# On the splitting conjecture in the hybrid model for the Riemann zeta
function

Winston Heap<EMAIL_ADDRESS>
###### Abstract.

We show that the splitting conjecture in the hybrid model of
Gonek–Hughes–Keating holds to order on the Riemann hypothesis. Our results are
valid in a larger range of the parameter $X$ which mediates between the
partial Euler and Hadamard products. We also show that the asymptotic
splitting conjecture holds for this larger range of $X$ in the cases of the
second and fourth moments.

##  1\. Introduction

The moments of the Riemann zeta function have been the subject of several
conjectural methods in recent years. Since the second and fourth moments of
Hardy–Littlewood [23] and Ingham [33], it is only relatively recently that a
full conjecture for all moments was given. This began with the work of
Keating–Snaith [34] who used the now famous connection with random matrix
theory to conjecture that for real $k>-1/2$,

$\frac{1}{T}\int_{T}^{2T}|\zeta(\tfrac{1}{2}+it)|^{2k}dt\sim a(k)g(k)(\log
T)^{k^{2}}$

where

(1) $a(k)=\prod_{p}\bigg{(}1-\frac{1}{p}\bigg{)}^{k^{2}}\sum_{m\geqslant
0}\frac{d_{k}(p^{m})^{2}}{p^{m}}$

and

(2) $g(k)=\frac{G(k+1)^{2}}{G(2k+1)}$

where $G$ is the Barnes $G$-function. This was preceded by conjectures for the
6th and 8th moments due to Conrey–Ghosh [14] and Conrey–Gonek [15],
respectively, using number theoretic methods. The Keating–Snaith conjecture
has since been derived with various different approaches [13, 18, 22].

A drawback of Keating and Snaith’s method was that the arithmetic factor
$a(k)$ had to be incorporated in an ad-hoc way since there was no input from
primes in their random matrix theory model. This was remedied in the method of
Gonek–Hughes–Keating (G–H–K) [22] which forms the main focus of this paper.

The first step of G–H–K’s method was to express the zeta function as the
product of partial Euler and Hadamard products. Precisely, Theorem 1 of [22]
states that for $2\leqslant X\leqslant t^{1/3}$ and large $t$,

(3)
$\zeta(\tfrac{1}{2}+it)=P_{X}(\tfrac{1}{2}+it)Z_{X}(\tfrac{1}{2}+it)\Big{(}1+O\Big{(}\frac{1}{\log
X}\Big{)}\Big{)}$

where

$P_{X}(s)=\exp\bigg{(}\sum_{n\leqslant X}\frac{\Lambda(n)}{n^{s}\log
n}\bigg{)},\,\,\,\,\,\,\,Z_{X}(s)=\exp\bigg{(}-\sum_{\rho}U((s-\rho)\log
X)\bigg{)}$

and

$U(z)=\int_{0}^{\infty}u(x)E_{1}(z\log x)dx$

with $E_{1}(z)=\int_{z}^{\infty}e^{-w}dw/w$ and $u(x)$ a smooth, non-negative
function of mass 1 with support in $[e^{1-1/X},e]$. To give a rough idea of
these objects, note that from the support conditions on $u$ we have
$U(z)\approx E_{1}(z)$. This has mass concentrated in the region $z\ll 1$
where we have the approximation $E_{1}(z)\approx-\gamma-\log z$. Thus, roughly
speaking, $Z_{X}(s)\approx\prod_{|\Im(s)-\Im(\rho)|\ll 1/\log
X}((s-\rho)e^{\gamma}\log X)$. Also, from the definition of the von Mangoldt
function and the Taylor series for the logarithm we find
$P_{X}(s)\approx\prod_{p\leqslant X}(1-p^{-s})^{-1}$. Therefore, we can indeed
view $P_{X}(s)$ and $Z_{X}(s)$ as partial Euler and Hadamard products.

G–H–K then proceeded to compute the moments of the Euler product, showing that
for $X\ll(\log T)^{2-\epsilon}$,

(4) $\frac{1}{T}\int_{T}^{2T}|P_{X}(\tfrac{1}{2}+it)|^{2k}dt\sim
a(k)(e^{\gamma}\log X)^{k^{2}},\qquad k\in\mathbb{R}.$

They conjectured with random matrix theory that

(5) $\,\,\,\,\frac{1}{T}\int_{T}^{2T}|Z_{X}(\tfrac{1}{2}+it)|^{2k}dt\sim
g(k)\Big{(}\frac{\log T}{e^{\gamma}\log
X}\Big{)}^{k^{2}},\,\,\,\,\,\,\,\,k>-1/2$

and then proved this in the cases $k=1,2$ for $X\ll(\log T)^{2-\epsilon}$. In
order to recover the Keating–Snaith conjecture they assumed that the moments
of the product of $P_{X}$ and $Z_{X}$ should split as the product of moments.

######  Conjecture 1 (Splitting conjecture, [22]).

Let $X,T\to\infty$ with $X\ll(\log T)^{2-\epsilon}$. Then for fixed $k>-1/2$
we have

$\frac{1}{T}\int_{T}^{2T}|P_{X}(\tfrac{1}{2}+it)Z_{X}(\tfrac{1}{2}+it)|^{2k}dt\sim\frac{1}{T}\int_{T}^{2T}|P_{X}(\tfrac{1}{2}+it)|^{2k}dt\cdot\frac{1}{T}\int_{T}^{2T}|Z_{X}(\tfrac{1}{2}+it)|^{2k}dt.$

Their reasoning behind this conjecture was that since $P_{X}$ and $Z_{X}$
oscillate at different scales ($1/\log X$ vs.$\,1/\log T$), their
contributions should act independently and hence the moment should split to
leading order. They verified this in the cases $k=1,2$ for $X\ll(\log
T)^{2-\epsilon}$. The methodology of the hybrid model has since been used in
various different settings to acquire conjectures for all sorts of
$L$-functions [1, 8, 9, 10, 11, 19, 25]. In all cases, an equivalent version
of the splitting conjecture plays a key role.

In this paper we prove that the splitting conjecture holds to order on the
Riemann hypothesis (RH). Furthermore, we can extend the range of $X$ past
$(\log T)^{2-\epsilon}$.

######  Theorem 1.

Assume RH. Let $\epsilon,k>0$ be fixed and suppose $X,T\to\infty$ with
$X\leqslant(\log T)^{\theta_{k}-\epsilon}$ where
$\theta_{k}=2\sqrt{1+{1}/{2|k|}}$. Then

$\frac{1}{T}\int_{T}^{2T}|P_{X}(\tfrac{1}{2}+it)Z_{X}(\tfrac{1}{2}+it)|^{2k}dt\asymp\frac{1}{T}\int_{T}^{2T}|P_{X}(\tfrac{1}{2}+it)|^{2k}dt\cdot\frac{1}{T}\int_{T}^{2T}|Z_{X}(\tfrac{1}{2}+it)|^{2k}dt.$

As mentioned, this holds in a range of $X$ larger than originally conjectured.
We can also extend the range of $X$ in the asymptotic results (4) and (5),
both unconditionally and on RH. This gives the following.

######  Theorem 2.

The Splitting conjecture holds for $k=1,2$ in the range

$X\leqslant\tfrac{1}{10^{4}}(\log T)^{2}(\log_{2}T)^{2}.$

Assuming RH, we may take

$X\leqslant\begin{cases}(\log T)^{\sqrt{6}-\epsilon}\,\,\text{ when }&k=1,\\\
(\log T)^{\sqrt{5}-\epsilon}\,\,\text{ when }&k=2.\end{cases}$

Our proofs utilise the recent developments in the theory of moments of
$L$-functions due to Soundararajan [43], Harper [24] and
Radziwiłł–Soundararajan [38]. These techniques were originally geared for
upper bounds although they can be brought to bear on lower bounds too [27]. We
highlight three main ideas.

The first is an innovation of Soundararajan [43]. This was to note that
$\log|\zeta(\tfrac{1}{2}+it)|$ can be bounded from above by a sum over primes
alone since the zeros contribute negatively to this quantity (see Lemma 6
below and c.f. formula (3)). With this, $|\zeta(\tfrac{1}{2}+it)|$ can be
bounded from above by an Euler product of flexible length.

The second idea can be found in a paper of Radziwiłł [37] and features heavily
in the later works of Harper [24] and Radziwiłł–Soundararajan [38]. It allows
one to compute moments of Euler products provided one can restrict to a
certain subset of $[T,2T]$. For the purposes of this discussion we consider
the example

$\exp\Big{(}\sum_{p\leqslant Y}p^{-1/2-it}\Big{)}$

with $Y=T^{1/(\log\log T)^{2}}$. On the face of it, this is a very long
Dirichlet polynomial. However, if we can restrict $t$ to a subset of $[T,2T]$
on which $|\sum_{p\leqslant Y}p^{-1/2-it}|\leqslant V$ for a given $V$, then
we can truncate the exponential series effectively using the fact that

(6) $e^{z}\sim\sum_{j=0}^{10V}\frac{z^{j}}{j!}$

for $|z|\leqslant V$ and large $V$. The choice of $V$ is naturally dictated by
the variance: setting $V=\sum_{p\leqslant Y}p^{-1}\sim\log\log T$ we get a
Dirichlet polynomial of length $Y^{10V}=T^{10/\log\log T}$. This is now short
and so the mean square is easily computed. Also, the exceptional set in this
case is of small measure.

