Convolution of triangle function with itself I'm trying to find the convolution $(A \ast A)(x)$, where 
$$A(x) = \begin{cases}
1 + x, & -1\leq x \leq 0,\\
1 - x, & 0\leq x \leq 1,\\
0, & \text{otherwise},
\end{cases}$$
so far without success. Can someone help?
 A: It is convenient to consider this problem in a probabilistic setting. Let $U_1,U_2,\ldots$ be independent uniform$[0,1]$ random variables. Then, your function $A(x)$ (supported on $[-1,1]$) is the probability density function (pdf) of the random variable $X = U_1 + U_2 - 1$; see, page 17 here (notes from Purdue university) for this easy fact. Thus, the convolution of $A$ with itself gives the pdf of the sum $X_1 + X_2$, where $X_1$ and $X_2$ are independent copies of $X$, hence the pdf of the random variable $U_1+U_2+U_3+U_4-2$ (which is supported on $[-2,2]$). According to MathWorld, the pdf of $U_1+\cdots+U_4$ is given by
$$
f_{U_1  +  \cdots  + U_4 } (x) = \frac{1}{{2(4 - 1)!}}\sum\limits_{k = 0}^4 {( - 1)^k {4 \choose k}(x - k)^{4 - 1} {\mathop{\rm sgn}} (x - k)} , \;\; x \in [0,4].
$$
(Replace $4$ with $n$ for the pdf of $U_1+\cdots+U_n$.) It thus follows that
$$
(A * A)(x) = f_{U_1  +  \cdots  + U_4 } (x+2),\;\; x \in [-2,2],
$$
where $*$ denotes convolution. (Note that $A$ is a symmetric function.) I have verified this result by comparing to  the right-hand side of
$$
(A * A)(x) = \int_{ - 1}^1 {A(x - y)A(y)\,{\rm d}y} , \;\; x \in [-2,2]
$$
(recall the definition of convolution, and note that $A * A$ is supported on $[-2,2]$). 
For example, the following values were obtained for $x=-1.65$ and $x=1.5$, respectively:
$$
0.0071458333333343435 , 0.007145833333334295
$$
and
$$
0.020833333333335497 , 0.020833333333333332,
$$
where the left-hand values are approximations of the above integral defining $(A*A)(x)$.
A: Hint: Write down what integral you need to compute, then split into four pieces, $x \geq 0$ and $y \geq x$, $\geq 0$ and $x < y$, $x < 0$ and $y \geq x$ and $x < 0$ and $y > x$.
A: if $-1\leq x-t\leq 0$, then $x\leq t\leq x+1$ and if $0\leq x-t\leq 1$, then $x-1\leq t\leq x$.  So
$$\int_{-\infty}^{\infty}A(t)A(x-t)dt=\int_{x-1}^xA(t)(1+x-t)dt+\int_x^{x+1}A(t)(1-x+t)dt$$
you can further split this up depending on where $-1,0,1$ fall wrt $x-1,x,x+1$.  for instance if $x\geq2$ or if $x\leq-2$ then both integrals are zero.  anyway, you can do the integral directly (although it is somewhat tedious).
