Let $\theta>0$ be irrational, and define
$$\zeta\_\theta(s) = \sum\_{n=1}^\infty \frac{\lfloor n \theta \rfloor}{n^s}.$$
This converges to an analytic function on $\Re(s)>2$. Does $\zeta\_\theta(s)$ have a meromorphic continuation to $\mathbb{C}$?


Some motivation: I'm interested in zeta functions associated to distance functions on $\mathbb{R}^n$: For a distance function $d$, consider 
$$\zeta\_d(s)=\sum\_{\mathbf{m}\in \mathbb{Z}^n}' d(\mathbf{m})^{-s}.$$ 
If $d$ is smooth outside $0$, one can show $\zeta\_d$ has a meromorphic continuation. If $d(x,y)=\max\{|y|,|x|/\theta\}$, then $\zeta\_\theta(s)$ is $d(x,y)^{-s}$ summed over the integral points in the sector $\{(x,y)\ |\ y\theta>x>0\}$. Meromorphic continuation of $\zeta\_\theta$ and $\zeta\_{\theta^{-1}}$ would then imply the meromorphic continuation of $\zeta\_d$.


Based on numerical evidence, I conjecture that for $\theta$ a quadratic irrational, $\zeta\_\theta(s)$ has a meromorphic continuation to (at least) a region $\Re(s)>\sigma\_0$, where $\sigma\_0<0$, and that its poles in $\Re(s)\geq 0$ occur at $s=2,1$ and a subset of $it\_0\mathbb{Z}$ for some $t\_0\in \mathbb{R}$. A more tentative conjecture is that this is true for all real algebraic numbers.


**EDIT**: George Lowther gave a nice answer below, but I've since learned that this question was answered by Hecke in 1922:


E. Hecke, *Uber Analytische Funktionen und die verteilung von zahlen mod
eins*, Hamburg. Math.Abh., 1(1922) 54 - 76



Meromorphic continuation of a Dirichlet series associated to an irrational number
Yes, the conjecture is true -- $\zeta\_\theta(s)$ extends to a meromorphic function on $\mathbb{C}$ for all real quadratic irrationals $\theta$ with poles at $s=2$, $s=1$ and in the infinite set of vertical lines $\lbrace -2n+i\mathbb{R}\colon n\in\mathbb{Z}\_{\ge0}\rbrace$. I'm not sure about the more tentative conjecture for arbitrary algebraic numbers. Also, more simply, it will extend to a meromorphic function over $\Re(s) > 0$ whenever $\theta$ cannot be approximated too well by rationals (in particular, if $\theta$ has irrationality measure 2 which, by the Thue-Siegel-Roth theorem, includes all algebraic irrationals). Conversely, if an irrational $\theta$ can be approximated too well by rationals then there will not be a meromorphic extension, which will be the case outside of a meagre subset of $\mathbb{R}$. I can give a rough proof of these statements now.


*[**Edit**: As Pieter pointed out in the comments, my initial answer carelessly missed out a rather important term, so was not correct. I've updated the proof sketch to deal with this term, and restricted to quadratic irrationals for the meromorphic extension. The proof is a bit sketchy, as I haven't justified that we can rearrange the order of the various summations/integrals. As they are only conditionally convergent, this would need to be fleshed out to make it more rigorous but, thinking about this a bit more, it looks like it all works out with no problems.]*


First, let $\lbrace x\rbrace = x - \lfloor x\rfloor$ denote the fractional part of $x$, and write
$$
\tilde\zeta\_\theta(s)=\sum\_{n\ge1}\frac{\lbrace n\theta\rbrace-1/2}{n^s},
$$
which converges, at least, on $\Re(s) > 1$.
Then, the function $\zeta\_\theta$ defined in the question is
$$
\zeta\_\theta(s)=\theta\zeta(s-1)-\frac12\zeta(s)-\tilde\zeta\_\theta(s),
$$
where $\zeta$ is the Riemann zeta function. So we have an extension to $\Re(s) > 1$. In the case where $\theta$ is rational then $\lbrace n\theta\rbrace-1/2$ is periodic in $n$, so $\tilde\zeta\_\theta$ is a linear combination of Dirichlet L-functions and has a meromorphic extension to $\mathbb{C}$. Henceforth, I will only consider irrational $\theta$. Setting $F(x)=\sum\_{1\le n\le x}(\lbrace n\theta\rbrace-1/2)$, the equidistribution theorem states that $F(x)/x\to0$ as $x\to\infty$. Rearranginging the expression for $\tilde\zeta\_\theta$ gives
$$
\tilde\zeta\_\theta(s)=\sum\_{n\ge1}\frac{F(n)}{n}n^{1-s}\left(1-(1+1/n)^{-s}\right)
$$
If $\lvert F(n)/n\rvert$ is bounded by some $\delta > 0$ then this expression will be bounded by $\delta\zeta(s)$ for all $s > 1$. So, by equidistribution, it follows that $(s-1)\tilde\zeta\_\theta(s)\to0$ as $s\to1$ (on $s > 1$). If a meromorphic extension existed in a neighbourhood of 1 then this would imply that $\tilde\zeta\_\theta$ is bounded near 1. However, if $\theta=p/q$ is rational ($p,q$ coprime) then $F(x)/x\to-1/(2q)$ so, if $x$ is too closely approximated by rationals then there will be a slight bias in $F(x)/x$ and $\tilde\zeta\_\theta(s)$ will not be bounded close to 1. I think that the existence of rational approximations $p\_n/q\_n$ with $q\_n^2\log\lvert\theta-p\_n/q\_n\rvert\to-\infty$ is enough for this to happen.


