In problem 4, we have the functions:

[ ÆŸ(x) = 0 \{x < 0\}, \space ÆŸ(x) = 1 \{x > 0\}] [ f(x) = 1 \{|x| > 1\}, f(x) = 0 \{|x| < 1\}]

and an idea to use the functional relation [f(x) = ÆŸ(x+1) - ÆŸ(x-1)]

But this identity does not hold. For instance, [f(0) = ÆŸ(1) - ÆŸ(-1) = 1 - 0 = 1 \ne 0]

An identity which I found to work instead for the integral is: [f(x) = ÆŸ(x - 1) + ÆŸ(-x - 1) ]

Have I made an error?