# Analytic continuation of a certain family of $L$ series

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?

• I don't think a density argument is ok. May 1, 2021 at 8:40
• Why is that? I think a uniform convergence argument might help out May 1, 2021 at 8:41
• It is ok if you can prove the "sum of two Hurwitz zeta" functional equation for $x\in \Bbb{Q}$, but otherwise I don't think it is, because you get a sequence of linear combinations of increasing many Dirichlet L-functions. Indeed $\zeta(s)$ has a pole at $s=1$ so the density doesn't prevent singularities. May 1, 2021 at 8:48
• Can you please elaborate @reuns ? May 1, 2021 at 8:49

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.
• About what? ${}{}$ May 1, 2021 at 8:24
• This is how Riemann obtained the functional equation for $\zeta(s)$ in his 1859 paper, search for "zeta functional equation contour integral" for example math.stackexchange.com/questions/2506151/… May 1, 2021 at 8:30