# I Integrating scaled and translated indicator function

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1. Nov 20, 2017

### Wuberdall

I am struggling to evaluate the following, relatively easy, integral (it might be because its early on a monday morning):
$$I_{jk}(a)=\int_0^a\chi_{[0,1)}(2^jx-k)\,dx,$$
where $\chi_{[0,1)}(x)$ denotes the indicator function on $[0,1)$ and $j,k$ are both integers.
My idea is to rewrite the indicator function as
$$\chi_{[0,1)}(2^jx-k) = \chi_{[2^{-j}k,2^{-j}(k+1))}(x).$$
Thus,
$$I_{jk}(a) = \int_0^a \chi_{[2^{-j}k,2^{-j}(k+1))}(x)\,dx.$$
And this is here I am stuck. I will welcome any ideas or advice with open arms.

Last edited by a moderator: Nov 21, 2017
2. Nov 20, 2017

### mathman

Your integral then has 3 cases: $$2^{-j}k > a,\ 2^{-j}k \le a<2^{-j}(k+1), \ and\ 2^{-j}(k+1) \le a$$. In each case, the integral is simply the length of the allowed interval. Note that the first case integral = 0, while $$2^{-j}k$$ is the lower limit for the other 2.