Selberg zeta functions have second moment at $\sigma=1$

Abstract.

In this paper, we demonstrate the existence of the second moment of the Selberg zeta function for a Fuchsian group of the first kind at $\sigma=1$. The prime geodesic theorem plays a crucial role in this context. The proof extends to Beurling zeta-functions satisfying a weak form of the Riemann hypothesis and to general Dirichlet series with positive coefficients, the partial sums of which are well-behaved. Note that by employing the recent approach of Broucke and Hilberdink in proving the second moment theorem, we can circumvent the separation condition introduced by Landau for general Dirichlet series.

1. Introduction

Let $s=\sigma+it$ be the complex variable. Let $\Gamma\subset\text{PSL}_{2}(\mathbb{R})$ be any Fuchsian group of the first kind. In this paper, we consider the Selberg zeta function for $\Gamma,$ which is defined for $\sigma>1$ by the absolutely convergent Euler product (see Iwaniec [iwaniec1995, Section 10.8])

$$Z(s)=\prod_{P}\prod_{k=0}^{\infty}(1-N(P)^{-s-k})$$

where $P$ runs through all the primitive hyperbolic conjugacy classes of $\Gamma$ and $N(P)=\alpha^{2}$ if the eigenvalues of $P$ are $\alpha$ and $\alpha^{-1}$, with $\alpha$ having the bigger modulus. Note that for $N(P)$, the prime geodesic theorem is valid, i.e. there exists $0<\theta<1$ such that

(1) $$\sum_{N(P)\leq x}1=\operatorname{Li}(x)+O(x^{\theta}).$$

The strongest version of the prime geodesic theorem can be found in Iwaniec [iwaniec1995, Theorem 10.5]. The function $Z(s)$ has a meromorphic continuation to the whole complex plane and satisfies the functional equation

$$Z(s)=\Psi(s)Z(1-s),$$

where $\Psi(s)$ is a meromorphic function of order 2. More about $Z(s)$ see, for example, Jorgenson and Smajlović [js2017].

The first author of the present paper, together with Drungilas and Novikas, have already investigated the existence of the second moment of $Z(s)$ for $\text{PSL}_{2}(\mathbb{Z})$ in [drungilas2021]. There a conditional result for the existence of the second moment of $Z(1+it)$ and $1/Z(1+it)$ is proven (see [drungilas2021, Theorem 3]). This paper demonstrates that they exist unconditionally for all $\Gamma$. The second moments of logarithm and the logarithmic derivative of Selberg-zeta functions were investigated in [dgk2013], Aoki [aoki2020], Hashimoto [hashimoto2023], [hashimoto2024].

For $\sigma>1$ we can write $Z(s)$ and its multiplicative inverse as a general Dirichlet series (compare to [drungilas2021, equations (11)])

$$Z(s)=\sum_{n=1}^{\infty}\frac{b_{n}}{y^{s}_{n}},\ \frac{1}{Z(s)}=\sum_{n=1}^{\infty}\frac{c_{n}}{y^{s}_{n}},$$

where $1=y_{1}<y_{2}<\dots$ and $b_{n},c_{n}\in\mathbb{Z}.$ Our main result is the following theorem.

Theorem 1.

Let $Z(s)$ be the Selberg zeta function for a Fuchsian group of the first kind. Then

(2) $$\lim_{T\rightarrow\infty}\frac{1}{T}\int_{0}^{T}|Z(1+it)|^{2}dt=\sum_{n=1}^{\infty}\frac{b_{n}^{2}}{y_{n}^{2}}$$

and

(3) $$\lim_{T\rightarrow\infty}\frac{1}{T}\int_{1}^{T}|Z(1+it)|^{-2}dt=\sum_{n=1}^{\infty}\frac{c_{n}^{2}}{y_{n}^{2}}.$$

Both series on the RHS converge.

For $\sigma>1,$ we can write

$$Z(s)=Z(s+1)\prod_{P}(1-N(P)^{-s}).$$

Theorem 1 will follow by considering the Ruelle zeta function, which is defined as

(4) $$Z_{1}(s)=Z(s)/Z(s+1)=\prod_{P}(1-N(P)^{-s}).$$

This function has a meromorphic continuation to $\mathbb{C}$. For $\sigma>1,$ we can write

$$Z_{1}(s)=\sum_{n=1}^{\infty}\frac{b^{\prime}_{n}}{x^{s}_{n}},\ \frac{1}{Z_{1}(s)}=\sum_{n=1}^{\infty}\frac{c^{\prime}_{n}}{x^{s}_{n}},$$

where $1=x_{1}<x_{2}<\dots$ and $b^{\prime}_{n},c^{\prime}_{n}\in\mathbb{Z}.$ Note that the absolute value of $b^{\prime}_{n}$ is bounded by how many expressions of the form

