Okay, let's start this over with more rigor. By which I'll work in steps.
We begin by denoting \(f(z) = e^{z}-1\) which has the neutral fixed point at \(0\) with multiplier \(1\). By which we can construct an Abel function in the left half plane \(\Re(z) < 0\), because this domain is within the attracting petal of \(0\). Which is to mean that \(\lim_{n\to\infty}f^{\circ n}(z) = 0\) for all \(\Re(z) < 0\). This is seen by just observing the orbit.
The abel function \(\alpha(z)\) is holomorphic in the Left half plane, and satisfies:
\[
\alpha(e^z-1) = \alpha(z) + 1\\
\]
By which we have the identity \(\alpha(z+2\pi i) = \alpha(z)\)--it must inherit \(f\)'s period. We can write:
\[
\alpha(z) = \alpha(f^{\circ n}(z)) - n\\
\]
Which ensures the expansion to the left half plane. This function is holomorphic in a petal near zero; by which it is non-singular; and we can take an inverse function, which we write as \(\alpha^{-1}(z)\), which satisfies \(f(\alpha^{-1}(z)) = \alpha^{-1}(z+1)\). We can construct a holomorphic squareroot function of \(f\), which we write:
\[
g(z) = \alpha^{-1}\left(\frac{1}{2} + \alpha(z)\right)\\
\]
As per David E. Speyer's comment, this function is extendable to \(\mathbb{C}/[0,\infty)\)--as per referenced here https://mathoverflow.net/questions/4347/...ar-and-exp Which is stated as a result of Baker (too lazy to find an English copy of the paper, the original is in German).
This function then has two properties we are interested in. Firstly, \(g : \mathbb{C}_{\Re(z) < 0} \to \mathbb{C}_{\Re(z) < 0}\), where additionally, when we assign the point at infinity in this space, we get that:
\[
g(\infty) = g^{-1}(-1)\\
\]
Because, on this domain, \(f(\infty) = -1\). The function \(g\) is non-singular, so a local inverse always exists. The second thing we need is that this function is bounded on the line \(i\mathbb{R}\). This can be found because \(g(g(i\mathbb{R})) = f(i\mathbb{R})\) is bounded; and by part so much be the first arc \(g(i\mathbb{R})\).
So we can strenghten our statement to \(g(z)\) is bounded for \(\Re(z) \le 0\).
Now the function \(g\) is not holomorphic at \(0\), but can be expanded into an asymptotic series we will write as:
\[
g(z) \sim \sum_{k=1}^\infty g_k z^k\\
\]
And the goal we are trying to show, is that there is some \(0 < c \) such that \(g_k = O(c^kk!)\). Or equivalently, that:
\[
\mathcal{B} g(z) = \sum_{k=1}^\infty g_k \frac{z^k}{k!}\\
\]
Has a non trivial radius of convergence.
----------------------------------------
The first theorem we'll write is as follows:
Theorem:
\[
F(z) = \int_0^\infty g(-x)x^{z-1}\,dx\\
\]
Converges for \( -1 < \Re(z) < 0\) and has decay like:
\[
|F(x + iy)| < M e^{-\frac{\pi}{2}|y|}\\
\]
For some constant \(M\).
Proof:
The integral:
\[
F_\theta(z) = \int_0^\infty g(-e^{i\theta} x) (e^{i\theta}x)^{z-1}e^{i\theta}\,dx\\
\]
converges for all \(-\pi /2 \le \theta \le \pi/2\) and \(-1 < \Re(z) < 0\). The integral converges at the endpoin \(0\), because \(g(z) \sim g_1 z\) at \(0\), and the integral converges at \(\infty\), because \(g(-e^{i\theta}x)\) is bounded and the integrand is bounded by \(|x|^{\Re(z) - 1}\).
