This question involves the Laplace transform of the convolution of two functions. The derivation in my textbook has a step that really confuses me. First I'll lay out their argument.
$$ f(t) = f_1(t) \otimes f_2(t) = \int_0^t f_1(t - \tau) f_2(\tau) \, d\tau $$ and it goes on to show that $$\mathscr L[f(t)] = F(s) = F_1(s)F_2(s)$$
Apparently their definition of convolution differs from other contexts as well (this is electrical engineering).
It starts with $$ \mathscr L[f(t) = \int_0^\infty \bigg[ \int_0^t f_1(t - \tau) f_2(\tau) \, d\tau \bigg]e^{-st} \, dt$$
and claims that by introducing $u(t - \tau)$ we can change the limits of the bracketed integral, becoming $$ \mathscr L[f(t) = \int_0^\infty \bigg[ \int_0^\infty f_1(t - \tau)u(t - \tau) f_2(\tau) \, d\tau \bigg]e^{-st} \, dt $$
My question is how does this allow us to go from $\int_0^t$ to $\int_0^\infty$? I have trouble understanding how we know to go from $0$ to $t$ to begin with (this pops up in solving RC differential equations, for example), but what exactly is the logic that allows us to achieve the last form?
I understand that, for example $\int_0^\infty u(t-2)dt = \int_2^\infty 1 dt$, so how can we change the upper limit? Thanks for your help.