$$N(P_{1})\dots N(P_{K})$$

the number $x_{n}$ has. While $c^{\prime}_{n}$ is positive and equal to how many expressions of the form

$$N(P_{1})^{\alpha_{1}}\dots N(P_{K})^{\alpha_{K}},$$

$\alpha_{k}\in\mathbb{N},$ the number $x_{n}$ has. Thus we have

(5) $$|b^{\prime}_{n}|\leq c^{\prime}_{n}.$$

In [drungilas2021], the conditional version of Theorem 1 was proved for $\Gamma=\text{PSL}_{2}(\mathbb{Z})$ assuming the conditions for coefficients

$$\sum_{x_{n}\leq x}b^{\prime}_{n},\sum_{x_{n}\leq x}c^{\prime}_{n}-\rho x\ll x\exp\left(-45\frac{\log x}{\log\log x}\right),$$

where $\rho>0$ is the residue of $Z_{1}^{-1}(s)$ at $s=1$, or the separation condition

(6) $$x_{n+1}-{x_{n}}\gg\exp\left(-\exp\left(c\sqrt{\log x_{n}\log\log x_{n}}\right)\right)$$

are valid for some specific $c>0$. The separation condition (6) was inherited from Landau [landau1909, Chapter 75], where he investigated the second moment for the general Dirichlet series.

The function $Z_{1}$ is a Beurling zeta-function (see also [dgn2019, Ending notes] and [drungilas2021, Introduction]). Notably, the proof that the second moment of $Z_{1}(1+it)$ exists does not rely on specific properties of the Selberg zeta-function, aside from the prime geodesic theorem (1) with $\theta<1$. Therefore, we will derive the moment of $Z_{1}(1+it)$ from the following two general propositions, which examine the cases of the Beurling zeta-function and general Dirichlet series.

We consider generalized prime numbers systems $\mathcal{P}$, often called $g$-primes,

$$1<p_{1}\leq p_{2}\leq\dots,$$

with $p_{j}\in\mathbb{R}$ tending to infinity and the associated Beurling zeta-function

(7) $$\zeta_{\mathcal{P}}(s)=\prod_{j=1}^{\infty}\left(1-\frac{1}{p_{j}^{s}}\right)^{-1}.$$

A formal multiplication of the product gives the Dirichlet series

(8) $$\zeta_{\mathcal{P}}(s)=\sum_{n=1}^{\infty}\frac{d_{n}}{\nu_{n}^{s}},$$

where $1=\nu_{1}<\nu_{2}<\nu_{3}<\dots$ is the increasing sequence of power-products of $g$-primes with corresponding multiplicities $1=d_{1},d_{2},d_{3},\dots$. Similarly, there are integers $1=e_{1},e_{2},e_{3},\dots,$ such that

(9) $$\frac{1}{\zeta_{\mathcal{P}}(s)}=\prod_{j=1}^{\infty}\left(1-\frac{1}{p_{j}^{s}}\right)=\sum_{n=1}^{\infty}\frac{e_{n}}{\nu_{n}^{s}}.$$

As in (5) we have

(10) $$|e_{n}|\leq d_{n}.$$

In this paper, we always assume that there is a fixed $0<\alpha<1$, such that

(11) $$\pi_{\mathcal{P}}(x)=\sum_{p_{n}\leq x}1=\operatorname{Li}(x)+O(x^{\alpha}).$$

Then Hilberdink and Lapidus [hilberdink2006, Theorem 2.1] obtained that $\zeta_{\mathcal{P}}(s)$ has an analytic continuation to the half-plane $\Re s>\alpha$ except for a simple pole at $s=1$ and $\zeta_{\mathcal{P}}(s)\neq 0$ in this region, With these considerations, we formulate the following proposition.

Proposition 2.

Under the assumption (11), we have that

(12) $$\lim_{T\rightarrow\infty}\frac{1}{T}\int_{1}^{T}|\zeta_{\mathcal{P}}(1+it)|^{2}dt=\sum_{n=1}^{\infty}\frac{d^{2}_{n}}{\nu_{n}^{2}}$$

and

(13) $$\lim_{T\rightarrow\infty}\frac{1}{T}\int_{1}^{T}|\zeta_{\mathcal{P}}(1+it)|^{-2}dt=\sum_{n=1}^{\infty}\frac{e^{2}_{n}}{\nu_{n}^{2}}.$$

Both series on the RHS converge.

Moreover in [hilberdink2006, Theorem 2.2] it is proved that

(14) $$\sum_{x_{n}\leq x}d_{n}=\eta x+D(x),$$

where $\eta>0$ is a constant and

(15) $$D(x)\ll x\exp\left(-c\sqrt{\log x\log\log x}\right),$$

with $c>0$ fixed. Note that Broucke, Debruyne, and Vindas [bdv20] constructed a Beurling generalized number system which satisfies (11) with $\alpha=1/2$ and (14) with $D(x)=\Omega\left(x\exp\left(-C\sqrt{\log x\log\log x}\right)\right)$ for some $C>0$. Proposition 2 allows to remove the separation condition of the type (6) in Theorem 1 in [drungilas2021], where the second moment of $\zeta_{\mathcal{P}}^{\pm 1}(1+it)$ was treated.