The final main input in the arguments of Harper and Radziwiłł–Soundararajan
allows one to push the length of the prime sum up to $Y=T^{\theta}$, for some
fixed $\theta>0$. This involves breaking the sum it into subsums of
progressively smaller variance. A similar splitting has appeared in the work
of Brun on the pure sieve (see Hooley’s refinement [29]).

This circle of ideas has been used in a wide variety of different contexts
recently. These include; short interval maxima of the Riemann zeta function
[2, 3, 4], unconditional bounds for the moments of zeta and $L$-functions [21,
26, 27], value distribution of $L$-functions [16, 30, 39], sign changes in
Fourier coefficients of modular forms [35], non-vanishing of central values of
$L$-functions [17] and equidistribution of lattice points on the sphere [32].
In our case, we use these ideas to prove the following.

######  Proposition 1.

Let $\epsilon>0$ and $k\in\mathbb{R}$ be fixed. Suppose
$X\leqslant\eta_{k}(\log T)^{2}(\log_{2}T)^{2}$ with
$\eta_{k}=\tfrac{1}{16k^{2}}-\epsilon$. Then

$\frac{1}{T}\int_{T}^{2T}|P_{X}(\tfrac{1}{2}+it)|^{2k}dt\sim
a(k)(e^{\gamma}\log X)^{k^{2}}$

where $a(k)$ is given by (1). Assuming RH, this holds for $X\leqslant(\log
T)^{\theta_{k}-\epsilon}$ with $\theta_{k}=2\sqrt{1+{1}/{2|k|}}.$

######  Proposition 2.

Suppose $X\leqslant\tfrac{1}{10^{4}}(\log T)^{2}(\log_{2}T)^{2}$. Then for
$k=1,2$ we have

$\frac{1}{T}\int_{T}^{2T}|Z_{X}(\tfrac{1}{2}+it)|^{2k}dt\sim
g(k)\bigg{(}\frac{\log T}{e^{\gamma}\log X}\bigg{)}^{k^{2}}$

where $g(k)$ is given by (2). Assuming RH we may take $X\leqslant(\log
T)^{\sqrt{6}-\epsilon}$ when $k=1$ and $X\leqslant(\log
T)^{\sqrt{5}-\epsilon}$ when $k=2$.

######  Proposition 3.

Assume RH and let $\epsilon,k>0$ be fixed. Suppose $X\leqslant(\log
T)^{\theta_{k}-\epsilon}$ with $\theta_{k}$ as above. Then

$\frac{1}{T}\int_{T}^{2T}|Z_{X}(\tfrac{1}{2}+it)|^{2k}dt\asymp\bigg{(}\frac{\log
T}{\log X}\bigg{)}^{k^{2}}.$

######  Remark.

The lower bound in Proposition 3 can be made unconditional provided
$X\leqslant\eta_{k}(\log T)^{2}(\log_{2}T)^{2}$. We say more on this in
section 7.

Since $\int_{T}^{2T}|\zeta(\tfrac{1}{2}+it)|^{2k}dt$ is $\ll T(\log
T)^{k^{2}}$ on RH [24] and $\gg T(\log T)^{k^{2}}$ unconditionally [27],
Theorem 1 follows from Propositions 1 and 3 when combined with (3). Likewise,
Theorem 2 follows on combining Propositions 1 and 2 and (3).

Using the case of $P_{X}$ as an example, we describe how the range of $X$ can
be increased past $(\log T)^{2-\epsilon}$. First of all, note that since

(7) $\Big{|}\sum_{n\leqslant X}\frac{\Lambda(n)}{n^{1/2+it}\log
n}\Big{|}\leqslant(1+o(1))\frac{2X^{1/2}}{\log X},$

we can approximate $P_{X}(1/2+it)^{k}$ with a Dirichlet polynomial of length
$X^{20|k|X^{1/2}/\log X}$ by using (6) to truncate the exponential. If
$X\ll(\log T)^{2-\epsilon}$ then this is $T^{o(1)}$ and so we have a short
Dirichlet polynomial. Note this holds for _all_ $t\in[T,2T]$ since the bound
(7) is pointwise. G–H–K computed a Dirichlet polynomial approximation in a
slightly different way, although in order for it to be short they required the
same bound on $X$, perhaps unsurprisingly.

If $X$ is larger, then in order to have a short Dirichlet polynomial we must
restrict to a subset of $[T,2T]$ and in this case we need good bounds on the
exceptional set. Typically, one would expect Gaussian bounds of the shape

(8) $\frac{1}{T}\mu\Big{(}\Big{\\{}t\in[T,2T]:\Big{|}\Re\sum_{n\leqslant
X}\frac{\Lambda(n)}{n^{1/2+it}\log n}\Big{|}\geqslant
V\Big{\\}}\Big{)}\ll\exp\Big{(}-\frac{V^{2}}{\log\log T}\Big{)}$

in a wide range of $V$. In practice we are limited to $V\ll\sqrt{(\log
T)(\log_{2}T)/\log X}$ which may be much smaller than the maximum
$2X^{1/2}/\log X$. For the remaining range of $V$ one must settle for weaker
bounds. For example, in [43] it is shown that the tails of
$\log|\zeta(1/2+it)|$ can be bounded by $e^{-V\log V}$ when
$V\gg\log_{2}T\log_{3}T$. We can show that the tails of our sum satisfy the
same bound in the range $\log_{2}T\log_{3}T\leqslant V\leqslant 2X^{1/2}/\log
X$ provided $X\ll(\log T)^{2}$. However, for our purposes the weaker bound of
$e^{-AV}$ with large $A$ is sufficient and this affords us slightly more room
in the size of $X$.

Another avenue for improvement is to reduce the trivial bound in (7). This
becomes a manageable task under RH and thus we are able to make further gains
in the size of $X$ under this assumption. We shall prove the following.

######  Theorem 3.

Assume RH. Then for large $t\in[T,2T]$ and $2(\log T)^{2}\leqslant X\leqslant
T$ we have

$\Big{|}\Re\sum_{n\leqslant X}\frac{\Lambda(n)}{n^{1/2+it}\log
n}\Big{|}\leqslant(\tfrac{1}{2}+o(1))\big{(}\log(\tfrac{X^{1/2}}{\log
T})+4\log\log X\big{)}\frac{\log T}{\log\log T}.$

For the imaginary part we can replace the factor $\tfrac{1}{2}+o(1)$ by
$\tfrac{1}{\pi}+o(1)$.

The factor of $1/2+o(1)$ here is related to the function $S(t)$ and can be
read as $2c$ where $c$ is a permissible constant in the bound
$S(t)\leqslant(c+o(1))\log t/\log\log t$. The current best is due to
Carneiro–Chandee–Milinovich [12] who give $c=1/4$. Our approach to Theorem 3
is to relate the sum with $S(t)$ via contour integrals and then input these
bounds. We have not made attempts to further optimise this argument but it
would be interesting to see if one could use the extremal function machinery
of Carneiro et al. in a more direct way.

Regarding further improvements in the size of $X$, if the conjectural bound
$S(t)\ll\sqrt{\log t\log_{2}t}$ of Farmer–Gonek–Hughes [20] holds, then one
could take $X\leqslant\exp($ $C(\log t)^{1/4})$. Also, assuming that the
bounds for the exceptional set in (8), or some minor variant of this111In
fact, anything of the form $e^{-AV}$ with large $A$ would be sufficient., hold
in the full range of $V$ for a given $X$, then our arguments can reproduce
Theorems 1 and 2 for $X$ as large as $T^{1/C\log\log T}$. This supports the
view of G–H–K that the splitting conjecture may hold as long as $X=o(T)$.

The paper is organised as follows. We first prove Theorem 3 in section 2 and
then the asymptotic results of Propositions 1 and 2 in sections 3 and 4,
respectively. In section 5 we describe some tools for later use. Then in
section 6 we prove the upper bound of Proposition 3 and in sections 7 and 8 we
prove the lower bound in the cases $0\leqslant k\leqslant 1$ and $k\geqslant
1$, respectively.

######  Acknowledgements.

The author would like to thank Jing Zhao and Junxian Li for their comments on
an early draft of this paper and Chris Hughes for some clarifying remarks.

##  2\. Bounds for prime sums: Proof of Theorem 3

We first give a lemma which relates our prime sum to $S(t)$. This shares some
similarities with previous convolution formulas from the literature [42, 45].

