2 Preliminaries

We first recall some useful facts about the Riemann zeta-function. Then we do the same for the ideal class zeta-function. Finally, we recall some techniques and state some lemmas that are useful in Section 2.3.

2.1 Riemann zeta-function

The Riemann zeta-function in the half-plane $\sigma>1$ is defined by

where the product runs through all primes. In this half-plane, the logarithmic derivative of the Riemann zeta-function can be written as a Dirichlet series,

$$ \begin{align*}\frac{\zeta'}{\zeta}(s)=-\sum_{j\ge 2}\frac{\Lambda( j)}{j^s},\end{align*} $$

where the von Mangoldt function is defined by

The Riemann zeta-function can be continued analytically to a meromorphic function on the whole complex plane with a single singularity at $s=1$ which is a simple pole. This continuation satisfies the functional equation

This implies that the logarithmic derivative of the Riemann zeta-function satisfies

$$ \begin{align*}\frac{\zeta'}\zeta({s})=\frac{\Delta'}{\Delta}({s})-\frac{\zeta'}{\zeta}({1-s}),\end{align*} $$

where by Stirling’s formula,

$$ \begin{align*}\frac{\Delta'}{\Delta}({1-s})=\frac{\Delta'}{\Delta}({s})=-\log\frac{t}{2\pi}+O\bigg({\frac 1 t}\bigg)\end{align*} $$

for $t>1$ . Since the Riemann zeta-function has a simple pole at $s=1$ , so does its logarithmic derivative and in the neighbourhood of $s=1$ ,

$$ \begin{align*}\frac{\zeta'}{\zeta}(s)=-\frac{1}{s-1}+O(1).\end{align*} $$

It follows from the functional equation that any negative even integer is a zero for the Riemann zeta-function; these zeros are called trivial zeros. All further zeros lie in the strip $0\le t\le 1$ , and are called nontrivial zeros. Let $N(T)$ be the number of nontrivial zeros of the Riemann zeta-function whose positive imaginary part does not exceed the number T, that is,

The Riemann–von Mangoldt formula provides asymptotic formulae:

$$ \begin{align*}N(T)=\frac T{2\pi}\log \frac T{2\pi e}+O({\log T})\end{align*} $$

and

$$ \begin{align*}N({T+h})-N(T)\ll \log T\end{align*} $$

for any fixed positive real number h. (Here, every multiple zero is counted according to its multiplicity. For more details, see [Reference Titchmarsh13, Ch. 9]). By an approximation formula for the logarithmic derivative of the Riemann zeta-function [Reference Titchmarsh13, Theorem 9.6] together with the Riemann–von Mangoldt formula, in the strip $-1\le \sigma \le 2$ ,

(2.1) $$ \begin{align} \frac{\zeta'}{\zeta}({\sigma+ it})\ll ({\log t})^2 \end{align} $$

for $\lvert t-\gamma \rvert \ge {c}/{\log t}$ , where c is a constant independent of t.

2.2 Ideal class zeta-function

Recall that the ideal class zeta-function associated to the ideal class $\mathfrak C$ in the half-plane $\sigma>1$ can be written as

$$ \begin{align*} \zeta_K({\mathfrak C;s})= \sum_{m\ge 1}\frac{c_m}{m^{s}}, \end{align*} $$

where $c_m$ is the number of integral ideals of norm m from the class $\mathfrak C$ . According to [Reference Paul and Sankaranarayanan10], one can show that $c_m \ll m^\varepsilon $ and

$$ \begin{align*}\sum_{m\le x}c_m =\kappa x+E({\mathfrak C;x}),\end{align*} $$

where

(2.2) $$ \begin{align} E({\mathfrak C;x})\ll \begin{cases} x^{{23}/{73}}({\log x})^{{315}/{146}} & \text{if } n=2,\\ x^{1-2/n+{8}/{n({5n+2})}} ({\log x})^{{10}/{(5n+2)}} & \text{if } 3\le n\le 6,\\ x^{1-2/n+{3}/{2n^2}}({\log x})^{{2}/{n}} & \text{if } n\ge 7,\\ \end{cases} \end{align} $$

and

Here, $r_1,r_2, R, w$ and d are the number of real embeddings, the number of pairs of complex conjugate embeddings, the regulator, the number of roots of unity and the absolute value of the discriminant of the number field K, respectively.

