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I Integrating scaled and translated indicator function

  1. Nov 20, 2017 #1
    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. jcsd
  3. Nov 20, 2017 #2

    mathman

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    Gold Member

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