Proposition 2 will follow from a statement for general Dirichlet series. Denote

$$f(s)=\sum_{j=1}^{\infty}\frac{a_{j}}{n_{j}^{s}},$$

where $(n_{j})$ is a strictly increasing sequence of positive reals which tends to infinity and $a_{j}>0$. Then the second moment of $f(s)$ at $\sigma=1$ exists even for a weaker bound than in (15).

Proposition 3.

Let $a_{j}>0$, $\varepsilon>0$ and assume that

$$A(x)=\sum_{n_{j}\leq x}a_{j}=\varrho x+R(x),$$

where $\varrho>0$ is a constant and

(16) $$R(x)\ll\frac{x}{(\log x)^{3/2+\varepsilon}}.$$

Then the sum defining $f(s)$ converges for $\sigma>1$ and the function has a continuous extension to $\sigma\geq 1$, $s\neq 1$, and

(17) $$\lim_{T\rightarrow\infty}\frac{1}{T}\int_{1}^{T}|f(1+it)|^{2}dt=\sum_{j=1}^{\infty}\frac{a^{2}_{j}}{n^{2}_{j}}.$$

The series on the RHS converges.

Proposition 3 extends the recent result of Broucke and Hilberdink [bh], where they obtained that the second moment of $f(s)$ exists for $\sigma>(1+\eta)/2$, if $\sum_{n_{j}\leq x}a_{j}=\varrho x+O(x^{\eta})$ for some fixed $\varrho>0$ and $\eta<1$. The significant feature of their approach is that they do not require any separation condition for $n_{j}$ in the proof.

We establish Proposition 3 in the second section. The third section is dedicated to proving Theorem 1 and Proposition 2.

We use notation $f(x)\ll g(x)$ or $f(x)=O(g(x))$ to mean that $|f(x)|\leq c|g(x)|$ for some constant $c>0,$ when $x$ is big enough. Additionally, we use $f(x)\prec g(x)$ or $f(x)=o(g(x))$ to signify that $f(x)/g(x)\rightarrow 0,$ when $x\rightarrow\infty.$ The notation $f(x)=\Omega(g(x))$ means that exists a constant $C$ such that the inequality $|f(x)|>C|g(x)|$ holds infinitely often for arbitrarily large values of $x$.

2. Proof of Proposition 3

For the proof of Proposition 3 we adapt the ideas of Broucke and Hilberdink [bh]. Let

$$f_{N}(s)=\sum_{n_{j}\leq N}\frac{a_{j}}{n_{j}^{s}}.$$

Lemma 4.

The following holds

$$\lim_{T\rightarrow\infty}\frac{1}{T}\int_{0}^{T}|f_{N}(1+it)|^{2}dt=\sum_{j=1}^{\infty}\frac{a_{j}^{2}}{n_{j}^{2}}$$

as $N,T\rightarrow\infty$ such that $(\log N)^{2}\prec T.$

Proof.

Assume that $N>e^{e}.$ We compute

(18) $$\begin{split}&\frac{1}{T}\int_{0}^{T}|f_{N}(1+it)|^{2}dt=\frac{1}{2T}\int_{-T}^{T}\left|\sum_{n_{j}\leq N}\frac{a_{j}}{n^{1+it}_{j}}\right|^{2}dt\\ &=\sum_{n_{j},n_{k}\leq N}\frac{a_{j}a_{k}}{2Tn_{j}n_{k}}\int_{-T}^{T}\left(\frac{n_{j}}{n_{k}}\right)^{it}dt\\ &=\sum_{n_{j}\leq N}\frac{a_{j}^{2}}{n_{j}^{2}}+\sum_{n_{j}<n_{k}\leq N}\frac{a_{j}a_{k}}{2Tn_{j}n_{k}}\int_{-T}^{T}\left(\frac{n_{j}}{n_{k}}\right)^{it}+\left(\frac{n_{k}}{n_{j}}\right)^{it}dt\\ &=\sum_{n_{j}\leq N}\frac{a_{j}^{2}}{n_{j}^{2}}+2\sum_{n_{j}<n_{k}\leq N}\frac{a_{j}a_{k}}{n_{j}n_{k}}s(T\log(n_{k}/n_{j})),\end{split}$$

where $s(x)=\sin(x)/x.$ We need to show that for every $\eta>0$ there exists a $T_{0}>0$ such that