The ideal class zeta-function can also be continued analytically to a meromorphic function on the whole complex plane with only a simple pole at $s=1$ . This continuation satisfies the functional equation

(2.3) $$ \begin{align} \zeta_K({\mathfrak C;s})=Z_K({s})\zeta_K({\mathfrak C';1-s}), \end{align} $$

where

$$ \begin{align*} Z_K(s)=d^{1/2-s}\big({\sqrt 2}\big)^n\bigg({\frac{\Gamma({1-s})}{({2\pi})^{1-s}}}\bigg)^n\bigg({\sqrt 2\sin\bigg({\pi\frac s2}}\bigg)\bigg)^{r_1}({\sin({\pi s})})^{r_2} \end{align*} $$

and $\mathfrak C'$ is the ideal class of K which is the complement of $\mathfrak C$ (with respect to the trace) (see [Reference Lang8, Ch. 13] for more details). In addition, in a neighbourhood of $s=1$ ,

$$ \begin{align*}\zeta_K({\mathfrak C;s})=\frac{\kappa}{s-1}+O(1).\end{align*} $$

The functional equation, together with Stirling’s formula and the Phragmén–Lindelöf principle [Reference Titchmarsh12, Section 5.65] implies that

(2.4) $$ \begin{align} \zeta_K({\mathfrak C;\sigma+it})\ll \begin{cases} t^\varepsilon & \text{if }\sigma>1,\\ t^{{n{(1-\sigma)}}/2+\varepsilon} & \text{if } -{1}/{\log t}\le \sigma \le 1+{1}/{\log t},\\ t^{n({1/2-\sigma})+\varepsilon} & \text{if }\sigma<0, \end{cases} \end{align} $$

as t tends to infinity.

2.3 Some useful lemmas

As a preparation for the proof of the main result (Theorem 1.1), we state three useful lemmas.

Lemma 2.1 (Abel’s summation formula)

Let $a_1,a_2,\dots $ be a sequence of real (or complex) numbers and suppose $f(x)$ has a continuous derivative on the interval $[y,x]$ , where $0<y<x$ . Define

$$ \begin{align*}A(x)=\sum_{m\le x}a_m,\end{align*} $$

where $A(x)=0$ if $x<1$ . Then,

$$ \begin{align*}\sum_{y<m\le x}a_mf(m)=A(x)f(x)-A(y)f(y)-\int_y^xA(t)f'(t)\,dt.\end{align*} $$

Proof. See [Reference Apostol1, Theorem 4.2].

Lemma 2.2 (The first derivative test)

Let $f(x)$ and $g(x)$ be real functions such that ${g(x)}/{f'(x)}$ is monotonic and ${f'(x)}/{g(x)}\ge m>0$ , or ${f'(x)}/{g(x)}\le -m<0$ . Then,

$$ \begin{align*}\bigg| {\int_a^bg(t)\exp({ if(t)})\,dt}\bigg| \le \frac 4m.\end{align*} $$

Proof. See [Reference Titchmarsh13, Lemma 4.3].

The following lemma is generalised from Gonek’s lemma [Reference Gonek7, Lemma 2].

