2.1 Standard lemmas

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

Lemma 2.1 [Reference Hildebrand and Tenenbaum13, 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}

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

Lemma 2.4 [Reference Montgomery and Vaughan15, 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 [Reference Titchmarsh19, 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 [Reference Hildebrand and Tenenbaum14, 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 [Reference Hildebrand and Tenenbaum13, Lem. 8], proved in the same way.

Lemma 2.7 [Reference Hildebrand and Tenenbaum13]

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 [Reference Gorodetsky9]

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 $T+1$ [Reference Montgomery and Vaughan15, 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 [Reference Montgomery and Vaughan15, 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 $\psi (y)-y=O(y^{1/2}\log ^2 y)$ [Reference von Koch20].

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 [Reference Gorodetsky9]

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).

3.1 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 [Reference Saias17] (cf. [Reference Tenenbaum18, p. 537]), which is not stated for general $T$ and $\sigma$ but contains the same ideas.

Proof. By [Reference Titchmarsh19, 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 [Reference Hildebrand and Tenenbaum14, 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 [Reference Saias17].

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 $x$ [Reference Hildebrand and Tenenbaum13], 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 [Reference Saias17, p. 98], which does not allow general $y$ and $T$ but contains the main ideas.

Proof. The truncated Perron's formula [Reference Hildebrand and Tenenbaum14, 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 [Reference Hildebrand and Tenenbaum13, Thm. 3]. To conclude, we recall Theorem 2.4 of [Reference de la Bretèche and Tenenbaum5] 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 $\alpha =\beta +O(1/\log y)$ [Reference Hildebrand and Tenenbaum13, 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 $\alpha =\beta +O(1/\log y)$ [Reference Hildebrand and Tenenbaum13, 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.

4.1 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 [Reference Gorodetsky9]), 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).