2·1. Notation and conventions

Throughout this paper, we use the following conventions:

1. (i) K is a number field;

2. (ii) $n_K = [K : {\mathbb {Q}}]$ ;

3. (iii) $r_1$ and $2r_2$ are the number of real and complex embeddings of K, respectively;

4. (iv) $\Delta_K$ is the discriminant of K;

5. (v) $\textrm{D}_K = \vert \Delta_K \vert$ is the absolute discriminant of K;

6. (vi) $\zeta_K(s)$ is the Dedekind zeta function of K.

Some theorems in this paper apply only to real or imaginary quadratic number fields. In these cases, we will specify any additional hypotheses on K. Otherwise, it is assumed that K is a general number field.

The Riemann hypothesis for $\zeta_K(s)$ will denote the conjecture that all nontrivial zeros of $\zeta_K(s)$ lie on the critical line $\textrm{Re}(s) = {1}/{2}$ . In light of this, $\rho$ will always denote a nontrivial zero of $\zeta_K(s)$ with imaginary part $\gamma$ .

2·2. Analytic properties of Dedekind zeta functions

In this section, we provide some preliminary statements about Dedekind zeta functions that will be used in the proofs of each of our results. We start with the functional equation for $\zeta_K(s)$ .

Theorem 2·1. For any number field K, the Dedekind zeta function $\zeta_K(s)$ satisfies

\begin{equation*} \zeta_K(1-s)= \zeta_K(s) \left(\frac{\textrm{D}_K}{\pi^{n_K}2^{n_K}}\right)^{s-\frac{1}{2}} \left( \frac{\pi}{2} \right)^{\frac{r_1}{2}}\frac{\Gamma(s)^{r_2}}{(\sin \frac{\pi s}{2})^{r_1} \Gamma(1-s)^{r_1+r_2}}.\end{equation*}

This functional equation extends $\zeta_K$ to a meromorphic function on $\mathbb C$ , which is analytic everywhere except for a simple pole at $s=1$ .

The Dedekind zeta function has trivial zeros of order $r_2$ at each negative odd integer and of order $r_1+r_2$ at each negative even integer, as well as a trivial zero of order $r_1+r_2-1$ at $s=0$ . It also possesses nontrivial zeros, each of which lies in the critical strip $0 < \textrm{Re}(s) < 1$ . By the above functional equation and the reflection principle, these zeros are symmetric about $\textrm{Re}(s) = {1}/{2}$ and the real line.

Next, we give a suitable upper bound, due to Chandrasekharan–Narasimhan, on the Dirichlet series coefficients for $\zeta_K(s)$ .

Lemma 2·2 (Chandrasekharan{Narasimhan [ Reference Chandrasekharan and NarasimhanCN63 , lemma 9]). Let K be any number field, and write in $\textrm{Re}(s) > 1$ ,

\begin{equation*} \zeta_K(s) = \sum_{n=1}^\infty \frac{a_n}{n^s}.\end{equation*}

Then $a_n$ is bounded by the coefficient $b_n$ of $n^{-s}$ in $\zeta(s)^{n_K}$ and there exists a constant C depending only on $n_K$ such that

\begin{equation*} a_n \leq b_n \ll_{n_K} n^{\frac{C}{\log \log n}}\ \text{ as } n \rightarrow \infty.\end{equation*}

The same holds for the Dirichlet series coefficients $a_n'$ of $1/\zeta_K(s)$ .

Useful for our purposes will be an estimate for the number of nontrivial zeros of $\zeta_K(s)$ in unit intervals in the critical strip. Define

\begin{equation*} N_K(T) \;:\!=\; \#\{\rho = \beta + i\gamma \mid \zeta_K(\rho) = 0,\; 0 < \beta < 1,\;0 \leq |\gamma|\;\leq T\}.\end{equation*}

It is well known that $N(T+1) - N(T) \ll \log T$ in the case of $K = \mathbb{Q}$ . An analogous result holds for any number field K.

In the explicit formulae to follow in Section 4, we will also require the following result which gives good upper bounds for $1/\zeta_K(s)$ on certain horizontal lines near the critical strip. We note that it is conditional on the Riemann hypothesis for $\zeta_K(s)$ ,

Lemma 2·4. Assume that $\zeta_K(s)$ satisfies the Riemann hypothesis. Then there exists a constant $C > 0$ such that, for each positive integer $n\geq 4$ , there exists some $n\leq T_n<n+1$ such that

\begin{equation*} \left|\zeta_K(\sigma+iT_n)\right|\geq \exp\left(-\frac{C\log n}{\log\log n}\right)\end{equation*}

for $-1\leq \sigma\leq 2$ . We denote by $\mathcal{T} = \{T_n\}_{n=4}^\infty$ the sequence so constructed.

This result is well known in the case $K = \mathbb{Q}$ (see [ Reference Montgomery and VaughanMV07 , theorem 13·22] or [ Reference TitchmarshTit86 , theorem 14·16]) and the proof for general K is mostly analogous. A complete proof can be found in Section 5.

2·3. $\Omega$ -type lemmas for $M_K(x)$ and disproofs of the naïve Mertens-type conjecture in degenerate cases

Fix any number field K. In this section, we provide various conditional $\Omega$ -type theorems for $M_K(x)$ which will reduce the unconditional disproofs in Section 3 to a number of conditional assumptions. The main result of this section is Theorem 2·10, which provides three cases in which the naïve Mertens-type conjecture is guaranteed to fail. Essential in the following proofs will be the following well-known method of Landau.

Lemma 2·5 (Landau [ Reference Montgomery and VaughanMV07 , theorem 15·1]). Suppose that A(x) is a bounded Riemann-integrable function in any finite interval $1 \leq x \leq X$ , and that $A(x) \geq 0$ for all $x > X_0$ . Let $\sigma_c$ denote the infimum of those $\sigma$ for which $\int_{X_0}^\infty A(x)x^{-\sigma}\, dx < \infty$ . Then the function

\begin{equation*} F(s) = \int_1^\infty A(x)x^{-s}\, dx\end{equation*}

is analytic in the half-plane $\textrm{Re}(s) > \sigma_c$ , but not at the point $s = \sigma_c$ .

We shall demonstrate that most cases in which zeros of $\zeta_K(s)$ behave contrary to conventional expectations (i.e. the Riemann hypothesis and simplicity of zeros) will degenerate into falsity of the naïve Mertens-type conjecture over K. One additional possible obstruction present in the case of Dedekind zeta functions, which is not so in the case of the Riemann zeta function, is the presence of nontrivial zeros on the real line. Let

\begin{align*} &\Theta = \sup_{\zeta_K(\rho) = 0} \{ \textrm{Re}(\rho)\} = \max\{\Theta', \Theta''\},\quad \\[5pt] &\text{ where } \Theta' = \sup_{\substack{\zeta_K(\rho) = 0 \\ \rho \in (0,1)}} \rho, \quad \Theta'' = \sup_{\substack{\zeta_K(\rho) = 0 \\ \textrm{Im}(\rho) \neq 0}} \{ \textrm{Re}(\rho)\}.\end{align*}

If no zero $\rho \in (0,1)$ exists, we set $\Theta' = 0$ . In particular, $\Theta \geq {1}/{2}$ and the Riemann hypothesis for $\zeta_K(s)$ is the statement $\Theta = {1}/{2}$ . The following is our first degenerate case.

Proof. Suppose that $M(x) < x^{\Theta - \epsilon}$ for all $x > X_0(\epsilon)$ . Consider, in view of (1·1), the function

\begin{equation*} \frac{1}{s - \Theta + \epsilon} - \frac{1}{s\zeta_K(s)} = \int_1^\infty (x^{\Theta - \epsilon} - M_K(x))x^{-s-1} \, dx.\end{equation*}

Here the left-hand side has a pole at $\Theta - \epsilon$ , but is analytic for real $s > \Theta - \epsilon$ , since $\zeta_K(s)$ has no zeros to the right of $\Theta - \epsilon$ on the real line. The integrand of the right-hand side is non-negative for all $x > X_0(\epsilon)$ . By an application of Lemma 2·5, the above identity holds for $\textrm{Re}(s) > \Theta - \epsilon$ , and both sides are analytic in this half-plane. But by definition of $\Theta$ , the function $1/\zeta_K$ has poles with real part $> \Theta - \epsilon$ , which is a contradiction. Hence, we deduce that $M(x) = \Omega_+(x^{\Theta - \epsilon})$ .

To obtain the $\Omega_-$ estimate, we argue similarly using the identity

\begin{equation*} \frac{1}{s - \Theta + \epsilon} + \frac{1}{s\zeta_K(s)} = \int_1^\infty (x^{\Theta - \epsilon} + M_K(x))x^{-s-1} \, dx.\end{equation*}

We then conclude that $M_K(x) = \Omega_{\pm}(x^{\Theta - \epsilon})$ . Specifically, this gives the falsity of the naïve Mertens-type conjecture over K, since $\Theta > {1}/{2}$ .

If $\Theta' = \Theta$ , then since real zeros must be isolated, there is a zero at $s = \Theta$ which prohibits the application of Lemma 2·5 in the proof above. In this case, we give a result in lieu of the previous proposition, which involves the elimination of the polar behaviour of $1/s\zeta_K(s)$ at the real nontrivial zeros of $\zeta_K(s)$ . More precisely, there exist constants $c_{k,\alpha} \in \mathbb{C}$ , $1 \leq k \leq m_{\alpha}$ , for each of the finitely many real zeros $\alpha \in (0,1)$ , such that the function

(2·1) \begin{equation} \frac{1}{s\zeta_K(s)} - \sum_{\substack{\zeta_K(\alpha) = 0 \\ \alpha \in (0,1)}} \sum_{k=1}^{m_\alpha} \frac{c_{k,\alpha}}{(s - \alpha)^k}\end{equation}

is analytic on the segment (0, 1) on the real line, where $m_\alpha$ is the multiplicity of the zero at $\alpha$ . Upon this simplification, we may provide the following result.

Proposition 2·7. Suppose that $\zeta_K(s)$ does not satisfy the Riemann hypothesis, and $\Theta' = \Theta$ , $\frac{1}{2} < \Theta'' \leq \Theta$ . Let $c_{k,\alpha}$ be as in the equation (2·1). Then for all $0 < \epsilon < \Theta'' - {1}/{2}$ ,

\begin{equation*} M_K(x) - \tilde M_K(x) = \Omega_{\pm}(x^{\Theta'' - \epsilon}),\end{equation*}

where

(2·2) \begin{equation} \tilde M_K(x) \;:\!=\; \sum_{\substack{\zeta_K(\alpha) = 0 \\ \alpha \in (0,1)}} \sum_{k=1}^{m_\alpha} \frac{c_{k,\alpha} x^\alpha (\log x)^{k-1}}{(k-1)!\!}.\end{equation}

In particular, the naïve Mertens-type conjecture over K is false.

