Borel summation, Mellin Transforms, Parabolic iteration

[bo198214]

So to begin, we'll get some concepts clear. A parabolic fixed point \(f(0) = 0\) and \(f'(0) = e^{2 \pi i m/q}\) for \(q,m \in \mathbb{Z}\). And the multiplier of this fixed point, as Milnor states it, or the \(\ell = \text{valit}\) as Bo, refers to it, is given in the expansion:

I think you mean "multiplicity", perhaps edit that in your original post - because it is called "multiplier" in the hyperbolic case for the first coefficient, right?
EDIT: After reading Milnor again: the multiplicity is valit+1, they are not the same. The multiplicity is the index of the first non-zero coefficient after the first coefficient. Valit is the multiplicity - 1. The number of petals is 2*valit.

Here is where things start to get difficult. We have an abel function on half plane, which is really just the abel function of our function \(f\), but fixated in a petal, and near \(0\). We have to from here, restrict (we don't have to in the final result, but we do now). This function \(\alpha(w)\) satisfies the asymptotic expansion:

\[
\alpha(w) = w + A \log(w) + \sum_{k=0}^\infty b_k w^{-k}\\
\]

Where the coefficients \(b_k\) are of order \(O(c^kk!)\) for some random value \(c\) which is unimportant to us. This asymptotic only holds for \(|\arg(w)|  < \kappa\), for some value \(\kappa>0\)--dependent on the multiplicity of this fixed point (Milnor), or the valit, as Bo calls it.

But I would like to remark here that asymptotics is for all petals the same, only the branch(cut) of the logarithm is what needs to be adjusted from petal to petal.

And so enters in our new change of variables \(u = 1/w\):
\[
\alpha(u) = \frac{1}{u} - A \log(u) + \sum_{k=0}^\infty b_k u^k\\
\]

Don't you need more terms \(\frac{1}{u^k}\) and also having coefficients, like:
\[ \alpha(u) = \left(\sum_{k=1}^{\ell} \frac{c_k}{u^k}\right) - A \log(u) + \sum_{k=0}^\infty b_k u^k \]

\[
T_c(u) = \int_{c-i\infty}^{c+i\infty} \frac{H(s)}{\Gamma(1-s)} u^{-s}\,ds\\
\]

We get the perfect version of:

\[
T_c(u) = \sum_{k=1}^\infty \frac{b_k}{k!}u^k
\]

Where we can choose the index \(k=1\) by moving \(c\) to the right or left. For \(c = \delta - N\) we get:

\[
T_c(u) = \sum_{k=N}^\infty \frac{b_k}{k!}u^k
\]

Where, (I'm not sure if this converges, if it doesn't, it's irrelevant), if \(c = 1+\delta\) we have the borel sum of the Alpha function.

Therefore:

\[
\int_0^\infty e^{-t/u}T_{1+\delta}(t) \,dt = \alpha(u)\\
\]

But now, in addition to everything above, this will prove A GLOBAL SOLUTION OF OUR BOREL SUM METHODOLOGY!

Where do \(\delta\) and N come from?