# Unit step function present in convolution result

We are currently learning about convolution in my signals and systems class, and one of our textbook problems is to compute the convolution of $x(t) = e^{-\alpha t} u(t)$ and $y(t) = e^{-\beta t} u(t)$. After working out the integral, I obtained the following result

$$x(t) * h(t) = \frac{e^{-\alpha t} - e^{-\beta t}}{\beta - \alpha}.$$

The solutions manual for my textbook has the same answer, except their result is multiplied by $u(t)$. I am unsure of where this is coming from, since I know that I computed the integral correctly. I used the unit step functions to simplify the bounds on the integral, so I do not see how they are still around in the final answer.

Can anyone explain what is going on here?

\begin{align} f(t) &= \int_{-\infty}^\infty x(\tau) h(t-\tau) \,d\tau \\ &= \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) &= \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 $t < 0$ 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) &= u(t) \int_0^t e^{-\alpha\tau} e^{-\beta(t-\tau)} \, d\tau \\ &= \cdots \\ &= u(t) \frac{e^{-\alpha t} - e^{-\beta t}}{\beta-\alpha} \end{align}