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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?

$$\left(\int_0^{+\infty}e^{-\lambda t}f(x,t)\,\mathrm dt\right)\times\left(\int_0^{+\infty}e^{-\lambda t}g(t)\,\mathrm dt\right)=\int_0^{+\infty}e^{-\lambda t}\left(\int_0^tf(x,t-s)g(s)\,\mathrm ds\right)\,\mathrm dt$$

$$\begin{align}A&=\int_0^{+\infty}e^{-\lambda t}f(x,t)\,\mathrm dt\\B&=\int_0^{+\infty}e^{-\lambda t}g(t)\,\mathrm dt\end{align}$$

$$A\times B=\int_0^{+\infty}e^{-\lambda t}C\,\mathrm dt$$

$$\begin{align}A\times B&=\left(\int_0^{+\infty}e^{-\lambda t}f(x,t)\,\mathrm dt\right)\times\left(\int_0^{+\infty}e^{-\lambda t}g(t)\,\mathrm dt\right)\\[5pt]&=\left[\int_0^{+\infty}e^{-\lambda t}\left(\int_0^tf(x,t-s)g(s)\,\mathrm ds\right)\,\mathrm dt\right]\end{align}$$

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1 Answer 1

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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.

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