Define $x_n (t) = 2^{j/2} H(2^j t - k)$, with $n = 2^j - k$, $j \geq 0$, $0 \leq k < 2^j $, and
$H(t) = \begin{cases} 1, \text{ for } 0 \leq t< 1/2\\ -1, \text{ for } 1/2 \leq t<1 \\ 0, \text{ otherwise } \end{cases}$
Show that $\int^t _{0} x_n(s) ds = y_n z_n (t)$, with
$y_n = \frac{1}{2} 2^{-j/2}$ for $n = 2^j + k$, $j \geq 0$, $0 \leq k < 2^j $, and
$z_n(t) = z(2^j t - k)$, with
$z(t) = \begin{cases} 2t, \text{ for } 0 \leq t< 1/2\\ -2(t-1), \text{ for } 1/2 \leq t<1 \\ 0, \text{ otherwise } \end{cases}$
I have
$\int^t _{0} x_n(s) ds = 2^{j/2} \int^t _{0} H(2^j s - k) ds = 2^{j/2} \cdot 2^{-j}\int^{2^j t - k} _{-k} H(a) da $
with substitution $a = 2^j s - k$.
The outcome of this integral is $2^{j/2} \cdot 2^{-j} \cdot 2^j t = 2^{j/2} t$, for $0 \leq 2^j t - k < \frac{1}{2}$. This is not what I should get. How to proceed?