# Convolution: characteristic-functions

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?

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 $$g\ast h_{\alpha}(t)=\frac{1}{\alpha} t \mathbf{1}_{[0,\alpha]}(t) + \mathbf{1}_{(\alpha,\infty)}(t).$$

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) $$g\ast h_{\alpha}(t)=\frac{t}{\alpha} \mathbf{1}_{[0,\alpha]}(t) + \mathbf{1}_{(\alpha,\infty]}(t).$$