######  Lemma 1.

Assume RH. For large $t\in[T,2T]$ and $2\leqslant X\leqslant T$, $Y\leqslant
T/2$, we have

$\sum_{n\leqslant X}\frac{\Lambda(n)}{n^{1/2+it}\log
n}=\int_{t-Y}^{t+Y}S(y)\frac{1-X^{-i(t-y)}}{t-y}dy+E(X,Y,T)$

where

$E(X,Y,T)\ll\frac{X^{1/2}(\log X+\tfrac{\log_{3}T}{\log X})}{Y}+\frac{\log
T}{\log_{2}T}\Big{(}\frac{1}{Y^{1/2}}+\frac{X^{{C}/{\log_{2}T}}}{Y}\mathds{1}_{X>(\log
T)^{A}}\Big{)}$

and $A$ is a large constant.

###### Proof.

By Perron’s formula (Lemma 3.19, [44]) we have

$\sum_{n\leqslant X}\frac{\Lambda(n)}{n^{1/2+it}\log n}=\frac{1}{2\pi
i}\int_{1/2+1/\log X-iY}^{1/2+1/\log
X+iY}\log\zeta(z+\tfrac{1}{2}+it)X^{z}\frac{dz}{z}+O\Big{(}\frac{X^{1/2}\log
X}{Y}\Big{)}.$

We shift the contour to the line with real part $\Re(z)=1/\log X$. Restricting
$Y\leqslant T/2$ we don’t encounter any poles. In the region
$1/2+C/\log\log\tau\leqslant\sigma\leqslant 1$, $\tau\gg 1$, we have

$\log\zeta(\sigma+i\tau)\ll\frac{(\log\tau)^{2-2\sigma}\log_{3}\tau}{\log_{2}\tau}.$

This follows from (14.14.5) of [44], the Phragmen–Lindelöf principle and the
bound $\log\zeta(\sigma+i\tau)\ll\log_{3}\tau$, $\sigma\geqslant 1$ (the
latter can be deduced from the proof of Theorem 14.8 of [44]). From Theorem
14.14 (B) of [44] we have

$\log\zeta(\sigma+i\tau)\ll\frac{\log\tau}{\log_{2}\tau}\log\Big{(}\frac{2}{(\sigma-1/2)\log_{2}\tau}\Big{)},\qquad\tfrac{1}{2}<\sigma\leqslant\tfrac{1}{2}+C/\log_{2}\tau.$

Therefore, the horizontal contours contribute

(9) $\ll\frac{X^{{C}/{\log_{2}T}}\log(\tfrac{\log X}{\log_{2}T})\log
T}{Y\log_{2}T\log X}\mathds{1}_{X>(\log
T)^{A}}+\frac{X^{1/2}\log_{3}T}{Y\log_{2}T\max(\log(\tfrac{X^{1/2}}{(\log
T)^{2}}),1)}+\frac{X^{1/2}\log_{3}T}{Y\log X}.$

By formula (14.10.5) of [44] (see also (14.12.4) there) we have

$\log\zeta(z+\tfrac{1}{2}+it)=i\int_{t/2-Y}^{2t+Y}\frac{S(y)}{z+i(t-y)}dy+O\Big{(}\frac{\log
T}{T}\Big{)}.$

This implies

(10) $\frac{1}{2\pi i}\int_{1/\log X-iY}^{1/\log
X+iY}\log\zeta(z+\tfrac{1}{2}+it)X^{z}\frac{dz}{z}=i\int_{t/2-Y}^{2t+Y}S(y)I(t-y)dy+O\Big{(}\frac{(\log
T)^{2}}{T}\Big{)}.$

where

$I(t-y)=\frac{1}{2\pi i}\int_{1/\log X-iY}^{1/\log
X+iY}\frac{X^{z}}{z(z+i(t-y))}dz.$

A trivial estimate gives

(11) $I(t-y)\ll\int_{-Y}^{Y}\frac{dx}{(1/\log X+|x|)(1/\log X+|x+t-y|)}\ll\log
X.$

On the other hand, shifting the contour to the left we find

(12) $I(t-y)=\frac{1-X^{-i(t-y)}}{i(t-y)}\mathds{1}_{|t-y|\leqslant
Y}+\frac{1}{i(t-y)}\mathds{1}_{|t-y|>Y}+O\Big{(}\frac{1}{Y|Y-|t-y||\log
X}\Big{)}.$

For a given $\delta>0$ to be chosen, we find by (11) that

$\int_{|t-y\pm Y|\leqslant\delta}S(y)I(t-y)dy\ll\delta\log X\log T/\log_{2}T$

since $S(\tau)\ll\log\tau/\log_{2}\tau$. In the integral over the remaining
region, the error term of (12) contributes

$\ll\frac{\log T}{\delta Y\log X\log_{2}T}+\frac{\log Y\log T}{Y\log
X\log_{2}T}$

after considering the regions $\delta<|Y-|t-y||\leqslant 1$ and $1<|Y-|t-y||$
separately. Therefore, on choosing $\delta=1/(Y^{1/2}\log X)$ we find that the
integral on the right of (10) is

$i\int_{\begin{subarray}{c}t/2-Y\leqslant y\leqslant 2t+Y\\\ |t-y\pm
Y|>\delta\end{subarray}}S(y)\bigg{(}\frac{1-X^{-i(t-y)}}{i(t-y)}\mathds{1}_{|t-y|\leqslant
Y}+\frac{1}{i(t-y)}\mathds{1}_{|t-y|>Y}\bigg{)}dy+O\Big{(}\frac{\log
T}{Y^{1/2}\log_{2}T}\Big{)}.$

Integrating by parts along with the bound
$S_{1}(\tau)\ll\log\tau/(\log_{2}\tau)^{2}$, we find that the second term in
this integral is $\ll\log T/(Y(\log_{2}T)^{2})$ which can be absorbed into the
error term immediately above. The range of integration of the first term can
be extended to $|t-y|\leqslant Y$ at the cost of an error $\ll\log
T/(Y\log_{2}T)$. Combining this in (10) along with the error terms of (9) the
result follows.

∎

###### Proof of Theorem 3.

We apply Lemma 1 with

$Y=\frac{X^{1/2}(\log X)^{2+\epsilon}}{\log T}.$

With this choice we have $E(X,Y,T)=o(\log T/\log\log T)$ since $\log
X\gg\log_{2}T$. Therefore, on taking real parts in Lemma 1 we find

$\Re\sum_{n\leqslant X}\frac{\Lambda(n)}{n^{1/2+it}\log
n}=-\int_{-Y}^{Y}S(t+y)\frac{1-\cos(y\log X)}{y}dy+o(\log T/\log_{2}T).$

From [12] we have

$|S(\tau)|\leqslant(1+o(1))\frac{1}{4}\frac{\log\tau}{\log\log\tau}$

for large $\tau$ and so

$\Big{|}\Re\sum_{n\leqslant X}\frac{\Lambda(n)}{n^{1/2+it}\log
n}\Big{|}\leqslant(1+o(1))\frac{\log T}{4\log\log T}\int_{-Y\log X}^{Y\log
X}\frac{1-\cos y}{|y|}dy.$

The integral here is

$\displaystyle 2\int_{\pi}^{Y\log X}\frac{1-\cos y}{y}dy+O(1)\leqslant$
$\displaystyle 2\sum_{n=1}^{Y\log X}\frac{1}{\pi n}\int_{\pi
n}^{\pi(n+1)}(1-\cos y)dy+O(1)$ $\displaystyle=$ $\displaystyle 2\log(Y\log
X)+O(1).$

Thus, we acquire

$\Big{|}\Re\sum_{n\leqslant X}\frac{\Lambda(n)}{n^{1/2+it}\log
n}\Big{|}\leqslant(1+o(1))\frac{\log(Y\log X)\log T}{2\log\log T}$

and the result follows on inputting our choice of $Y$.

In the case of the imaginary part, we acquire the integral
$\int_{-Y}^{Y}|\sin(y\log X)/y|dy$ which, on following the same argument, is
$\leqslant(4/\pi)\log(Y\log X)+O(1)$. ∎

From the proof we see that the factor $\log(X^{1/2}/\log T)$ comes from the
divergent integral $\int_{-Y}^{Y}|1-X^{-iy}|/|y|dy$. One may then wonder if
smoothing would help here, that is, if the problem could be modified so that
we consider a smoothed sum instead; say

$\sum_{n\leqslant X}\frac{\Lambda(n)}{n^{1/2+it}\log n}\Big{(}1-\frac{\log
n}{\log X}\Big{)}.$

The imaginary part responds well to this procedure and the above proof
recovers the formulas of Selberg [42] and Tsang [45] which have the convergent
integrand $\sin^{2}(y\log X)/y^{2}$. Unfortunately, for the real part the
integrand is again of the form $\approx 1/|y|$ owing to large negative values
of $\Re\log\zeta(1/2+it)$ (cf. Lemma 5 of [45]).

