Analytic continuation of a certain family of $L$ series

Question

Consider the function

$$ G(s,x) : = \sum_{n=1}^\infty \frac{\exp(2\pi i nx)}{n^s} $$ where $x\in[0,1)$ and $s\in\mathbb{C}$. The series is absolutely convergent for $Re(s) > 1$.

If $x\in\mathbb{Q}\cap[0,1)$, then it is not difficult to see that $G(s,x)$ can be written as a finite sum of certain Dirichlet $L$ series which have analytic continuation, functional equation and other nice properties. Now, can we use the fact that $\mathbb{Q}\cap [0,1)$ is dense in $[0,1)$ and show that $G(s,x)$ has analytic continuation and functional equation for all $x\in[0,1)$.

The intuition behind the question is that $G$ is continuous in the $x$ variable and holomorphic in the $s$ variable in the region of convergence. Can we use these two facts and the existence of analytic continuation on a dense subset to conclude the existence of analytic continuation everywhere?

Answer 1

For $x\in (0,1)$, $\Re(s) > 1$ $$\Gamma(s)\sum_{n\ge 1} e^{2i\pi nx}n^{-s}=\int_0^\infty \frac{t^{s-1}}{e^{t-2i\pi x}-1}dt=\int_r^\infty \frac{t^{s-1}}{e^{t-2i\pi x}-1}dt+\sum_{k\ge 0} \frac{c_k(x) r^{s+k}}{s+k} $$ where $\sum_{k\ge 0} c_k(x)t^k=\frac1{e^{t-2i\pi x}-1}$ for $|t|\le r$ and the RHS gives the analytic continuation to $\Bbb{C}$.

The functional equation (in term of Hurwitz zeta) is obtained from applying the residue theorem to $$2i\sin(\pi s)\Gamma(s)\sum_{n\ge 1} e^{2i\pi nx}n^{-s} = \int_C \frac{(-z)^{s-1}}{e^{z-2i\pi x}-1}dz$$ where $C$ is a Bromwich contour.