Now, by contour integration, we have that \(F_\theta\) is constant in \(\theta\). To see this, consider the contour:
\[
C = [0,R] + \gamma_R - [0,Re^{i\theta}]\\
\]
Then:
\[
\int_C g(x) x^{z-1}\,dx\\ = 0\\
\]
The integral along the arc \(\gamma_R\) is bounded by \(|R|^z\), and therefore the two integrals:
\[
\int_0^\infty g(x)x^{z-1}\,dx - \int_0^\infty g(-e^{i\theta} x)(e^{i\theta}x)^{z-1} e^{i\theta}\,dx = 0\\
\]
Now consider \(y = \Im(z) > 0\), and \(\theta = \pi/2\). We have that:
\[
F(x+iy) = e^{i\frac{\pi}{2}(x+iy)}\int_0^\infty g(ix)x^{z-1}\,dx\\
\]
But this is bounded as \(|F(x+iy) | \le M^+ e^{-\frac{\pi}{2} y}\). A similar procedure can be done for \(\Im(z) < 0\) and \(\theta = - \pi/2\), and we are given:
\[
|F(z)| \le M e^{-\frac{\pi}{2} |y|}\\
\]
Where \(M\) depends on the real part of \(z\) only. This is a Stein & Shakarchi argument, which they develop using the Fourier transform; this is just a change of variables of this. We point to Stein & Shakarchi Complex Analysis.
QED
---------------------------------------
Our second theorem will be that \(F(z)\) is actually meromorphic for \(\Re(z) < 0\). This is a bit more tricky.
Theorem:
The function \(F(z)\) is meromorphic for \(\Re(z) < 0\) with simple poles at \(k \in \mathbb{Z}_{<0}\), with residues \((-1)^k g_k\).
Proof:
This is more of a just check the result. I could go by induction but I want to be quick.
Begin by expanding the integral defining \(F\) as follows:
\[
F(z) = \int_0^1 g(-x)x^{z-1}\,dx + \int_1^\infty g(-x)x^{z-1}\,dx\\
\]
You will note instantly that the second integral is holomorphic for \(\Re(z) < 0\)--while the first integral is only holomorphic for \(\Re(z) > -1\). So let's add in our asymptotic expansion, which is given as:
\[
\int_0^1 \left(g(-x) - \sum_{k=1}^N g_k (-x)^k + \sum_{k=1}^N g_k (-x)^k\right) x^{z-1}\,dx\\
\]
Where the term by term integral of this can be written:
\[
\int_0^1 g(-x)x^{z-1} \,dx = \sum_{k=1}^N g_k\frac{(-1)^k}{k+z} + \int_0^1\left(g(-x) - \sum_{k=1}^N g_k (-x)^k\right) x^{z-1}\,dx\\
\]
But now the integral on the right has an \(N+1\)'th order zero at \(0\), and therefore the integral on the left is meromorphic for \(\Re(z) > -N\).
Combining this together, and since \(N\) is arbitrary. We have that \(F(z)\) is holomorphic for \(\Re(z) < 0\), and has simple poles at \(z = -k\), with residues:
\[
\mathop{\mathrm{Res}}_{z=-k} F(z) = (-1)^k g_k\\
\]
QED
-----------------------------------------------
Now we will place here a theorem on Gamma functions asymptotics in the complex plane. Partially attributed to Stirling, and named for him.
\[
\Gamma(z) \sim \sqrt{2\pi}z^{z-1/2}e^{-z}\,\,\text{as}\,\,|z| \to \infty\,\,\text{while}\,\,|\arg(z)| < \pi\\
\]
Which, for our purposes, in the imaginary asymptotic, can be written:
\[
|\Gamma(x+iy)| \sim \sqrt{2\pi} |y|^{x-1/2} e^{-\frac{\pi}{2}|y|}\,\,\text{as}\,\,|y|\to\infty\\
\]
Of which, we get that:
\[
\frac{|F(x+iy)|}{|\Gamma(1-x-iy)|} \le M |y|^{x-1+1/2}\\
\]
Which gives our third theorem:
Theorem: Let \(-1 < c <-1/2\), then the integral:
\[
h(x) = \frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty} \frac{F(z)}{\Gamma(1-z)}x^{-z}\,dz\\
\]
Converges absolutely, and is continuous for \(x > 0\).