##  3\. Moments of the Euler product: Proof of Proposition 1

Recall that

$P_{X}(s)=\exp\Big{(}\sum_{n\leqslant X}\frac{\Lambda(n)}{n^{s}\log
n}\Big{)}.$

Here and throughout the paper we consider the following subsets of $[T,2T]$ on
which the sum in the exponential attains typical values. For $V\geqslant 0$
set

$\qquad\qquad\mathcal{S}_{\Re}(V)=\Big{\\{}t\in[T,2T]:\Big{|}\Re\sum_{n\leqslant
X}\frac{\Lambda(n)}{n^{1/2+it}\log n}\Big{|}\leqslant V\Big{\\}},$

and define the set relating to the imaginary part $\mathcal{S}_{\Im}(V)$
similarly. Let

$\mathcal{S}(V)=\mathcal{S}_{\Re}(V)\cap\mathcal{S}_{\Im}(V)$

and denote

$V_{0}=\log_{2}T\log_{3}T,\qquad\,\,\,\,\,\,\mathcal{S}=\mathcal{S}(V_{0}).$

Define the complementary sets by

$\mathcal{E}_{\Re}(V)=\overline{\mathcal{S}_{\Re}(V)},\,\,\mathcal{E}_{\Im}(V)=\overline{\mathcal{S}_{\Im}(V)},\,\,\mathcal{E}(V)=\overline{\mathcal{S}(V)},\,\,\mathcal{E}=\overline{\mathcal{S}}\,\,$

where $\overline{A}=[T,2T]\backslash A$ for a given set $A$. Also, let

$V_{\max}=\max_{t\in[T,2T]}\bigg{(}\Big{|}\Re\sum_{n\leqslant
X}\frac{\Lambda(n)}{n^{1/2+it}\log n}\Big{|},\Big{|}\Im\sum_{n\leqslant
X}\frac{\Lambda(n)}{n^{1/2+it}\log n}\Big{|}\bigg{)}.$

Note that unconditionally $V_{\max}\leqslant(2+o(1))X^{1/2}/\log X$ and that
on the Riemann hypothesis $V_{\max}\leqslant(1/2+o(1))(\log(X^{1/2}/\log
T)+O(\log_{2}X))\log T/\log_{2}T$ by Theorem 3. The reason we work with the
real and imaginary parts (as opposed to working with the modulus directly) is
so that we have slightly better conditional bounds for $V_{\max}$ whilst still
maintaining control over the modulus (which is important for Lemma 5 below).
This gives better exponents for our logarithms in the conditional results but
entails slightly more work.

To the estimate the measure of the complementary sets we use the following.

######  Lemma 2 ([43]).

Let $T$ be large and let $2\leqslant x\leqslant T$. Let $m$ be a natural
number such that $x^{m}\leqslant T/\log T$. Then for any complex numbers
$a(p)$ we have

$\frac{1}{T}\int_{T}^{2T}\bigg{|}\sum_{p\leqslant
x}\frac{a(p)}{p^{1/2+it}}\bigg{|}^{2m}dt\ll m!\bigg{(}\sum_{p\leqslant
x}\frac{|a(p)|^{2}}{p}\bigg{)}^{m}.$

######  Lemma 3.

Let $\epsilon,\kappa>0$. Then for $X\leqslant\eta_{\kappa}(\log
T)^{2}(\log_{2}T)^{2}$ with $\eta_{\kappa}=\tfrac{1}{16\kappa^{2}}-\epsilon$
we have

$\qquad\qquad\qquad\qquad\qquad\qquad\mu(\mathcal{E}_{\Re}(V))\ll
Te^{-(2\kappa+o(1))V},\qquad\qquad V_{0}\leqslant V\leqslant V_{\max}$

where $\mu$ denotes Lebesgue measure. Assuming RH, the same bound holds
provided $X\leqslant(\log T)^{\theta_{\kappa}-\epsilon}$ where

(13) $\theta_{\kappa}=2\sqrt{1+\tfrac{1}{2\kappa}}.$

The same results hold for $\mathcal{E}_{\Im}(V)$ also.

###### Proof.

We first prove the unconditional result. Write

$\sum_{n\leqslant X}\frac{\Lambda(n)}{n^{1/2+it}\log n}=\sum_{p\leqslant
X}\frac{1}{p^{1/2+it}}+\sum_{p\leqslant\sqrt{X}}\frac{1}{2p^{1+2it}}+O(1).$

Then from Jensen’s inequality in the form $(a+b+c)^{2m}\leqslant
3^{2m-1}(a^{2m}+b^{2m}+c^{2m})$ with $m\geqslant 1$, we have

$\mu(\mathcal{E}_{\Re}(V))\leqslant\frac{3^{2m-1}}{V^{2m}}\bigg{(}\int_{T}^{2T}\Big{|}\sum_{p\leqslant
X}\frac{1}{p^{1/2+it}}\Big{|}^{2m}dt+\int_{T}^{2T}\Big{|}\sum_{p\leqslant\sqrt{X}}\frac{1}{2p^{1+2it}}\Big{|}^{2m}dt+O(TC^{2m})\bigg{)}.$

By Lemma 2 the right hand side is

(14) $\ll T\frac{3^{2m}m!}{V^{2m}}\Big{(}\sum_{p\leqslant
X}\frac{1}{p}\Big{)}^{m}\ll Tm^{1/2}\Big{(}\frac{9m\log\log
X}{eV^{2}}\Big{)}^{m}$

provided $m\leqslant(\log T-\log_{2}T)/\log X$. Choosing

$m=\frac{2\kappa V}{\log V}$

the bound $\ll Te^{-(2\kappa+o(1))V}$ follows since $\log\log X\leqslant
V^{o(1)}$. Note that our choice of $m$ is legal since

$m\leqslant\frac{2\kappa V_{\max}}{\log V_{\max}}\lesssim\frac{8\kappa
X^{1/2}}{(\log X)^{2}}\lesssim\frac{4\kappa\eta_{k}^{1/2}\log T}{\log
X}\leqslant(1-\epsilon)\frac{\log T}{\log X}$

where $\lesssim$ means $\leqslant$ times a constant of the form $1+o(1)$.

Assuming RH, then on writing $\theta_{\kappa}=2+\nu_{\kappa}$ we find

$V_{\max}\leqslant(\tfrac{1}{2}+o(1))\big{(}\log(\tfrac{X^{1/2}}{\log
T})+4\log\log X\big{)}\frac{\log T}{\log\log
T}\lesssim\frac{\nu_{\kappa}-\epsilon}{4}\log T$

by Theorem 3. Choosing $m$ as before gives the desired bound for
$\mu(\mathcal{E}_{\Re}(V))$ and, again, our choice of $m$ is legal since in
this case we have

$m\lesssim\frac{\kappa(\nu_{\kappa}-\epsilon)\log T}{2\log\log T}$

which is $\leqslant(\log T-\log_{2}T)/\log X$ provided

$2+\nu_{\kappa}\leqslant\frac{2}{\kappa\nu_{\kappa}}\,\,\iff\,\,\nu_{\kappa}\leqslant
2\sqrt{1+\tfrac{1}{2\kappa}}-2.$

∎

######  Lemma 4.

Let $\epsilon>0$, $\nu\in\mathbb{R}$ and suppose $X\leqslant\eta_{\nu}(\log
T)^{2}(\log_{2}T)^{2}$ with $\eta_{\nu}=\tfrac{1}{16\nu^{2}}-\epsilon$. Then

$\frac{1}{T}\int_{\mathcal{E}}|P_{X}(\tfrac{1}{2}+it)|^{2\nu}dt\ll e^{-\delta
V_{0}}$

for some $\delta>0$ dependent on $\epsilon$. Assuming RH, this holds provided
$X\leqslant(\log T)^{\theta_{\nu}-\epsilon}$ where $\theta_{\nu}$ is given by
(13).

###### Proof.

Since $A\cup B=A\cup(\overline{A}\cap B)$ we may write

(15)
$\mathcal{E}=\mathcal{E}_{\Re}(V_{0})\cup\mathcal{E}_{\Im}(V_{0})=\mathcal{E}_{\Re}(V_{0})\cup(\mathcal{S}_{\Re}(V_{0})\cap\mathcal{E}_{\Im}(V_{0})).$

The integral over $\mathcal{E}_{\Re}(V_{0})$ is

(16)
$\begin{split}\leqslant&\frac{1}{T}\int_{\mathcal{E}_{\Re}(V_{0})}\exp\Big{(}2|\nu|\cdot\Big{|}\Re\sum_{n\leqslant
X}\frac{\Lambda(n)}{n^{1/2+it}\log n}\Big{|}\Big{)}dt\\\
=&\frac{1}{T}e^{2|\nu|V_{0}}\mu(\mathcal{E}_{\Re}(V_{0}))+\frac{2|\nu|}{T}\int_{V_{0}}^{V_{\max}}e^{2|\nu|V}\mu(\mathcal{E}_{\Re}(V))dV.\end{split}$

Since $\eta_{\nu}\leqslant 1/16(|\nu|+\delta/2)^{2}-\epsilon/2$ for some
$\delta>0$ dependent on $\epsilon$, Lemma 3 gives
$\mu(\mathcal{E}_{\Re}(V))\ll Te^{-(2|\nu|+\delta)V}$. Hence the quantity in
(16) is $\ll e^{-\delta V_{0}}$. Similarly, the integral over
$\mathcal{S}_{\Re}(V_{0})\cap\mathcal{E}_{\Im}(V_{0})$ is

$\leqslant\frac{1}{T}e^{2|\nu|V_{0}}\mu(\mathcal{E}_{\Im}(V_{0}))\leqslant
e^{-\delta V_{0}}.$

The result then follows by the union bound. Likewise,
$\theta_{\nu}-\epsilon\leqslant\theta_{\nu+\delta/2}-\epsilon/2$ and so the
same argument gives the conditional case. ∎

We now show that $P_{X}(1/2+it)^{k}$ can be approximated by a short Dirichlet
polynomial provided $t\in\mathcal{S}$. To do this we note that for
$|z|\leqslant Z$ we have

(17) $\Big{|}e^{z}-\sum_{j=0}^{10Z}\frac{z^{j}}{j!}\Big{|}\leqslant e^{-10Z}$

by a Taylor expansion and Stirling’s formula.

