There is asked in an example to do convolution $ h_1(t)*h_2(t) + h_3(t)*h_4(t) $ where
$h_1(t) = e^{-2t}u(t)$
$h_2(t) = 2e^{-t}u(t) $
$h_3(t) = e^{-3t}u(t) $
$h_4(t) = 4\delta(t) $
and then the answer is written $ [e^{-t}-e^{-2t}]u(t) + 12e^{-3t}u(t)$.
But for second part $$ h_3(t)*h_4(t) = h_3(t)*4\delta(t) = 4h_3(t)*\delta(t) = 4h_3(t) =4e^{-3t}u(t) $$
as $ x(t)*\delta(t)=x(t) $ and for first part:
as $ u(\tau)u(t-\tau) = 1 $ for $ 0<t<\tau $ and $0$ for otherwise
$$ h_1(t)*h_2(t) = \int_{0}^{t}{e^{-2\tau}2e^{-t+\tau}}d\tau =2e^{-t}\int_{0}^{t}e^{-\tau}d\tau =-2e^{-t}[e^{-t}-1] =-2[e^{-t}-e^{-2t}] $$
Where am I wrong? How did they get $u(t)$ though we use value for $ u(\tau)u(t-\tau)$?
Am I missing something ?
Both $h_1(t)$ and $h_2(t)$ are left-sided signals, therefore their ROC will also be left-sided. We know that the Laplace transform of two signals convolved in time domain is the product of their individual Laplace transform, and the resulting ROC is (at least) the intersection of their individual ROCs. So, the ROC of $h_1(t) \ast h_2(t)$ is left-sided. Therefore, the time-domain signal $h_1(t) \ast h_2(t)$ has to be left-sided. As a result, after doing the integration, we have to add $u(t)$ to the result to make it left-sided.