Lemma 2.3 (Generalised Gonek lemma)

Let a be a fixed real number. For large T, let $\mathcal J$ be the unique positive integer such that $2^{-\mathcal J}T<1\le 2^{1-\mathcal J}T$ . Then,

$$ \begin{align*} &\int_1^T\bigg({\bigg({\frac{t}{2\pi}}\bigg)^{a-1/2}\exp\bigg({ it\log \frac{t}{r e}}}\bigg)\bigg)^n\, dt\\ &\quad = \frac{2\pi}{\sqrt n}\bigg({\frac{2\pi}{r}}\bigg)^{{(n-1)}/2-an}\exp\bigg({- i\bigg({nr-\frac \pi 4}}\bigg)\bigg)\,\chi_{(2^{-\mathcal J}T,T]}(r)+\sum_{j=1}^{\mathcal J}E({r;2^{-j}T,2^{1-j}T}) \end{align*} $$

with the characteristic function

$$ \begin{align*} \chi_I(x)=\begin{cases} 1 &\text{if } x\in I,\\ 0 &\text{otherwise}, \end{cases} \end{align*} $$

and the error term

$$ \begin{align*}E({r;A,B})\ll A^{n({a-1/2})}+\frac{A^{n({a-1/2})+1}}{|{A-r}|+A^{1/2}}+\frac{B^{n\,({a-1/2})+1}}{|B-r|+B^{1/2}}.\end{align*} $$

Proof. Substituting $t\mapsto t/n$ together with Gonek’s lemma [Reference Gonek7, Lemma 2] yields

$$ \begin{align*} &\int_A^B\bigg({\bigg({\frac{t}{2\pi}}\bigg)^{a-1/2}\exp\bigg({ it\log \frac{t}{r e}}}\bigg)\bigg)^n\,dt\\ &\quad=\frac{1}{n^{n({a-1/2})+1}}\int_{nA}^{nB}\bigg({\frac{t}{2\pi}}\bigg)^{n({a-1/2})}\exp\bigg({ it\log \frac{t}{nr e}}\bigg)\,dt\\ &\quad= \frac{2\pi}{\sqrt n}\bigg({\frac{2\pi}{r}}\bigg)^{{(n-1)}/2-an}\exp\bigg({- i\bigg({nr-\frac \pi 4}}\bigg)\bigg)\kern2pt\chi_{(A,B]}(r)+E({r;A,B}) \end{align*} $$

for large A and $A<B\le 2A$ . Then,

$$ \begin{align*} &\int_1^T\bigg({\bigg({\frac{t}{2\pi}}\bigg)^{a-1/2}\exp\bigg({ it\log \frac{t}{r e}}}\bigg)\bigg)^n\, dt\\ &\quad= \sum_{j=1}^{\mathcal J}\int_{2^{-j}T}^{2^{1-j}T}\bigg({\bigg({\frac{t}{2\pi}}\bigg)^{a- 1/2}\exp\bigg({ it\log \frac{t}{r e}}}\bigg)\bigg)^n\, dt+O({1})\\ &\quad=\frac{2\pi}{\sqrt n}\bigg({\frac{2\pi}{r}}\bigg)^{{(n-1)}/2-an}\exp\bigg({- i\bigg({nr-\frac \pi 4}}\bigg)\bigg)\kern2pt\chi_{(2^{-\mathcal J}T,T]}(r)+\sum_{j=1}^{\mathcal J}E({r;2^{-j}T,2^{1-j}T}).\\[-3.8pc] \end{align*} $$

3 Proof of Theorem 1.1

In this section, let $\mathfrak C$ be an ideal class of the class group, $\mathfrak C'$ the class containing the complements of the members of the class $\mathfrak C$ , and $c_m$ and $c^{\prime }_m$ the number of integral ideals of norm m from the class $\mathfrak C$ and $\mathfrak C'$ , respectively.

