# Friable integers and the Dickman $\rho$ function

Ofir Gorodetsky

###### Abstract.

This paper is concerned with the relationship of $y$-friable (i.e. $y$-smooth)
integers and the Dickman function. Under the Riemann Hypothesis (RH), an
asymptotic formula for the count of $y$-friable integers up to $x$,
$\Psi(x,y)$, in terms of the Dickman function was previously available only
for $y\geq(\log x)^{2+\varepsilon}$, thanks to works of Hildebrand and Saias.
Unconditionally we establish an asymptotic formula for $\Psi(x,y)$ in the
wider range $y\geq(1+\varepsilon)\log x$, whose shape is $x\rho(\log x/\log
y)$ times correction factors. These factors take into account the
contributions of zeta zeros and of prime powers. With this formula at hand, we
resolve two questions of Hildebrand and Pomerance.

Hildebrand conjectured that $\Psi(x,y)$ is not $\asymp x\rho(\log x/\log y)$
once $y$ is smaller than $(\log x)^{2+\varepsilon}$, and we show
unconditionally he was correct.

Pomerance asked whether the inequality $\Psi(x,y)\geq x\rho(\log x/\log y)$
holds for $x/2\geq y\geq 2$. If RH is false we show this fails infinitely
often. When RH is true, the inequality holds for $x/2\geq y\geq 2$, $x\gg 1$
except possibly for $y$ close to the critical point $y=(\log x)^{2}$. Near
this point, the question is essentially equivalent to

$\liminf_{y\to\infty}\frac{\psi(y)-y}{\log y\sqrt{y}}>L$

for a constant $L\approx-0.666217$, where $\psi$ is the Chebyshev function. It
is expected that this limit is $0$, but even under RH we cannot rule out that
it is $-\infty$.

As another consequence of our formula, we show that under RH $\Psi(x,y)$
exhibits an unexpected phase transition when $y\approx(\log x)^{3/2}$.

##  1\. Introduction

A positive integer is called $y$-friable (or $y$-smooth) if all its prime
factors do not exceed $y$. We denote the number of $y$-friable integers up to
$x$ by $\Psi(x,y)$. We assume throughout $x\geq y\geq 2$.

We denote by $\rho\colon[0,\infty)\to(0,\infty)$ the Dickman function, defined
as $\rho(t)=1$ for $t\in[0,1]$, while for larger values it is defined via the
delay differential equation $\rho^{\prime}(t)=-\rho(t-1)/t$. Dickman [Dic30]
showed that

(1.1) $\Psi(x,y)\sim x\rho\left(\log x/\log y\right)\qquad(x\to\infty)$

holds as long as $\log x/\log y$ is bounded from above, that is, $y\geq
x^{\varepsilon}$. For this reason, it is useful to introduce the parameter

$u:=\log x/\log y.$

###  1.1. Hildebrand’s work

The range of validity of (1.1) was considerably improved by de Bruijn [dB51b].
He stated his result in terms of the error term in the prime number theorem.
Define $R(x)$ via

$\pi(x)=\mathrm{Li}(x)\left(1+O(R(x))\right)$

where $\pi$ is the prime counting function and $\mathrm{Li}$ is the
logarithmic integral. He used Buchstab’s identity to show (essentially) that
if $R(x)\ll_{\varepsilon}\exp(-(\log x)^{a-\varepsilon})$ ($a\in(0,1)$) then

(1.2) $\Psi(x,y)=x\rho(u)\left(1+O_{\varepsilon}\left(\log(u+1)/\log
y\right)\right)$

holds uniformly in the range $\log y\geq(\log x)^{1/(1+a)+\varepsilon}$ for
every $\varepsilon>0$. Using the Korobov–Vinogradov zero-free region for the
Riemann zeta function one may take $a=3/5$. Additionally, in the range of his
results, he proved an asymptotic expansion for $\Psi(x,y)$ in (roughly) powers
of $\log(u+1)/\log y$.

In [Hil86], Hildebrand extended de Bruijn’s range qualitatively, showing that
(1.2) holds uniformly in the range

(1.3) $\log y\geq(\log\log x)^{\frac{1}{a}+\varepsilon}$

for every $\varepsilon>0$. Here $a$ is the same as defined above, so we may
take $a=3/5$. In an earlier paper [Hil84], Hildebrand showed that RH implies a
further qualitative improvement, namely that (1.2) holds in the wider range

(1.4) $y\geq(\log x)^{2+\varepsilon}.$

The reverse implication is also true: if even the weaker estimate
$\Psi(x,y)=x\rho(u)\exp(O_{\varepsilon}(y^{\varepsilon}))$ holds in the range
(1.4) then RH must be true. Hildebrand’s proofs rely on a beautiful identity
[Hil84, p. 261].

######  Remark 1.

Hildebrand’s conditional result does not give an asymptotic result when

(1.5) $(\log x)^{A}\geq\log y\geq(\log x)^{2+\varepsilon},$

$A$ being an arbitrary number. Indeed, the error term $\log(u+1)/\log y$ is
bounded away from $0$ when (1.5) holds. Hildebrand’s result only gives an
upper bound in this regime, and if $y\geq(\log x)^{C}$ for sufficiently large
$C$ then also a lower bound is implied.

###  1.2. Hildebrand’s conjecture

In [Hil86, p. 290], Hildebrand speculates that $\Psi(x,y)\sim x\rho(u)$ for
$y\geq(\log x)^{2+\varepsilon}$ but not for $y\leq(\log x)^{2-\varepsilon}$.
Specifically, he writes

> If the Riemann hypothesis is assumed, the range for $u$ can be further
> extended to $1\leq u\leq\log x/(2+\varepsilon)\log\log x$, but it seems
> likely that then the critical limit is attained: it may be conjectured that
> for $\log y<(2-\varepsilon)\log\log x$, the relation $\Psi(x,x^{1/u})\sim
> x\rho(u)$ no longer holds.

This conjecture is repeated by Granville in [Gra89], and in [Gra93, p. 258] he
writes

> … and Hildebrand has even shown that (2.3) holds for all
> $y\geq\log^{2+\varepsilon}x$ if and only if the Riemann Hypothesis is true.
> However we do not believe that (2.1) can hold uniformly for
> $y=\log^{2-\varepsilon}x$ for any fixed $\varepsilon>0$.

We confirm these speculations:

######  Theorem 1.1.

Fix $\varepsilon\in(0,2)$. There are sequences $x_{n}\to\infty$,
$y_{n}\to\infty$ with $y_{n}=(\log x_{n})^{2-\varepsilon+o(1)}$ and

$\Psi(x_{n},y_{n})>x_{n}\rho\left(\log x_{n}/\log
y_{n}\right)\exp\left(y_{n}^{\varepsilon/(2-\varepsilon)+o(1)}\right).$

The theorem is proved in the next section, as part of the stronger Proposition
2.12.

###  1.3. Pomerance’s question

In [Gra08, LP18], Pomerance asked whether

(1.6) $\Psi(x,y)\geq x\rho(u)$

holds for all $x/2\geq y\geq 1$. The motivation is related to de Bruijn’s
approximation to $\Psi(x,y)$, called $\Lambda(x,y)$, which in some ranges is
strictly larger than $x\rho(u)$, see §2.2.

If RH is false, we show Pomerance’s inequality fails infinitely often. If RH
is true, we show it is true when $y\geq(\log x)^{2+\varepsilon}$ or
$y\leq(\log x)^{2-\varepsilon}$ (at least for $x\gg_{\varepsilon}1$). Near
$y=(\log x)^{2}$, the question lies beyond RH in a precise sense, but we
indicate that a positive answer follows from a conjecture of Montgomery and
Vaughan on the size of the remainder term in the prime number theorem.

######  Theorem 1.2.

Fix $\varepsilon>0$.

  1. (1)

Suppose $x\gg_{\varepsilon}1$. Unconditionally, (1.6) holds in
$(1-\varepsilon)x\geq y\geq\exp((\log\log x)^{5/3+\varepsilon})$.

  2. (2)

Suppose RH is not true. Fix $\sigma_{0}\in(1-\Theta,\Theta)$ where
$\Theta\in(1/2,1]$ is the supremum of the real parts of the zeros of $\zeta$.
Then, there are sequences $x_{n}\to\infty$, $y_{n}\to\infty$ satisfying
$y_{n}=(\log x_{n})^{1/(1-\sigma_{0})+o(1)}$ and

$\Psi(x_{n},y_{n})<x_{n}\rho\left(\log x_{n}/\log
y_{n}\right)\exp(-y_{n}^{\Theta-\sigma_{0}-\varepsilon}).$

  3. (3)

If RH is true, (1.6) holds when $x(1-\varepsilon)\geq y\geq(\log
x)^{2+\varepsilon}$ and $2\leq y\leq(\log x)^{2-\varepsilon}$, as long as
$x\gg_{\varepsilon}1$.

  4. (4)

Suppose RH is true. Let $L\in\mathbb{R}$ be the following constant:

$L=\max_{v\in\mathbb{R}}e^{v}\left(-\log(-\zeta(1/2))-\frac{1}{2}\int_{v}^{2v}e^{-r}r^{-1}dr\right)\approx-0.666217.$

Let $\psi$ be the Chebyshev function. A necessary condition for (1.6) to hold
in $(\log x)^{3}\geq y\geq(\log x)^{3/2}$ is

(1.7) $\liminf_{y\to\infty}\frac{\psi(y)-y}{\sqrt{y}\log y}\geq L.$

A sufficient condition for (1.6) to hold in $(\log x)^{3}\geq y\geq(\log
x)^{3/2}$ if $y\gg 1$ is that (1.7) holds with strict inequality.

The theorem is proved in §2.8. Note that RH implies

(1.8) $\psi(y)-y=O(\sqrt{y}(\log y)^{2})$

as shown by von Koch in 1901 [MV07, Thm. 13.1], and this has not been improved
since. It is believed that

(1.9) $\liminf_{y\to\infty}\frac{\psi(y)-y}{\sqrt{y}(\log\log\log
y)^{2}}=-\frac{1}{2\pi},$

see the discussion in [MV07, p. 484]; (1.9) implies that the limit considered
in the last part of Theorem 1.2 is $0$. Goldston and Suriajaya showed that
sufficiently uniform versions of Montgomery’s Pair Correlation lead to
improvements on (1.8) which would also show the limit is $0$.

### Structure of paper

In §2.3 we give some intuition for the behavior of $\Psi(x,y)/(x\rho(u))$ and
then go on to prove Theorems 1.1 and 1.2 as well as a phase transition result
(Theorem 2.14), new inequalities (Corollary 2.11, Theorem 2.16) and a simple
formula for $\Psi(x,y)/(x\rho(u))$ in a wide range given a zero-free strip for
$\zeta$ (Theorem 2.13).

In Theorem 2.15 we study under RH in what range does
$\Psi(x,y)\sim\Lambda(x,y)$ hold where $\Lambda$ is the de Bruijn
approximation [dB51b], and as in Theorem 1.2 the answer relates to the size of
$(\psi(y)-y)/(\sqrt{y}\log y)$.

In §3 and §4 we develop some standard material (including properties of the
saddle point for $\zeta(s,y)$ introduced first in [HT86], and a variant of the
truncated explicit formula for $\psi(y)$) that is needed for a subset of our
results.

### Conventions

We use the convention where $C,c$ denote absolute positive constants which may
change between different occurrences. The notation $A\ll B$ means $|A|\leq CB$
for some absolute constant $C$, and $A\ll_{\varepsilon}B$ means $C$ may depend
on $\varepsilon$. We write $A\asymp B$ to mean $C_{1}B\leq A\leq C_{2}B$ for
some absolute positive constants $C_{i}$, and $A\asymp_{\varepsilon}B$ means
$C_{i}$ may depend on $\varepsilon$. We write $A=\Theta(B)$ and
$A=\Theta_{\varepsilon}(B)$ to mean $A\asymp B$ and $A\asymp_{\varepsilon}B$,
respectively. If we differentiate a bivariate function, we always do so with
respect to the first variable. Throughout, $L(x)=\exp((\log
x)^{\frac{3}{5}}(\log\log x)^{-\frac{1}{5}})$.

### Acknowledgements

We are grateful to Sacha Mangerel for asking as about integer analogues of
[Gor22]. We thank Zeev Rudnick and James Maynard for comments on exposition.
This project has received funding from the European Research Council (ERC)
under the European Union’s Horizon 2020 research and innovation programme
(grant agreement No 851318).

##  2\. A new approximation

###  2.1. Definitions and standard results

We define $\xi\colon[1,\infty)\to[0,\infty)$, a function of $u>1$, via
$e^{\xi(u)}=1+u\xi(u)$.

######  Lemma 2.1.

[HT86, Lem. 1][Hil84, Lem. 1] For $u\geq 3$ we have

(2.1) $\displaystyle\xi(u)$ $\displaystyle=\log u+\log\log u+O\left(\log\log
u/\log u\right),$ (2.2) $\displaystyle\xi^{\prime}(u)^{-1}$
$\displaystyle=u\left(1+O\left((\log u)^{-1}\right)\right).$

The first part of Lemma 2.1 shows

######  Lemma 2.2.

Let $\sigma=1-\xi(u)/\log y$. Fix $\varepsilon>0$. If $x\geq
y\geq(1+\varepsilon)\log x$ and $x\gg_{\varepsilon}1$ then

(2.3) $\sigma=\frac{\log\left(\frac{y}{\log x}\right)}{\log
y}\left(1+O_{\varepsilon}\left(\frac{\log\log y}{\log y}\right)\right).$

######  Lemma 2.3.

Let $\sigma=1-\xi(u)/\log y$, where $u=\log x/\log y$. We have $\sigma\leq 1$
with equality if and only if $u=1$. We have $\sigma\geq 0$ if and only if
$y\geq 1+\log x$, and $\sigma=0$ if and only if $y=1+\log x$.