######  Lemma 5.

Suppose $t\in\mathcal{S}$. Let $k\in\mathbb{R}$ and define

$W_{0}=W_{0}(k,T)=20|k|V_{0}.$

Then as $X\to\infty$,

(18)
$\begin{split}P_{X}(\tfrac{1}{2}+it)^{k}=&\,\,\big{(}1+O(e^{-19|k|V_{0}})\big{)}D(t,k)\end{split}$

where

(19) $D(t,k)=\sum_{\begin{subarray}{c}n\in
S(X)\end{subarray}}\frac{\alpha_{k}(n)}{n^{1/2+it}}$

with $S(X)=\big{\\{}n\in\mathbb{N}:p|n\implies p\leqslant X\big{\\}}$ and
where the coefficients $\alpha_{k}(n)$ satisfy the following properties:

  * •

$\alpha_{k}(n)$ is supported on integers $n\leqslant X^{W_{0}}$ and

$|\alpha_{k}(n)|\leqslant d_{|k|}(n).$

  * •

If $\,\,\Omega(n)\leqslant W_{0}$ then $\alpha_{k}(n)=\beta_{k}(n)$ where
$\beta_{k}(n)$ is a multiplicative function satisfying

$\beta_{k}(n)=d_{k}(n)$

if $p^{m}|n\implies p^{m}\leqslant X$ and $|\beta_{k}(n)|\leqslant d_{|k|}(n)$
in general.

###### Proof.

Since $t\in\mathcal{S}$ we have

$\Big{|}\sum_{\ell\leqslant
X}\frac{\Lambda(\ell)}{\ell^{1/2+it}\log\ell}\Big{|}\leqslant
2\log_{2}T\log_{3}T,$

and so by (17) we acquire

(20)
$P_{X}(\tfrac{1}{2}+it)^{k}=\big{(}1+O(e^{-19|k|V_{0}})\big{)}\sum_{j=0}^{W_{0}}\frac{k^{j}}{j!}\bigg{(}\sum_{\ell\leqslant
X}\frac{\Lambda(\ell)}{\ell^{1/2+it}\log\ell}\bigg{)}^{j}.$

Writing the sum on the right as the Dirichlet polynomial $D(t,k)$ it remains
to deduce the properties of the coefficients $\alpha_{k}(n)$.

Clearly, this is a Dirichlet polynomial of length $X^{W_{0}}$ over the
$X$-smooth numbers $S(X)$. For the remaining properties, first note that we
may write

(21) $\exp\Big{(}k\sum_{\ell\leqslant
X}\frac{\Lambda(\ell)}{\ell^{s}\log\ell}\Big{)}=\prod_{p\leqslant
X}\exp\Big{(}-k\log(1-p^{-s})-k\sum_{m:\,\,p^{m}>X}\frac{1}{mp^{ms}}\Big{)}.$

Note also that after performing a Taylor expansion of the left hand side and
collecting like terms for $n^{s}$, the coefficients are a sum of positive
terms if $k>0$, whilst they can be bounded from above by the same sum but
involving $|k|$ if $k<0$. From these observations it is clear that
$|\alpha_{k}(n)|\leqslant d_{|k|}(n)$ since the right hand side is the
generating function for the divisor functions $d_{k}(n)$ with some terms
removed.

Moreover, if we form the product on the right hand side of (21) into a series
$\sum_{n\in S(X)}\beta_{k}(n)n^{-s}$, then we see that the coefficients are
multiplicative and satisfy $\beta_{k}(n)=d_{k}(n)$ if $p^{m}|n\implies
p^{m}\leqslant X$. Since the highest power $j$ in (20) is $W_{0}$ we see that,
certainly, if $\Omega(n)\leqslant W_{0}$ then $\alpha_{k}(n)=\beta_{k}(n)$. ∎

###### Proof of Proposition 1.

Write

$\frac{1}{T}\int_{T}^{2T}|P_{X}(\tfrac{1}{2}+it)|^{2k}dt=\frac{1}{T}\int_{\mathcal{S}}|P_{X}(\tfrac{1}{2}+it)|^{2k}dt+\frac{1}{T}\int_{\mathcal{E}}|P_{X}(\tfrac{1}{2}+it)|^{2k}dt.$

By Lemma 4, the second integral here is $o(1)$. For the first integral, Lemma
5 gives

$\displaystyle\frac{1}{T}\int_{\mathcal{S}}|P_{X}(\tfrac{1}{2}+it)|^{2k}dt\sim$
$\displaystyle\frac{1}{T}\int_{\mathcal{S}}|D(t,k)|^{2}dt$ $\displaystyle=$
$\displaystyle\frac{1}{T}\int_{T}^{2T}|D(t,k)|^{2}dt+O\bigg{(}\frac{1}{T}\int_{\mathcal{E}}|D(t,k)|^{2}dt\bigg{)}$

By the Cauchy–Schwarz inequality the error term here is

(22)
$\ll\Big{(}\frac{\mu(\mathcal{E})}{T}\Big{)}^{1/2}\bigg{(}\frac{1}{T}\int_{T}^{2T}|D(t,k)|^{4}dt\bigg{)}^{1/2}.$

Since $D(t,k)$ is of length $X^{W_{0}}\leqslant
e^{C(\log_{2}T)^{2}\log_{3}T}$, the Montgomery–Vaughan mean value Theorem (see
(39) below) and the coefficient bounds of Lemma 5 give

$\displaystyle\frac{1}{T}\int_{T}^{2T}|D(t,k)|^{4}dt=$
$\displaystyle\,(1+o(1))\sum_{\begin{subarray}{c}n_{1}n_{2}=n_{3}n_{4}\\\
n_{j}\in
S(X)\end{subarray}}\frac{\alpha_{k}(n_{1})\alpha_{k}(n_{2})\alpha_{k}(n_{3})\alpha_{k}(n_{4})}{(n_{1}n_{2}n_{3}n_{4})^{1/2}}$
(23) $\displaystyle\ll$ $\displaystyle\prod_{p\leqslant
X}\bigg{(}1+\frac{4k^{2}}{p}+O(p^{-2})\bigg{)}\ll(\log X)^{4k^{2}}.$

More generally, for fixed $m\in\mathbb{N}$ we have

(24) $\frac{1}{T}\int_{T}^{2T}|D(t,k)|^{2m}dt\ll(\log X)^{m^{2}k^{2}}.$

Therefore, by Lemma 3 the expression in (22) is

$\ll e^{-|k|(\log_{2}T)(\log_{3}T)}(\log X)^{2k^{2}}=o(1)$

and hence we may concentrate on the integral of $|D(t,k)|^{2}$ over the full
set $[T,2T]$.

Applying the Montgomery–Vaughan mean value theorem again gives

$\frac{1}{T}\int_{T}^{2T}|D(t,k)|^{2}dt=(1+o(1))\sum_{n\in
S(X)}\frac{\alpha_{k}(n)^{2}}{n}.$

Since $|\alpha_{k}(n)|\leqslant d_{|k|}(n)$, the sum over terms with
$\Omega(n)>W_{0}$ is, for any $1<r<2$,

(25) $\sum_{\begin{subarray}{c}n\in S(X)\\\
\Omega(n)>W_{0}\end{subarray}}\frac{\alpha_{k}(n)^{2}}{n}\ll
r^{-W_{0}}\sum_{\begin{subarray}{c}n\in
S(X)\end{subarray}}\frac{d_{|k|}(n)^{2}r^{\Omega(n)}}{n}\ll r^{-W_{0}}(\log
X)^{rk^{2}}=o(1)$

where in the first inequality we have applied Rankin’s trick in the form
$r^{\Omega(n)-W_{0}}\geqslant 1$.

