Let $x(t) = \sigma(t)$ and $h(t) = e^{-4t} \cdot \sigma(t)$, where $\sigma(t)$ is the unit step function. When I try to solve the convolution $x(t) \ast h(t)$, I get
$$ x(t) \ast h(t) = \int_{-\infty}^{\infty}\sigma(\tau) \cdot \sigma(t-\tau) \cdot e^{-4(t-\tau)} d\tau $$ $$ = \int_{0}^{\infty}\sigma(t-\tau) \cdot e^{-4(t-\tau)} d\tau $$ $$ = \int_{0}^{t}e^{-4(t-\tau)} d\tau = \frac{1}{4} - \frac{1}{4} \cdot e^{-4t}$$
According to my exercise sheet, the correct solution should be
$$ x(t) \ast h(t) = \left(\frac{1}{4} - \frac{1}{4} \cdot e^{-4t}\right) \cdot \sigma(t) $$
Why is that? In which step did I forget something so that there is a step function in the solution?