Calculate integral $\int_0^1 K(x)dK^2(x)$ where $K(x)$ is Cantor ternary function.
I tried to use Cantor function's property that $K(1-x) = 1 - K(x)$ So:
$\int_0^1K(x)dK^2(x)=/x = 1-x/=\int_1^0K(1-x)dK^2(1-x)=\int_1^0(1-K(x))d(1 - 2K(x) + K^2(x))=2\int_0^1dK(x) - 2\int_0^1K(x)dK(x) + \int_0^1K(x)dK^2(x) - \int_0^1dK^2(x)$
The first and the last integrals equal $1$, the third integral is the one we started from. So we have $I = -2\int_0^1K(x)dK(x) + I + 1$ and therefore $I$ is reduced and I'm stuck. Am I going the wrong way or simply made some mistakes during the calculations?
Divide integral on domains: $[0, \frac 1 3], [\frac 1 3, \frac 2 3], [\frac 2 3, 1]$. Let's start with $[0, \frac 1 3]$, where $K(x)=\frac {K(3x)} 2$ $\int_{0}^{1/3} K(x) dK^{2}(x) = \int_{0}^{1/3} \frac {K(3x)} {2}dK^{2}(x) = \int_{0}^{1} \frac {K(t)} {2} dK^{2}(\frac t 3). K(\frac t 3) = \frac {K(t)} 2$. So $\int_{0}^{1} \frac {K(t)} {2} dK^{2}(\frac t 3)$$=\int_{0}^{1} \frac {K(t)} {2} d(\frac {K^{2}(t)} 4) = \frac 1 8 \int_{0}^{1} K(t)dK^{2}(t)$. The last is the same thing we want to calculate. So u should next calculate integral at $[\frac 1 3,\frac 2 3]$ and $[\frac 2 3, 1] $. And u got an equation on our integral.