2 Background

Let $K/{\mathbb {Q}}$ be a non-Galois cubic field, and let $\widehat {K}$ be the Galois closure of K over $\mathbb {Q}$ . Then, the Dedekind zeta function of the field K has the decomposition

$$ \begin{align*} \zeta_K(s) = \zeta(s) L(s,\rho,\widehat{K}/\mathbb{Q}), \end{align*} $$

where $L(s,\rho ,\widehat {K}/\mathbb {Q})$ is the Artin L-function associated to the two-dimensional representation $\rho $ of $\mathrm {Gal}(\widehat K/{\mathbb {Q}}) \simeq S_3$ . The strong Artin conjecture is known for such representations; in this particular case, we have an explicit underlying cuspidal representation $\tau $ of $GL_2/\mathbb {Q}$ , such that $L(s,\rho ,\widehat {K}/\mathbb {Q})=L(s,\tau )$ . For the sake of completeness, let us describe $\tau $ in more detail. Let $F=\mathbb {Q}[\sqrt {D_K}]$ , and let $\chi $ be a nontrivial character of Gal $(\widehat {K}/F)\simeq C_3$ , considered as a Hecke character of F. Then $\tau =\text {Ind}^{\mathbb Q}_{F} \chi $ is a dihedral representation of central character $\chi _{D_K}=\big (\frac {D_K}\cdot \big )$ . When $D_K <0$ , $\tau $ corresponds to a weight one newform of level $\left | D_K \right |$ and nebentypus $\chi _{D_K}$ , and when $D_K>0$ , it corresponds to a weight zero Maass form (see [Reference Duke, Friedlander and IwaniecDFI, Introduction]). In both cases, we will denote the corresponding automorphic form by $f_K$ , and, in particular, we have the equality

$$ \begin{align*}L(s,\rho,\widehat{K}/\mathbb{Q}) = L(s,f_K).\end{align*} $$

We are interested in the analytic properties of $\zeta _K(s)/\zeta (s)=L(s,f_K)$ . We have the functional equation

(2.1) $$ \begin{align} \Lambda(s,f_K)=\Lambda(1-s,f_K). \end{align} $$

Here, $\Lambda (s,f_K):= |D_K|^{\frac s2} \Gamma _{f_K}(s) L(s,f_K)$ is the completed L-function, with the gamma factor

$$ \begin{align*}\Gamma_{f_K}(s)=\begin{cases} \Gamma_+(s) & \text{ if } D_K>0 \text{ } (\text{that is } K \text{ has signature } (3,0) ); \\ \Gamma_-(s) & \text{ if } D_K < 0 \text{ } (\text{that is } K \text{ has signature } (1,1) ) , \end{cases}\end{align*} $$

where $\Gamma _+(s):= \pi ^{-s}\Gamma (\frac s2)^2$ and $\Gamma _-(s):= \pi ^{-s}\Gamma (\frac s2)\Gamma (\frac {s+1}2)$ .

The coefficients of $L(s,f_K)$ have an explicit description in terms of the splitting type of the prime ideal $(p)\mathcal O_{K}$ . Writing

$$ \begin{align*}L(s,f_K) = \sum_{n =1}^\infty \frac{\lambda_K (n)}{n^s},\end{align*} $$

we have that

where

$$ \begin{align*}\tau_e:= \begin{cases} 1 &\text{ if } e\equiv 0 \bmod 3; \\ -1 &\text{ if } e\equiv 1 \bmod 3; \\ 0 &\text{ if } e\equiv 2 \bmod 3. \end{cases}\end{align*} $$

Furthermore, we find that the coefficients of the reciprocal

(2.2) $$ \begin{align} \frac{1}{L(s,f_K)} = \sum_{n=1}^\infty \frac{ \mu_K (n)}{n^s} \end{align} $$

are given by

$$ \begin{align*} \mu_{K}(p^k) = \begin{cases} -\lambda_K(p) & \mbox{ if } k = 1;\\ \Big( \frac{D_K} p\Big) & \mbox{ if } k = 2;\\ 0 & \mbox{ if } k> 2. \end{cases} \end{align*} $$

The remaining values of $\lambda _K(n)$ and $\mu _K(n)$ are determined by multiplicativity. Finally, the coefficients of the logarithmic derivative

$$ \begin{align*}-\frac{L'}{L}(s,f_K)=\sum_{n\geq 1} \frac{\Lambda(n)a_K(n)}{n^s}\end{align*} $$

are given by

where

$$ \begin{align*}\eta_e:= \begin{cases} 2 &\text{ if } e \equiv 0 \bmod 3; \\ -1 &\text{ if } e \equiv \pm 1 \bmod 3. \end{cases}\end{align*} $$

We now describe explicitly the constants $A_p^\pm (T)$ and $B_p^\pm (T)$ that appear in (1.2). More generally, let $\mathbf {p} = ( p_1 , \ldots , p_J)$ be a vector of primes and let $\mathbf {k} = ( k_1 ,\ldots , k_J )\in \{1,2,3,4,5\}^J $ (when $J=1$ , $\mathbf {p}=(p)$ is a scalar, and we will abbreviate by writing $\mathbf {p}=p$ and similarly for $\mathbf {k}$ ). We expect that

(2.3) $$ \begin{align} N^\pm_{\mathbf{p}} (X,T_{\mathbf{k}}) : &= \#\{ K \in \mathcal F^\pm(X) : p_j \text{ has splitting type }T_{k_j} \text{ in }K \; (1\leq j\leq J)\} \notag \\& = A^\pm_{\mathbf{p}} (T_{\mathbf{k}} ) X +B^\pm_{\mathbf{p}}(T_{\mathbf{k}}) X^{\frac 56} + O_{\varepsilon}((p_1 \cdots p_J )^{ \omega}X^{\theta+\varepsilon}), \end{align} $$

for some $\omega \geq 0$ and with the same $\theta $ as in (1.1). Here,

$$ \begin{align*}A^\pm_{\mathbf{p}} ( T_{\mathbf{k}} ) = C_1^\pm \prod_{j = 1}^J ( x_{p_j} c_{k_j} (p_j ) ),\hspace{1cm} B^\pm_{\mathbf{p}} ( T_{\mathbf{k}} ) = C_2^\pm \prod_{j = 1}^J ( y_{p_j} d_{k_j} (p_j ) ) ,\end{align*} $$

$$ \begin{align*}x_p:=\Big(1+\frac 1p+\frac 1{p^2}\Big)^{-1}, \hspace{2cm} y_p:=\frac{1-p^{-\frac 13}}{(1-p^{-\frac 53})(1+p^{-1})},\end{align*} $$

and $c_k(p)$ and $d_k(p)$ are defined in the following table:

Recently, Bhargava, Taniguchi and Thorne [Reference Bhargava, Taniguchi and ThorneBTT] have shown that the values $\theta =\omega =\frac {2}3$ are admissible in (2.3).

3 New lower-order terms in the one-level density

In this section, we shall estimate the one-level density

$$ \begin{align*}\frac 1{N^\pm(X)} \sum_{K \in \mathcal{F}^\pm(X)} \mathfrak{D}_{\phi}(K)\end{align*} $$

assuming the cubic field count (1.2) for some fixed parameters $\frac 12\leq \theta <\frac 56$ and $ \omega \geq 0$ . Throughout the paper, we will use the shorthand

$$ \begin{align*}L =\log \Big(\frac X{(2\pi e)^2}\Big).\end{align*} $$

The starting point of this section is the explicit formula.

Lemma 3.1. Let $\phi $ be a real even Schwartz function whose Fourier transform is compactly supported, and let $K\in \mathcal {F}^\pm (X)$ . We have the formula

(3.1) $$ \begin{align} \mathfrak{D}_\phi(K) = \sum_{\gamma_K}\phi\Big(\frac{L\gamma_K}{2\pi}\Big)=& \frac{\widehat\phi(0)}{L } \log{|D_K|} + \frac1{\pi}\int_{-\infty}^{\infty}\phi\Big(\frac{Lr}{2\pi}\Big)\operatorname{Re}\Big(\frac{\Gamma'_{\pm}}{\Gamma_{\pm}}(\tfrac12+ir)\Big)dr \nonumber\\ & -\frac2{L }\sum_{n=1}^{\infty}\frac{\Lambda(n)}{\sqrt n}\widehat\phi\Big(\frac{\log n}{L}\Big) a_K(n), \end{align} $$

where $\rho _K=\frac 12+i\gamma _K$ runs over the nontrivial zeros of $L(s,f_K)$ .

Proof. This follows from, for example, [Reference Rudnick and SarnakRS, Proposition 2.1], but for the sake of completeness, we reproduce the proof here. By Cauchy’s integral formula, we have the identity

$$ \begin{align*} \sum_{\gamma_K}\phi\Big(\frac{L\gamma_K}{2\pi}\Big) &=\frac{1}{2\pi i} \int_{(\frac 32)}\phi\left( \frac{L}{2\pi i} \left( s - \frac{1}{2} \right)\right) \frac{\Lambda'}{\Lambda}(s,f_K)ds \\ &\quad-\frac{1}{2\pi i} \int_{(-\frac 12)}\phi\left( \frac{L}{2\pi i} \left( s - \frac{1}{2} \right)\right) \frac{\Lambda'}{\Lambda}(s,f_K) ds. \end{align*} $$

These integrals converge since $\phi \left (\frac {L}{2 \pi i}\left ( s-\frac {1}{2} \right ) \right )$ is rapidly decreasing in vertical strips. For the second integral, we apply the change of variables $s \rightarrow 1-s$ . Then, by the functional equation in the form $ \frac {\Lambda '}{\Lambda }(1-s,f_K)=-\frac {\Lambda '}{\Lambda }(s,f_K)$ and since $\phi (-s)=\phi (s)$ , we deduce that

$$ \begin{align*} \sum_{\gamma_K}\phi\Big(\frac{L\gamma_K}{2\pi}\Big)=\frac{1}{\pi i} \int_{(\frac 32)} \phi\left( \frac{L}{2\pi i} \left( s - \frac{1}{2} \right)\right) \frac{\Lambda'}{\Lambda}(s,f_K)ds. \end{align*} $$

Next, we insert the identity

$$ \begin{align*}\frac{\Lambda'}{\Lambda}(s,f_K)=\frac{1}{2}\log |D_K| + \frac{\Gamma_{f_K}'}{\Gamma_{f_K}}(s) - \sum_{n\geq 1}\frac{\Lambda(n)a_K(n)}{n^s}\end{align*} $$

and separate into three integrals. By shifting the contour of integration to $\operatorname {Re}(s)=\frac 12$ in the first two integrals, we obtain the first two terms on the right-hand side of (3.1). The third integral is equal to

$$ \begin{align*} -2\sum_{n\geq 1} \frac{\Lambda(n)a_K(n)}{\sqrt{n}}\frac{1}{2\pi i} \int_{(\frac 32)} \phi\left( \frac{L}{2\pi i} \left( s - \frac{1}{2} \right)\right)n^{-(s-\frac 12)}ds. \end{align*} $$

By moving the contour to $\operatorname {Re}(s)=\frac 12$ and applying Fourier inversion, we find the third term on the right-hand side of (3.1) and the claim follows.

Our goal is to average (3.1) over $K \in \mathcal {F}^\pm (X)$ . We begin with the first term.

Lemma 3.2. Assume that (1.1) holds for some $0\leq \theta <\frac 56$ . Then, we have the estimate

$$ \begin{align*}\frac{1}{ N^\pm(X)} \sum_{K \in \mathcal{F}^\pm(X)} \log{|D_K|} = \log X - 1 -\frac{C_2^\pm}{5C_1^\pm} X^{-\frac 16} + \frac{(C_2^\pm)^2}{ 5 (C_1^\pm)^2 } X^{- \frac13} +O_{\varepsilon}(X^{\theta-1+\varepsilon}+X^{-\frac 12}).\end{align*} $$

Proof. Applying partial summation, we find that

$$ \begin{align*} \sum_{K \in \mathcal{F}^\pm(X)} \log |D_K| = \int_1^X (\log t) dN^\pm(t) = N^\pm(X)\log X-N^\pm(X)-\frac 15 C_2^\pm X^{\frac 56} +O_{\varepsilon}(X^{\theta+\varepsilon}). \end{align*} $$

The claimed estimate follows from applying (1.1).