(19) $$\left|\sum_{n_{j}<n_{k}\leq N}\frac{a_{j}a_{k}}{n_{j}n_{k}}s(T\log(n_{k}/n_{j}))\right|<\eta,$$

when $T>T_{0}$ and $(\log N)^{2}\prec T.$

Denote

$$R_{k}=\frac{n_{k}}{(\log n_{k})^{3/2+\varepsilon}}.$$

Let $n_{0}>0$. We divide the sum in (19) into three parts

$$\left(\sum_{n_{j}<n_{k}\leq n_{0}}+\sum_{{n_{0}<n_{j}<n_{k}\leq N\atop n_{j}\leq n_{k}-R_{k}}}+\sum_{{n_{0}<n_{j}<n_{k}\leq N\atop n_{j}>n_{k}-R_{k}}}\right)\frac{a_{j}a_{k}}{n_{j}n_{k}}s(T\log(n_{k}/n_{j}))=:S_{1}+S_{2}+S_{3}.$$

To get the bound (19) we will prove the following three statements.

(i) There is $n_{0}=n_{0}(\eta)$ such that $|S_{3}|<\eta/3$ for all $T$;

(ii) there is $T_{1}=T_{1}(\eta)$ such that $|S_{2}|\leq\eta/3$ for $T>T_{1}$ and all $n_{0}$;

(iii) for any given $n_{0}$ there is $T_{2}=T_{2}(\eta,n_{0})$ such that $|S_{1}|\leq\eta/3$ for $T>T_{2}$.

Proof of (i). For $S_{3}$ use $|s(x)|\leq 1.$ Then, for all $T$,

$$\begin{split}&|S_{3}|\leq\sum_{n_{0}\leq n_{k}\leq N}\frac{a_{k}}{n_{k}}\sum_{n_{k}-R_{k}\leq n_{j}<n_{k}}\frac{a_{j}}{n_{j}}\\ &\ll\sum_{n_{0}\leq n_{k}<N}\frac{a_{k}(A(n_{k})-A(n_{k}-R_{k}))}{n_{k}(n_{k}-R_{k})}\\ &\ll\sum_{n_{0}\leq n_{k}}\frac{a_{k}}{(n_{k}-R_{k})(\log n_{k})^{3/2+\varepsilon}}.\end{split}$$

The last sum converges if and only if the sum

(20) $$\sum_{n_{0}\leq n_{k}}\frac{a_{k}}{n_{k}(\log n_{k})^{3/2+\varepsilon}}$$

converges. Using Abel’s summation formula we obtain

$$\begin{split}&\sum_{n_{0}\leq n_{k}\leq N}\frac{a_{k}}{n_{k}(\log n_{k})^{3/2+\varepsilon}}\ll\frac{1}{(\log N)^{3/2+\varepsilon}}+\frac{1}{(\log n_{0})^{3/2+\varepsilon}}\\ &+\int_{n_{0}}^{N}\frac{1}{u(\log(u))^{3/2+\varepsilon}}du\ll 1.\end{split}$$

Thus, the sum (20) converges. Hence, we can pick $n_{0}$ large enough such that, for all $T$,

$$|S_{3}|<\frac{\eta}{3}.$$

This proves (i).

Proof of (ii). For $S_{2}$ use $|s(x)|\leq x^{-1}$ and $\log(n_{k}/n_{j})\geq(n_{k}-n_{j})/n_{k}.$ Then

$$\begin{split}&|S_{2}|\leq\frac{1}{T}\sum_{n_{k}\leq N}\frac{a_{k}}{n_{k}}\sum_{n_{j}\leq n_{k}-R_{k}}\frac{a_{j}n_{k}}{n_{j}(n_{k}-n_{j})}\\ &\ll\frac{1}{T}\sum_{n_{k}\leq N}\frac{a_{k}}{n_{k}}\left(\sum_{n_{j}\leq n_{k}/2}\frac{a_{j}}{n_{j}}+\sum_{n_{k}/2<n_{j}\leq n_{k}-R_{k}}\frac{a_{j}}{n_{k}-n_{j}}\right).\end{split}$$

Using Abel’s summation formula twice we obtain

$$\begin{split}&\frac{1}{T}\sum_{n_{k}\leq N}\frac{a_{k}}{n_{k}}\sum_{n_{j}\leq n_{k}/2}\frac{a_{j}}{n_{j}}\ll\frac{1}{T}\sum_{n_{k}\leq N}\frac{a_{k}}{n_{k}}\left(\frac{2A(n_{k}/2)}{n_{k}}+\int_{1}^{n_{k}}\frac{A(u)}{u^{2}}du\right)\\ &\ll\frac{1}{T}\sum_{n_{k}\leq N}\frac{a_{k}\log n_{k}}{n_{k}}\ll\frac{\log N}{T}\left(\frac{A(N)}{N}+\int_{1}^{N}\frac{A(u)}{u^{2}}du\right)\ll\frac{(\log N)^{2}}{T}.\end{split}$$

