So, let's write:
\[
f_\theta^{\circ s}(z) = h^{\circ \frac{s}{q} + \frac{2 \pi s}{q^2\theta}}\\
\]
This can be writ, as:
\[
f_\theta^{\circ -s}(z) = \frac{1}{\Gamma(s)} \int_0^\infty \vartheta[h](x,z) x^{\frac{s}{q} + \frac{2 \pi s}{q^2\theta} -1}\,dx\\
\]
Let \(u = x^{\frac{1}{q} + \frac{2 \pi}{q^2\theta}}\) where by \(du = \left(\frac{1}{q} + \frac{2 \pi}{q^2\theta}\right)x^{\frac{1}{q} + \frac{2 \pi}{q^2\theta}-1}\,dx\). Whereby we now have the equation:
\[
f_\theta^{\circ -s}(z) = \frac{1}{(\frac{1}{q} + \frac{2 \pi}{q^2\theta})\Gamma(s)}\int_0^\infty \vartheta[h](u^{q + \frac{q^2}{2\pi\theta}},z) u^{s-1}\,du\\
\]
Thereby, we can write:
\[
H(x,z) = \frac{1}{\frac{1}{q} + \frac{2 \pi}{q^2\theta}}\sum_{k=0}^\infty f_\theta^{\circ kq}(z) \frac{(-1)^kx^{(q + \frac{q^2}{2\pi\theta})k}}{k!}\\
\]
And we should be able to derive that:
\[
\frac{d^s}{dx^s}\Big{|}_{x=0} H(x,z) = f_\theta^{\circ s}(z)\\
\]
.......... Provided I didn't screw up my elementary calculus.... We may have to take this integral along the arc \([0, \lim_{x\to\infty}x^{1/\theta}]\), rather than just along the real line....
God, I haven't had to keep track of this many variable substitutions since like 2nd or 3rd year undergrad. Getting frustrated....
Either way, it's something like this, lol.
Oddly enough; the value:
\[
\frac{1}{\frac{1}{q} + \frac{2 \pi}{q^2\theta}} \approx \frac{1}{m}\\
\]
where \(m\to\infty\) as \(q \to \infty\). But additionally:
\[
f_\theta^{\circ kq} \frac{(-1)^kx^{(q + \frac{q^2}{2\pi\theta})k}}{k!} \approx f_\theta^{kq} \frac{(-1)^k x^{kq}}{k!}\\
\]
So that:
\[
\frac{1}{m} \sum_{k=0}^\infty f_\theta^{kq/m} \frac{(-1)^k x^{kq/m}}{k!} \approx \int_0^\infty f^{\circ kq}(z) \frac{e^{\pi i k}x^{kq}}{k!}\,dk\\
\]
I smell a fourier transform somewhere in here. Too tired tn, but I think we can solve for when \(q \to \infty\) using integral calculus. Which would allow us to talk about the neutral case... when \(\Im\theta \to 0\) and \(q = \infty\)...
Recalling that all the work of this post, only works for the set \(\theta \in 2 \pi \mathbb{Q}(i) \cap \Im(\theta) > 0\), and nowhere else. We have to take limits to get it to work for \(\Im(\theta) \ge 0\) for arbitrary functions.
\[
f_\theta^{\circ s}(z) = h^{\circ \frac{s}{q} + \frac{2 \pi s}{q^2\theta}}\\
\]
This can be writ, as:
\[
f_\theta^{\circ -s}(z) = \frac{1}{\Gamma(s)} \int_0^\infty \vartheta[h](x,z) x^{\frac{s}{q} + \frac{2 \pi s}{q^2\theta} -1}\,dx\\
\]
Let \(u = x^{\frac{1}{q} + \frac{2 \pi}{q^2\theta}}\) where by \(du = \left(\frac{1}{q} + \frac{2 \pi}{q^2\theta}\right)x^{\frac{1}{q} + \frac{2 \pi}{q^2\theta}-1}\,dx\). Whereby we now have the equation:
\[
f_\theta^{\circ -s}(z) = \frac{1}{(\frac{1}{q} + \frac{2 \pi}{q^2\theta})\Gamma(s)}\int_0^\infty \vartheta[h](u^{q + \frac{q^2}{2\pi\theta}},z) u^{s-1}\,du\\
\]
Thereby, we can write:
\[
H(x,z) = \frac{1}{\frac{1}{q} + \frac{2 \pi}{q^2\theta}}\sum_{k=0}^\infty f_\theta^{\circ kq}(z) \frac{(-1)^kx^{(q + \frac{q^2}{2\pi\theta})k}}{k!}\\
\]
And we should be able to derive that:
\[
\frac{d^s}{dx^s}\Big{|}_{x=0} H(x,z) = f_\theta^{\circ s}(z)\\
\]
.......... Provided I didn't screw up my elementary calculus.... We may have to take this integral along the arc \([0, \lim_{x\to\infty}x^{1/\theta}]\), rather than just along the real line....
God, I haven't had to keep track of this many variable substitutions since like 2nd or 3rd year undergrad. Getting frustrated....
Either way, it's something like this, lol.Oddly enough; the value:
\[
\frac{1}{\frac{1}{q} + \frac{2 \pi}{q^2\theta}} \approx \frac{1}{m}\\
\]
where \(m\to\infty\) as \(q \to \infty\). But additionally:
\[
f_\theta^{\circ kq} \frac{(-1)^kx^{(q + \frac{q^2}{2\pi\theta})k}}{k!} \approx f_\theta^{kq} \frac{(-1)^k x^{kq}}{k!}\\
\]
So that:
\[
\frac{1}{m} \sum_{k=0}^\infty f_\theta^{kq/m} \frac{(-1)^k x^{kq/m}}{k!} \approx \int_0^\infty f^{\circ kq}(z) \frac{e^{\pi i k}x^{kq}}{k!}\,dk\\
\]
I smell a fourier transform somewhere in here. Too tired tn, but I think we can solve for when \(q \to \infty\) using integral calculus. Which would allow us to talk about the neutral case... when \(\Im\theta \to 0\) and \(q = \infty\)...
Recalling that all the work of this post, only works for the set \(\theta \in 2 \pi \mathbb{Q}(i) \cap \Im(\theta) > 0\), and nowhere else. We have to take limits to get it to work for \(\Im(\theta) \ge 0\) for arbitrary functions.