Proof:
Just take absolute values and use the above asymptotics, and note that \(|x^{-z}| < |x|^{-\Re(z)}\).
QED
-----------------------------
And now we enter in the hard part!!! Essentially, we wish to show that \(h(x) = \mathcal{B} g(-x)\). And to do that we need a lemma that is deceptively simple, but may be confusing.
\[
h(x) = \sum_{k=1}^\infty \mathop{\mathrm{Res}}_{z=-k}\frac{F(z)}{\Gamma(1-z)}x^{-z}\\
\]
To calculate these residues isn't very hard. We know that \(F(z)\) has a simple pole at each \(k\), with a residue of \((-1)^kg_k\), so by running Cauchy's integral formula we get:
\[
\mathop{\mathrm{Res}}_{z=-k}\frac{F(z)}{\Gamma(1-z)}x^{-z} = g_k \frac{(-x)^k}{k!}\\
\]
Which is indeed the expansion of \(\mathcal{B}g(-x)\). The trouble is, we need to show that the contour integral satisfies equalling these residues. We write this below:
Theorem:
The function \(h(x)\) from the previous theorem, satisfies:
\[
h(x) = \sum_{k=1}^\infty g_k \frac{(-x)^k}{k!}\\
\]
For a value \(\delta > 0\) and \(0<x < \delta\)
Proof:
Take a contour \(C_R\) such that:
\[
C_R = [c-iR,c+iR] - \gamma_R\\
\]
Where \(\gamma_R\) is the semi-circle to the left of this vertical line. Then we write:
\[
\frac{1}{2\pi i} \int_{C_R} \frac{F(z)}{\Gamma(1-z)} x^{-z}\,dz = \sum \mathop{\mathrm{Res}}\\
\]
Which means it's the sum of the residues of the poles inside the contour. If we can show that:
\[
\lim_{R \to \infty} \int_{\gamma_R} \frac{F(z)}{\Gamma(1-z)} x^{-z}\,dz \to 0\\
\]
By which we get that the integral equals the sum of the residues; and additionally shows that the series converges. To show this, all we need is to show that:
\[
\lim_{|z|\to\infty} \frac{F(z)}{\Gamma(1-z)}x^{-z} \to 0\,\,\text{as}\,\,|z| \to \infty\,\,\text{while}\,\,\Re(z) < 0\\
\]
This needs to be said now, that \(0 < x < \delta\) for arbitrary small, by which the limit has exponential decay as \(\Re(z) \to -\infty\). The limit is immediately true as \(\Im(z) \to \pm \infty\), as we've already shown when discussing the validity of this integral. So all that's left is to show that:
\[
|\frac{F(z)}{\Gamma(1-z)}| = o(x^{z})\,\, \text{as}\,\, \Re(z) \to - \infty\\
\]
Now to solve this we're going to use integration by parts
\[
F(z-1) = \int_0^\infty g(-x)x^{z-2}\,dx = \frac{1}{z-1}\int_0^\infty g'(x)x^{z-1}\,dx\\
\]
And iterating this, we get:
\[
F(z-N) = \frac{1}{(z-1)(z-2)\cdots (z-N)}\int_0^\infty g^{(N)}(x)x^{z-1}\,dx\\
\]
And this clearly has Gamma function growth, where by we must have the appropriate decay.
QED.
--------------------------
Thereby, we've shown that:
\[
\frac{1}{2\pi i} \int_{c-i\infty}^{c+i\infty} \frac{F(z)}{\Gamma(1-z)} x^{-z}\,dz = \sum_{k=1}^\infty g_k \frac{(-x)^k}{k!}\\
\]
For small enough \(0 < x < \delta\). And additionally, since the Residue theorem can be rigorously applied, and the integral converges, we have that the series on the right has a non-trivial radius of convergence.
This is enough to state that \(g_k\) looks like \(O(c^kk!)\) for \(c = \frac{1}{\delta}\).
Which doesn't give as tight a bound as Gottfried wanted, but does supply a bound.
I tried not to muddy any of the details. If you have any questions please ask! I knew Fractional Calculus was good for something, lol.