###### Proof.

The first part follows from $\xi$ being $0$ at $u=1$ and being strictly
increasing. Next, we need to solve $\sigma\geq 0$, or $\log y\geq\xi(u)$.
Again, since $\xi$ is strictly increasing, it actually suffices to solve $\log
y=\xi(u)$. Exponentiating, this implies

$y=e^{\xi(u)}=1+u\xi(u)=1+u\log y=1+\log x$

as needed. ∎

We define the entire function $I(s)$ by $I(s)=\int_{0}^{s}(e^{v}-1)dv/v$. The
following are standard identities.

######  Lemma 2.4.

[Ten15, Eq. (III.5.56)] For $u\geq 1$ we have $I^{\prime}(\xi(u))=u$ and
$I^{\prime\prime}(\xi(u))=1/\xi^{\prime}(u)$.

######  Lemma 2.5.

[Ten15, Thm. III.5.10] Let $\gamma$ be the Euler-Mascheroni constant. For all
$s\in\mathbb{C}$,

(2.4)
$\hat{\rho}(s):=\int_{0}^{\infty}e^{-sv}\rho(v)\,dv=\exp\left(\gamma+I(-s)\right).$

###  2.2. De Bruijn’s approximation

Our results are based on a new approximation for $\Psi(x,y)$. It has
consequences beyond resolving Hildebrand’s and Pomerance’s questions. To put
our approximation in context, we shall briefly discuss de Bruijn’s
approximation $\Lambda(x,y)$ [dB51b]. For $x\not\in\mathbb{Z}$ it is given by

$\Lambda(x,y):=x\int_{\mathbb{R}}\rho\left(u-v\right)d\left(\lfloor
y^{v}\rfloor/y^{v}\right),$

otherwise $\Lambda(x,y)=\Lambda(x+,y)$ (one has
$\Lambda(x,y)=\Lambda(x-,y)+O(1)$ if $x\in\mathbb{Z}$ [dB51b, p. 54]). We
refer the reader to de Bruijn’s original paper for the motivations for this
definition. Integrating the definition by parts gives

(2.5)
$\Lambda(x,y)=x\rho(u)-\\{x\\}+\int_{0}^{u-1}(-\rho^{\prime}(u-v))\\{y^{v}\\}y^{-v}dv$

when $x\notin\mathbb{Z}$. Due to $\rho$ being decreasing, the integral in the
right-hand side of (2.5) is non-negative, which motivates Pomerance’s
question. Saias [Sai89, Lem. 4], improving on de Bruijn [dB51b], proved

(2.6) $\Lambda(x,y)=x\rho(u)\left(1+O_{\varepsilon}(\log(u+1)/\log y)\right)$

holds uniformly in $y\geq(\log x)^{1+\varepsilon}$, and that [Sai89, Thm.]

(2.7) $\Psi(x,y)=\Lambda(x,y)\left(1+O_{\varepsilon}\left(\exp(-(\log
y)^{a-\varepsilon}\right)\right)$

holds uniformly in the range (1.3). In particular, one recovers Hildebrand’s
unconditional result. Saias indicates in [Sai89, p. 79] that if one assumes RH
then his proof gives

(2.8)
$\Psi(x,y)=\Lambda(x,y)\left(1+O_{\varepsilon}\left(y^{\varepsilon-1/2}\log
x\right)\right)$

in the range $y\geq(\log x)^{2+\varepsilon}$. Implicit in the proof of
Proposition 4.1 of La Bretèche and Tenenbaum [dlBT05] is the estimate

(2.9) $\Lambda(x,y)=x\rho(u)Z(1-\xi(u)/\log y)\left(1+O_{\varepsilon}(1/\log
x)\right)$

uniformly for $y\geq(\log x)^{1+\varepsilon}$ where

$Z(t):=\zeta(t)(t-1)/t,\qquad Z(1)=1,$

$\zeta$ is the Riemann zeta function and $\xi$ is defined in §2.1. The
function $Z$ originates in de Bruijn’s work [dB51b, Eq. (2.8)], where it is
denoted by $K(t+1)$. It is evident that $\lim_{t\to 0^{+}}Z(t)=\infty$.
Moreover,

######  Lemma 2.6.

The function $Z$ is strictly decreasing in $(0,1]$.

###### Proof.

We have

$Z^{\prime}(t)=((\zeta(t)(t-1))^{\prime}-\zeta(t)(t-1))/t^{2}.$

The integral representation
$\zeta(s)=s/(s-1)-s\int_{1}^{\infty}\\{x\\}dx/x^{1+s}$ for $\Re s>0$ [MV07,
Thm. 1.2] implies
$Z^{\prime}(t)=-\int_{1}^{\infty}(x+1-t^{2})\\{x\\}x^{-2-t}dx<0$. ∎

It follows from Saias’ work and (2.9) that under RH, the quantities
$\Psi(x,y)$ and $x\rho(u)$ are not asymptotic in the regime (1.5), but still
of the same order of magnitude.

###  2.3. The function $G$ and informal discussion

Our approximations will be given in terms of the function

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

where

$\zeta(s,y):=\prod_{p\leq y}(1-p^{-s})^{-1}=\sum_{n\text{ is
}y\text{-friable}}n^{-s}\qquad(\Re s>0)$

is the partial zeta function, and

(2.10) $F(s,y):=\zeta(s)(s-1)\log y\hat{\rho}((s-1)\log
y)\qquad(s\in\mathbb{C})$

where $\hat{\rho}$ is the Laplace transform of the Dickman function, which is
never zero by Lemma 2.5. Hence the function $G$ is defined for every
$s\in\mathbb{C}$ with $\Re s>0$ which is not a zero of $\zeta$.

The ratio $G$ arises naturally: $\zeta(s,y)/s$ appears as the Mellin transform
of $x\mapsto\Psi(x,y)$ while $F(s,y)/s$ appears as the Mellin transform of
$x\mapsto\Lambda(x,y)$. This latter fact is due to de Bruijn [dB51b, p. 54]
(cf. [Sai89]). The ratio $G$ contains information on the ratio
$\Psi(x,y)/\Lambda(x,y)\sim\Psi(x,y)/(x\rho(u)Z(\sigma))$.

We choose the following logarithm of $\zeta(s,y)$:

$\log\zeta(s,y)=\sum_{p\leq y}(-\log(1-p^{-s}))=\sum_{n\text{ is
}y\text{-friable}}\Lambda(n)/(n^{s}\log n),$

so we can write $G$ as $G_{1}G_{2}$ where

(2.11) $G_{1}(s,y)=\exp\big{(}\sum_{n\leq y}\frac{\Lambda(n)}{n^{s}\log
n}\big{)}/F(s,y),\,\log G_{2}(s,y)=\sum_{k\geq 2}\sum_{y^{1/k}<p\leq
y}p^{-ks}/k.$

We use the decomposition into $G_{1}$ and $G_{2}$ throughout. Informally, for
real $s\in(0,1)$ we show in §4 that

$\log G_{1}(s,y)\approx-\sum_{\rho}\frac{y^{\rho-s}}{(\rho-s)\log y}$

where the sum is over zeros of $\zeta$, and

$\log G_{2}(s,y)\approx y^{\max\\{1-2s,1/2-s\\}}/\log y.$

The relevant value of $s$ when studying $\Psi(x,y)$ and $\Lambda(x,y)$ using
their Mellin transforms is known to be close to $1-\log\log x/\log y$. In
particular, if we fix $A>1$ and consider $y\approx(\log x)^{A}$, then for the
purposes of studying $\Psi(x,y)$ we care mostly about $s\approx 1-1/A$. At
this point

(2.12) $\displaystyle\log G_{1}\left(1-1/A,y\right)$ $\displaystyle\ll
y^{\Theta-1+1/A+o(1)},$ (2.13) $\displaystyle\log G_{2}\left(1-1/A,y\right)$
$\displaystyle\approx y^{\max\left\\{1/A-1/2,2/A-1\right\\}+o(1)}$

where $\Theta$ is the supremum of the real parts of the zeros of $\zeta$.

Our understanding of $\log G_{1}$ relies on our understanding of the zeros.
For instance, let us suppose RH holds. Then $\Theta=1/2$ and $\log
G_{1}(1-1/A,y)$ is $o(1)$ if $A>2$. In the other direction we end up using
Landau’s oscillation theorem to show that not only is it bounded by
$y^{-1/A-1/2+o(1)}$, but that it can reach this order of magnitude _with both
signs_ infinitely often, so $\log G_{1}(1-1/A,y)\neq o(1)$ once $A<2$. See
§4.2.

The term $\log G_{2}(1-1/A,y)$ is elementary. For $A>2$, $\log G_{2}(1-1/A,y)$
is $o(1)$, while for $A<2$ it has a large positive contribution.

In summary, $A=2$ is a critical point for two different reasons: zeros and
prime powers.

###  2.4. Main formula

Hildebrand and Tenenbaum proved the following asymptotic formula.

######  Theorem 2.7.

[HT86, Thms. 1, 2] Uniformly for $x\geq y\geq\log x$ we have

(2.14)
$\Psi(x,y)=\frac{x^{\alpha}\zeta(\alpha,y)}{\alpha\sqrt{2\pi\phi_{2}(\alpha,y)}}\left(1+O\left(u^{-1}\right)\right)$

where $\alpha>0$ is defined as the minimum of $s\mapsto\zeta(s,y)x^{s}$, and

(2.15) $\phi_{2}(\alpha,y):=\sum_{p\leq y}\frac{p^{\alpha}(\log
p)^{2}}{(p^{\alpha}-1)^{2}}=\big{(}1+\frac{\log x}{y}\big{)}\log x\log
y\left(1+O\left((\log(1+u))^{-1}\right)\right).$

The saddle point $\alpha$ satisfies, uniformly for $x\geq y\geq 2$,

(2.16) $\alpha=\frac{\log\left(1+\frac{y}{\log x}\right)}{\log
y}\left(1+O\left(\frac{\log\log y}{\log y}\right)\right).$

The saddle point proof of Theorem 2.7 adapts to the study of $\rho$:

######  Theorem 2.8.

[Ten15, Thm. III.5.13] For $u\geq 1$ we have

(2.17) $\rho(u)=(\sqrt{2\pi
I^{\prime\prime}(\xi(u))})^{-1}e^{\gamma-u\xi+I(\xi)}(1+O(u^{-1})).$

By the definition of $F$, we can restate Theorem 2.8 as

######  Corollary 2.9.

Suppose $y\neq 1+\log x$. We have, for $\sigma=1-\xi(u)/\log y$,

(2.18) $x\rho(u)Z(\sigma)=\frac{x^{\sigma}F(\sigma,y)}{\sigma\sqrt{2\pi
I^{\prime\prime}(\xi)(\log y)^{2}}}\left(1+O\left(u^{-1}\right)\right).$

Throughout we use $\alpha$ and $\sigma$ as in Theorem 2.7 and Corollary 2.9.
For $t\in(0,1]$ let

(2.19) $f(t):=t\log x+\log F(t,y),\qquad g(t):=t\log x+\log\zeta(t,y),$

and

$B(x,y):=\frac{\sigma}{\alpha}\frac{\sqrt{I^{\prime\prime}(\xi)(\log
y)^{2}}}{\sqrt{\phi_{2}(\alpha,y)}}.$

Observe also that

(2.20) $\displaystyle g^{\prime\prime}(t)$ $\displaystyle=\sum_{p\leq
y}\frac{p^{t}(\log p)^{2}}{(p^{t}-1)^{2}}>0,$ (2.21) $\displaystyle
f^{\prime\prime}(t)$ $\displaystyle=(\log(\zeta(t)(t-1)))^{\prime\prime}+(\log
y)^{2}I^{\prime\prime}((1-t)\log y).$

######  Proposition 2.10 (Main formula).

If $x\geq y>1+\log x$ then

(2.22) $\displaystyle\frac{\Psi(x,y)}{x\rho(u)Z(\sigma)}$
$\displaystyle=G(\sigma,y)\exp\left(g(\alpha)-g(\sigma)\right)B(x,y)\left(1+O\left(u^{-1}\right)\right)$
(2.23)
$\displaystyle=G(\alpha,y)\exp\left(f(\alpha)-f(\sigma)\right)B(x,y)\left(1+O\left(u^{-1}\right)\right).$

###### Proof.

We divide the left-hand side of (2.14) by the left-hand side of (2.18), and
equate with the right-hand side of (2.14) divided by the right-hand side of
(2.18). We then rearrange in two different ways via

$\frac{x^{\alpha}\zeta(\alpha,y)}{x^{\sigma}F(\sigma,y)}=\frac{\zeta(\alpha,y)}{F(\alpha,y)}\frac{x^{\alpha}F(\alpha,y)}{x^{\sigma}F(\sigma,y)}=\frac{\zeta(\sigma,y)}{F(\sigma,y)}\frac{x^{\alpha}\zeta(\alpha,y)}{x^{\sigma}\zeta(\sigma,y)}.$

Finally, recall $G=\zeta/F$. ∎

If $u\to\infty$, Lemmas 2.1 and 2.4 show $I^{\prime\prime}(\xi(u))\sim u$. If
$y/\log x\to\infty$ and $u\to\infty$ then $\phi_{2}(\alpha,y)\sim\log x\log y$
by (2.15). Moreover, if $y/\log x\to\infty$ and $u\to\infty$ then
$\alpha\sim\sigma$ by (2.16) and Lemma 2.1. Hence

$B(x,y)\sim 1$

if $u\to\infty$ and $y/\log x\to\infty$. In §3 we study $B(x,y)$ in depth,
which is not needed for Theorem 1.1 but is needed for other results such as
Theorem 2.16.