Proof. Suppose that $M_K(x) - \tilde M_K(x) < x^{\Theta - \epsilon}$ for all $x > X_0(\epsilon)$ . Now

\begin{align*} \frac{1}{s-\Theta +\epsilon} - \frac{1}{s\zeta_K(s)} + \sum_{\substack{\zeta_K(\alpha) = 0 \\ \alpha \in (0,1)}} \sum_{k=1}^{m_\alpha} \frac{c_{k,\alpha}}{(s - \alpha)^k} = \int_1^\infty ( x^{\Theta - \epsilon} - M_K(x) + \tilde M_K(x))x^{-s-1}\, dx\end{align*}

holds for all real $s > \Theta - \epsilon$ , and by Lemma 2·5 extends to $\textrm{Re}(s) > \Theta - \epsilon$ . From here, the proof is similar to that of Proposition 2·6. For the last statement, note that $s = \Theta$ is one of the $\alpha$ in the sum $\tilde M_K(x)$ . Since $\Theta > {1}/{2}$ , this sum surpasses the order of $x^{1/2}$ , and since $M_K(x) - \tilde M_K(x)$ is oscillatory we deduce the falsity of the naïve Mertens-type conjecture over K.

It remains to address the case where ${1}/{2} = \Theta'' < \Theta' = \Theta$ . We give the following auxiliary result, which is independent of RH.

Proposition 2·8. Suppose $\Theta$ , $\Theta'$ , $\Theta''$ are as above, and that there is a non-real zero $\rho$ of $\zeta_K$ of multiplicity $m \geq 1$ with $\textrm{Re}(\rho) = \Theta''$ , say $\rho = \Theta'' + i\gamma$ . Then

\begin{equation*} M_K(x) - \tilde M_K(x) = \Omega_{\pm}(x^{\Theta''}(\log x)^{m-1}),\end{equation*}

where $\tilde M_K(x)$ is defined as in (2·2). More precisely,

\begin{equation*} \liminf_{x \rightarrow \infty} \frac{M_K(x) - \tilde M_K(x)}{x^{\Theta''}(\log x)^{m-1}} \leq -\frac{m}{|\rho \zeta_K^{(m)}(\rho)|}<\frac{m}{|\rho \zeta_K^{(m)}(\rho)|}\leq \limsup_{x \rightarrow \infty} \frac{M_K(x) - \tilde M_K(x)}{x^{\Theta''}(\log x)^{m-1}}.\end{equation*}

Proof. Suppose that $M_K(x) - \tilde M_K(x) \leq cx^{\Theta''}(\log x)^{m-1}$ for all $x > X_0$ . It suffices to prove that $c \geq m/|\rho \zeta_K^{(m)}(\rho)|$ . Consider the function

\begin{align*} &\frac{c(m-1)!\!}{(s-\Theta'')^m} - \frac{1}{s \zeta_K(s)} + \sum_{\substack{\zeta_K(\alpha) = 0 \\ \alpha \in (0,1)}} \frac{c_{k,\alpha} x^{\alpha} (\log x)^{k-1}}{(k-1)!\!}\\[5pt] &\quad= \int_1^\infty (cx^{\Theta''}(\log x)^{m-1} - M_K(x) + \tilde M_K(x))x^{-s-1} \, dx \end{align*}

for real $s > \Theta''$ . By Lemma 2·5, this can be extended to $\textrm{Re}(s) > \Theta''$ . Denote the function F(s). Then

\begin{align*} &F(s) + \frac{1}{2}e^{i\phi}F(s + i\gamma) + \frac{1}{2}e^{-i\phi}F(s - i \gamma) \\[5pt] &\quad = {\int_1^\infty (cx^{\Theta''}(\log x)^{m-1} - M_K(x) + \tilde M_K(x))(1 + \cos(\phi - \gamma \log x) x^{-s-1})\, dx}\end{align*}

for $\textrm{Re}(s) > \Theta''$ . On the right-hand side, the integral from 1 to $X_0$ is uniformly bounded, while the integral from $X_0$ to $\infty$ is non-negative. Thus the $\liminf$ of the right-hand side is bounded below as $s \rightarrow \Theta^+$ . As a result, the coefficient of $(s - \Theta'')^{-m}$ in its Laurent series must be non-negative. On the other hand, the left-hand side has a pole of multiplicity m at $s = \Theta''$ , at which the Laurent series expansion contains a term $(s - \Theta'')^{-m}$ with coefficient equal to

\begin{equation*} c(m-1)!\! - \frac{m!\! e^{i\phi}}{2\rho \zeta_K^{(m)}(\rho)} - \frac{m!\! e^{-i\phi}}{2\bar \rho \zeta_K^{(m)}(\bar \rho)}.\end{equation*}

Choosing $\phi$ so that

\begin{equation*} e^{i\phi} = \frac{\rho \zeta_K^{(m)}(\rho)}{|\rho \zeta_K^{(m)}(\rho)|}.\end{equation*}

Then the above is $(m-1)!\!(c - m/|\rho \zeta_K'(\rho)|)$ . This quantity must be non-negative, otherwise the left-hand side would tend to $-\infty$ as $s \rightarrow \Theta^+$ . Hence $c \geq m/|\rho \zeta_K^{(m)}(\rho)|$ . The $\Omega_-$ case is similar.

In the scenario where ${1}/{2} = \Theta'' < \Theta' = \Theta$ , we deduce from the previous proposition that the naïve Mertens-type conjecture is false over K. This completes the discussion under falsity of the Riemann hypothesis for $\zeta_K(s)$ .

Turning to results that are instead conditional on the Riemann hypothesis for K, Proposition 2·8 in fact yields the following immediate corollary.

Corollary 2·9. Assume the Riemann hypothesis for $\zeta_K(s)$ , and that $\zeta_K(s)$ has a zero $\rho = {1}/{2} + i\gamma$ , $\gamma > 0$ , of multiplicity $m \geq 1$ . Then

\begin{equation*} M_K(x) - \tilde M_K(x) = \Omega_{\pm}(x^{1/2}(\log x)^{m-1}),\end{equation*}

where $\tilde M_K(x)$ defined as in (2·2) is the part corresponding to a possible zero at $s = {1}/{2}$ .

Corollary 2·9 implies in particular that if $\zeta_K(s)$ satisfies the Riemann hypothesis, but possesses nontrivial zeros with multiplicity $m \geq 2$ (at $s = {1}/{2}$ or otherwise), then the naïve Mertens-type conjecture is false. Therefore, the only non-degenerate case of the naïve Mertens-type conjecture occurs when $\zeta_K(s)$ satisfies the Riemann hypothesis and has only simple nontrivial zeros (a priori none at $s = {1}/{2}$ ).

The following theorem summarizes the results of this section.

In particular, if $M_K(x)/\sqrt x$ is bounded (both above and below), then the Riemann hypothesis for $\zeta_K(s)$ holds, and $\zeta_K(s)$ has no multiple zero in the critical strip, thereby implying that $\zeta_K(\frac12)\neq 0$ . We expand on the possibility of growth in only one direction in Section 6.

2·4. The explicit formula for $M_K(x)$

We now state the explicit formula for the Mertens function, which, in its compatibility with analytic methods, is essential to our applications. We must assume the Riemann hypothesis for $\zeta_K(s)$ and that it has no nontrivial multiple zeros. The error terms in our explicit formula are presented in different cases depending on their applicability to $x > 0$ ; the generality of $x>0$ is necessary for the unconditional results of Section 3, while more specific bounds for $x>1$ are necessary to prove the existence of a limiting distribution in Section 4.

Proposition 2·12. Assume the Riemann hypothesis for $\zeta_K(s)$ , and that all nontrivial zeros of $\zeta_K(s)$ are simple. For any real $x>0$ and any $T\in \mathcal T$ , where $\mathcal T$ is as in Lemma 2·4,

$$M_K(x)=\sum_{|\gamma|\leq T} \frac{x^\rho}{\rho\zeta_K'(\rho)}+\sum_{k=0}^\infty \textrm{Res}_{s=-k}\frac{x^s}{s\zeta_K(s)}+E(x,T),$$

where for any $x>0$ , $E(x,T)=O_x(T^{1-\epsilon})$ for any $\epsilon$ , and for $x > 1$ we have

$$E(x,T)\ll\frac{x^{1+C/\log \log x}\log x}T+\frac{x}{T^{1-\epsilon}}+x^{C/\log\log x},$$

where C is as in Lemma 2·2. In addition, for $x > 1$ ,

$$\sum_{k=0}^\infty \textrm{Res}_{s=-k}\frac{x^s}{s\zeta_K(s)}\ll (\log x)^{O(1)},$$

which in this case can be subsumed into the error term E(x,T).

We defer the proof of this formula to Section 5.

For the previous result and others to follow, we also require the following bound on the residues of ${x^s}/{s\zeta_K(s)}$ at the trivial zeros $s = -k$ . The purpose of this bound is two-fold: it reveals dependence on k and x, which suffices already to prove the last statement in Proposition 2.12, and it clarifies dependence on $r_1$ , $r_2$ , and $D_K$ , which is crucial in our proof of Theorem 1·2.

Lemma 2·13. Fix non-negative integers $r_1$ and $r_2$ , and let K be a number field with $r_1$ real embeddings and $r_2$ complex conjugate pairs of embeddings. Then there exists some constant $c=c_{r_1,2r_2}$ such that, if $\zeta_K(s)$ has a zero at $s=-k$ for a positive integer k, then

$$\left|\textrm{Res}_{s=-k}\frac{x^s}{s\zeta_K(s)}\right|\leq \frac{c(2\pi)^{kn_K}(\log (kx\textrm{D}_K))^{r_1 + r_2}}{kx^k\textrm{D}_K^{k+1/2}(k!\!)^{n_K}}.$$

In addition,

$$\left|\textrm{Res}_{s=0}\frac{x^s}{s\zeta_K(s)}\right|\ll_K (\log x)^{r_1+r_2-1}.$$

See Section 5 for the proof. Heavier consideration of the residue at $s = 0$ is necessary to prove Theorems 1·1 and 1·2, wherein dependence on $n_K$ and $D_K$ will be significant.

3·1. The general result

In this section, we give a generic result about the limiting behaviour of $x^{-1/2}M_K(x)$ which holds unconditionally. We follow a method of Jurkat from [ Reference JurkatJur73 ], using his main theorem directly in our application. In particular, Jurkat has identified a wide class of trigonometric series which are seen to share a key limiting property enjoyed by Bohr-almost periodic functions. By exploiting this property, we show $x^{-1/2}M_K(x)$ approaches certain explicitly computable values infinitely often and arbitrarily closely.

Formally, this means f admits an expansion of the form

(3·1) \begin{equation} f(x) \sim \sum_{n=1}^\infty \textrm{Re}(a_n e^{i \lambda_n x}),\end{equation}

where $0 < \lambda_n \nearrow \infty$ and $a_n$ are complex constants such that

\begin{equation*} \sum_{n=1}^\infty \frac{|a_n|}{\lambda_n^k} < \infty\end{equation*}

for some $k \in \mathbb{N}$ . When referring to an APD-function, we shall specify k.

The notion of almost periodicity in a distributional sense is useful because such functions have easily-calculable limit points, as shown by the following main theorem of Jurkat.

Proposition 3·2 (Jurkat [ Reference JurkatJur73 , pp. 151[152]). Suppose $f\;:\;{\mathbb {R}}\to{\mathbb {R}}$ is an APD-function, and let L(f) denote the set of Lebesgue points of f. Then for any $t\in L(f)$ ,

$$\liminf_{\substack{x\to\infty\\ x\in L(f)}}f(x)\leq f(t)\leq \limsup_{\substack{x\to\infty\\ x\in L(f)}}f(x).$$

Moreover, all values of f at Lebesgue points are limit points of f.

The technical condition on Lebesgue points can be safely ignored for the remainder of this paper. The functions we work with are sufficiently well-behaved that the inclusion of their Lebesgue points does not change either limit.

