Let's write:
\[
f_\theta^{\circ -s}(z) = \frac{1}{\Gamma(s)} \int_\gamma H(x,z)x^{s-1}\,dx\\
\]
When:
\[
f_\theta(z) = e^{e^{i\theta}z}-1\\
\]
We assign the following linear relationships:
\[
\begin{align}
e^{i\theta q} &\in (0,1)\\
\theta q &\in 2 \pi \mathbb{Z}(i)\\
q \,&\text{is the minimal value}\, q \in \mathbb{N}\,\text{of all solutions}\\
\Im(\theta) &> 0\\
\theta &\in 2\pi \mathbb{Q}(i)\\
\end{align}
\]
Then...
\[
H(x,z) = \frac{1}{\frac{1}{q} + \frac{2\pi}{q^2\theta}} \sum_{k=0}^\infty f_\theta^{\circ kq}(z) \frac{(-1)^k x^{(q+\frac{q^2\theta}{2\pi})k}}{k!}\\
\]
Here enters the integral...
Let \(h = \frac{1}{q} + \frac{2\pi}{q^2\theta}\).
As \(h \to 0\) we have \(q \to \infty\). This happens exactly as \(\theta \to \mathbb{C}(i)\), rather than \(\theta \in 2\pi \mathbb{Q}(i)\). As we hit limit points, the value \(q \to \infty\)...
This means the expression:
\[
x^{(q+\frac{q^2\theta}{2\pi})k} = x^{k/h}\\
\]
This is close enough to pull a proper inspection--and only be a linear factor off...
From here we have:
\[
f_\theta^{\circ kq} = f_\theta^{\circ \frac{k}{h}}\\
\]
This is because we only have added a second order error to \(h\), and the second order error will make no demonstration here.
Let's run the limit formula using these constraints...
\[
H(x,z) = \lim_{h\to 0} \frac{1}{h} \sum_{k=0}^\infty f_\theta^{\circ \frac{k}{h}} \frac{(-1)^k x^{\frac{k}{h}}}{k!}\\
\]
This is close to the integral we actually want. It will not be the integral. But it's reallllllyyyyyyyy close.
I CAN FUCKING SMELL FOURIER BUT I KNOW I'M MISSING SOMETHING
Okay, so I know that :
\[
H(x,z) = \int_0^\infty f_\theta^{\circ t} x^t\,d\mu\\
\]
Where \(\mu(t)\) is a measure on \([0,\infty]\). It's something like this this.... Fuck boys! I may have dug a hole too deep!!!!
\[
f_\theta^{\circ -s}(z) = \frac{1}{\Gamma(s)} \int_\gamma H(x,z)x^{s-1}\,dx\\
\]
When:
\[
f_\theta(z) = e^{e^{i\theta}z}-1\\
\]
We assign the following linear relationships:
\[
\begin{align}
e^{i\theta q} &\in (0,1)\\
\theta q &\in 2 \pi \mathbb{Z}(i)\\
q \,&\text{is the minimal value}\, q \in \mathbb{N}\,\text{of all solutions}\\
\Im(\theta) &> 0\\
\theta &\in 2\pi \mathbb{Q}(i)\\
\end{align}
\]
Then...
\[
H(x,z) = \frac{1}{\frac{1}{q} + \frac{2\pi}{q^2\theta}} \sum_{k=0}^\infty f_\theta^{\circ kq}(z) \frac{(-1)^k x^{(q+\frac{q^2\theta}{2\pi})k}}{k!}\\
\]
Here enters the integral...
Let \(h = \frac{1}{q} + \frac{2\pi}{q^2\theta}\).
As \(h \to 0\) we have \(q \to \infty\). This happens exactly as \(\theta \to \mathbb{C}(i)\), rather than \(\theta \in 2\pi \mathbb{Q}(i)\). As we hit limit points, the value \(q \to \infty\)...
This means the expression:
\[
x^{(q+\frac{q^2\theta}{2\pi})k} = x^{k/h}\\
\]
This is close enough to pull a proper inspection--and only be a linear factor off...
From here we have:
\[
f_\theta^{\circ kq} = f_\theta^{\circ \frac{k}{h}}\\
\]
This is because we only have added a second order error to \(h\), and the second order error will make no demonstration here.
Let's run the limit formula using these constraints...
\[
H(x,z) = \lim_{h\to 0} \frac{1}{h} \sum_{k=0}^\infty f_\theta^{\circ \frac{k}{h}} \frac{(-1)^k x^{\frac{k}{h}}}{k!}\\
\]
This is close to the integral we actually want. It will not be the integral. But it's reallllllyyyyyyyy close.
I CAN FUCKING SMELL FOURIER BUT I KNOW I'M MISSING SOMETHING
Okay, so I know that :
\[
H(x,z) = \int_0^\infty f_\theta^{\circ t} x^t\,d\mu\\
\]
Where \(\mu(t)\) is a measure on \([0,\infty]\). It's something like this this.... Fuck boys! I may have dug a hole too deep!!!!