For the second term of (3.1), we note that it is constant on $\mathcal {F}^\pm (X)$ . We can now concentrate our efforts on the average of the third (and most crucial) term

(3.2) $$ \begin{align} I^{\pm}(X;\phi) := -\frac2{LN^\pm(X) }\sum_{p} \sum_{e=1}^\infty \frac{\log p }{p^{e/2}} \widehat\phi\Big(\frac{e \log p}{L}\Big) \sum_{K \in \mathcal{F}^\pm(X)} a_K(p^e ). \end{align} $$

It follows from (1.2) that

(3.3) $$ \begin{align} \notag \sum_{K \in \mathcal{F}^\pm(X)} a_K(p^e)&=2N^\pm_p(X,T_1)+(1+(-1)^e)N^\pm_p(X,T_2)+ \eta_e N^\pm_p(X,T_3)+N^\pm_p(X,T_4)\\ &= C_1^\pm X (\theta_e+\tfrac 1p)x_p+ C_2^\pm X^{\frac 56} (1+p^{-\frac 13})(\kappa_e(p)+p^{-1}+p^{-\frac 43})y_p +O_{\varepsilon}(p^{\omega }X^{\theta +\varepsilon}), \end{align} $$

where

(3.4) $$ \begin{align} \theta_e:=\delta_{2\mid e}+\delta_{3\mid e}=\begin{cases} 2 &\text{ if } e\equiv 0 \bmod 6 \\ 0 &\text{ if } e\equiv 1 \bmod 6 \\ 1 &\text{ if } e\equiv 2 \bmod 6 \\ 1 &\text{ if } e\equiv 3 \bmod 6 \\ 1 &\text{ if } e\equiv 4 \bmod 6 \\ 0 &\text{ if } e\equiv 5 \bmod 6, \end{cases} \end{align} $$

and

(3.5) $$ \begin{align} \kappa_e(p):=(\delta_{2\mid e}+\delta_{3\mid e})(1+p^{-\frac 23}) + (1-\delta_{3\mid e})p^{-\frac 13}=\begin{cases} 2+2p^{-\frac 23} &\text{ if } e\equiv 0 \bmod 6 \\ p^{-\frac 13} &\text{ if } e\equiv 1 \bmod 6 \\ 1+p^{-\frac 13}+p^{-\frac 23} &\text{ if } e\equiv 2 \bmod 6 \\ 1+p^{-\frac 23} &\text{ if } e\equiv 3 \bmod 6 \\ 1+p^{-\frac 13}+p^{-\frac 23} &\text{ if } e\equiv 4 \bmod 6 \\ p^{-\frac 13} &\text{ if } e\equiv 5 \bmod 6. \end{cases} \end{align} $$

Here, $ \delta _{\mathcal P} $ is equal to $1$ if $ \mathcal P$ is true and is equal to $0$ otherwise. Note that we have the symmetries $\theta _{-e} = \theta _e$ and $\kappa _{-e}(p) = \kappa _e(p)$ . With this notation, we prove the following proposition.

Proposition 3.3. Let $\phi $ be a real even Schwartz function for which $\widehat \phi $ has compact support, and let $\sigma :=\sup (\mathrm {supp}(\widehat \phi ))$ . Assume that (1.2) holds for some fixed parameters $0\leq \theta <\frac 56$ and $\omega \geq 0$ . Then we have the estimate

$$ \begin{align*} &I^{\pm}(X;\phi)= -\frac{2}{L}\sum_{p,e}\frac{x_p\log p}{p^{\frac e2}}\widehat\phi\Big(\frac{\log p^e}{L}\Big) (\theta_e+\tfrac 1p) \\ &\quad + \frac{2}{L}\bigg( - \frac{ C_2^\pm }{C_1^\pm }X^{-\frac 16} + \frac{ ( C_2^\pm )^2 }{(C_1^\pm )^2 }X^{-\frac 13} \bigg) \sum_{p,e}\frac{\log p}{p^{\frac e2}}\widehat\phi\Big(\frac{\log p^e}{L}\Big) \beta_e(p) +O_{\varepsilon}(X^{\theta - 1 + \sigma(\omega + \frac12 ) +\varepsilon}+X^{-\frac 12+\frac{\sigma}6}), \end{align*} $$

where

(3.6) $$ \begin{align} \beta_e(p):=y_p (1+p^{-\frac 13})(\kappa_e(p) +p^{-1}+p^{-\frac 43})-x_p(\theta_e+\tfrac 1p). \end{align} $$

Proof. Applying (3.3), we see that

$$ \begin{align*} I^{\pm}(X;\phi) &= -\frac{2C_1^\pm X}{LN^\pm(X)}\sum_p \sum_{e=1}^\infty \frac{x_p\log p}{p^{\frac e2}}\widehat\phi\Big(\frac{\log p^e}{L}\Big) (\theta_e+\tfrac 1p)\\ &\quad- \frac{2C_2^\pm X^{\frac 56}}{LN^\pm(X)} \sum_p \sum_{e=1}^\infty \frac{y_p(1+p^{-\frac 13})\log p}{p^{\frac e2}}\widehat\phi\Big(\frac{\log p^e}{L}\Big) (\kappa_e(p) +p^{-1}+p^{-\frac 43}) \\ &\quad+O_{\varepsilon}\Big(X^{\theta - 1 +\varepsilon}\sum_{\substack{p^e \leq X^\sigma \\ e\geq 1}}p^{\omega -\frac e2}\log p \Big)\\ & = -\frac{2 }{L }\bigg( 1 - \frac{ C_2^\pm}{C_1^\pm} X^{- \frac16} + \frac{ (C_2^\pm)^2 }{(C_1^\pm)^2 } X^{- \frac13} \bigg) \sum_p \sum_{e=1}^\infty \frac{x_p\log p}{p^{\frac e2}}\widehat\phi\Big(\frac{\log p^e}{L}\Big) (\theta_e+\tfrac 1p)\\ &\quad - \frac{2}{L} \bigg( \frac{C_2^\pm }{C_1^\pm}X^{-\frac 16} - \frac{ (C_2^\pm)^2 }{(C_1^\pm)^2 } X^{- \frac13} \bigg) \sum_p \sum_{e=1}^\infty \frac{y_p(1+p^{-\frac 13})\log p}{p^{\frac e2}}\widehat\phi\Big(\frac{\log p^e}{L}\Big) (\kappa_e(p) +p^{-1}+p^{-\frac 43}) \\ &\quad +O_{\varepsilon}\big(X^{\theta - 1 + \sigma(\omega + \frac12 ) +\varepsilon} +X^{-\frac 12+\frac{\sigma}6}\big). \end{align*} $$

Note, in particular, that the error term $O(X^{-\frac 12+\frac {\sigma }6})$ bounds the size of the contribution of the first omitted term in the expansion of $X^{\frac 56}/N^{\pm }(X)$ appearing in the second double sum above. Indeed, this follows since $\kappa _1(p)=p^{-\frac 13}$ and

$$ \begin{align*} X^{-\frac12}\sum_{p\leq X^{\sigma}}\frac{\log p}{p^{\frac56}}=O\big(X^{-\frac 12+\frac{\sigma}6}\big). \end{align*} $$

The claimed estimate follows.

We shall estimate $ I^{\pm }(X;\phi )$ further and find asymptotic expansions for the double sums in Proposition 3.3.

Lemma 3.4. Let $\phi $ be a real even Schwartz function whose Fourier transform is compactly supported, define $\sigma :=\sup (\mathrm {supp}(\widehat \phi ))$ , and let $\ell $ be a positive integer. Define

$$ \begin{align*}I_1(X;\phi) := \sum_{p}\sum_{e=1}^\infty \frac{x_p\log p}{p^{\frac e2}}\widehat\phi\Big(\frac{\log p^e}{L}\Big) (\theta_e+\tfrac 1p) , \qquad I_2(X;\phi) := \sum_{p}\sum_{e=1}^\infty \frac{\log p}{p^{\frac e2}}\widehat\phi\Big(\frac{\log p^e}{L}\Big) \beta_e(p) .\end{align*} $$

Then, we have the asymptotic expansion

$$ \begin{align*}I_1(X;\phi) = \frac{\phi (0)}{4} L + \sum_{n=0 }^\ell \frac{ \widehat{\phi}^{(n)} (0) \nu_1 (n) }{ n!} \frac{ 1 }{L^n } + O_{\ell} \Big( \frac{1 }{L^{\ell+1} } \Big), \end{align*} $$

where

$$ \begin{align*} \nu_1 (n ) &:= \delta_{n=0} + \sum_{p}\sum_{ e \neq 2 } \frac{x_p e^n ( \log p)^{n+1} }{p^{\frac e2}} (\theta_e+\tfrac 1p) + \sum_p \frac{ 2^n ( \log p)^{n+1} }{p} \Big( x_p\Big( 1 + \frac1p\Big) -1 \Big) \\ &\quad + \int_1^\infty \frac{2^n ( \log u)^{n-1} ( \log u - n ) }{u^2} \mathcal{R} (u) du \end{align*} $$

with $ \mathcal {R}(u) := \sum _{ p \leq u } \log p - u $ . Moreover, we have the estimate

$$ \begin{align*}I_2(X;\phi) = L \int_0^\infty \widehat{\phi} ( u) e^{\frac{Lu}6} du + O\Big( X^{\frac \sigma 6} e^{-c_0(\sigma)\sqrt{\log X}}\Big),\end{align*} $$

where $c_0(\sigma )>0$ is a constant. Under the Riemann Hypothesis, we have the more precise expansion

$$ \begin{align*}I_2(X;\phi) = L \int_0^\infty \widehat{\phi} ( u) e^{\frac{Lu}{6}} du + \sum_{n=0}^\ell \frac{ \widehat{\phi}^{(n)} (0) \nu_2 (n) }{ n!} \frac{ 1 }{L^n } + O_{\ell} \Big( \frac{1 }{L^{\ell+1} } \Big),\end{align*} $$

where

$$ \begin{align*} \nu_2 (n ) &:= \delta_{n=0} + \sum_{p}\sum_{e=2}^\infty \frac{e^n (\log p )^{n+1} \beta_e(p) }{p^{\frac e2}} + \sum_{p} \frac{(\log p)^{n+1}}{p^{\frac 12}} \Big( \beta_1 (p) - \frac{1}{p^{\frac13}} \Big)\\ &\quad + \int_1^{\infty} \frac{( \log u)^{n-1} ( 5 \log u - 6n) }{6 u^{\frac{11}{6} }} \mathcal{R} (u) du. \end{align*} $$

Proof. We first split the sums as

(3.7) $$ \begin{align} I_1(X;\phi) = \sum_p \frac{\log p }{p} \widehat{\phi} \Big( \frac{ 2 \log p }{L} \Big) + I'_1(X;\phi), \qquad I_2(X;\phi) =\sum_p \frac{ \log p }{ p^{\frac56}}\widehat\phi\Big(\frac{\log p }{L}\Big) +I'_2(X;\phi) , \end{align} $$

where

(3.8) $$ \begin{align} I'_1(X;\phi) & := \sum_{p}\sum_{e \neq 2 } \frac{x_p\log p}{p^{\frac e2}}\widehat\phi\Big(\frac{\log p^e}{L}\Big) (\theta_e+\tfrac 1p) + \sum_p \frac{ \log p }{p} \Big( x_p\Big( 1 + \frac1p\Big) -1 \Big) \widehat{\phi} \Big( \frac{ 2 \log p }{L} \Big), \nonumber\\ I'_2(X;\phi) & := \sum_{p}\sum_{e=2}^\infty \frac{\log p}{p^{\frac e2}}\widehat\phi\Big(\frac{\log p^e}{L}\Big) \beta_e(p) + \sum_{p} \frac{\log p}{p^{\frac 12}}\widehat\phi\Big(\frac{\log p }{L}\Big) \Big( \beta_1 (p) - \frac{1}{p^{\frac13}} \Big). \end{align} $$

We may also rewrite the sums in (3.7) using partial summation as follows:

(3.9) $$ \begin{align} \sum_p \frac{\log p }{p} \widehat{\phi} \Big( \frac{ 2 \log p }{L} \Big) & = \int_1^\infty \frac{1 }{u} \widehat{\phi} \Big( \frac{ 2 \log u }{L} \Big)d( u+ \mathcal{R}(u)) \nonumber\\ & = \frac{ \phi(0) }{4} L + \widehat{\phi}(0) - \int_1^\infty \Big(\frac{-1}{u^2} \widehat{\phi} \Big( \frac{2 \log u }{L} \Big) + \frac{2}{u^2L} \widehat{\phi}' \Big( \frac{2 \log u }{L} \Big) \Big) \mathcal{R} (u) du , \nonumber\\ \sum_p \frac{\log p }{p^{\frac56} } \widehat{\phi} \Big( \frac{ \log p }{L} \Big) & = L \int_0^\infty \widehat{\phi} ( u) e^{Lu/6} du + \widehat{\phi}(0) \nonumber\\ & \quad - \int_1^{X^{\sigma}} \Big(\frac{-5}{6 u^2} \widehat{\phi} \Big( \frac{ \log u }{L} \Big) + \frac{ 1 }{u^2L} \widehat{\phi}' \Big( \frac{ \log u }{L} \Big) \Big) u^{\frac16} \mathcal{R} (u) du. \end{align} $$

Next, for any $\ell \geq 1$ and $ |t| \leq \sigma $ , Taylor’s theorem reads

(3.10) $$ \begin{align} \widehat{\phi} ( t ) = \sum_{n=0}^\ell \frac{ \widehat{\phi}^{(n)} (0) }{ n!} t^n + O_{\ell} ( |t|^{\ell+1}), \end{align} $$

and one has a similar expansion for $\widehat \phi '$ . The claimed estimates follow from substituting this expression into (3.8) and (3.9) and evaluating the error term using the prime number theorem $ \mathcal {R}(u)\ll u e^{-c \sqrt { \log u }} $ .

We end this section by proving Theorem 1.1.

Proof of Theorem 1.1

Assume that $\theta ,\omega \geq 0$ are admissible values in (1.2) and are such that $\theta +\omega <\frac 12$ . Let $\phi $ be any real even Schwartz function, such that $\widehat \phi \geq 0$ and $1< \sup (\mathrm {supp}(\widehat \phi )) < (\frac 56-\theta )/(\frac 13+\omega )$ ; this is possible thanks to the restriction $\theta +\omega <\frac 12$ . Combining Lemmas 3.1 and 3.2 with Proposition 3.3, we obtain the estimateFootnote ³

(3.11) $$ \begin{align} &\frac 1{N^\pm(X)} \sum_{K \in \mathcal{F}^\pm(X)} \mathfrak{D}_{\phi}(K)=\widehat \phi(0)\Big(1 + \frac{ \log (4 \pi^2 e) }{L} -\frac{C_2^\pm}{5C_1^\pm} \frac{X^{-\frac 16}}{L} + \frac{(C_2^\pm)^2 }{5(C_1^\pm)^2 } \frac{X^{-\frac 13}}{L} \Big) \nonumber\\ &\qquad\quad+ \frac1{\pi}\int_{-\infty}^{\infty}\phi\Big(\frac{Lr}{2\pi}\Big)\operatorname{Re}\Big(\frac{\Gamma'_{\pm}}{\Gamma_{\pm}}(\tfrac12+ir)\Big)dr -\frac{2}{L}\sum_{p,e}\frac{x_p\log p}{p^{\frac e2}}\widehat\phi\Big(\frac{\log p^e}{L}\Big) (\theta_e+\tfrac 1p)\nonumber\\ & -\frac{2C_2^\pm X^{-\frac 16}}{C_1^\pm L}\Big( 1-\frac{C_2^\pm}{C_1^\pm}X^{-\frac{1}{6}}\Big) \sum_{p,e}\frac{\log p}{p^{\frac e2}}\widehat\phi\Big(\frac{\log p^e}{L}\Big) \beta_e(p) +O_{\varepsilon}(X^{\theta - 1 + \sigma(\omega + \frac12 ) +\varepsilon}+X^{-\frac 12+\frac \sigma 6}), \end{align} $$

where $\sigma =\sup (\mathrm {supp}(\widehat \phi ))$ .

