H(x-1) = 0 for x< 1 otherwise it is 1.
This tells us that,
H(x+1) - H(x-1) = 0 - 0 = 0 for x<1
H(x+1) - H(x-1) = 1 - 0 = 1 for -1<x<1
H(x+1) - H(x-1) = 1 - 1 = 0 for 1<x
Thus, the function H(x+1)-H(x-1) is non-zero on the interval -1<x<1
So when you integrate on the interval (-oo,+oo) you can just do it on (-1,1)
Because it is zero everywhere else and it does not matter.
But furthermore, on the interval (-1,1) the Heaviside functions add up to 1.
Which makes no difference when you multiply a number by them.
1/sqrt(2*pi) * INT(-1,1) (1-|x|)*exp(-ist) dx
Is the required Fourier Transform.