MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $N,a\in\mathbf{Z}_{\geq 1}$. Define a partial $\zeta$-function as $$ \zeta(s;N,a):=\sum_{\substack{n\geq 1\newline n\equiv a\pmod{N}}} \frac{1}{n^s} $$ where $Re(s)>1$. Let $\omega$ be either the trivial character or the sign character i.e. $x\mapsto sign(x)$. Define a partial $\Psi$-function as $$ \Psi(s,\omega;N,a):=\sum_{\substack{0\neq n\in\mathbf{Z}\newline n\equiv a\pmod{N}}}\frac{\omega(n)}{|n|^s}, $$ for $Re(s)>1$. Then it is well known that $\zeta(s;N,a)$ and $\Psi(s,\omega;N,a)$ admit a meromorphic continuation to $\mathbf{C}$ with at most of pole of order $1$ at $s=1$. If $\omega$ is the sign character then $$ \Psi(s,\omega;N,a)+\Psi(s,1;N,a)=2\zeta(s;N,a) $$ and $$ \zeta(s;N,a)-\zeta(s;N,-a)=\Psi(s,\omega;N,a). $$ Note that when $N>1$, the functions $\Psi$ and $\zeta$ do not have an Euler product. So here are 2 natural questions:

Q1: For a fixed $N$ do we know if there exists a constant $C_N>1$ such that if $Re(s)>C_N$ then $\Psi$ and $\zeta$ do not vanish (if the answer is yes then how to prove it)?

Q2: What do we know in general about the nontrivial zeros of $\Psi$ and $\zeta$?

share|cite|improve this question
up vote 10 down vote accepted

The answer to question 1 is classical: Any Dirichlet series which has a finite abscissa of absolute convergence has a zero-free half-plane.

Suppose the Dirichlet series $$ A(s)=\sum_{n=1}^\infty \frac{a_n}{n^s}$$ has an abscissa of absolute convergence $\sigma_a$. If $a_m$ is the first non-zero coefficient, then for $\Re(s)>\sigma_a$ $$ m^s A(s) = a_m + a_{m+1} \Big(\frac{m}{m+1}\Big)^s + a_{m+2} \Big(\frac{m}{m+2}\Big)^s +\cdots \to a_m \ \text{ as } \ \Re(s) \to +\infty. $$ Hence, there exists an absolute constant $C$ such that for $\Re(s)>C$ we have $$ \left|a_{m+1} \Big(\frac{m}{m+1}\Big)^s + a_{m+2} \Big(\frac{m}{m+2}\Big)^s +\cdots \right| \leq \frac{|a_m|}{2}.$$ Consequently, $m^s A(s)$ has no zeros in the half-plane $\Re(s)>C$.

share|cite|improve this answer
Thanks Micah, do you think it is possible to give a sharp approximation of $C_N$? So for example, do you think that $C_N=1$ should work? – Hugo Chapdelaine May 17 '12 at 2:48
With a little work, one should be able to prove an explicit value of for $C_N$. Proving "sharpness" may not be so easy. It is unclear to me that $C_N=1$ will work. There are plenty of fairly basic Dirichlet series with zeros to the right of $\Re(s)=1$. – Micah Milinovich May 17 '12 at 14:40

Micah's answer answers your Q1. My reply gives some information on your Q2. Let us assume that $1 \leq a\leq N$ for simplicity (this condition can be removed). Your zeta-functions $\zeta$ and $\Psi$ can be expressed by the Hurwitz zeta-function For your first function $$\zeta(s;N,a)=N^{-s} \zeta \left(s,\frac a N \right).$$ Davenport and Heilbronn proved that for any $c>0$, $a/N \neq 1/2,1$ there exists $\gg T$ zeros of this function in the strip $1<\Re(s)<1+c, | \Im (s)| < T $. By Voronin universality for the Hurwitz zeta-function the same can be said in any strip $1/2<\sigma_0< \Re (s)<\sigma_1<1$ (this can be found for example in Jörn Steuding's SLN or Garunkstis-Laurincikas book on the Lerch zeta-function).

