Convolution of $f(x)={1 \over 2} \chi_{[-1,1]}*\chi_{[-5,5]}$ Convolution of $f(x)={1 \over 2} \chi_{[-1,1]}*\chi_{[-5,5]}$. This is my first exercise, it's basically an interpolation between the two functions. So here my results:
Calling $u={1 \over 2} \chi_{[-1,1]}$ and $v=\chi_{[-5,5]}$ I can calculate the different values:
$u*v(0)={ 3 \over 2}$
$u*v({1 \over 2})={ 3 \over 2}$
$u*v(-{1 \over 2})={ 3 \over 2}$
So for the borders, we have:


*

*$0$ for $x<-5,x>5$

*$1$ for $x \geq -5,x \leq -{ 3\over 4}$

*$-x+{ 3 \over 4}$ for $-{ 3\over 4} \leq x \leq -{ 1 \over 4}$

*${ 3 \over 2}$ for $-{ 1 \over 4} \leq x \leq { 1 \over 4}$

*$x-{ 3 \over 4}$ for ${ 1 \over 4} \leq x \leq { 3 \over 4}$

*$1$ for $x \geq { 3 \over 4},x\leq5$
I hope it's right (I can't draw the graphic here); if there is some error or not rigorous passage, please let me know I want to be capable to solve this at the best of possibilities

$f(x)=\frac{1}{2}\int_{-\infty}^{+\infty} X_{[-5,5]}(y)X_{[-1,1]}(x-y)dy$
$$X_{[-1,1]}(x-y)=1 \Longleftrightarrow -1 \leq x-y \leq 1 $$$$\Longleftrightarrow x-1\leq y \leq x+1 \Longleftrightarrow y \in[x-1,x+1]\Longleftrightarrow X_{[x-1,x+1]}(y)=1$$
Thus $X_{[-1,1]}(x-y)=X_{[x-1,x+1]}(y)$
Also $G(y)=X_{[-5,5]}(y)X_{[x-1,x+1]}(y)=X_{[-5,5] \cap [x-1,x+1]}(y)$
$1.$ If $|x| \leq 4$ then $[x-1,x+1] \cap [-5,5]=[x-1,x+1]$ thus $G(y)=X_{[x-1,x+1]}(y) \Longrightarrow f(x)=1$
$2.$ If $x-1>5$ the the intervals are disjoint so the integral is zero,so $f(x)=0$
$3.$ If $x+1<-5$ again as in $2$ we have $f(x)=0$
$4.$ If $x=-6,6$ the intersection of the intervals is a singleton set so $f(x)=0$
$5.$ If $-6<x<-4$ then $x+1<-3$ and $x-1<-5$ and $-5<x+1<-3 $ and the intersection of the intervals is $[-5,x+1]$ thus  $f(x)=\int_{-5}^{x+1}dy=\frac{x+6}{2}$
$6.$ If $4<x<6$ then $x+1>5$ and $3<x-1<5$  then  the intersection is the interval $[x-1,5]$ thus $f(x)=\int_{x-1}^{5}dy=\frac{6-x}{2}$