By the Riemann–von Mangoldt formula, for a given number $t_o\ge 3$ , there is a positive integer $t\in [t_o,t_o+1)$ such that

$$ \begin{align*}\lvert t-\gamma \rvert\ge \frac{c}{\log t},\end{align*} $$

where $\gamma $ is any imaginary value of a nontrivial zero $\rho =\beta + i\gamma $ of the Riemann zeta-function and c is a constant (independent of t). Now, let $\varepsilon>0$ and $T\ge 3$ be such that

$$ \begin{align*}\lvert T-\gamma \rvert\ge \frac{c}{\log T}.\end{align*} $$

Note that the least imaginary value $\gamma $ of a nontrivial zero of the Riemann zeta-function $\rho =\beta + i\gamma $ in the upper half-plane is a little larger than $14$ and the logarithmic derivative ${\zeta '}/{\zeta }$ of the Riemann zeta-function has simple poles at the zeros of the Riemann zeta-function and is analytic elsewhere except for a simple pole at $s=1$ . By Cauchy’s theorem,

$$ \begin{align*}\sum_{0<\gamma<T}\zeta_{K}({\mathfrak C;\rho}) = \frac{1}{2\pi i}\int_{\mathcal C}\frac{\zeta'}{\zeta}(s)\zeta_K({\mathfrak C;s})\, ds,\end{align*} $$

where the counter-clockwise oriented contour $\mathcal C$ is a rectangle with vertices $a+ i$ , $a+ iT, 1-a+ iT, 1-a+ i$ with

. We rewrite the contour integral as

First, we start with the lower horizontal integral. We obtain

(3.1) $$ \begin{align} \mathcal I_4=\frac{1}{2\pi i}\int_{1-a+ i}^{a+ i}\frac{\zeta'}\zeta(s)\zeta_K({\mathfrak{C};s})\, ds\ll 1 \end{align} $$

since this integral is independent of T.

Next, we consider the vertical line segment on the right, which lies inside the half-plane of absolute convergence for the Dirichlet series. By interchanging integration and summation, it follows that

$$ \begin{align*} \mathcal I_1 & =\frac{1}{2\pi i}\int_{a+ i}^{a+ iT}\frac{\zeta'}\zeta(s)\zeta_K({\mathfrak{C};s})\, ds\\ &=\frac{1}{2\pi}\int_1^T\bigg({-\sum_{j\ge 2}\frac{\Lambda(\,j)}{j^{a+ it}}}\bigg)\bigg({\sum_{m\ge 1}\frac{c_m}{m^{a+{it}}}}\bigg)\, dt\\ &=-\frac 1{2\pi}\sum_{j\ge 2}\frac{\Lambda(\,j)}{j^{a}}\sum_{m\ge 1}\frac{c_m}{m^{a}}\int_1^T(\,{jm})^{- it}\, dt. \end{align*} $$

It is easy to see that the latter integral is bounded. By the Laurent expansion of the ideal class zeta-function and the logarithmic derivative of the Riemann zeta-function at $s=1$ ,

(3.2) $$ \begin{align} \mathcal I_1 \ll \frac{\zeta'}{\zeta}(a) \zeta_K({\mathfrak C;a}) \ll ({\log T})^2. \end{align} $$

In view of the estimates for the logarithmic derivative of the Riemann zeta-function and the ideal class zeta-function, (2.1) and (2.4), we obtain the upper horizontal integral

(3.3) $$ \begin{align} \mathcal I_2=\frac{1}{2\pi i}\int_{a+ iT}^{1-a+ iT}\frac{\zeta'}\zeta(s)\zeta_K({\mathfrak{C};s})\,ds\ll T^{n/2+\varepsilon} \end{align} $$

by the trivial estimation.

It remains to consider the vertical integral on the left. By the functional equation (2.3) and substituting $s\mapsto 1-s$ ,

$$ \begin{align*} \mathcal I_3 &=\frac{1}{2\pi i}\int_{1-a+ iT}^{1-a+ i}\frac{\zeta'}\zeta(s)\zeta_K({\mathfrak{C};s})\, ds\\ &=\frac{1}{2\pi i}\int_{1-a+ iT}^{1-a+ i}\frac{\zeta'}\zeta(s) Z_K(s)\zeta_K({\mathfrak{C}';1-s})\, ds\\ &=\frac{1}{2\pi i}\int_{a- i}^{a- iT}\frac{\zeta'}\zeta({1-s}) Z_K({1-s})\zeta_K({\mathfrak{C}';s})\, ds. \end{align*} $$