We begin by denoting \(f(z) = e^{z}-1\) which has the neutral fixed point at \(0\) with multiplier \(1\). By which we can construct an Abel function in the left half plane \(\Re(z) < 0\), because this domain is within the attracting petal of \(0\). Which is to mean that \(\lim_{n\to\infty}f^{\circ n}(z) = 0\) for all \(\Re(z) < 0\). This is seen by just observing the orbit.
The abel function \(\alpha(z)\) is holomorphic in the Left half plane, and satisfies:
\[
\alpha(e^z-1) = \alpha(z) + 1\\
\]
By which we have the identity \(\alpha(z+2\pi i) = \alpha(z)\)--it must inherit \(f\)'s period. We can write:
\[
\alpha(z) = \alpha(f^{\circ n}(z)) - n\\
\]
Which ensures the expansion to the left half plane. This function is holomorphic in a petal near zero; by which it is non-singular; and we can take an inverse function, which we write as \(\alpha^{-1}(z)\), which satisfies \(f(\alpha^{-1}(z)) = \alpha^{-1}(z+1)\). We can construct a holomorphic squareroot function of \(f\), which we write:
\[
g(z) = \alpha^{-1}\left(\frac{1}{2} + \alpha(z)\right)\\
\]
As per David E. Speyer's comment, this function is extendable to \(\mathbb{C}/[0,\infty)\)--as per referenced here https://mathoverflow.net/questions/4347/...ar-and-exp Which is stated as a result of Baker (too lazy to find an English copy of the paper, the original is in German).
This function then has two properties we are interested in. Firstly, \(g : \mathbb{C}_{\Re(z) < 0} \to \mathbb{C}_{\Re(z) < 0}\), where additionally, when we assign the point at infinity in this space, we get that:
\[
g(\infty) = g^{-1}(-1)\\
\]
Because, on this domain, \(f(\infty) = -1\). The function \(g\) is non-singular, so a local inverse always exists. The second thing we need is that this function is bounded on the line \(i\mathbb{R}\). This can be found because \(g(g(i\mathbb{R})) = f(i\mathbb{R})\) is bounded; and by part so much be the first arc \(g(i\mathbb{R})\).
So we can strenghten our statement to \(g(z)\) is bounded for \(\Re(z) \le 0\).
Now the function \(g\) is not holomorphic at \(0\), but can be expanded into an asymptotic series we will write as:
\[
g(z) \sim \sum_{k=1}^\infty g_k z^k\\
\]
And the goal we are trying to show, is that there is some \(0 < c \) such that \(g_k = O(c^kk!)\). Or equivalently, that:
\[
\mathcal{B} g(z) = \sum_{k=1}^\infty g_k \frac{z^k}{k!}\\
\]
Has a non trivial radius of convergence.
----------------------------------------
The first theorem we'll write is as follows:
Theorem:
\[
F(z) = \int_0^\infty g(-x)x^{z-1}\,dx\\
\]
Converges for \( -1 < \Re(z) < 0\) and has decay like:
\[
|F(x + iy)| < M e^{-\frac{\pi}{2}|y|}\\
\]
For some constant \(M\).
Proof:
The integral:
\[
F_\theta(z) = \int_0^\infty g(-e^{i\theta} x) (e^{i\theta}x)^{z-1}e^{i\theta}\,dx\\
\]
converges for all \(-\pi /2 \le \theta \le \pi/2\) and \(-1 < \Re(z) < 0\). The integral converges at the endpoin \(0\), because \(g(z) \sim g_1 z\) at \(0\), and the integral converges at \(\infty\), because \(g(-e^{i\theta}x)\) is bounded and the integrand is bounded by \(|x|^{\Re(z) - 1}\).