For any real $x > 0$ , define

(3·2) \begin{equation}M_K^*(x)= - \sum_{k=0}^\infty \textrm{Res}_{s=-k}\frac{x^s}{s\zeta_K(s)}.\end{equation}

(We remark that this carries a slightly different normalisation than Jurkat’s $M^*(x)$ , as he makes a few simplifications particular to the case $K={\mathbb {Q}}$ .) Motivated by Proposition 2·12, this is a “trivial completion” of $M_K(x)$ , in the sense that, for $T\in\mathcal T$ (where $\mathcal T$ is given by Lemma 2·4),

(3·3) \begin{equation}M_K(x)+M_K^*(x)=\sum_{|\gamma|\leq T}\frac{x^\rho}{\rho\zeta_K'(\rho)}+O_x\left(\frac1{T^{1-\epsilon}}\right)\end{equation}

for all $x>0$ , assuming the preconditions of Proposition 2·12 hold. We first show an analogue of Jurkat’s main theorem for number fields under these assumptions, as well as some more stringent assumptions.

The function $f\;:\; \mathbb{R} \rightarrow \mathbb{R}$ defined by

\begin{equation*} f(y) \;:\!=\; e^{-y/2}(M_K(e^y) + M_K^*(e^y))\end{equation*}

is locally integrable. If we assume the Riemann hypothesis for $\zeta_K(s)$ and that all nontrivial zeros are simple, and let $T \rightarrow \infty$ along $\mathcal{T}$ in the equation (3·3) above (with the necessary modifications for $M_K(e^y)$ if y is the logarithm of an integer), it furthermore has an expansion of the form in (3·1) with $a_n = 2(\rho_n \zeta_K'(\rho_n))^{-1}$ , $\lambda_n = \gamma_n$ , where $\gamma_n$ is the ordinate of the nth highest nontrivial zero $\rho_n = {1}/{2} + i \gamma_n$ in the upper half plane. Here, the sum converges boundedly given this grouping of the terms. Assuming for now that there exists some constant $C > 0$ such that

\begin{equation*} \frac{1}{|\rho \zeta_K'(\rho)|} \leq C,\end{equation*}

the result of Lemma 2·3 implies that f is an APD-function with $k = 2$ . Hence, Proposition 3·2 functions produces the following.

Proposition 3·3. Let K be a number field. Assume the Riemann hypothesis for $\zeta_K(s)$ , that all nontrivial zeros of $\zeta_K(s)$ are simple, and that there exists some constant $C > 0$ for which

$$\frac{1}{|\rho\zeta_K'(\rho)|}\leq C$$

for every nontrivial zero $\rho$ of $\zeta_K(s)$ . Then, for any $x>0$ ,

$$\frac{M_K(x)+M_K^*(x)}{x^{1/2}}$$

is a limit point of the function $x^{-1/2}M_K(x)$ .

We now describe our major application of this result. Recall the definition

$$M_K^-=\liminf_{x\to\infty}\frac{M_K(x)}{x^{1/2}}, \quad M_K^+=\limsup_{x\to\infty}\frac{M_K(x)}{x^{1/2}}.$$

In the case where the Riemann hypothesis holds and all nontrivial zeros are simple, we may apply Proposition 2·8 with $\Theta''=1/2$ and $m=1$ to obtain

$$M_K^+,-M_K^-\geq \frac1{|\rho\zeta_K'(\rho)|}$$

for any nontrivial zero $\rho$ of $\zeta_K(s)$ . If one of $M_K^\pm$ is infinite, then (a) and (b) hold as before. Otherwise, if $M_K^\pm$ are both finite, then there is some constant C for which

$$\frac1{|\rho\zeta_K'(\rho)|}\leq C$$

for every zero $\rho$ of $\zeta_K(s)$ . So, the preconditions of Proposition 3·3 hold, and we have for any x that

$$\frac{M_K(x)+M_K^*(x)}{x^{1/2}}$$

is between $M_K^-$ and $M_K^+$ . In particular,

$$M_K^-\leq \lim_{x\to 1^-}\frac{M_K(x)+M_K^*(x)}{x^{1/2}}=M_K^*(1)<M_K^*(1)+1=\lim_{x\to 1^+}\frac{M_K(x)+M_K^*(x)}{x^{1/2}}\leq M_K^+,$$

where limits to $1^-$ and $1^+$ denote approaching 1 from below and above, respectively. This inequality chain shows both (a) and (b).

3·2. Imaginary quadratic fields

In this subsection, we calculate and bound $M_K^*(x)$ explicitly for imaginary quadratic fields. We pick $D>0$ such that $-D$ is a negative fundamental discriminant and let $K={\mathbb {Q}}(\sqrt{-D})$ be the imaginary quadratic field with discriminant $-D$ . Let $\chi_{-D}$ denote the odd quadratic character $\chi_{-D}(n)=\left({-D}/n\right)$ of conductor D, so that $\zeta_K(s)=\zeta(s)L(s,\chi_{-D})$ . Using Theorem 2·1, we may compute

(3·4) \begin{equation} M_K^*(x) \;:\!=\; \frac{2\pi}{L(1,\chi_{-D})D^{1/2}} + \sum_{k=1}^\infty \left( \frac{D}{4\pi^2} \right)^{-k - \frac{1}{2}}\frac{({-}1)^k x^{-k}}{k (k!\!)^2\zeta_K(k+1)}.\end{equation}

Via the class number formula of Dirichlet, the leading term of (3·4) is always positive. We show that, for $x = 1$ and sufficiently large D, it exceeds the other terms in absolute value.

Lemma 3·5. If $D>0$ and $-D$ is a fundamental discriminant, and $K={\mathbb {Q}}(\sqrt{-D})$ , then

$$M_K^*(1)\geq \frac{2\pi}{D^{1/2}}\left(\frac1{\frac12\log D+\log\log D+2+\log 2}-\exp\left(\frac{4\pi^2}D\right)+1\right).$$

In particular, if $D>307$ , then $M_K^*(1)>0$ .

Proof. We begin by bounding from below the leading term of (3·4). The Euler product for $L(s,\chi_{-D})$ implies that it is positive for all real $s > 1$ , and hence $L(1,\chi_{-D}) > 0$ , knowing that it is nonzero. On the other hand,

$$L(1,\chi_{-D}) = \sum_{n\leq N} \frac{\chi_{-D}(n)}{n} + \sum_{n > N} \frac{\chi_{-D}(n)}{n} \leq 1 + \log N + \frac{2}{N} \sqrt{D} \log D,$$

where we have used partial summation and the Pólya–Vinogradov bound $\sqrt{D} \log D$ for the partial sums of the primitive character $\chi_{-D}$ modulo D, as well as the bound $1 + \log x$ for $\sum_{n \leq x} \frac{1}{n}$ . Setting $N = 2\sqrt{D} \log D$ gives that

(3·5) \begin{equation}\frac{2\pi}{D^{1/2}L(1,\chi_{-D})}\geq \frac{2\pi}{D^{1/2}\left(\frac12\log D+\log\log D+2+\log 2\right)}.\end{equation}

Now, for any real $\sigma>1$ , we have $\zeta_K(\sigma)>1$ by definition, so $\zeta_K(k+1)^{-1}<1$ for every $k \geq 1$ . So, we have the bound

\begin{align*} \left| \sum_{k=1}^\infty \left( \frac{D}{4\pi^2} \right)^{-k - \frac{1}{2}} \frac{(-1)^k}{k(k!\!)^2 \zeta_K(k+1)} \right| &\leq \frac{2\pi}{D^{1/2}} \sum_{k=1}^\infty \left( \frac{4\pi^2}{D} \right)^k \frac{1}{k (k!)^2\zeta_K(k+1)} \\[5pt] &\leq \frac{2\pi}{D^{1/2}} \sum_{k=1}^\infty \left( \frac{4\pi^2}{D} \right)^k \frac{1}{k!} \\[5pt] &\leq \frac{2\pi}{D^{1/2}} \left( \exp\left( \frac{4\pi^2}{D} \right) - 1 \right).\end{align*}

Combining this bound with (3·5) gives the desired result.

Lemma 3·5 has the following immediate consequence via Lemma 3·4.

By direct computation, we may address the smaller discriminants.

Table 1.

Values of $M_K^*(1)$ for imaginary quadratic fields $K = \mathbb{Q}(\sqrt{-D})$ of discriminant $-D$ , $D \leq 307$ .

Table 2.

Given $K={\mathbb {Q}}(\sqrt{-D})$ for a fundamental discriminant $-D$ , the smallest positive integer n for which $M_K(n^+)+M_K^*(n)>n^{1/2}$ .

Proof. The values of $M_K^*(1)$ are tabulated in Table 1, where the sum has been computed up to threshold $k = 50$ . This implies the falsity of the naïve Mertens-type conjecture over K for all $D \leq 307$ where

\begin{equation*} -D \notin \{-3,-4,-7,-8,-11,-15,-20,-23\},\end{equation*}

since, via Lemma 3·4, we deduce from $M_K^*(1) > 0$ that $M_K^+ > 1$ . The procedure can be remedied for most of those exceptional values of $-D$ by letting $x \rightarrow n^+$ for different choices of positive integers n where

\begin{equation*} \frac{M_K(n^{+}) + M_K^*(n)}{n^{1/2}} > 1,\end{equation*}

as represented in Table 2 (we have taken the right-hand limits, as $n^{-1/2}M_K^*(n)$ is generally decreasing). This suffices to resolve all cases except for $D = 3$ (the field $K = \mathbb{Q}(\sqrt{-3})$ ) and $D = 4$ (the field $K = \mathbb{Q}(i)$ ).

In order to resolve the case $D = 4$ , we appeal to a method originally due to Ingham [ Reference InghamIng42 , theorem 1] in examining the smoothed sums of $\sum_{|\gamma| <T} {e^{i\gamma t}}/{\rho \zeta_K'(\rho)}$ by a suitable kernel function f(y). In particular, we select the kernel function of Jurkat–Peyerimhoff (see [ Reference Jurkat and PeyerimhoffJP76, Reference Odlyzko and te RieleOtR85 ]),

\begin{align*} f(y) = \begin{cases} (1 - y) \cos \pi y + \frac{1}{\pi} \sin \pi y &\text{ if } 0 \leq y \leq 1, \\[5pt] 0 &\text{ if } y \geq 1, \end{cases}\end{align*}

which produces the smoothed sum

(3·6) \begin{align} & h_{K,T}^*(t) = \sum_{|\gamma| < T} f\left( \frac{|\gamma|}{T} \right) \frac{e^{i \gamma y}}{\rho \zeta_K'(\rho)}\nonumber \\[5pt] &\quad = \sum_{|\gamma| < T} \left( \left( 1 - \frac{|\gamma|}{T} \right) \cos \left( \frac{\pi |\gamma|}{T} \right) + \frac{1}{\pi} \sin \left(\frac{\pi |\gamma|}{T} \right) \right) \frac{e^{i \gamma y}}{\rho \zeta_K'(\rho)}.\end{align}

It holds for this sum that for $T > 0$ fixed,

\begin{equation*} M_K^- \leq \liminf_{t \rightarrow \infty} h_{K,T}^*(t) \leq \limsup_{t \rightarrow \infty} h_{K,T}^*(t) \leq M_K^+,\end{equation*}

and it follows from the fact that $h_{K,T}^*(t)$ is almost periodic that

\begin{equation*} \liminf_{t \rightarrow \infty} h_{K,T}^*(t) \leq h_{K,T}^*(t) \leq \limsup_{t \rightarrow \infty} h_{K,T}^*(t),\end{equation*}

so it suffices to exhibit $T > 0$ and $t \in \mathbb{R}$ for which $h_{K,T}^*(t) > 1$ or $h_{K,T}^*(t) < -1$ .

