Does this function have a holomorphic continuation in $\sigma > \frac{1}{2}$?

Question

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 $2\sigma>1$.

Answer 1

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 $\sigma\geq 1$.