Define
\begin{equation}
F(\sigma) = \Re \sum\_{h=1}^{\infty} \sum\_{n=1}^{\infty} \frac{b\_{n, h}}{n^{2\sigma}}(1+h/n)^{-\sigma}\Bigg( \frac{e^{i\log(1+h/n)}-1}{i\log(1+h/n)} \Bigg)
\end{equation} where $\sigma \in \mathbb{R}$ and $|b\_{n, h}|\ll \log n$. *What is the minimal real number $c$ such that $F(\sigma)$ has a holomorphic continuation in $\sigma>c$ ?*


Heuristically, it seems to me that $c \leq 1/2$, considering the fact that the inner sum is dominated by terms with small $h/n$, so that convergence of the whole series is assured for $2\sigma>1$.



Does this function have a holomorphic continuation in $\sigma > \frac{1}{2}$?
Observe first that
$$\lim\_{x\to 0}\frac{e^{i\log(1+x)}-1}{i\log(1+x)}=1,$$
hence there exists an absolute constant $C>0$ such that
$$\Re\frac{e^{i\log(1+h/n)}-1}{i\log(1+h/n)}>\frac{1}{2}\qquad\text{for}\qquad n>Ch.$$
Now assume that $b\_{n,h}$ is the indicator function of $n\in(Ch,2Ch)$. Then for $\sigma\in[1,2]$ we have
$$F(\sigma)>\frac{1}{2}\left(1+\frac{1}{C}\right)^{-2}\sum\_{h=1}^\infty\,\sum\_{n\in(Ch,2Ch)}\frac{1}{n^{2\sigma}}\gg\sum\_{h=1}^\infty\frac{1}{h^{2\sigma-1}}.$$
It follows that $F(1)$ diverges, and $F(\sigma)\to\infty$ under $\sigma\to 1+$. The second statement shows that $F(\sigma)$ has no continuous extension to $\sigma\geq 1$.