The differences $g(\alpha)-g(\sigma)$ and $f(\alpha)-f(\sigma)$ are more
delicate, but it is easy to determine their signs. Since
$g^{\prime}(\alpha)=0$ by definition and $g^{\prime\prime}(t)>0$, it follows
that $g(\alpha)-g(\sigma)\leq 0$. A similar argument works for
$f(\alpha)-f(\sigma)$, but more care is needed because $f^{\prime}(\sigma)$ is
not $0$. A Taylor approximation shows

$f(\alpha)-f(\sigma)=(\alpha-\sigma)f^{\prime}(\sigma)+(\alpha-\sigma)^{2}f^{\prime\prime}(t)/2$

for some $t$ between $\alpha$ and $\sigma$. We have

$f^{\prime}(\sigma)=\log x+(\log\zeta(\sigma)(\sigma-1))^{\prime}-\log
yI^{\prime}(\xi)=(\log\zeta(\sigma)(\sigma-1))^{\prime}=O(1).$

Moreover, $f^{\prime\prime}(t)>0$ by Lemma 3.2 and $\alpha=\sigma+o(1)$ by
Lemma 3.1. Hence $f(\alpha)-f(\sigma)\geq o(1)$. We just established

######  Corollary 2.11.

If $y/\log x\to\infty$ and $u\to\infty$ when $x\to\infty$ then

$(1+o(1))G(\alpha,y)\leq\frac{\Psi(x,y)}{x\rho(u)Z(\sigma)}\leq(1+o(1))G(\sigma,y).$

Fix $\varepsilon>0$. If $y\geq(1+\varepsilon)\log x$ and $x\gg_{\varepsilon}1$
then

$G(\alpha,y)\ll_{\varepsilon}\frac{\Psi(x,y)}{x\rho(u)Z(\sigma)}\ll_{\varepsilon}G(\sigma,y).$

######  Remark 2.

There is a variant of Proposition 2.10 proved in exactly the same way. Letting

$\tilde{f}(t):=t\log x+\hat{\rho}((s-1)\log
y),\quad\tilde{G}(s,y):=G(s,y)Z(s),$

one has

(2.24) $\displaystyle\frac{\Psi(x,y)}{x\rho(u)}$
$\displaystyle=\tilde{G}(\sigma,y)\exp\left(g(\alpha)-g(\sigma)\right)B(x,y)\left(1+O\left(u^{-1}\right)\right)$
(2.25)
$\displaystyle=\tilde{G}(\alpha,y)\exp\big{(}\tilde{f}(\alpha)-\tilde{f}(\sigma)\big{)}B(x,y)\left(1+O\left(u^{-1}\right)\right).$

On the one hand, $\tilde{f}^{\prime}(t)$ is _exactly_ $0$ at $t=\sigma$ and
$\tilde{f}(\alpha)-\tilde{f}(\sigma)$ is non-negative. On the other hand,
$\tilde{G}$ is more complicated than $G$.

In §3 we study $g(\alpha)-g(\sigma)$ and $f(\alpha)-f(\sigma)$ which is needed
for some of our later results, such as our phase transition result (Theorem
2.14), but not for Theorem 1.1. In Lemma 3.9 we improve the error in
Proposition 2.10 to $O((\alpha\log x)^{-1})$ which is needed e.g. in Theorem
2.16, but not for Theorem 1.1. In view of Lemma 3.9 and the estimate for $B$
in Lemma 3.8, we can drop the condition $u\to\infty$ in Corollary 2.11.

###  2.5. Oscillations

Recall $\sigma=\sigma(x,y)=1-\xi(u)/\log y$ and $\alpha=\alpha(x,y)$ are
functions of $x$ and $y$. Given $y\geq 2$ and fixed $\sigma_{0}\in(0,1)$,
there is a unique $x$ with $\sigma(x,y)=\sigma_{0}$. It is determined by the
relation

$(y^{1-\sigma_{0}}-1)/(1-\sigma_{0})=\log x.$

By Lemma 2.1 $\sigma_{0}=1-\log\log x/\log y+o(1)$ as $y\to\infty$, hence

$y=(\log x)^{1/(1-\sigma_{0})+o(1)}.$

Similarly, given $y\geq 2$ and fixed $\alpha_{0}>0$, there is a unique $x$
with $\alpha(x,y)=\alpha_{0}$, determined by

$-\log^{\prime}(\alpha_{0},y)=\sum_{p\leq y}(p^{\alpha_{0}}-1)^{-1}=\log x.$

We have the relation $\alpha_{0}=1-\log\log x/\log y+o(1)$ by (2.16), so that
$y=(\log x)^{1/(1-\alpha_{0})+o(1)}$.

######  Proposition 2.12.

Let $\Theta\in[1/2,1]$ be the supremum of the real parts of zeros of $\zeta$.
Fix $\varepsilon>0$.

  1. (1)

Assume RH fails and fix $\sigma_{0}\in(1-\Theta,\Theta)$. Given $y>2$, let
$x=x(y)$ be the solution to $\sigma(x,y)=\sigma_{0}$. Then

$\Psi(x(y),y)\ll
x\rho(u)\exp\left(\Omega_{-}\left(y^{\Theta-\sigma_{0}-\varepsilon}\right)\right).$

  2. (2)

Fix $\alpha_{0}\in(0,\Theta)$. Given $y>2$, let $x=x(y)$ be the solution to
$\alpha(x,y)=\alpha_{0}$. Then (regardless of the truth of RH), for some
$c_{\alpha_{0}}>0$,

$\Psi(x(y),y)\gg
x\rho(u)\exp\left(c_{\alpha_{0}}y^{\max\left\\{1-2\alpha_{0},\frac{1}{2}-\alpha_{0}\right\\}}/\log
y\right)\exp\left(\Omega_{+}\left(y^{\Theta-\alpha_{0}-\varepsilon}\right)\right).$

###### Proof.

Let as assume RH fails. Let us fix $\sigma_{0}\in(1-\Theta,\Theta)$, and given
$y$ let $x=x(y)$ be the solution to $\sigma(x,y)=\sigma_{0}$. We have
$\Psi(x(y),y)\ll x\rho(u)G(\sigma_{0},y)$ by Corollary 2.11. We have $\log
G(\sigma_{0},y)=\log G_{1}(\sigma_{0},y)+\log G_{2}(\sigma_{0},y)$. For our
fixed $\sigma_{0}$, Corollary 4.12 tells us that the function $\log
G_{2}(\sigma_{0},y)$ is

$\log G_{2}(\sigma_{0},y)\asymp
y^{\max\left\\{1-2\sigma_{0},\frac{1}{2}-\sigma_{0}\right\\}}/\log y$

if $\sigma_{0}\neq 1/2$, and otherwise $\log G_{2}(\sigma_{0},y)\asymp 1$. For
$\log G_{1}(\sigma_{0},y)$ we have $\log
G_{1}(\sigma_{0},y)=\Omega_{\pm}(y^{\Theta-\sigma_{0}-\varepsilon})$ by
Proposition 4.8. Since $y^{\Theta-\sigma-\varepsilon}$ dominates $\log
G_{2}(\sigma,y)$ by our choice of $\sigma_{0}$, the first result follows.

We now fix $\alpha_{0}\in(0,1)$ and assume nothing about RH. We argue as
before, except that now we exploit the fact that $\log G_{2}$ is positive. ∎

Applying the second part of the proposition with
$1/(1-\alpha_{0})=2-\varepsilon$ proves Theorem 1.1. The reader who is only
interested in understanding Proposition 2.12 in depth can go directly to the
short proofs of Corollary 4.12 and Proposition 4.8 which do not require
material from the rest of this section or the next one.

###  2.6. $\Psi(x,y)$ under a zero-free strip

######  Theorem 2.13.

Let $\Theta$ be the supremum of the real parts of zeros of $\zeta$ and suppose
$\Theta<1$. Fix $\varepsilon>0$. If

$x\geq y\geq(\log
x)^{\frac{1}{2}\max\left\\{3,(1-\Theta)^{-1}\right\\}+\varepsilon}$

then, as $x\to\infty$,

(2.26) $\Psi(x,y)\sim x\rho(u)Z(\sigma)G(\sigma,y)\sim
x\rho(u)Z(\sigma)G(\alpha,y).$

###### Proof.

Our starting point is Lemma 3.9. It suffices to show

$g(\alpha)-g(\sigma),\,f(\alpha)-f(\sigma)=o(1).$

By Lemma 3.7,

(2.27) $\displaystyle g(\sigma)-g(\alpha)$
$\displaystyle\asymp_{\varepsilon}(\sigma-\alpha)^{2}\log x\log y,$ (2.28)
$\displaystyle f(\alpha)-f(\sigma)+o(1)$
$\displaystyle\asymp_{\varepsilon}(\sigma-\alpha)^{2}\log x\log y.$

By Lemma 3.3,

(2.29) $(\log x\log
y)(\sigma-\alpha)\ll_{\varepsilon}\left|G^{\prime}(\alpha,y)/G(\alpha,y)\right|+1.$

By Proposition 4.6,

(2.30) $G_{1}^{\prime}(\alpha,y)/G_{1}(\alpha,y)\ll y^{\Theta-\alpha}(\log
y)^{2}.$

By Lemma 4.13,

(2.31)
$G^{\prime}_{2}(\alpha,y)/G_{2}(\alpha,y)\ll_{\varepsilon}\int_{\sqrt{y}}^{y}t^{-2\alpha}dt\ll
y^{\max\left\\{1-2\alpha,\frac{1}{2}-\alpha\right\\}}\log y.$

Since $y^{1-\alpha}\ll u\log(u+1)$ by Lemma 3.1, these estimates give the
result. ∎

Suppose we knew that $\Theta\leq 3/4$. Then, if $x\to\infty$ and $y\geq(\log
x)^{2+\varepsilon}$, Theorem 2.13 would tell us that

$\Psi(x,y)\sim x\rho(u)Z(\sigma)G(\sigma,y).$

###  2.7. Phase transition

Under RH, Theorem 2.13 implies that, as $x\to\infty$,

$\Psi(x,y)\sim x\rho(u)Z(\sigma)G(\sigma,y)$

for $x\geq y\geq(\log x)^{3/2+\varepsilon}$. The next theorem shows a
different behavior emerges once

$y\asymp(\log x)^{3/2}(\log\log x)^{-1/2}.$

######  Theorem 2.14.

Assume $RH$. Fix $\varepsilon>0$. If $(1+\varepsilon)\log x\leq y\leq(\log
x)^{2-\varepsilon}$ and $x\gg_{\varepsilon}1$ then

$\frac{\Psi(x,y)}{x\rho(u)Z(\sigma)}=G(\sigma,y)\exp\big{(}-\Theta_{\varepsilon}\big{(}\frac{(\log
x)^{3}}{y^{2}\log
y}\big{)}\big{)}=G(\alpha,y)\exp\big{(}\Theta_{\varepsilon}\big{(}\frac{(\log
x)^{3}}{y^{2}\log y}\big{)}\big{)}.$

###### Proof.

Our starting point is Proposition 2.10. In the considered range, $\alpha\leq
1/2-c_{\varepsilon}$ by (2.16). According to Lemma 3.8, RH implies in our
range that

$\log B(x,y)\ll_{\varepsilon}\frac{(\log y)^{2}}{\sqrt{y}}+\frac{\log
x}{y}=o\left(\frac{(\log x)^{3}}{y^{2}\log y}\right).$

It remains to study the differences $g(\alpha)-g(\sigma)$,
$f(\alpha)-f(\sigma)$. By Lemma 3.7,

(2.32) $\displaystyle g(\alpha)-g(\sigma)$
$\displaystyle\asymp_{\varepsilon}-(\sigma-\alpha)^{2}\log x\log y,$ (2.33)
$\displaystyle f(\alpha)-f(\sigma)$
$\displaystyle\asymp_{\varepsilon}(\sigma-\alpha)^{2}\log x\log
y+O(|\alpha-\sigma|).$

By (3.9), RH implies

$\alpha-\sigma\ll_{\varepsilon}\frac{\log(u+1)}{\sqrt{y}}+\frac{1}{\log x\log
y}+\frac{\log x}{y\log y}=o\left(\frac{(\log x)^{3}}{y^{2}\log y}\right)$

in this range, so the term $O(|\alpha-\sigma|)$ is negligible. It remains to
understand $(\sigma-\alpha)^{2}\log x\log y$. By Lemma 3.3,

$\sigma-\alpha\asymp_{\varepsilon}\big{(}\frac{G^{\prime}}{G}(\alpha,y)+O(1)\big{)}/(\log
x\log y).$

By Proposition 4.6,

(2.34) $G^{\prime}_{1}(\alpha,y)/G_{1}(\alpha,y)\ll
y^{\frac{1}{2}-\alpha}(\log y)^{2}.$

By Lemma 4.13,

(2.35)
$G_{2}^{\prime}(\alpha,y)/G_{2}(\alpha,y)\asymp_{\varepsilon}-y^{1-2\alpha}.$

Since $y^{1-\alpha}\asymp u\log(u+1)\asymp\log x$ in this range by Lemma 3.1,
these estimates show $(\sigma-\alpha)^{2}\log x\log y$ is of order of
magnitude $(\log x)^{3}/(y^{2}\log y)$. ∎

###  2.8. Pomerance’s question

Here we prove Theorem 1.2. The first part is essentially due to Saias. We
claim

(2.36) $\Lambda(x,y)\geq x\rho(u)(1+c\log(u+1)/\log y)$

holds in $(1-\varepsilon)x\geq y\geq(\log x)^{1+\varepsilon}$ for sufficiently
small $c>0$. By (2.9), (2.36) holds if $u\gg 1$. For bounded $u$ with
$(1-\varepsilon)x\geq y$, we consider the contribution of $0\leq v\leq c/\log
y$ to the integral in the right-hand side of (2.5) to get (2.36). Now observe
that the error term in Saias’ estimate, (2.7), is smaller than $\log(u+1)/\log
y$. This finishes the first part.

The second part of the theorem is just the first part of Proposition 2.12.

