I'm struggling in demonstrating that the relation in the first quoted equation.
I've met this problem in finding the inverse Laplace Transform of an equation in the form $W=A\times B$, there $A$ and $B$ have the form written in the second quote.
To find the inverse Laplace transform of $W$, I have to obtain something like the thing in the third image, where $C$ is the inverse transform I'm looking for. I know that the relation in the fourth quote holds, but I don't know how to arrive at something like the equation in the first quote.
What am I missing?
$$\left(\int_0^{+\infty}e^{-\lambda t}f(x,t)\,\mathrm dt\right)\times\left(\int_0^{+\infty}e^{-\lambda t}g(t)\,\mathrm dt\right)=\int_0^{+\infty}e^{-\lambda t}\left(\int_0^tf(x,t-s)g(s)\,\mathrm ds\right)\,\mathrm dt$$
$$\begin{align}A&=\int_0^{+\infty}e^{-\lambda t}f(x,t)\,\mathrm dt\\B&=\int_0^{+\infty}e^{-\lambda t}g(t)\,\mathrm dt\end{align}$$
$$A\times B=\int_0^{+\infty}e^{-\lambda t}C\,\mathrm dt$$
$$\begin{align}A\times B&=\left(\int_0^{+\infty}e^{-\lambda t}f(x,t)\,\mathrm dt\right)\times\left(\int_0^{+\infty}e^{-\lambda t}g(t)\,\mathrm dt\right)\\[5pt]&=\left[\int_0^{+\infty}e^{-\lambda t}\left(\int_0^tf(x,t-s)g(s)\,\mathrm ds\right)\,\mathrm dt\right]\end{align}$$
This is an application of the Laplace Convolution Formula.
Let $h(t) = f(x,t)$.
Let $L[h] = \int_0^\infty h(t)e^{-st} dt$, the Laplace transform of h.
Then the Laplace Convolution Formula states that:
$L[h]L[g] = L[h * g]$, where $h * g = \int_o^t h(x)g(t - x) dx$.
Hence $L[h * g] = \int_0^\infty e^{-st}\int_o^t h(x)g(t - x) dx dt $
This is exactly the first equality you have above.