# Cantor function integral.

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

$$\mathcal{I}=\int_{0}^{1}C\left(\sqrt{1-x^2}\right)dx$$

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?

• For what it is worth, my attempt numerically suggested about $0.735718$ though it is possible I have rounding errors Commented Oct 19, 2021 at 18:06
• 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. Commented Oct 19, 2021 at 18:08

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