I'm trying to solve the following integral for a problem about fractals involving Cantor set:


Where $C(x)$ denotes the Cantor ternary function.

I don't know a lot about measure theory and Lebesgue integrals, so I don't know what to do here. I tried the substitution $u=\sqrt{1-x^2}$, but it led me nowhere.

I've made a small program to numerically evaluate it, and it yielded $\mathcal{I}\approx0.7357895383$.

Is it possible to find a closed-form expression or a series expansion for this integral?

  • 2
    $\begingroup$ For what it is worth, my attempt numerically suggested about $0.735718$ though it is possible I have rounding errors $\endgroup$
    – Henry
    Commented Oct 19, 2021 at 18:06
  • 3
    $\begingroup$ For what it's worth, with that substitution you should end up with $$ \mathcal I = \int_0^1 C\left(\sqrt{1 - x^2}\right) = \int_0^1 C(u) \cdot \frac{u}{\sqrt{1-u^2}}\,du. $$ At the very least, combining this with the fact that $\int \frac u{\sqrt{1 - u^2}}\,du = -\sqrt{1 - u^2} + C$ leads to an expression of this integral as an infinite sum. $\endgroup$ Commented Oct 19, 2021 at 18:08

1 Answer 1


Partial solution.

Put $t = \sqrt{1-x^2}$. Hence $x = \sqrt{1-t^2}$ and $\mathcal{I} = \int_0^1 \frac{C(t) tdt}{\sqrt{1-t^2}}$.

Open Cantor set $A = \bigsqcup_{n \ge1, 1 \le k \le 2^{n-1}} (a_{nk}, b_{nk})$, where $(a_{11}, b_{11}) = (\frac13, \frac23)$, $(a_{21}, b_{21}) = (\frac19, \frac29)$, $(a_{22}, b_{22}) = (\frac79, \frac89)$, $(a_{31}, b_{31}) = (\frac1{27}, \frac2{27})$, $(a_{32}, b_{32}) = (\frac7{27}, \frac8{27})$, $(a_{33}, b_{33}) = (\frac{19}{27}, \frac{20}{27})$, $(a_{34}, b_{34}) = (\frac{25}{27}, \frac{26}{27})$ etc. Denote by $C_{ij}$ the value of $C$ on $(a_{nk}, b_{nk})$. It's easily seen that $C_{11} = \frac{1}2, C_{21} = \frac{1}4, C_{22} = \frac{3}4, C_{31} = \frac{1}8, C_{32} = \frac{3}8, C_{33} = \frac{5}8, C_{34} = \frac{7}8$ and $C(t) = \frac{2k-1}{2^n}$ for $t \in (a_{nk}, b_{nk})$. Thus

$$\mathcal{I} = \int_A \frac{C(t)tdt}{\sqrt{1-t^2}} = \sum_{n \ge 1, 1 \le k \le 2^{n-1}} \int_{a_{nk}}^{b_{nk}} \frac{C(t)tdt}{\sqrt{1-t^2}} = \sum_{n \ge 1, 1 \le k \le 2^{n-1}} \int_{a_{nk}}^{b_{nk}} \frac{2k-1}{2^n}\frac{tdt}{\sqrt{1-t^2}} = $$ $$ = \sum_{n \ge 1, 1 \le k \le 2^{n-1}} \frac{2k-1}{2^n} \int_{a_{nk}}^{b_{nk}} \frac{tdt}{\sqrt{1-t^2}}$$

It's sufficient to paste formulas for $a_{nk}$ and $b_{nk}$ and find the sum.

  • 1
    $\begingroup$ Notably, we can simplify the integral in the summation to $$ \int_{a_{nk}}^{b_{nk}} \frac{t}{\sqrt{1-t}}\, dt = \sqrt{1 - a_{nk}^2} - \sqrt{1 - b_{nk}^2}. $$ $\endgroup$ Commented Oct 20, 2021 at 18:33

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