A computation demonstrates that $h^*_{{\mathbb {Q}}(i), 600}(72.85) \leq -1.008$ and $h^*_{{\mathbb {Q}}(i), 600}(-85.15) \geq 1.029$ . That is, $M_K^- \leq -1.008 < 1.029 \leq M_K^+$ . This completes the proof.

Combining Theorem 3·6 and Theorem 3·7 concludes the proof of Theorem 1·1(a).

3·3. Real quadratic fields

Next, we calculate and bound $M_K^*(x)$ explicitly for real quadratic fields. Pick a positive fundamental discriminant $D>0$ and let $K={\mathbb {Q}}(\sqrt{D})$ be the real quadratic field with discriminant D. Then $\zeta_K(s)=\zeta(s)L(s,\chi_D)$ , where $\chi_D$ is the even quadratic character $\chi_D(n)=\left(\frac Dn\right)$ of conductor D. We compute

(3·7) \begin{align} M_K^*(x) &= \frac{4}{L(1,\chi_D)D^{1/2}} \left( \log \left( \frac{xD}{4\pi^2} \right) - \gamma + \frac{L'}{L}(1,\chi_D) \right)\nonumber \\[5pt] &\quad + \frac{2}{D^{1/2}} {\sum_{k=1}^\infty \frac{\left(\frac{4\pi^2}{xD}\right)^{2k}}{k((2k)!\!)^2 \zeta_K(2k+1)} \left( \log \left( \frac{xD}{4\pi^2} \right) + \frac{1}{2k} + \frac{\zeta_K'}{\zeta_K}(2k+1) + 2 \frac{\Gamma'}{\Gamma}(2k+1)\! \right)\!.}\end{align}

Here, $\gamma = 0.5772\ldots$ is Euler’s constant. The $L'/L$ , $\zeta_K'/\zeta_K$ , and $\Gamma'/\Gamma$ terms arise (in contrast to (3·4)) since $\zeta_K(s)$ has zeros of multiplicity two when K is real quadratic. We now only need to identify which D satisfy the estimate $M_K^*(1)>0$ . We first provide a bound on the $L'/L$ term that is particular to the case of real characters.

Lemma 3·8. Let $\chi$ be a non-principal real primitive character of conductor D. Then

$$\frac{L'}{L}(1,\chi)> -\frac12\left(\log\frac D{2\pi}-\gamma\right)+\frac{\chi(-1)}2\log 2.$$

Proof. Recall [ Reference Montgomery and VaughanMV07 , corollary 10·18] for any Dirichlet character $\chi$ of conductor D the identity

$$\frac{L'}{L}(s,\chi)=-\frac{L'}{L}(1,\overline\chi)-\frac12\frac{\Gamma'}{\Gamma}\left(\frac{s+\kappa}2\right)-\log\frac D\pi+\sum_\rho\left(\frac1{s-\rho}+\frac1\rho\right)+\frac\gamma2+(1-\kappa)\log 2,$$

where $\kappa=({1-\chi(-1)})/2$ and $\rho$ runs over all nontrivial zeros of $L(s,\chi)$ . Specifying to $s=1$ and $\chi = \overline{\chi}$ gives

$$2\frac{L'}{L}(1,\chi)=-\frac12\frac{\Gamma'}{\Gamma}\left(\frac{1+\kappa}2\right)-\log\frac D\pi+\sum_\rho\frac1{\rho(1-\rho)}+\frac{\gamma}2+(1-\kappa)\log 2.$$

Since

$$\frac{\Gamma'}{\Gamma}\left(\frac12\right)=-\gamma-2\log 2\qquad \frac{\Gamma'}{\Gamma}(1)=-\gamma,$$

this gives

$$2\frac{L'}{L}(1,\chi)=\gamma+2(1-\kappa)\log 2-\log\frac D\pi+\sum_\rho\frac1{\rho(1-\rho)}.$$

By grouping $\rho$ and $\overline\rho$ in the sum we have

\begin{align*}\sum_\rho\frac1{\rho(1-\rho)}&=\frac12\sum_\rho\left(\frac1{\rho(1-\rho)}+\frac1{\overline\rho(1-\overline\rho)}\right)\\[5pt] &=\sum_\rho\frac{\textrm{Re}(\rho-\rho^2)}{|\rho|^2|1-\rho|^2}\\[5pt] &=\sum_\rho\frac{\beta(1-\beta)+\gamma^2}{|\rho|^2|1-\rho|^2}>0,\end{align*}

where $\rho=\beta+i\gamma$ and we have used $0<\beta<1$ . This gives

\begin{align*}2\frac{L'}{L}(1,\chi)&=\gamma+2(1-\kappa)\log 2-\log\frac D\pi+\sum_\rho\frac1{\rho(1-\rho)}> \gamma+\log2\\[5pt] &\quad +\chi(-1)\log 2-\log\frac D\pi,\end{align*}

as desired.

This allows us to establish the following result on $M_K^*(1)$ .

Proof. We begin by bounding the contribution of the infinite series in $M_K^*(1)$ . After dividing out by a factor of $2/D^{1/2}$ , the series under consideration takes the form

(3·8) \begin{equation} \Sigma = \sum_{k=1}^\infty \left( \frac{4\pi^2}{D} \right)^{2k} \frac{1}{k ((2k)!\!)^2 \zeta_K(2k+1)} \left( \log\left( \frac{D}{4\pi^2} \right) + \frac{1}{2k} + \frac{\zeta_K'}{\zeta_K}(2k+1) + 2 \frac{\Gamma'}{\Gamma}(2k+1) \right).\end{equation}

We note that $\zeta_K(2k + 1) \geq 1$ for all $k \geq 1$ , and by the triangle inequality,

\begin{align*} \left| \frac{\zeta_K'}{\zeta_K}(2k+1) \right| &= \left| \frac{\zeta'}{\zeta}(2k+1) + \frac{L'}{L}(2k+1,\chi_D) \right| \leq 2\sum_{n=1}^\infty \frac{\Lambda(n)}{n^{2k+1}} \\[5pt] &\leq 2\int_1^\infty \frac{\log x}{x^{2k+1}}\, dx = \frac{1}{2k^2}.\end{align*}

We also have the following known identity for the digamma function:

\begin{equation*} 2\frac{\Gamma'}{\Gamma}(2k + 1) = 2 \left( \sum_{n=1}^{2k} \frac{1}{n} - \gamma \right) \leq 2 \log(2k) + (2 - 2\gamma),\end{equation*}

where $\gamma$ is Euler’s constant. Finally, we use these bounds to see that, for $D > 4\pi^2$ ,

\begin{align*} \Sigma &\leq \sum_{k=1}^\infty \left( \frac{4\pi^2}{D} \right)^{2k} \frac{1}{k ((2k)!\!)^2} \left( \log\left( \frac{D}{4\pi^2} \right) + \frac{1}{2k} + \frac{1}{2k^2} + 2 \log(2k) + (2 - 2\gamma) \right) \\[5pt] &\leq \left( \cosh\left( \frac{4\pi^2}{D} \right) - 1 \right) \left( \frac{1}{2}\log \left( \frac{D}{4\pi^2} \right) + \frac{1}{4} + \frac{1}{4} + \log 2 + (1 - \gamma) \right)\\[5pt] &= \left( \cosh \left( \frac{4\pi^2}{D} \right) - 1 \right) \left( \frac{1}{2} \log D +\frac 32-\log\pi -\gamma\right).\end{align*}

We now turn our attention to the first term of $M_K^*(1)$ in (3·7). Upon dividing out by a factor of $2/D^{1/2}$ , we have

\begin{equation*} \frac{2}{L(1,\chi_D)} \left( \log \left( \frac{D}{4\pi^2} \right) - \gamma + \frac{L'}{L}(1,\chi_D) \right).\end{equation*}

To this end, we recall the previously obtained bound from (3·5)

\begin{equation*} \frac{1}{L(1,\chi_D)} \geq \frac{1}{\frac{1}{2} \log D + \log \log D + 2 + \log 2}.\end{equation*}

Finally, using Lemma 3·8 in the case of an even character, the problem reduces to finding the values D at which the inequality

\begin{equation*} \frac{2 \left(\frac12\log D - \frac32\log\pi-\log2-\frac\gamma2\right)}{\frac{1}{2} \log D + \log \log D + 2 + \log 2} > \left( \cosh\left( \frac{4\pi^2}{D} \right) - 1 \right) \left( \frac{1}{2} \log D + \frac32-\gamma-\log\pi \right)\end{equation*}

holds. This indeed holds for all $D > 269$ .

By direct computation, we may address the smaller discriminants.

Table 3.

Values of $M_K^*(1)$ for imaginary quadratic fields $K = \mathbb{Q}(\sqrt{D})$ of discriminant D, $D \leq 269$ .

Table 4.

Given $K={\mathbb {Q}}(\sqrt{-D})$ for a fundamental discriminant $-D$ , the smallest positive integer n for which $M_K(n^+)+M_K^*(n)>n^{1/2}$ .

Proof. As in the proof of Theorem 3·7, the values of $M_K^*(1)$ are tabulated in Table 3, where the sum has been computed up to threshold $k = 50$ . This implies the falsity of the naïve Mertens-type conjecture over K for all $D \leq 269$ where

\begin{equation*} D \notin \{5,8,12,13,17,21,24,28,29,33,37,40,41,44,56,57,60,61,65,73\}.\end{equation*}

Again, this can be remedied for most of the above discriminants by letting $x \rightarrow n^{+}$ for other positive integers n for which $n^{-1/2}(M_K(n^{+})+ M_K^*(n)) > 1$ . These findings are presented in Table 4, which suffices to resolve all cases except for $D = 5$ (the field ${\mathbb {Q}}(\sqrt{5})$ ), $D = 8$ (the field ${\mathbb {Q}}(\sqrt{2})$ ), and $D = 12$ (the field ${\mathbb {Q}}(\sqrt{3})$ ).

For these fields, we return to the means $h_{K,T}^*(t)$ as defined in (3·6). A computation shows that $t = 17.32$ at $T = 200$ suffices for $K = {\mathbb {Q}}(\sqrt{3})$ with a bound of $M_K^+ > 1.027$ . Moreover, $t = -24.64$ at $T = 150$ suffices for $K = {\mathbb {Q}}(\sqrt{2})$ with a bound of $M_K^+ > 1.049$ .

This concludes the proof of Theorem 1·1(b).

3·4. General number fields

In this subsection, we show that $M_K^*(1)>0$ for all number fields K with sufficiently large discriminant and fixed signature. In light of Lemma 3·4, this implies that the naïve Mertens-type conjecture is false for such K. We first show that the contribution of the residues at strictly negative integers is relatively small.