From now on we assume that RH holds. We have

$\rho(u)=\left(u\log u/e(1+o(1))\right)^{-u}$

as $u\to\infty$ by [dB51a], which implies $\Psi(x,y)\geq 1>x\rho(u)$ for
$y\leq e\log x(1-\varepsilon)$ and $x\gg_{\varepsilon}1$. This observation is
due to Granville [Gra08, p. 270]. So we may assume $y\geq 2\log x$. In the
range $2\log x\leq y\leq(\log x)^{2-\varepsilon}$,

$1/2-c_{\varepsilon}\geq\alpha,\sigma\geq c/\log y$

by (2.16) and Lemma 2.2. Pomerance’s inequality follows from Theorem 2.14 in
this range. Indeed, the theorem shows

$\Psi(x,y)\geq
x\rho(u)Z(\sigma)G(\alpha,y)\exp\bigg{(}c_{\varepsilon}\frac{(\log
x)^{3}}{y^{2}\log y}\bigg{)}$

for $x\gg_{\varepsilon}1$. All the terms to the right of $\rho(u)$ are $>1$.
For the last one this is obvious. For $Z(\sigma)$ this follows by
monotonicity:

$Z(\sigma)\geq Z(1/2)>Z(1)=1.$

For $G(\alpha,y)$, we use Proposition 4.6,

$\log G_{1}(\alpha,y)\ll_{\varepsilon}y^{\frac{1}{2}-\alpha}\log y$

and Corollary 4.12,

$\log G_{2}(\alpha,y)\geq c_{\varepsilon}y^{1-2\alpha}/\log y$

to find

$\log G(\alpha,y)\geq c_{\varepsilon}y^{1-2\alpha}/\log y>0.$

We now consider $x(1-\varepsilon)\geq y\geq(\log x)^{2+\varepsilon}$. By
Saias’ RH result (2.8) and (2.36),

$\frac{\Psi(x,y)}{x\rho(u)}=\frac{\Psi(x,y)}{\Lambda(x,y)}\frac{\Lambda(x,y)}{x\rho(u)}\geq\left(1+c\frac{\log(u+1)}{\log
y}\right)\left(1+O\left(\frac{\log x}{y^{1/3}}\right)\right)>1$

if $x(1-\varepsilon)\geq y\geq(\log x)^{4}$ and $x\gg_{\varepsilon}1$. If
$(\log x)^{2+\varepsilon}\leq y\leq(\log x)^{4}$, we use Theorem 2.13 and the
monotonicity of $Z$ to find

$\frac{\Psi(x,y)}{x\rho(u)}=Z(\sigma)G(\alpha,y)(1+o(1))\geq
Z(4/5)G(\alpha,y)(1+o(1))$

as $x\to\infty$. We have $Z(4/5)>1$, and $G(\alpha,y)\sim 1$ by Corollary 4.12
and (4.13), implying

$\Psi(x,y)>x\rho(u)$

if $(\log x)^{2+\varepsilon}\leq y\leq(\log x)^{4}$ and $x\gg_{\varepsilon}1$.
We now prove the last parts of the theorem, which deal with

$(\log x)^{2+\varepsilon}\geq y\geq(\log x)^{2-\varepsilon}.$

In this range, Theorem 2.13 tells us

$\Psi(x,y)\sim x\rho(u)Z(\sigma)G(\sigma,y).$

The asymptotic estimate for $\log G_{2}$ given in Corollary 4.12, and the
estimate for $\log G_{1}$ given in (4.15) yield

$\log G(\sigma,y)=\frac{1+o(1)}{2}\int_{\sqrt{y}}^{y}\frac{dt}{t^{2\sigma}\log
t}dt-\frac{1}{\log y}\sum_{|\Im\rho|\leq
T}\frac{y^{\rho-\sigma}}{\rho-\sigma}+E$

where

$E\ll y^{-\sigma}\left(\frac{y\log^{2}(yT)}{T}+\log y+\sum_{|\Im\rho|\leq
T}\left|\frac{y^{\rho}}{(\rho-\sigma)^{2}\log^{2}y}\right|\right)$

for any choice of $T\geq 2$. Here the summations are over non-trivial zeros of
$\zeta$ up to height $T$. We take $T=y$. Recall $\sum_{\rho}1/|\rho|^{2}$
converges. It follows that

$\log G(\sigma,y)=\frac{1+o(1)}{2}\int_{\sqrt{y}}^{y}\frac{dt}{t^{2\sigma}\log
t}dt-\frac{1}{\log y}\sum_{|\Im\rho|\leq
y}\frac{y^{\rho-\sigma}}{\rho}+O\left(\frac{y^{\frac{1}{2}-\sigma}}{\log
y}\right).$

We now recognize $\sum_{|\Im\rho|\leq y}y^{\rho}/\rho$ as the error in the
prime number theorem. Specifically,

(2.37) $-\sum_{|\Im\rho|\leq y}y^{\rho}/\rho=\psi(y)-y+O(\log^{2}y)$

by the truncated explicit formula [MV07, Thm. 12.5], where $\psi$ is the
Chebyshev function. Hence,

(2.38) $\log
G(\sigma,y)=\frac{1+o(1)}{2}\int_{\sqrt{y}}^{y}\frac{dt}{t^{2\sigma}\log
t}dt+\frac{\psi(y)-y}{y^{\sigma}\log
y}+O\bigg{(}\frac{y^{\frac{1}{2}-\sigma}}{\log y}\bigg{)}.$

In summary, we want

$\log Z(\sigma)+\frac{y^{\frac{1}{2}-\sigma}}{\log
y}\left(\frac{\psi(y)-y}{\sqrt{y}}+O(1)\right)+\frac{1+o(1)}{2}\int_{\sqrt{y}}^{y}\frac{dt}{t^{2\sigma}\log
t}dt+o(1)$

to be non-negative. We show that

(2.39) $\liminf_{y\to\infty}\frac{\psi(y)-y}{\sqrt{y}\log y}>L$

is a sufficient condition, if $x\gg 1$. We consider three separate cases. If
$(2\sigma-1)\log y$ tends to $\infty$ then

$\int_{\sqrt{y}}^{y}\frac{dt}{t^{2\sigma}\log
t}\sim\frac{y^{\frac{1}{2}-\sigma}}{(\sigma-1/2)\log y}$

by Lemma 4.10. Thus,

$\log\left(\frac{\Psi(x,y)}{x\rho(u)}\right)=\log
Z(\sigma)+\frac{y^{\frac{1}{2}-\sigma}}{\log
y}\frac{\psi(y)-y}{\sqrt{y}}+o(1).$

If $x\gg 1$, (2.39) implies this is positive. If $(2\sigma-1)\log y$ tends to
$-\infty$, a similar argument works, using Lemma 4.10 again. The most delicate
range is $(2\sigma-1)\log y=O(1)$. Here $Z(\sigma)\sim Z(1/2)$. Set

$\sigma=\frac{1}{2}+\frac{v}{\log y}$

so that $v$ is bounded. We express $\log(\Psi(x,y)/(x\rho(u))$ as a function
of $y$ and $v$:

$\log\left(\frac{\Psi(x,y)}{x\rho(u)}\right)=\log
Z(1/2)+e^{-v}\frac{\psi(y)-y}{\sqrt{y}\log
y}+\frac{1}{2}\int_{v}^{2v}\frac{e^{-r}}{r}dr+o(1).$

If (2.39) holds, we find by the definition of $L$ that the last expression is
$\geq c$ for some positive $c$, if $y$ is sufficiently large. If instead

(2.40) $\liminf_{y\to\infty}\frac{\psi(y)-y}{\sqrt{y}\log y}<L$

then, by definition, we can find $v\in\mathbb{R}$ such that if
$\sigma=1/2+v/\log y$ then

$\log\left(\frac{\Psi(x,y)}{x\rho(u)}\right)<-c$

for some $c>0$, if $y$ is sufficiently large.

We record (2.38) below.

######  Theorem 2.15.

Assume RH. In the range $x\geq y\geq(\log x)^{1+\varepsilon}$ we have

(2.41) $\log
G(\sigma,y)=\frac{1+o(1)}{2}\int_{\sqrt{y}}^{y}\frac{dt}{t^{2\sigma}\log
t}dt+\frac{\psi(y)-y}{y^{\sigma}\log
y}+O\left(\frac{y^{\frac{1}{2}-\sigma}}{\log y}\right)$

as $x\to\infty$ where $\sigma=1-\xi(u)/\log y$. In particular,

$\psi(x,y)\sim x\rho(u)Z(\sigma)\sim\Lambda(x,y)$

holds when $y/(\log x\log\log x)^{2}\to\infty$, and if

(2.42) $\psi(y)-y=o(\sqrt{y}\log y)$

is true then

$\psi(x,y)\sim x\rho(u)Z(\sigma)\sim\Lambda(x,y)$

holds when $y/(\log x)^{2}\to\infty$ and this range is optimal.

###### Proof.

By (2.9), $x\rho(u)Z(\sigma)\sim\Lambda(x,y)$. By Theorem 2.13, in the range
considered, $\psi(x,y)\sim x\rho(u)Z(\sigma)$ is equivalent to
$G(\sigma,y)\sim 1$. It remains to understand when $\log G(\sigma,y)=o(1)$ and
we use (2.41) to do so. By definition of $\sigma$, $y^{1-\sigma}\asymp
u\log(u+1)$, so

$\frac{y^{\frac{1}{2}-\sigma}}{\log y}\asymp\frac{u\log(u+1)}{y\log y}$

is $o(1)$ when $y=O((\log x)^{2})$. Von Koch’s estimate (1.8) implies

$\frac{\psi(y)-y}{y^{\sigma}\log y}\asymp\frac{u\log(u+1)}{y\log
y}(\psi(y)-y)=O\left(\frac{\log x\log(u+1)}{\sqrt{y}}\right)$

is $o(1)$ if $y/(\log x\log\log x)^{2}\to\infty$, and (2.42) implies

$\frac{\psi(y)-y}{y^{\sigma}\log y}=o(1)$

if $y/(\log x)^{2}\to\infty$. Using $y^{1-\sigma}\asymp u\log(u+1)$ and
Corollary 4.12, we see that if $y/(\log x)^{2}=\Theta(1)$ then
$\sigma=1/2+O(1/\log y)$ and the integral in (2.41) is $\Theta(1)$, while if
$y/(\log x)^{2}$ tends to $0$ then $(\sigma-1/2)\log y\to\infty$ integral goes
to $0$. ∎

###  2.9. Inequality

In [Ten15, Thm. III.5.21], Hildebrand and Tenenbaum showed

(2.43)
$\log\left(\frac{\Psi(x,y)}{x}\right)=\log\rho(u)\left(1+O_{\varepsilon}\left(\exp(-(\log
y)^{\frac{3}{5}-\varepsilon})\right)\right)+O_{\varepsilon}\left(\frac{\log(u+1)}{\log
y}\right)$

holds for $y\geq(\log x)^{1+\varepsilon}$. We offer an improvement in terms of
range and error, which also shows (2.43) does not hold for $y\asymp\log x$.

######  Theorem 2.16.

Fix $\varepsilon>0$. Uniformly for $x\geq y\geq(1+\varepsilon)\log x$,

(2.44)
$\log\left(\frac{\Psi(x,y)}{xZ(\sigma)}\right)=\log\rho(u)\big{(}1+O_{\varepsilon}\big{(}L(y)^{-c}\big{)}\big{)}+O_{\varepsilon}\left(\frac{1}{\alpha\log
x}\right)+\log G_{2}(\sigma,y).$

If $y\geq\log x\cdot L(\log x)^{C}$ the term $\log G_{2}(\sigma,y)$ is
absorbed in the existing errors. Otherwise, it contributes

(2.45) $\displaystyle\log G_{2}(\sigma,y)$
$\displaystyle\asymp_{\varepsilon}(\log x)^{2}/(y\log y).$

###### Proof.

Taking logs in Lemma 3.9 we see

(2.46) $\displaystyle\log\big{(}\frac{\Psi(x,y)}{x\rho(u)Z(\sigma)}\big{)}$
$\displaystyle=\log G(\sigma,y)+g(\alpha)-g(\sigma)+\log
B(x,y)+O\left(\frac{1}{\alpha\log x}\right).$

Recall $G=G_{1}G_{2}$. By Proposition 4.6,

$\log G_{1}(\sigma,y)\ll y^{1-\sigma}L(y)^{-c}.$

We have $y^{1-\sigma}\asymp u\log(u+1)$ which implies that

$\log G_{1}(\sigma,y)\ll(-\log\rho(u))L(y)^{-c}.$

By Lemma 3.8, $\log B(x,y)$ can be absorbed in the error term $O(1/(\alpha\log
x))$ and in the bound for $\log G_{1}(\sigma,y)$. We have the estimate
$g(\sigma)-g(\alpha)\ll_{\varepsilon}(\sigma-\alpha)^{2}\log x\log y$ by Lemma
3.7 and the size of $\alpha-\sigma$ is studied in Lemma 3.3. We see that
$g(\sigma)-g(\alpha)$ is also absorbed in the existing error terms. The term
$\log G_{2}(\sigma,y)$ is studied in Corollary 4.12, and the error term there
is simplified using the definition of $\sigma$. ∎

De Bruijn found the asymptotics for $\log\Psi(x,y)$ uniformly for $x\geq y\geq
2$ [dB66]. In the range $y\geq(1+\varepsilon)\log x$, Theorem 2.16 strengthens
his result when combined with the asymptotics for $\log G_{2}(\sigma,y)$ given
in Corollary 4.12 and Lemma 4.14.

##  3\. Study of the main formula

Recall $\alpha$ and $\sigma$ were defined in Theorem 2.7 and Corollary 2.9.

######  Lemma 3.1.

We have $y^{1-\sigma}\asymp u\log(u+1)$ uniformly for $x\geq y\geq 2$.
Uniformly for $x\geq y>\log x$ we have $\sigma-\alpha=O(1/\log y)$ and so
$y^{1-\alpha}\asymp u\log(u+1)$ as well.

###### Proof.

The first part follows from the definitions of $\sigma$ and $\xi$ and from
Lemma 2.1. The second part follows from [HT86, Eq. (3.5)]. ∎

######  Lemma 3.2.

Fix $2\leq k\leq 5$. Let $I$ be the interval with endpoints $\alpha$ and
$\sigma=1-\xi(u)/\log y$. Fix $\varepsilon>0$. Suppose $x\geq
y\geq(1+\varepsilon)\log x$ and $x\gg_{\varepsilon}1$. Uniformly for $t\in I$
we have

(3.1) $\displaystyle f^{(k)}(t),\,g^{(k)}(t)$
$\displaystyle\asymp_{\varepsilon}(-1)^{k}\log x(\log y)^{k-1}.$

Additionally, $f^{(2)}(t)$ and $g^{(2)}(t)$ are positive for $t\in(0,1]$.

