Restrucutred this question, as it felt more natural to be asked in a different matter than I first wrote.
Given the functions
$$f(t) = \frac{e^{-2|t|}}{4},\quad g(t) = \frac{e^{-3|t|}}{6}$$
Find the convolution $f(t) * g(t)$
$$f(t)*g(t) = \int_{-\infty}^\infty f(t-\tau)g(\tau)\,d\tau =\frac{1}{24} \int_{-\infty}^\infty e^{-2|t-\tau|}e^{-3|\tau|}\,d\tau$$
Because $t$ is a variable, how do I deal with this? It seems like I can't just divide it into two integrands, because $|\tau|$ won't always have the same discontinuity point as $|t-\tau|$. In fact, this only happens when $t=0$.
Assuming $t<0\,:$
$$\frac{1}{24} \int_{-\infty}^\infty e^{-2|t-\tau|}e^{-3|\tau|}\,d\tau = \frac{1}{24}\left(\int_{-\infty}^0 e^{-2|t-\tau|}e^{3\tau}\,d\tau + \int_{0}^\infty e^{2(t-\tau)}e^{-3\tau}\,d\tau\right)$$
I don't know how to apply the sign replacement in the case of $e^{-2|t-\tau|}$ when integrating $(-\infty,0]$ due to the fact that if $|\tau| \ge |t|$, then we'd have a positive sign replace the absolute value function; but if $|\tau| < |t|$, a negative sign would replace the absolute value function.