Now we estimate

$$\sum_{n_{k}/2<n_{j}\leq n_{k}-R_{k}}\frac{a_{j}}{n_{k}-n_{j}}.$$

Let $K(k)$ be the unique integer such that $n_{k}/4\leq 2^{K(k)}R_{k}<n_{k}/2.$ Thus

$$K(k)=\frac{3/2+\varepsilon}{\log 2}\log\log n_{k}+O(1)\ll\log\log n_{k}.$$

Then

$$\begin{split}&\sum_{n_{k}/2<n_{j}\leq n_{k}-R_{k}}\frac{a_{j}}{n_{k}-n_{j}}\\ &=\sum_{r=1}^{K(k)}\sum_{2^{r-1}R_{k}\leq n_{k}-n_{j}<2^{r}R_{k}}\frac{a_{j}}{n_{k}-n_{j}}+\sum_{2^{K(k)}R_{k}\leq n_{k}-n_{j}\leq n_{k}/2}\frac{a_{j}}{n_{k}-n_{j}}\\ &\ll\sum_{r=1}^{K(k)}\frac{A(n_{k}-2^{r-1}R_{k})-A(n_{k}-2^{r}R_{k})}{2^{r-1}R_{k}}+\frac{A(3n_{k}/4)-A(n_{k}/2)}{2^{K(k)}R_{k}}\\ &\ll\sum_{r=1}^{K(k)}\frac{2^{r}R_{k}}{2^{r-1}R_{k}}+1\ll K(k)\ll\log\log n_{k}.\end{split}$$

Thus,

(21) $$|S_{2}|\ll\frac{(\log N)^{2}}{T}+\frac{1}{T}\sum_{n_{k}\leq N}\frac{a_{k}\log\log n_{k}}{n_{k}}\ll\frac{(\log N)^{2}}{T}$$

uniformly in $n_{0}$. This proves the statement (ii), since $(\log N)^{2}\prec T$.

The statement (iii) follows by the bound $S_{1}\ll_{n_{0}}T^{-1}.$

We have proven (19) and this proves the lemma. ∎

Lemma 5.

We have

$$\frac{1}{T}\int_{1}^{T}|f(1+it)-f_{N}(1+it)|^{2}dt\ll\frac{1}{T^{2}}+\frac{1}{(\log N)^{3+2\varepsilon}}+\frac{T}{(\log N)^{2(1+\varepsilon)}}$$

uniformly in $N>e^{e}$.

Proof.

Assume that $N>e^{e}.$ For $\sigma>1$

(22) $$f(s)-f_{N}(s)=\int_{N}^{\infty}\frac{1}{x^{s}}dA(x)=-\frac{\varrho N^{1-s}}{1-s}-\frac{R(N)}{N^{s}}+s\int_{N}^{\infty}\frac{R(x)}{x^{s+1}}dx.$$

The integral

$$\int_{N}^{\infty}\frac{dx}{x(\log x)^{3/2+\varepsilon}}\ll\frac{1}{(\log N)^{1/2+\varepsilon}}$$

converges. Thus, the equation (22) holds when $\sigma=1$ and gives the continuous extension of $f(s)$ to $\sigma\geq 1$. Then, for $t\in[1,2T]$ we have

(23) $$|f(1+it)-f_{N}(1+it)|\ll\frac{1}{t}+\frac{1}{(\log N)^{3/2+\varepsilon}}+T\left|\int_{N}^{\infty}\frac{R(x)}{x^{2+it}}dx\right|.$$

Using the inequality $(a+b+c)^{2}\leq 3(a^{2}+b^{2}+c^{2})$ we get

$$\frac{1}{T}\int_{1}^{T}|f(1+it)-f_{N}(1+it)|^{2}dt\ll\frac{1}{T^{2}}+\frac{1}{(\log N)^{3+2\varepsilon}}+T\int_{1}^{T}\left|\int_{N}^{\infty}\frac{R(x)}{x^{2+it}}dx\right|^{2}dt.$$

Therefore, it is enough to evaluate

$$I(T)=\int_{1}^{2T}\left|\int_{N}^{\infty}\frac{R(x)}{x^{2+it}}dx\right|^{2}dt.$$

To do so we use the observation that the Fourier transform of

$$\frac{1}{4T}\left(1-\frac{|t|}{4T}\right)_{+}=\frac{1}{4T}\max\left\{1-\frac{|t|}{4T},0\right\}$$

is

(24) $$s^{2}(2Tu)=\left(\frac{\sin(2Tu)}{2Tu}\right)^{2}.$$

Define

$$J(T)=\frac{1}{4T}\int_{-4T}^{4T}\left(1-\frac{|t|}{4T}\right)_{+}\left|\int_{N}^{\infty}\frac{R(x)}{x^{2+it}}dx\right|^{2}dt.$$