###### Proof.

As shown in [HT86, Lem. 4],

$g^{(k)}(t)=(-1)^{k}\sum_{p\leq y}(\log p)(p^{t}-1)^{-k}Q_{k-1}(p^{t}\log p)$

for a polynomial $Q_{k-1}$ of degree $k-1$ and non-negative coefficients, so
$(-1)^{k}g^{(k)}(t)$ is positive and monotone for $t>0$. Moreover, by the same
lemma,

$g^{(k)}(\alpha)\asymp(-1)^{k}\log x(\log y)^{k-1}$

uniformly for $x\geq y\geq\log x$. It remains to show $g^{(k)}(\sigma)$ is
also of order $(-1)^{k}\log x(\log y)^{k-1}$. Since
$\sigma\gg_{\varepsilon}1/\log y$, the same lemma shows, as is, that

(3.2) $(\log
y)^{k-1}\frac{y^{1-\sigma}-1}{1-\sigma}\ll_{\varepsilon}(-1)^{k}g^{(k)}(\sigma)\ll_{\varepsilon}(\log
y)^{k-1}\sum_{p\leq y}\frac{\log p}{p^{\sigma}-1}.$

The lower bound is, by definition of $\sigma$, $\log x(\log y)^{k-1}$. The sum
in the upper bound is estimated in [Ten15, p. 552] as

$\sum_{p\leq y}\frac{\log
p}{p^{\sigma}-1}\ll\frac{1}{1-y^{-\sigma}}\int_{1}^{y}t^{-\sigma}dt+O(1)=\frac{\log
x}{1-y^{-\sigma}}+O(1)\ll_{\varepsilon}\log x$

by definition of $\sigma$ and Lemma 2.2. This finishes the bounds needed for
$g$. For $f$,

(3.3) $\displaystyle f^{(k)}(t)$
$\displaystyle=(\log(\zeta(t)(t-1)))^{(k)}+(-\log y)^{k}I^{(k)}((1-t)\log y).$

The function $I^{(k)}(r)=\sum_{i\geq 0}r^{i}/(i!(i+k))$ is monotone increasing
for $r\geq 0$ and $I^{(k)}(0)=1/k$. The expression
$(\log(\zeta(t)(t-1)))^{(k)}$ is $O(1)$ for $t\in[0,1]$, and a computer
calculation shows that for $k=2$ it is in $[-0.4,-0.1]$. This already shows
$f^{(2)}(t)>0$. To obtain the order of magnitude for $(-1)^{k}f^{(k)}(t)$,
observe $I^{(k)}(v)\asymp_{k}e^{v}/(v+1)$ and $e^{v}/(v+1)\asymp u$ as long as
$v=\xi(u)+O(1)$, so we want to show $(1-t)\log y=\xi(u)+O(1)$ for $t=\sigma$
and $t=\alpha$. For $t=\sigma$ it is trivial while for $t=\alpha$ it follows
from Lemma 3.1. ∎

We use Lemma 3.2 to study $\sigma-\alpha$, unconditionally and conditionally.
Our estimates will benefit from introducing

(3.4)
$H(y,\alpha):=\frac{y^{1-2\alpha}-y^{\frac{1}{2}-\alpha}}{(1-2\alpha)\log
y}>0$

which at $\alpha=1/2$ is defined as the limit at $1/2$. For $\alpha\leq
1/2-\varepsilon$,

$H(y,\alpha)\ll_{\varepsilon}y^{1-2\alpha}/\log y\asymp(\log x)^{2}/(y\log y)$

by Lemma 3.1.

######  Lemma 3.3.

Fix $\varepsilon>0$. Suppose $x\geq y\geq(1+\varepsilon)\log x$ and
$x\gg_{\varepsilon}1$. We have

(3.5) $\displaystyle\sigma-\alpha$
$\displaystyle\asymp_{\varepsilon}\frac{\frac{G^{\prime}}{G}(\alpha,y)+C_{\sigma}}{\log
x\log y},$ (3.6) $\displaystyle C_{\sigma}$
$\displaystyle:=(\log(\zeta(\sigma)(\sigma-1)))^{\prime}=\Theta(1).$

Unconditionally,

(3.7) $\displaystyle\sigma-\alpha$
$\displaystyle\asymp_{\varepsilon}\frac{H(y,\alpha)}{\log
x}+O\left(\frac{1}{\log x\log y}+L(y)^{-c}\right)$ (3.8)
$\displaystyle\ll_{\varepsilon}\frac{\log x}{y\log y}+\frac{1}{\log x\log
y}+L(y)^{-c}.$

Under RH,

(3.9) $\sigma-\alpha\ll_{\varepsilon}\frac{\log(u+1)}{\sqrt{y}}+\frac{1}{\log
x\log y}+\frac{H(y,\alpha)}{\log x}.$

###### Proof.

We have

(3.10) $-\frac{\zeta^{\prime}(\alpha,y)}{\zeta(\alpha,y)}=\log
x,\qquad-\frac{F^{\prime}(\sigma,y)}{F(\sigma,y)}=\log x-C_{\sigma}.$

Writing $\zeta(s,y)$ as $F(s,y)$ times $G(s,y)$ we find that

(3.11)
$-\frac{F^{\prime}(\alpha,y)}{F(\alpha,y)}+\frac{F^{\prime}(\sigma,y)}{F(\sigma,y)}=\log
x+\frac{G^{\prime}(\alpha,y)}{G(\alpha,y)}-(\log
x-C_{\sigma})=\frac{G^{\prime}(\alpha,y)}{G(\alpha,y)}+C_{\sigma}.$

By the mean value theorem, for some $t$ between $\alpha$ and $\sigma$ we have

(3.12)
$-\frac{F^{\prime}(\alpha,y)}{F(\alpha,y)}+\frac{F^{\prime}(\sigma,y)}{F(\sigma,y)}=-(\alpha-\sigma)\left(\frac{F^{\prime}}{F}\right)^{\prime}(t).$

We have $(F^{\prime}/F)^{\prime}=f^{\prime\prime}$, and by Lemma 3.2,
$f^{\prime\prime}(t)\asymp_{\varepsilon}\log x\log y$. To conclude (3.5), we
compare (3.11) and (3.12). We now show (3.7). By (2.16) and Lemma 2.2,
$\sigma,\alpha\gg_{\varepsilon}1/\log y$. By (4.14) and Lemma 4.13,

(3.13) $\displaystyle\frac{G^{\prime}_{1}}{G_{1}}(\alpha,y)$
$\displaystyle\ll_{\varepsilon}y^{1-\alpha}L(y)^{-c},$ (3.14)
$\displaystyle\frac{G^{\prime}_{2}}{G_{2}}(\alpha,y)$
$\displaystyle\asymp_{\varepsilon}-\frac{y^{1-2\alpha}-y^{\frac{1}{2}-\alpha}}{1-2\alpha}.$

Hence, by (3.5),

$\sigma-\alpha\asymp_{\varepsilon}-\frac{\frac{y^{1-2\alpha}-y^{\frac{1}{2}-\alpha}}{1-2\alpha}+O\big{(}1+y^{1-\alpha}L(y)^{-c}\big{)}}{\log
x\log y}.$

By Lemma 3.1 this implies (3.7). If RH holds, we use
$(G^{\prime}_{1}/G_{1})(\alpha,y)\ll_{\varepsilon}y^{1/2-\alpha}(\log y)^{2}$
which is shown in (4.13). ∎

We use Lemma 3.3 to improve on Lemma 3.2.

######  Lemma 3.4.

Fix $\varepsilon>0$. Suppose $x\geq y\geq(1+\varepsilon)\log x$ and
$x\gg_{\varepsilon}1$. Let $I$ be the interval with endpoints
$\sigma=1-\xi(u)/\log y$ and $\alpha$. Then, uniformly for $t\in I$,

(3.15) $\displaystyle g^{\prime\prime}(t)$
$\displaystyle=g^{\prime\prime}(\alpha)\left(1+O_{\varepsilon}\left(L(y)^{-c}+\frac{1}{\log
x}+\frac{\log x}{y}\right)\right),$ (3.16) $\displaystyle f^{\prime\prime}(t)$
$\displaystyle=f^{\prime\prime}(\sigma)\left(1+O_{\varepsilon}\left(L(y)^{-c}+\frac{1}{\log
x}+\frac{\log x}{y}\right)\right).$

###### Proof.

For any $t\in I$,

$g^{\prime\prime}(t)=g^{\prime\prime}(\alpha)+(t-\alpha)g^{(3)}(t_{2})$

for some $t_{2}\in I$. The estimates for $g^{\prime\prime}$ and $g^{(3)}$ in
Lemma 3.2 imply

$g^{\prime\prime}(t)=g^{\prime\prime}(\alpha)\left(1+O_{\varepsilon}(|\alpha-\sigma|\log
y)\right).$

Plugging (3.7) here concludes the estimate for $g^{\prime\prime}$. As for
$f^{\prime\prime}$, we write

$f^{\prime\prime}(t)=f^{\prime\prime}(\sigma)+(t-\sigma)f^{(3)}(t_{3})$

for some $t_{3}\in I$ and argue as before. ∎

We use Lemma 3.4 to improve on Lemma 3.3.

######  Corollary 3.5.

Fix $\varepsilon>0$. Suppose $x\geq y\geq(1+\varepsilon)\log x$ and
$x\gg_{\varepsilon}1$. We have

(3.17)
$\sigma-\alpha=\frac{\frac{G^{\prime}}{G}(\alpha,y)+C_{\sigma}}{f^{\prime\prime}(\sigma)}\big{(}1+O_{\varepsilon}\big{(}L(y)^{-c}+\frac{1}{\log
x}+\frac{\log x}{y}\big{)}\big{)}$

where $C_{\sigma}$ is as defined in (3.6). We have

(3.18) $\displaystyle\frac{G^{\prime}}{G}(\alpha,y)$
$\displaystyle=-H(y,\alpha)\log
y\left(1+O_{\varepsilon}\left(L(y)^{-c}+y^{-\alpha}\right)\right)$ (3.19)
$\displaystyle\qquad+O_{\varepsilon}\left(u\log(u+1)L(y)^{-c}\right).$

Under RH, the right-most error in (3.18) can be replaced with
$y^{1/2-\alpha}(\log y)^{2}$.

###### Proof.

The equality (3.17) follows by repeating the proof of Lemma 3.3 and inputting
the bounds for $f^{\prime\prime}$ given in Lemma 3.4. The estimates for
$G^{\prime}/G$ follow from (4.13) and Lemma 4.13. ∎

######  Lemma 3.6.

Fix $\varepsilon>0$. Suppose $x\geq y\geq(1+\varepsilon)\log x$ and
$x\gg_{\varepsilon}1$. We have

(3.20) $g^{(k)}(\alpha)-f^{(k)}(\sigma)\ll_{\varepsilon}|(\log
G)^{(k)}(\alpha,y)|+|\alpha-\sigma|\log x(\log y)^{k}.$

for $2\leq k\leq 4$. In particular,

(3.21) $\frac{g^{(k)}(\alpha)-f^{(k)}(\sigma)}{\log x(\log
y)^{k-1}}\ll_{\varepsilon}L(y)^{-c}+\frac{1}{\log x}+\frac{\log x}{y}.$

Under RH,

(3.22) $\frac{g^{(k)}(\alpha)-f^{(k)}(\sigma)}{\log x(\log
y)^{k-1}}\ll_{\varepsilon}\frac{\log y\log(u+1)}{\sqrt{y}}+\frac{1}{\log
x}+\frac{H(y,\alpha)}{u}.$

###### Proof.

We write

$g^{(k)}(\alpha)=(f^{(k)}(\alpha)-f^{(k)}(\sigma))+f^{(k)}(\sigma)+\left(\log
G\right)^{(k)}(\alpha,y)$

and replace $f^{(k)}(\alpha)-f^{(k)}(\sigma)$ by $(\alpha-\sigma)f^{(k+1)}(t)$
for $t$ between $\alpha$ and $\sigma$. Lemma 3.2 bounds $f^{(k+1)}(t)$ by
$\ll_{\varepsilon}\log x(\log y)^{k}$. This yields (3.20). To deduce (3.21)
from (3.20) we estimate $\alpha-\sigma$ using Lemma 3.3 and $(\log G)^{(k)}$
using (4.14), (4.13) and Lemma 4.13. We simplify the resulting bounds using
$y^{1-\alpha}\ll u\log(u+1)$ from Lemma 3.1. ∎

Applying Lemma 3.6 with $k=2$, we find

$g^{\prime\prime}(\alpha)=\log^{2}yI^{\prime\prime}(\xi)\big{(}1+O_{\varepsilon}\big{(}L(y)^{-c}+\frac{1}{\log
x}+\frac{\log x}{y}\big{)}\big{)}$

uniformly for $x\geq y\geq(1+\varepsilon)\log x$, which improves on (2.15) if
$y\gg\log x\log\log x$. We now prove asymptotics for $B(x,y)$,
$g(\sigma)-g(\alpha)$ and $f(\sigma)-f(\alpha)$.

######  Lemma 3.7.

Fix $\varepsilon>0$. If $x\geq y\geq(1+\varepsilon)\log x$ and
$x\gg_{\varepsilon}1$ then

(3.23) $\displaystyle g(\sigma)-g(\alpha)$
$\displaystyle\asymp_{\varepsilon}(\sigma-\alpha)^{2}\log x\log y,$ (3.24)
$\displaystyle f(\alpha)-f(\sigma)+o(1)$
$\displaystyle\asymp_{\varepsilon}(\sigma-\alpha)^{2}\log x\log y+o(1)$

where the $o(1)$ term is
$(\log(\zeta(\sigma)(\sigma-1)))^{\prime}(\alpha-\sigma)$. More accurately,

(3.25) $\displaystyle g(\sigma)-g(\alpha)$
$\displaystyle=g^{\prime\prime}(\alpha)(\sigma-\alpha)^{2}\left(1+E_{1}\right)/2,$
(3.26) $\displaystyle f(\alpha)-f(\sigma)$
$\displaystyle=(\log((\zeta(\sigma)(\sigma-1)))^{\prime}(\alpha-\sigma)+f^{\prime\prime}(\sigma)(\alpha-\sigma)^{2}\left(1+E_{2}\right)/2,$

where

(3.27) $E_{1},\,E_{2}\ll L(y)^{-c}+\frac{1}{\log x}+\frac{\log x}{y}$

and, under RH,

(3.28) $E_{1},\,E_{2}\ll\frac{\log y\log(u+1)}{\sqrt{y}}+\frac{1}{\log
x}+\frac{H(y,\alpha)}{u}$

where $H(y,\alpha)$ was defined (3.4).