By the Schwarz reflection principle, the conjugate of this integral is

$$ \begin{align*} \overline{\mathcal I}_3 &=-\frac{1}{2\pi i}\int_{a+ i}^{a+ iT}\frac{\zeta'}\zeta({1-s}) Z_K({1-s})\zeta_K({\mathfrak{C}';s})\, ds\\ &=-\frac{1}{2\pi i^{r_1+r_2+1}}\int_{a+ i}^{a+ iT}\bigg({\frac{\Delta'}{\Delta}({s})-\frac{\zeta'}\zeta({s})}\bigg) \bigg({\frac{\Gamma(s)}{({2\pi})^s}}\bigg)^n d^{s-1/2}E_K({s})\zeta_K({\mathfrak{C}';s})\, ds, \end{align*} $$

where the function $E_K({s})$ is defined by

(3.4)

for some integers $e_k$ with $e_n=1$ and $e_{-n}=({-1})^{r_1+r_2}$ . Observe that the ideal class zeta-function and logarithmic derivative of the Riemann zeta-function can be represented as absolutely convergent Dirichlet series. By Stirling’s formula, for $s=\sigma + it$ and a real number $\ell $ ,

$$ \begin{align*}\frac{\Gamma(s)}{({2\pi})^s}\exp\bigg({ i\pi\ell\frac s2}\bigg)=\bigg({\frac t{2\pi}}\bigg)^{\sigma- 1/2+ it}\exp\bigg({-\frac{\pi t({\ell+1})}2- i\bigg({t-\frac{\pi({\ell+1})\sigma}2+\frac \pi 4}}\bigg)\bigg)\bigg({1+O\bigg({\frac 1t}}\bigg)\bigg).\end{align*} $$

Now, based on (3.4), we split $\overline {\mathcal I_{3}}$ into a sum of integrals and estimate each integral as follows.

Case I:  $-n\le k< n$ . The integrals with $\exp ({ i\pi k{(1-s)}/2})$ contribute to the error term. More precisely,

$$ \begin{align*} &\int_{a+ i}^{a+ iT}\frac{\Delta'}{\Delta}({s})\bigg({\frac{\Gamma(s)}{({2\pi})^s}}\bigg)^n d^{s-1/2}e_k\exp\bigg({ i\pi k\frac{1-s}2}\bigg)\zeta_K({\mathfrak C';s})\, ds\\ &\quad \ll \sum_{m\ge 1}\frac{c_m'}{m^a}\int_1^T\bigg({-\log \frac t{2\pi}+O\bigg({\frac 1t}}\bigg)\bigg)t^{n({a- 1/2})}\exp\bigg({-\frac{\pi t({n-k})}{2n}}\bigg)\, dt \ll \log T \end{align*} $$

since the integral is bounded and the last asymptotic formula follows from the Laurent expansion of the ideal class zeta-function at $s=1$ . In a similar way,

$$ \begin{align*}\int_{a+ i}^{a+ iT}\frac{\zeta'}{\zeta}(s)\bigg({\frac{\Gamma(s)}{({2\pi})^s}}\bigg)^n d^{s- 1/2}e_k\exp\bigg({ i\pi k\frac{1-s}2}\bigg)\zeta_K({\mathfrak C';s})\, ds\ll ({\log T})^2\end{align*} $$

since the ideal class zeta-function and the logarithmic derivative of the Riemann zeta-function have a simple pole at $s=1$ .

Case II: $k=n$ . It remains to evaluate the integral with $\exp ({ i\pi n{(1-s)}/2})$ . We write the integral as the difference of one with ${\Delta '}/{\Delta }(s)$ and the other with ${\zeta '}/{\zeta }(s)$ , and estimate each of those integrals as follows.

