How do I find the convolution of two functions with different domains?

[bobred]

Homework Statement

I have the two functions below and have to find the convolution \beta * L

Homework Equations

Assume a<1
<br /> \beta(x)=\begin{cases}<br /> \frac{\pi}{4a}\cos\left(\frac{\pi x}{2a}\right) &amp; \left|x\right|&lt;a\\<br /> 0 &amp; \left|x\right|\geq a<br /> \end{cases}<br />
<br /> L(x)=\begin{cases}<br /> 1 &amp; \left|x\right|&lt;1\\<br /> 0 &amp; \left|x\right|\geq 1<br /> \end{cases}<br />

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.

 

[Ray Vickson]

Homework Statement

I have the two functions below and have to find the convolution \beta * L

Homework Equations

Assume a<1
<br /> \beta(x)=\begin{cases}<br /> \frac{\pi}{4a}\cos\left(\frac{\pi x}{2a}\right) &amp; \left|x\right|&lt;a\\<br /> 0 &amp; \left|x\right|\geq a<br /> \end{cases}<br />
<br /> L(x)=\begin{cases}<br /> 1 &amp; \left|x\right|&lt;1\\<br /> 0 &amp; \left|x\right|\geq 1<br /> \end{cases}<br />

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.

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 $y_2(x)$.