Note that $I(T)\leq 4TJ(T).$ Now square out the integral in $J(T),$ interchange the order of integration and use the observation to see that

$$J(T)\ll\int_{N}^{\infty}\int_{N}^{\infty}\frac{|R(x)R(y)|}{(xy)^{2}}s^{2}(2T\log\frac{x}{y})dydx.$$

By symmetry, it suffices to bound the integral over the domain $N\leq y\leq x.$ Split the integral over $y$ in three parts: $N\leq y\leq\max\{N,x/2\},$ $\max\{N,x/2\}\leq y\leq\max\{N,x-x/T\}$, and $\max\{N,x-x/T\}\leq y\leq x.$ Use $s^{2}(u)\leq\min\{1,u^{-2}\}$ and $\log(x/y)\geq(x-y)/x.$ Then the contribution of the first range is bounded by

$$\begin{split}&\ll\frac{1}{T^{2}}\int_{N}^{\infty}\frac{1}{x(\log x)^{3/2+\varepsilon}}\int_{N}^{x/2}\frac{x^{2}}{y(x-y)^{2}(\log y)^{3/2+\varepsilon}}dydx\\ &\ll\frac{1}{T^{2}}\int_{N}^{\infty}\frac{1}{x(\log x)^{3/2+\varepsilon}}\int_{N}^{x/2}\frac{1}{y(\log y)^{3/2+\varepsilon}}dydx\ll\frac{1}{T^{2}(\log N)^{1+2\varepsilon}}.\end{split}$$

The second range gives

$$\begin{split}&\ll\frac{1}{T^{2}}\int_{N}^{\infty}\frac{1}{x(\log x)^{3/2+\varepsilon}}\int_{x/2}^{x-x/T}\frac{x^{2}}{y(x-y)^{2}(\log y)^{3/2+\varepsilon}}dydx\\ &\ll\frac{1}{T}\int_{N}^{\infty}\frac{dx}{x(\log x)^{3+2\varepsilon}}\ll\frac{1}{T(\log N)^{2(1+\varepsilon)}},\end{split}$$

here we used the substitution $u=x/(x-y).$ The last range yields

$$\ll\int_{N}^{\infty}\frac{1}{x(\log x)^{3/2+\varepsilon}}\int_{x-x/T}^{x}\frac{1}{y(\log y)^{3/2+\varepsilon}}dydx\ll\frac{1}{T}\int_{N}^{\infty}\frac{dx}{x(\log x)^{3+2\varepsilon}}$$

$$\ll\frac{1}{T(\log N)^{2(1+\varepsilon)}}.$$

Combining everything we get the desired bound. ∎

Proof of Proposition 3.

The continuous extension of $f(s)$ is obtained at the beginning of the proof of Lemma 5.

Next, we rewrite the LHS of (17) as

$$\frac{1}{T}\int_{0}^{T}|f(1+it)|^{2}dt= \frac{1}{T}\int_{0}^{T}|f_{N}(1+it)|^{2}dt+\int_{0}^{T}|f(1+it)-f_{N}(1+it)|^{2}dt$$

(25) $$+2\Re\left(\frac{1}{T}\int_{0}^{T}f_{N}(1+it)(\overline{f(1+it)-f_{N}(1+it)})dt\right),$$

where the last term is bounded by

$$2\left(\frac{1}{T^{2}}\int_{0}^{T}|f_{N}(1+it)|^{2}dt\int_{0}^{T}|f(1+it)-f_{N}(1+it)|^{2}dt\right)^{1/2}.$$

Then choosing $N$ such that $(\log N)^{2}\prec T\prec(\log N)^{2(1+\varepsilon)}$ by Lemmas 4 and 5 we obtain the moment (17).

The series on the RHS of (17) converges because

$$\sum_{j=1}^{\infty}\frac{a^{2}_{j}}{n_{j}^{2}}\ll\sum_{j=1}^{\infty}\frac{a_{j}(A(n_{j})-A(n_{j}-1))}{n_{j}^{2}}\ll\sum_{n=1}^{\infty}\frac{a_{j}}{n_{j}(\log n_{j})^{3/2+\varepsilon}}.$$

This concludes the proposition. ∎

3. Proofs of Theorem 1 and Proposition 2

Proof of Proposition 2.

Equality (12) follows from Proposition 3.