Subcase I: the logarithmic derivative  ${\Delta '}/{\Delta }({s})$ . By the fundamental theorem of calculus,

(3.5)

where the function $J(\tau )$ is defined by

for some real numbers $b_t$ (which are bounded as functions of t). Now, we consider the function $J(\tau )$ . It can be rewritten as

$$ \begin{align*}J(\tau) = d^{a- 1/2}\exp\bigg({- i\pi\frac{n}4}\bigg)\sum_{m\ge 1}\frac{c_m'}{m^{a}} \int_{1}^{\tau}\bigg({\bigg({\frac{t}{2\pi}}\bigg)^{a- 1/2}\exp\bigg({ it\log\frac{d^{ 1/n}t}{2\pi m^{ 1/n} e}}\bigg)\bigg({1+\frac{b_t}t}}\bigg)\bigg)^n \, dt. \end{align*} $$

Applying the generalised Gonek lemma (Lemma 2.3), the function $J( \tau )$ is equal to

$$ \begin{align*} &\frac{2\pi}{d^{ 1/{2n}}n^{ 1/2}}\exp\bigg({ i\pi\frac{1-n} 4}\bigg)\sum_{m\ge 1}\frac{c^{\prime}_m}{m^{{(n-1)}/{2n}}}\exp\bigg({-2\pi i n\bigg({\frac{m}d}\bigg)^{ 1/n}}\bigg)\kern1pt\chi_{(2^{-\mathcal J}\tau, \tau]}\bigg({2\pi\bigg({\frac m d}\bigg)^{ 1/n}}\bigg) + O \\ &\quad =\frac{2\pi}{d^{ 1/{2n}}n^{ 1/2}}\exp\bigg({ i\pi\frac{1-n} 4}\bigg)\sum_{d({(2^{-\mathcal J}\tau)}/{2\pi})^n< m\le d({\tau}/{2\pi})^n}\frac{c^{\prime}_m}{m^{{(n-1)}/{2n}}}\exp\bigg({-2\pi i n\bigg({\frac{m}d}\bigg)^{1/n}}\bigg) + O, \end{align*} $$

where $\mathcal J$ is the unique positive integer such that $2^{-\mathcal J}\tau <1\le 2^{1-\mathcal J}\tau $ , and the error term $O=O({\tau ^{n({a- 1/2)}}})$ can be estimated similarly to [Reference Gonek7, Lemma 4].

Now, the method of partial summation and the asymptotic formula for the number of nonzero integral ideals in the class $\mathfrak C'$ with norm up to a given number can be applied to evaluate the above series. For convenience, let $x=d({ \tau /{2\pi }})^n$ , $y=d({2^{-\mathcal J}\tau }/{2\pi })^n$ and

$$ \begin{align*}\sum_{m\le X}c^{\prime}_m=\kappa X+O({X^{\alpha}({\log X})^\beta}),\end{align*} $$

where $\alpha ,\beta $ are determined in (2.2). By Abel’s partial summation (Lemma 2.1),

$$ \begin{align*} &\sum_{y<m\le x}\frac{c^{\prime}_m}{m^{{(n-1)}/{2n}}}\exp\bigg({-2\pi i n\bigg({\frac{m}d}\bigg)^{ 1/n}}\bigg)\\ &\quad =\frac{\kappa\exp({-2\pi in({ x/d})^{ 1/n}})}{x^{{-(n+1)}/{2n}}}-\frac{\kappa\exp({-2\pi in({ y/d})^{ 1/n}})}{y^{{-(n+1)}/{2n}}} \\ &\qquad+\int_y^x \frac{\kappa\exp({-2\pi in({ t/d})^{ 1/n}})}{t^{{(n-1)}/{2n}}}\bigg({2\pi i\bigg({\frac td}\bigg)^{ 1/n}+\frac{n-1}{2n}}\bigg)\, dt+ O({x^{\alpha-{(n-3)}/{2n}}({\log x})^\beta}). \end{align*} $$