Your second function can also be expressed in terms of the Hurwitz zeta-function $$\Psi(s,\omega;N,a)=N^{-s} \left(\zeta\left(s,\frac a N\right)+\omega(-1)\zeta \left( s,1-\frac a N \right) \right).$$ In this case joint universality of the Hurwitz zeta-function (see same references as above) can be used to show that except when $\Psi$ has an Euler-product (which only occurs for some small special cases when $ a / N$ equals 1/6, 1/4, 1/3, 1/2, 2/3, 3/4, 5/6, or 1.) there are $\gg T $ zeros in any strip $1/2<\sigma_0< \Re (s) < \sigma_1 < 1$ and $|\Im(s)| < T$. Davenport-Heilbronn's result for $\Re(s)>1$ can be obtained in this case as well.

For the special cases where your function has an Euler-product it will essentially be a Dirichlet L-function or the Riemann zeta-function, and its zeros will be the same.

share|cite|improve this answer
Dear Johan, this is quite an interesting result! It seems to me that this result suggests that given $c>0$, if we look for zeros with $1<Re(s)<1+c$ up to some large imaginary part $T$ then "most of them" will have a real part close to $1$. Are analytic number theorists close to prove anything like that? – Hugo Chapdelaine May 19 '12 at 16:23
No, I do not think so. In fact if $\sup_{\zeta(s,N,a)=0}\Re(s)=c_1$ then $c_1>1$ and given any $c_0>1$ there will be $\gg T$ zeros in $c_0<\Re(s)<c_1$. These results follows from results of Bohr and Kronecker's theorem. – Johan Andersson May 19 '12 at 16:41
I see, thanks for setting me on the right path! – Hugo Chapdelaine May 20 '12 at 2:26

EDIT Experimentally the zeros of $\zeta(s;5,1)$ and $\zeta(s;6,1)$ are those of zeta (the previous revision incorrectly included wrong zeros caused by insufficient precision).

The zeros of $\zeta(s;6,2)$ (if computed correctly) don't appear related to those of zeta.

sage code:

import mpmath
def L(x):
    return mpmath.dirichlet(x,an)

def search1():
    for k in xrange(1,40):
        rr += [0.1 + mpmath.j*k]
    for x in rr:

        except:  continue
        print r
        if ks in cac:  continue
        P += [(RR(r.real),RR(r.imag))]
    return P

share|cite|improve this answer
Thanks a lot joro for the data. So I find it quite interesting that all the zeroes you computed are located in the vertical strip $-1<Re(s)<1$. It would be interesting to find a zero $\rho$ with $Re(\rho)>1$. – Hugo Chapdelaine May 17 '12 at 14:29
Also @Joro, note that $\zeta(s;6,2)$ is not primitive in the sense that $gcd(6,2)=2$ and therefore $\zeta(s;6,2)=2^{-s}\zeta(s;3,1)$ where $\zeta(s;3,1)$ is now primitive. – Hugo Chapdelaine May 17 '12 at 14:38

This is only a comment regarding the functional equation of the $\Psi$-function. So define $$ \Psi^*(s,\omega;N,a):=\sum_{0\neq n\in\mathbf{Z}}\frac{\omega(n)}{|n|^s}e^{-2\pi i\frac{a}{N}n} $$ Then one can show using Riemann's classical idea (see for example 1) that $$ (-1)^p N^s\cdot\Gamma_{\infty}(s)\Psi(s,\omega;N,a)=\Gamma_{\infty}(1-s)\Psi^*(1-s,\omega;N,a), $$ where $p=0$ (resp. p=$1$) if $\omega=1$ (resp. $\omega=sign$) and $\Gamma_{\infty}(s):=\pi^{-s/2}\Gamma(\frac{s+p}{2})$.

So it seems that many of the non-trivial zeros of the $\Psi$-functions are located along the symmetry axis of the functional equation. Is there some heuristic argument that could explain this phenomenon?

share|cite|improve this answer
That zeroes cluster around the cricital line is well-known phenomena, see for example my answer…, where I cite a paper of Steuding that proves this result in the case of the periodic zeta-function (Your zeta-function $\Psi^*$ is such a function). – Johan Andersson May 19 '12 at 16:19
Thanks a lot for this reference of Steuding! – Hugo Chapdelaine May 19 '12 at 16:31

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.