Since $\alpha_{k}(n)=\beta_{k}(n)$ if $\Omega(n)\leqslant W_{0}$ the main term
is

$\displaystyle\sum_{\begin{subarray}{c}n\in S(X)\\\ \Omega(n)\leqslant
W_{0}\end{subarray}}\frac{\alpha_{k}(n)^{2}}{n}=$
$\displaystyle\sum_{\begin{subarray}{c}n\in
S(X)\end{subarray}}\frac{\beta_{k}(n)^{2}}{n}+O\bigg{(}\sum_{\begin{subarray}{c}n\in
S(X)\\\ \Omega(n)>W_{0}\end{subarray}}\frac{|\beta_{k}(n)|^{2}}{n}\bigg{)}.$

The bound $|\beta_{k}(n)|\leqslant d_{|k|}(n)$ and the same analysis as in
(25) shows that this error term is $o(1)$. From the properties of
$\beta_{k}(n)$ we get

$\displaystyle\sum_{\begin{subarray}{c}n\in
S(X)\end{subarray}}\frac{\beta_{k}(n)^{2}}{n}=$
$\displaystyle\prod_{p\leqslant X}\bigg{(}\sum_{m:\,p^{m}\leqslant
X}\frac{d_{k}(p^{m})^{2}}{p^{m}}+\sum_{m:\,p^{m}>X}\frac{\beta_{k}(p^{m})^{2}}{p^{m}}\bigg{)}$
$\displaystyle=$ $\displaystyle\prod_{p\leqslant X}\sum_{m\geqslant
0}\frac{d_{k}(p^{m})^{2}}{p^{m}}\prod_{p\leqslant
X}\bigg{(}1+O\Big{(}\sum_{m:\,p^{m}>X}\frac{d_{|k|}(p^{m})^{2}}{p^{m}}\Big{)}\bigg{)}.$

We split the second product at $\sqrt{X}$ and apply the bound $d_{k}(n)\ll
n^{\epsilon}$ to find that it is

(26)
$\prod_{p\leqslant\sqrt{X}}\Big{(}1+O\Big{(}\frac{1}{X^{1-\epsilon}}\Big{)}\Big{)}\prod_{\sqrt{X}<p\leqslant
X}\Big{(}1+O\Big{(}\frac{1}{p^{2-\epsilon}}\Big{)}\Big{)}=1+O(X^{-1/2+\epsilon})$

by the prime number theorem. Then by Mertens’ theorem we have

$\prod_{p\leqslant X}\sum_{m\geqslant 0}\frac{d_{k}(p^{m})^{2}}{p^{m}}\sim
a(k)(e^{\gamma}\log X)^{k^{2}}$

since $a(k)$ is an absolutely convergent product. ∎

##  4\. Asymptotics for the 2nd and 4th moments of the Hadamard product: Proof
of Proposition 2

From (3) we have

$Z_{X}(\tfrac{1}{2}+it)=\zeta(\tfrac{1}{2}+it)P_{X}(\tfrac{1}{2}+it)^{-1}\big{(}1+O(1/\log
X)\big{)}$

and thus it suffices to consider the second and fourth moment of the object on
the right. Our aim is to first replace $P_{X}$ by its Dirichlet polynomial
approximation and then apply formulas for the twisted second and fourth
moments of the zeta function.

###  4.1. The second moment

As before, we decompose the integral as

(27)
$\frac{1}{T}\int_{T}^{2T}|Z_{X}(\tfrac{1}{2}+it)|^{2}dt=\frac{1}{T}\int_{\mathcal{S}}|Z_{X}(\tfrac{1}{2}+it)|^{2}dt+\frac{1}{T}\int_{\mathcal{E}}|Z_{X}(\tfrac{1}{2}+it)|^{2}dt.$

Working unconditionally first, we apply the Cauchy–Schwarz inequality to the
integral over $\mathcal{E}$ to find it is

$\ll(\log
T)^{2}\bigg{(}\frac{1}{T}\int_{\mathcal{E}}|P_{X}(\tfrac{1}{2}+it)^{-1}|^{4}dt\bigg{)}^{1/2}$

using Ingham’s asymptotic for the fourth moment. Since
$X\leqslant\tfrac{1}{10^{4}}(\log T)^{2}(\log_{2}T)^{2}$ and
$1/10^{4}<\eta_{2}$ we find that this is $\ll(\log T)^{2}e^{-\delta
V_{0}}=o(1)$ by Lemma 4.

If we assume RH we can apply Hölder’s inequality in the form

(28)
$\begin{split}\frac{1}{T}\int_{\mathcal{E}}&|Z_{X}(\tfrac{1}{2}+it)|^{2}dt\\\
\ll&\bigg{(}\frac{1}{T}\int_{T}^{2T}|\zeta(\tfrac{1}{2}+it)|^{2(1+{1}/{\epsilon})}dt\bigg{)}^{\tfrac{\epsilon}{1+\epsilon}}\bigg{(}\frac{1}{T}\int_{\mathcal{E}}|P_{X}(\tfrac{1}{2}+it)|^{-2(1+{\epsilon})}dt\bigg{)}^{\tfrac{1}{1+\epsilon}}\end{split}$

for some $\epsilon>0$. The first term on the right is $\ll(\log
T)^{4(1+1/\epsilon)}$ by Harper’s [24] conditional bound
$\int_{T}^{2T}|\zeta(1/2+it)|^{2k}\ll T(\log T)^{k^{2}}$. Since
$X\leqslant(\log T)^{\sqrt{6}-\epsilon^{\prime}}$ and
$\sqrt{6}-\epsilon^{\prime}\leqslant\theta_{1+\epsilon}-\epsilon^{\prime}/2$
on choosing $\epsilon$ small enough, the second term on the right is $\ll
e^{-\delta V_{0}}$ by Lemma 4. Therefore, the quantity in (28) is $o(1)$.

By Lemma 5 we have

$\displaystyle\frac{1}{T}\int_{\mathcal{S}}|Z_{X}(\tfrac{1}{2}+it)|^{2}dt\sim$
$\displaystyle\frac{1}{T}\int_{\mathcal{S}}|\zeta(\tfrac{1}{2}+it)|^{2}|D(t,-1)|^{2}dt$

which we write as

$\frac{1}{T}\int_{T}^{2T}|\zeta(\tfrac{1}{2}+it)|^{2}|D(t,-1)|^{2}dt+O\bigg{(}\frac{1}{T}\int_{\mathcal{E}}|\zeta(\tfrac{1}{2}+it)|^{2}|D(t,-1)|^{2}dt\bigg{)}.$

Applying the Cauchy–Schwarz inequality twice along with Ingham’s fourth moment
bound, Lemma 3 and (24) we find that the error term here is $o(1)$.

It remains to show that

$I_{1}:=\frac{1}{T}\int_{T}^{2T}|\zeta(\tfrac{1}{2}+it)|^{2}|D(t,-1)|^{2}dt\sim\frac{\log
T}{e^{\gamma}\log X}.$

The mean square of the zeta function times an arbitrary Dirichlet polynomial
has been computed before e.g. see [5, 7]. From there we see that

$I_{1}=\sum_{m,n\in
S(X)}\frac{\alpha_{-1}(m)\alpha_{-1}(n)(m,n)}{mn}\log\Big{(}\frac{BT(m,n)^{2}}{mn}\Big{)}+o(1)$

for some constant $B$. On applying the bound $\alpha_{-1}(n)\ll d_{1}(n)\ll 1$
and following the argument in the proof of Theorem 3, pg. 530, of [22] we
easily see that

$\sum_{m,n\in
S(X)}\frac{\alpha_{-1}(m)\alpha_{-1}(n)(m,n)}{mn}\log\Big{(}\frac{B(m,n)^{2}}{mn}\Big{)}\ll(\log
X)^{10}$

and thus we can consider the remaining sum.

As in the previous section, we first estimate the sum over integers for which
$\Omega(m),\Omega(n)>W_{0}$ (which equals $10\log_{2}T\log_{3}T$ in this
case). Applying the bound $\alpha_{-1}(n)\ll 1$ leads to an error of size

$\ll\log T\sum_{\begin{subarray}{c}m,n\in S(X)\\\
\Omega(m)>W_{0}\end{subarray}}\frac{(m,n)}{mn}\ll(\log
T)r^{-W_{0}}\sum_{\begin{subarray}{c}m,n\in
S(X)\end{subarray}}\frac{(m,n)r^{\Omega(m)}}{mn}$

for any $1<r<2$. A short computation shows this last sum is

$\prod_{p\leqslant X}\Big{(}1+{(2r+1)}{p^{-1}}+O(p^{-2})\Big{)}\ll(\log
X)^{2r+1}$

and so the terms with $\Omega(m),\Omega(n)>W_{0}$ contribute an error of size
$o(1)$.

