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?

  • $\begingroup$ I don't think a density argument is ok. $\endgroup$
    – reuns
    Commented May 1, 2021 at 8:40
  • $\begingroup$ Why is that? I think a uniform convergence argument might help out $\endgroup$ Commented May 1, 2021 at 8:41
  • $\begingroup$ 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. $\endgroup$
    – reuns
    Commented May 1, 2021 at 8:48
  • $\begingroup$ Can you please elaborate @reuns ? $\endgroup$ Commented May 1, 2021 at 8:49

1 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.

  • $\begingroup$ Can you include a bit more details... $\endgroup$ Commented May 1, 2021 at 8:24
  • $\begingroup$ About what? ${}{}$ $\endgroup$
    – reuns
    Commented May 1, 2021 at 8:24
  • $\begingroup$ Also, some details about the Bromwich contour would be appreciated. I am hearing the word for the first time... $\endgroup$ Commented May 1, 2021 at 8:27
  • 1
    $\begingroup$ 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/… $\endgroup$
    – reuns
    Commented May 1, 2021 at 8:30

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