Convolution of two functions

1. Oct 6, 2016

bobred

1. The problem statement, all variables and given/known data
I have the two functions below and have to find the convolution $\beta * L$

2. Relevant equations
Assume a<1
$$\beta(x)=\begin{cases} \frac{\pi}{4a}\cos\left(\frac{\pi x}{2a}\right) & \left|x\right|<a\\ 0 & \left|x\right|\geq a \end{cases}$$
$$L(x)=\begin{cases} 1 & \left|x\right|<1\\ 0 & \left|x\right|\geq 1 \end{cases}$$

3. The attempt at a solution
$$\beta * L=\int^\infty_{-\infty}\beta(y)L(x-y)dy=\int^\infty_{-\infty}\beta(x-y)L(y)dy$$
What I am unsure about is the limits and which integral to use. Any hints would be greatly appreciated.

2. Oct 6, 2016

Ray Vickson

Start by drawing the two-dimensional $A$ region for the second form:
$$A_a = \{ (x,y) : -1 \leq y \leq 1, |x-y| \leq a \}.$$
There are three cases: (1) $a < 1$; (2) $a = 1$; and (3) $a > 1$. The region will look slightly different in cases (1) and (3).

For some values of $x$ the line vertical line through $x$ will miss region $A_a$ completely, so the $y$integral will equal 0. For other values of $x$ the vertical line through $x$ will intersect region $A_a$ in a vertical line segment, so will have the form $y_1(x) \leq y \leq y_2(x)$, and your $y$-integration will be from $y_1(x)$ to $y_2(x)$.