In the main term we replace $\alpha_{-1}(n)$ with $\beta_{-1}(n)$ and then re-
extend the sum to include those integers for which
$\Omega(m),\Omega(n)>W_{0}$. By the bounds $\beta_{-1}(n)\ll d_{1}(n)\ll 1$,
the same argument shows that this introduces an error of $o(1)$. Thus,

$I_{1}=\log T\sum_{m,n\in
S(X)}\frac{\beta_{-1}(m)\beta_{-1}(n)(m,n)}{mn}+O((\log X)^{10}).$

By symmetry and the properties of $\beta_{k}(n)$ we find that the sum is

$\displaystyle\prod_{p\leqslant
X}\bigg{(}\sum_{\begin{subarray}{c}m,n\geqslant 0\\\ p^{m},\,\,p^{n}\leqslant
X\end{subarray}}\frac{\mu(p^{m})\mu(p^{n})}{p^{m+n-\min(m,n)}}+O\Big{(}\sum_{\begin{subarray}{c}m,n\geqslant
0\\\ p^{m}>X\end{subarray}}\frac{1}{p^{m+n-\min(m,n)}}\Big{)}\bigg{)}$
$\displaystyle=$ $\displaystyle\prod_{p\leqslant
X}\Big{(}1-\frac{1}{p}\Big{)}\prod_{p\leqslant
X}\bigg{(}1+O\Big{(}\sum_{\begin{subarray}{c}m,n\geqslant 0\\\
p^{m}>X\end{subarray}}\frac{1}{p^{m+n-\min(m,n)}}\Big{)}.\bigg{)}$

To estimate the second product we note that the sum in the error term is
$\ll\sum_{p^{m}>X}(m+1)p^{-m}\ll(\log X)p^{-\lceil\log X/\log p\rceil}$ and
then split the product at $\sqrt{X}$, as before. In this way we find it is
$1+O(X^{-1/2+\epsilon})$ and therefore by Mertens’ Theorem

$I_{1}\sim\frac{\log T}{e^{\gamma}\log X}$

as desired.

###  4.2. The fourth moment: Initial clearing

Not surprisingly, the fourth moment requires more work in both the initial
stages and the arithmetic computations. Our aim is to show that

$I_{2}:=\frac{1}{T}\int_{T}^{2T}|\zeta(\tfrac{1}{2}+it)|^{4}|P_{X}(\tfrac{1}{2}+it)|^{-4}dt\sim\frac{1}{12}\bigg{(}\frac{\log
T}{e^{\gamma}\log X}\bigg{)}^{4}.$

In this subsection the goal is to replace $P_{X}(1/2+it)^{-2}$ by $D(t,-2)$.

Splitting the integral as in (27) we see that our first task is to bound

(29)
$\frac{1}{T}\int_{\mathcal{E}}|\zeta(\tfrac{1}{2}+it)|^{4}|P_{X}(\tfrac{1}{2}+it)|^{-4}dt.$

On RH we can deal with this by applying Hölder’s inequality as in (28).
Following the same argument and using the fact that
$\sqrt{5}-\epsilon^{\prime}\leqslant\theta_{2(1+\epsilon)}-\epsilon^{\prime}/2$
shows that this is $o(1)$.

To bound this unconditionally requires more work. First note that since

$\mathcal{E}\subset\mathcal{E}^{\prime}:=\Big{\\{}t\in[T,2T]:\Big{|}\sum_{n\leqslant
X}\frac{\Lambda(n)}{n^{1/2+it}\log n}\Big{|}\geqslant V_{0}\Big{\\}}$

we can upper bound by the integral over $\mathcal{E}^{\prime}$. Let
$V_{j}=e^{j}V_{0}$ and define $\mathcal{J}$ to be the maximal $j$ such that
$V_{\mathcal{J}}\leqslant V_{\max}$. Let

$\mathcal{E}_{j}=\Big{\\{}t\in[T,2T]:V_{j}\leqslant\Big{|}\sum_{n\leqslant
X}\frac{\Lambda(n)}{n^{1/2+it}\log n}\Big{|}\leqslant V_{j+1}\Big{\\}}$

so that

$\mathcal{E}^{\prime}=\cup_{j=0}^{\mathcal{J}}\mathcal{E}_{j}.$

Then

(30)
$\begin{split}&\frac{1}{T}\int_{\mathcal{E}^{\prime}}|\zeta(\tfrac{1}{2}+it)|^{4}|P_{X}(\tfrac{1}{2}+it)|^{-4}dt\leqslant\sum_{j=0}^{\mathcal{J}}e^{4V_{j+1}}\frac{1}{T}\int_{\mathcal{E}_{j}}|\zeta(\tfrac{1}{2}+it)|^{4}dt\\\
\leqslant&\sum_{j=0}^{\mathcal{J}}e^{4V_{j+1}}V_{j}^{-2r_{j}}\frac{1}{T}\int_{T}^{2T}|\zeta(\tfrac{1}{2}+it)|^{4}\Big{|}\sum_{n\leqslant
X}\frac{\Lambda(n)}{n^{1/2+it}\log n}\Big{|}^{2r_{j}}dt\end{split}$

for any given integer $r_{j}\geqslant 0$. The combinatorics are simplified if
we focus on the prime sums so we apply Jensen’s inequality in the form

$\Big{|}\sum_{n\leqslant X}\frac{\Lambda(n)}{n^{1/2+it}\log
n}\Big{|}^{2r_{j}}\leqslant 9^{r_{j}}\bigg{(}\Big{|}\sum_{p\leqslant
X}\frac{1}{p^{1/2+it}}\Big{|}^{2r_{j}}+\Big{|}\sum_{p\leqslant\sqrt{X}}\frac{1}{2p^{1+2it}}\Big{|}^{2r_{j}}+O(C^{2r_{j}})\bigg{)}.$

It will be clear after the computations that the first sum here gives the
dominant contribution and so we focus on this. Note that by the multinomial
theorem,

$\Big{(}\sum_{p\leqslant
X}\frac{1}{p^{1/2+it}}\Big{)}^{r_{j}}=r_{j}!\sum_{\begin{subarray}{c}n\leqslant
X^{r_{j}}\\\ \Omega(n)=r_{j}\end{subarray}}\frac{\mathfrak{g}(n)}{n^{1/2+it}}$

where $\mathfrak{g}(n)$ is the multiplicative function satisfying
$\mathfrak{g}(p^{\alpha})=1/\alpha!$. Accordingly, (30) is

(31)
$\ll\sum_{j=0}^{\mathcal{J}}e^{4V_{j+1}}9^{r_{j}}V_{j}^{-2r_{j}}\cdot\frac{1}{T}\int_{T}^{2T}|\zeta(\tfrac{1}{2}+it)|^{4}\Big{|}r_{j}!\sum_{\begin{subarray}{c}n\leqslant
X^{r_{j}}\\\
\Omega(n)=r_{j}\end{subarray}}\frac{\mathfrak{g}(n)}{n^{1/2+it}}\Big{|}^{2}dt$

The twisted fourth moment of the zeta function has been computed before [6,
31] and has been applied in similar situations [26]. Provided
$X^{r_{j}}\leqslant T^{1/4-\epsilon}$ i.e.

(32) $r_{j}\leqslant(1/4-\epsilon)\frac{\log T}{\log X},$

Proposition 4 and formula (8) of [26] (see section 6 there) give

$\frac{1}{T}\int_{T}^{2T}|\zeta(\tfrac{1}{2}+it)|^{4}\Big{|}r_{j}!\sum_{\begin{subarray}{c}n\leqslant
X^{r_{j}}\\\
\Omega(n)=r_{j}\end{subarray}}\frac{\mathfrak{g}(n)}{n^{1/2+it}}\Big{|}^{2}dt\ll(\log
T)^{4}9^{r_{j}}\sum_{\begin{subarray}{c}m,n\leqslant X^{r_{j}}\\\
\Omega(m)=\Omega(n)=r_{j}\end{subarray}}\frac{r_{j}!^{2}\mathfrak{g}(n)\mathfrak{g}(m)}{[n,m]}.$

Following the arguments of [26] which lead to formula (9) there, we find that
this is

(33) $\ll(\log T)^{4}9^{r_{j}}r_{j}!\Big{(}\sum_{p\leqslant
X}\frac{1}{p}\Big{)}^{r_{j}}\exp\Big{(}\sum_{p\leqslant
X}\frac{1}{p}\Big{)}\ll(\log T)^{5}r_{j}^{1/2}\Big{(}\frac{9r_{j}\log\log
X}{e}\Big{)}^{r_{j}}.$

We choose

$r_{j}=\frac{12V_{j}}{\log V_{j}}\leqslant\frac{12V_{\max}}{\log
V_{\max}}\lesssim\frac{48X^{1/2}}{(\log X)^{2}}\lesssim\frac{24\log T}{100\log
X}\qquad\forall j$

since $X\leqslant\tfrac{1}{10^{4}}(\log T)^{2}(\log_{2}T)^{2}$. Clearly, this
satisfies (32). Then applying (33) in (31) we find that

$\displaystyle\frac{1}{T}\int_{\mathcal{E}^{\prime}}$
$\displaystyle|\zeta(\tfrac{1}{2}+it)|^{4}|P_{X}(\tfrac{1}{2}+it)|^{-4}dt$
$\displaystyle\ll$ $\displaystyle(\log
T)^{5}\sum_{j=0}^{\mathcal{J}}e^{4V_{j+1}}\Big{(}\frac{C\log\log X}{V_{j}\log
V_{j}}\Big{)}^{12V_{j}/\log V_{j}}$ $\displaystyle\ll$ $\displaystyle(\log
T)^{5}\sum_{j=0}^{\mathcal{J}}e^{4V_{j+1}-(12-o(1))V_{j}}\ll(\log
T)^{5}\sum_{j=0}^{\mathcal{J}}e^{-e^{j}V_{0}(12-4e+o(1))}=o(1).$