Lemma 3·11. Fix a signature $(r_1,r_2)$ , and consider a number field K with $r_1$ real embeddings, $2r_2$ complex embeddings, and varying discriminant $\Delta_K$ of magnitude $\textrm{D}_K$ . Then for all $x \geq 1$ ,

$$\sum_{k=1}^\infty \textrm{Res}_{s=-k}\frac{x^s}{s\zeta_K(s)}\ll \frac{(\log(x\textrm{D}_K))^{n_K}}{x\textrm{D}_K^{3/2}}.$$

The remainder of this section will involve bounding the residue at $s=0$ . In our application, we are taking $x = 1$ , and it is easy to see that

\begin{equation*} \textrm{Res}_{s = 0} \frac{x^s}{s\zeta_K(s)} \bigg|_{x = 1} = \textrm{Res}_{s = 0} \frac{1}{s\zeta_K(s)}.\end{equation*}

We first give a result about the Laurent series coefficients of $\zeta_K'(s)/\zeta_K(s)$ at $s = 1$ .

Lemma 3·12. Let K be a number field of fixed degree, and assume the Riemann hypothesis for $\zeta_K(s)$ . Then, for a fixed integer $j\geq 0$ ,

$$\lim_{s\to 1}\frac{d^j}{ds^j}\left(\frac1{s-1}+\frac{\zeta_K'(s)}{\zeta_K(s)}\right)\ll(\log\log\textrm{D}_K)^{j+1}$$

Proof. We mimic the proof of [ Reference Montgomery and VaughanMV07 , theorem 13·13]. Fix some real $x,y\geq 2$ , to be chosen later. Define $\Lambda^K(n)$ to be the coefficient of $n^{-s}$ in the Dirichlet series expansion of $\zeta_K'(s)/\zeta_K(s)$ , and define

$$w(u)=w(x, y; u) \;:\!=\; \begin{cases} 1 & 1 \leq u \leq x \\[5pt] 1 - \frac{\log u/x}{\log y} & x \leq u \leq xy\\[5pt] 0 & u \geq xy.\end{cases}$$

By Riesz summation ([ Reference Montgomery and VaughanMV07 , (5·20)]; cf. [ Reference Montgomery and VaughanMV07 , theorem 13·13]), we have

(3·9) \begin{equation}\sum_{n\leq xy}\frac{w(n)\Lambda^K(n)}{n^s}=-\frac{\zeta_K'(s)}{\zeta_K(s)}+\frac{(xy)^{1-s}-x^{1-s}}{(1-s)^2\log y}-\sum_\rho\frac{(xy)^{\rho-s}-x^{\rho-s}}{(\rho-s)^2\log y},\end{equation}

where the sum runs over all zeros of $\zeta_K(s)$ , both trivial and nontrivial, and a zero of multiplicity m appears m times. Now, as the nontrivial zeros occur only at non-positive integers and with multiplicity bounded by $n_K$

(3·10) \begin{equation}\left|\sum_{\rho\text{ trivial}}\frac{(xy)^{\rho-s}-x^{\rho-s}}{(\rho-s)^2\log y}\right|\leq n_K\sum_{k=0}^\infty \left|\frac{(xy)^{-k-s}-x^{-k-s}}{(k+s)^2\log y}\right|\ll \frac{2x^{-3/4}}{\log y},\end{equation}

where the last bound follows from $|s-1|\;\leq {1}/{4}$ . Also,

\begin{align*}\left|\sum_{\rho\text{ nontrivial}}\frac{(xy)^{\rho-s}-x^{\rho-s}}{(\rho-s)^2\log y}\right|&\leq 2\sum_{T=0}^\infty \sum_{T\leq \gamma<T+1}\frac{x^{1/2-\sigma}\left|y^{\rho-s}-1\right|}{\left|\frac12-\sigma+i(\gamma-t)\right|^2\log y}\\[5pt] &\leq \frac{4x^{-1/4}}{\log y}\sum_{T=0}^\infty \frac{N(T+1)-N(T)}{\frac1{16}+T^2}\ll \frac{x^{-1/4}\log\textrm{D}_K}{\log y},\end{align*}

where we have used Lemma 2·3 in the last simplification. As a result,

$$f(s)\;:\!=\; \frac{\zeta_K'(s)}{\zeta_K(s)}-\left[-\sum_{n\leq xy}\frac{w(n)\Lambda^K(n)}{n^s}+\frac{(xy)^{1-s}-x^{1-s}}{(1-s)^2\log y}\right]\ll O\left(\frac{x^{-1/4}\log\textrm{D}_K}{\log y}\right)$$

for $|s-1|\;\leq {1}/{4}$ . Now, the Taylor series expansion

$$\frac{\zeta_K'(s)}{\zeta_K(s)}=-\frac1{s-1}+\sum_{j=0}^\infty \frac{a_j(s-1)^j}{j!\!},\qquad a_j=\lim_{s\to 1}\frac{d^j}{ds^j}\left(\frac1{s-1}+\frac{\zeta_K'(s)}{\zeta_K(s)}\right)$$

converges for $|s-1|\; < {1}/{2}$ , since $\zeta_K(s)$ has no zeros strictly within distance ${1}/{2}$ of $s=1$ . Furthermore, we have the expansions

$$\sum_{n\leq xy}\frac{w(n)\Lambda^K(n)}{n^s}=\sum_{j=0}^\infty \frac{(s-1)^j}{j!\!}\sum_{n\leq xy}\frac{w(n)\Lambda^K(n)(-\log n)^j}{n}$$

and

$$\frac{(xy)^{1-s}-x^{1-s}}{(1-s)^2\log y}=-\frac1{s-1}+\sum_{j=0}^\infty \frac{(s-1)^j}{j!\!}\left(\frac{({-}1)^j\left(\log(xy)^{j+2}-(\log x)^{j+2}\right)}{(j+1)(j+2)\log y}\right).$$

Now, select $x=(\log\textrm{D}_K)^4$ and $y=e$ , so that $f(s)=O(1)$ . Writing

$$f(s)=\sum_{j=0}^\infty \frac{b_j(s-1)^j}{j!\!},$$

we have

$$b_j=j!\! \oint_{\Gamma_r} f(s)(s-1)^{-j-1}ds=\frac1{2\pi}\int_0^{2\pi}f\left(1+re^{i\theta}\right)r^{-j-1}e^{-i(j+1)\theta}d\theta$$

for any $r\leq {1}/{4}$ , where $\Gamma_r$ is a counterclockwise circular contour of radius r around 1. For fixed j and r, bounding the integrand in the second integral gives $b_j=O(1)$ . So, recalling our choices of x and y and using partial summation gives

\begin{align*}a_j&=\sum_{n\leq xy}\frac{w(n)\Lambda^K(n)(-\log n)^j}{n}+\frac{({-}1)^j\left(\log(xy)^{j+2}-(\log x)^{j+2}\right)}{(j+1)(j+2)\log y}\\[5pt] &\quad +O(1)\ll(\log\log\textrm{D}_K)^{j+1}.\end{align*}

This completes the proof.

We are now ready to bound the residue at 0.

Lemma 3·13. In the same setting as Lemma 3·11,

$$-\textrm{Res}_{s=0}\frac1{s\zeta_K(s)}=\frac{2^{r_1+r_2}\pi^{r_2}(\log\textrm{D}_K)^{r_1+r_2-1}}{\textrm{D}_K^{1/2}\lim_{s\to 1}{(s-1)\zeta_K(s)}}(1+o(1))\gg_{r_1,r_2} \frac{(\log\textrm{D}_K)^{-r_2}}{D_K^{1/2}}.$$

Proof. Write

$$\textrm{Res}_{s=0}\frac1{s\zeta_K(s)}=\lim_{s\to 0}\frac{d^{r_1+r_2-1}}{ds^{r_1+r_2-1}}\left(\frac{s^{r_1+r_2-1}}{\zeta_K(s)}\right).$$

Using the functional equation, we may write

$$\frac{s^{r_1+r_2-1}}{\zeta_K(s)}=\left(\frac{2^{r_2}\pi^{n_K}}{\textrm{D}_K^{1/2}}\right)\left(\frac{\textrm{D}_K}{(2\pi)^{n_K}}\right)^s\left(\frac1{s\zeta_K(1-s)}\right)\left(\frac{s}{\sin\frac{\pi s}2}\right)^{r_1}\left(\frac s{\sin\pi s}\right)^{r_2}\left(\frac1{\Gamma(1-s)}\right)^{n_K}.$$

Define

\begin{align*}f(s) &=\frac{s^{r_1+r_2-1}}{\zeta_K(s)}\\[5pt] g_1(s)&=s\log\left(\frac{\textrm{D}_K}{(2\pi)^{n_K}}\right)\\[5pt] g_2(s)&=\log(s\zeta_K(1-s))\\[5pt] g_3(s)&=\log s-\log\sin\frac{\pi s}2\\[5pt] g_4(s)&=\log s-\log\sin\pi s\\[5pt] g_5(s)&=\log\Gamma(1-s)\end{align*}

so that

$$g(s)\;:\!=\;\log f(s)=\log\left(\frac{2^{r_2}\pi^{n_K}}{\textrm{D}_K^{1/2}}\right)+g_1(s)-g_2(s)+r_1g_3(s)+r_2g_4(s)-n_Kg_5(s).$$

Since $f'(s)=g'(s)f(s)$ , we have via induction on j that

$$f^{(j)}(s)=f(s)P_j\left(g^{(1)}(s),g^{(2)}(s),\ldots,g^{(j)}(s)\right)$$

for some polynomial $P_j(t_1,\ldots,t_j)$ in j variables with positive integral coefficients, which is homogeneous of degree j when $t_i$ is assigned degree i. In addition, the coefficient of $t_1^j$ in $P_j(t_1,\ldots,t_j)$ is 1. We now bound the derivatives of g at 1.

For each $1\leq j\leq r_1+r_2-1$ , the derivatives $g_i^{(j)}(s)$ for $i\in\{3,4,5\}$ near $s=0$ depend only on $r_1$ and $r_2$ , so we can treat the terms corresponding to these $g_i$ as constants. This gives that, for $j\geq 1$ ,

$$g^{(j)}(s)=-g_2^{(j)}(s)+O(1)+\begin{cases}\log\textrm{D}_K&j=1\\[5pt] 0&j> 1.\end{cases}$$

Now, note that

$$g_2^{(j)}(s)=\frac{d^{j-1}}{ds^{j-1}}\left(\frac1s-\frac{\zeta_K'(1-s)}{\zeta_K(1-s)}\right),$$

so Lemma 3·12 shows that $g_2^{(j)}(0)\ll(\log\log\textrm{D}_K)^{j}$ for all $j\geq 1$ . This implies that, using the properties of $P_j$ ,

$$\lim_{s\to 0}P_{r_1+r_2-1}\left(g^{(1)}(s),g^{(2)}(s),\ldots,g^{(r_1+r_2-1)}(s)\right)=(\log \textrm{D}_K)^{r_1+r_2-1}(1+o(1)).$$

Also,

$$\lim_{s\to 0}f(s)=\left(\frac{2^{r_2}\pi^{n_K}}{\textrm{D}_K^{1/2}}\right)\left(\lim_{s\to 0}\frac1{s\zeta_K(1-s)}\right)\left(\frac2\pi\right)^{r_1}\left(\frac1\pi\right)^{r_2}=-\frac{2^{r_1+r_2}\pi^{r_2}}{\textrm{D}_K^{1/2}}\lim_{s\to 1}\frac1{(s-1)\zeta_K(s)}.$$

By a theorem of Louboutin [ Reference LouboutinLou00a , theorem 1],

$$\lim_{s\to 1}(s-1)\zeta_K(s)\ll (\log \textrm{D}_K)^{n_K-1},$$

and is clearly positive. This gives the desired result.