Now, by contour integration, we have that \(F_\theta\) is constant in \(\theta\). To see this, consider the contour:
\[
C = [0,R] + \gamma_R - [0,Re^{i\theta}]\\
\]
Then:
\[
\int_C g(x) x^{z-1}\,dx\\ = 0\\
\]
The integral along the arc \(\gamma_R\) is bounded by \(|R|^z\), and therefore the two integrals:
\[
\int_0^\infty g(x)x^{z-1}\,dx - \int_0^\infty g(-e^{i\theta} x)(e^{i\theta}x)^{z-1} e^{i\theta}\,dx = 0\\
\]
Now consider \(y = \Im(z) > 0\), and \(\theta = \pi/2\). We have that:
\[
F(x+iy) = e^{i\frac{\pi}{2}(x+iy)}\int_0^\infty g(ix)x^{z-1}\,dx\\
\]
But this is bounded as \(|F(x+iy) | \le M^+ e^{-\frac{\pi}{2} y}\). A similar procedure can be done for \(\Im(z) < 0\) and \(\theta = - \pi/2\), and we are given:
\[
|F(z)| \le M e^{-\frac{\pi}{2} |y|}\\
\]
Where \(M\) depends on the real part of \(z\) only. This is a Stein & Shakarchi argument, which they develop using the Fourier transform; this is just a change of variables of this. We point to Stein & Shakarchi Complex Analysis.
QED
---------------------------------------
Our second theorem will be that \(F(z)\) is actually meromorphic for \(\Re(z) < 0\). This is a bit more tricky.
Theorem:
The function \(F(z)\) is meromorphic for \(\Re(z) < 0\) with simple poles at \(k \in \mathbb{Z}_{<0}\), with residues \((-1)^k g_k\).
Proof:
This is more of a just check the result. I could go by induction but I want to be quick.
Begin by expanding the integral defining \(F\) as follows:
\[
F(z) = \int_0^1 g(-x)x^{z-1}\,dx + \int_1^\infty g(-x)x^{z-1}\,dx\\
\]
You will note instantly that the second integral is holomorphic for \(\Re(z) < 0\)--while the first integral is only holomorphic for \(\Re(z) > -1\). So let's add in our asymptotic expansion, which is given as:
\[
\int_0^1 \left(g(-x) - \sum_{k=1}^N g_k (-x)^k + \sum_{k=1}^N g_k (-x)^k\right) x^{z-1}\,dx\\
\]
Where the term by term integral of this can be written:
\[
\int_0^1 g(-x)x^{z-1} \,dx = \sum_{k=1}^N g_k\frac{(-1)^k}{k+z} + \int_0^1\left(g(-x) - \sum_{k=1}^N g_k (-x)^k\right) x^{z-1}\,dx\\
\]
But now the integral on the right has an \(N+1\)'th order zero at \(0\), and therefore the integral on the left is meromorphic for \(\Re(z) > -N\).
Combining this together, and since \(N\) is arbitrary. We have that \(F(z)\) is holomorphic for \(\Re(z) < 0\), and has simple poles at \(z = -k\), with residues:
\[
\mathop{\mathrm{Res}}_{z=-k} F(z) = (-1)^k g_k\\
\]
QED
-----------------------------------------------
Now we will place here a theorem on Gamma functions asymptotics in the complex plane. Partially attributed to Stirling, and named for him.
\[
\Gamma(z) \sim \sqrt{2\pi}z^{z-1/2}e^{-z}\,\,\text{as}\,\,|z| \to \infty\,\,\text{while}\,\,|\arg(z)| < \pi\\
\]
Which, for our purposes, in the imaginary asymptotic, can be written:
\[
|\Gamma(x+iy)| \sim \sqrt{2\pi} |y|^{x-1/2} e^{-\frac{\pi}{2}|y|}\,\,\text{as}\,\,|y|\to\infty\\
\]
Of which, we get that:
\[
\frac{|F(x+iy)|}{|\Gamma(1-x-iy)|} \le M |y|^{x-1+1/2}\\
\]
Which gives our third theorem:
Theorem: Let \(-1 < c <-1/2\), then the integral:
\[
h(x) = \frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty} \frac{F(z)}{\Gamma(1-z)}x^{-z}\,dz\\
\]
Converges absolutely, and is continuous for \(x > 0\).
