What's about $\sum_{n=1}^\infty\frac{\mu(n)}{n}f(X^{\frac{1}{n}})$, where $\mu(n)$ is the Möbius function? Let $X=\sigma+it$ the complex variable, and $\mu(n)$ the Möbius function. 
Inspired in Riemann function $R(X)$ I would like to ask you

Question. What conditions are required to be satisfied by a function $f(X)$ with the purpose to presume that $$F(X):=\sum_{n=1}^\infty\frac{\mu(n)}{n}f(X^{\frac{1}{n}})$$ is well defined? I say conditions for good functions (an ample set of complex functions for which is easy answer the question, for a half-plane of convergence). Thanks in advance.

I know that Riemann function $R(X)$ is smooth, then if you want to provide us conditions for which $F(X)$ be diffentiable in the complex sense, you are welcome. I know that it it possible write $$|F(X)|\leq \sum_{n=1}^\infty\frac{|f(X^{\frac{1}{n}})|}{n},$$
and that could be useful define $F(X)$ as $\lim_{N\to\infty}\sum_{n=1}^N\frac{\mu(n)}{n}f(X^{\frac{1}{n}})$, but I have no idea to solve and deduce where is defined $F(X)$.
Please if there are inaccurancies in my questions add a comment to improve this post.
 A: Short answer: If $f$ is holomorphic on $\mathbb{C} \setminus \{0\}$, $f(1) = 0$, and $\text{Log}$ is your favorite branch of the complex logarithm, then, defining $X^{1/n} = e^{(\text{Log} X)/n}$, $F$ is defined except at the branch of the complex logarithm where $\text{Log}$ was undefined. In fact, $F$ is holomorphic on this region.
To state this a little more cleanly, let me instead consider the functions $G(Y) = F(e^Y)$ and $g(y) = f(e^y)$, so that
$$
G(Y)
  = \sum_{n=1}^\infty\frac{\mu(n)}{n}f((e^Y)^{\frac{1}{n}})
  = \sum_{n=1}^\infty\frac{\mu(n)}{n}g\left(\frac{Y}{n}\right).
$$
If $f(1) = 0$, then $g(0) = 0$.
Below, I will show that given this, if $g$ is holomorphic in an open disc of radius $R$ around $0$, then so is $G$. We may then recover a (not unique) definition for $F$ by picking a branch of the complex logarithm $\text{Log}$ as described above, thus giving a unique definition to $X^{(1/n)}$ and letting us recover $F$ by the formula $F(x) = G(\text{Log}(x))$.
$F$ will thus be holomorphic on the set of $x$ such that $\text{Log}(x)$ lies in the domain of $G$, i.e. the set of $x$ such that $|\text{Log}(x)| < R$.
In particular, to justify my first sentence, if $f$ is holomorphic on $\mathbb{C} \setminus \{0\}$, then $g$ is holomorphic everywhere, so $G$ is holomorphic everywhere, so $F$ is holomorphic on the domain of $\text{Log}$.

Details:
It remains to prove the statement above, that is that assuming $g(0) = 0$, if $g$ is holomorphic in an open disc of radius $R \ge 0$ around $0$, then $G$ is defined and also holomorphic in that open disc. We specifically allow for $R = \infty$.
Note that $g(Y/n)$ is holomorphic for $|Y| < R$, for each $n$. So we have an infinite sum of holomorphic functions in this disc. Referring to Soarer's comment here:

[an] infinite sum of holomorphic functions (on a domain) that converge uniformly on every compact subset is holomorphic

we just need to show that series for $G(Y)$ converges uniformly on the set $|Y| \le r$, for any finite radius $r < R$ (again this allows for $R = \infty$). It suffices to bound the tail end of the series uniformly and with a bound that goes to $0$. So let $N \in \mathbb{N}$ and let $|Y| \le r$.
Expand $g$ as a power series around $0$ as
$$
g(z) = \sum_{i \ge 1} a_i \frac{z^i}{i!}
$$
with $i \ge 1$ since $g(0) = 0$.
Note that $g$ converges absolutely for $|z| \le r$, so that we may define a finite constant $M_r = \sum_{i \ge 1} |a_i| \frac{r^i}{i!}$.
Now we have
\begin{align*}
\left|\sum_{n=N}^\infty\frac{\mu(n)}{n}g\left(\tfrac{Y}{n}\right)\right|
&\le \sum_{n=N}^\infty \frac{1}{n} \left| g\left(\tfrac{Y}{n}\right)\right| \\
&= \sum_{n=N}^\infty \frac{1}{n} \left| \sum_{i = 1}^\infty a_i \frac{(Y/n)^i}{i!}\right| \\
&\le \sum_{n=N}^\infty \sum_{i = 1}^\infty \frac{1}{n^{i+1} i!} |a_i| |Y|^i \\
&= \sum_{i = 1}^\infty |a_i| \frac{|Y|^i}{i!} \sum_{n=N}^\infty \frac{1}{n^{i+1}} \\
&\le \sum_{i = 1}^\infty |a_i| \frac{|Y|^i}{i!} \sum_{n=N}^\infty \frac{1}{n^2} \\
&= \left( \sum_{i = 1}^\infty |a_i| \frac{|Y|^i}{i!} \right)
\left( \sum_{n=N}^\infty \frac{1}{n^2} \right) \\
&\le M_r \sum_{n=N}^\infty \frac{1}{n^2},
\end{align*}
which approaches $0$ as $N \to \infty$ since the series for $\zeta(2)$ converges.
Therefore, the series converges uniformly on any closed discs $|z| \le r$. Thus it converges uniformly on all compact subsets of $|z| < R$, and so the sum of these infinitely many holomorphic functions is holomorphic. So $G$ is holomorphic on $|z| < R$.