Using this bound, we finish the proof that $M_K^*(1)>0$ for all sufficiently large $D_K$ .

Proof of Theorem 1·2. By Lemma 3·11,

$$M_K^*(1)=-\textrm{Res}_{s=0}\frac1{s\zeta_K(s)}+O\left(\frac{(\log \textrm{D}_K)^{n_K}}{\textrm{D}_K^{3/2}}\right).$$

Now, Lemma 3·13 implies that the first term is positive, and that it grows at least as quickly as

$$\frac{(\log\textrm{D}_K)^{-r_2}}{\textrm{D}_K^{1/2}}.$$

For large $\textrm{D}_K$ depending on $r_1$ and $r_2$ , this outpaces the other terms, and so $M_K^*(1)>0$ . An application of Lemma 3·4 finishes the proof.

4·1. Support for the conjecture (1·3)

By the discussion following the statement of Theorem 1·4, the picture is most transparent when $K/\mathbb{Q}$ is an abelian Galois extension. In this case, the irreducible characters on $\textrm{Gal}(K/\mathbb{Q})$ are all 1-dimensional, and therefore by abelian reciprocity the functions appearing in the Artin factorisation of $\zeta_K(s)$ are moreover distinct primitive Dirichlet L-functions, among them $\zeta(s)$ . Label the functions $L_1(s),\ldots,L_n(s)$ , where $L_1(s) = \zeta(s)$ , say, and $n = [K:\mathbb{Q}]$ . We assume the generalised Riemann hypothesis for the Dirichlet L-functions, simplicity of all nontrivial zeros (a priori none at $s = {1}/{2}$ ), and that no two Dirichlet L-functions share a nontrivial zero. Let $k \in \mathbb{R}$ . Then,

\begin{equation*} \sum_{\substack{0 < \gamma \leq T \\ \zeta_K(\frac{1}{2} + i\gamma) = 0}} |\zeta_K'(\rho)|^{2k} = \sum_{i=1}^n \sum_{\substack{0 < \gamma_i \leq T \\ L_i(\frac{1}{2} + i\gamma_i) = 0}} |L_i'(\rho)|^{2k} \,\prod_{j \neq i} |L_j(\rho)|^{2k},\end{equation*}

where we maintain the convention $\rho = {1}/{2} + i\gamma_i$ . Upon normalising the ith summand by the factor $N_i(T) = \#\{0 \leq \gamma_i \leq T \mid L_i({1}/{2} + i \gamma_i) = 0 \}$ , we could anticipate that

\begin{align*} & \frac{1}{N_i(T)} \sum_{\substack{0 < \gamma_i \leq T \\ L_i(\frac{1}{2} + i\gamma_i) = 0}} |L_i'(\rho)|^{2k} \,\prod_{j \neq i} |L_j(\rho)|^{2k} \\[5pt] &\phantom{\,\,\,\,}\asymp \left( \frac{1}{N_i(T)} \sum_{\substack{0 < \gamma_i \leq T \\ L_i(\frac{1}{2} + i \gamma_i) = 0}} |L_i'(\rho)|^{2k} \right) \prod_{j \neq i} \left(\frac{1}{ N_i(T)} \sum_{\substack{0 < \gamma_i \leq T \\ L_i(\frac{1}{2} + i\gamma_i) = 0}} |L_j(\rho)|^{2k} \right).\end{align*}

Noting its similarity to the sum $J_{k}(T)$ , and recalling the known asymptotic $N_i(T) \asymp T(\log T)$ (see [ Reference Montgomery and VaughanMV07 , theorem 14·5]), we may posit that

\begin{equation*} \frac{1}{N_i(T)} \sum_{0 < \gamma_i \leq T} |L_i'(\rho)|^{2k}\;\asymp (\log T)^{k(k+2)},\end{equation*}

as in the case of the Riemann zeta function (cf. [ Reference GonekGon89 ]). Meanwhile, to address the inner sums

\begin{equation*} \frac{1}{N_i(T)} \sum_{\substack{0 < \gamma_i \leq T \\ L_i(\frac{1}{2} + i\gamma_i) = 0}} |L_j(\rho)|^{2k},\end{equation*}

where $j \neq i$ , it would be most favorable to impose some restriction on the proximity of the zeros $\gamma_i$ of $L_i$ to lower values of the $L_j$ ; this is because we are taking $k = -1$ . In this case, we expect that these normalised sums are $O_{\varepsilon}(T^\varepsilon)$ for all $\varepsilon > 0$ . This would suggest that $J_{-1}^K(T) \ll_{\alpha} T^{1 + \alpha}$ for all $\alpha > 0$ , which would be sufficient for Theorem 1·4. In fact, we believe that $J_{-1}^K(T) \ll_K T$ holds for all abelian number fields K.

5·1. Bounds in horizontal strips

In this section, we outline a proof of Lemma 2·4. This proof is heavily based on the proof of [ Reference Montgomery and VaughanMV07 , theorem 13·22], the corresponding statement for $K={\mathbb {Q}}$ . We will run through the intermediate statements needed for this result as it is proven in [ Reference Montgomery and VaughanMV07 ], and in each step explain the necessary modifications in order to generalise to the setting of Dedekind zeta functions.

Lemma 5·1 (cf. [ Reference Montgomery and VaughanMV07 , corollary 10·5]). Suppose $-1 \leq \sigma \leq \frac{1}{2}$ . Then

$$\left \vert \frac{\zeta_K(\sigma + iT)}{\zeta_K(1 - \sigma - iT)} \right \vert \asymp \vert T \vert^{(1/2-\sigma)n_K^2}.$$

This follows from the reflection formula for Dedekind zeta functions and Stirling’s approximation. By the Hadamard product for the completed Dedekind zeta function, one gets the following formula for the logarithmic derivative of $\zeta_K$ .

Lemma 5·2 (cf. [ Reference Montgomery and VaughanMV07 , lemma 12·1]). We have

$$\frac{\zeta_K'(s)}{\zeta_K(s)} = -\left(\frac{1}{s} + \frac{1}{s-1}\right)+ \sum_{\vert \gamma - t \vert \leq 1}\left(\frac{1}{\rho} + \frac{1}{s - \rho}\right) + O(\log\tau)$$

uniformly for $-1 \leq \sigma \leq 2$ , where $\tau = |t| + 4$ .

Define $\Lambda^K(n)$ to be, as in the proof of Lemma 3·12, such that

$$\frac{\zeta_K'(s)}{\zeta_K(s)}=\sum_{n=1}^\infty \Lambda^K(n)n^{-s}.$$

We will use the bound

(5·1) \begin{equation}0\leq \Lambda^K(n)\leq n_K\Lambda(n)\end{equation}

for all positive integers n in many of the following results. It may be proven by a simple consideration of the powers of prime ideals of $\mathcal{O}_K$ with given norm; for the upper bound, equality is reached for n a power of a prime p which splits completely in K. The first result needed for Lemma 2·4 is the following.

Lemma 5·3 (cf. [ Reference Montgomery and VaughanMV07 , theorem 13·13]). Assume the Riemann hypothesis for K. Then

$$\left \vert \frac{\zeta_K'(s)}{\zeta_K(s)}\right \vert \leq \sum_{n \leq (\log \tau)^2} \frac{\Lambda^K(n)}{n^\sigma} + O((\log \tau)^{2 - 2\sigma})$$

uniformly for ${1}/{2} + {1}/{\log \log \tau} \leq \sigma \leq {3}/{2}$ , $\vert t \vert \;\geq 1$ .

This lemma is proven essentially identically to the corresponding result from [ Reference Montgomery and VaughanMV07 ], and a proof along very similar lines is given in this paper for Lemma 3·12. In particular, the only modification to the proof in [ Reference Montgomery and VaughanMV07 ] that is needed results from the location and multiplicity of trivial zeros (see (3·10)). We remark that, in light of (5·1), this lemma implies the weaker bound

(5·2) \begin{equation}\left \vert \frac{\zeta_K'(s)}{\zeta_K(s)}\right \vert \leq n_K\sum_{n \leq (\log \tau)^2} \frac{\Lambda(n)}{n^\sigma} + O((\log \tau)^{2 - 2\sigma}),\end{equation}

in which a multiplicative constant is sacrificed for conceptual simplicity.

Lemma 5·4 (cf. [ Reference Montgomery and VaughanMV07 , corollary 13·16]). Assume the Riemann hypothesis for $\zeta_K(s)$ . Then, for $|t|\;\geq 1$ ,

$$|\log\zeta_K(s)|\;\leq n_K\cdot\begin{cases} \log\frac1{\sigma-1}+O(\sigma-1) &\text{if } 1+\frac1{\log\log\tau}\leq\sigma\leq\frac32\\[5pt] \log\log\log\tau+O(1) &\text{if } 1-\frac1{\log\log\tau}\leq\sigma\leq1+\frac1{\log\log\tau}\\[5pt] \log\frac1{1-\sigma}+O\left(\frac{(\log \tau)^{2-2\sigma}}{(1-\sigma)\log\log\tau}\right) &\text{if } \frac12+\frac1{\log\log\tau}\leq\sigma\leq1-\frac1{\log\log\tau}. \end{cases}$$

The proof of this corollary is identical to the proof of [ Reference Montgomery and VaughanMV07 , corollary 13·16], with (5·2) used in the place of [ Reference Montgomery and VaughanMV07 , theorem 13·13]. We also need two additional results about the zeros of $\zeta_K(s)$ , direct analogues of [ Reference Montgomery and VaughanMV07 , lemma 13·19] and [ Reference Montgomery and VaughanMV07 , lemma 13·20], with analogous proofs. Firstly, the bound in Lemma 5·3 can be used to show that

(5·3) \begin{equation}N_K\left(T+\frac1{\log\log T}\right)-N_K(T)\ll \frac{\log T}{\log\log T},\end{equation}

a refinement of Lemma 2·3. Secondly, an application of Lemma 5·2 combined with the counting result from the preceding equation, shows that, for $|t|\;\geq 1$ and $|\sigma-{1}/{2}|\;\leq {1}/{\log \log \tau}$ , then

(5·4) \begin{equation}\frac{\zeta_K'(s)}{\zeta_K(s)}=\sum_{|\gamma-t|\leq 1/\log\log\tau}\frac1{s-\rho}+O(\log\tau).\end{equation}

These results are enough to show Lemma 2·4 following the proof of [ Reference Montgomery and VaughanMV07 , theorem 13·22].

Proof of Lemma 2·4. Lemma 5·1 implies that we need only to consider ${1}/{2}<\sigma<2$ . From here, Lemma 5·4 shows the result for $\sigma> {1}/{2}+{1}/{\log\log T}$ , and integrating (5·4) over a short horizontal segment gives a bound of the desired form, but in terms of the distances between T and various ordinates of the zeros of $\zeta_K(s)$ . Integrating this bound over $n\leq T<n+1$ for any n gives that the result holds on average, and thus for some fixed $T=T_n$ , as desired.

5·2. Bounds on residues at the trivial zeros

We now prove Lemma 2·13, following the method of the proof of Lemma 2·13.

