On a doubt about a simple exercise concerning Nachbin resummation

Question

One can read here what is Nachbin resummation. This is an exercise that I've created (first I tried get simple examples about the $ \Psi $ type, but it was a failure):

I want to solve $$\sum_{n=0}^{\infty}\frac{\mu(n+1)}{s^n}=s\int_0^\infty e^{-(st)^k}f(t)dt,\tag{1}$$ where $\mu(n)$ is the Möbius function and $k\geq 1$ a fixed integer.

Question. Since from previous Wikipedia article one can write the explicit solution, I want write it, and I am asking for some details to fill the reasoning (in these questions, but feel free to add more details if it is neccesary):

A) Can you provide me hints or the proof to the deduce the Mellin transform, and its domain, of the kernel $$K(u)=e^{-u^{k}}$$ for a fixed integer $k\geq 1$, with the purpose to show that $$f(x)=k\sum_{n=0}^\infty\frac{\mu(n+1)}{\Gamma\left(\frac{n+1}{k}\right)}x^n\tag{2}$$ is the solution of previous $(1)$? My calculations were with an online calculator$^{\dagger}$.

B) Can you justify if $(2)$ holds for $x\geq 1$? I am asking it doing by similarity with Gram series, what I am asking if how do you deduce the domain of $f(x)$. THen, is it obvious from the integral $(1)$ that the domain is $0\leq x$? Thanks in advance.

$\dagger$ At this moment with standard time of computation I can not get one more time this code to show here a closed-form.

Answer 1

The Mellin transform of $K(u) = e^{-u^k}$ is by definition $$ M(s) = \int_0^\infty u^{s-1} e^{-u^k} \, du $$ Using the substitution $t = u^k$ in the integral, this turns into $$ M(s) = \frac1k \int_0^\infty t^{s/k-1} e^{-t} \, dt = \frac1k \Gamma\left(\frac{s}{k}\right) $$ which together with the formula from Wikipedia gives the desired series solution.

For your question about the domain of the solution you can just observe that it is given by a power series, and from the fact that the $\Gamma$ function grows faster than any geometric sequence and the fact that $|\mu| \le 1$ you get easily that it has an infinite radius of convergence, so it represents an entire function.