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

Define $$$$F(\sigma) = \Re \sum_{h=1}^{\infty} \sum_{n=1}^{\infty} \frac{b_{n, h}}{n^{2\sigma}}(1+h/n)^{-\sigma}\Bigg( \frac{e^{i\log(1+h/n)}-1}{i\log(1+h/n)} \Bigg)$$$$ where $$\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 $$2\sigma>1$$.

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 $$\sigma\geq 1$$.
• Thanks ! I i guess will be interesting to see what happends modulo $\sigma=1$...That is, whether or not $\sigma=1$ is an isolated singulaity... – Pres10 Aug 9 '20 at 7:28