If $f(t)$ an $g(t)$ are piecewise continuous functions on $[\ 0, \infty)$ then the convolution integral of $f(t)$ and $g(t)$ is,
$$(f*g)(t) = \int_{0}^{t}f(t-\tau)g(\tau) \text{d} \tau.$$
The text then gives a 'fact':
$\mathcal{L} \{f*g \} =F(s)G(s),$ where $\mathcal{L} \{ f(t) \} = F(s)$.
I tried to show this, but I'm not sure if it's correct. First
$$ \mathcal{L} \{f*g \} = \int_{0}^{\infty} e^{-st} \left [\ \int_{0}^{t}f(t-\tau)g(\tau) \text{d} \tau \right]\ \text{d}t. $$
I'm weary of swapping integral signs because don't exactly know when it is valid, but
$$ \mathcal{L} \{f*g \} = \int_{0}^{\infty} g(\tau) \left [\ \int_{\tau}^{\infty}f(t-\tau) e^{-st} \text{d} t \right]\ \text{d} \tau. $$ Using the substitution $u = t - \tau,$
$$ \mathcal{L} \{f*g \} = \int_{0}^{\infty} g(\tau) e^{-s \tau} \left [\ \int_{0}^{\infty}f(u) e^{-su} \text{d} u \right]\ \text{d} \tau = \left [\ \int_{0}^{\infty}f(u) e^{-su} \text{d} u \right]\ \left [\ \int_{0}^{\infty} g(\tau) e^{-s \tau} \text{d} \tau \right] $$
Does this make any sense?
Edit: $e^{-st}$ replaced with $e^{-su}$ and corrections to errors
As you know we have $$\mathcal{L} \{f*g \} = \int_{0}^{\infty} e^{-st} \left [\ \int_{0}^{t}f(t-\tau)g(\tau) \text{d} \tau \right]\ \text{d}t.$$ Note that the exponential has the following property $$ e^{-st} = e^{-s(t - \tau )} e^{-s\tau} $$ We may exploit this by sneaking the exponential into the $\tau$ integral with Fubini's to obtain $$\int_0^\infty \left [\int_0^t e^{-s ( t- \tau)} f( t - \tau) e^{-s \tau} g ( \tau) d \tau \right ] dt $$ Now let's change variables, say $u = t-\tau $ and $v = \tau$. What happens to the integral under this change of variables?