###### Proof.

Approximating $g$ at $\alpha$ using a linear Taylor polynomial shows

$g(\sigma)=g(\alpha)+g^{\prime}(\alpha)(\sigma-\alpha)+g^{\prime\prime}(t)(\sigma-\alpha)^{2}/2$

for some $t$ between $\sigma$ and $\alpha$. By definition,
$g^{\prime}(\alpha)=0$, and $g^{\prime\prime}(t)\asymp_{\varepsilon}\log x\log
y$ is shown in Lemma 3.2. For a more accurate result, we use a quadratic
approximation:

$g(\sigma)-g(\alpha)=\frac{g^{\prime\prime}(\alpha)(\sigma-\alpha)^{2}}{2}\bigg{(}1+O\bigg{(}\frac{g^{(3)}(t)|\sigma-\alpha|}{\log
x\log y}\bigg{)}\bigg{)}$

for some $t$ between $\sigma$ and $\alpha$. By Lemma 3.2,
$g^{(3)}(t)\ll_{\varepsilon}\log x(\log y)^{2}$ and we can bound
$|\sigma-\alpha|$ using the estimates in (3.7) and (3.9).

The same argument works for $f(\alpha)-f(\sigma)$ by Taylor-expanding $f$ at
$\sigma$ and using
$f^{\prime}(\sigma)=(\log(\zeta(\sigma)(\sigma-1))^{\prime}$. Since
$\alpha-\sigma$ goes to $0$, the term $(\alpha-\sigma)f^{\prime}(\sigma)$
contributes $o(1)$ to $f(\alpha)-f(\sigma)$. ∎

Let

$h(t):=\frac{\log t}{\sqrt{1+t^{-1}}\log(1+t)}.$

######  Lemma 3.8.

If $y/\log x\to\infty$ then $B(x,y)\sim 1$. More precisely, if $x\geq
y\geq(1+\varepsilon)\log x$ and $x\gg_{\varepsilon}1$ then

(3.29) $B(x,y)=1+O_{\varepsilon}\left(L(y)^{-c}+\frac{1}{\log x}+\frac{\log
x}{y}\right).$

Recall $H(y,\alpha)$ was defined (3.4). Under RH,

(3.30) $B(x,y)=1+O_{\varepsilon}\left(\frac{\log
y\log(u+1)}{\sqrt{y}}+\frac{1}{\log x}+\frac{H(y,\alpha)}{u}\right).$

If $y/\log x\to t\in(1,\infty)$ then $B(x,y)\sim h(t)$. More precisely, if
$x\geq y\geq(1+\varepsilon)\log x$ and $x\gg_{\varepsilon}1$ then

$B(x,y)=h\left(\frac{y}{\log
x}\right)\left(1+O_{\varepsilon}\left(\frac{1}{\log(1+u)}+\frac{\log\log
y}{\log y}\right)\right).$

###### Proof.

We first study $\sigma/\alpha$. Writing this ratio as
$1+(\sigma-\alpha)/\alpha$, inputting the estimate for $\alpha$ from (2.16)
the estimates of $\sigma-\alpha$ given in Lemma 3.3 yield

(3.31) $\frac{\sigma}{\alpha}=1+O_{\varepsilon}\left(L(y)^{-c}+\frac{1}{\log
x}+\frac{\log x}{y}\right)$

unconditionally and

(3.32) $\frac{\sigma}{\alpha}=1+O_{\varepsilon}\left(\frac{\log
y\log(u+1)}{\sqrt{y}}+\frac{1}{\log x}+\frac{H(y,\alpha)}{u}\right)$

under RH. We may also simply divide the expression for $\sigma$ and $\alpha$
given in the (2.16) and in Lemma 2.2 to deduce an alternative estimate:

$\frac{\sigma}{\alpha}=\frac{\log\left(\frac{y}{\log
x}\right)}{\log\left(1+\frac{y}{\log
x}\right)}\left(1+O_{\varepsilon}\left(\frac{\log\log y}{\log
y}\right)\right)$

when $y\geq(1+\varepsilon)\log x$. We turn to $(I^{\prime\prime}(\xi)(\log
y)^{2})/\phi_{2}(\alpha,y)$. This ratio can also be written as

(3.33) $\frac{I^{\prime\prime}(\xi)(\log
y)^{2}}{\phi_{2}(\alpha,y)}=\frac{f^{\prime\prime}(\sigma)+O(1)}{g^{\prime\prime}(\alpha)}=1+\frac{f^{\prime\prime}(\sigma)-g^{\prime\prime}(\alpha)+O(1)}{g^{\prime\prime}(\alpha)}.$

The denominator is $\asymp_{\varepsilon}\log x\log y$ according to Lemma 3.2,
and the numerator is estimated in Lemma 3.6, giving

(3.34) $\frac{I^{\prime\prime}(\xi)(\log
y)^{2}}{\phi_{2}(\alpha,y)}=1+O_{\varepsilon}\left(L(y)^{-c}+\frac{1}{\log
x}+\frac{\log x}{y}\right)$

unconditionally and

(3.35) $\frac{I^{\prime\prime}(\xi)(\log
y)^{2}}{\phi_{2}(\alpha,y)}=1+O_{\varepsilon}\left(\frac{\log
y\log(u+1)}{\sqrt{y}}+\frac{1}{\log x}+\frac{H(y,\alpha)}{u}\right)$

under RH. We can get an alternative estimate for $(I^{\prime\prime}(\xi)(\log
y)^{2})/\phi_{2}(\alpha,y)$ using (2.15) for the denominator and

$I^{\prime\prime}(\xi)=\xi^{\prime}(u)^{-1}=u\left(1+O\left((\log(1+u))^{-1}\right)\right)$

for the numerator, see Lemmas 2.4 and 2.1. Hence

(3.36) $\frac{I^{\prime\prime}(\xi)(\log
y)^{2}}{\phi_{2}(\alpha,y)}=\bigg{(}1+\frac{\log
x}{y}\bigg{)}^{-1}\left(1+O\left(\log(1+u)^{-1}\right)\right).$

Combining the estimates for $\sigma/\alpha$ and $(I^{\prime\prime}(\xi)(\log
y)^{2})/\phi_{2}(\alpha,y)$ finishes the proof. ∎

######  Lemma 3.9 (Sharp formula).

Suppose $y>1+\log x$. The error term in Proposition 2.10 can be taken to be
$O(1/(\alpha\log x))$.

###### Proof.

The range $2\log x\geq y\geq\log x$ is already in Proposition 2.10 because
$\alpha\asymp 1/\log y$ if $y\asymp\log x$. We assume $y>2\log x$ from now on.
Next we consider $\log y>\sqrt{\log x}$. By Lemmas 3.7 and 3.3,

$g(\alpha)-g(\sigma),f(\alpha)-f(\sigma)\ll 1/\log x.$

By (4.14) and Corollary 4.12,

$\log G(\sigma,y),\log G(\alpha,y)\ll 1/\log x.$

We conclude by appealing to (2.9). If $\log y\leq\sqrt{\log x}$ and $y\geq
2\log x$, we make use of the Main Theorem of Saha, Sankaranarayanan and Suzuki
[SSS20], which in the current range gives

$\Psi(x,y)=\frac{x^{\alpha}\zeta(\alpha,y)}{\alpha\sqrt{2\pi\phi_{2}(\alpha,y)}}\left(1+\frac{g^{(4)}(\alpha)}{8g^{(2)}(\alpha)^{2}}-\frac{5g^{(3)}(\alpha)^{2}}{24g^{(2)}(\alpha)^{3}}+O\left(\frac{1}{\alpha\log
x}\right)\right),$

which strengthens Theorem 2.7. We also use Smida’s result [Smi91, Thm. 1]

$\rho(u)=\frac{e^{\gamma-u\xi+I(\xi)}}{\sqrt{2\pi
I^{\prime\prime}(\xi(u))}}\left(1+\frac{\tilde{f}^{(4)}(\sigma)}{8\tilde{f}^{(2)}(\sigma)^{2}}-\frac{5\tilde{f}^{(3)}(\sigma)^{2}}{24\tilde{f}^{(2)}(\sigma)^{3}}+O(u^{-2})\right)$

which strengthens Theorem 2.8. Here $\tilde{f}(t)$ is as in Remark 2, so
$\tilde{f}^{(k)}(t)-f^{(k)}(t)=O(1)$ for $k=2,3,4$. We divide these two
estimates to get the formulas in Proposition 2.10, with the term $1+O(1/u)$
replaced with

$1+\frac{1}{8}\left(\frac{g^{(4)}(\alpha)}{g^{(2)}(\alpha)^{2}}-\frac{f^{(4)}(\sigma)}{f^{(2)}(\sigma)^{2}}\right)-\frac{5}{24}\left(\frac{g^{(3)}(\alpha)^{2}}{g^{(2)}(\alpha)^{3}}-\frac{f^{(3)}(\sigma)^{2}}{f^{(2)}(\sigma)^{3}}\right)+O\left(\frac{1}{\alpha\log
x}\right).$

This is estimated in Lemma 3.6. ∎

##  4\. Study of $G$

###  4.1. Formulas and bounds for $G_{1}$

Given $x>0$, $s\in\mathbb{C}$, let

$S_{1}(x,s):={\sum_{n\leq x}}^{\prime}\Lambda(n)/n^{s},$

where the prime on the summation indicates that if $x$ is a prime power, the
last term of the sum should be multiplied by $1/2$. Landau [Lan11, p. 353]
established an explicit formula for $S_{1}(x,s)$ in terms of zeros of $\zeta$.
We need a truncated version of it, which we give below and is surely known to
experts. It can be proved e.g. by adapting [MV07, Thm. 12.5] and such a proof
is given in the appendix. Throughout, $\langle x\rangle$ is the distance of
$x$ to the nearest prime power not equal to $x$.

######  Lemma 4.1.

Suppose $\Re s\geq 0$. Uniformly for $x\geq 4$ and $T\geq 2+3|\Im s|$ we have

(4.1)
$S_{1}(x,s)=\frac{x^{1-s}}{1-s}-\frac{\zeta^{\prime}(s)}{\zeta(s)}-\sum_{\begin{subarray}{c}\rho\\\
|\Im(\rho+s)|\leq
T\end{subarray}}\frac{x^{\rho-s}}{\rho-s}+\sum_{k=1}^{\infty}\frac{x^{-2k-s}}{2k+s}+R_{1}(x,T,s)$

where the sum is over non-trivial zeros of $\zeta$ and

(4.2) $R_{1}(x,T,s)\ll(\log x)(x-1)^{-\Re s}\min\left\\{1,\frac{x}{T\langle
x\rangle}\right\\}+\frac{\log^{2}(xT)}{T}\big{(}2^{\Re s}x^{1-\Re
s}+\frac{2^{-\Re s}}{\log x}\big{)}.$

If $s=1$, the ‘main term’ $x^{1-s}/(1-s)-\zeta^{\prime}(s)/\zeta(s)$ should be
interpreted as its limit at $s=1$.

Let

$S_{2}(x,s):={\sum_{n\leq x}}^{\prime}\Lambda(n)/(n^{s}\log n).$

######  Corollary 4.2.

Let $s\in[0,1]$. Uniformly for $x\geq 4$ and $T\geq 2$ we have

(4.3) $\displaystyle S_{2}(x,s)$ $\displaystyle=I((1-s)\log x)+\gamma+\log\log
x+\log(\zeta(s)(s-1))$ (4.4)
$\displaystyle\quad-\sum_{\begin{subarray}{c}\rho\\\ |\Im\rho|\leq
T\end{subarray}}\int_{0}^{\infty}\frac{x^{\rho-s-t}}{\rho-
s-t}dt+\sum_{k=1}^{\infty}\int_{0}^{\infty}\frac{x^{-2k-s-t}}{2k+s+t}dt+R_{2}(x,T,s),$
(4.5) $\displaystyle R_{2}(x,T,s)$ $\displaystyle\ll
x^{-s}\min\left\\{1,\frac{x}{T\langle
x\rangle}\right\\}+\frac{\log^{2}(xT)}{T\log x}x^{1-s}.$

###### Proof.

The starting observation for the proof, as in similar results [Sai89, Prop.
1], is

$S_{2}(x,s)=\int_{0}^{\infty}{\sum_{n\leq
x}}^{\prime}\frac{\Lambda(n)}{n^{s+t}}dt=\int_{0}^{\infty}S_{1}(x,s+t)dt.$

We integrate both sides of (4.1) along $\\{s+t:t\geq 0\\}$. We may interchange
sum and integral because the sum over $\rho$ is finite, while the integral of
the $k$-sum converges absolutely. It remains to show

$I((1-s)\log x)+\gamma+\log(\log
x\zeta(s)(s-1))=\int_{0}^{\infty}\bigg{(}\frac{x^{1-s-t}}{1-s-t}-\frac{\zeta^{\prime}(s+t)}{\zeta(s+t)}\bigg{)}dt.$

If $s\neq 1$, this is Lemma III.5.9 of [Ten15] with $(s-1)\log x$ in place of
$s$. If $s=1$, this follows by a continuity argument from the $s\neq 1$ case.
∎

The following are direct consequences of the definitions, Lemma 4.1 and
Corollary 4.2.