Proof:
Just take absolute values and use the above asymptotics, and note that \(|x^{-z}| < |x|^{-\Re(z)}\).
QED
-----------------------------
And now we enter in the hard part!!! Essentially, we wish to show that \(h(x) = \mathcal{B} g(-x)\). And to do that we need a lemma that is deceptively simple, but may be confusing.
\[
h(x) = \sum_{k=1}^\infty \mathop{\mathrm{Res}}_{z=-k}\frac{F(z)}{\Gamma(1-z)}x^{-z}\\
\]
To calculate these residues isn't very hard. We know that \(F(z)\) has a simple pole at each \(k\), with a residue of \((-1)^kg_k\), so by running Cauchy's integral formula we get:
\[
\mathop{\mathrm{Res}}_{z=-k}\frac{F(z)}{\Gamma(1-z)}x^{-z} = g_k \frac{(-x)^k}{k!}\\
\]
Which is indeed the expansion of \(\mathcal{B}g(-x)\). The trouble is, we need to show that the contour integral satisfies equalling these residues. We write this below:
Theorem:
The function \(h(x)\) from the previous theorem, satisfies:
\[
h(x) = \sum_{k=1}^\infty g_k \frac{(-x)^k}{k!}\\
\]
For a value \(\delta > 0\) and \(0<x < \delta\)
Proof:
Take a contour \(C_R\) such that:
\[
C_R = [c-iR,c+iR] - \gamma_R\\
\]
Where \(\gamma_R\) is the semi-circle to the left of this vertical line. Then we write:
\[
\frac{1}{2\pi i} \int_{C_R} \frac{F(z)}{\Gamma(1-z)} x^{-z}\,dz = \sum \mathop{\mathrm{Res}}\\
\]
Which means it's the sum of the residues of the poles inside the contour. If we can show that:
\[
\lim_{R \to \infty} \int_{\gamma_R} \frac{F(z)}{\Gamma(1-z)} x^{-z}\,dz \to 0\\
\]
By which we get that the integral equals the sum of the residues; and additionally shows that the series converges. To show this, all we need is to show that:
\[
\lim_{|z|\to\infty} \frac{F(z)}{\Gamma(1-z)}x^{-z} \to 0\,\,\text{as}\,\,|z| \to \infty\,\,\text{while}\,\,\Re(z) < 0\\
\]
This needs to be said now, that \(0 < x < \delta\) for arbitrary small, by which the limit has exponential decay as \(\Re(z) \to -\infty\). The limit is immediately true as \(\Im(z) \to \pm \infty\), as we've already shown when discussing the validity of this integral. So all that's left is to show that:
\[
|\frac{F(z)}{\Gamma(1-z)}| = o(x^{z})\,\, \text{as}\,\, \Re(z) \to - \infty\\
\]
Now to solve this we're going to use integration by parts
\[
F(z-1) = \int_0^\infty g(-x)x^{z-2}\,dx = \frac{1}{z-1}\int_0^\infty g'(x)x^{z-1}\,dx\\
\]
And iterating this, we get:
\[
F(z-N) = \frac{1}{(z-1)(z-2)\cdots (z-N)}\int_0^\infty g^{(N)}(x)x^{z-1}\,dx\\
\]
And this clearly has Gamma function growth, where by we must have the appropriate decay.
QED.
--------------------------
Thereby, we've shown that:
\[
\frac{1}{2\pi i} \int_{c-i\infty}^{c+i\infty} \frac{F(z)}{\Gamma(1-z)} x^{-z}\,dz = \sum_{k=1}^\infty g_k \frac{(-x)^k}{k!}\\
\]
For small enough \(0 < x < \delta\). And additionally, since the Residue theorem can be rigorously applied, and the integral converges, we have that the series on the right has a non-trivial radius of convergence.
This is enough to state that \(g_k\) looks like \(O(c^kk!)\) for \(c = \frac{1}{\delta}\).
Which doesn't give as tight a bound as Gottfried wanted, but does supply a bound.
I tried not to muddy any of the details. If you have any questions please ask! I knew Fractional Calculus was good for something, lol.