The series on the RHS of (13) converges because, using (10), we get

$$\sum_{n=1}^{\infty}\frac{e^{2}_{n}}{x_{n}^{2}}\leq\sum_{n=1}^{\infty}\frac{d^{2}_{n}}{x_{n}^{2}}.$$

Next, we prove equality (13). Let

$$\zeta_{\mathcal{P}N}(s)=\sum_{x_{n}\leq N}\frac{d_{n}}{x^{s}_{n}},\ (\zeta_{\mathcal{P}}^{-1})_{N}(s)=\sum_{x_{n}\leq N}\frac{e_{n}}{x^{s}_{n}},$$

where $(\zeta_{\mathcal{P}}^{-1})_{N}(s)$ denotes the partial Dirichlet sum of $1/\zeta_{\mathcal{P}}(s)$. To prove (13) we will show that for $(\zeta_{\mathcal{P}}^{-1})_{N}(1+it)$ and $1/\zeta_{\mathcal{P}}(1+it)-(\zeta_{\mathcal{P}}^{-1})_{N}(1+it)$ the statements analogous to Lemmas 4 and 5 hold.

To get the analog of Lemma 4 we use the equality

$$\frac{1}{T}\int_{0}^{T}|(\zeta_{\mathcal{P}}^{-1})_{N}(1+it)|^{2}dt=\sum_{\nu_{j}\leq N}\frac{e^{2}_{j}}{\nu_{j}^{2}}+2\sum_{\nu_{j}<\nu_{k}\leq N}\frac{e_{j}e_{k}}{\nu_{j}\nu_{k}}s(T\log(\nu_{k}/\nu_{j}))$$

and by (10) we get

$$\left|\sum_{\nu_{j}<\nu_{k}\leq N}\frac{e_{j}e_{k}}{\nu_{j}\nu_{k}}s(T\log(\nu_{k}/\nu_{j}))\right|\leq\sum_{x_{j}<x_{k}\leq N}\frac{d_{j}d_{k}}{\nu_{j}\nu_{k}}|s(T\log(\nu_{k}/\nu_{j}))|.$$

Then in the same way as in Lemma 4 we get that

$$\sum_{x_{j}<x_{k}\leq N}\frac{d_{j}d_{k}}{\nu_{j}\nu_{k}}|s(T\log(\nu_{k}/\nu_{j}))|=\sum_{x_{j}<x_{k}\leq N}\frac{d_{j}d_{k}}{\nu_{j}\nu_{k}}s(T\log(\nu_{k}/\nu_{j}))\rightarrow 0,$$

if $(\log N)^{2}\prec T.$ Hence,

(26) $$\lim_{T\rightarrow\infty}\frac{1}{T}\int_{0}^{T}|(\zeta_{\mathcal{P}}^{-1})_{N}(1+it)|^{2}dt=\sum_{n=1}^{\infty}\frac{e^{2}_{n}}{\nu_{n}^{2}}$$

as $N,T\rightarrow\infty$ such that $(\log N)^{2}\prec T.$

We consider the statement analogous to Lemma 5. By the discussion below Theorem 1 in [dgn2019] we have that

(27) $$E(x):=\sum_{\nu_{n}\leq x}e_{n}\ll x\exp\left(-c\sqrt{\log x\log\log x}\right)\quad(c>0).$$

Clearly, $x\exp\left(-c\sqrt{\log x\log\log x}\right)\ll x(\log x)^{-3/2-\varepsilon}$. Similarly to (23) we obtain

(28) $$\left|\zeta_{\mathcal{P}}(1+it)^{-1}-(\zeta_{\mathcal{P}}^{-1})_{N}(1+it)\right|=-\frac{E(N)}{N^{1+it}}+T\left|\int_{N}^{\infty}\frac{E(x)}{x^{2+it}}dx\right|.$$

Further, following the proof of Lemma 5 (replace $R(x)$ by $E(x)$) we arrive at

(29) $$\frac{1}{T}\int_{1}^{T}|\zeta_{\mathcal{P}}(1+it)^{-1}-(\zeta_{\mathcal{P}}^{-1})_{N}(1+it)|^{2}dt\ll\frac{T}{(\log N)^{2(1+\varepsilon)}}.$$

By (26) and (29) similarly as in the proof of Theorem 3 we derive (13).

Before concluding the proof, we note that the sharper bound (27) (compared to (16)) enables us to derive a bound of the type (29) without utilizing the Fourier transform (24). Indeed, from (28) by (27) we obtain

$$|\zeta_{\mathcal{P}}(1+it)^{-1}-(\zeta_{\mathcal{P}}^{-1})_{N}(1+it)|\ll T\exp\left(-c\sqrt{\log N}\right)\sqrt{\log N}$$

and

$$\frac{1}{T}\int_{1}^{T}|\zeta_{\mathcal{P}}(1+it)^{-1}-(\zeta_{\mathcal{P}}^{-1})_{N}(1+it)|^{2}dt\ll T^{2}\exp\left(-2c\sqrt{\log N}\right)\log N.$$

Then choosing $T=(\log N)^{3}$ we again derive (13). Similarly, the moment (12) can be proved without using the Fourier transform. ∎

Proof of Theorem 1.