To bound the integral involving the gamma function in (3.11), we note that Stirling’s formula implies that for s in any fixed vertical strip minus discs centred at the poles of $\Gamma _\pm (s)$ , we have the estimate

$$ \begin{align*}\operatorname{Re} \left( \frac{ \Gamma'_{\pm} }{ \Gamma_{\pm}} ( s) \right) = \log | s | + O(1).\end{align*} $$

Now, $ \phi (x) \ll |x|^{-2}$ , and thus,

$$ \begin{align*}\frac{1 }{ \pi} \int_{-\infty}^\infty \phi \Big( \frac{ Lr}{ 2 \pi} \Big) \operatorname{Re} \Big( \frac{ \Gamma'_{\pm} }{ \Gamma_{\pm}} ( \tfrac12 + ir ) \Big) dr \ll \int_{-1}^1 \left| \phi \Big( \frac{ Lr}{ 2 \pi} \Big) \right| dr + \int_{ |r|\geq 1 } \frac{ \log (1+|r|)}{ (Lr)^2} dr \ll \frac1L.\end{align*} $$

Moreover, Lemma 3.4 implies the estimates

$$ \begin{align*}- \frac{2}{L}\sum_{p,e}\frac{x_p\log p}{p^{\frac e2}}\widehat\phi\Big(\frac{\log p^e}{L}\Big) (\theta_e+\tfrac 1p) \ll 1\end{align*} $$

and

$$ \begin{align*} &-\frac{2C_2^\pm X^{-\frac 16}}{C_1^\pm L}\Big( 1-\frac{C_2^\pm}{C_1^\pm}X^{-\frac{1}{6}}\Big) \sum_{p,e}\frac{\log p}{p^{\frac e2}}\widehat\phi\Big(\frac{\log p^e}{L}\Big) \beta_e(p) \\ &\qquad = -\frac{2C_2^\pm X^{-\frac 16}}{C_1^\pm } \int_0^\infty \widehat \phi(u) e^{\frac{Lu}{6}} du +O\big( X^{-\frac { 1}6}+X^{\frac{\sigma -2}{6}} \big), \end{align*} $$

since the Riemann Hypothesis for $\zeta _K(s)$ implies the Riemann Hypothesis for $\zeta (s)$ . Combining these estimates, we deduce that the right-hand side of (3.11) is

$$ \begin{align*}\leq -C_\varepsilon X^{\frac{\sigma-1}6-\varepsilon}+O_\varepsilon(1+X^{\frac{\sigma-1}6-\delta+\varepsilon}+X^{-\frac 13+\frac \sigma 6}),\end{align*} $$

where $\varepsilon>0$ is arbitrary, $C_\varepsilon $ is a positive constant and $\delta :=\frac {\sigma -1}6 -( \theta - 1 + \sigma (\omega + \frac 12 ))>0 $ . However, for small enough $\varepsilon $ , this contradicts the bound

$$ \begin{align*}\frac 1{N^\pm(X)} \sum_{K \in \mathcal{F}^\pm(X)} \mathfrak{D}_{\phi}(K) = O( \log X),\end{align*} $$

which is a direct consequence of the Riemann Hypothesis for $\zeta _K(s)$ and the Riemann-von Mangoldt formula [Reference Iwaniec and KowalskiIK, Theorem 5.31].

4 A refined Ratios Conjecture

The celebrated L-functions Ratios Conjecture [Reference Conrey, Farmer and ZirnbauerCFZ] predicts precise formulas for estimates of averages of ratios of (products of) L-functions evaluated at points close to the critical line. The conjecture is presented in the form of a recipe with instructions on how to produce predictions of a certain type in any family of L-functions. In order to follow the recipe, it is of fundamental importance to have control of counting functions of the type (1.1) and (2.3) related to the family. The connections between counting functions, low-lying zeros and the Ratios Conjecture are central in the present investigation.

The Ratios Conjecture has a large variety of applications. Applications to problems about low-lying zeros first appeared in the work of Conrey and Snaith [Reference Conrey and SnaithCS], where they study the one-level density of families of quadratic Dirichlet L-functions and quadratic twists of a holomorphic modular form. The investigation in [Reference Conrey and SnaithCS] has inspired a large amount of work on low-lying zeros in different families (see, e.g. [Reference MillerM2, Reference MillerM3, Reference Huynh, Keating and SnaithHKS, Reference Fiorilli and MillerFM, Reference David, Huynh and ParksDHP, Reference Fiorilli, Parks and SödergrenFPS1, Reference Fiorilli, Parks and SödergrenFPS3, Reference Mason and SnaithMS, Reference Cho and ParkCP, Reference WaxmanW]).

As part of this project, we went through the steps of the Ratios Conjecture recipe with the goal of estimating the one-level density. We noticed that the resulting estimate does not predict certain terms in Theorem 1.2. To fix this, we modified [Reference Conrey, Farmer and ZirnbauerCFZ, Step 4], which is the evaluation of the average of the coefficients appearing in the approximation of the expression

(4.1) $$ \begin{align} R(\alpha,\gamma;X):=\frac 1{N^\pm(X)} \sum_{K \in \mathcal{F}^\pm(X)} \frac{L\left(\frac{1}{2}+\alpha,f_K\right)}{L\left(\frac{1}{2}+\gamma,f_K\right)}. \end{align} $$

More precisely, instead of only considering the main term, we kept track of the secondary term in Lemma 4.1.

We now describe more precisely the steps in the Ratios Conjecture recipe. The first step involves the approximate functional equation for $L(s,f_K)$ , which reads

(4.2) $$ \begin{align} L\left(s, f_K\right) = \sum_{n<x} \frac{\lambda_K(n)}{n^s}+|D_K|^{\frac 12-s}\frac{\Gamma_{\pm}(1-s)}{\Gamma_{\pm}(s)} \sum_{n<y} \frac{\lambda_K(n)}{n^{1-s}}+\:\mathrm{Error}, \end{align} $$

where $x,y$ are such that $xy\asymp |D_K| (1+|t|)^2 $ (this is in analogy with [Reference Conrey and SnaithCS]; see [Reference Iwaniec and KowalskiIK, Theorem 5.3] for a description of the approximate functional equation of a general L-function). The analysis will be carried out assuming that the error term can be neglected and that the sums can be completed.

Following [Reference Conrey, Farmer and ZirnbauerCFZ], we replace the numerator of (4.1) with the approximate functional equation (4.2) and the denominator of (4.1) with (2.2). We will need to estimate the first sum in (4.2) evaluated at $s=\frac {1}{2}+\alpha $ , where $|\operatorname {Re}(\alpha )|$ is sufficiently small. This gives the contribution

(4.3) $$ \begin{align} R_1(\alpha, \gamma;X):=\frac1{N^\pm(X)} \sum_{K \in \mathcal{F}^\pm(X)} \sum_{h,m} \frac{\lambda_K(m)\mu_{K}(h)}{m^{\frac{1}{2} + \alpha}h^{\frac{1}{2}+\gamma}} \end{align} $$

to (4.1). This infinite sum converges absolutely in the region $\operatorname {Re}(\alpha )>\frac 12$ and $\operatorname {Re}(\gamma )>\frac 12$ , however, later in this section, we will provide an analytic continuation to a wider domain. We will also need to evaluate the contribution of the second sum in (4.2), which is given by

(4.4) $$ \begin{align} R_2(\alpha, \gamma;X):=\frac1{N^\pm(X)} \sum_{K \in \mathcal{F}^\pm(X)} |D_K|^{-\alpha}\frac{\Gamma_{\pm}\left(\frac 12-\alpha \right)}{\Gamma_{\pm}\left(\frac 12+\alpha\right)}\sum_{h,m} \frac{\lambda_K(m)\mu_{K}(h)}{m^{\frac{1}{2}-\alpha}h^{\frac{1}{2}+\gamma}} \end{align} $$

(Once more, the series converges absolutely for $\operatorname {Re}(\alpha )<-\frac 12$ and $\operatorname {Re}(\gamma )>\frac 12$ , but we will later provide an analytic continuation to a wider domain).

A first step in the understanding of the $R_j(\alpha , \gamma ;X)$ will be achieved using the following precise evaluation of the expected value of $\lambda _K(m)\mu _K(h) $ .

Lemma 4.1. Let $m,h\in \mathbb N$ , and let $\frac 12 \leq \theta <\frac 56$ and $\omega \geq 0$ be, such that (2.3) holds. Assume that h is cubefree. We have the estimate

$$ \begin{align*} &\frac{1}{N^\pm(X)} \sum_{K \in \mathcal{F}^\pm(X)}\lambda_K(m)\mu_K(h) \\ &\quad\!\! = \prod_{p^e\parallel m, p^s \parallel h}f(e,s,p)x_p + \Bigg( \prod_{p^e\parallel m, p^s \parallel h}g(e,s,p)y_p- \prod_{p^e\parallel m, p^s \parallel h}f(e,s,p)x_p\Bigg) \frac{C_2^\pm}{C_1^\pm}X^{-\frac{1}{6}}\Big( 1-\frac{C_2^\pm}{C_1^\pm}X^{-\frac{1}{6}}\Big) \\ &\qquad +O_\varepsilon\bigg( \prod_{ p \mid hm , p^e \parallel m } \big( (2 e + 5)p^{\omega}\big) X^{\theta-1 +\varepsilon}\bigg), \end{align*} $$

where

$$ \begin{align*} f(e,0,p)&:= \frac{e+1}{6}+\frac{1+(-1)^e}{4}+\frac{\tau_e}{3}+\frac{1}{p}; \\ f(e,1,p)&:= - \frac{e+1}{3}+\frac{\tau_e}{3}- \frac{1}{p}; \\ f(e,2,p)&:=\frac{e+1}{6}-\frac{1+(-1)^e}{4}+\frac{\tau_e}{3}; \\ g(e,0,p)&:= \frac{(e+1)(1+p^{-\frac 13})^3}{6} +\frac{(1+(-1)^e)(1+p^{-\frac 13})(1+p^{-\frac 23})}{4} \\ & \quad + \frac{\tau_e(1+p^{-1})}{3} + \frac{(1+p^{-\frac 13})^2}{p}; \\ g(e,1,p)&:= -\frac{(e+1)(1+p^{-\frac 13})^3}{3}+ \frac{\tau_e(1+p^{-1})}{3}-\frac{(1+p^{-\frac 13})^2}{p}; \\ g(e,2,p)&:= \frac{(e+1)(1+p^{-\frac 13})^3}{6} -\frac{(1+(-1)^e)(1+p^{-\frac 13})(1+p^{-\frac 23})}{4} + \frac{\tau_e(1+p^{-1})}{3}. \end{align*} $$

Proof. We may write $ m = \prod _{j=1}^J p_j^{e_j}$ and $ h = \prod _{j=1}^J p_j^{s_j }$ , where $ p_1 , \ldots , p_J$ are distinct primes and for each $ j $ , $ e_j $ and $ s_j $ are nonnegative integers but not both zero. Then we see that

$$ \begin{align*} \sum_{K \in \mathcal{F}^\pm(X)} \lambda_K (m) \mu_K (h) = & \sum_{K \in \mathcal{F}^\pm(X)} \prod_{j=1}^J \bigg( \lambda_K \big( p_j ^{e_j} \big) \mu_K \big( p_j^{s_j} \big) \bigg) = \sum_{\mathbf{k}} \sum_{ \substack{ K \in \mathcal{F}^{\pm}(X) \\ \mathbf{p} : ~type ~T_{\mathbf{k}} }} \prod_{j=1}^J \bigg( \lambda_K \big( p_j ^{e_j} \big) \mu_K \big( p_j^{s_j} \big) \bigg) , \end{align*} $$

where $\mathbf {k} = (k_1 , \ldots , k_J ) $ runs over $\{ 1, 2, 3, 4, 5 \}^J $ and $\mathbf {p} = ( p_1 , \ldots , p_J)$ . When each $p_j$ has splitting type $T_{k_j}$ in K, the values $\lambda _K (p_j^{e_j})$ and $ \mu _K ( p_j^{s_j})$ depend on $p_j$ , $k_j $ , $e_j$ and $ s_j$ . Define

$$ \begin{align*} \eta_{1, p_j}(k_j, e_j ) := \lambda_K (p_j^{e_j}) , \qquad \eta_{2, p_j} (k_j, s_j ) := \mu_K (p_j^{s_j}) \end{align*} $$

for each $ j \leq J$ with $p_j $ of splitting type $T_{k_j}$ in K, as well as

(4.5) $$ \begin{align} \eta_{1, \mathbf{p}} ( \mathbf{k}, \mathbf{e}) := \prod_{j=1}^J \eta_{1, p_j} (k_j, e_j ), \qquad \eta_{2, \mathbf{p}} ( \mathbf{k}, \mathbf{s}) := \prod_{j=1}^J \eta_{2, p_j} (k_j, s_j ). \end{align} $$

We see that

$$ \begin{align*} \sum_{K \in \mathcal{F}^\pm(X)} \lambda_K (m) \mu_K (h) & = \sum_{\mathbf{k}} \eta_{1, \mathbf{p}} ( \mathbf{k}, \mathbf{e}) \eta_{2,\mathbf{p}} ( \mathbf{k}, \mathbf{s}) \sum_{ \substack{ K \in \mathcal{F}^{\pm}(X) \\ \mathbf{p} : ~type ~T_{\mathbf{k}} }} 1 \\ & = \sum_{\mathbf{k}} \eta_{1, \mathbf{p}} ( \mathbf{k}, \mathbf{e}) \eta_{2,\mathbf{p}} ( \mathbf{k}, \mathbf{s}) N^{\pm}_{\mathbf{p}} ( X , T_{\mathbf{k}}), \end{align*} $$

which by (2.3) is equal to

$$ \begin{align*} & \sum_{\mathbf{k}} \eta_{1, \mathbf{p}} ( \mathbf{k}, \mathbf{e}) \eta_{2,\mathbf{p}} ( \mathbf{k}, \mathbf{s}) \bigg(C_1^\pm \prod_{j = 1}^J ( x_{p_j} c_{k_j} (p_j ) ) X + C_2^\pm \prod_{j = 1}^J ( y_{p_j} d_{k_j} (p_j ) ) X^{\frac 56} + O_{\varepsilon} \bigg(\prod_{j =1}^J p_j^{\omega}X^{\theta+\varepsilon} \bigg) \bigg) \\ &= C_1^\pm X \bigg( \sum_{\mathbf{k}} \eta_{1, \mathbf{p}} ( \mathbf{k}, \mathbf{e}) \eta_{2,\mathbf{p}} ( \mathbf{k}, \mathbf{s}) \prod_{j = 1}^J ( x_{p_j} c_{k_j} (p_j )) \bigg) + C_2^\pm X^{\frac 56} \bigg( \sum_{\mathbf{k}} \eta_{1, \mathbf{p}} ( \mathbf{k}, \mathbf{e}) \eta_{2,\mathbf{p}} ( \mathbf{k}, \mathbf{s}) \prod_{j = 1}^J ( y_{p_j} d_{k_j} (p_j )) \bigg) \\ &\quad + O_{\varepsilon} \bigg( \sum_{\mathbf{k}} | \eta_{1, \mathbf{p}} ( \mathbf{k}, \mathbf{e}) \eta_{2,\mathbf{p}} ( \mathbf{k}, \mathbf{s}) | \prod_{j =1}^J p_j^{\omega}X^{\theta+\varepsilon} \bigg). \end{align*} $$

We can change the last three $\mathbf {k}$ -sums into products by (4.5). Doing so, we obtain that the above is equal to

$$ \begin{align*} &C_1^\pm X \prod_{j =1}^J \bigg( x_{p_j} \widetilde f(e_j, s_j, p_j ) \bigg) + C_2^\pm X^{\frac 56} \prod_{j =1}^J \bigg( y_{p_j} \widetilde g(e_j, s_j, p_j ) \bigg) + O_{\varepsilon} \bigg( \prod_{j =1}^J \bigg( p_j^{\omega} (2e_j +5 ) \bigg) X^{\theta+\varepsilon} \bigg) \\ &\quad = C_1^\pm X \prod_{p^e\parallel m, p^s \parallel h}\widetilde f(e,s,p)x_p + C_2^\pm X^{\frac 56} \prod_{p^e\parallel m, p^s \parallel h}\widetilde g(e,s,p)y_p + O_{\varepsilon} \bigg( \prod_{j =1}^J \bigg( p_j^{\omega} ( 2 e_j + 5 ) \bigg) X^{\theta+\varepsilon} \bigg) , \end{align*} $$

where

$$ \begin{align*}\widetilde f( e, s, p) := \sum_{k=1}^5 \eta_{1, p } (k , e ) \eta_{2,p } (k , s ) c_{k } (p ) , \qquad \widetilde g(e,s,p) := \sum_{k =1}^5 \eta_{1, p } ( k , e ) \eta_{2,p } ( k , s ) d_{k } (p ) .\end{align*} $$

A straightforward calculation shows that $\widetilde f( e, s, p) = f( e, s, p) $ and $\widetilde g( e, s, p)=g( e, s, p) $ (see the explicit description of the coefficients in Section 2; note that $ \eta _{2,p}(k,0) = 1 $ ) and the lemma follows.

We now proceed with the estimation of $R_1(\alpha , \gamma; X )$ . Taking into account the two main terms in Lemma 4.1, we expect that

(4.6) $$ \begin{align} R_1(\alpha, \gamma;X)=R_1^M(\alpha, \gamma)+\frac{C_2^\pm}{C_1^\pm}X^{-\frac 16}\Big( 1-\frac{C_2^\pm}{C_1^\pm}X^{-\frac{1}{6}}\Big) (R_1^S(\alpha, \gamma)-R_1^M(\alpha, \gamma))+\text{Error}, \end{align} $$

where

(4.7) $$ \begin{align} R_1^M(\alpha, \gamma):=&\prod_p \Big( 1 + \sum_{ e\geq 1} \frac{x_p f(e,0,p)}{p^{e(\frac 12+\alpha)}} + \sum_{ e\geq 0} \frac{x_p f(e,1,p)}{p^{e(\frac 12+\alpha) +(\frac 12+\gamma) }} + \sum_{ e\geq 0} \frac{x_p f(e,2,p)}{p^{e(\frac 12+\alpha) + 2(\frac 12+\gamma) }}\Big),\nonumber\\ R_1^S(\alpha, \gamma):=&\prod_p \Big( 1 + \sum_{ e\geq 1} \frac{y_p g(e,0,p)}{p^{e(\frac 12+\alpha)}} + \sum_{ e\geq 0} \frac{y_p g(e,1,p)}{p^{e(\frac 12+\alpha) +(\frac 12+\gamma) }} + \sum_{ e\geq 0} \frac{y_p g(e,2,p)}{p^{e(\frac 12+\alpha) + 2(\frac 12+\gamma) }}\Big) \end{align} $$

for $ \operatorname {Re} ( \alpha ) , \operatorname {Re} ( \gamma )> \frac 12$ . Since

$$ \begin{align*} R_1^M(\alpha, \gamma) &= \prod_p \Bigg( 1 + \frac{1}{p^{1+2\alpha}} - \frac{1}{p^{1+\alpha+\gamma}}\\ &\quad+ O \Big(\frac{1}{p^{\frac 32 +\operatorname{Re}(\alpha)}}+\frac{1}{p^{\frac 32 +3\operatorname{Re}(\alpha)}}+\frac{1}{p^{\frac 32 +\operatorname{Re}(\gamma)}} + \frac{1}{p^{\frac 32+ \operatorname{Re} ( 2 \alpha + \gamma) }} + \frac{1}{p^{\frac 52 + \operatorname{Re} (3 \alpha+ 2 \gamma) }} \Big) \Bigg), \end{align*} $$

we see that

(4.8) $$ \begin{align} A_3(\alpha, \gamma) := \frac{\zeta(1+\alpha+\gamma)}{\zeta(1+2\alpha) } R_1^M(\alpha, \gamma) \end{align} $$

is analytically continued to the regionFootnote ⁴ $ \operatorname {Re} (\alpha ), \operatorname {Re} (\gamma )> - \frac 16 $ . Similarly, from the estimates

$$ \begin{align*} & \sum_{ e\geq 1} \frac{y_p g(e,0,p)}{p^{e(\frac 12+\alpha)}} = \frac 1{p^{\frac 56+\alpha}}+\frac 1{p^{1+2\alpha}} +O\Big( \frac 1{p^{\operatorname{Re}(\alpha) + \frac 32}}+\frac 1{p^{2\operatorname{Re}(\alpha) + \frac 43}}+ \frac 1{p^{3\operatorname{Re}(\alpha) + \frac 32}}\Big),\\ &\sum_{ e\geq 0} \frac{y_p g(e,1,p)}{p^{e(\frac 12+\alpha) +(\frac 12+\gamma) }}=-\frac 1{p^{\frac 56+\gamma}} - \frac 1{p^{1+\alpha+\gamma}}+O\Big(\frac 1{p^{\frac 43+\operatorname{Re}(\alpha+\gamma)}} \Big),\\ &\sum_{ e\geq 0} \frac{y_p g(e,2,p)}{p^{e(\frac 12+\alpha) + 2(\frac 12+\gamma) }} = O \Big( \frac 1{p^{\frac 32+\operatorname{Re}(\alpha+2\gamma)}} \Big) , \end{align*} $$

we deduce that

(4.9) $$ \begin{align} A_4(\alpha,\gamma):= \frac{ \zeta(\tfrac 56+\gamma) \zeta(1+\alpha+\gamma)}{\zeta(\tfrac 56+\alpha) \zeta(1+2\alpha) }R_1^S(\alpha, \gamma) \end{align} $$

is analytic in the region $\operatorname {Re}(\alpha ),\operatorname {Re}(\gamma )>-\frac 16$ . Note that by their defining product formulas, we have the bounds

(4.10) $$ \begin{align} A_3 ( \alpha, \gamma)=O_\varepsilon (1) ,\quad A_4 ( \alpha, \gamma) = O_\varepsilon (1) \end{align} $$

for $ \operatorname {Re} ( \alpha ) , \operatorname {Re} ( \gamma ) \geq - \frac 16 + \varepsilon> - \frac 16$ . Using this notation, (4.6) takes the form

$$ \begin{align*} & R_1(\alpha,\gamma;X) = \frac{\zeta(1+2\alpha)}{\zeta(1+\alpha+\gamma)} \Big( A_3(\alpha,\gamma)+\frac{C_2^\pm}{C_1^\pm} X^{-\frac 16} \Big( 1-\frac{C_2^\pm}{C_1^\pm}X^{-\frac{1}{6}}\Big) \Big( \frac{\zeta(\tfrac 56+\alpha)}{\zeta(\tfrac 56+\gamma)} A_4(\alpha,\gamma) - A_3(\alpha,\gamma) \Big)\Big)\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad +\text{Error}. \end{align*} $$

The above computation is sufficient in order to obtain a conjectural evaluation of the average (4.3). However, our goal is to evaluate the one-level density through the average of $\frac {L'}{L} (\frac 12+r ,f_K ) $ ; therefore, it is necessary to also compute the partial derivative $\frac {\partial }{ \partial \alpha }R_1(\alpha , \gamma ;X)|_{\alpha =\gamma =r} $ . To do so, we need to make sure that the error term stays small after a differentiation. This is achieved by applying Cauchy’s integral formula for the derivative

$$ \begin{align*}f'(a)=\frac{1}{2\pi i}\int_{|z-a|=\kappa} \frac{f(z)}{(z-a)^2}dz\end{align*} $$

(valid for all small enough $\kappa>0$ ) and bounding the integrand using the approximation for $R_1(\alpha ,\gamma ;X) $ above. As for the main terms, one can differentiate them term by term and obtain the expected approximation

(4.11) $$ \begin{align} & \frac{\partial }{ \partial \alpha}R_1(\alpha, \gamma;X)\Big|_{\alpha=\gamma=r}=A_{3,\alpha}(r,r)+ \frac{\zeta'}{\zeta}(1+2r) A_3(r,r) + \frac{C_2^\pm}{C_1^\pm} X^{-\frac 16}\Big( 1-\frac{C_2^\pm}{C_1^\pm}X^{-\frac{1}{6}}\Big)\nonumber\\ & \quad \times \Big( A_{4,\alpha}(r,r) +\frac{\zeta'}{\zeta} (\tfrac 56+r) A_4 (r,r) - A_{3,\alpha}(r,r) +\frac{\zeta'}{\zeta}(1+2r) ( A_4(r,r)-A_3(r,r)) \Big) \nonumber\\ &\quad +\text{Error}, \end{align} $$

where $A_{3,\alpha }(r,r)=\frac {\partial }{ \partial \alpha }A_3(\alpha ,\gamma )\Big |_{\alpha =\gamma =r}$ and $A_{4,\alpha }(r,r)=\frac {\partial }{ \partial \alpha }A_4(\alpha ,\gamma )\Big |_{\alpha =\gamma =r}.$

Now, from the definition of $f(e,j,p) $ and $g(e,j,p)$ (see Lemma 4.1) as well as (3.4) and (3.5), we have

$$ \begin{align*} f(1,0,p)+f(0,1,p) & =g(1,0,p)+g(0,1,p)=0,\\ f(e,0,p)+f(e-1,1,p) +f(e-2,2,p) & =g(e,0,p)+g(e-1,1,p) +g(e-2,2,p)=0, \\ f(e, 0 , p ) - f(e-2 , 2, p ) & = \theta_e + p^{-1} , \\ g(e, 0 , p ) - g(e-2 , 2, p ) & = ( 1 + p^{- \frac13} ) ( \kappa_e ( p ) + p^{-1} + p^{- \frac43} ). \end{align*} $$

By the above identities and the definition (4.7), we deduce that

$$ \begin{align*}R_1^M(r,r)=A_3(r,r)=R_1^S(r,r)=A_4(r,r)=1.\end{align*} $$

It follows that for $\operatorname {Re}(r)>\frac 12$ ,

$$ \begin{align*} R^M_{1,\alpha}(r,r) &= \frac{R^M_{1,\alpha}(r,r)}{R^M_1(r,r)} = \frac{ \partial}{\partial \alpha} \log R^M_{1} ( \alpha, \gamma) \bigg|_{ \alpha=\gamma = r}\\ &= \sum_p \left( - \frac{ x_p \log p }{p^{\frac12 + r}}f(1,0,p) - \sum_{ e \geq 2} \frac{ x_p \log p }{ p^{ e( \frac12 + r ) } } \big( f(e,0,p ) - f(e-2,2,p)\big) \right)\\ &\quad + \sum_p \left( - \sum_{ e \geq 2} \frac{ x_p \log p }{ p^{ e( \frac12 + r ) } } (e-1) \big( f( e,0,p) + f(e-1,1,p) + f( e-2,2,p) \big) \right) \\ & =- \sum_p \sum_{ e \geq 1} \frac{ x_p \log p }{ p^{ e( \frac12 + r ) } } \Big( \theta_e + \frac1p \Big) \end{align*} $$

and

$$ \begin{align*} R^S_{1,\alpha}(r,r) &= \sum_p \left( - \frac{ y_p \log p }{p^{\frac12 + r}}g(1,0,p) - \sum_{ e \geq 2} \frac{ y_p \log p }{ p^{ e( \frac12 + r ) } } \big( g(e,0,p ) - g(e-2,2,p)\big) \right)\\ &\quad + \sum_p \left( - \sum_{ e \geq 2} \frac{ y_p \log p }{ p^{ e( \frac12 + r ) } } (e-1) \big( g( e,0,p) + g(e-1,1,p) + g( e-2,2,p) \big) \right) \\ & =- \sum_p \sum_{ e \geq 1} \frac{y_p \log p }{ p^{ e( \frac12 + r ) } } \big( 1 + p^{- \frac13} \big) \big( \kappa_e ( p ) + p^{-1} + p^{- \frac43} \big) \\ & =- \sum_p \sum_{ e \geq 1} \frac{ \log p }{ p^{ e( \frac12 + r ) } } \Big( \beta_e (p)+ x_p \Big( \theta_e+ \frac1p \Big)\Big) , \end{align*} $$

by (3.6). Thus, we have

$$ \begin{align*} A_{3,\alpha}(r,r)&=R^M_{1,\alpha}(r,r) - \frac{\zeta'}{\zeta}(1+2r) = -\sum_{p,e\geq 1} \Big(\theta_e+\frac 1p\Big) \frac{x_p\log p }{p^{e(\frac 12+r)}}- \frac{\zeta'}{\zeta}(1+2r) \end{align*} $$

and

(4.12) $$ \begin{align} A_{4,\alpha}(r,r)-A_{3,\alpha}(r,r) = -\sum_{p,e\geq 1} \frac{(\beta_e(p)-p^{-\frac e3}) \log p }{p^{e(\frac 12+r)}}, \end{align} $$

which are now valid in the extended region $ \operatorname {Re}(r)>0 $ . Coming back to (4.11), we deduce that

$$ \begin{align*}\notag \frac{\partial }{ \partial \alpha}R_1(\alpha, \gamma;X)\Big|_{\alpha=\gamma=r} & =A_{3,\alpha}(r,r)+ \frac{\zeta'}{\zeta}(1+2r) \\&\notag\quad + \frac{C_2^\pm}{C_1^\pm} X^{-\frac 16}\Big( 1-\frac{C_2^\pm}{C_1^\pm}X^{-\frac{1}{6}}\Big) \Big( A_{4,\alpha}(r,r) - A_{3,\alpha}(r,r) +\frac{\zeta'}{\zeta} (\tfrac 56+r) \Big) +\text{Error}\\ \notag &=-\sum_{p,e\geq 1} \Big(\theta_e+\frac 1p\Big) \frac{x_p\log p }{p^{e(\frac 12+r)}} -\frac{C_2^\pm}{C_1^\pm} X^{-\frac 16}\Big( 1-\frac{C_2^\pm}{C_1^\pm}X^{-\frac{1}{6}}\Big) \sum_{p,e\geq 1} \frac{(\beta_e(p)-p^{-\frac e3} )\log p }{p^{e(\frac 12+r)}}\\ &\quad +\frac{C_2^\pm}{C_1^\pm} X^{-\frac 16}\Big( 1-\frac{C_2^\pm}{C_1^\pm}X^{-\frac{1}{6}}\Big)\frac{\zeta'}{\zeta} (\tfrac 56+r) +\text{Error}, \end{align*} $$

where the second equality is valid in the region $\operatorname {Re}(r)>0$ .

We now move to $R_2(\alpha , \gamma ;X).$ We recall that

(4.13) $$ \begin{align} R_2(\alpha, \gamma;X)=\frac1{N^\pm(X)} \sum_{K \in \mathcal{F}^\pm(X)} |D_K|^{-\alpha}\frac{\Gamma_{\pm}(\frac 12-\alpha)}{\Gamma_{\pm}(\frac 12+\alpha)} \sum_{h,m} \frac{\lambda_K(m)\mu_{K}(h)}{m^{\frac{1}{2}-\alpha}h^{\frac{1}{2}+\gamma}}, \end{align} $$

and the Ratios Conjecture recipe tells us that we should replace $\lambda _K(m)\mu _K(h)$ with its average. However, a calculation involving Lemma 4.1 suggests that the terms $ |D_K|^{-\alpha }$ and $\lambda _K(m)\mu _{K}(h)$ have nonnegligible covariance. To take this into account, we substitute this step with the use of the following corollary of Lemma 4.1.

Corollary 4.2. Let $m,h\in \mathbb N$ , and let $\frac 12 \leq \theta <\frac 56$ and $\omega \geq 0$ be, such that (2.3) holds. For $\alpha \in \mathbb {C}$ with $0<\operatorname {Re}(\alpha ) < \frac 12$ , we have the estimate

$$ \begin{align*} &\frac{1}{N^\pm(X)} \sum_{K \in \mathcal{F}^\pm(X)} |D_K|^{-\alpha} \lambda_K(m)\mu_K(h) = \frac{X^{-\alpha}}{1-\alpha} \prod_{p^e\parallel m, p^s \parallel h}f(e,s,p)x_p \\ &\quad + X^{-\frac16-\alpha} \Bigg( \frac{1}{1- \frac{6\alpha}5} \prod_{p^e\parallel m, p^s \parallel h}g(e,s,p)y_p- \frac{1}{1-\alpha} \prod_{p^e\parallel m, p^s \parallel h}f(e,s,p)x_p\Bigg) \frac{C_2^\pm}{C_1^\pm} \Big( 1-\frac{C_2^\pm}{C_1^\pm}X^{-\frac{1}{6}}\Big) \\ &\quad +O_{\varepsilon}\bigg((1+|\alpha|)\prod_{ p \mid hm, p^e \parallel m } \big( (2 e + 5)p^{\omega}\big)X^{\theta-1-\operatorname{Re}(\alpha)+\varepsilon}\bigg). \end{align*} $$

Proof. This follows from applying Lemma 4.1 and (1.1) to the identity

$$ \begin{align*} & \sum_{K \in \mathcal{F}^\pm(X)} |D_K|^{-\alpha} \lambda_K(m)\mu_K(h) = \int_1^X u^{-\alpha} d\bigg( \sum_{K \in \mathcal{F}^\pm(u)} \lambda_K(m)\mu_K(h) \bigg) \\ &\qquad\qquad\qquad\qquad= X^{-\alpha} \sum_{K \in \mathcal{F}^\pm(X)} \lambda_K(m)\mu_K(h) + \alpha \int_1^X u^{-\alpha -1 } \bigg( \sum_{K \in \mathcal{F}^\pm(u)} \lambda_K(m)\mu_K(h)\bigg) du. \end{align*} $$

Applying this lemma, we deduce the following heuristic approximation of $R_2 ( \alpha , \gamma; X)$ :

$$ \begin{align*} & \frac{\Gamma_{\pm}(\frac 12-\alpha)}{\Gamma_{\pm}(\frac 12+\alpha)} \sum_{h,m} \frac{1}{m^{\frac{1}{2}-\alpha}h^{\frac{1}{2}+\gamma}} \bigg\{ \frac{X^{-\alpha}}{1-\alpha} \prod_{p^e\parallel m, p^s \parallel h}f(e,s,p)x_p\\ & + X^{-\frac 16-\alpha} \frac{C_2^\pm}{C_1^\pm} \Big( 1-\frac{C_2^\pm}{C_1^\pm}X^{-\frac{1}{6}}\Big) \Big( \frac{1}{1-\frac{6\alpha}5} \prod_{p^e\parallel m, p^s \parallel h}g(e,s,p)y_p- \frac{1}{1-\alpha} \prod_{p^e\parallel m, p^s \parallel h}f(e,s,p)x_p\Big) \bigg\} \\ = & \frac{\Gamma_{\pm}(\frac 12-\alpha)}{\Gamma_{\pm}(\frac 12+\alpha)} \bigg\{ X^{-\alpha} \frac{ R_1^M ( - \alpha, \gamma)}{1-\alpha} + X^{-\frac 16-\alpha}\frac{C_2^\pm}{C_1^\pm} \Big( 1-\frac{C_2^\pm}{C_1^\pm}X^{-\frac{1}{6}}\Big) \Big( \frac{R_1^S ( - \alpha, \gamma) }{1-\frac{6\alpha}5} - \frac{R_1^M ( - \alpha, \gamma)}{1-\alpha} \Big) \bigg\} \\ = & \frac{\Gamma_{\pm}(\frac 12-\alpha)}{\Gamma_{\pm}(\frac 12+\alpha)} \frac{\zeta(1-2\alpha)}{ \zeta( 1 - \alpha + \gamma) } \bigg\{ X^{-\alpha} \frac{A_3 ( - \alpha, \gamma)}{1-\alpha} \\ & + X^{-\frac 16-\alpha}\frac{C_2^\pm}{C_1^\pm} \Big( 1-\frac{C_2^\pm}{C_1^\pm}X^{-\frac{1}{6}}\Big) \Big( \frac{A_4 ( - \alpha, \gamma)}{1- \frac{6\alpha}5} \frac{ \zeta(\frac 56-\alpha)}{ \zeta( \frac 56+\gamma)} - \frac{A_3 ( - \alpha, \gamma) }{1-\alpha} \Big) \bigg\}. \end{align*} $$

If $\operatorname {Re}( r) $ is positive and small enough, then we expect that

$$ \begin{align*} \frac{\partial }{ \partial \alpha}R_2(\alpha, \gamma;X)\Big|_{\alpha=\gamma=r} = & - \frac{\Gamma_{\pm}(\frac 12-r)}{\Gamma_{\pm}(\frac 12+r)} \zeta(1-2r) \bigg\{ X^{-r} \frac{A_3 ( - r,r)}{1-r} \\ & + X^{-\frac 16-r} \frac{C_2^\pm}{C_1^\pm} \Big( 1-\frac{C_2^\pm}{C_1^\pm}X^{-\frac{1}{6}}\Big) \Big( \frac{\zeta(\tfrac 56-r)}{\zeta(\tfrac 56+r)} \frac{A_4(-r,r)}{ 1- \frac{6r}5} - \frac{ A_3( -r, r)}{1-r} \Big) \bigg\}+\text{Error}. \end{align*} $$

We arrive at the following conjecture.

Conjecture 4.3. Let $\frac 12 \leq \theta <\frac 56$ and $\omega \geq 0$ be, such that (2.3) holds. There exists $0<\delta <\frac 1{6}$ , such that for any fixed $\varepsilon>0$ and for $r\in \mathbb C$ with $ \frac 1{L} \ll \operatorname {Re}(r) <\delta $ and $|r|\leq X^{\frac \varepsilon 2}$ ,

(4.14) $$ \begin{align} &\frac 1{N^\pm(X)} \sum_{K \in \mathcal{F}^\pm(X)} \frac{L'(\frac 12+r,f_K)}{L(\frac 12+r,f_K)}\nonumber\\ &\quad = -\sum_{p,e\geq 1} \Big(\theta_e+\frac 1p\Big) \frac{x_p\log p }{p^{e(\frac 12+r)}} -\frac{C_2^\pm}{C_1^\pm} X^{-\frac 16} \Big( 1-\frac{C_2^\pm}{C_1^\pm}X^{-\frac{1}{6}}\Big) \sum_{p,e\geq 1} \frac{(\beta_e(p)-p^{-\frac e3}) \log p }{p^{e(\frac 12+r)}} \nonumber\\ &\qquad +\frac{C_2^\pm}{C_1^\pm} X^{-\frac 16}\Big( 1-\frac{C_2^\pm}{C_1^\pm}X^{-\frac{1}{6}}\Big) \frac{\zeta'}{\zeta}(\tfrac 56+r) - X^{-r} \frac{\Gamma_{\pm}(\frac 12-r)}{\Gamma_{\pm}(\frac 12+r)}\zeta(1-2r) \frac{A_3( -r, r)}{1-r} \nonumber\\ &\qquad - \frac{C_2^\pm }{C_1^\pm } X^{-r-\frac 16} \Big( 1-\frac{C_2^\pm}{C_1^\pm}X^{-\frac{1}{6}}\Big) \frac{\Gamma_{\pm}(\frac 12-r)}{\Gamma_{\pm}(\frac 12+r)}\zeta(1-2r) \Big( \frac{\zeta(\tfrac 56-r)}{\zeta(\tfrac 56+r)} \frac{A_4(-r,r)}{ 1- \frac{6r}5} - \frac{ A_3( -r, r)}{1-r} \Big) +O_\varepsilon(X^{\theta-1+\varepsilon}). \end{align} $$

Note that the two sums on the right-hand side are absolutely convergent.

Traditionally, when applying the Ratios Conjecture recipe, one has to restrict the real part of the variable r to small enough positive values. For example, in the family of quadratic Dirichlet L-functions [Reference Conrey and SnaithCS, Reference Fiorilli, Parks and SödergrenFPS3], one requires that $\frac 1{\log X}\ll \operatorname {Re}(r)<\frac 14$ . This ensures that one is far enough from a pole for the expression in the right-hand side. In the current situation, we will see that the term involving $X^{-r-\frac 16}$ has a pole at $s=\frac 16$ .

Proposition 4.4. Assume Conjecture 4.3 and the Riemann Hypothesis for $\zeta _K(s)$ for all $K\in \mathcal {F}^{\pm }(X)$ , and let $\phi $ be a real even Schwartz function, such that $ \widehat \phi $ is compactly supported. For any constant $ 0< c < \frac 16$ , we have that

(4.15) $$ \begin{align} &\frac 1{N^\pm(X)} \sum_{K \in \mathcal{F}^\pm(X)} \sum_{\gamma_K}\phi \Big(\frac{L\gamma_K}{2\pi }\Big)=\widehat \phi(0)\Big(1+ \frac{\log(4 \pi^2 e)}{L} -\frac{C_2^\pm}{5C_1^\pm} \frac{X^{-\frac 16}}{L} + \frac{(C_2^\pm)^2 }{5(C_1^\pm)^2 } \frac{X^{-\frac 13}}{L} \Big)\nonumber\\ &\qquad + \frac1{\pi}\int_{-\infty}^{\infty}\phi\left(\frac{Lr}{2\pi}\right)\operatorname{Re}\Big(\frac{\Gamma'_{\pm}}{\Gamma_{\pm}}(\tfrac12+ir)\Big)dr -\frac{2}{L}\sum_{p,e}\frac{x_p\log p}{p^{\frac e2}}\widehat\phi\left(\frac{\log p^e}{L}\right) (\theta_e+\tfrac 1p)\nonumber\\ &\qquad -\frac{2C_2^\pm X^{-\frac 16}}{C_1^\pm L}\Big( 1-\frac{C_2^\pm}{C_1^\pm}X^{-\frac{1}{6}}\Big) \sum_{p,e}\frac{\log p}{p^{\frac e2}}\widehat\phi\left(\frac{\log p^e}{L}\right) (\beta_e(p)-p^{-\frac e3})\nonumber\\ &\qquad -\frac{1}{\pi i} \int_{(c )} \phi \Big(\frac{Ls}{2\pi i}\Big) \Big\{ -\frac{C_2^\pm}{C_1^\pm} X^{-\frac 16}\Big( 1-\frac{C_2^\pm}{C_1^\pm}X^{-\frac{1}{6}}\Big) \frac{\zeta'}{\zeta}(\tfrac 56+s) + X^{-s} \frac{\Gamma_{\pm}(\frac 12-s)}{\Gamma_{\pm}(\frac 12+s)}\zeta(1-2s) \frac{A_3( -s, s) }{1-s} \nonumber\\ &\qquad +\frac{C_2^\pm }{C_1^\pm} X^{-s-\frac 16}\Big( 1-\frac{C_2^\pm}{C_1^\pm}X^{-\frac{1}{6}}\Big) \frac{\Gamma_{\pm}(\frac 12-s)}{\Gamma_{\pm}(\frac 12+s)}\zeta(1-2s) \Big( \frac{\zeta(\tfrac 56-s)}{\zeta(\tfrac 56+s)} \frac{A_4(-s,s)}{ 1- \frac{6s}5} - \frac{ A_3( -s, s)}{1-s} \Big) \Big\}ds \nonumber\\ &\qquad + O_\varepsilon(X^{\theta-1+\varepsilon}). \end{align} $$

Proof. By the residue theorem, we have the identity

(4.16) $$ \begin{align} \frac 1{N^\pm(X)} \sum_{K \in \mathcal{F}^\pm(X)} \mathfrak{D}_\phi(K)= \frac{1}{2 \pi i} \left( \int_{(\frac1L)} - \int_{(-\frac1L)} \right) \frac 1{N^\pm(X)} \sum_{K \in \mathcal{F}^\pm(X)} \frac{L'(s+\frac 12,f_K)}{L(s+\frac 12,f_K)}\phi \Big(\frac{Ls}{2\pi i}\Big)ds. \end{align} $$

Under Conjecture 4.3 and well-known arguments (see, e.g. [Reference Fiorilli, Parks and SödergrenFPS3, Section 3.2]), the part of this sum involving the first integral is equal to

$$ \begin{align*} &-\frac{1}{2\pi i} \int_{(\frac1L)} \phi \Big(\frac{Ls}{2\pi i}\Big) \bigg\{\sum_{p,e\geq 1} \Big(\theta_e+\frac 1p\Big) \frac{x_p\log p }{p^{e(\frac 12+s)}} +\frac{C_2^\pm}{C_1^\pm} X^{-\frac 16} \Big( 1-\frac{C_2^\pm}{C_1^\pm}X^{-\frac{1}{6}}\Big) \sum_{p,e\geq 1} \frac{(\beta_e(p)-p^{-\frac e3}) \log p }{p^{e(\frac 12+s)}} \\ &\qquad -\frac{C_2^\pm}{C_1^\pm} X^{-\frac 16}\Big( 1-\frac{C_2^\pm}{C_1^\pm}X^{-\frac{1}{6}}\Big) \frac{\zeta'}{\zeta}(\tfrac 56+s) + X^{-s} \frac{\Gamma_{\pm}(\frac 12-s)}{\Gamma_{\pm}(\frac 12+s)}\zeta(1-2s) \frac{A_3( -s, s) }{1-s} \\ &\qquad +\frac{C_2^\pm }{C_1^\pm}X^{-s-\frac 16}\Big( 1-\frac{C_2^\pm}{C_1^\pm}X^{-\frac{1}{6}}\Big) \frac{\Gamma_{\pm}(\frac 12-s)}{\Gamma_{\pm}(\frac 12+s)}\zeta(1-2s) \Big( \frac{\zeta(\tfrac 56-s)}{\zeta(\tfrac 56+s)} \frac{A_4(-s,s)}{ 1- \frac{6s}5} - \frac{ A_3( -s, s)}{1-s} \Big) \bigg\} ds \\ &\qquad +O_\varepsilon(X^{\theta-1+\varepsilon}), \end{align*} $$

where we used the bounds (4.10) and

(4.17) $$ \begin{align} \phi\Big(\frac{Ls}{2\pi i}\Big) = \frac { (-1)^\ell} {L^\ell s^\ell }\int_{\mathbb R} e^{ L \operatorname{Re}(s) x} e^{ i L \operatorname{Im}(s) x} \widehat \phi^{(\ell)}(x) dx \ll_\ell \frac{e^{L|\operatorname{Re}(s)| \sup(\mathrm{supp}(\widehat \phi)) } }{L^\ell |s|^\ell } \end{align} $$

for every integer $\ell>0 $ , which is decaying on the line $\operatorname {Re}(s) = \frac 1L$ . We may also shift the contour of integration to the line $\operatorname {Re}(s)=c$ with $ 0 < c < \frac 16$ .

For the second integral in (4.16) (over the line $\operatorname {Re}(s)=-\frac {1}{L}$ ), we treat it as follows. By the functional equation (2.1), we have

$$ \begin{align*} - \frac{1}{2 \pi i} & \int_{(-\frac1L)} \frac 1{N^\pm(X)} \sum_{K \in \mathcal{F}^\pm(X)} \frac{L'(s+\frac 12,f_K)}{L(s+\frac 12,f_K)}\phi \Big(\frac{Ls}{2\pi i}\Big)ds \\ = & \frac{1}{2 \pi i} \int_{(\frac1L)} \frac 1{N^\pm(X)} \sum_{K \in \mathcal{F}^\pm(X)} \frac{L'(s+\frac 12,f_K)}{L(s+\frac 12,f_K)}\phi \Big(\frac{Ls}{2\pi i}\Big)ds \\ & + \frac{1}{2 \pi i} \int_{(-\frac1L)} \frac 1{N^\pm(X)} \sum_{K \in \mathcal{F}^\pm(X)} \left( \log |D_K| + \frac{ \Gamma'_{\pm}}{\Gamma_\pm} ( \tfrac12 + s )+ \frac{ \Gamma'_{\pm}}{\Gamma_\pm} ( \tfrac12 - s ) \right)\phi \Big(\frac{Ls}{2\pi i}\Big)ds. \end{align*} $$

The first integral on the right-hand side is identically equal to the integral that was just evaluated in the first part of this proof. As for the second, by shifting the contour to the line $\operatorname {Re} (s)=0$ , we find that it equals

$$ \begin{align*} & \left( \frac 1{N^\pm(X)} \sum_{K \in \mathcal{F}^\pm(X)} \log |D_K| \right) \frac{1}{2 \pi i} \int_{(0)} \phi \Big(\frac{Ls}{2\pi i}\Big)ds + \frac{1}{2 \pi i} \int_{(0)} \left( \frac{ \Gamma'_{\pm}}{\Gamma_\pm} ( \tfrac12 + s )+ \frac{ \Gamma'_{\pm}}{\Gamma_\pm} ( \tfrac12 - s ) \right) \phi \Big(\frac{Ls}{2\pi i}\Big)ds \\ & = \left( \frac 1{N^\pm(X)} \sum_{K \in \mathcal{F}^\pm(X)} \log |D_K| \right) \frac{ \widehat{\phi} (0)}{L} + \frac{1}{ \pi} \int_{- \infty}^\infty \phi \Big(\frac{Lr}{2\pi }\Big) \operatorname{Re} \left( \frac{ \Gamma'_{\pm}}{\Gamma_\pm} ( \tfrac12 + ir ) \right)dr. \end{align*} $$

By applying Lemma 3.2 to the first term, we find the leading terms on the right-hand side of (4.15).

Finally, by absolute convergence, we have the identity

$$ \begin{align*} \frac{1}{2\pi i} \int_{(c)}\phi \Big(\frac{Ls}{2\pi i}\Big)\sum_{p,e\geq 1} \Big(\theta_e+\frac 1p\Big) \frac{x_p\log p }{p^{e(\frac 12+s)}} ds &= \sum_{p,e\geq 1} \Big(\theta_e+\frac 1p\Big) \frac{x_p\log p } {p^{\frac e2}} \frac 1{2\pi i} \int_{(c)}\phi \Big(\frac{Ls}{2\pi i}\Big) p^{-es}ds\\ &=\frac{1}{L}\sum_{p,e\geq 1} \Big(\theta_e+\frac 1p\Big) \frac{x_p\log p }{p^{\frac e2}}\widehat \phi\Big( \frac{e\log p}{L} \Big), \end{align*} $$

since the contour of the inner integral can be shifted to the line $\operatorname {Re}(s)=0$ . The same argument works for the term involving $\beta _e(p)-p^{-\frac e3}$ . Hence, the proposition follows.

5 Analytic continuation of $A_3 ( -s, s ) $ and $A_4 ( -s , s )$

The goal of this section is to prove Theorem 1.4. To do so, we will need to estimate some of the terms in (4.15), namely,

(5.1) $$ \begin{align} J^\pm & (X) := \frac{2C_2^\pm X^{-\frac 16}}{C_1^\pm L}\Big( 1-\frac{C_2^\pm}{C_1^\pm}X^{-\frac{1}{6}}\Big) \sum_{p,e}\frac{\log p}{p^{\frac {5e}{6}}}\widehat\phi\left(\frac{\log p^e}{L}\right) \nonumber\\ & -\frac{1}{\pi i} \int_{(c )} \phi \Big(\frac{Ls}{2\pi i}\Big) \Big\{ -\frac{C_2^\pm}{C_1^\pm} X^{-\frac 16}\Big( 1-\frac{C_2^\pm}{C_1^\pm}X^{-\frac{1}{6}}\Big) \frac{\zeta'}{\zeta}(\tfrac 56+s) + X^{-s} \frac{\Gamma_{\pm}(\frac 12-s)}{\Gamma_{\pm}(\frac 12+s)}\zeta(1-2s) \frac{ A_3( -s, s) }{1-s}\nonumber\\ & +\frac{C_2^\pm }{C_1^\pm} X^{-s-\frac 16}\Big( 1-\frac{C_2^\pm}{C_1^\pm}X^{-\frac{1}{6}}\Big) \frac{\Gamma_{\pm}(\frac 12-s)}{\Gamma_{\pm}(\frac 12+s)}\zeta(1-2s) \Big( \frac{\zeta(\tfrac 56-s)}{\zeta(\tfrac 56+s)} \frac{A_4(-s,s)}{ 1- \frac{6s}5} - \frac{ A_3( -s, s)}{1-s} \Big) \Big\}ds, \end{align} $$

for $ 0 < c < \frac 16 $ . The idea is to provide an analytic continuation to the Dirichlet series $A_3(-s,s)$ and $A_4(-s,s)$ in the strip $ 0<\operatorname {Re}(s) < \frac 12$ and to shift the contour of integration to the right.

Lemma 5.1. The product formula

(5.2) $$ \begin{align} A_3 ( -s, s) = \zeta(3) \zeta( \tfrac32 - 3s ) \prod_p \bigg( 1 - \frac{1}{ p^{\frac32 +s}} + \frac{1}{p^{\frac52 - s}} - \frac{1}{p^{\frac52 -3s}} - \frac{1}{p^{3-4s}} + \frac{1}{p^{\frac92 - 5s}} \bigg) \end{align} $$

provides an analytic continuation of $A_3 (-s, s)$ to $|\operatorname {Re} ( s )| < \frac 12$ except for a simple pole at $s= \frac 16$ with residue

$$ \begin{align*}-\frac{\zeta(3)}{3\zeta(\frac53)\zeta(2)}.\end{align*} $$

Proof. From (4.7) and (4.8), we see that in the region $ | \operatorname {Re} (s) | < \frac 16$ ,

(5.3) $$ \begin{align} \notag A_3 ( -s, s) & = \prod_p \bigg( 1 - \frac1{p^3} \bigg)^{-1} \bigg( 1 - \frac1{p^{1-2s}}\bigg) \\ &\quad \times\bigg( 1+ \frac1p + \frac1{p^2} + \sum_{e \geq 1 } \frac{ f(e,0,p)}{p^{e(\frac 12-s)}} + \sum_{ e \geq 0 } \frac{f(e,1,p)}{p^{e(\frac 12-s)+\frac12+s}} + \sum_{e \geq 0 } \frac{f(e,2,p)}{ p^{e(\frac 12-s)+1+2s }} \bigg) \notag \\ & = \zeta(3) \prod_p \bigg( 1 - \frac1{p^{1-2s}}\bigg) \bigg( \frac1{p^2} + \sum_{e \geq 0 } \frac{1}{p^{e(\frac 12-s)}} \bigg( f(e,0,p) + \frac{f(e,1,p)}{p^{\frac 12+s}} + \frac{f(e,2,p)}{ p^{1+2s }}\bigg) \bigg). \end{align} $$

The sum over $e\geq 0$ on the right-hand side is equal to

(5.4) $$ \begin{align} & \frac16 \bigg( 1 - \frac{1}{ p^{\frac 12+s}} \bigg)^2 \sum_{e \geq 0 } (e+1) \frac{1}{p^{e(\frac 12-s)}} + \frac12 \bigg( 1 - \frac{1}{p^{1+2s}} \bigg) \sum_{e \geq 0 } \frac{ 1+(-1)^e}{2} \frac{1}{p^{e(\frac 12-s)}} \nonumber\\ &\qquad + \frac13 \bigg( 1 + \frac{1}{p^{\frac 12+s}}+ \frac{1}{p^{1+2s}} \bigg) \sum_{e \geq 0 }\tau_e \frac{1}{p^{e(\frac 12-s)}} + \frac1p \bigg( 1- \frac{1}{p^{\frac 12+s}}\bigg) \sum_{e \geq 0 } \frac{1}{p^{e(\frac 12-s)}}\nonumber\\ &\quad = \frac16 \cdot \frac{ \Big(1 - \frac{1}{ p^{\frac 12+s}} \Big)^2}{ \Big(1 - \frac{1}{ p^{\frac 12-s}}\Big)^2} + \frac12 \cdot \frac{ 1 - \frac{1}{p^{1+2s}} }{ 1 - \frac{1}{p^{1-2s}} } + \frac13 \cdot \frac{ 1 + \frac{1}{p^{\frac 12+s}}+ \frac{1}{p^{1+2s}}}{ 1 + \frac{1}{p^{\frac 12-s}}+ \frac{1}{p^{1-2s}} } + \frac1p \cdot \frac{ 1- \frac{1}{p^{\frac 12+s}}}{ 1- \frac{1}{p^{\frac 12-s}} }. \end{align} $$

Here, we have used geometric sum identities, for example,

$$ \begin{align*} \sum_{k=0}^\infty \tau_k x^k & = \sum_{k=0}^\infty x^{3k} - \sum_{k=0}^\infty x^{3k+1} = \frac{1-x}{1-x^3 } = \frac{1}{1+x+x^2 } \qquad (|x| < 1). \end{align*} $$

Inserting the expression (5.4) in (5.3) and simplifying, we obtain the identity

$$ \begin{align*} A_3 ( -s, s) = \zeta(3) \zeta( \tfrac32 - 3s ) \prod_p \bigg( 1 - \frac{1}{ p^{\frac 32+s}} + \frac{1}{p^{\frac 52 - s}} - \frac{1}{p^{\frac 52-3s}} - \frac{1}{p^{3-4s}} + \frac{1}{p^{\frac 92 - 5s}} \bigg) \end{align*} $$

in the region $|\operatorname {Re}(s)|<1/6$ . Now, this clearly extends to $ | \operatorname {Re} ( s ) | < 1/2$ except for a simple pole at $ s= 1/6 $ with residue equal to

$$ \begin{align*}-\frac{\zeta(3)}{3} \prod_p \big( 1 - p^{-\frac53} - p^{-2} + p^{-\frac{11}{3}}\big) = - \frac{ \zeta(3)}{3} \frac{ 1}{ \zeta(\frac53) \zeta(2)} ,\end{align*} $$

as desired.

Lemma 5.2. Assuming the Riemann Hypothesis, the function $A_4(-s,s)$ admits an analytic continuation to the region $|\operatorname {Re}(s)|< \frac 12$ , except for a double pole at $s=\frac 16$ . Furthermore, for any $0<\varepsilon <\frac 14$ and in the region $|\operatorname {Re}(s)|< \frac 12-\varepsilon $ , we have the bound

$$ \begin{align*}A_4(-s,s) \ll_\varepsilon (|\operatorname{Im}(s)|+1)^{ \frac23} .\end{align*} $$

Proof. By (4.7) and (4.9), for $ | \operatorname {Re} (s) | < \frac 16$ , we have that

$$ \begin{align*} A_4 ( -s, s) = & \prod_p \frac{ \Big(1- \frac1{p^{1-2s}}\Big)\Big( 1- \frac1{p^{\frac56-s}}\Big)\Big( 1 - \frac1{p^{\frac13}} \Big) }{ \Big(1- \frac1{p^2}\Big)\Big( 1- \frac1{p^{\frac56+s}}\Big) \Big(1- \frac1{p^{\frac53}}\Big) } \Bigg( \frac1{p^2} \Big( 1+ \frac1{p^{\frac13}}\Big) + \sum_{e \geq 0 } \frac{ g(e,0,p) + \frac{g(e,1,p)}{p^{\frac12+s}} + \frac{g(e,2,p)}{ p^{1+2s }} }{p^{e(\frac 12-s)}} \Bigg), \end{align*} $$

since $ y_p^{-1} - g(0,0,p) = \frac 1{p^2} \Big ( 1+ \frac 1{p^{\frac 13}}\Big ) $ . Recalling the definition of $g(e,j,p)$ (see Lemma 4.1), a straightforward evaluation of the infinite sum over $e\geq 0$ yields the expression

$$ \begin{align*} & A_4 ( -s, s) = \zeta(2) \zeta(\tfrac53) \prod_p \frac{ \Big(1- \frac1{p^{1-2s}}\Big)\Big( 1- \frac1{p^{\frac56 -s}}\Big)\Big( 1 - \frac1{p^{\frac13}}\Big) }{ \Big( 1- \frac1{p^{\frac56 +s}}\Big) } \Bigg( \frac{ \Big(1+\frac{1}{p^{ \frac13}} \Big)^3\Big( 1 - \frac{1}{ p^{\frac12+s}} \Big)^2 }{ 6 \Big( 1 - \frac{1}{ p^{\frac12-s}} \Big)^2} \\ & + \frac{ \Big(1+\frac{1}{p^{ \frac13}} \Big)\Big(1+\frac{1}{p^{ \frac23}} \Big) \Big( 1 - \frac{1}{p^{1+2s}}\Big) }{2 \Big( 1 - \frac{1}{p^{1-2s}}\Big)} + \frac{ \Big( 1+\frac1p \Big) \Big( 1 + \frac{1}{p^{\frac12 +s}}+ \frac{1}{p^{1+2s}}\Big) }{ 3 \Big( 1 + \frac{1}{p^{\frac12-s}} + \frac{1}{p^{1-2s}} \Big) } + \frac{\Big(1+\frac{1}{p^{ \frac13}} \Big)^2 \Big( 1- \frac{1}{p^{\frac12+s}}\Big) }{ p\Big( 1- \frac{1}{p^{\frac12-s}}\Big) } + \frac{ 1+\frac{1}{p^{ \frac13}} }{p^2} \Bigg). \end{align*} $$

Isolating the ‘divergent terms’ leads us to the identity

$$ \begin{align*} A_4 ( -s, s) = & \zeta(2) \zeta(\tfrac53) \prod_p (D_{4,p,1}(s) + A_{4,p,1}(s) ), \end{align*} $$

where

$$ \begin{align*} D_{4,p,1}(s) & := \frac{ 1- \frac1{p^{\frac56-s}} }{ 1- \frac1{p^{\frac56+s}} } \Bigg( \frac{ \Big(1+ \frac2{p^{\frac13}} - \frac2p \Big) \Big(1 - \frac{1}{ p^{\frac12+s}} \Big)^2 \Big(1 + \frac1{p^{\frac12-s}}\Big) }{ 6\Big( 1 - \frac{1}{ p^{\frac12-s}}\Big)} \\ &\quad + \frac{ 1 - \frac{1}{p^{1+2s}} }{2} + \frac{\Big(1 - \frac1{p^{\frac13}} +\frac1p \Big)\Big( 1 + \frac{1}{p^{\frac12+s}}+ \frac{1}{p^{1+2s}} \Big) \Big(1- \frac1{p^{1-2s}}\Big) }{3\Big( 1 + \frac{1}{p^{\frac12-s}} + \frac{1}{p^{1-2s}} \Big) } + \frac1p \Bigg) \end{align*} $$

and

$$ \begin{align*} A_{4,p,1}(s) &:= \frac{ 1- \frac1{p^{\frac56-s}} }{ 1- \frac1{p^{\frac56+s}} } \Bigg( - \frac{ \Big(1 - \frac{1}{ p^{\frac12+s}} \Big)^2 \Big(1 + \frac1{p^{\frac12-s}}\Big) }{ 6 p^{\frac43} \Big( 1 - \frac{1}{ p^{\frac12-s}}\Big)} - \frac{ 1 - \frac{1}{p^{1+2s}} }{2p^{\frac43}} - \frac{ \Big( 1 + \frac{1}{p^{\frac12+s}}+\frac{1}{p^{1+2s}} \Big) \Big(1- \frac1{p^{1-2s}}\Big) }{3 p^{\frac43} \Big( 1 + \frac{1}{p^{\frac12-s}} + \frac{1}{p^{1-2s}} \Big) } \\ & \quad + \frac{\Big(1+ \frac1{p^{\frac13}} - \frac1{p^{\frac23}}- \frac1p \Big) \Big( 1- \frac{1}{p^{\frac12+s}}\Big)\Big( 1+ \frac{1}{p^{\frac12-s}} \Big) -1 }{p} + \frac1{p^2} \Big( 1- \frac1{p^{\frac23}}\Big) \Big(1- \frac1{p^{1-2s}}\Big) \Bigg). \end{align*} $$

The term $A_{4,p,1}(s)$ is ‘small’ for $ | \operatorname {Re} (s)|< \frac 12$ , hence, we will concentrate our attention on $D_{4,p,1}(s)$ . We see that

$$ \begin{align*}D_{4,p,1}(s) = \frac{ 1- \frac1{p^{\frac56-s}} }{ 1- \frac1{p^{\frac56+s}} } D_{4,p,2}(s) + \frac1p + A_{4,p,2}(s) , \end{align*} $$

where

$$ \begin{align*}D_{4,p,2}(s) & := \frac{ \Big(1+ \frac2{p^{\frac13}} \Big) \Big(1 - \frac{1}{ p^{\frac12+s}} \Big)^2 \Big(1 + \frac1{p^{\frac12-s}}\Big) }{ 6\Big( 1 - \frac{1}{ p^{\frac12-s}}\Big)} + \frac{ 1 - \frac{1}{p^{1+2s}} }{2} \\ &\quad + \frac{\Big(1 - \frac1{p^{\frac13}} \Big)\Big( 1 + \frac{1}{p^{\frac12+s}}+ \frac{1}{p^{1+2s}} \Big) \Big(1- \frac1{p^{1-2s}}\Big) }{3\Big( 1 + \frac{1}{p^{\frac12-s}} + \frac{1}{p^{1-2s}} \Big) }\end{align*} $$

and

$$ \begin{align*} A_{4,p,2}(s):= \frac{ \Big( 1- \frac1{p^{\frac56-s}} \Big) }{ p \Big( 1- \frac1{p^{\frac56+s}} \Big) } \Bigg( & - \frac{ \Big(1 - \frac{1}{ p^{\frac12+s}} \Big)^2 \Big(1 + \frac1{p^{\frac12-s}}\Big) }{ 3\Big( 1 - \frac{1}{ p^{\frac12 -s}}\Big)} + \frac{ \Big( 1 + \frac{1}{p^{\frac12+s}}+ \frac{1}{p^{1+2s}} \Big) \Big(1- \frac1{p^{1-2s}}\Big) }{3\Big( 1 + \frac{1}{p^{\frac12-s}} + \frac{1}{p^{1-2s}} \Big) } +1 \Bigg) - \frac1p, \end{align*} $$

which is also ‘small’. Taking common denominators and expanding out shows that

$$ \begin{align*}D_{4,p,2}(s)=\frac{1}{ 1- \frac{1}{p^{\frac32 -3s}} } \bigg( 1 - \frac1p - \frac{1}{p^{\frac56+s}} + \frac{1}{p^{\frac56-s}} + \frac1{p^{\frac43-2s}} + A_{4,p,3}(s) \bigg),\end{align*} $$

where

$$ \begin{align*}A_{4,p,3}(s) := - \frac{1}{p^{\frac32-s}} + \frac{1}{ p^{\frac52-s}} - \frac1{p^{\frac43}} + \frac{1}{p^{\frac{11}6+s}} - \frac1{p^{\frac{11}6 -s}} + \frac1{p^{\frac73} } - \frac{1}{p^{\frac73-2s}}\end{align*} $$

is ‘small’. More precisely, for $ | \operatorname {Re} (s) |\leq \frac 12 - \varepsilon < \frac 12 $ and $ j = 1,2,3$ , we have the bound $ A_{4,p,j } (s) = O_\varepsilon \Big ( \frac {1}{p^{1+\varepsilon }}\Big ) $ . Therefore,

(5.5) $$ \begin{align} A_4 ( -s, s) = \zeta(2) \zeta(\tfrac53) \zeta( \tfrac32-3s) \widetilde{A}_4 (s) \prod_p \Bigg( \frac{\Big( 1- \frac1{p^{\frac56-s}}\Big) }{ \Big( 1- \frac1{p^{\frac56+s}}\Big) } \bigg( 1 - \frac{1}{p^{\frac56+s}} + \frac{1}{p^{\frac56-s}} + \frac1{p^{\frac43-2s}} \bigg) \Bigg) , \end{align} $$

where

$$ \begin{align*} \widetilde{A}_4 (s) := & \prod_p \left( 1 + \frac{ \frac1p \bigg( 1- \frac{1}{p^{\frac32 -3s}} - \frac{ 1- p^{-\frac56+s} }{ 1- p^{- \frac56-s} } \bigg) + \frac{ 1- p^{-\frac56+s} }{ 1- p^{- \frac56-s} } A_{4,p,3}(s) + \Big( 1- \frac{1}{p^{\frac32-3s}}\Big)(A_{4,p,2}(s) + A_{4,p,1}(s)) }{ \frac{ 1- p^{-\frac56+s} }{ 1- p^{- \frac56-s} } \Big( 1 - \frac{1}{p^{\frac56+s}} + \frac{1}{p^{\frac56-s}} + \frac1{p^{\frac43-2s}} \Big) } \right) \end{align*} $$

is absolutely convergent for $|\operatorname {Re} (s) | < \frac 12 $ . Hence, the final step is to find a meromorphic continuation for the infinite product on the right-hand side of (5.5), which we will denote by $D_3(s)$ . However, it is straightforward to show that

(5.6) $$ \begin{align} A_{4,4}(s):= D_3(s) \frac{ \zeta(\tfrac83-4s) \zeta(\tfrac53-2s) \zeta(\tfrac{13}6-3s) } { \zeta(\tfrac43-2s)} \end{align} $$

converges absolutely for $|\operatorname {Re}(s)|<\frac 12$ . This finishes the proof of the first claim in the lemma.

Finally, the growth estimate

$$ \begin{align*}A_4(-s,s)\ll_\varepsilon (|\operatorname{Im}(s)|+1)^{\varepsilon} | \zeta( \tfrac32 - 3s ) \zeta( \tfrac43 - 2s)| \ll_\varepsilon (|\operatorname{Im}(s)|+1)^{ \frac23}\end{align*} $$

follows from (5.5), (5.6), as well as [Reference Montgomery and VaughanMV, Theorems 13.18 and 13.23] and the functional equation for $\zeta (s)$ .

Now that we have a meromorphic continuation of $A_4(-s,s)$ , we will calculate the leading Laurent coefficient at $s=\frac 16$ .

Lemma 5.3. We have the formula

$$ \begin{align*}\lim_{s\rightarrow \frac 16} (s-\tfrac 16)^2 A_4(-s,s) = \frac16 \frac{ \zeta(2)\zeta(\tfrac 53)}{\zeta( \tfrac43)} \prod_p \Big(1-\frac 1{p^{\frac 23}}\Big)^2 \Big(1- \frac1p \Big)\Big(1+\frac 2{p^{\frac 23}}+\frac 1p+\frac 1{p^{\frac 43}}\Big) .\end{align*} $$

Proof. By Lemma 5.2, $A_4 ( -s, s ) $ has a double pole at $ s= \frac 16 $ . Moreover, by (5.5) and (5.6), we find that $ \frac { A_4 ( -s, s)}{ \zeta ( \frac 32 - 3s ) \zeta ( \frac 43 - 2s ) } $ has a convergent Euler product in the region $|\operatorname {Re}(s)| < \frac 13$ (this allows us to interchange the order of the limit and the product in the calculation below), so that

$$ \begin{align*} \lim_{s \to \frac16 } & (s-\tfrac16 )^2 A_4 ( -s , s) = \frac{1}{6} \lim_{s \to \frac16} \frac{ A_4 ( -s, s)}{ \zeta( \frac32 - 3s ) \zeta ( \frac43 - 2s ) } \\ = & \frac{ \zeta(2)\zeta(\tfrac 53) }{6} \prod_p \Big(1-\frac1p \Big) \Big(1-\frac 1{p^{\frac 23}}\Big)^2 \Big(1-\frac 1{p^{\frac 13}}\Big) \Bigg\{ \frac{ \Big(1+\frac 1{p^{\frac 13}}\Big)^3 \Big(1 - \frac{1}{ p^{\frac23}}\Big)^2 }{ 6 \Big( 1 - \frac{1}{ p^{\frac13}}\Big)^2 } \\ & + \frac{ \Big(1+ \frac1{p^{ \frac13}} \Big) \Big( 1+ \frac1{p^{ \frac23}} \Big) \Big( 1 - \frac{1}{p^{\frac43}}\Big) }{2\Big( 1 - \frac{1}{p^{\frac23}}\Big) } + \frac{ \Big(1+ \frac1p \Big)\Big(1 + \frac{1}{p^{\frac23}}+ \frac{1}{p^{\frac43}}\Big) }{3 \Big( 1 + \frac{1}{p^{\frac13}} + \frac{1}{p^{\frac23}} \Big) } + \frac{ \Big( 1 + \frac1{p^{ \frac13}} \Big)^2 \Big( 1- \frac{1}{p^{\frac23}} \Big) }{p\Big( 1- \frac{1}{p^{\frac13}} \Big) } + \frac{ 1+ \frac1{p^{\frac13}} }{p^2} \Bigg\}. \end{align*} $$

The claim follows.

We are now ready to estimate $J^\pm (X)$ when the support of $\widehat \phi $ is small.

Lemma 5.4. Let $\phi $ be a real even Schwartz function, such that $\sigma =\sup (\mathrm {supp} (\widehat \phi )) <1$ . Let $J^\pm (X)$ be defined by (5.1). Then we have the estimate

$$ \begin{align*}J^\pm(X) = C^\pm \phi \Big( \frac{L }{12 \pi i}\Big) X^{-\frac 13}+O_{\varepsilon} \Big(X^{\frac {\sigma-1}2+\varepsilon }\Big),\end{align*} $$

where

(5.7) $$ \begin{align} C^\pm := \frac{5}{12} \frac{C_2^\pm }{C_1^\pm} \frac{\Gamma_{\pm}(\frac 13)}{\Gamma_{\pm}(\frac 23)} \frac{ \zeta(\tfrac 23)^2 \zeta(\tfrac 53) \zeta(2) }{ \zeta( \tfrac43)} \prod_p \Big(1-\frac 1{p^{\frac 23}}\Big)^2 \Big(1- \frac1p \Big)\Big(1+\frac 2{p^{\frac 23}}+\frac 1p+\frac 1{p^{\frac 43}}\Big). \end{align} $$

Proof. We rewrite the integral in $J^\pm (X)$ as

(5.8) $$ \begin{align} &\frac{1}{2 \pi i} \int_{(c)} (-2) \phi \Big(\frac{Ls}{2\pi i}\Big)\Big\{ \Big( 1-\frac{C_2^\pm}{C_1^\pm}X^{-\frac{1}{6}}\Big)\Big(-\frac{C_2^\pm}{C_1^\pm} X^{-\frac 16} \frac{\zeta'}{\zeta}(\tfrac 56+s)+ X^{-s} \frac{\Gamma_{\pm}(\frac 12-s)}{\Gamma_{\pm}(\frac 12+s)}\zeta(1-2s) \frac{A_3( -s, s)}{1-s} \Big)\nonumber\\ &\qquad +\frac{C_2^\pm }{C_1^\pm} X^{-s-\frac 16} \Big( 1-\frac{C_2^\pm}{C_1^\pm}X^{-\frac{1}{6}}\Big) \frac{\Gamma_{\pm}(\frac 12-s)}{\Gamma_{\pm}(\frac 12+s)} \zeta(1-2s) \frac{\zeta(\tfrac 56-s)}{\zeta(\tfrac 56+s)} \frac{A_4(-s,s)}{ 1- \frac{6s}5} \nonumber\\ &\qquad +\Big(\frac{C_2^\pm}{C_1^\pm}\Big)^2 X^{-s-\frac 13} \frac{\Gamma_{\pm}(\frac 12-s)}{\Gamma_{\pm}(\frac 12+s)}\zeta(1-2s) \frac{ A_3( -s, s)}{1-s} \Big\}ds \end{align} $$

for $ 0 < c < \frac 16$ . The integrand has a simple pole at $s=\frac 16$ with residue

(5.9) $$ \begin{align} & -2\phi \Big( \frac{L }{12 \pi i}\Big) \Big( 1-\frac{C_2^\pm}{C_1^\pm}X^{-\frac{1}{6}}\Big) X^{-\frac 16} \Big(\frac{C_2^\pm}{C_1^\pm} - \frac25 \frac{\Gamma_{\pm}(\frac 13)}{\Gamma_{\pm}(\frac 23)} \frac{ \zeta(\tfrac23 ) \zeta(3)}{ \zeta( \tfrac53 ) \zeta(2)} \Big) \notag \\ & -2 \phi \Big( \frac{L }{12 \pi i}\Big) \frac{C_2^\pm }{C_1^\pm} X^{-\frac 13} \frac{\Gamma_{\pm}(\frac 13)}{\Gamma_{\pm}(\frac 23)} \frac{5\zeta(\tfrac 23)^2}{4} \lim_{s\rightarrow \frac 16} (s-\tfrac 16)^2A_4(-s,s) +O\Big(\phi\Big( \frac{L}{12 \pi i}\Big) X^{-\frac 12} \Big) \\ &\qquad = -{C^\pm} \phi \Big( \frac{L }{12 \pi i}\Big) X^{- \frac13} + O( X^{\frac{\sigma}{6}- \frac12}) \notag \end{align} $$

by Lemma 5.3, as well as the fact that the first line vanishes. Due to Lemmas 5.1 and 5.2, we can shift the contour of integration to the line $\operatorname {Re}(s)=\frac 12 - \frac {\varepsilon }{2}$ , at the cost of $-1$ times the residue (5.9).

We now estimate the shifted integral. The term involving $\frac {\zeta '}{\zeta }(\tfrac 56+s)$ can be evaluated by interchanging sum and integral; we obtain the identity

(5.10) $$ \begin{align} \frac{1}{\pi i} \int_{(\frac 12-\frac \varepsilon 2)} \phi \Big(\frac{Ls}{2\pi i}\Big) \frac{\zeta'}{\zeta}(\tfrac 56+s)ds = - \frac 2 L\sum_{p,e}\frac{\log p}{p^{\frac {5e}6}}\widehat\phi\Big(\frac{\log p^e}{L}\Big). \end{align} $$

The last step is to bound the remaining terms, which is carried out by combining (4.17) with Lemmas 5.1 and 5.2.

Finally, we complete the Proof of Theorem 1.4.

Proof of Theorem 1.4

Given Proposition 4.4 and Lemma 5.4, the only thing remaining to prove is (1.7). Applying (5.8) with $c=\frac 1{20}$ and splitting the integral into two parts, we obtain the identity

$$ \begin{align*} &J^\pm (X) = \frac{2C_2^\pm X^{-\frac 16}}{C_1^\pm L}\Big( 1-\frac{C_2^\pm}{C_1^\pm}X^{-\frac{1}{6}}\Big) \sum_{p,e}\frac{\log p}{p^{\frac {5e}{6}}}\widehat\phi\left(\frac{\log p^e}{L}\right) \\ &\quad -\frac{1}{\pi i} \int_{(\frac{1}{20})} \phi \Big(\frac{Ls}{2\pi i}\Big)\Big\{ \Big( 1-\frac{C_2^\pm}{C_1^\pm}X^{-\frac{1}{6}}\Big)\Big(-\frac{C_2^\pm}{C_1^\pm} X^{-\frac 16} \frac{\zeta'}{\zeta}(\tfrac 56+s)+ X^{-s} \frac{\Gamma_{\pm}(\frac 12-s)}{\Gamma_{\pm}(\frac 12+s)}\zeta(1-2s) \frac{ A_3( -s, s)}{1-s} \Big) \Big\} ds\\ &\quad-\frac{1}{\pi i} \int_{(\frac{1}{20})} \phi \Big(\frac{Ls}{2\pi i}\Big)\Big\{ \frac{C_2^\pm }{C_1^\pm} X^{-s-\frac 16} \Big( 1-\frac{C_2^\pm}{C_1^\pm}X^{-\frac{1}{6}}\Big) \frac{\Gamma_{\pm}(\frac 12-s)}{\Gamma_{\pm}(\frac 12+s)} \zeta(1-2s) \frac{\zeta(\tfrac 56-s)}{\zeta(\tfrac 56+s)} \frac{A_4(-s,s)}{ 1- \frac{6s}5} \\ &\quad +\Big(\frac{C_2^\pm}{C_1^\pm}\Big)^2X^{-s-\frac 13} \frac{\Gamma_{\pm}(\frac 12-s)}{\Gamma_{\pm}(\frac 12+s)}\zeta(1-2s) \frac{ A_3( -s, s) }{1-s} \Big\}ds. \end{align*} $$

By shifting the first integral to the line $\operatorname {Re}(s) = \frac 15$ and applying (5.10), we derive (1.7). Note that the residue at $s=\frac 16$ is the first line of (5.9), which is equal to zero.

Figure 2 A plot of $(p,f_{p}(10^4,T_j))$ for $p<10^4$ and $j=1,2,3$ .