Proof of Lemma 2·13. First, the residue at 0 can be written as

$$\lim_{s\to 0}\frac{d^{r_1+r_2-1}}{ds^{r_1+r_2-1}}\frac{s^{r_1+r_2-1}x^s}{\zeta_K(s)}=\lim_{s\to 0}\frac{d^{r_1+r_2-1}}{ds^{r_1+r_2-1}}x^sf(s)$$

for some f independent of x with neither a pole nor a zero at $s=0$ . Evaluating the derivative explicitly yields that this residue is polylogarithmic in x, with exponent $r_1 + r_2 - 1$ .

For $k>0$ , by Theorem 2·1, we write

(5·5) \begin{equation} \frac{x^s}{s\zeta_K(s)} = \frac{2^{r_2}\pi^{n_K}}{D_K^{1/2}} \left(\frac{x D_K}{(2\pi)^{n_K}}\right)^s\left( \frac{1}{s\zeta_K(1-s)} \right) \left( \frac{1}{\sin \frac{\pi s}2} \right)^{r_1} \left( \frac{1}{\sin \pi s} \right)^{r_2} \left( \frac{1}{\Gamma(1-s)} \right)^{n_K},\end{equation}

where we have used the reflection formula $\Gamma(s)\Gamma(1-s) = \pi / \sin \pi s$ . At $s = -k$ , none of these terms are zero. The fourth term has a pole of order $r_1$ if k is even, and the fifth term has a pole of order $r_2$ regardless. Define $\epsilon$ to be 1 if k is even and 0 otherwise, so that ${x^s}/{s\zeta_K(s)}$ has a pole of order $r_1\epsilon+r_2$ at $s=-k$ . Define

\begin{align*}f(s) &=\frac{(s+k)^{r_1\epsilon+r_2}x^s}{s\zeta_K(s)}\\[5pt] g_1(s)&=s\log\left(\frac{x\textrm{D}_K}{(2\pi)^{n_K}}\right)\\[5pt] g_2(s)&=\log(s\zeta_K(1-s))\\[5pt] g_3(s)&=\epsilon\log(s+k)-\log\sin\frac{\pi s}2\\[5pt] g_4(s)&=\log(s+k)-\log\sin\pi s\\[5pt] g_5(s)&=\log\Gamma(1-s)\end{align*}

so that

$$g(s)\;:\!=\;\log f(s)=\log\left(\frac{2^{r_2}\pi^{n_K}}{\textrm{D}_K^{1/2}}\right)+g_1(s)-g_2(s)+r_1g_3(s)+r_2g_4(s)-n_Kg_5(s),$$

and so that

$$\textrm{Res}_{s=-k}\frac{x^s}{s\zeta_K(s)}=\lim_{s\to -k}\frac{d^{r_1\epsilon+r_2-1}}{d^{r_1\epsilon+r_2-1}s}f(s).$$

We may show that, for $\ell\leq n_K$ ,

\begin{align*}\left|g_1^{(\ell)}(k)\right|&\leq \log\left(\frac{xD_K}{(2\pi)^{n_K}}\right)\\[5pt] \left|g_2^{(\ell)}(k)\right|,\left|g_3^{(\ell)}(k)\right|,\left|g_4^{(\ell)}(k)\right|&\ll_{r_1,r_2} 1\\[5pt] \left|g_5^{(\ell)}(k)\right|&\ll \log k.\end{align*}

This shows that $g^{(\ell)}(s)\ll_{r_1,r_2} \log(kx\textrm{D}_K)$ . In addition,

\begin{align*}\left|\lim_{s\to -k}f(s)\right|&=\left|\left(\frac{2^{r_2}\pi^{n_K}}{\textrm{D}_K^{1/2}}\right)\frac{\left(\frac{x\textrm{D}_K}{(2\pi)^{n_K}}\right)^{-k}}{\lim_{s\to -k}s\zeta_K(1-s)}\left(\frac 2\pi\right)^{\epsilon r_1}\left(\frac 1\pi\right)^{r_2}\frac1{k!^{n_K}}\right|\\[5pt] &\ll_{r_1,r_2}\frac{(2\pi)^{kn_K}}{x^k\textrm{D}_K^{k+1/2}k!^{n_K}}\cdot \frac{1}{\left|\lim_{s\to -k}s\zeta_K(1-s)\right|}\ll \frac{(2\pi)^{kn_K}}{kx^k\textrm{D}_K^{k+1/2}k!^{n_K}}.\end{align*}

As in the proof of Lemma 3·13, since $f'(s)=g'(s)f(s)$ , we have that $f^{(j)}(s)/f(s)$ is a polynomial of degree at most j in the first j derivatives of g. This finishes the proof.

5·3. Proof of the explicit formula

To prove Proposition 2·12, the explicit formula for $M_K(x)$ , we need a rather precise truncated form of Perron’s formula. This can be found by combining the statements of [ Reference TitchmarshTit86 , lemma 3·12] and [ Reference NgNg04 , lemma 2], and a similar result can be found in [ Reference Montgomery and VaughanMV07 , corollary 5·3].

Lemma 5·5. Suppose $f(s)=\sum_{n=1}^\infty a_nn^{-s}$ is absolutely convergent for $\textrm{Re}(s)>1$ , and let $\Phi(n)$ be a positive and non-decreasing function such that $a_n\ll\Phi(n)$ . Assume that

$$\sum_{n=1}^\infty \frac{|a_n|}{n^\sigma}=O\left(\frac1{(\sigma-1)^\alpha}\right)$$

as $\sigma$ tends to 1 from above. Pick $w=u+iv$ and $c>0$ such that $u+c>1$ , and pick $T>0$ . Then, for $x>0$ ,

\begin{equation*} \sum_{n \leq x}\frac{a_n}{n^w}=\frac1{2\pi i}\int_{c-iT}^{c+iT}f(w+s)\frac{x^s}sds+O\left(\frac{x^c}{T(u+c-1)^\alpha}+\frac{\Phi(2x)x^{1-u}\log x}{T}+E_1(x,T)\right),\end{equation*}

where

$$E_1(x,T)\ll \Phi(2x)x^{-u}$$

if $x > 1$ , and, if N is the nearest integer to x (besides possibly x itself),

$$E_1(x,T)\ll \frac{\Phi(N)x^{1-u}}{T|x-N|}$$

for any $x>0$ .

We now give a bound on the integral found in our application of Perron’s formula. This is done separately from the main proof of the explicit formula in order to elucidate the locations where the appropriate error terms arise.

Lemma 5·6. Fix a number field K, and let $\mathcal T$ be as in Lemma 2·4. For any $x>0$ , $0<T\in\mathcal T$ , and $1<c \leq 2$ , write

\begin{equation*}\frac1{2\pi i}\int_{c-iT}^{c+iT}\frac{x^s}{s\zeta_K(s)}ds= \sum_{|\gamma| < T} \frac{x^{\rho}}{\rho \zeta_K'(\rho)} + \sum_{k=0}^{\infty}\textrm{Res}_{s=-k}\frac{x^s}{s\zeta_K(s)}+E_2(x,c,T).\end{equation*}

Then, for any $\epsilon>0$ , $E_2(x,c,T)=O_x(1/T^{1-\epsilon})$ . When $x>1$ , $E_2(x,c,T)=O(x^c/T^{1-\epsilon})$ uniformly.

Proof. Pick a sufficiently large positive half-integer U (we will let $U\to \infty$ ), and integrate around the rectangular contour with vertices $c\pm iT$ and $-U\pm iT$ . This gives

(5·6) \begin{align}&\frac1{2\pi i}\int_{c-iT}^{c+iT}\frac{x^s}{s\zeta_K(s)}ds\nonumber \\[5pt] &=\sum_{|\gamma|<T}\frac{x^\rho}{\rho\zeta_K'(\rho)}+\sum_{k=0}^{\lfloor U\rfloor} \textrm{Res}_{s=-k}\frac{x^s}{s\zeta_K(s)}+\frac1{2\pi i}\left(\int_{c-iT}^{-U-iT}+\int_{-U-iT}^{-U+iT}+\int_{-U+iT}^{c+iT}\right)\frac{x^s}{s\zeta_K(s)}ds.\end{align}

We first bound these integrals using the functional equation for $\zeta_K(s)$ (Theorem 2·1). First, we recall the identities and bounds

\begin{align*}\Gamma(s)\Gamma(1-s) &= \frac{\pi}{\sin \pi s}\\[5pt] \Gamma(s)^{-1}&\ll e^{\sigma-(\sigma-1/2)\log\sigma+\pi|t|/2}\\[5pt] |\sin\pi s|^{-1}&\ll e^{-\pi |t|}\\[5pt] \left|\sin\frac{\pi s}2\right|^{\pm 1}&\ll e^{\pm\pi |t|/2}\\[5pt] \zeta_K(s)^{-1}&\ll 1\text{ for }\sigma>2,\end{align*}

which hold for s bounded away from integers. As the paths of the three integrals in (5·6) always stay a distance of ${1}/{4}$ away from each integer, we have for s on such a path, and with real part not between $-1$ and 2, that

(5·7) \begin{equation}\frac{x^s}{s\zeta_K(s)}\ll_K \frac{\left(x\textrm{D}_K/(2\pi)^{n_K}\right)^\sigma}{T}e^{n_K\left[1-\sigma-(1/2-\sigma)\log(1-\sigma)\right]}.\end{equation}

Thus, the middle integral is asymptotically at most $x^Ue^{-U\log U+O(U)}$ uniformly in x and T as $U\to\infty$ . For the two horizontal integrals, we first bound the portion where $\textrm{Re}(s)$ runs from $-1$ to $-U$ . Here, we may use (5·7). Since the integral in the right-hand side of

$$\int_{-1 + iT}^{-U + iT}\frac{x^s}{s\zeta_K(s)}\,ds \ll \frac1T\int_{-1}^{-U}\left(\frac{x\textrm{D}_K}{(2\pi)^{n_K}}\right)^\sigma e^{n_K\left[1-\sigma-(1/2-\sigma)\log(1-\sigma)\right]}\,d\sigma$$

converges as $U\to\infty$ , this is $O(1/T)$ for fixed x, and for $x>1$ the dependence is $O(x^{-1})$ . As a result, we may write

\begin{align*}&\frac1{2\pi i}\int_{c-iT}^{c+iT}\frac{x^s}{s\zeta_K(s)}ds\\[5pt] &=\sum_{|\gamma|<T}\frac{x^\rho}{\rho\zeta_K'(\rho)}+\sum_{k=0}^\infty \textrm{Res}_{s=-k}\frac{x^s}{s\zeta_K(s)}+\frac1{2\pi i}\left(\int_{c-iT}^{-1-iT}+\int_{-1+iT}^{c+iT}\right)\frac{x^s}{s\zeta_K(s)}ds+E_3(x,T),\end{align*}

where $E_3(x,T)=O_x(1/T)$ for $x > 0$ , and $E_3(x,T)=O(1/xT)$ for $x>1$ . Finally, for the remaining integrals, we use the definition of $\mathcal T$ . Since $|\zeta_K(\sigma\pm iT)|^{-1}\ll T^\epsilon$ uniformly in $-1\leq \sigma \leq 2$ for $T\in\mathcal T$ and any $\epsilon>0$ , we have

$$\frac{x^s}{s\zeta_K(s)}\ll \frac{\max(x^c,1/x)}{T^{1-\epsilon}}$$

for any s on one of our remaining line segments. This bounds our integrals by $\max(x^c,1/x)/T^{1-\epsilon}$ , which is enough.

