How to integrate moving Heaviside block?

Question

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.

Answer 1

Write the integrand as $$ f(t,\tau ) = H\left( {{1 \over 2} - \tau } \right)H_{\,1/2} \left( \tau \right)H_{\,1} \left( {t - \tau } \right) $$ then according to your definitions, the integrand will be one when all the three terms are so, and null otherwise.
That means to impose the contemporaneous validity of the following inequalities: $$ \eqalign{ & f(t,\tau ) = 1\;:\;\left\{ \matrix{ 0 \le {1 \over 2} - \tau \hfill \cr - {1 \over 2} \le \tau \le {1 \over 2} \hfill \cr - 1 \le t - \tau \le 1 \hfill \cr} \right.\quad \Rightarrow \cr & \Rightarrow \quad \left\{ \matrix{ - {1 \over 2} \le \tau \le {1 \over 2} \hfill \cr - 1 + t \le \tau \le 1 + t \hfill \cr} \right.\quad \Rightarrow \cr & \Rightarrow \quad \left\{ {\matrix{ { - {1 \over 2} \le \tau \le 1 + t} & {\left| {\; - {3 \over 2} \le t < - {1 \over 2}} \right.} \cr { - {1 \over 2} \le \tau \le {1 \over 2}} & {\left| {\; - {1 \over 2} \le t < {1 \over 2}} \right.} \cr { - 1 + t \le \tau \le {1 \over 2}} & {\left| {\;{1 \over 2} \le t < {3 \over 2}} \right.} \cr } } \right. \cr} $$

Thus the integral will be $$ \int_{ - \infty }^\infty {f(t,\tau )d\tau } = \left\{ {\matrix{ {{3 \over 2} + t} & {\left| {\; - {3 \over 2} \le t < - {1 \over 2}} \right.} \cr 1 & {\left| {\; - {1 \over 2} \le t < {1 \over 2}} \right.} \cr {{3 \over 2} - t} & {\left| {\;{1 \over 2} \le t < {3 \over 2}} \right.} \cr } } \right. $$

Answer 2

Let $f(t) = \int_{-\infty}^{\infty} H(\frac{1}{2}-t) H_\frac{1}{2}(\tau)H_1(t-\tau) d\tau$ $$ \int_{-\infty}^{\infty} H(\frac{1}{2}-t) H_\frac{1}{2}(\tau)H_1(t-\tau) d\tau =$$ \begin{multline*} =\int_{-\infty}^{-\frac{1}{2}} H(\frac{1}{2}-t) H_\frac{1}{2}(\tau)H_1(t-\tau) d\tau\ + \\ + \int_{-\frac{1}{2}}^{\frac{1}{2}} H(\frac{1}{2}-t) H_\frac{1}{2}(\tau)H_1(t-\tau) d\tau\ + \\ + \int_{\frac{1}{2}}^{\infty} H(\frac{1}{2}-t) H_\frac{1}{2}(\tau)H_1(t-\tau) d\tau \end{multline*}

First and third integral equal to zero, because $H_{\frac{1}{2}}(\tau) = 0,\ \tau \in (-\infty; -\dfrac{1}{2}) \cup (\dfrac{1}{2}; \infty)$

$$f(t) = \int_{-\frac{1}{2}}^{\frac{1}{2}} H(\frac{1}{2}-t)H_1(t-\tau) d\tau$$

If ($\dfrac{1}{2} - t) < 0 \text{, or } t > \dfrac{1}{2} \Rightarrow H(\frac{1}{2}-t) = 0 \Rightarrow f(t) = 0$

Otherwise: $$ f(t) = \int_{-\frac{1}{2}}^{\frac{1}{2}} H_1(t-\tau) d\tau\qquad t \in [\dfrac{1}{2}; \infty)$$

Answer 3

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 $t>0$: