# Inverse Laplace transform of a product of integrals

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.