Integration by parts yields

$$ \begin{align*} &\int_y^x \frac{\kappa\exp({-2\pi in({ t/d})^{ 1/n}})}{t^{{(n-1)}/{2n}}} {2\pi i\bigg({\frac td}\bigg)^{ 1/n}}\, dt = -\frac{\kappa\exp({-2\pi in({ x/d})^{ 1/n}})}{x^{{-(n+1)}/{2n}}} \\ &\quad +\frac{\kappa\exp({-2\pi in({ y/d})^{ 1/n}})}{y^{{-(n+1)}/{2n}}}+\frac{\kappa{(n+1)}}{2n} \int_y^x \frac{\exp({-2\pi in({ t/d})^{ 1/n}})}{t^{{(n-1)}/{2n}}}\, dt. \end{align*} $$

This implies that

$$ \begin{align*} &\sum_{y<m\le x}\frac{c^{\prime}_m}{m^{{(n-1)}/{2n}}}\exp\bigg({-2\pi i n\bigg({\frac{m}d}\bigg)^{ 1/n}}\bigg) \\ &\qquad =\int_y^{x} \frac{\kappa\exp({-2\pi in( {t/d})^{ 1/n}})}{t^{{(n-1)}/{2n}}}\, d t+O({x^{\alpha-{(n-3)}/{2n}}({\log x})^\beta}) \ll \tau^{n\alpha-{(n-3)}/2}({\log \tau})^{\beta}, \end{align*} $$

where the last asymptotic estimate follows from the integral term by applying the first derivative test (Lemma 2.2). More precisely, let $f(t)=-2\pi n({ t/d})^{ 1/n}$ and $g(t)={1}/{t^{{(n-1)}/{2n}}}$ . Then for $y\le t\le x$ ,

$$ \begin{align*} \bigg|{\frac{f'(t)}{g(t)}}\bigg|=\bigg({\frac{2\pi }{d^{ 1/n}}}\bigg)t^{{(1-n)}/{2n}}\ge \bigg({\frac{2\pi}{d^{ 1/n}}}\bigg)x^{{(1-n)}/{2n}}. \end{align*} $$

By the first derivative test (Lemma 2.2),

$$ \begin{align*} \bigg|{\int_y^xg(t)\exp({ if(t)})\, dt}\,\bigg| \le \frac {2d^{ 1/n}}{\pi}x^{{(n-1)}/{2n}}\ll x^{{(n-1)}/{2n}}\ll x^{\alpha -{(n-3)}/{2n}}.\end{align*} $$

Therefore,

$$ \begin{align*}J(\tau) \ll \max\{\tau^{n({a- 1/2})},\tau^{{165}/{146}}({\log \tau})^{{315}/{146}}\}.\end{align*} $$

Substituting $J({\tau })$ into (3.5),

$$ \begin{align*} \mathcal I_{3,\Delta} &\ll\max\{T^{ n/2}\log T,T^{{165}/{146}}({\log T})^{{461}/{146}}\}. \end{align*} $$

Subcase II: the logarithmic derivative  ${\zeta '}/{\zeta }({s})$ . By Stirling’s formula and the generalised Gonek lemma (Lemma 2.3),

(3.6)

where $\mathcal J$ is the unique positive integer such that $2^{-\mathcal J}T<1\le 2^{1-\mathcal J}T$ and the error term follows as in [Reference Gonek7, Lemma 4] together with the Laurent expansion of the logarithmic derivative of the Riemann zeta-function and the ideal class zeta-function at $s=1$ . After conjugation, we obtain the first desired result.

