How to invert this relationship (Möbius function involved)?

Question

Background

Let,

$$ S(x,a) = \sum_{r=1}^\infty \sum_{d|r} \frac{x^r}{a^d d!}$$

Another representation of this series for all derivatives around $0$ is given by:

$$ S(x,a) = \sum_{r=1}^\infty (e^{\frac{x^r}{a}} -1 ) $$

where $|x| < 1$ and $a \neq 0$

$$ \frac{1}{2 \pi i} \oint \frac{f(a)}{a} \lim_{x \to 0 } \frac{\partial^r S}{\partial x^{r}} da = \delta_{1,r}$$

where $r$ is a positive non-zero integer and $\delta_{1,r}$ is only $1$ when $r = 1$

With, $f$ being analytic then we claim:

$$f^{'n}(0) = \mu(n)$$

where $\mu$ is the Möbius function.

Question

Are there any simplifications which can be done to:

$$ \frac{1}{2 \pi i} \oint \frac{f(a)}{a} \lim_{x \to 0 } \frac{\partial^r S}{\partial x^{r}} da = \delta_{1,r}$$

To possibly invert it?

The below are examples of the claim:

Example $1$

Consider, the following equation:

$$ S(x,a) = \sum_{r=1}^\infty (e^{\frac{x^r}{a}} -1 )$$

Differentiating both sides with respect to $x$:

$$ S'(x,a) = \frac{1}{a}e^{\frac{x}{a}} + \frac{2x}{a}e^{\frac{x^2}{a}} + \dots$$

Multiplying by $\frac{f(a)}{a}$ and taking $x \to 0$:

$$ \frac{f(a)}{a} \lim_{x \to 0}S'(x,a) = \frac{f(a)}{a^2}$$

Taking the contour integral with respect to $a$:

$$\oint \frac{f(a)}{a^2} da = f'(0) = 1$$

Example $2$

Consider, the following equation:

$$ S(x,a) = \sum_{r=1}^\infty (e^{\frac{x^r}{a}} -1 ) $$

Differentiating both sides with respect to $x$ twice:

$$ S''(x,a) = \frac{1}{a^2}e^{\frac{x}{a}} + \frac{2}{a}e^{\frac{x^2}{a}} + \dots$$

Multiplying by $\frac{f(a)}{a}$ and taking $x \to 0$:

$$ \frac{f(a)}{a} \lim_{x \to 0} S''(x,a) = \frac{f(a)}{a^3} + 2 \frac{f(a)}{a^2}$$

Taking the contour integral with respect to $a$:

$$\oint (\frac{f(a)}{a^3} + 2 \frac{f(a)}{a^2}) da = 2! f''(0) + 2f'(0) \implies f''(0) = -1$$

(Using example $1$ in the last step)

Answer 1

I thought about this for a bit and actually I arrive at a different result. So probably I made a mistake.

Let me change the variables and indices a bit. For $a ∈ ℂ^×$ and $z$ ranging around zero let $$S(z;a) = \sum_{n=1}^∞ \sum_{d \mid n} \frac{z^n}{a^d d!} = \sum_{ν=1}^∞ (\mathrm e^{\frac {z^ν} a} - 1).$$ Then for any $a ∈ ℂ^×$, $S(z; a)$ is analytical in $z$ around zero and furthermore for $n ≥ 1$ $$S_{n,0}(a) := \lim_{z → 0} \frac {∂^n} {∂z^n} S(z; a) = \sum_{d \mid n} \frac {n!} {a^d d!}.$$ By Cauchy’s integral formula we have for any analytical function $f$ around zero $$f^{(d)}(0) = \frac {d!} {2π \mathrm i} \oint \frac{f(a)} {a^{d+1}} \mathrm d a,$$ and so $$\frac 1 {2π \mathrm i} \oint \frac{f(a)} a S_{n,0}(a) \mathrm d a = \sum_{d \mid n} \frac {n!} {d!} \frac 1 {2π \mathrm i}\oint \frac {f(a)} {a^{d+1}} \mathrm d a= \sum_{d \mid n} \frac {n!} {d!^2} f^{(d)} (0).$$ Now if for every $n ≥ 1$ $$\frac 1 {2π \mathrm i} \oint \frac{f(a)} a S_{n,0}(a) \mathrm d a = ε(n)$$ with $ε(n) = δ_{n,1}$, then by canceling $n!$, we also get $$\sum_{d \mid n} \frac{f^{(d)}(0)} {d!^2} = ε(n),$$ and since also $\sum_{d \mid n} μ(d) = ε(n)$, by a simple induction on $n$ we conclude $μ(n) = \frac{f^{(n)}(0)}{n!^2}$.