Convolution of a Convolution I am confusing myself more and more with this question and require some assistance. The question reads:

Let:
  $$g(x) = \begin{cases}1 & \text{ if } |x|<\frac{1}{2}\\
0 & \text{ otherwise }\end{cases}$$
  Compute $(g*g)(x)$ and $(g*g*g)$ explicitly. 

Now I have found that:
$$(g*g)(x) = \begin{cases}
1-x & 0\leq x\leq 1\\
x+1 & -1\leq x\leq 0\\
0 & \text{ otherwise}\end{cases}$$
But now I am trying to compute $g*g*g$. I have started:
$$g*(g*g) = \int_{-\infty}^{\infty}g(x-y)\cdot (g*g)(y)dy$$
Thus, we require $-1\leq y\leq 1$ and $x-\frac{1}{2}\leq y\leq x+\frac{1}{2}$ for the integrand to not vanish. This says that $-\frac{3}{2}\leq x\leq \frac{3}{2}$. But now I'm confusing the hell out of myself. How do I divide this part into cases like I did with $(g*g)$? Do I separate it into two intervals $[-\frac{3}{2},0]$ and $[0,\frac{3}{2}]$? And treat each case separately?
 A: Hint:
You want to do the convolution the hard way (using the first principles i.e., the integral form), so here it is made simpler with crude images:
$g$ looks like 
$g*g$ looks like 
Due to symmetry for $g*g*g$ it suffices to look only in the domain of $[-\frac{3}{2},0]$
$g*g*g$ in the domain $-3/2$ to $-1/2$ increases exponentially (in power of $2$ to be precise) and is depicted in the  where the area of the shaded region is what you are after. 
Similarly, $g*g*g$ in the domain $-1/2$ to $0$ increases in quadratic sense and is depicted in the  where the area of the shaded region is what you are after.
A: Using Heaviside step function $g(x)={\bf H}(x+\dfrac12)-{\bf H}(x-\dfrac12)$ then with Laplace transform
$${\mathcal L}(g)=\dfrac1s(e^{-\frac12s}-e^{\frac12s})=G(s)$$
\begin{align}
g*g
&= {\mathcal L}^{-1}(G^2(s)) \\
&= {\mathcal L}^{-1}\left(\dfrac1s(e^{-\frac12s}-e^{\frac12s})\right)^2 \\
&= {\mathcal L}^{-1}\left(\dfrac{1}{s^2}(e^{-s}+e^{s}-e^{0s})\right) \\
&= (x+1){\bf H}(x+1)+(x-1){\bf H}(x-1)-2x{\bf H}(x)
\end{align}
\begin{align}
g*g*g
&= {\mathcal L}^{-1}(G^3(s)) \\
&= {\mathcal L}^{-1}\left(\dfrac1s(e^{-\frac12s}-e^{\frac12s})\right)^3 \\
&= {\mathcal L}^{-1}\left(\dfrac{1}{s^3}(e^{-\frac32s}+3e^{-\frac12s}+3e^{\frac12s}+e^{\frac32s})\right) \\
&= (x+\dfrac32)^2{\bf H}(x+\dfrac32)+3(x+\dfrac12){\bf H}(x+\dfrac12)+3(x-\dfrac12)^2{\bf H}(x-\dfrac12)+(x-\dfrac32)^2{\bf H}(x-\dfrac32)
\end{align}
A: First, notice that convolving two even functions results in an even function. Your $g(x)$ is an even function and hence all of its convolution powers are even functions. This saves us half of the work.
Second, notice that convolving piecewise polynomial functions results in a piecewise polynomial function whose degree is one more than the sum of the degrees of the polynomials we are convolving. Your $g(x)$ is a piecewise degree $0$ polynomial, $g\star g$ is a piecewise degree $1$ polynomial, and $g^{\star 3}:=g\star g\star g$ is a piecewise degree $2$ polynomial. 
Third, notice that convolving two functions with finite support results in a function with finite support. Your $g(x)$ has support on the interval $(-1/2,1/2)$, $g\star g$ has support on the interval $(-1,1)$, and $g^{\star 3}$ has support on the interval $(-3/2,3/2)$.
Putting these three observations together we conclude that $g^{\star 3}$ is an even function which is a quadratic polynomial on $(-1/2,1/2)$ while it is a different quadratic polynomial on both $(1/2,3/2)$ and $(-3/2,-1/2)$. Notice that a quadratic polynomial is determined by its values at three different points so our work is greatly eased by this fact.