Now, suppose that $\theta$ has irrationality measure no greater than some finite $\gamma$, so that there are only finitely many rational approximations $\lvert\theta-p/q\rvert\le q^{-\gamma-\delta}$ for any $\delta > 0$. Equivalently, there is a constant $C\_\delta > 0$ such that $n^{\gamma-1+\delta}\lvert n\theta-m\rvert\ge C$ for all positive integer $n$ and integer $m$. Then, the discrepancy of the set $\left\lbrace \lbrace k\theta\rbrace\colon 1\le k\le n\right\rbrace$ is bounded by $Kn^{-1/(\gamma-1)+\delta}$ (for a constant $K$). This means that $F(n)/n$ is bounded by $2Kn^{-1/(\gamma-1)+\delta}$ and, hence, the expression above for $\tilde\zeta\_\theta(s)$ has summand going to zero at rate $O(n^{-s-1/(\gamma-1)+\delta})$. So, $\tilde\zeta\_\theta(s)$ is an analytic function on $\Re(s) > (\gamma-2)/(\gamma-1)$. In particular, for algebraic irrationals, $\gamma=2$ and we get a meromorphic extension of $\zeta\_\theta(s)$ to $\Re(s) > 0$ with simple poles at $s=2$ and $s=1$.


I now attempt to extend to the whole of $\mathbb{C}$ via Hurwitz's formula which, for $\Re(s) > 0$ and irrational $x$ in the unit interval, gives
$$
\begin{array}{l}
&\sum\_{n\ge1}e^{2\pi inx}n^{-s}=(2\pi)^{s-1}\Gamma(1-s)\left(e^{i\pi(1-s)/2}\zeta(1-s,x)+e^{i\pi(s-1)/2}\zeta(1-s,1-x)\right)
\end{array}
$$
where $\zeta(s,x)=\sum\_{n\ge0}(n+x)^{-s}$ is the Hurwitz Zeta function.
Plugging in the Fourier series $\lbrace x\rbrace=1/2+\sum\_{k\not=0}(i/(2\pi k))e^{2\pi ikx}$, applying Hurwitz's formula, and being careless about when the sums converge/commute (it can be made more rigorous) gives,
$$
\begin{array}{rl}
\tilde\zeta\_\theta(s)&=\sum\_{n\ge1}\sum\_{k\not=0}\frac{ie^{2\pi ink\theta}}{2\pi kn^s}\cr
&=(2\pi)^{s-2}\Gamma(1-s)\sum\_{k\not=0}\frac{i}{k}\left(e^{i\pi(1-s)/2}\zeta(1-s,\lbrace k\theta\rbrace)+e^{i\pi(s-1)/2}\zeta(1-s,1-\lbrace k\theta\rbrace)\right)\cr
&=(2\pi)^{s-2}\Gamma(1-s)\sum\_{k\not=0}\frac{i}{k}\left(e^{i\pi(1-s)/2}\zeta(1-s,\lbrace k\theta\rbrace)-e^{i\pi(s-1)/2}\zeta(1-s,\lbrace k\theta\rbrace)\right)\cr
&=2\sin((s-1)\pi/2)(2\pi)^{s-2}\Gamma(1-s)\sum\_{k\not=0}k^{-1}\zeta(1-s,\lbrace k\theta\rbrace)\cr
&=2\sin((s-1)\pi/2)(2\pi)^{s-2}\Gamma(1-s)\sum\_{n\ge0}\sum\_{k\not=0}k^{-1}(n+\lbrace k\theta\rbrace)^{s-1}.
\end{array}
$$
Split up the summation as
$$
\sum\_{n\ge1}n^{s-1}\sum\_{k\not=0}k^{-1}\left(1+\lbrace k\theta\rbrace/n\right)^{s-1}+\sum\_{k\not=0}k^{-1}\lbrace k\theta\rbrace^{s-1}.
$$
The first term can be handled easily. Use the fact that $\theta$ has irrationality measure 2, so that $\left\lbrace\lbrace k\theta\rbrace\colon k=1,\ldots,n\right\rbrace$ has discrepancy $O(n^{-1+\delta})$. As the function $x\mapsto(1+x/n)^{s-1}$ has bounded variation of size $O(1/n)$ over the unit interval, the summation over $k$ converges and is of size $O(1/n)$. Therefore, the sum over $n$ converges on $\Re(s) < 1$, showing that the first term converges to an analytic function on $\Re(s) < 1$.


