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## Homework Statement

I have found the following exercise just solved, but I haven't understood some steps...

[tex]f(t)=\begin {cases}1, t\in [-1,1] \\ 0, t\not \in [-1,1]\end{cases}[/tex]

We have to do the convolution [tex]f*f=\int^{+\infty}_{-\infty}f(\tau)f(\tau)f(t- \tau) d\tau=\int^{+1}_{-1}f(t- \tau) d\tau [/tex]

**2. The suggested solution is:**

We know that [tex] -1\ge t-\tau \le 1[/tex] and [tex] -1\ge \tau \le 1[/tex]

so we can write: [tex]f*f=\begin {cases}0, t<-2\\?, t \in[-2,0]\\?,

> t\in[0,2]\\0, t>2 \end {cases}[/tex]

1) If [tex]t<0 \rightarrow t+1>-1 \rightarrow t>-2[/tex]

Thus [tex]f*f=\int_{-1}^{t+1}dt=t+2[/tex]

2) while if [tex]t>0 \rightarrow t+1>-1 \rightarrow t>-2[/tex]

Thus [tex]f*f=\int_{t-1}^{1}dt=2-t[/tex]

**3. My doubts:**

1) I don't understand why the last two integrals are in "dt", while in the first integral the integration variable is [tex]\tau[/tex].. I have thought to a typing error, but I'm not sure

2) I haven't understood in which way the condition on t are chosen (i.e. [tex]t<0 \rightarrow t+1>-1, if t>0 \rightarrow t-1<1[/tex]). They are "true", but I haven't understood why they are put in this way and not in another one.

3) I haven't understood in which way the "extremes" of integration are chosen. I suspect that there is a trick, but I'm not sure..

4) There is another way to carry on the correct result?

A lot of thanks!