u\ge0,t\ge u\}$ you can write explicetly the integral
$$\int_{\{t\ge0,u\ge0,t\ge u\}}f(t,u)dudt=\int_{\{t\ge0\}}\int_{\{0\le u\le t\}}f(t,u)dudt=\int_0^\infty\int_0^tf(t,u)dudt$$