Next, we focus on (3.6). Applying the fact that $c^{\prime }_k\ll k^\varepsilon ,\Lambda (k)\ll k^\varepsilon $ and the number-of-divisors function $\sigma _o(k)\ll k^\varepsilon $ for any $\varepsilon>0$ ,

$$ \begin{align*}\sum_{\ell\,\mid\, k}c^{\prime}_\ell \Lambda\bigg({\frac k\ell}\bigg)\ll k^\varepsilon,\end{align*} $$

and then

$$ \begin{align*}\mathcal I_{3,\zeta}\ll \sum_{d({{2^{-\mathcal J}T}/{2\pi}})^n<k\le d({{T}/{2\pi}})^n} \frac{1}{k^{{(n-1)}/{2n}-\varepsilon}} +O({T^{ n/2}\log T}) \ll T^{{(n+1)}/{2}+\varepsilon}\end{align*} $$

by applying Abel’s summation formula (Lemma 2.1). Hence,

(3.7) $$ \begin{align} \overline{\mathcal I_3} \ll T^{{(n+1)}/2+\varepsilon}. \end{align} $$

The same estimate holds for $\mathcal I_3$ . Summing up (3.1), (3.2), (3.3) and (3.7), we obtain

$$ \begin{align*}\sum_{0<\gamma<T}\zeta_K({\mathfrak{C};\rho})\ll T^{{(n+1)}/2+\varepsilon} .\end{align*} $$

4 Concluding remarks

The results we have obtained are not trivial. To see this, we first deduce from Theorem 1.1 that

(4.1) $$ \begin{align} \sum_{T<\gamma\leq 2T}\zeta_K(\mathfrak{C};\rho)\ll T^{{(n+1)}/{2}+\varepsilon}. \end{align} $$

In fact, this estimate also implies the one given in Theorem 1.1. Note that the hypothetical zeros of the Riemann zeta-function off the critical line appear in pairs, that is, if $\rho =\beta + i\gamma $ is one, then $1-\beta + i\gamma $ is another (as follows from the functional equation). If now $\frac 12<\beta <1$ , then, in view of the bound (2.4), the contribution of this zero to the sum in question would be of size

$$ \begin{align*} \zeta_K({\mathfrak{C};\rho})\ll T^{{ n}\beta/2+\varepsilon} \end{align*} $$

for $\gamma \in (T,2T]$ . Multiplying this by the number of all zeros $\rho =\frac 12+ i\gamma $ with $T<\gamma \leq 2T$ yields

$$ \begin{align*} \sum_{T<\gamma\leq 2T}\zeta_K(\mathfrak{C};\rho)\ll T^{1+ n/2+\varepsilon}\log T. \end{align*} $$

Comparing this with (4.1) shows that Theorem 1.1 gives a better bound. One should mention that there are techniques available to find unconditional bounds for $\zeta _K({\mathfrak {C};\frac 12+ it})$ which may improve (4.1); however, the authors are not aware of any such result in the literature and their derivation would not be too easy.

The proof of Theorem 1.1 is of interest also for another reason. It should be mentioned that, at least to the best knowledge of the authors, the above reasoning is the first application of the method of Conrey et al. to a zeta-function of degree $n>1$ (in the sense of the extended Selberg class). Note that Conrey et al. [Reference Conrey, Ghosh and Gonek3–Reference Conrey, Ghosh and Gonek5] rewrote the first derivative of a Dedekind zeta-function $\zeta _K(s)=\zeta (s)L({s,\psi })$ of a quadratic number field at $s=\rho $ as

$$ \begin{align*} \zeta_K'(\rho)=\zeta'(\rho) L({1-\rho,\psi}), \end{align*} $$

where the symmetry of the zeros of the Dirichlet L-function $L({s,\psi })$ for a real character $\psi $ (by the functional equation) is used. This allowed them to evaluate the sum over the values $\zeta _K'({\rho })$ . Their reasoning is then based on an evaluation of the integral

$$ \begin{align*} \int \zeta'(s)L({1-s,\psi})\, d s, \end{align*} $$

or rather a variation of it after applying the functional equation. On both vertical sides of the path of integration, the integrand is of order $O({t^{1/2+\varepsilon }})$ (as for a degree-one element of the Selberg class on the left). The underlying symmetry, however, does not apply to our case in general.

In principle, the method can be extended further, for example, to consider the mean square. It is to be expected that, as in our note, the bound comes from the integral over the left vertical line segment.