We are now ready to show the explicit formula for $M_K(x)$ .

Proof of Proposition 2·12. First, note that $|\mu_K(n)|$ is bounded by the coefficient of $n^{-s}$ in the Dirichlet series expansion of $\zeta_K(s)$ . As a result,

$$\sum_{n=1}^\infty \frac{|\mu_K(n)|}{n^\sigma}=O\left(\frac{1}{\sigma - 1}\right) \text{ as } \sigma \rightarrow 1^+,$$

since the pole of $\zeta_K(s)$ at $s = 1$ is simple. Define

$$\Phi(x)=x^{\frac{C}{\log\log(x+10)}},$$

where C is as in Lemma 2·2, so that $\Phi(x)$ is positive, non-decreasing, and satisfies $\mu_K(n)\ll \Phi(n)$ . Now, apply 5.5 to $f(s)=1/{\zeta_K(s)}$ , $w=0$ , $\alpha=1$ , $T>0$ , and with $1<c\leq 2$ to be chosen later. This gives

$$M_K(x)=\frac1{2\pi i}\int_{c-iT}^{c+iT}\frac{x^s}{s\zeta_K(s)}ds+O\left(\frac{x^c}{T(c-1)}+\frac{\Phi(2x)x\log x}{T}+E_1(x,T)\right),$$

where $E_1(x,T)$ is as in the statement of Lemma 5·5. By Lemma 5·6, we have

\begin{align*}M_K(x)&={\sum_{|\gamma|\leq T} \frac{x^\rho}{\rho\zeta_K'(\rho)}+\sum_{k=0}^{\infty}\textrm{Res}_{s=-k}\frac{x^s}{s\zeta_K(s)}}\\[5pt] &+O\left(\frac{x^c}{T(c-1)}+\frac{\Phi(2x)x\log x}{T}+E_1(x,T)+E_2(x,c,T)\right).\end{align*}

This is enough to show our inequality on the error term when $x > 0$ , as each term in the above is asymptotically bounded by $T^{-(1-\epsilon)}$ (any $1<c\leq2$ will suffice). When $x > 1$ , we have

$$E(x,T)\ll \frac{x^c}{T(c-1)}+\frac{\Phi(2x)x\log x}{T}+\frac{x^c}{T^{1-\epsilon}}+\Phi(2x).$$

Setting $c=1+(\log x)^{-1}$ gives

$$E(x,T)\ll\frac{x^{1+C/\log \log x}\log x}T+\frac{x}{T^{1-\epsilon}}+x^{C/\log\log x}.$$

Finally, Lemma 2·13 implies that the sum of the residues at negative integers is convergent as a series in k, and its dependence on x is $O((\log x)^{r_1 + r_2})$ depending on K.

6·1. Complexities of non-abelian number fields

A main tool in studying the limiting distribution of $M_K(x)/\sqrt{x}$ is the explicit formula (Proposition 2·12). This formula was derived via Perron’s formula, which allows one to express $M_K(x)$ as a sum of residues of ${x^s}/{s\zeta_K(s)}$ . If $\rho$ is a simple zero of $\zeta_K(s)$ , then this residue has a particularly simple formula and is of order $x^{1/2}$ as a function of x. However, if $\textrm{ord}_{s = \rho}\zeta_K(s) = n > 1$ , then this residue is a linear combination of terms $x^{1/2}(\log x)^k$ where $k < n$ . In particular, the coefficients of these terms involve large positive and negative powers of higher derivatives of $\zeta_K(x)$ , quantities whose moments are not well understood.

Assuming some standard conjectures, the zeros of $\zeta_K(s)$ for a non-abelian number field K (with the possible exception of a zero at $s = {1}/{2}$ ) have multiplicity at most the largest dimension of an irreducible complex representation of $\textrm{Gal}(K/{\mathbb {Q}})$ . This motivates the following question.

Moreover, assume that the $\zeta_K(s)$ has a zero at $s = {1}/{2}$ of multiplicity larger than that of any other nontrivial zero. As described in Proposition 2·7, this term does not oscillate between large positive and negative values (like terms associated with non-real zeros), and will dominate all others in the explicit formula for $M_K(x)$ . Hence, one may ask the following question.

One may ask if such number fields satisfying the hypothesis of this question even exist. This is answered in the affirmative, assuming appropriate standard conjectures, by the work of Louboutin [ Reference LouboutinLou00b ].

It now remains to identify number fields K of interest, as well as give heuristic bounds on certain sums over zeros, which we turn to now. In what follows, it will be relevant to state the following assumptions on Artin L-functions, for a Galois extension of number fields $L/K$ :

6·2. Candidate number fields

The first appearance of a number field K satisfying $\zeta_K(\frac{1}{2}) = 0$ occurred in a paper of Armitage [ Reference ArmitageArm72 ], who studied a degree 48 extension of $\mathbb{Q}$ which had first appeared in a work of Serre [ Reference SerreSer71 ]. Later, Serre (unpublished, see [ Reference NgNg00 , chapter 2]) discovered the simpler example

$$K={\mathbb {Q}}\left(\sqrt{(5+\sqrt 5)(41+\sqrt{205})}\right),$$

which is Galois with quaternionic Galois group $Q_8$ .

In emulation of this latter construction, we identify as candidates for $\textrm{Gal}(K/\mathbb{Q})$ the following well-known class of groups to which $Q_8$ belongs. These groups possess convenient representation-theoretic properties, which allow control over the behaviour of $\zeta_K(s)$ at $s={1}/{2}$ .

Definition 6·7. Let $n \geq 2$ an integer. We define the dicyclic group $Q_{4n}$ by the presentation

\begin{equation*} Q_{4n} = \langle a, b: a^{2p} = 1, a^{p} = b^{2}, b^{-1} ab = a^{-1} \rangle.\end{equation*}

This is a solvable non-abelian group of order 4n.

In what follows, let $p \geq 3$ be an odd prime, and let $N/\mathbb{Q}$ be a normal extension with Galois group $G = Q_{4p}$ , called a dicyclic number field of degree 4p. We recall the Galois theory of such fields, which in be found in the introduction to [ Reference LouboutinLou00b ].

If p is not totally ramified in $N/\mathbb{Q}$ , then by [ Reference LouboutinLou00b , theorem 1], the root number $W_{N/\mathbb{Q}}(\psi) = W_{N/L}(\chi)$ does not depend on $\chi$ but just on the field N, so either there are $(p-1)/2$ two-dimensional characters with root number $-1$ or there are none. Hence, assuming $\text{SH}_{N/{\mathbb {Q}}}$ , the order of the zero at $s = {1}/{2}$ of the Dedekind zeta function is either $p-1$ or zero. Moreover, there exist infinitely many examples of dicyclic fields N which produce either case. If p is totally ramified in $N/\mathbb{Q}$ , which can only occur when $p \equiv 1 \pmod{4}$ , and if $L = \mathbb{Q}(\sqrt{p})$ , then half of the $(p-1)/2$ quaternionic characters have root number $-1$ and half have root number $+1$ . Hence, assuming $\text{SH}_{N/{\mathbb {Q}}}$ , the order of the zero at $s = {1}/{2}$ is $(p-1)/2$ .

Now that we have built up the multiplicity of the zero at $s = {1}/{2}$ , we invoke the conditions $\text{SH}_{N/{\mathbb {Q}}}$ and $\text{IC}_{N/{\mathbb {Q}}}$ in order to tame the multiplicities of all the other nontrivial zeros. Equipped with these assumptions, the largest dimension of an irreducible representation of a dicyclic number field would be 2, so that all other zeroes besides $s = {1}/{2}$ should have order at most 2.

Assuming moreover $\text{IC}_{N/{\mathbb {Q}}}$ , and for p sufficiently large, this order is higher than that of any other nontrivial zero.

Moreover, there are infinitely many examples of N as in case (a) assuming $\text{SH}_{N/{\mathbb {Q}}}$ .

6·3. Heuristics in support of one-sided growth of $M_K(x)$

From now on, we shall specialise to the scenario of Theorem 6·8, where $K \;:\!=\; N$ . Hence, K is a dicyclic number field satisfying $\text{SH}_{K/{\mathbb {Q}}}$ and $\text{IC}_{K/{\mathbb {Q}}}$ , with the function $\zeta_K(s)$ having a zero of multiplicity $\mu \geq 3$ at $s = 1/2$ and all other nontrivial zeros of multiplicity equal to either 1 (simple) or 2 (double). In the notation of Proposition 2·12, for all $T_n \in \mathcal{T}$ and $x > 1$ we have, under the Riemann hypothesis for $\zeta_K(s)$ ,

(6·1) \begin{equation} M_K(x) = \textrm{Res}_{s = 1/2} \frac{x^s}{s\zeta_K(s)} + \sum_{\substack{|\gamma| \leq T_n \\ \rho \text{ simple}}} \textrm{Res}_{s = \rho} \frac{x^s}{s\zeta_K(s)} +\sum_{\substack{|\gamma| \leq T_n \\ \rho \text{ double}}} \textrm{Res}_{s = \rho} \frac{x^s}{s\zeta_K(s)} + E(x,T_n).\end{equation}

As noted already, the leading term above grows one-sidedly in x as $x^{1/2}(\log x)^{\mu - 1}$ , so we seek to restrain the growth of the remaining terms. We have

\begin{align*} \textrm{Res}_{s = \rho} \frac{x^s}{s\zeta_K(s)} = \frac{x^{\rho}}{\rho \zeta_K'(\rho)} \text{ and } \textrm{Res}_{s = \rho} \frac{x^s}{s\zeta_K(s)} = \frac{2x^{\rho}}{\rho \zeta_K''(\rho)} \left( \log x - \frac{1}{\rho} - \frac{\zeta_K'''(\rho)}{\zeta_K''(\rho)}\right),\end{align*}

when $s = \rho$ is a simple and double zero, respectively. In the former case, one suitable bound would be a generalised form of (1·3) which is applicable to Dedekind zeta functions having nontrivial zeros of multiplicity, but which targets only the simple ones, namely

\begin{equation*} \sum_{\substack{0 < \gamma \leq T \\ \rho \text{ simple}}} \frac{1}{|\zeta_K'(\rho)|^2} \ll T,\end{equation*}

give or take the addition of a low power of $\log T$ to the right-hand side. In this case, we may replace the $T_n$ in the second term of (6·1) with any $T > 0$ , upon inserting an additional error term of

\begin{equation*} \left( \frac{x \log T}{T} \right)^{1/2}.\end{equation*}

A similar estimate pertaining to the double zeros,

\begin{equation*} \sum_{\substack{0 < \gamma \leq T \\ \rho \text{ double}}} \frac{1}{|\zeta_K''(\rho)|^2} \ll T,\end{equation*}

and an estimate of the form

\begin{equation*} \sum_{\substack{0 < \gamma \leq T \\ \rho \text{ double}}} \frac{|\zeta_K'''(\rho)|}{|\rho||\zeta_K''(\rho)|^2} \ll (\log T)^a,\end{equation*}

with a a reasonably small integer, would also come in handy, once again allowing for the replacement of $T_n$ with any $T > 0$ in (6·1) in exchange for residual error terms. Finally, following the arguments in the proof of [ Reference NgNg04 , theorem 1] would produce the conclusion of Conjecture 6·1, assuming that the resulting contribution of extra powers of $\log x$ has been smaller than $(\log x)^{\mu - 1}$ .