######  Lemma 4.3.

Let $s\in[0,1]$. Uniformly for $x\geq 4$ and $T\geq 2$ we have, for $R_{2}$
estimated in (4.5),

(4.6) $\displaystyle\log G_{1}(s,x)$
$\displaystyle=\mathbf{1}_{x\in\mathbb{N},\,\Lambda(x)\neq
0}\frac{\Lambda(x)}{2x^{s}\log x}-\sum_{\begin{subarray}{c}\rho\\\
|\Im\rho|\leq T\end{subarray}}\int_{0}^{\infty}\frac{x^{\rho-s-t}}{\rho-
s-t}dt$ (4.7)
$\displaystyle\quad+\sum_{k=1}^{\infty}\int_{0}^{\infty}\frac{x^{-2k-s-t}}{2k+s+t}dt+R_{2}(x,T,s).$

Let $s\in\mathbb{C}$ with $\Re s\geq 0$. Uniformly for $x\geq 4$ and $T\geq
2+3|\Im s|$ we have

(4.8) $\displaystyle-\frac{G^{\prime}_{1}(s,x)}{G_{1}(s,x)}$
$\displaystyle=\mathbf{1}_{x\in\mathbb{N},\,\Lambda(x)\neq
0}\frac{\Lambda(x)}{2x^{s}}-\sum_{\begin{subarray}{c}\rho\\\ |\Im(\rho+s)|\leq
T\end{subarray}}\frac{x^{\rho-s}}{\rho-s}+\sum_{k=1}^{\infty}\frac{x^{-2k-s}}{2k+s}+R_{1}(x,T,s)$

for $R_{1}$ estimated in (4.5).

Applying Cauchy’s integral formula to the second part of Lemma 4.3 in the form

$f^{(j)}(s)=\frac{j!}{2\pi i}\int_{|z|=\frac{\varepsilon}{\log
x}}\frac{f(z)}{(z-s)^{j+1}}dz$

we get for free

######  Corollary 4.4.

Let $s\in[0,1]$ and fix $0\leq i\leq 5$. Uniformly for $x\geq 4$ and $T\geq 2$
we have

(4.9) $\displaystyle-\frac{\partial^{i}}{\partial
s^{i}}\frac{G_{1}^{\prime}(s,x)}{G_{1}(s,x)}$
$\displaystyle=\mathbf{1}_{x\in\mathbb{N},\,\Lambda(x)\neq
0}\frac{\Lambda(x)(-\log x)^{i}}{2x^{s}}-\sum_{\begin{subarray}{c}\rho\\\
|\Im\rho|\leq T\end{subarray}}\frac{\partial^{i}}{\partial
s^{i}}\frac{x^{\rho-s}}{\rho-s}$ (4.10)
$\displaystyle\quad+\frac{\partial^{i}}{\partial
s^{i}}\sum_{k=1}^{\infty}\frac{x^{-2k-s}}{2k+s}+R_{1,i}(x,T,s),$ (4.11)
$\displaystyle R_{1,i}(x,T,s)$ $\displaystyle\ll(\log
x)^{i+1}x^{-s}\min\left\\{1,\frac{x}{T\langle
x\rangle}\right\\}+\frac{\log^{2}(xT)(\log x)^{i}}{T}x^{1-s}.$

######  Lemma 4.5.

Let $s\geq 0$ and let $\rho$ be a non-trivial zero of $\zeta$. Let
$d:=\min_{t\geq 0}|\rho-s-t|$. Uniformly for $x\geq 2$ we have

(4.12) $\int_{0}^{\infty}\frac{x^{\rho-s-t}}{\rho-
s-t}dt=\frac{x^{\rho-s}}{(\rho-s)\log
x}+O\left(\frac{x^{\Re\rho-s}}{d|\rho-s|(\log x)^{2}}\right).$

###### Proof.

We write the integrand as

$\frac{x^{\rho-s-t}}{\rho-s}\frac{1}{1-\frac{t}{\rho-s}}=\frac{x^{\rho-
s-t}}{\rho-s}\left(1+O\left(\frac{t}{|\rho-s-t|}\right)\right)$

and integrate. ∎

######  Proposition 4.6.

Let $s\in[0,1]$ and $x\geq 4$. If $\Theta\in[1/2,1]$ denotes the supremum of
the real parts of zeros of $\zeta$ then, for $0\leq i\leq 6$,

(4.13) $\displaystyle(\log G_{1}(s,x))^{(i)}$ $\displaystyle\ll
x^{\Theta-s}(\log x)^{i+1},$ (4.14) $\displaystyle(\log G_{1}(s,x))^{(i)}$
$\displaystyle\ll x^{1-s}L(x)^{-c}.$

Also

(4.15) $\displaystyle\log G_{1}(s,x)$ $\displaystyle=-(\log
x)^{-1}x^{-s}\left(\sum_{|\Im\rho|\leq
T}\frac{x^{\rho}}{\rho-s}\left(1+O\left(\frac{1}{|\rho-s|(\log
x)}\right)\right)\right.$ (4.16)
$\displaystyle\qquad\qquad\left.+O\left(\frac{x\log^{2}(xT)}{T}+\log
x\right)\right).$

###### Proof.

The estimates in (4.13) follow by taking $T=\sqrt{x}$ in Lemma 4.3 and
Corollary 4.4, and using

$\sum_{|\Im\rho|\leq T}1/|\rho-s|\ll(\log T)^{2}$

coming from the fact that between height $N$ and $N+1$, $\zeta$ has $\ll\log
N$ zeros. To prove (4.14) we take $T=L(x)^{c}$ and use the Vinogradov–Korobov
zero-free region. The last part follows by simplifying Lemma 4.3. ∎

###  4.2. Oscillations of $G_{1}$

Given a function $A(x)$, its Mellin transform is

$\\{\mathcal{M}A\\}(s)=\int_{0}^{\infty}A(x)x^{s-1}dx.$

The following proposition computes the Mellin transforms of

(4.17) $A_{1}(x):=\sum_{n\leq x}\Lambda(n)/(n^{s_{0}}\log n),\qquad
A_{2}(x):=I((1-s_{0})\log x)$

for fixed $s_{0}\in(0,1)$. It is inspired by Mellin transform computations of
Diamond and Pintz [DP09], who studied the transform of

$\sum_{p\leq x}-\log\left(1-1/p\right)-\left(\log\log x+\gamma\right)$

in order to show oscillation of this difference.

######  Proposition 4.7.

Fix $s_{0}\in(0,1)$. We have, for $\Re s>1-s_{0}$,

(4.18)
$\\{\mathcal{M}A_{1}\\}(-s)=\frac{1}{s}\log\zeta(s+s_{0}),\qquad\\{\mathcal{M}A_{2}\\}(-s)=\frac{1}{s}\log\frac{s}{s+s_{0}-1}.$

###### Proof.

For $A_{1}$ this is [MV07, Thm. 1.3]. For $A_{2}$, we start with [DP09, Eq.
(2.3)]:

(4.19) $\log\frac{z+1}{z}=\int_{1}^{\infty}t^{-z}\frac{1-t^{-1}}{t\log t}dx$

for $\Re z>0$. Applying (4.19) with $z=(s_{0}+s-1)/(1-s_{0})$ and performing
the change of variables $t=x^{1-s_{0}}$ in (4.19) we obtain

$\log\frac{s}{s+s_{0}-1}=\int_{1}^{\infty}x^{-s}\frac{x^{1-s_{0}}-1}{x\log
x}dx$

for $\Re s>1-s_{0}$. Integration by part, inspired by [DP09, p. 526], shows

$\log\frac{s}{s+s_{0}-1}=s\int_{1}^{\infty}x^{-s-1}\int_{1}^{x}\frac{t^{1-s_{0}}-1}{t\log
t}dtdx.$

The change of variables $t=e^{v/(1-s_{0})}$ shows that the inner integral
equals $I((1-s_{0})\log x)$. ∎

######  Proposition 4.8.

Let $\Theta\in[1/2,1]$ be the supremum of the real parts of zeros of $\zeta$.
Fix $s_{0}\in(0,\Theta)$. For any $\varepsilon>0$ we have

$\log G_{1}(s_{0},x)=\Omega_{\pm}(x^{\Theta-s_{0}-\varepsilon}).$

###### Proof.

By Lemma 4.3,

$\log G_{1}(s_{0},x)=\sum_{n\leq x}\frac{\Lambda(n)}{n^{s_{0}}\log
n}-I((1-s_{0})\log x)+O_{s_{0}}(\log\log x).$

Letting

$\Delta(x)=\sum_{n\leq x}\frac{\Lambda(n)}{n^{s_{0}}\log n}-I((1-s_{0})\log
x)$

it suffices to show that $\Delta(x)=\Omega_{\pm}(x^{\Theta-
s_{0}-\varepsilon})$. We show one direction, namely that $\Delta(x)\geq
x^{\Theta-s_{0}-\varepsilon}$ occurs infinitely often, the other inequality is
established similarly. By Proposition 4.7,

$\\{\mathcal{M}\Delta\\}(-s)=\frac{1}{s}\log\frac{\zeta(s+s_{0})(s+s_{0}-1)}{s}$

for $\Re s>1-s_{0}$, and it is easily seen that, if we let $B_{a}(x):=x^{a}$,

$\\{\mathcal{M}B_{a}\\}(-s)=\frac{1}{s-a}.$

Suppose for contradiction sake that $\Delta(x)<x^{\Theta-s_{0}-\varepsilon}$
holds once $x\geq X$ ($X$ may depend on $\varepsilon)$. Let

$F(s):=\\{\mathcal{M}(B_{\Theta-
s_{0}-\varepsilon}-\Delta)\\}(-s)=\int_{1}^{\infty}\left(x^{\Theta-
s_{0}-\varepsilon}-\Delta(x)\right)x^{-s-1}dx.$

Let $\sigma_{c}$ be the infimum of $\sigma$ for which $F(\sigma)$ converges.
Then Lemma 15.1 of [MV07] (Landau’s Oscillation Theorem) says that $F(s)$ is
analytic in the half-plane $\Re s>\sigma_{c}$, but not at $s=\sigma_{c}$.
However,

$F(s)=\frac{1}{s-(\Theta-
s_{0}-\varepsilon)}-\frac{1}{s}\log\frac{\zeta(s+s_{0})(s+s_{0}-1)}{s}.$

This function has a simple pole at $s=\Theta-s_{0}-\varepsilon$, and is
analytic for real $s>\Theta-s_{0}-\varepsilon$ through the inequalities
$\zeta(\sigma)(\sigma-1)\in(1,\sigma)$ for all $\sigma>0$ [MV07, Cor. 1.14].
So $\sigma_{c}$ must be $\Theta-s_{0}-\varepsilon$, implying $F(s)$ is
analytic in the half-plane $\Re s>\Theta-s_{0}-\varepsilon$. However, by
definition of $\Theta$, this gives a contradiction. ∎

###  4.3. Estimates for $G_{2}$

We have $\log G_{2}=\log G_{2,1}+\log G_{2,2}$ for

(4.20) $\log G_{2,1}(s,x)=\sum_{\sqrt{x}<p\leq x}p^{-2s}/2,\qquad\log
G_{2,2}(s,x)=\sum_{k\geq 3}\sum_{x^{1/k}<p\leq x}p^{-ks}/k.$

The Prime Number Theorem (PNT) with error term shows

######  Lemma 4.9.

Uniformly for $x\geq 2$ and $s\in[0,1]$ we have

(4.21) $\displaystyle\log G_{2,1}(s,x)$
$\displaystyle=\frac{1+O(L(x)^{-c})}{2}\int_{\sqrt{x}}^{x}\frac{dt}{t^{2s}\log
t}$ (4.22) $\displaystyle\asymp\frac{1}{\log
x}\int_{\sqrt{x}}^{x}\frac{dt}{t^{2s}}=\frac{x^{1-2s}-x^{\frac{1}{2}-s}}{(1-2s)\log
x}.$

For $x\neq 0$ let $\operatorname{Ei}(x)$ be the exponential integral, to be
understood in principal value sense:

$\operatorname{Ei}(x)=-\int_{-x}^{\infty}e^{-t}t^{-1}dt=\int_{-\infty}^{x}e^{t}t^{-1}dt=e^{x}x^{-1}\left(1+O(x^{-1})\right).$

######  Lemma 4.10.

For $1/2\neq s\in[0,1]$ we have

(4.23) $\displaystyle\int_{\sqrt{x}}^{x}\frac{dt}{t^{2s}\log t}$
$\displaystyle=\operatorname{Ei}\left(\log
x\left(1-2s\right)\right)-\operatorname{Ei}\left(\log
x\left(\frac{1}{2}-s\right)\right)$ (4.24)
$\displaystyle\sim\begin{cases}\frac{x^{\frac{1}{2}-s}}{\log
x\left(s-\frac{1}{2}\right)}&\text{if }(2s-1)\log x\to\infty,\\\
\frac{x^{1-2s}}{\log x\left(1-2s\right)}&\text{if }(2s-1)\log
x\to-\infty.\end{cases}$

When $s=1/2$, the integral is $\log 2$. When $s=1/2+O(1/\log x)$ the integral
is $\Theta(1)$.