Theorem 1 follows from Proposition 2. The proof is similar to that of Theorem 3 in [drungilas2021], but we include a shortened version here for the reader’s convenience.

We write (see (4))

$$Z(s)=Z_{1}(s)Z_{2}(s),$$

where $Z_{1}$ is the Ruelle zeta function and $Z_{2}(s)=Z(s+1).$

The function $Z_{2}$ is bounded on the line $\sigma=1,$ thus there exists a constant $C>0$ such that

$$|Z_{2}(1+it)|^{\pm 2}\leq C,$$

for all $t\in\mathbb{R}.$ Also, the Dirichlet series representation of this function is absolutely convergent at $\sigma=1.$

We are going to show that the series

$$\sum_{n=1}^{\infty}\frac{b^{2}_{n}}{y_{n}^{2}}\quad\text{and}\quad\sum_{n=1}^{\infty}\frac{c^{2}_{n}}{y_{n}^{2}}$$

on the right-hand side of (2) and (3) converge. We start with the second one.

For $\sigma>1,$ write

$$\frac{1}{Z_{1}(s)}=\sum_{n=1}^{\infty}\frac{c^{\prime}_{n}}{x_{n}^{s}}\quad\text{and}\quad\frac{1}{Z_{2}(s)}=\sum_{n=1}^{\infty}\frac{c^{\prime\prime}_{n}}{y_{n}^{s}},$$

Also let

$$(Z^{-1}_{1})_{N}(s)=\sum_{x_{n}\leq N}\frac{c^{\prime}_{n}}{x_{n}^{s}}$$

be the partial sum of $1/Z_{1}(s)$.

Let

$$c_{n}(N)=\sum_{x_{j}y_{k}=y_{n}\atop x_{j}\leq N}c^{\prime}_{j}c^{\prime\prime}_{k}.$$

Then,

$$0\leq c_{n}(N)\leq\sum_{x_{j}y_{k}=y_{n}}c^{\prime}_{j}c^{\prime\prime}_{k}=c_{n}.$$

In view of [landau1909, Satz 29 in §221], we obtain

(30) $$\begin{split}&\sum_{y_{n}\leq N}\frac{c^{2}_{n}}{y_{n}^{2}}+\sum_{y_{n}>N}\frac{c^{2}_{n}(N)}{y_{n}^{2}}=\lim_{T\rightarrow\infty}\frac{1}{T}\int_{1}^{T}|Z_{1N}(1+it)Z_{2}(1+it)|^{-2}dt\leq\\ &C\sum_{n=1}^{\infty}\frac{c^{\prime 2}_{n}}{x_{n}^{2}}.\end{split}$$

By Proposition 2, the series on the far right-hand side above converges, therefore the series on the left-hand side also converges. Thus

$$\lim_{N\rightarrow\infty}\lim_{T\rightarrow\infty}\frac{1}{T}\int_{1}^{T}|Z_{1N}(1+it)Z_{2}(1+it)|^{-2}dt=\sum_{n=1}^{\infty}\frac{c^{2}_{n}}{y_{n}^{2}}$$

and the series on the right-hand side converges. Using inequalities (5), $|b_{n}|\leq c_{n}$ and reasoning as in (30) we have that

$$\lim_{N\rightarrow\infty}\lim_{T\rightarrow\infty}\frac{1}{T}\int_{1}^{T}|Z_{1N}(1+it)Z_{2}(1+it)|^{2}dt=\sum_{n=1}^{\infty}\frac{b^{2}_{n}}{y_{n}^{2}}$$

and that the series on the right-hand side converges.

Similarly as in (2), we get

$$\begin{split}&\frac{1}{T}\left(\int_{1}^{T}|Z(1+it)|^{-2}dt-\int_{1}^{T}|Z_{1N}(1+it)Z_{2}(1+it)|^{-2}dt\right)\leq\\ &\frac{C}{T}\int_{1}^{T}|Z(1+it)^{-1}-(Z^{-1}_{1})_{N}(1+it)|dt+\\ &\frac{2C}{T}\Re\int_{1}^{T}\left(Z(1+it)^{-1}-(Z^{-1}_{1})_{N}(1+it))\overline{(Z^{-1}_{1})_{N}(1+it)}\right)dt.\end{split}$$

Using the Cauchy-Schwarz inequality, Lemmas 4 and 5, and choosing $N$ such that $(\log N)^{2}\prec T\prec(\log N)^{2(1+\varepsilon)}$, we see that the left-hand side of the above inequality goes to 0 as $T\rightarrow\infty.$ Therefore, (3) is proven. To prove (2), we argue likewise, using (26) and (29) instead of Lemmas 4 and 5. ∎

Acknowledgment. This work is funded by the Research Council of Lithuania (LMTLT), agreement No. S-MIP-22-81.