Unit step function present in convolution result
A step by step derivation of this convolution would start with the following:

$$
\begin{align}
f(t)
  &amp;= \int_{-\infty}^\infty x(\tau) h(t-\tau) \,d\tau \\
  &amp;= \int_{-\infty}^\infty e^{-\alpha\tau} u(\tau) e^{-\beta(t-\tau)} u(t-\tau) \,d\tau
\end{align}
$$
And this is probably where you probably went on and tried to simplified the bounds on the integral with:
$$
\begin{align}
  f(t) &amp;= \int_0^t e^{-\alpha\tau} e^{-\beta(t-\tau)} \, d\tau \\
\end{align}
$$
forgetting that this is only true if $t \geq 0$. 
Indeed for $t &lt; 0$ the product $u(\tau) u(t-\tau)$ is 0 for all values of $\tau$, and the result of the integral is thus 0.

So, the correct expression would be:
$$
\begin{align}
f(t)
  &amp;= u(t) \int_0^t e^{-\alpha\tau} e^{-\beta(t-\tau)} \, d\tau \\
  &amp;= \cdots \\
  &amp;= u(t) \frac{e^{-\alpha t} - e^{-\beta t}}{\beta-\alpha}
\end{align} 
$$