I spent a lot of time figuring out how to integrate a convolution of a heaviside function with another heaviside function, but so far I couldn't find any closed form.
$$\int_{-\infty}^{\infty} H\left(\frac{1}{2}-\tau\right) H_\frac{1}{2}(\tau)H_1(t-\tau) d\tau$$ and by $H_a$ I mean:
$$
H_a(x)=
\begin{cases}
1 & -a\leq x\leq a \\
0 & \;\text{ otherwise}
\end{cases}
$$
I know the integral is equal to:
$$\int_{-\frac{1}{2}}^{\frac{1}{2}}H_1(t-\tau) d\tau$$
But I can't work this off any further.
Best Answer
I have changed a little bit the notation by putting $H_a=\chi_a$, meaning this as the characteristic function. of the interval $[-a,a]$: therefore we have $$ \begin{split} \int\limits_{-\infty}^{+\infty} H\left(\frac{1}{2}-\tau\right) \chi_\frac{1}{2}(\tau)\,\chi_1(t-\tau) \mathrm{d}\tau &= \int\limits_{-\infty}^{+\frac{1}{2}} \chi_\frac{1}{2}(\tau)\,\chi_1(t-\tau) \mathrm{d}\tau \\ & = \int\limits_{-\frac{1}{2}}^{+\frac{1}{2}} \chi_1(t-\tau) \mathrm{d}\tau \\ \text{ and by the change of variables $t-\tau =z$ (and } z=\tau &\text{) we have that}\\ & = \int\limits_{t-\frac{1}{2}}^{t+\frac{1}{2}} \chi_1(\tau) \mathrm{d}\tau\\ \end{split}\label{1}\tag{1} $$ Now we have to analyze the different values of the integral for $t$ varying in $\mathbb{R}$ (I will consider only the values $t>0$ and $|t|\le \frac{1}{2}$ since the method is identical for $t<0$):
if $-\frac{1}{2}\le t\le \frac{1}{2}$, then $|\tau|\le 1$ thus $\chi_1(\tau)$ is equal to $1$. The last integral at the right member of \eqref{1} is then equal to the measure of the interval $\left[t-\frac{1}{2},t+ \frac{1}{2}\right]$ i.e. $1$: $$ \int\limits_{t-\frac{1}{2}}^{t+\frac{1}{2}} \chi_1(\tau) \mathrm{d}\tau = \int\limits_{t-\frac{1}{2}}^{t+\frac{1}{2}}\mathrm{d}\tau =\left[ \tau\right]^{t+\frac{1}{2}}_{t-\frac{1}{2}}=1 $$
if $\frac{1}{2} < t \le \frac{3}{2}$, then $\tau \in [a,b]$ where $0 < a=t-\frac{1}{2}\le 1$ and $1 < b =t+\frac{1}{2}\le 2$. Then the function $\chi_1(\tau)$ is equal to $1$ only on the interval $[a,1]$ and the integral \eqref{1} is equal to $$ \int\limits_{t-\frac{1}{2}}^{t+\frac{1}{2}} \chi_1(\tau) \mathrm{d}\tau = \int\limits_{t-\frac{1}{2}}^{1}\mathrm{d}\tau = \left[ \tau\right]^{1}_{t-\frac{1}{2}}=\frac{3}{2}-t $$
if $\frac{3}{2} < t $, then $\tau$ is always larger than $1$, thus the function $\chi_1(\tau)$ and as a consequence the integral \eqref{1} is always $0$.
In sum, to evaluate integral \eqref{1} we have to compare the support of the indicator (characteristic function) of $[-1,+1]$ and the domain of integration as $t$ varies in $\mathbb{R}$, as shown in the picture below for $t>0$: