Smooth integers and de Bruijn's approximation Ʌ

Abstract

This paper is concerned with the relationship of $y$-smooth integers and de Bruijn's approximation $\Lambda (x,\,y)$. Under the Riemann hypothesis, Saias proved that the count of $y$-smooth integers up to $x$, $\Psi (x,\,y)$, is asymptotic to $\Lambda (x,\,y)$ when $y \ge (\log x)^{2+\varepsilon }$. We extend the range to $y \ge (\log x)^{3/2+\varepsilon }$ by introducing a correction factor that takes into account the contributions of zeta zeros and prime powers. We use this correction term to uncover a lower order term in the asymptotics of $\Psi (x,\,y)/\Lambda (x,\,y)$. The term relates to the error term in the prime number theorem, and implies that large positive (resp. negative) values of $\sum _{n \le y} \Lambda (n)-y$ lead to large positive (resp. negative) values of $\Psi (x,\,y)-\Lambda (x,\,y)$, and vice versa. Under the Linear Independence hypothesis, we show a Chebyshev's bias in $\Psi (x,\,y)-\Lambda (x,\,y)$.

1. Introduction

A positive integer is called $y$-smooth if each of its prime factors does not exceed $y$. We denote the number of $y$-smooth integers not exceeding $x$ by $\Psi (x,\,y)$. We assume throughout $x \ge y \ge 2$. Let $\rho \colon [0,\,\infty ) \to (0,\,\infty )$ be the Dickman function, defined as $\rho (t)=1$ for $t \in [0,\,1]$ and via the delay differential equation $t \rho '(t) =-\rho (t-1)$ for $t>1$. Dickman [7] showed that

(1.1)\begin{equation} \Psi(x,y) \sim x \rho ( \log x/\log y) \quad (x \to \infty) \end{equation}

holds when $y \ge x^{\varepsilon }$. For this reason, it is useful to introduce

\[ u:=\log x/\log y. \]

De Bruijn [3, Eqs. (1.3), (4.6)] showed that

(1.2)\begin{equation} \Psi(x,y)- x\rho(u) \sim (1-\gamma)\frac{x\rho(u-1)}{\log x}>0 \end{equation}

when $x \to \infty$ and $(\log x)/2 >\log y > (\log x)^{5/8}$. Here and later $\gamma$ is the Euler–Mascheroni constant. As we see, there is no arithmetic information in the leading behaviour of the error term $\Psi (x,\,y) - x \rho (u)$, and in particular it does not oscillate. Moreover, the error term is large: the saving (1.2) gives over the main term is merely $\asymp \log (u+1)/\log y$ [3, p. 56].

This begs the question, what is the correct main term for $\Psi (x,\,y)$ that leads to a small and arithmetically rich error term? De Bruijn [3, Eq. (2.9)] introduced a refinement of $\rho$, often denoted $\lambda _y$:

\[ \lambda_y(u) := \int_{0}^{\infty} \rho\left(u-\frac{\log t}{\log y}\right)d\left(\frac{\lfloor t \rfloor}{t}\right) = \int_{\mathbb{R}} \rho(u-v)d\left(\frac{\lfloor y^v \rfloor}{y^v}\right) \]

if $y^u \notin \mathbb {Z}$; otherwise $\lambda _y(u)=\lambda _y(u+)$ (one has $\lambda _y(u)= \lambda _y(u-)+O(1/x)$ if $y^u \in \mathbb {Z}$ [3, p. 54]). The count $\Psi (x,\,y)$ should be compared to

\[ \Lambda(x,y):= x \lambda_y(u). \]

We refer the reader to de Bruijn's original paper for the motivation for this definition. In particular, $\Lambda$ satisfies the following continuous variant of Buchstab's identity:

\[ \Lambda(x,y)=\Lambda(x,z)-\int_{y}^{z}\Lambda\left(\frac{x}{t},t\right)\frac{{\rm d}t}{\log t} \]

for $y \le z$, to be compared with $\Psi (x,\,y)=\Psi (x,\,z)-\sum _{y < p \le z}\Psi (x/p,\,p)$. De Bruijn proved [3, Eq. (1.4)]

(1.3)\begin{equation} \Lambda(x,y)=x\rho(u)\left(1+O_{\varepsilon}\left(\frac{\log(u+1)}{\log y}\right)\right) \end{equation}

holds for $\log y > \sqrt {\log x}$. Saias [17, Lem. 4] improved the range to $y \ge (\log x)^{1+\varepsilon }$. De Bruijn and Saias also provided asymptotic series expansion for $\lambda _y(u)$ in (roughly) powers of $\log (u+1)/\log y$. Hildebrand and Tenenbaum [14, Lem. 3.1] showed that for $y \ge (\log x)^{1+\varepsilon }$,

(1.4)\begin{equation} \Lambda(x,y) \asymp_{\varepsilon} x\rho(u) \end{equation}

for $y \ge (\log x)^{1+\varepsilon }$. Implicit in the proof of proposition 4.1 of La Bretèche and Tenenbaum [5] is the estimate

(1.5)\begin{equation} \Lambda(x,y) = x \rho(u) K\left( - \frac{\xi(u)}{\log y}\right) \left(1 + O_{\varepsilon}\left( \frac{1}{\log x}\right)\right), \quad K(t):=\frac{t\zeta(t+1)}{t+1}, \end{equation}

for $y \ge (\log x)^{1+\varepsilon }$ where $\zeta$ is the Riemann zeta function and $\xi \colon [1,\,\infty ) \to [0,\,\infty )$ is defined via

\[ e^{\xi(u)}=1+u\xi(u). \]

We include as an appendix a proof in English of (1.5). The function $K$ originates in de Bruijn's work [3, Eq. (2.8)]. Evidently, $K(0)=1$ and $\lim _{t \to -1^+} K(t)= \infty$. Moreover, $K$ is strictly decreasing in $(-1,\,0]$ [9].

Suppose $\pi (x)=\mathrm {Li}(x)(1+O(\exp (-(\log x)^{a})))$ for some $a \in (0,\,1)$. Saias [17, Thm.], improving on De Bruijn [3], proved that

(1.6)\begin{equation} \Psi(x,y) = \Lambda(x,y)(1 + O_{\varepsilon}(\exp(-(\log y)^{a-\varepsilon}))) \end{equation}

holds in the range $\log y \ge (\log \log x)^{{1}/{a}+\varepsilon }$. By the Vinogradov–Korobov zero-free region, we may take $a=3/5$. Saias writes without proof [17, p. 81] that under the Riemann hypothesis (RH) his methods give

(1.7)\begin{equation} \Psi(x,y) = \Lambda(x,y) (1+O_{\varepsilon}(y^{\varepsilon-1/2}\log x)) \end{equation}

in the range $y \ge (\log x)^{2+\varepsilon }$, which recovers a conditional result of Hildebrand [11].

1.1 $G$

Define the entire function $I(s)=\int _{0}^{s} \tfrac {e^v-1}{v}\,{\rm d}v$. As shown in [14, Lem. 2.6], the Laplace transform of $\rho$ is

(1.8)\begin{equation} \hat{\rho}(s) := \int_{0}^{\infty} e^{{-}sv}\rho(v)\, \,{\rm d}v = \exp( \gamma + I({-}s)) \end{equation}

for all $s \in \mathbb {C}$. In [9] we studied in detail the ratio

\[ G(s,y) := \zeta(s,y) / F(s,y) \]

where

\[ \zeta(s,y):=\prod_{p \le y} (1-p^{{-}s})^{{-}1}=\sum_{n \text{ is }y\text{-smooth}} n^{{-}s} \quad (\Re s >0) \]

is the partial zeta function and

(1.9)\begin{equation} F(s,y):= \hat{\rho}((s-1)\log y)\zeta(s)(s-1)\log y. \end{equation}

The function $G(s,\,y)$ is defined for $\Re s>0$ such that $\zeta (s) \neq 0$. Informally, $G$ carries information about the ratio $\Psi (x,\,y)/\Lambda (x,\,y)$, since $s\mapsto \zeta (s,\,y)/s$ is the Mellin transform of $x\mapsto \Psi (x,\,y)$ while $s\mapsto F(s,\,y)/s$ is the Mellin transform of $x\mapsto \Lambda (x,\,y)$ [3, p. 54]. As in [9], it is essential to write $G$ as $G_1 G_2$ where

\begin{align*} \log G_1(s,y) & = \sum_{n \le y} \frac{\Lambda(n)}{n^{s}\log n}-(\log (\zeta(s)(s-1))+\log \log y+\gamma+ I((1-s)\log y)),\\ \log G_2(s,y) & = \sum_{k \ge 2} \sum_{y^{1/k} < p \le y} \frac{p^{{-}ks}}{k}. \end{align*}

We assume $\log \zeta (s)$ is chosen to be real when $s>1$.

1.2 Main results

Let $\psi (y)=\sum _{n \le y}\Lambda (n)$ and

(1.10)\begin{equation} \beta:=1-\frac{\xi(u)}{\log y}. \end{equation}

Theorem 1.1 Assume RH. Fix $\varepsilon \in (0,\,1)$. Suppose that $x \ge C_{\varepsilon}$ and $x^{1-\varepsilon } \ge y \ge (\log x)^{2+\varepsilon }$. Then

(1.11)\begin{equation} \Psi(x,y) = \Lambda(x,y) G(\beta,y) \left( 1 + O_{\varepsilon}\left( \frac{\log (u+1)}{y \log y} \left( |\psi(y)-y| + y^{{1}/{2}}\right)\right)\right). \end{equation}

The following theorem gives an asymptotic formula for $\Psi (x,\,y)$ for $y$ smaller than $(\log x)^{2}$.

Theorem 1.2 Assume RH. Fix $\varepsilon \in (0,\,1/3)$. Suppose that $x \ge C_{\varepsilon}$ and $(\log x)^{3} \ge y \ge (\log x)^{4/3+\varepsilon }$. Then

(1.12)\begin{equation} \Psi(x,y) = \Lambda(x,y) G(\beta,y) \left( 1 + O_{\varepsilon}\left( \frac{(\log y)^3}{y^{{1}/{2}}} + \frac{(\log x)^3 (\log y)^3}{y^2}\right)\right). \end{equation}

If $y \le (\log x)^{2-\varepsilon }$ then the error term can be improved to $O_{\varepsilon } ((\log x)^3/(y^2 \log y))$.

Theorems 1.1 and 1.2, proved in § 4, show that

\[ \Psi(x,y) \sim \Lambda(x,y) G(\beta,y) \]

holds when $y/((\log x)^{3/2}(\log \log x)^{-1/2}) \to \infty$. This range is shown to be optimal in Theorem 2.14 of [9]. The same theorem also supplies an alternative proof of theorem 1.2 when $y \le (\log x)^{2-\varepsilon }$ (the proof can be adapted to cover $(\log x)^{2-\varepsilon } \le y \le (\log x)^3$ as well).

Hildebrand showed that RH is equivalent to $\Psi (x,\,y) \asymp _{\varepsilon } x\rho (u)$ for $y \ge (\log x)^{2+\varepsilon }$ [11]. He conjectured that $\Psi (x,\,y)$ is not of size $\asymp x\rho (u)$ when $y \le (\log x)^{2-\varepsilon }$ [12]. This was recently confirmed by the author [9]. This also follows (under RH) from theorem 1.2, since $\Lambda (x,\,y) \asymp _{\varepsilon } x\rho (u)$ for $y \ge (\log x)^{1+\varepsilon }$ while (under RH) $G(\beta,\,y) \to \infty$ when $y \le (\log x)^{2-\varepsilon }$ and $x \to \infty$ (this follows from the estimates for $G$ in [9], see § 2).

Theorems 1.1 and 1.2 and their proofs have their origin in our work in the polynomial setting [10], where $\Psi (x,\,y)$ corresponds to the number of $m$-smooth polynomials of degree $n$ over a finite field, while $\Lambda (x,\,y)$ is analogous to the number of $m$-smooth permutations of $S_n$ (multiplied by $q^n/n!$). In that setting, the analogue of $G_1(s,\,y)$ is identically $1$ (the relevant zeta function has no zeros) which makes the analysis unconditional.

1.3 Applications: sign changes and biases

From theorem 1.1 we deduce in § 2.2 the following

Corollary 1.3 Assume RH. Fix $\varepsilon \in (0,\,1)$. Suppose that $x\ge C_{\varepsilon}$ and $x^{1-\varepsilon } \ge y \ge (\log x)^{2+\varepsilon }$. Then

\begin{align*} \Psi(x,y)/\Lambda(x,y) & = 1 +\frac{y^{-\beta}}{\log y} \bigg(- \sum_{|\rho| \le T} \frac{y^{\rho}}{\rho-\beta}\\ & \quad + \frac{y^{{1}/{2}}}{2\beta-1}+ O_{\varepsilon}\big(\frac{y^{{1}/{2}}}{\log y } +\frac{y\log^2(yT)}{T}+ \frac{|\psi(y)-y|+y^{{1}/{2}}}{u}\big)\bigg)\\ & = 1 + \frac{y^{-\beta}}{\log y} ( (\psi(y)-y)(1+O_{\varepsilon}(u^{{-}1})) + O_{\varepsilon}(y^{{1}/{2}}))\\ & =1 + O_{\varepsilon}((\log (u+1))(\log x)y^{-{1}/{2}}) \end{align*}

holds for $T \ge 4$, where the sum is over zeros of $\zeta$.

Corollary 1.3 implies that large positive (resp. negative) values of $\psi (y)-y$ lead to large positive (resp. negative) values of $\Psi (x,\,y) -\Lambda (x,\,y)$ and vice versa. Large and small values of $\psi (y)-y$ were exhibited by Littlewood [15, Thm. 15.11]. Note that corollary 1.3 sharpens (1.7) if $y \le x^{1-\varepsilon }$.¹

Let $\pi (x)$ be the count of primes up to $x$ and $\mathrm {Li}(x)$ be the logarithmic integral. It is known that $\pi (x)-\mathrm {Li}(x)$ is biased towards positive values in the following sense. Assuming RH and the Linear Independence hypothesis (LI) for zeros of $\zeta$, Rubinstein and Sarnak [16] showed that the set

\[ \{ x\ge 2 : \pi(x) > \mathrm{Li}(x)\} \]

has logarithmic density $\approx 0.999997$. This is an Archimedean analogue of the classical Chebyshev's bias on primes in arithmetic progressions. We use corollary 1.3 to exhibit a similar bias for smooth integers. Let us fix the value of $\beta =1-\xi (u)/\log y$ to be

\[ \beta=\beta_0 \]

where $\beta _0 \in (1/2,\,1)$. This amounts to restricting $x$ to be a function $x=x(y)$ of $y$ defined by

(1.13)\begin{equation} x= \exp\left(\frac{y^{1-\beta_0}-1}{1-\beta_0} \right). \end{equation}

In particular, $y=(\log x)^{1/(1-\beta _0)+o(1)}$. Then corollary 1.3 shows

(1.14)\begin{align} \frac{\Psi(x(y),y)-\Lambda(x(y),y)}{\Lambda(x(y),y)} y^{\beta_0-\frac{1}{2}}\log y & ={-}\sum_{|\rho| \le T} \frac{y^{\rho-{1}/{2}}}{\rho-\beta_0} + \frac{1}{2\beta_0-1}\nonumber\\ & \quad + O_{\beta_0}\left( \frac{y^{{1}/{2}}\log^2 (yT)}{T}+\frac{1}{\log y}\right). \end{align}

Applying the formalism of Akbary et al. [1] to the right-hand side of (1.14) we deduce immediately

Corollary 1.4 Assume RH. Assume LI for $\zeta$. Fix $\beta _0 \in (1/2,\,1)$ and let $x$ be a function of $y$ defined as in (1.13). Then the set

\[ \{ y \ge 2: \Psi(x(y),y) > \Lambda(x(y),x) \} \]

has logarithmic density greater than $1/2$, and the left-hand side of (1.14) has a limiting distribution in logarithmic sense.

In the same way that Chebyshev's bias for primes relates to the contribution of prime squares, this is also the case for smooth integers. Writing $G$ as $G_1 G_2$ as in § 1.1, $G_2$ captures the contribution of proper powers of primes. When $\beta _0 \in (1/2,\,1)$, the only significant term in $G_2(\beta _0,\,y)$ is $k=2$, which corresponds to squares of primes. The squares lead to the term $y^{1/2}/(2\beta _0-1)$ in (1.14) which creates the bias.

Remark 1.5 Consider the arithmetic function $\alpha _y(n)$ defined implicitly via

\[ \sum_{n\ge 1} \frac{\alpha_y(n)}{n^s} = \exp\bigg(\sum_{m \le y} \frac{\Lambda(m)}{ \log m}\frac{1}{m^s}\bigg). \]

This function is supported on $y$-smooth numbers and coincides with the indicator of $y$-smooth numbers on squarefree integers. Working with the summatory function of $\alpha _y$ instead of $\Psi (x,\,y)$, the bias discussed above disappears. This is because, modifying the proof of theorem 1.1, one finds that

\[ \sum_{n \le y} \alpha_y(n) = \Lambda(x,y) G_1(\beta,y)\left( 1 + O_{\varepsilon}\left( \frac{\log (u+1)}{y \log y} ( |\psi(y)-y|+y^{{1}/{2}})\right)\right) \]

holds in $x^{1-\varepsilon } \ge y \ge (\log x)^{2+\varepsilon }$, meaning the bias-causing factor $G_2(\beta,\,y)$ does not arise. This is analogous to how the indicator function of primes is biased, while $\Lambda (n)/\log n$ is not.

Remark 1.6 It is interesting to see if one can formulate and prove variants of corollaries 1.3 and 1.4 in the range $y \le (\log x)^{1-\varepsilon }$. In this range, an accurate main term for $\Psi (x,\,y)$ was established in [6].

1.4 Strategy behind theorems 1.1 and 1.2

We write $\Psi (x,\,y)$ as a Perron integral, at least for non-integer $x$:

\[ \Psi(x,y) = \frac{1}{2\pi i} \int_{(\sigma)} \zeta(s,y) \frac{x^s}{s} \,{\rm d}s \]

where $\sigma$ can be any positive real. For non-integer $x$ we also have

(1.15)\begin{equation} \Lambda(x,y)=\frac{1 }{2\pi i}\int_{(\sigma)} F(s,y) \frac{x^s}{s}\,{\rm d}s \end{equation}

whenever $\sigma >\varepsilon$ and $y \ge C_{\varepsilon}$. Indeed, the Laplace inversion formula expresses $\Lambda (x,\,y)$ as

(1.16)\begin{align} \Lambda(x,y)& =x\lambda_y(u)=\frac{x}{2\pi i}\int_{(c)}\hat{\lambda}_y(s)e^{us}\,{\rm d}s \nonumber\\ & =\frac{1 }{2\pi i}\int_{(1+{c}/{\log y})}( \hat{\lambda}_y((s-1)\log y) \log y) x^s \,{\rm d}s \end{align}

for any $c$ such that

(1.17)\begin{equation} \hat{\lambda}_y(s):=\int_{0}^{\infty} e^{{-}sv} \lambda_y(v)\,{\rm d}v, \end{equation}

converges absolutely for $\Re s \ge c$. In particular, we may take $c>-(\log y)/(1+\varepsilon )$ if we assume $y\ge C_{\varepsilon}$, as Saias showed, see corollary A.2. As shown by de Bruijn [3, Eq. (2.6)] (cf. [17, Lem. 6]),

\[ \hat{\lambda}_y(s) = \hat{\rho}(s) K(s/\log y). \]

By definition of $F$, (1.9), we can rewrite (1.16) as (1.15). As Saias does, we choose to work with $\sigma =\beta$, which is essentially a saddle point for $F(s,\,y)x^s$. If $x \ge y \ge (\log x)^{1+\varepsilon }$ and $x \ge C_{\varepsilon}$ then lemma 2.1 implies

\[ \beta \ge c_{\varepsilon}>0. \]

Saias proved (1.6) by showing that $\zeta (s,\,y)$ and $F(s,\,y)$ are close and so if we subtract

\[ \Psi(x,y) -\Lambda(x,y) = \frac{1}{2\pi i} \int_{(\beta)} (\zeta(s,y)-F(s,y)) \frac{x^s}{s}\,{\rm d}s \]

then we can bound the integral by using pointwise bounds for the integrand. Instead of subtracting $\Lambda (x,\,y)$, we subtract $\Lambda (x,\,y)$ times $G(\beta,\,y)$, which leads to

(1.18)\begin{equation} \Psi(x,y) =\Lambda(x,y) G(\beta,y) \left(1 + \frac{\Lambda(x,y)^{{-}1}}{2\pi i} \int_{(\beta)} \frac{G(s,y) - G(\beta,y)}{G(\beta,y)}F(s,y) \frac{x^s}{s} \,{\rm d}s\right). \end{equation}

We want to bound the integral in (1.18). The proof of theorem 1.1 considers separately the range

(1.19)\begin{equation} u \ge (\log y) ( \log \log y)^3 \end{equation}

and its complement. When $u$ satisfies (1.19), then in (1.18) one needs only small values of $\Re s$ to estimate the integral ($|\Re s| \le 1/\log y$) with arbitrary power saving in $y$. This is an unconditional observation established in proposition 3.1. However, for smaller $u$, one needs $|\Re s|$ going up to a power of $y$ if one desires power saving in $y$, which makes the proof more involved.

In our proofs, RH is only invoked at the very end to estimate $G_1$ and its derivatives. For instance, in the range where (1.19) and $y \ge (\log x)^{2+\varepsilon }$ hold, we prove in (4.12) the unconditional estimate

(1.20)\begin{align} \Psi(x,y) & = \Lambda(x,y) G(\beta,y) \left( 1+O_{\varepsilon}\left( \frac{\max_{|v|\le 1} |G'(\beta+iv,y)|}{G(\beta,y) \log x}\right.\right.\nonumber\\ & \quad + \left.\left. \frac{ \max_{|v|\le 1}|G''(\beta+iv,y)|}{G(\beta,y) (\log x)(\log y)} + \frac{1}{y}\right)\right). \end{align}

See (4.16) for a similar estimate for $u \le (\log y)(\log \log y)^3$. In particular, our proofs are easily modified to recover (1.6).

Conventions

The letters $C,\,c$ denote absolute positive constants that may change between different occurrences. We denote by $C_{\varepsilon },\,c_{\varepsilon }$ positive constants depending only on $\varepsilon$, which may also change between different occurrences. The notation $A \ll B$ means $|A| \le C B$ for some absolute constant $C$, and $A\ll _{\varepsilon } B$ means $|A| \le C_{\varepsilon } B$. We write $A \asymp B$ to mean $C_1 B \le A \le C_2 B$ for some absolute positive constants $C_i$, and $A \asymp _{\varepsilon } B$ means $C_i$ may depend on $\varepsilon$. The letter $\rho$ will always indicate a non-trivial zero of $\zeta$. When we differentiate a bivariate function, we always do so with respect to the first variable. We set

\[ L(y):=\exp((\log y)^{{3}/{5}}(\log \log y)^{-{1}/{5}}). \]

2. Preliminaries

2.1 Standard lemmas

Recall $\beta$ was defined in (1.10).

Lemma 2.1 [13, Lem. 1] For $u \ge 3$ we have $\xi (u) = \log u + \log \log u + O( (\log \log u) / \log u)$. In particular,

(2.1)\begin{equation} y^{1-\beta} \asymp u \log(u+1), \quad u \ge 1. \end{equation}

Lemma 2.2 [2]

For $u \ge 1$ we have $\rho (u) \asymp e^{-u\xi +I(\xi )} u^{-1/2} = x^{\beta -1} e^{I(\xi )}u^{-1/2}$.

In the next lemmas we write $s \in \mathbb {C}$ as $s=\sigma + it$.

Lemma 2.3 [15, Cor. 10.5] For $|\sigma | \le A$ and $|t| \ge 1$, $|\zeta (s)| \asymp _A (|t|+4)^{1/2-\sigma }|\zeta (1-s)|$.

Lemma 2.4 [15, Cor. 1.17] Fix $\varepsilon >0$. For $\sigma \in [\varepsilon,\,2]$ and $|t| \ge 1$ we have

\[ \zeta(s) \ll_{\varepsilon} (1+(|t|+4)^{1-\sigma})\min\left\{\frac{1}{|\sigma-1|},\log (|t|+4)\right\}. \]

Lemma 2.5 [19, Thm. 7.2(A)]

We have, for $\sigma \in [1/2,\,2]$ and $T \ge 2$,

\[ \int_{1}^{T} |\zeta(\sigma+it)|^2 \,{\rm d}t \ll T \min\left\{ \log T, \frac{1}{\sigma-\frac{1}{2}}\right\}. \]

Lemma 2.6 [14, Lem. 2.7] The following bounds hold for $s=-\xi (u)+it$:

(2.2)\begin{equation} \hat{\rho}(s) =e^{\gamma+I({-}s)}= \begin{cases} O\left(\exp\left(I(\xi)-\frac{t^2u}{2\pi^2}\right)\right) & \mbox{if }|t| \le \pi,\\ O\left(\exp\left(I(\xi)-\frac{u}{\pi^2+\xi^2}\right)\right) & \mbox{if }|t| \ge \pi,\\ \frac{1}{s} + O\left( \frac{1+u \xi }{|s|^2} \right) & \mbox{if }1+u\xi=O(|t|).\end{cases} \end{equation}

The third case of lemma 2.6 is usually stated in the range $1+u\xi \le |t|$, but the same proof works for $1+u\xi = O(|t|)$. Since $1+u\xi =e^{\xi }$, the third case can also be written as

(2.3)\begin{equation} s\hat{\rho}(s) = 1+O( e^{-\sigma}/|t|) \end{equation}

for $s=\sigma +it$, assuming $\sigma <0$ and $e^{-\sigma } =O(|t|)$. The following lemma is a variant of [13, Lem. 8], proved in the same way.

Lemma 2.7 [13]

Fix $\varepsilon >0$. Suppose $x \ge y \ge (\log x)^{1+\varepsilon }$ and $x \ge C_{\varepsilon}$. For $|t| \le 1/\log y$,

\[ \left|\frac{\zeta(\beta+it,y)}{\zeta(\beta,y)}\right| \le \exp({-}ct^2 (\log x )(\log y) ). \]

For $1/\log y \le |t| \le \exp ((\log y)^{3/2-\varepsilon })$,

(2.4)\begin{equation} \frac{\zeta(\beta+it,y)}{\zeta(\beta,y)} \ll_{\varepsilon} \exp\left(-\frac{c ut^2}{(1-\beta)^2+t^2}\right). \end{equation}

2.2 More on $G$

Lemma 2.8 [9]

Fix $0 \le i \le 4$. Let $y \ge 4$. Let $s \in \mathbb {C}$ with $\Re s \in [0,\,1]$ and the property that

(2.5)\begin{equation} \min_{\zeta(\rho)=0,\quad t \ge 0} |\rho-s-t| \gg 1. \end{equation}

Then for $T \ge 3+|\Im s|$ we have

(2.6)\begin{align} (\log G_1)^{(i)}(s,y)& ={-}\sum_{|\Im (\rho-s)| \le T}\frac{{\rm d}^{i}}{{\rm d}s^{i}} \int_{0}^{\infty} \frac{y^{\rho-s-t}}{\rho-s-t}\,{\rm d}t\nonumber\\ & \quad +O\left((\log y)^{i} y^{-\Re s}+ \frac{\log^2 (yT)(\log y)^{i-1} }{T} y^{1-\Re s}\right). \end{align}

Corollary 2.9 Fix $0 \le i \le 4$. Let $y \ge 4$. Let $s \in \mathbb {C}$ with $\Re s \in [0,\,1]$. If $|\Im s| \le 1$ we have $(\log G_1)^{(i)}(s,\,y) \ll L(y)^{-c} y^{1-\Re s}$ unconditionally. Under RH, if $T \ge 4$ and $|\Im s| \le 1$ then

(2.7)\begin{align} (\log G_1)^{(i)}(s,y) & =(-\log y)^{i-1}y^{- s}\bigg( \sum_{|\Im (\rho-s)| \le T} \frac{y^{\rho}}{\rho-s} + O \bigg( \frac{y^{{1}/{2}}}{\log y}+ \frac{y\log^2(yT)}{T}\bigg) \bigg) \nonumber\\ & =({-}1)^i(\log y)^{i-1} y^{- s} (\psi(y)-y+ O(y^{{1}/{2}})) \ll y^{{1}/{2}-\Re s}(\log y)^{i+1}. \end{align}

Under RH, if $T \ge 4$, $\Re s \in [3/4,\,1]$ and $|\Im s| \le y^{9/10}$ then

(2.8)\begin{align} (\log G_1)^{(i)}(s,y) & = ({-}1)^i(\log y)^{i-1} y^{- s} (\psi(y)-y+ O(y^{{1}/{2}}\log^2 (|\Im s|+2)))\nonumber\\ & \quad \ll y^{{1}/{2}-\Re s}(\log y)^{i+1}. \end{align}

Proof. If $|\Im s| \le 1$ then (2.5) holds. It is easily seen that, for any zero $\rho$ of $\zeta$,

(2.9)\begin{equation} \frac{{\rm d}^{i}}{{\rm d}s^{i}}\int_{0}^{\infty} \frac{y^{\rho-s-t}}{\rho-s-t}\,{\rm d}t ={-} \frac{(-\log y)^{i-1}y^{\rho-s}}{\rho-s} \left( 1 +O\left( \frac{1}{\min_{t \ge 0} |\rho-s-t|\log y}\right)\right) \end{equation}

if (2.5) holds. We apply lemma 2.8 with $T=L(y)^c$ and use the Vinogradov–Korobov zero-free region and (2.9) to simplify. Now assume RH, i.e. $|y^{\rho }|=y^{1/2}$. We demonstrate (2.7), and (2.8) is proved along similar lines. We apply lemma 2.8 with $T\ge 4$ and simplify it using (2.9). We bound the resulting error using the facts $\min _{t \ge 0}|\rho -s-t| \asymp |\rho -s|$ and $\sum _{\rho } 1/|\rho -s|^2 \ll 1$ for $|s|\le 2$, since there are $\ll \log T$ zeros of $\zeta$ between height $T$ and $T+1$ [15, Thm. 10.13]. This gives the first equality in (2.7). The second equality in (2.7) follows by taking $T=y$, recalling the classical estimate

(2.10)\begin{equation} \psi(y)-y ={-}\sum_{|\rho| \le y} \frac{y^{\rho}}{\rho} + O(\log^2 y) \end{equation}

given in [15, Thm. 12.5] (it also follows from lemma 2.8 with $(i,\,s,\,T)=(1,\,0,\,y)$), and the bound $\sum _{\rho } 1/(|\rho -s||\rho |) \ll 1$. The last inequality in (2.7) is von Koch's bound $\psi (y)-y=O(y^{1/2}\log ^2 y)$ [20].

We turn to $G_2$. By the non-negativity of the coefficients of $\log G_2$, for $i \ge 0$ and $\Re s>0$ we have

(2.11)\begin{equation} |(\log G_{2})^{(i)}(s,y)| \le ({-}1)^i\log G_{2}^{(i)}(\Re s ,y ). \end{equation}

Lemma 2.10 [9]

Fix $\varepsilon >0$ and $0 \le i \le 4$. For $y \ge 2$ and $1 \ge s \ge \varepsilon$,

(2.12)\begin{align} (\log G_2)^{(i)}(s,y) & =( 1+O_{\varepsilon}( L(y)^{{-}c})) \frac{({-}2)^i}{2}\int_{y^{1/2}}^{y}(\log t)^{i-1} t^{{-}2s} \,{\rm d}t\nonumber\\ & \quad \asymp_{\varepsilon} \frac{(-\log y)^{i}y^{\max\{1-2s,\frac{1}{2}-s\}}}{\max\{1,|s-1/2|\log y\}}. \end{align}

Corollary 2.9 and lemma 2.10, applied with $i=0$, imply the following

Lemma 2.11 Assume RH. Fix $\varepsilon >0$. If $1 \ge s \ge 1/2+\varepsilon$ and $T \ge 4$ then

\begin{align*} G(s,y) & = 1 +\frac{y^{{-}s}}{\log y} \bigg(- \sum_{|\rho| \le T} \frac{y^{\rho}}{\rho-s} + \frac{y^{{1}/{2}}}{2s-1}+ O_{\varepsilon}\bigg(\frac{y^{{1}/{2}}}{\log y} + \frac{y \log^2( yT)}{T}\bigg)\bigg)\\ & =1 + \frac{y^{{-}s}}{\log y} ( \psi(y)-y + O_{\varepsilon}(y^{{1}/{2}}))=1 + O_{\varepsilon}( y^{{1}/{2}-s} \log y). \end{align*}

Corollary 1.3 follows from theorem 1.1 by simplifying $G(\beta,\,y)$ using lemma 2.11 and (2.1).

3. Truncation estimates for $\Psi$ and $\Lambda$

The purpose of this section is to prove the following two propositions.

Proposition 3.1 Medium $u$

Suppose $x \ge y \ge 2$ satisfy

\[ u \ge (\log y) ( \log \log y)^3. \]

Fix $\varepsilon >0$. Suppose $y \ge (\log x)^{1+\varepsilon }$ and $x \ge C_{\varepsilon}$. Then

(3.1)\begin{align} \Psi(x,y) & = \frac{1}{2\pi i} \int_{\beta-\frac{i}{\log y}}^{\beta+\frac{i}{\log y}} \zeta(s,y) \frac{x^s}{s}\,{\rm d}s\nonumber\\ & \quad + O_{\varepsilon}\left( \frac{\Psi(x,y)+x\rho(u) G(\beta,y)}{\exp(c_{\varepsilon}\min\{u/\log^2(u+1),(\log y)^{4/3}\})}\right), \end{align}

(3.2)\begin{align} \Lambda(x,y) & = \frac{1}{2\pi i} \int_{\beta-\frac{i}{\log y}}^{\beta+{i}/{\log y}} F(s,y) \frac{x^s}{s}\,{\rm d}s + O_{\varepsilon}\left(\frac{x\rho(u)}{\exp( c u/\log^2(u+1))}\right). \end{align}

Proposition 3.2 Small $u$

Suppose $x \ge y \ge 2$ satisfy

\[ u \le (\log y) ( \log \log y)^3. \]

Suppose $x \ge C$ and let $T \in [(\log x)^5,\,x\rho (u)]$. Then

\begin{align*} \Psi(x,y) & = \frac{1}{2\pi i} \int_{\beta-i T}^{\beta+iT} \zeta(s,y) \frac{x^s}{s}\,{\rm d}s + O\left( \frac{\Psi(x,y)+x\rho(u) G(\beta,y)}{T^{4/5}}\right),\\ \Lambda(x,y) & = \frac{1}{2\pi i} \int_{\beta-iT}^{\beta+iT} F(s,y) \frac{x^s}{s}\,{\rm d}s + O\left(\frac{x\rho(u)}{T^{4/5}}\right). \end{align*}

3.1 Preparation

Lemma 3.3 Fix $\varepsilon \in (0,\,1)$. For $\sigma \in [\varepsilon,\,1]$ and $x \ge T \ge 2$ we have

(3.3)\begin{equation} \frac{1}{2\pi i}\int_{\sigma+it: \, |t|>T} \zeta(s) \frac{x^s}{s} \,{\rm d}s \ll_{\varepsilon} \frac{x^{\sigma}}{T^{\sigma}}\log T + \log x. \end{equation}

The integral should be understood in principal value sense. Lemma 3.3 makes more precise a computation done in p. 96 of Saias’ paper [17] (cf. [18, p. 537]), which is not stated for general $T$ and $\sigma$ but contains the same ideas.

Proof. By [19, Thm. 4.11], for every $r>0$ we have

\[ \zeta(s) =\sum_{n \le r} n^{{-}s} - \frac{r^{1-s}}{1-s} + O_{\varepsilon}(r^{-\Re s}) \]

as long as $s \neq 1$, $\Re s \ge \varepsilon$ and $|\Im s|\le 2r$. Suppose $s=\sigma +it$ with $|t| \ge 1$. We apply this estimate with $r=|t|$, obtaining

(3.4)\begin{equation} \zeta(s) =\sum_{n \le |t|} n^{{-}s} - \frac{|t|^{1-s}}{1-s} + O_{\varepsilon}(|t|^{-\sigma})= \sum_{n \le |t|} n^{{-}s} + O_{\varepsilon}(|t|^{-\sigma}). \end{equation}

We now plug (3.4) in the left-hand side of (3.3). The contribution of the error term to the integral is acceptable:

\[ \int_{\sigma+it: \, |t|>T} O(|t|^{-\sigma}) \frac{x^s}{s} \,{\rm d}s \ll x^{\sigma}\int_{T}^{\infty} |t|^{-\sigma-1}\,{\rm d}t \ll_{\varepsilon} \frac{x^{\sigma}}{T^{\sigma}}. \]

The contribution of $n^{-s} \mathbf {1}_{n\le |t|}$ in (3.4) to the left-hand side of (3.3) is

(3.5)\begin{equation} \frac{1}{2\pi i} \int_{\sigma+it: |t|>\max\{n,T\}} n^{{-}s} \frac{x^s}{s}\,{\rm d}s. \end{equation}

Since

\[ \frac{1}{2\pi i} \int_{\sigma+it:\, |t| \le S} n^{{-}s} \frac{x^s}{s}\,{\rm d}s =\mathbf{1}_{x>n} + \frac{\mathbf{1}_{x=n}}{2} +O\left( \frac{(x/n)^{\sigma}}{1+S|\log(x/n)|}\right), \quad S \ge 1, \]

by the truncated Perron's formula [14, p. 435], and

\[ \frac{1}{2\pi i} \int_{(\sigma)} n^{{-}s} \frac{x^s}{s}\,{\rm d}s =\mathbf{1}_{x>n} + \frac{\mathbf{1}_{x=n}}{2} \]

by Perron's formula, it follows that the integral in (3.5) is bounded by

\[{\ll} \frac{(x/n)^{\sigma}}{1+\max\{n,T\}|\log(x/n)|} \]

and so the total contribution of the $n$-sum in (3.4) to the left-hand side of (3.3) is

(3.6)\begin{equation} \ll x^{\sigma}\sum_{n \ge 1} \frac{n^{-\sigma}}{1+\max\{n,T\}|\log(x/n)|}. \end{equation}

It remains to estimate (3.6), which we do according to the size of $n$. The contribution of $n \ge 2x$ is

\[{\ll} x^{\sigma} \sum_{n \ge 2x} n^{-\sigma-1} \ll_{\varepsilon} 1. \]

The contribution of $n \in (x/2,\,2x)$ can be bounded by considering separately the $n$ closest to $x$, and partitioning the rest of the $n$s according to the value of $k\ge 0$ for which $|\log (x/n)| \in [2^{-k},\,2^{1-k})$:

\[{\ll} x^{\sigma}\sum_{n \in (x/2,2x)} \frac{n^{-\sigma}}{1+x|\log (x/n)|}\ll 1+ \sum_{k \ge 0:\, 2^k \le 2x} \frac{x}{2^k} \frac{1}{1+x/2^k} \ll \log x. \]

The contribution of $n \le T/2$ is

\[{\ll} \frac{x^{\sigma}}{T} \sum_{n \le T/2}n^{-\sigma} \ll \frac{x^{\sigma}}{T^{\sigma}} \log T. \]

Finally, the contribution of $T/2< n\le x/2$ is

\[{\ll} x^{\sigma} \sum_{n>T/2} n^{{-}1-\sigma} \ll_{\varepsilon} \frac{x^{\sigma}}{T^{\sigma}}, \]

acceptable as well.

Corollary 3.4 Fix $\varepsilon \in (0,\,1)$. Suppose $x \ge y \ge C_{\varepsilon}$. For $\sigma \in [\varepsilon,\,1]$ and $x \ge T \ge \max \{2,\,y^{1-\sigma }/\log y\}$ we have

\begin{align*} \Lambda(x,y) & = \frac{1}{2\pi i}\int_{\sigma-iT}^{\sigma+iT} F(s,y) \frac{x^s}{s} \,{\rm d}s \\ & \quad + O_{\varepsilon}\left( \frac{x^{\sigma}}{T^{\sigma}}\log T+\log x+ x^{\sigma} \frac{y^{1-\sigma}}{\log y}\frac{\log^{1/2} T}{T^{\min\{1,1/2+\sigma\}}} \right). \end{align*}

Corollary 3.4 rests on lemma 3.3, and makes more precise Proposition 2 of Saias [17].

Proof. Our starting point is the identity (1.15). (If $x \in \mathbb {Z}$ it still holds with an error term of $O(1)$, since the integral converges to the average $(\Lambda (x+,\,y)+\Lambda (x-,\,y))/2 = \Lambda (x,\,y)+O(1)$.) From that identity it follows that our task is equivalent to upper bounding

\[ \left|\int_{\sigma+it:\, |t|>T} F(s,y) \frac{x^s}{s} \,{\rm d}s\right|. \]

Recall $F(s,\,y) = \hat {\rho }((s-1)\log y)\zeta (s)(s-1)\log y$. By (2.3) with $(s-1)\log y$ instead of $s$ we find

\[ F(s,y)=\zeta(s)\left(1+O\left(\frac{y^{1-\sigma}}{|t|\log y}\right)\right) \]

if $y^{1-\sigma } = O(|t| \log y)$, which holds by our assumptions on $T$. By the triangle inequality,

(3.7)\begin{align} & \left|\int_{\sigma+it:\, |t|>T} F(s,y) \frac{x^s}{s} \,{\rm d}s\right| \ll \left|\int_{\sigma+it:\, |t|>T} \frac{\zeta(s)}{s}x^s \,{\rm d}s\right|\nonumber\\ & \quad + x^{\sigma}\frac{y^{1-\sigma}}{\log y} \int_{\sigma+it:\,|t|>T} \frac{|\zeta(s)|}{|t|^2} |\,{\rm d}s| . \end{align}

The first integral in the right-hand side of (3.7) is estimated in lemma 3.3. To bound the second integral we apply the second moment estimate for $\zeta$ given in lemma 2.5. We first suppose that $\sigma \ge 1/2$. Using Cauchy–Schwarz, the second integral in the right-hand side of (3.7) is at most

(3.8)\begin{align} & \int_{\sigma+it:\,|t|>T} \frac{|\zeta(s)|}{|t|^2} |\,{\rm d}s| \ll \sum_{2^k \ge T/2}4^{{-}k} \int_{2^k}^{2^{k+1}} |\zeta(\sigma+it)|\,{\rm d}t\ll \sum_{2^k \ge T/2}2^{{-}k} k^{1/2} \nonumber\\ & \quad\ll \frac{\log^{1/2} T}{T}. \end{align}

Multiplying this by the prefactor $x^{\sigma }y^{1-\sigma }/\log y$, we see that this is acceptable. If $\varepsilon \le \sigma \le 1/2$ we use lemma 2.3. We obtain that the second integral in the right-hand side of (3.7) is at most

(3.9)\begin{align} & \int_{\sigma+it:\,|t|>T} \frac{|\zeta(s)|}{|t|^2} |\,{\rm d}s| \ll \int_{1-\sigma+it:\,|t|>T} \frac{|\zeta(s)|}{|t|^{2+\sigma-1/2}} |\,{\rm d}s|\nonumber\\ & \quad \ll \sum_{2^k \ge T/2} 2^{{-}k(\sigma+1/2)}k^{1/2} \ll \frac{\log^{1/2} T}{T^{{1}/{2}+\sigma}}, \end{align}

concluding the proof.

Let $\alpha =\alpha (x,\,y)$ be the saddle point associated with $y$-smooth numbers up to $x$ [13], that is, the minimizer of the convex function $s\mapsto x^s \zeta (s,\,y)$ ($s>0$).

Lemma 3.5 For $\sigma \in (0,\,1]$, $x\ge y \ge C$ and $T \ge 2$ we have

(3.10)\begin{equation} \Psi(x,y)=\frac{1}{2\pi i}\int_{\sigma-iT}^{\sigma+iT} \zeta(s,y)\frac{x^s}{s}\,{\rm d}s+O\left( \frac{x^{\sigma}\zeta(\sigma,y)}{T} + \frac{\Psi(x,y) \log T}{T^{\alpha}}+1\right). \end{equation}

Our proof makes more precise a similar estimate appearing in Saias [17, p. 98], which does not allow general $y$ and $T$ but contains the main ideas.

Proof. The truncated Perron's formula [14, p. 435] bounds the error in (3.10) by

\[{\ll} x^{\sigma}\sum_{\substack{n\ge 1\\ n\text{ is }y\text{-smooth}}}\frac{1}{n^{\sigma}(1+T|\log(x/n)|)}. \]

The contribution of the terms with $|\log (x/n)|\ge 1$ is

\[{\ll} \frac{x^{\sigma}}{T} \sum_{\substack{n\ge 1\\n\text{ is } y\text{-smooth}}} \frac{1}{n^{\sigma}} = \frac{x^{\sigma}\zeta(\sigma,y)}{T}. \]

We now study the terms with $|\log (x/n)|<1$. These contribute

(3.11)\begin{equation} \ll \sum_{\substack{e^{{-}1}x< n< ex\\n\text{ is }y\text{-smooth}}} \frac{1}{1+T|\log (x/n)|}. \end{equation}

The subset of terms with $|\log (x/n)| \le 1/T$ contributes to (3.11)

(3.12)\begin{equation} \ll \sum_{\substack{|n-x| \le Cx/T \\ n\, y\text{-smooth}}} 1 \ll \Psi\left(x+ \frac{Cx}{T},y\right)-\Psi\left(x-\frac{Cx}{T},y\right). \end{equation}

The contribution of the rest of the terms to (3.11), namely, those terms with $1/T<|\log (x/n)| <1$, can be dyadically dissected to terms with $|\log (x/n)| \in [2^{-k},\,2^{1-k})$ for each integer $k\ge 1$ such that $2^k<2\,T$ holds. Their total contribution is

(3.13)\begin{equation} \ll \frac{1}{T}\sum_{1 \le k \le \log_2\,T + 1} 2^k\left(\Psi\left(x+ \frac{Cx}{2^k},y\right)-\Psi\left(x-\frac{Cx}{2^k},y\right)\right), \end{equation}

where $\log _2$ is the base-2 logarithm. (We interpret $\Psi (a,\,y)$ for negative $a$ as equal to $0$.) Note that the sum in (3.13) dominates the right-hand side of (3.12). We shall make use of Hildebrand's inequality $\Psi (a+b,\,y)-\Psi (a,\,y) \le \Psi (b,\,y)$, valid for $y \ge C$ and $a,\,b \ge y$. It implies

(3.14)\begin{equation} \Psi(a+b,y)-\Psi(a,y) \le \Psi(b,y)+1 \end{equation}

for $y\ge C$ and all $a,\,b$. We apply (3.14) with $a=x-Cx/2^k$ and $b=2Cx/2^k$ to find that (3.13) is bounded by

(3.15)\begin{equation} \ll \frac{1}{T}\sum_{1 \le k \le \log_2\,T + 1} 2^k\left(\Psi\left(\frac{Cx}{2^k},y\right)+1\right) \ll \frac{1}{T}\sum_{1 \le k \le \log_2\,T + 1} 2^k\left(\Psi\left(\frac{x}{2^k},y\right)+1\right) \end{equation}

where in the second inequality we replaced $\Psi (Cx,\,y)$ with $\Psi (x,\,y)$ using [13, Thm. 3]. To conclude, we recall Theorem 2.4 of [5] says $\Psi (x/d,\,y) \ll \Psi (x,\,y)/d^{\alpha }$ holds for $x \ge y \ge 2$ and $1 \le d \le x$. We apply this inequality with $d=2^k$ and obtain

(3.16)\begin{align} & \frac{1}{T}\sum_{1 \le k \le \log_2\,T + 1} 2^k\left(\Psi\left(\frac{x}{2^k},y\right)+1\right) \ll 1+\frac{\Psi(x,y)}{T}\sum_{1 \le k \le \log_2\,T + 1} 2^{(1-\alpha)k}\nonumber\\ & \quad\ll 1+\frac{\Psi(x,y)\log T}{T^{\alpha}} \end{align}

as needed.

3.2 Proof of proposition 3.1

We first truncate the Perron integral for $\Psi (x,\,y)$. We apply lemma 3.5 with $\sigma = \beta$ and $T=\exp ((\log y)^{4/3})$. The assumption $y \ge (\log x)^{1+\varepsilon }$ implies $\beta \gg _{\varepsilon } 1$ and $\Psi (x,\,y) \ge x^{c_{\varepsilon }}$. Since $\alpha =\beta +O(1/\log y)$ [13, Lem. 2] it follows that $\alpha \gg _{\varepsilon } 1$ and so

(3.17)\begin{equation} \Psi(x,y)=\frac{1}{2\pi i}\int_{\beta-iT}^{\beta+iT} \zeta(s,y)\frac{x^s}{s}\,{\rm d}s+O_{\varepsilon}\left( \frac{x^{\beta}\zeta(\beta,y)+\Psi(x,y)}{T^{c_{\varepsilon}}}\right). \end{equation}

We use lemma 2.7 to bound the contribution of $1/\log y \le |\Im s| \le T$:

\begin{align*} \int_{\beta+i/\log y}^{\beta+iT} \zeta(s,y)\frac{x^s}{s}\,{\rm d}s & \ll x^{\beta} \zeta(\beta,y) \int_{1/\log y}^{T} \left|\frac{\zeta(\beta+it,y)}{\zeta(\beta,y)}\right|\frac{{\rm d}t}{\beta+t}\\ & \ll x^{\beta} \zeta(\beta,y) \int_{1/\log y}^{T} \exp\left(-\frac{c ut^2}{(1-\beta)^2+t^2}\right)\frac{{\rm d}t}{\beta+t}\\ & \ll x^{\beta} \zeta(\beta,y) \left( \exp({-}c u)\log T + \int_{1/\log y}^{\xi(u)/\log y}\right.\nonumber\\ & \quad\left. \exp\left(-\frac{c (\log x)( \log y)}{\log^2 (u+1)}t^2\right)\,{\rm d}t \right)\\ & \ll x^{\beta} \zeta(\beta,y)\exp\left(-\frac{c u}{\log^2(u+1)}\right). \end{align*}

We estimate $x^{\beta } \zeta (\beta,\,y)$:

(3.18)\begin{align} ^{\beta}\zeta(\beta,y) & = \frac{x}{e^{u\xi(u)}}F(\beta,y) G(\beta,y) \nonumber\\ & =\zeta(\beta)(\beta-1) \frac{x e^{I(\xi)+\gamma}\log y}{e^{u\xi(u)}} G(\beta,y) \ll_{\varepsilon} x\rho(u)\sqrt{(\log x)( \log y)} G(\beta,y) \end{align}

using (1.9) and lemma 2.2. Finally, note that both $T$ and $\exp (u/\log ^2(u+1))$ grow faster than any power of $\log x$. We turn to $\Lambda (x,\,y)$. We apply corollary 3.4 with $\sigma =\beta$ and

\[ T = \frac{y^{1-\beta}}{\log y} = \frac{e^{\xi(u)}}{\log y} \asymp \frac{u \log (u+1)}{\log y} \gg (\log \log y)^4. \]

We obtain

\[ \Lambda(x,y) = \frac{1}{2\pi i}\int_{\beta-iT}^{\beta+iT} F(s,y) \frac{x^s}{s} \,{\rm d}s+ O_{\varepsilon}\left( \frac{ux}{\exp(u\xi)}\right). \]

We now treat the range $1/\log y\le | \Im s| \le T$. By the definition of $F$,

(3.19)\begin{equation} \int_{\beta+\frac{i}{\log y}}^{\beta+iT} F(s,y) \frac{x^s}{s} \,{\rm d}s\ll_{\varepsilon} \frac{x\log y}{\exp(u\xi)} \int_{1/\log y}^{T} |\zeta(\beta+it)| |\hat{\rho}(-\xi(u)+it \log y)| \,{\rm d}t. \end{equation}

First suppose $t \ge \pi /\log y$. By the second case of lemma 2.6, this range contributes

(3.20)\begin{align} & \ll_{\varepsilon} \frac{x \exp(I(\xi))\log y}{\exp(u\xi)} \exp\left( - \frac{u}{\pi^2 +\xi^2}\right) \int_{\pi/\log y}^{T} |\zeta(\beta+it)|\,{\rm d}t \nonumber\\ & \ll x\rho(u) \sqrt{(\log x)(\log y)}\exp\left(- \frac{u}{\pi^2+\xi^2}\right)\int_{\pi/\log y}^{T} |\zeta(\beta+it)|\,{\rm d}t \end{align}

using lemma 2.2 in the second inequality. Recall the second moment estimate for $\zeta$ given in lemma 2.5. It shows that right-hand side of (3.20) is bounded by

\[{\ll} x\rho(u) \sqrt{(\log x )(\log y)}\exp\left(- \frac{u}{\pi^2+\xi^2}\right) T^{\max\{1,3/2-\beta\}}\sqrt{\log T} \]

where we used the functional equation if $\beta < 1/2$ (lemma 2.3). The contribution of $1/\log y \le t \le \pi /\log y$ to the right-hand side of (3.19) is treated using the first part of lemma 2.6, and we find that it is at most

(3.21)\begin{equation} \ll_{\varepsilon} \frac{x \exp(I(\xi))\log y}{\exp(u\xi)} \int_{1/\log y}^{\pi/\log y} \exp\bigg(- \frac{(\log x)( \log y)}{2\pi^2}t^2\bigg)\,{\rm d}t \ll_{\varepsilon} x\rho(u) \exp({-}cu), \end{equation}

using lemma 2.2 in the second inequality. In conclusion,

\[ \Lambda(x,y) = \frac{1}{2\pi i}\int_{\beta-i/\log y}^{\beta+i/\log y} F(s,y) \frac{x^s}{s} \,{\rm d}s+ E \]

where

\begin{align*} & E \ll_{\varepsilon} \frac{ux}{\exp(u\xi)} + x\rho(u) \left( \sqrt{(\log x)( \log y)}\right.\\ & \left.\quad \exp\left(- \frac{u}{\pi^2+\xi^2}\right) T^{\max\{1,3/2-\beta\}}\sqrt{\log T} + \exp({-}cu)\right). \end{align*}

By our choice of $T$ and assumptions on $u$ and $y$, this can be absorbed in the error term of (3.2).

3.3 Proof of proposition 3.2

We first truncate the Perron integral for $\Psi (x,\,y)$. We apply lemma 3.5 with $\sigma =\beta$ and our $T$, finding

(3.22)\begin{equation} \Psi(x,y)=\frac{1}{2\pi i}\int_{\beta-iT}^{\beta+iT} \zeta(s,y)\frac{x^s}{s}\,{\rm d}s+O\left( 1+ \frac{\Psi(x,y)\log T}{T^{\alpha}}+\frac{x^{\beta}\zeta(\beta,y)}{T}\right). \end{equation}

In the considered range, $\Psi (x,\,y) \asymp x \rho (u)$. In particular, the error term $O(1)$ is acceptable since our $T$ is $\ll x\rho (u) \ll \Psi (x,\,y)$ and so $1 \ll \Psi (x,\,y) /T^{4/5}$. Additionally, $\beta \sim 1$ as $x \to \infty$ by lemma 2.1 and $\alpha =\beta +O(1/\log y)$ [13, Lem. 2], so $\alpha \sim 1$. This implies that $(\log T)/T^{\alpha } \ll 1/T^{4/5}$ and the error term $O(\Psi (x,\,y)(\log T)/T^{\alpha })$ is also acceptable. The estimate (3.18) treats the last error term and finishes the estimation. We turn to $\Lambda (x,\,y)$. We apply corollary 3.4 with our $T$, obtaining

(3.23)\begin{equation} \Lambda(x,y) = \frac{1}{2\pi i}\int_{\beta-iT}^{\beta+iT} F(s,y) \frac{x^s}{s} \,{\rm d}s+ O\left(\log x +x \exp({-}u\xi) u \log (u+1) \frac{\log T}{T^{\sigma}}\right). \end{equation}

In our range $x\rho (u) \asymp x^{1+o(1)}$, so the term $\log x$ is acceptable. We have $\exp (-u\xi ) u\log (u+1) \ll \rho (u)$ by lemma 2.2, so the second term in the error term of (3.23) is also acceptable.

4. Proofs of theorems 1.1 and 1.2

Proposition 4.1 Medium $u$

Suppose $x \ge y \ge 2$ satisfy

\[ u \ge (\log y) ( \log \log y)^3. \]

Fix $\varepsilon >0$ and suppose $y \ge (\log x)^{1+\varepsilon }$ and $x \ge C_{\varepsilon}$. Let

\[ t_0:= (\log x)^{{-}1/3}(\log y)^{{-}2/3}, \quad T := \exp(\min\{u/\log^2(u+1),(\log y)^{4/3}\}). \]

Then $\Psi (x,\,y) = \Lambda (x,\,y) G(\beta,\,y) ( 1 + E)$ for

(4.1)\begin{align} & E \ll_{\varepsilon} \frac{ |G'(\beta,y)|}{G(\beta,y)\log x} + \frac{ \max_{|v|\le t_0}|G''(\beta+iv,y)|}{G(\beta,y)(\log x)(\log y)}\nonumber\\ & \quad +\frac{\max_{|v|\le \frac{1}{\log y}} |G'(\beta+iv,y)|\exp({-}u^{1/3}/20)}{G(\beta,y)\log x} + \frac{1}{T^{c_{\varepsilon}}}. \end{align}

Proof. Our strategy is to establish $\Psi (x,\,y) = \Lambda (x,\,y) G(\beta,\,y) (1 + E_1 + E_2) + E_3$ for

\begin{align*} E_1 & \ll_{\varepsilon} \frac{ |G'(\beta,y)|}{G(\beta,y)\log x} + \frac{ \max_{|v|\le t_0}|G''(\beta+iv,y)|}{G(\beta,y)(\log x) (\log y)},\\ E_2 & \ll_{\varepsilon} \frac{\max_{|v|\le \frac{1}{\log y}} |G'(\beta+iv,y)|\exp({-}u^{1/3}/20)}{G(\beta,y)\log x},\\ E_3 & \ll_{\varepsilon} \frac{\Psi(x,y)+x\rho(u)G(\beta,y)}{T^{c_{\varepsilon}}}. \end{align*}

The theorem will then follow by rearranging, once we recall that $x\rho (u) \asymp _{\varepsilon } \Lambda (x,\,y)$. From proposition 3.1,

(4.2)\begin{align} & \Psi(x,y) -\Lambda(x,y)G(\beta,y) \nonumber\\ & \quad =\frac{1}{2\pi i} \int_{\beta-\frac{i}{\log y}}^{\beta+\frac{i}{\log y}} ( G(s,y)-G(\beta,y))F(s,y) \frac{x^s}{s}\,{\rm d}s + O_{\varepsilon}\left( \frac{\Psi(x,y)+x\rho(u) G(\beta,y)}{T^{c_{\varepsilon}}} \right), \end{align}

which explains $E_3$. Let $t_0$ be as in the statement of the proposition. We upper bound the contribution of $t_0 \le |\Im s| \le 1/\log y$ to the integral in the right-hand side of (4.2). We have

\[ |G(s,y)-G(\beta,y)| \le |\Im s| \max_{ |t|\le |\Im s|} |G'(\beta+it,y)|. \]

The triangle inequality shows, by definition of $F$, that

(4.3)\begin{align} & \int_{\beta+it_0}^{\beta+\frac{i}{\log y}} ( G(s,y)-G(\beta,y))F(s,y) \frac{x^s}{s}\,{\rm d}s \nonumber\\ & \quad \ll_{\varepsilon} \max_{|t| \le \frac{1}{\log y}} |G'(\beta+it,y)| x^{\beta} \log y \int_{t_0}^{{1}/{\log y}} t |e^{I(\xi-it\log y)}| \,{\rm d}t. \end{align}

Since $-e^{-v^2/2}$ is the antiderivative of $e^{-v^2/2}v$, the first part of lemma 2.6 shows

\begin{align*} \int_{t_0}^{{1}/{\log y}} t |e^{ I(\xi-it\log y)}| \,{\rm d}t & \ll \exp(I(\xi))\int_{t_0}^{{1}/{\log y}} t \exp(-(\log x) (\log y)t^2/(2\pi^2)) \,{\rm d}t\\ & \ll \exp(I(\xi)) \frac{\exp({-}u^{1/3}/(2\pi^2))}{(\log x)( \log y)}. \end{align*}

Hence, $t_0 \le |\Im s| \le 1/\log y$ contributes in total

\[ \ll_{\varepsilon} \max_{|t| \le 1/\log y}|G'(\beta+it,y)| x\rho(u)\exp({-}u^{1/3}/20)/\log x \]

where we used lemma 2.2 to simplify. Once we divide this by $\Lambda (x,\,y)G(\beta,\,y) \asymp _{\varepsilon } x\rho (u)G(\beta,\,y)$ we obtain the error term $E_2$. It remains to study the contribution of $|\Im s|\le t_0$ to the integral in the right-hand side of (4.2), which will yield $E_1$. We Taylor-expand the integrand at $s=\beta$. We write $s=\beta +it$, $|t|\le t_0$. We first simplify the integrand using the definition of $F$:

\begin{align*} \frac{F(s,y)x^{s}}{s} & = (\log y) K(s-1) e^{\gamma+I(\xi)}x^{\beta+it} \exp(I(\xi-it \log y)-I(\xi))\\ & = (\log y) K(s-1) x^{\beta}e^{\gamma+I(\xi)} \exp(I(\xi-it \log y)-I(\xi)+it \log x). \end{align*}

We Taylor-expand $\log K(s-1)$ and $G(s,\,y)-G(\beta,\,y)$:

\begin{align*} K(s-1) & = K(\beta-1) (1+O_{\varepsilon}(t)),\\ G(s,y)-G(\beta,y) & = itG'(\beta,y)+ O(t^2 \max_{|v|\le t} |G''(\beta+iv,y)|). \end{align*}

We expand $I(\xi -it \log y)-I(\xi )+it \log x$:

(4.4)\begin{equation} I(\xi-it \log y)-I(\xi)+it \log x ={-}\frac{t^2}{2}I''(\xi)\log^2 y + O( |t|^3 (\log x) (\log y)^2), \end{equation}

where we used $I'(\xi (u))=u$ and $I^{(3)}(\xi (u)+it) \ll e^{\xi (u)}/(1+\xi (u)) \asymp u$. This implies

(4.5)\begin{align} & \exp(I(\xi-it \log y)-I(\xi)-it\log y)\nonumber\\ & \quad = \exp\left(-\frac{t^2}{2}I''(\xi)\log^2 y\right)( 1+ O( |t|^3 (\log x) (\log y)^2)) \end{align}

for $|t| \le t_0$. By two basic properties of moments of the Gaussian,

\begin{align*} & \int_{{-}t_0}^{t_0} t\exp\left(-\frac{t^2}{2}I''(\xi)\log^2 y\right) \,{\rm d}t = 0,\\ & \int_{{-}t_0}^{t_0} |t|^{k}\exp\left(-\frac{t^2}{2}I''(\xi)\log^2 y\right) \,{\rm d}t\nonumber\\ & \quad \ll_k (I''(\xi)\log^2 y)^{-{k+1}/{2}} \ll_k ((\log x)(\log y))^{-{k+1}/{2}}, \end{align*}

we find

(4.6)\begin{align} & \int_{\beta-it_0}^{\beta+it_0} ( G(s,y)-G(\beta,y))F(s,y) \frac{x^s}{s}\,{\rm d}s\nonumber\\ & \quad\ll_{\varepsilon} x^{\beta}e^{I(\xi)} \left( \frac{|G'(\beta,y)|\sqrt{\log y}}{(\log x)^{3/2} } + \frac{\max_{|v|\le t_0}|G''(\beta+iv)|}{(\log x)^{3/2}(\log y)^{1/2}}\right). \end{align}

By lemma 2.2, we can replace $x^{\beta } e^{I(\xi )}$ with $x\rho (u)\sqrt {u}$, to obtain

(4.7)\begin{align} & \int_{\beta-it_0}^{\beta+it_0} ( G(s,y)-G(\beta,y))F(s,y) \frac{x^s}{s}\,{\rm d}s\nonumber\\ & \quad \ll_{\varepsilon} x\rho(u)\left( \frac{|G'(\beta,y)|}{\log x} + \frac{\max_{|v|\le t_0}|G''(\beta+iv)|}{(\log x)(\log y)}\right). \end{align}

Dividing by $G(\beta,\,y)\Lambda (x,\,y) \asymp _{\varepsilon } G(\beta,\,y) x\rho (u)$ gives the error term $E_1$.

Proposition 4.2 Small $u$

Suppose $x \ge y \ge C$ satisfy

(4.8)\begin{equation} u \le (\log y) ( \log \log y)^3. \end{equation}

Let

(4.9)\begin{equation} t_0:= (\log x)^{{-}1/3}(\log y)^{{-}2/3}, \quad t_1: = \frac{u\log(u+1)}{\log y},\quad t_2 \in [(\log x)^{5}, y^{4/5}]. \end{equation}

Then $\Psi (x,\,y) = \Lambda (x,\,y) G(\beta,\,y) ( 1 + E)$ for

\begin{align*} E & \ll \frac{ |G'( \beta,y)|}{\log x} + \frac{ \max_{|v|\le t_0}|G''( \beta+iv,y)|}{(\log x)(\log y)}\\ & \quad + \frac{\max_{|v|\le t_1} |G'( \beta+iv,y)|\exp({-}u^{1/3}/20)}{\log x}+t_2^{{-}4/5} \\ & \quad + \exp({-}u/2)\left( \max_{|t|\le t_2}\left| \frac{G( \beta+it,y)}{G( \beta,y)}-1\right|\right.\\ & \quad +\left. \left|\int_{t_1 \le |t| \le t_2} K( \beta+it-1)x^{it} \frac{G( \beta+it,y)-G( \beta,y)}{G( \beta,y)}\frac{{\rm d}t}{t}\right|\right). \end{align*}

Proof. Our strategy is to establish $\Psi (x,\,y) = \Lambda (x,\,y) G( \beta,\,y) (1 + E_1 + E_2 + E_3 + E_4) + E_5$ for

(4.10)\begin{align} E_1 & \ll \frac{ |G'( \beta,y)|}{G( \beta,y)\log x} + \frac{ \max_{|v|\le t_0}|G''( \beta+iv,y)|}{G( \beta,y)(\log x)( \log y)}, \nonumber\\ E_2 & \ll \frac{\max_{|v|\le t_1} |G'( \beta+iv,y)|\exp({-}u^{1/3}/20)}{G( \beta,y)\log x},\nonumber\\ E_3 & \ll \frac{\exp({-}u/2)}{\log y} \int_{t_1 \le |t| \le t_2} \left| \frac{G( \beta+it)-G( \beta,y)}{G( \beta,y)}\right| \frac{\log(|t|+2)}{t^2} \,{\rm d}t, \nonumber\\ E_4 & \ll \exp({-}u/2) \left|\int_{t_1 \le |t| \le t_2} K( \beta+it-1)x^{it} \frac{G( \beta+it,y)-G( \beta,y)}{G( \beta,y)}\frac{{\rm d}t}{t}\right|,\nonumber\\ E_5 & \ll t_2^{{-}4/5} (\Psi(x,y)+x\rho(u)G( \beta,y)). \end{align}

The proposition will then follow by rearranging and the fact that $G(\beta,\,y) \asymp 1$ in the considered range, unconditionally, as follows from corollary 2.9 and lemma 2.10. From proposition 3.2 with $T=t_2$,

\begin{align*} & \Psi(x,y) -\Lambda(x,y)G( \beta,y) \\ & \quad =\frac{1}{2\pi i} \int_{ \beta-it_2}^{ \beta+it_2} ( G(s,y)-G( \beta,y))F(s,y) \frac{x^s}{s}\,{\rm d}s\\ & \qquad + O( t_2^{{-}4/5}(\Psi(x,y)+x\rho(u) G( \beta,y)) ), \end{align*}

which explains $E_5$. For $|\Im s| \le t_0$, we Taylor-expand $I(\xi -it\log y)$ as in the medium $u$ range and obtain the contribution of $E_1$ (see (4.7)) We treat the contribution of $|\Im s| \in [t_0,\,t_1]$. We replace $G(s,\,y)-G( \beta,\,y)$ with

\[ |G(s,y)-G( \beta,y)| \le |\Im s| \max_{0 \le |t|\le |\Im s|} |G'( \beta+it,y)|. \]

The first two parts of lemma 2.6 show

\begin{align*} & \int_{ \beta+it, \, |t| \in [t_0,t_1]} ( G(s,y)-G( \beta,y))F(s,y) \frac{x^s}{s}\,{\rm d}s \\ & \quad \ll \max_{|t|\le t_1} |G'( \beta+it,y)| x\rho(u)(\log y)\sqrt{u}\\ & \qquad \int_{|t| \in [t_0,t_1]} |t|\left( \exp\left(-\frac{t^2 (\log x)( \log y)}{2\pi^2}\right) + \exp({-}u/(\pi^2+\xi^2))\right) \,{\rm d}t\\ & \quad \ll \max_{|t|\le t_1} |G'( \beta+it,y)| x\rho(u)\sqrt{u}\frac{\exp({-}u^{1/3}/2\pi^2)}{\log x}. \end{align*}

This explains $E_2$. It remains to consider $t_2 \ge |\Im s| \ge t_1$. We use the third part of lemma 2.6 to replace $\hat {\rho }((s-1)\log y)$, appearing in $F(s,\,y)$, with its approximation:

(4.11)\begin{align} & \int_{ \beta+it, \, |t| \in [t_1,t_2]} ( G(s,y)-G( \beta,y))F(s,y) \frac{x^s}{s}\,{\rm d}s \nonumber\\ & \quad = (\log y) x^{ \beta} \int_{s= \beta+it, \, |t| \in [t_1,t_2]} K(s-1)x^{it} ( G(s,y)\nonumber\\ & \qquad -G( \beta,y))\left( \frac{i}{t\log y} + O\left( \frac{u\log (u+1)}{t^2 \log^2 y}\right)\right) \,{\rm d}s. \end{align}

Recall $x^{ \beta } \ll x\rho (u) \sqrt {u} \exp (-I(\xi (u)))$ by lemma 2.2, and that $I(\xi (u)) \sim u$ since a change of variables shows $I(r)= \mathrm {Li}(e^r)+O(\log r)\sim e^r/r$. The contribution of the error term in the right-hand side of (4.11) is

\begin{align*} & \ll x^{ \beta} \log y \int_{s= \beta+it, \, |t| \in [t_1,t_2]} |K(s-1)x^{it} ( G(s,y)-G( \beta,y))| \frac{u\log (u+1)}{t^2 \log^2 y} |\,{\rm d}s| \\ & \ll \frac{x\rho(u)\exp({-}2u/3)}{\log y} \int_{|t|\in[t_1,t_2]}| G( \beta+it)-G( \beta,y)| \frac{|\zeta( \beta+it)| | \beta+it-1|} {t^2 | \beta+it|} \,{\rm d}t. \end{align*}

If $|t|\le 2$ we use $|\zeta ( \beta +it)( \beta +it-1)| \ll 1$ while if $|t| \ge 2$ we use lemma 2.4, to obtain an error term of size $E_3$. The main term of (4.11) gives $E_4$.

4.1 Proof of theorem 1.1: medium $u$

Here we prove theorem 1.1 in the range (1.19). We obtain from proposition 4.1 that unconditionally

(4.12)\begin{equation} \Psi(x,y) = \Lambda(x,y) G(\beta,y) ( 1+E) \end{equation}

for

(4.13)\begin{equation} E \ll_{\varepsilon} \frac{\max_{|v|\le 1} |G'(\beta+iv,y)|}{G(\beta,y)\log x} + \frac{ \max_{|v|\le 1}|G''(\beta+iv,y)|}{G(\beta,y)(\log x )(\log y)}+ \frac{1}{y}. \end{equation}

Because we assume $y \ge (\log x)^{2+\varepsilon }$, we have $\beta \ge 1/2+ c_{\varepsilon }$. Under RH, $\log G(\beta,\,y) = O_{\varepsilon }(1)$ by lemma 2.11. To bound the quantities appearing in $E$, we write $G(\beta +it,\,y)$ as $G_1(\beta +it,\,y)$ times $G_2(\beta +it,\,y)$. Lemma 2.10 and equation (2.11) tell us that

(4.14)\begin{equation} (\log G_2)^{(i)}(\beta+it,y) \ll_{\varepsilon} (\log y)^{i-1} y^{{1}/{2}-\beta} \end{equation}

for $i=0,\,1,\,2$ and $t \in \mathbb {R}$. Corollary 2.9 says that under RH

(4.15)\begin{align} (\log G_1)^{(i)}(\beta+it,y) & = ({-}1)^i(\log y)^{i-1}y^{-\beta-it} ( \psi(y)-y+ O_{\varepsilon}(y^{{1}/{2}}))\nonumber\\ & \quad \ll_{\varepsilon} (\log y)^{i+1} y^{{1}/{2}-\beta} \end{align}

for all $i=0,\,1,\,2$ and $|t|\le 1$. Putting these two together, one obtains (1.11).

4.2 Proof of theorem 1.1: small $u$

Here we prove theorem 1.1 for $u$ in the range (4.8). In this range, $\beta =1+o(1)$ and $\Psi (x,\,y)=x^{1+o(1)}$. Moreover, $\log G(\beta,\,y) = O(1)$ unconditionally by corollary 2.9 and lemma 2.10. The hardest range of the proof will be $u\asymp 1$. Before proceeding with the actual proof, note that from proposition 4.2 and the triangle inequality, it follows that

(4.16)\begin{align} \Psi(x,y) & = \Lambda(x,y) G(\beta,y) \left(1+O\left(t_2^{{-}4/5} + t_2 \max_{|t|\le t_2}|G'(\beta+it,y)|\right.\right.\nonumber\\ & \left.\left.\quad+\max_{|t|\le 1}|G''(\beta+it,y)|\right)\right) \end{align}

holds unconditionally for $t_2 \in [(\log x)^5,\, y^{4/5}]$ and the range $x \ge y \ge C$, $u \le (\log y)(\log \log y)^3$.

We obtain from proposition 4.2 with $t_2=y^{4/5}$ that

\[ \Psi(x,y) = \Lambda(x,y) G(\beta,y) ( 1+E_1 + E_2 + E_3 + E_4 + y^{{-}3/5}) \]

for $E_i$ bounded in (4.10). We write $G(\beta +it,\,y)$ as $G_1(\beta +it,\,y)$ times $G_2(\beta +it,\,y)$. By lemma 2.10 and (2.11),

(4.17)\begin{equation} (\log G_2)^{(i)}(\beta+it,y) \ll (\log y)^{i-1}u\log (u+1) y^{-{1}/{2}} \end{equation}

for $i=0,\,1,\,2$ and $t \in \mathbb {R}$ where we simplified $y^{-\beta }$ using (2.1). From now on we assume RH. Corollary 2.9 implies

(4.18)\begin{equation} (\log G_1)^{(i)}(\beta+it,y) \ll \frac{ (\log y)^{i-1}u \log (u+1)}{y}(|\psi(y)-y|+y^{{1}/{2}}) \end{equation}

for $i=0,\,1,\,2$ when $|t|\le 1$. As in the medium $u$ case, one can bound $E_1$ by an acceptable quantity using our estimates for $(\log G_1)^{(i)}$ and $(\log G_2)^{(i)}$. Recall

\[ E_2 \ll \frac{\max_{|v|\le t_1} |G'( \beta+iv,y)|\exp({-}u^{1/3}/20)}{G( \beta,y)\log x} \]

where $t_1 = u\log (u+1)/\log y$. If $t_1 \le 1$ we bound $E_2$ in the same way we bounded $E_1$. Otherwise we use (2.8), which implies that

(4.19)\begin{equation} (\log G_1)^{(i)}(\beta+it,y) \ll (\log y)^{i+1}u \log (u+1) y^{-{1}/{2}} \end{equation}

holds for $i=0,\,1,\,2$ and $|t|\le y^{9/10}$. This shows that, if $t_1>1$, i.e. $u\log (u+1)\ge \log y$,

\[ E_2 \ll \frac{(\log y)^2 u \log(u+1)\exp({-}u^{1/3}/20)}{y^{{1}/{2}}\log x} \ll \log (u+1) y^{-{1}/{2}}. \]

This is an acceptable contribution when $u\log (u+1)> \log y$. We now study $E_3$ and $E_4$. Due to $G(\beta +it,\,y)/G(\beta,\,y)$ being very close to $1$ in our considered range by (4.17) and (4.19), we may replace

\[ G(\beta+it,y)/G(\beta,y)-1 \]

by

\[ \log G(\beta+it,y)-\log G(\beta,y) \]

and incur a negligible error, in both $E_3$ and $E_4$. So to show $E_3$ is acceptable we need to prove

(4.20)\begin{equation} \int_{t_1 \le |t| \le y^{4/5}} | \log G( \beta+it,y)\!-\!\log G( \beta,y)| \frac{\log(|t|+2)}{t^2} \,{\rm d}t \!\ll\! \frac{e^{u/3}}{y} (|\psi(y)-y|\!+\!y^{{1}/{2}}). \end{equation}

This is shown using the bound

(4.21)\begin{equation} \log G( \beta+it,y) \ll \frac{u \log (u+1)}{y \log y} (|\psi(y)-y| + y^{{1}/{2}}\log^2(|t|+2)), \quad |t| \le y^{9/10}, \end{equation}

which is a consequence of (2.8) and (4.17). To handle $E_4$ it remains to prove

(4.22)\begin{align} & \int_{t_1 \le |t| \le y^{4/5}} K(\beta+it-1)x^{it}( \log G( \beta+it,y)-\log G( \beta,y)) \frac{{\rm d}t}{t}\nonumber\\ & \quad \ll_{\varepsilon} \frac{e^{u/2}}{y \log y} (|\psi(y)-y|+y^{{1}/{2}}). \end{align}

Here we cannot use the triangle inequality and put absolute value inside the integral. Indeed, if we use the pointwise bound (4.21), along with our bounds for $\zeta$ (lemmas 2.4 and 2.5), we get a bound which falls short by a factor of $(\log y)^3$. We shall overcome this by several integrations by parts as we now describe.

To deal with the contribution of $\log G(\beta,\,y)$ to (4.22) we use (4.21) with $t=0$ along with the bound

\[ \int_{t_1 \le |t| \le y^{4/5}} K(\beta+it-1)x^{it}\frac{{\rm d}t}{t} \ll u^2 \]

which follows by integration by parts, where we replace $x^{it}$ by its antiderivative $x^{it}/\log x$.

Note that due to integration by parts, derivatives of $\zeta$ arise. This means that in addition to lemmas 2.4 and 2.5 we need the bounds $\zeta ^{(k)}(s) \ll _k (1+(|t|+4)^{1-\sigma })\log ^{k+1}(|t|+4)$ and $\int _{1}^{T} |\zeta ^{(k)}(\sigma +it)|^2 \,{\rm d}t \ll _k T$ for $\sigma \in [2/3,\,1]$ and $T,\, |t| \ge 1$. These bounds follow from lemmas 2.4 and 2.5 through Cauchy's integral formula.

To deal with the contribution of $\log G(\beta +it,\,y)$ to (4.22) we write it $\log G_1( \beta +it,\,y)+\log G_2( \beta +it,\,y)$ and obtain two integrals which we bound separately.

4.2.1 Treatment of $\log G_1$

Recall we assume $y \le x^{1-\varepsilon }$. We want to show

(4.23)\begin{equation} \int_{t_1 \le |t| \le y^{4/5}} K(\beta+it-1)x^{it} \log G_1( \beta+it,y) \frac{{\rm d}t}{t} \ll_{\varepsilon} \frac{e^{u/2}}{y \log y} (|\psi(y)-y|+y^{{1}/{2}}). \end{equation}

We integrate by parts, replacing $x^{it}$ by its antiderivative, reducing matters to showing

(4.24)\begin{equation} \frac{1}{\log x}\int_{t_1\le |t| \le y^{4/5}} K(\beta+it-1) x^{it} \frac{G_1'}{G_1}(\beta+it,y) \frac{{\rm d}t}{t} \ll_{\varepsilon} \frac{e^{u/2}}{y \log y} (|\psi(y)-y|+y^{{1}/{2}}). \end{equation}

We divide and multiply the integrand by $y^{it}$, so the left-hand side of (4.23) is now

(4.25)\begin{equation} \frac{1}{\log x}\int_{t_1\le |t| \le y^{4/5}} K(\beta+it-1) (x/y)^{it} H(t)\frac{{\rm d}t}{t} \end{equation}

where $H(t):= y^{it}(G'_1/G_1)(\beta +it,\,y)$. From lemma 2.8,

\[ y^{\beta} \cdot H(t) = \sum_{|\Im (\rho)-t| \le 2y^{4/5}}\frac{y^{\rho}}{\rho-\beta-it}+O(y^{{2}/{5}}) \ll |\psi(y)-y|+ y^{{1}/{2}}\log^2 (|t|+2) \]

and, for $k=1,\,2,\,3$,

\[ y^{\beta} \cdot H^{(k)}(t) \!=\! (k+1)!i^{k} \sum_{|\Im (\rho)-t| \le 2y^{4/5}}\frac{y^{\rho}}{(\rho-\beta-it)^{k+1}}+O(y^{{2}/{5}})\ll y^{{1}/{2}}\log (|t|\!+\!2). \]

We integrate by parts 3 times, replacing $(x/y)^{it}$ by its antiderivative. We are guaranteed to get enough saving since $\log (x/y) \gg _{\varepsilon } \log x$.

4.2.2 Treatment of $\log G_2$

The function $\log G_2(\beta +it,\,y)$ is given as a sum over proper primes powers. As the cubes and higher powers contribute at most $\ll y^{-2/3+o(1)}$ to it by the prime number theorem (see [9]), we can replace $\log G_2(\beta +it,\,y)$ with the prime sum $\sum _{y^{1/2}< p \le y} p^{-2(\beta +it)}/2$, so we are left to show

\[ \sum_{y^{1/2}< p \le y} p^{{-}2\beta}\int_{t_1 \le |t| \le y^{4/5}} K(\beta+it-1)(x/p^2)^{it} \frac{{\rm d}t}{t} \ll \frac{e^{u/2}}{y^{{1}/{2}} \log y}. \]

For a given $p$, the pointwise bound $(x/p^2)^{it} \ll 1$ leads to the above integral being bounded by $\ll \log y$. This is good enough for the primes $p \in [y^{1/2}\log y,\, y]$, since

\[ \sum_{y^{{1}/{2}}\log y \le p\le y}p^{{-}2\beta} \log y \asymp \frac{u\log (u+1)}{y^{{1}/{2}}\log y}. \]

For the primes $p \in (y^{1/2},\,y^{1/2}\log y)$ we integrate by parts, replacing $(x/p^2)^{it}$ by its antiderivatives.

4.3 Proof of theorem 1.2

Suppose $(\log x)^{3} \ge y \ge (\log x)^{4/3+\varepsilon }$. It follows from proposition 4.1 that $\Psi (x,\,y) = \Lambda (x,\,y) G(\beta,\,y) ( 1 + E)$ holds unconditionally for

(4.26)\begin{equation} E \ll_{\varepsilon} \frac{ |G'(\beta,y)|}{G(\beta,y)\log x} + \frac{ \max_{|v|\le t_0}|G''(\beta+iv,y)|}{G(\beta,y)(\log x)(\log y)} + \frac{\max_{|v|\le \frac{1}{\log y}} |G'(\beta+iv,y)|}{G(\beta,y)\exp(u^{1/3}/20)} +\frac{1}{y} \end{equation}

where $t_0$ is given in the proposition. It remains to bound the quantities appearing in $E$. From now on we assume RH. Let $A:= (\log x) / y^{1/2}$. We will prove the stronger bound

(4.27)\begin{align} E & \ll_{\varepsilon} \frac{|\psi(y)-y|+y^{{1}/{2}}}{y} \bigg( 1+ u \frac{|\psi(y)-y|+y^{\frac{1}{2}}}{y}\bigg)\nonumber\\ & \quad + \frac{\max\{A,A^2\}}{u\max\{1,|\log A|\}} \bigg( 1+\frac{\max\{A,A^2\}}{\max\{1,|\log A|\}}\bigg), \end{align}

which implies the theorem using $\psi (y)-y \ll y^{1/2} \log ^2 y$. Recall we can always simplify $y^{-\beta }$ using (2.1) as $\asymp _{\varepsilon } (\log x)/y$. In particular, $y^{1/2-\beta } \asymp _{\varepsilon } A$. Recall $G=G_1 G_2$. Lemma 2.10 and equation (2.11) tell us that

(4.28)\begin{equation} (\log G_2)^{(i)}(\beta+it,y) \ll (\log y)^{i} \frac{\max\{A, A^2\}}{\max\{1,|\log A|\}} \end{equation}

for $i=0,\,1,\,2$ and $t \in \mathbb {R}$. Corollary 2.9 says that under RH

(4.29)\begin{equation} (\log G_1)^{(i)}(\beta+it,y) \ll (\log y)^{i-1}\frac{\log x}{y} ( |\psi(y)-y| + y^{{1}/{2}}) \end{equation}

for $i=0,\,1,\,2$ and $|t|\le 1$. Applying (4.28) and (4.29) with $i=1$ shows

\[ \frac{ |G'(\beta,y)|}{G(\beta,y)} \frac{1}{\log x} \ll \frac{|\psi(y)-y|+y^{{1}/{2}}}{y}+ \frac{\max\{A,A^2\}}{u\max\{1,|\log A|\}} \]

which treats the first quantity in (4.26). We now consider the third term in (4.26). Observe

(4.30)\begin{align} & \frac{ \max_{|v|\le 1/\log y}|G'(\beta+iv,y)|}{G(\beta,y)\exp(u^{1/3}/20)} \le \frac{ \max_{|v|\le 1/\log y}|G(\beta+iv,y)|}{G(\beta,y)\exp(u^{1/3}/20)}\nonumber\\ & \quad \cdot \max_{|v|\le 1 }|(\log G)'(\beta+iv,y)|. \end{align}

From (4.28) and (4.29) we have

(4.31)\begin{equation} \max_{|v|\le 1 }|(\log G)'(\beta+iv,y)| \ll (\log x)^4, \end{equation}

say, and, by (2.11) and (4.29),

(4.32)\begin{equation} \frac{ \max_{|v|\le 1/\log y}|G(\beta+iv,y)|}{G(\beta,y)}\le \exp(C_{\varepsilon} (\log y)^2 (\log x) /y^{1/2}), \end{equation}

so that (4.30) leads to

\[ \frac{\max_{|v|\le 1/\log y} |G'(\beta+iv,y)|}{G(\beta,y)\exp(u^{1/3}/20)} \ll_{\varepsilon} \frac{\exp(C_{\varepsilon} (\log y)^2 (\log x) /y^{1/2})}{ \exp(u^{1/3}/40)} \ll_{\varepsilon} \frac{1}{y}. \]

It remains to bound the second term in (4.26). Observe

(4.33)\begin{align} & \frac{ \max_{|v|\le t_0}|G''(\beta+iv,y)|}{G(\beta,y)(\log x) (\log y)} \le \frac{ \max_{|v|\le t_0}|G(\beta+iv,y)|}{G(\beta,y)(\log x) (\log y)} \nonumber\\ & \quad \cdot ( \max_{|v|\le 1}|(\log G)''(\beta+iv,y)| + \max_{|v|\le 1}|(\log G)'(\beta+iv,y)|^2). \end{align}

By (2.11) we can bound the fraction in the right-hand side of (4.33) by $O_{\varepsilon }(1)$:

\begin{align*} & \frac{ \max_{|v|\le t_0}|G(\beta+iv,y)|}{G(\beta,y)}\le \frac{ \max_{|v|\le t_0}|G_1(\beta+iv,y)|}{G_1(\beta,y)}\\ & \quad \le \exp\big( \int_{{-}t_0}^{t_0} |G_1'/G_1|(\beta+iv,y) \,{\rm d}v\big)\le \exp(C_{\varepsilon}t_0 (\log y)^2(\log x)/y^{1/2}) \ll_{\varepsilon} 1. \end{align*}

The derivatives of $\log G$ in the right-hand side of (4.33) are handled by (4.28) and (4.29), giving

\begin{align*} & \max_{|v|\le 1}|(\log G)''(\beta+iv,y)| + \max_{|v|\le 1}|(\log G)'(\beta+iv,y)|^2 \\ & \quad \ll \frac{(\log y )(\log x)}{y}(|\psi(y)-y|+y^{{1}/{2}}|) + \frac{(\log x)^2}{y^2} (|\psi(y)-y|+y^{{1}/{2}})^2\\ & \quad + (\log y)^{2} \left( \frac{\max\{A, A^2\}}{\max\{1,|\log A|\}} + \frac{\max\{A, A^2\}^2}{\max\{1,|\log A|\}^2}\right). \end{align*}

Dividing this by $(\log x) (\log y)$ gives a bound for the second term in (4.26).