Let $\alpha>0$ and $t\in\mathbb{R}$.
$g(t):=\mathbf{1}_{[0,\infty)}(t)$ and $h_{\alpha}(t):=\frac{1}{\alpha}\mathbf{1}_{[0,\alpha]}(t)$.
The convolution is:
$g\ast h_{\alpha}(t)=\int g(x)h_{\alpha}(t-x)dx$ $=\int \mathbf{1}_{[0,\infty)}(x)\frac{1}{\alpha}\mathbf{1}_{[0,\alpha]}(t-x)dx$ $=\frac{1}{\alpha}\int_{0}^{t}1 dx$ $=\frac{1}{\alpha} t$
But actually the solution is: $\frac{1}{\alpha} t \mathbf{1}_{[0,\alpha)}(t)$.
Does anyone see where I made a mistake?
On
Following your computation, just being a bit more cautious about the connection between the upper limit of the integral and the definition of $h_\alpha$ yields $$ g\ast h_{\alpha}(t)=\int_{x=0}^t g(x)h_{\alpha}(t-x)dx =\int_{x=0}^t \mathbf{1}_{[0,\infty)}(x)\frac{1}{\alpha}\mathbf{1}_{[0,\alpha]}(t-x)dx\\ =\frac{1}{\alpha} \int_{x=0}^t \mathbf{1}_{[0,\alpha]}(t-x)dx =\frac{1}{\alpha}\int_{y=0}^{t} \mathbf{1}_{[0,\alpha]}(y)dy = \frac{t}\alpha \quad \mathrm{for}\; t\leq\alpha $$ where we used the simple substitution $y=t-x$. When $t$ goes over $\alpha$, the integrand becomes zero so for $t>\alpha$ we can write the last integral above as $$ \frac{1}{\alpha}\int_{y=0}^{t} \mathbf{1}_{[0,\alpha]}(y)dy =\frac{1}{\alpha}\int_{y=0}^{\alpha} \mathbf{1}_{[0,\alpha]}(y)dy +\frac{1}{\alpha}\int_{y=\alpha}^{t} \mathbf{1}_{[0,\alpha]}(y)dy =1 \quad \mathrm{for}\; t>\alpha $$ the last integral being zero. If you want to put the above into one formula, you can write (as in the answer by Dosidis) $$ g\ast h_{\alpha}(t)=\frac{t}{\alpha} \mathbf{1}_{[0,\alpha]}(t) + \mathbf{1}_{(\alpha,\infty]}(t). $$
Starting from the integral you wrote $$\int \mathbf{1}_{[0,\infty)}(x)\frac{1}{\alpha}\mathbf{1}_{[0,\alpha]}(t-x)dx$$
we have two restrictions
$$\begin{cases}x\geq 0\\0\leq t-x\leq a\end{cases}\Leftrightarrow \begin{cases}x\geq 0\\t-a\leq x\leq t.\end{cases} $$
There are 3 cases. When $t<0$ there are no solutions and the integrant is zero.
If $0\leq t\leq \alpha $, then $t-\alpha\leq 0 $ and the system above gives $0\leq x\leq t$ and the intergal becomes $\frac{1}{\alpha}\int_{0}^{t}1 dx = \frac{1}{\alpha} t$ as you wrote.
If $t>\alpha$ then $t-\alpha>0$ and the system gives $t-a\leq x\leq t$. The integral then becomes $$\frac{1}{\alpha}\int_{t-a}^{t}1 dx = \frac{1}{\alpha} (t - (t-\alpha)) = 1$$ Thus $$g\ast h_{\alpha}(t)=\frac{1}{\alpha} t \mathbf{1}_{[0,\alpha]}(t) + \mathbf{1}_{(\alpha,\infty)}(t).$$