Evaluate $\int_{-\infty}^{\infty} \chi_{[0,1]}(x-y) \chi_{[0,1]}(y) \, \mathrm{d}y$ I'm trying to evaluate the integral $$\int_{-\infty}^{\infty} \chi_{[0,1]}(x-y) \chi_{[0,1]}(y) \, \mathrm{d}y$$
where $\chi_{[0,1]}(x)=1$ is the characteristic function, i.e. equals $1$ for $x \in [0,1]$ and $0$ otherwise. How do I evaluate this integral? So far I've done
$$\int_{-\infty}^{\infty} \chi_{[0,1]}(x-y) \chi_{[0,1]}(y) \, \mathrm{d}y = \int_{0}^1 \chi_{[0,1]} (x-y) \, \mathrm{d}y.$$
I've tried substitution but I don't find it yielding anything useful. The answer should be the function $$f(x)= \left\{ 
  \begin{array}{l l}
    x & \quad \text{if $x \in [0,1]$}\\
    2-x & \quad \text{if $x \in [1,2]$}\\
    0 & \quad \text{otherwise.}
  \end{array} \right.$$
 A: Change variables $x-y=s$, thus the new domain is $(x,x-1)$. You have
$$
\int_0^1\Xi(x-y)dy=\int_{x-1}^x\Xi(s)ds.
$$
Now, if $x-1>1$ and if $x<0$ the integral vanishes. Moreover, you have
$$
\int_{x-1}^x\Xi(s)ds=\text{length}\left((x-1,x)\cap(0,1)\right).
$$
If $0<x<1$
$$
\text{length}\left((x-1,x)\cap(0,1)\right)=x.
$$
Otherwise, if $1<x<2$
$$
\text{length}\left((x-1,x)\cap(0,1)\right)=1-(x-1)=2-x
$$
A: You're off to a good start. Next note that
$$
\chi_{[0,1]}(x-y) = \begin{cases} 1, & \text{for $x-1 \le y \le x$} \\ 0, & \text{otherwise}\end{cases}.
$$
Draw some graphs!
Hence, if $0< x < 1$, then $\chi_{[0,1]}(x-y)$ on the whole of $[0,1]$. Thus the integral is $0$.
If $0 < x < 1$, then $\chi_{[0,1]}(x-y) = 1$ on $[0,x]$ and $=0$ on the rest of $[0,1]$. In this case, the integral will be
$$ \int_0^x 1\,dy = x. $$
Third case, if $1 < x < 2$, then $\chi_{[0,1]}(x-y) = 1$ on $[x-1,1]$ and $=0$ on the rest of $[0,1]$, so the integral will be
$$ \int_{x-1}^1 1\,dy = 2-x.$$
Finally, if $x > 0$, then $\chi_{[0,1]}(x-y) = 0$ on $[0,1]$ again.
