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
 A: Is known that

$$\mathcal L(h(x-b)) = \dfrac1s e^{-bs},$$
$$\mathcal L((x-b)h(x-b)) = \dfrac1{s^2} e^{-bs},$$
$$\mathcal L(u(x)*v(x)) = \mathcal L(u(x)) \mathcal L(v(x)),$$
where $\mathcal L(f(x))$ is Laplace transform,
$$h(t)=
\begin{cases}
0,\quad\text{if}\quad t\in(-\infty,0]\\
1,\quad\text{otherwize}
\end{cases}$$
is the Heaviside function.

So
$$u(x)=\dfrac12(h(x-1)-h(x+1)),\quad v(x)=h(x-5)-h(x+5),$$
$$\mathcal L(u(x)*v(x)) = \dfrac1{2s^2}(e^{-5s}-e^{5s})(e^{-s}-e^s)
 = \dfrac1{2s^2}(e^{-6s}-e^{-4s}-e^{4s}+e^{6s}),$$
$$u(x)*v(x) = \dfrac12\left((t+6)h(t+6)- (t+4)h(t+4)- (t-4)h(t-4)+ (t-6)h(t-6)\right),$$
$$u(x)*v(x)=\begin{cases}
0,\quad\text{if}\quad x \in(-\infty, -6]\\
\dfrac t2+3,\quad\text{if}\quad x \in(-6, -4]\\
1,\quad\text{if}\quad x \in(-4, 4]\\
3- \dfrac t2,\quad\text{if}\quad x \in(4, 6]\\
0,\quad\text{if}\quad x \in(6,\infty).
\end{cases}$$
A: $f(x)=\frac{1}{2}\int_{-\infty}^{+\infty} X_{[-5,5]}(y)X_{[-1,1]}(x-y)dy$
$$X_{[-1,1]}(x-y)=1 \Longleftrightarrow -1 \leq x-y \leq 1 $$$$\Longleftrightarrow x-1\leq y \leq x+1 \Longleftrightarrow y \in[x-1,x+1]\Longleftrightarrow X_{[x-1,x+1]}(y)=1$$
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 $f(x)=\int_{x-1}^{5}dy=\frac{6-x}{2}$
