- #1
Wuberdall
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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.
$$I_{jk}(a)=\int_0^a\chi_{[0,1)}(2^jx-k)\,dx,$$
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.
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