Problem with piecewise functions, translational invariance (CNN)

Question

I don't understand how to do this question:

Compute the convolution $(f*g)(x)$ for the functions $f(x)$ and $g(x)$ defined as,

$f(x)$=\begin{cases} 0&\text{if}\, x\ < a\\ \exp(-x)&\text{if}\, a \leq x \ \ \end{cases}

$g(x)$=\begin{cases} 1&\text{if} \, \frac{-1}{2} \leq x \leq \frac{1}{2}\\ 0&\text{otherwise}\ \ \end{cases}

where $(f*g)(x) = \int^{\infty}_{-\infty}f(y)g(x-y)dy$

The answer should be a piecewise function like $f$ and $g$, with a as parameter, like the $f(x)$ function.

The problem I have is that I don't understand how to do the calculations with the intervals given for $f(x)$ and $g(x)$. Any help is appreciated :)

Kind regards, Pontus

Answer 1

1. Can you first of all see that
   $$(f*g)(x) = \int^{\infty}_{a}e^{-y}g(x-y)dy$$ by the definition of $f$
2. Make things easier by changing variables $x-y = p$. Can you see that $$(f*g)(x) = \int^{a-x}_{-\infty}e^{-(x-p)}g(p)dp=e^{-x}\int^{a-x}_{-\infty}e^{p}g(p)dp$$
3. Now we can consider values of $a-x$. If $a-x<-\frac{1}{2}$ what is the $x$ range and the value of $f*g(x)$? If $a-x>\frac{1}{2}$ then
   $$(f*g)(x) =e^{-x}\int^{\frac{1}{2}}_{-\frac{1}{2}}e^{p}dp$$ which leaves the range $-\frac{1}{2}\le a-x \le \frac{1}{2}$ and
   $$(f*g)(x) =e^{-x}\int^{a-x}_{-\frac{1}{2}}e^{p}dp$$