Double integral and variable change problem

Question

I need to prove this following double integral
$$\int_{0}^{\infty} \left(\int_{0}^{t}f(s)\:g(t-s)\:ds \right)dt$$
can be rewritten like this:
$$\int_{0}^{\infty}f(s)\:ds\:\int_{s}^{\infty}g(t-s)\:ds$$
I tried a variable change but I didn't go too far. $f(x)$ and $g(x)$ are continuous functions.
Thank you in advance.

Answer 1

So, we start with the integral

$$I=\int_{0}^{\infty} \left(\int_{0}^{t}f(s)\:g(t-s)\:ds \right)dt$$

and by changing the order of integration, we obtain

$$I=\int_{0}^{\infty} \left(\int_{s}^{\infty}f(s)\:g(t-s)\:dt \right)ds$$

Now, since $f$ is only a function of the dummy integration variable $s$, we can write

$$I=\int_{0}^{\infty} f(s)\left(\int_{s}^{\infty}\:g(t-s)\:dt \right)ds$$

Now, for the inner integral, substitute $t-s=x$ so that $dt=dx$ and the limits of integration become $0$ to $\infty$. Then, we have

$$\begin{align} I&=\int_{0}^{\infty} f(s)\left(\int_{s}^{\infty}\:g(t-s)\:dt \right)ds\\\\ &=\int_{0}^{\infty} f(s)\left(\int_{0}^{\infty}\:g(x)\:dx \right)ds\\\\ &=\int_{0}^{\infty} f(s)ds\int_{0}^{\infty}\:g(x)\:dx \end{align}$$