So I was answering this question about whether or not the Julia Set was self-similar in a known way. Of course it is, and that got me thinking. Even though the self similarity is nonlinear, what if you could exploit it to integrate over a normalized haussdorf...so I started small, I just wanted to see if I could get,
$$(1) \quad \int_J z \ \mu(dz)$$
The Julia Set, $J$, is the attractor of the dynamical system (IFS),
$$(2) \quad F \rightarrow w_1(F) \cup w_2(F)$$ $$w_1(z)=\sqrt{z+\lambda} \ , \quad w_1(z)=-\sqrt{z+\lambda}$$
So, the integral can be rewritten as,
$$(3) \quad \int_J z \ \mu(dz)=\int_{w_1(J)} z \ \mu(dz)+\int_{w_2(J)} z \ \mu(dz)$$
The first an second integrals on the right can be written after the substitutions $z \rightarrow w_1(z)$ and $z \rightarrow w_2(t)$ as,
$$(4) \quad \int_J z \ \mu(dz)=\int_{J} \sqrt{z+\lambda} \ \mu \left(\cfrac{dz}{\sqrt{z+\lambda}} \right)+\int_{J} -\sqrt{z+\lambda} \ \mu \left(-\cfrac{dz}{\sqrt{z+\lambda}} \right)=0$$
Which is rather boring. So, next I tried,
$$(5) \quad \int_J z^2 \ \mu(dz)$$
After applying the self-similarity, and applying a change of variables, I get,
$$(6) \quad \int_J z^2 \ \mu(dz)=\int_J z+\lambda \ \mu \left(\cfrac{dz}{\sqrt{z+\lambda}} \right)+\int_J z+\lambda \ \mu \left(-\cfrac{dz}{\sqrt{z+\lambda}} \right)$$
I remember the value of $(1)$ so combining the integral into one with that in it is a good choice. I also remember that the negative sign inside the measure is irrelevant, so I get,
$$(7) \quad \int_J z^2 \ \mu(dz)=2 \int_J z+\mu \left(\cfrac{dz}{\sqrt{z+\lambda}} \right)+2 \lambda \int_J \mu \left(\cfrac{dz}{\sqrt{z+\lambda}} \right)=2 \lambda \int_J \mu \left(\cfrac{dz}{\sqrt{z+\lambda}} \right)$$
Then changing variables back using $z \rightarrow w^{-1}_1(z)$, I get,
$$(8) \quad \int_J z^2 \ \mu(dz)=2 \lambda \int_{w_1(J)} \mu \left( dz \right)=\lambda$$
So my result for $(5)$ is,
$$(9) \quad \int_J z^2 \ \mu(dz)=\lambda$$
Is this correct? If not, what's the correct way? Anyone see any tricks I could've used to speed this up?