###### Proof.

When $s\neq 1/2$, we perform the change of variables $v=(1-2s)\log t$ and use
the asymptotics for $\operatorname{Ei}$. When $s=1/2$, we use the fact that
$\log\log t$ is an antiderivative of $1/(t\log t)$. ∎

######  Lemma 4.11.

Fix $\varepsilon>0$. For $x\geq 2$ and $1\geq s\geq\varepsilon/\log x$,

$\log
G_{2,2}(s,x)\ll_{\varepsilon}\frac{x^{1-3s}-x^{\frac{1}{3}-s}}{(1-3s)\log
x}\asymp\begin{cases}\frac{x^{\frac{1}{3}-s}}{(3s-1)\log x}&\text{if
}(3s-1)\log x\geq 1,\\\ 1&\text{if }|(3s-1)\log x|\leq 1,\\\
\frac{x^{1-3s}}{(1-3s)\log x}&\text{if }(3s-1)\log x\leq-1.\end{cases}$

###### Proof.

The same argument as in Lemma 4.9 shows that the contribution of $k=3$ to
$\log G_{2,2}(s,x)$ is acceptable, so we omit this case from now on. We
consider the contribution of $k\geq\max\\{2/s,\log_{2}x\\}$ (base-$2$
logarithm). For such $k$,

$\sum_{x^{1/k}<p\leq x}p^{-ks}\leq 2^{-ks}+\sum_{p\geq 3}p^{-ks}\ll
2^{-ks}+\int_{2}^{\infty}t^{-ks}dt\ll 2^{-ks}.$

Hence

$\sum_{k\geq\max\\{2/s,\log_{2}x\\}}\sum_{x^{1/k}<p\leq
x}p^{-ks}/k\ll\sum_{k\geq\max\\{2/s,\log_{2}x\\}}2^{-ks}/k\ll x^{-s},$

which is sufficiently small. It remains to consider the contribution of $4\leq
k\leq\max\\{2/s,\log_{2}x\\}$ to $\log G_{2,2}$. We show that primes
$p\in(x^{1/4},x]\subseteq(x^{1/k},x]$ have an acceptable contribution. The
assumption $s\geq\varepsilon/\log x$ implies $1/(1-t^{-s})\ll_{\varepsilon}1$
when $t\geq x^{1/4}$, and so

$\sum_{\max\\{2/s,\log_{2}x\\}\geq k\geq 4}\sum_{x^{1/4}<p\leq
x}\frac{1}{p^{ks}k}\ll\sum_{k\geq 4}\int_{x^{1/4}}^{x}\frac{dt}{t^{ks}k\log
t}\ll_{\varepsilon}\int_{x^{1/4}}^{x}\frac{dt}{t^{4s}\log t}$

which is smaller than the bound we give. For the smaller primes,
$x^{1/k}<p\leq x^{1/4}$, we use

$\sum_{x^{1/k}<p\leq x^{1/4}}p^{-ks}\ll x^{\frac{1}{4}-s}/\log x$

which implies

$\sum_{\max\\{2/s,\log_{2}x\\}\geq k\geq 4}\sum_{x^{1/k}<p\leq
x^{1/4}}p^{-ks}/k\ll x^{\frac{1}{4}-s}\sum_{\max\\{2/s,\log_{2}x\\}\geq k\geq
4}1/\log x$

which is $\ll_{\varepsilon}x^{1/4-s}$. ∎

######  Corollary 4.12.

Fix $\varepsilon>0$. Suppose $x\geq 2$ and $1\geq s\geq\varepsilon/\log x$.
Then

(4.25) $\displaystyle\log G_{2}(s,x)$
$\displaystyle=\frac{1}{2}\int_{\sqrt{x}}^{x}\frac{dt}{t^{2s}\log
t}\left(1+O_{\varepsilon}\left(L(x)^{-c}+x^{-s}\right)\right)$ (4.26)
$\displaystyle\asymp_{\varepsilon}(\log
x)^{-1}\int_{\sqrt{x}}^{x}t^{-2s}dt=\frac{x^{1-2s}-x^{\frac{1}{2}-s}}{(1-2s)\log
x}.$

In the same way we establish

######  Lemma 4.13.

Fix $\varepsilon>0$ and $0\leq i\leq 6$. For $x\geq 2$ and $1\geq
s\geq\varepsilon/\log x$,

(4.27)
$\displaystyle\big{(}\frac{G_{2}^{\prime}(s,x)}{G_{2}(s,x)}\big{)}^{(i)}$
$\displaystyle=(-1)^{i-1}2^{i}\int_{\sqrt{x}}^{x}(\log
t)^{i}t^{-2s}dt\left(1+O_{\varepsilon}\left(L(x)^{-c}+x^{-s}\right)\right)$
(4.28) $\displaystyle\asymp_{\varepsilon}(-\log x)^{i+1}\log G_{2}(s,x).$

######  Lemma 4.14.

Fix $\varepsilon>0$. Suppose $1/10\geq s\geq\varepsilon/\log x$. Then

$\log G_{2}(s,x)=\int_{\sqrt{x}}^{x}(-\log(1-t^{-s})-t^{-s})\frac{dt}{\log
t}(1+O_{\varepsilon}(L(x)^{-c}))$

###### Proof.

The contribution of $k\geq\max\\{2,s/\log x\\}$ is $\ll x^{-s}$ as in Lemma
4.11. We now consider $2\leq k<\max\\{2,s/\log x\\}$. If
$x^{1/k}<p\leq\sqrt{x}$, we get a contribution of $\ll_{\varepsilon}x^{1/2-s}$
as in Lemma 4.11. We handle $2\leq k<\max\\{2,s/\log x\\}$ and $\sqrt{x}<p\leq
x$ by the PNT, obtaining

$\int_{\sqrt{x}}^{x}\sum_{2\leq
k\leq\max\\{2/s,\log_{2}x\\}}\frac{t^{-ks}}{k}\frac{dt}{\log
t}(1+O_{\varepsilon}(L(x)^{-c})).$

We use $t^{s}-1\gg_{\varepsilon}1$ when $t\in[\sqrt{x},x]$ to complete the sum
over $k$ at a negligible cost. ∎

##  Appendix A Proof of Lemma 4.1

We apply [MV07, Cor. 5.3] with $\sigma_{0}=\max\\{0,1-\Re s\\}+1/\log x$ and
the sequence $a_{n}=\Lambda(n)/n^{s}$ to obtain

(A.1) $\displaystyle S_{1}(x,s)$ $\displaystyle=\frac{1}{2\pi
i}\int_{\sigma_{0}-iT}^{\sigma_{0}+iT}-\frac{\zeta^{\prime}(s+w)}{\zeta(s+w)}x^{w}\frac{dw}{w}+O(E_{s}),$
(A.2) $\displaystyle E_{s}$
$\displaystyle=\sum_{\begin{subarray}{c}x/2<n<2x\\\ n\neq
x\end{subarray}}\frac{\Lambda(n)}{n^{\Re
s}}\min\left\\{1,\frac{x}{T|x-n|}\right\\}+\frac{x^{\sigma_{0}}}{T}\sum_{n=1}^{\infty}\frac{\Lambda(n)}{n^{\sigma_{0}+\Re
s}}.$

If $0\leq\Re s\leq 2$, we handle separately $n\in(x/2,x-1]$, $n\in[x+1,2x)$
and $n\in(x-1,x+1)$ in the first sum of $E_{s}$, and use
$\sum_{n=1}^{\infty}\Lambda(n)/n^{t}=-\zeta^{\prime}(t)/\zeta(t)\asymp
1/(t-1)$ for $t\in(1,10]$ in the second sum, to find

$E_{s}\ll(\log x)x^{-\Re s}\min\left\\{1,\frac{x}{T\langle
x\rangle}\right\\}+\frac{x^{1-\Re s}\log^{2}x}{T}+\frac{x^{\sigma_{0}}\log
x}{T}.$

If $\Re s\geq 2$ we argue similarly but this time make use of
$-\zeta^{\prime}(t)/\zeta(t)\ll 2^{-t}$ for $t\geq 2$ to deduce

$E_{s}\ll(\log x)(x-1)^{-\Re s}\min\left\\{1,\frac{x}{T\langle
x\rangle}\right\\}+2^{\Re s}\frac{x^{1-\Re s}\log^{2}x}{T}+\frac{2^{-\Re
s}}{T}.$

Recall $T\geq 2+3|\Im s|$. By [MV07, Lem. 12.2], there is $T_{1}\in[T,T+1]$
such that

(A.3) $\frac{\zeta^{\prime}(\sigma+i\Im s+iT_{1})}{\zeta(\sigma+i\Im
s+iT_{1})}\ll(\log T)^{2}$

holds uniformly for $-1\leq\sigma\leq 2$. The lemma also guarantees the
existence of $T_{2}\in[T,T+1]$ such that

(A.4) $\frac{\zeta^{\prime}(\sigma+i\Im s-iT_{2})}{\zeta(\sigma+i\Im
s-iT_{2})}\ll(\log T)^{2}$

uniformly for $-1\leq\sigma\leq 2$. We now change the range of integration in
(A.1) from $|\Im w|\leq T$ to $-T_{2}\leq\Im w\leq T_{1}$. The error we incur
is at most

$\ll\frac{x^{\sigma_{0}}}{T}\left(-\frac{\zeta^{\prime}(\sigma_{0}+\Re
s)}{\zeta(\sigma_{0}+\Re s)}\right)$

which can be absorbed in $O(E_{s})$. Let $K$ denote an odd positive integer,
and let $\mathcal{C}$ denote the contour consisting of three line segments,
connecting $\sigma_{0}-iT_{2}$, $-K-\Re s-iT_{2}$, $-K-\Re s+iT_{1}$,
$\sigma_{0}+iT_{1}$, in this order. Cauchy’s residue theorem shows that, if
$s\neq 1$, then

(A.5) $\frac{1}{2\pi
i}\int_{\sigma_{0}-iT_{2}}^{\sigma_{0}+iT_{1}}-\frac{\zeta^{\prime}(s+w)}{\zeta(s+w)}x^{w}\frac{dw}{w}=\frac{x^{1-s}}{1-s}-\frac{\zeta^{\prime}(s)}{\zeta(s)}\\\
-\sum_{\begin{subarray}{c}\rho\\\
-T_{2}<\Im(\rho+s)<T_{1}\end{subarray}}\frac{x^{\rho-s}}{\rho-s}+\sum_{1\leq
k<K/2}\frac{x^{-2k-s}}{2k+s}+\frac{1}{2\pi
i}\int_{\mathcal{C}}-\frac{\zeta^{\prime}(s+w)}{\zeta(s+w)}x^{w}\frac{dw}{w}.$

If $s=1$, the integrand has a double pole at $w=0$ and the main term
$x^{1-s}/(1-s)-\zeta^{\prime}(s)/\zeta(s)$ should be replaced with $\log
x-\gamma$, the residue at $0$ (which is also the limit of
$x^{1-s}/(1-s)-\zeta^{\prime}(s)/\zeta(s)$ as $s$ tends to $1$).

We replace the sum over $-T_{2}<\Im(\rho+s)<T_{1}$ with one over
$-T\leq\Im(\rho+s)\leq T$, and the incurred error is at most

$\ll\sum_{\begin{subarray}{c}\rho\\\
\Im(\rho+s)\in[T_{1},T]\cup[-T_{2},-T]\end{subarray}}\frac{x^{1-\Re
s}}{|\rho-s|}\ll\frac{x^{1-\Re s}\log T}{T}$

which is acceptable (i.e. can be absorbed in $R(x,T,s)$). It remains to bound
the integral over $\mathcal{C}$.

To bound the horizontal parts of the integral, we consider separately three
ranges of $\Re w\in[-K-\Re s,\sigma_{0}]$. The contribution of $\Re
w\in[-1-\Re s,\min\\{2-\Re s,\sigma_{0}\\}]$ can be bounded using (A.3) and
(A.4):

$\frac{1}{2\pi i}\int_{-1-\Re s+iT_{1}}^{\min\\{2-\Re
s,\sigma_{0}\\}+iT_{1}}-\frac{\zeta^{\prime}(s+w)}{\zeta(s+w)}x^{w}\frac{dw}{w}\ll\frac{\log^{2}T}{T}\frac{x^{\min\\{2-\Re
s,\sigma_{0}\\}}}{\log x},$

and the same bound holds if $T_{1}$ is replaced with $-T_{2}$. This error is
acceptable.

Next, the contribution of $\Re w\in(2-\Re s,\sigma_{0}]$ should only be
considered it this is a non-empty interval, which happens exactly when $\Re
s\geq 2-1/\log x$. In this case, we use $-\zeta^{\prime}(t)/\zeta(t)\ll
2^{-t}$ ($t\geq 2$) to estimate the integral as follows:

$\frac{1}{2\pi i}\int_{2-\Re
s+iT_{1}}^{\sigma_{0}+iT_{1}}-\frac{\zeta^{\prime}(s+w)}{\zeta(s+w)}x^{w}\frac{dw}{w}\ll\frac{1}{T2^{\Re
s}}\frac{\left(\frac{x}{2}\right)^{\sigma_{0}}}{\log x}\ll\frac{1}{T2^{\Re
s}}$

which is acceptable. The same bound holds if $T_{1}$ is replaced with
$-T_{2}$. To bound the contribution of $\Re w\in[-K-\Re s,-1-\Re s]$ we make
use of [MV07, Lem. 12.4] which says that

(A.6) $\zeta^{\prime}(s)/\zeta(s)\ll\log(|s|+1)$

holds uniformly for all $s$ with $\Re s\leq-1$ and $\min_{k\geq 1}|s+2k|\geq
1/4$. This gives

(A.7) $\displaystyle\frac{1}{2\pi i}\int_{-K-\Re s+iT_{1}}^{-1-\Re s+iT_{1}}$
$\displaystyle-\frac{\zeta^{\prime}(s+w)}{\zeta(s+w)}x^{w}\frac{dw}{w}$ (A.8)
$\displaystyle\ll\int_{-K}^{-1}\log(T+|a|)\frac{x^{a-\Re
s}}{T+|a|}da\ll\frac{\log T}{T}\frac{x^{-1-\Re s}}{\log x}$

which is acceptable. The same bound holds if $T_{1}$ is replaced with
$-T_{2}$. To bound the integral over the vertical part of the $\mathcal{C}$,
we use (A.6) again to obtain

(A.9) $\frac{1}{2\pi i}\int_{-K-\Re s-iT_{2}}^{-K-\Re
s+iT_{1}}-\frac{\zeta^{\prime}(s+w)}{\zeta(s+w)}x^{w}\frac{dw}{w}\\\ \ll
x^{-K-\Re s}\int_{-K-\Re s-iT_{2}}^{-K-\Re
s+iT_{1}}\log(|s+w|+1)\frac{|dw|}{|w|}\ll x^{-K-\Re s}\log^{2}(K+T).$

As $K$ tends to $\infty$ the size of the last integral goes to $0$, which
concludes the proof.

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Mathematical Institute, University of Oxford, Oxford, OX2 6GG, UK

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