The final summation over $k$ is more problematic. Assuming $\theta$ is a quadratic irrational, and using $\bar z$ to denote the conjugate of $z\in\mathbb{Q}(\theta)$, it can be rewritten (up to order of summation) as
$$
(\bar\theta-\theta)\sum\_{0 < z < 1}\frac{z^{s-1}}{\bar z-z}=(\bar\theta-\theta)\sum\_{j=0}^\infty\sum\_{0 < z < 1}{\bar z}^{-j-1}z^{s+j-1}.
$$
Here, I have substituted in $z=\lbrace k\theta\rbrace$, and the summation is over all $z\in\mathbb{Z}+\mathbb{Z}\theta$ lying in the unit interval. For each bounded region for $s$ on which we want to extend we only actually have to expand out a finite number of terms in the sum over $j$ above, up until $\Re(s)+j-1$ is positive. In that case, the remainder is of order $\bar z^{-j-1}z^{s+j-1}=O(k^{-1-j})$ and has uniformly convergent sum. So, we just need to show that each term in the sum over $j$ extends to a meromorphic function.
For the moment, I'll restrict to the case where $\mathbb{Z}+\mathbb{Z}\theta$ is a number ring. For example, $\theta=\sqrt{d}$ for squarefree $d\not\cong1$ (mod 4) or $\theta=(1+\sqrt{d})/2$ for squarefree $d\cong1$ (mod 4). Let $\eta$ be the fundamental unit of $\mathbb{Z}[\theta]$ lying in the unit interval. Then, each $z$ in the unit interval can be written uniquely as $\eta^ny$ for $y$ in $[\eta,1)$ and nonnegative integer $n$. Writing $\epsilon=\bar\eta\eta=\pm1$, the term for a fixed $j$ in the expansion above is (again, being a bit careless about order of summation),
$$
\begin{array}{rl}
\sum\_{0 < z < 1}{\bar z}^{-j-1}z^{s+j-1}&=\sum\_{\eta\le z < 1}\sum\_{n=0}^{\infty}(\bar\eta ^n\bar z)^{-j-1}(\eta^nz)^{s+j-1}\cr
&=\sum\_{\eta\le z\le1}\sum\_{n=0}^\infty{\bar z}^{-j-1}z^{s+j-1}\epsilon^{n(j+1)}\eta^{n(s+2j)}\cr
&=\left(1-\epsilon^{j+1}\eta^{s+2j}\right)^{-1}\sum\_{\eta\le z < 1}{\bar z}^{-j-1}z^{s+j-1}.
\end{array}
$$
Now that the summation has being restricted to $z\ge\eta$, the term $z^{s+j-1}$ will be bounded, and the sum is guarateed to converge. To see this, substitute back $z=\lbrace k\theta\rbrace$ and $\bar z=k(\bar\theta-\theta)+\lbrace k\theta\rbrace$. The summand is of size $O(k^{-1-j})$, so the sum converges absolutely for $j\ge1$. For $j=0$ it is $(\bar\theta-\theta)^{-1}\lbrace k\theta\rbrace^{s-1}/k + O(k^{-2})$ for $\lbrace k\theta\rbrace\ge\eta$. As $x^{s-1}$ has finite variation over the interval $[\eta,1)$, the sum for $j=0$ also converges.


Finally, the term $(1-\epsilon^{j+1}\eta^{s+2j})^{-1}$ is meromorphic on $\mathbb{C}$ with zeros on the vertical line $-2j+i\mathbb{R}$. So, $\zeta\_\theta$ does extend with poles in the claimed set.


In the argument above, it was assumed for simplicity that the module $M\equiv\mathbb{Z}+\mathbb{Z}\theta$ is a number ring. The generalization to arbitrary quadratic irrationals is easy enough. The only use of the fact that $M$ is a number ring was to ensure that $\eta M\subseteq M$ and $\eta^{-1}M\subseteq M$. However, there will always exist some $r > 0$ such that $\eta^{\pm r}M\subseteq M$ and the argument above follows through unchanged with $\eta^r$ in place of $\eta$. To see that such an $r$ does exist, let $R$ be the ring of algebraic integers in $\mathbb{Q}(\theta)$, and let $a,b$ be nonzero integers with $bR\subseteq aM\subseteq R$. Then, multiplication by $\eta$ gives a permutation of the finite quotient $R/bR$, which has finite order $r > 0$. So, $\eta^{\pm r}x-x\in bR\subseteq aM$ for all $x\in R$, from which $\eta^{\pm r}aM\subseteq aM$ follows.