We have therefore arrived at

$I_{2}=(1+o(1))\frac{1}{T}\int_{\mathcal{S}}|\zeta(\tfrac{1}{2}+it)|^{4}|D(t,-2)|^{2}dt+o(1)$

on applying Lemma 5 in the integral over $\mathcal{S}$. After extending to the
full range of integration $[T,2T]$ it remains to estimate

$\frac{1}{T}\int_{\mathcal{E}}|\zeta(\tfrac{1}{2}+it)|^{4}|D(t,-2)|^{2}dt.$

However, from the definition of $D(t,k)$ we have

$|D(t,k)|\leqslant\sum_{j=0}^{W_{0}}\frac{|k|^{j}}{j!}\Big{|}\sum_{n\leqslant
X}\frac{\Lambda(n)}{n^{1/2+it}\log
n}\Big{|}^{j}\leqslant\exp\Big{(}\Big{|}k\sum_{n\leqslant
X}\frac{\Lambda(n)}{n^{1/2+it}\log n}\Big{|}\Big{)}.$

and so we can apply the same argument as above to acquire the bound $o(1)$ for
this integral. Thus we have

$I_{2}\sim J_{2}$

where

$J_{2}:=\frac{1}{T}\int_{T}^{2T}|\zeta(\tfrac{1}{2}+it)|^{4}|D(t,-2)|^{2}dt.$

###  4.3. The fourth moment: Arithmetic computations

In this section our aim is to show that

$J_{2}\sim\frac{1}{12}\bigg{(}\frac{\log T}{e^{\gamma}\log X}\bigg{)}^{4}.$

The formulas for the twisted fourth moment of the zeta function given in the
literature [6, 31] apply to smoothed integrals and accordingly we must first
smooth $J_{2}$. Let $\Phi_{-},\Phi_{+}$ be smooth approximations of compact
support satisfying

(34) $\Phi_{-}(t)\leqslant\mathds{1}_{t\in[T,2T]}\leqslant\Phi_{+}(t)$

with derivatives $\Phi_{\pm}^{(j)}(t)\ll T^{\epsilon}$. For example, we may
take $\Phi_{-}$ to be compactly supported on $[1,2]$ and equal to $1$ on the
interval $[1+T^{-\epsilon},2-T^{-\epsilon}]$ with smooth, monotonic decay to
zero at each endpoint. Then, on letting $\Phi$ be either $\Phi_{-}$ or
$\Phi_{+}$ we consider the smoothed integral

$J_{2,\Phi}:=\frac{1}{T}\int_{\mathbb{R}}\Phi\Big{(}\frac{t}{T}\Big{)}|\zeta(\tfrac{1}{2}+it)|^{4}|D(t,-2)|^{2}dt.$

We note that the error incurred from these approximations will be $\ll
T^{1-\epsilon}$ which is tolerable given the asymptotic we seek.

######  Theorem 4 (Theorem 1.2 of [6]).

Let $\Phi$ be as above and $\alpha_{1},\alpha_{2},\alpha_{3},\alpha_{4}\ll
1/\log T$. Let $\Xi$ be the subgroup of $S_{4}$ consisting of the identity,
those permutations which swap just one element of $\\{1,2\\}$ with $\\{3,4\\}$
and the permutation satisfying $\tau(1)=3,\tau(2)=4$. Then for any Dirichlet
polynomial $\sum_{n\leqslant y}a(n)n^{-s}$ satisfying $y\leqslant
T^{1/4-\epsilon}$ and $a(n)\ll n^{\epsilon}$ we have

$\int_{\mathbb{R}}\zeta(\tfrac{1}{2}+\alpha_{1}+it)\zeta(\tfrac{1}{2}+\alpha_{2}+it)\zeta(\tfrac{1}{2}-\alpha_{3}-it)\zeta(\tfrac{1}{2}-\alpha_{4}-it)\Big{|}\sum_{n\leqslant
y}\frac{a(m)}{m^{1/2+it}}\Big{|}^{2}\Phi\Big{(}\frac{t}{T}\Big{)}dt\\\
=\sum_{m_{1},m_{2}\leqslant
y}a(m_{1})\overline{a(m_{2})}\int_{\mathbb{R}}\Phi\Big{(}\frac{t}{T}\Big{)}\sum_{\tau\in\Xi}\Big{(}\frac{t}{2\pi}\Big{)}^{\sum_{j=1}^{4}\alpha_{\tau(j)}-\alpha_{j}}Z_{\tau(\alpha_{1}),\tau(\alpha_{2}),\tau(\alpha_{3}),\tau(\alpha_{4}),m_{1},m_{2}}dt\\\
+O(T^{1-\epsilon})$

where

$Z_{\alpha_{1},\alpha_{2},\alpha_{3},\alpha_{4},m_{1},m_{2}}=\sum_{m_{1}n_{1}n_{2}=m_{2}n_{3}n_{4}}\frac{1}{(m_{1}m_{2})^{1/2}n_{1}^{1/2+\alpha_{1}}n_{2}^{1/2+\alpha_{2}}n_{3}^{1/2-\alpha_{3}}n_{4}^{1/2-\alpha_{4}}}V\Big{(}\frac{n_{1}n_{2}n_{3}n_{4}}{t^{2}}\Big{)}$

and

$\qquad V(x)=\frac{1}{2\pi
i}\int_{c-i\infty}^{c+i\infty}\frac{G(s)}{s}(4\pi^{2}x)^{-s}ds,\qquad c>0$

with $G(s)$ an even function of rapid decay in vertical strips satisfying
$G(0)=1$.

######  Remark.

We remark that the choice of function $G(s)$ is flexible and it can be
prescribed to have zeros at linear combinations of the shifts. This is fairly
typical and is used to cancel unnecessary poles later on. We will take
$G(s)=Q_{{\alpha}}(s)\exp(s^{2})$ where $Q_{{\alpha}}(s)$ is an even
polynomial which is 1 at $s=0$ and zero at
$2s=\alpha_{3}-\alpha_{1},\alpha_{4}-\alpha_{1},\alpha_{3}-\alpha_{2}$ and
$\alpha_{4}-\alpha_{2}$. Note these conditions on $Q$ imply that for fixed
$\Re(s)$,

(35) $G(s)\ll(\log T)^{4}e^{-\Im(s)^{2}}$

since $\alpha_{j}\ll 1/\log T$.

Let us compute term corresponding to the identity: $\tau=\text{id}$. Denote
this by

$\displaystyle\mathcal{K}=\mathcal{K}_{\alpha}(t,X):=$
$\displaystyle\sum_{m_{1},m_{2}\in
S(X)}\alpha_{-2}(m_{1})\alpha_{-2}(m_{2})Z_{\alpha_{1},\alpha_{2},\alpha_{3},\alpha_{4},m_{1},m_{2}}$
$\displaystyle=$
$\displaystyle\sum_{\begin{subarray}{c}m_{1}n_{1}n_{2}=m_{2}n_{3}n_{4}\\\
m_{j}\in
S(X)\end{subarray}}\frac{\alpha_{-2}(m_{1})\alpha_{-2}(m_{2})}{(m_{1}m_{2})^{1/2}n_{1}^{1/2+\alpha_{1}}n_{2}^{1/2+\alpha_{2}}n_{3}^{1/2-\alpha_{3}}n_{4}^{1/2-\alpha_{4}}}V\Big{(}\frac{n_{1}n_{2}n_{3}n_{4}}{t^{2}}\Big{)}.$

By shifting the contour in $V$ to the either the left or right depending on
whether $x\ll 1$ or $x\gg 1$, respectively, we find that $V(x)\ll(1+|x|)^{-A}$
for any $A>0$. Accordingly, on applying the bounds $\alpha_{k}(m)\ll
m^{\epsilon}\cdot\mathds{1}_{m\leqslant T^{\delta}}$ we may restrict the above
sum to those $n_{j}$ satisfying $n_{1}n_{2}n_{3}n_{4}\ll t^{2+\epsilon}\ll
T^{2+\epsilon}$ at the cost of an error of size $o(1)$. Then the contribution
from those $m_{1}$ with $\Omega(m_{1})>W_{0}$ is for $1<r<2$

$\ll r^{-W_{0}}\sum_{m\leqslant
T^{2+\epsilon}}\frac{r^{\Omega(n)}d_{4}(m)^{2}}{m}\ll r^{-W_{0}}(\log
T)^{16r}=o(1)$

where for the first inequality we have applied Rankin’s trick in the form
$r^{\Omega(m)-W_{0}}\geqslant 1$ along with the bound $\alpha_{-2}(n)\ll
d(n)$. The same bound holds for the sum over $\Omega(m_{2})>W_{0}$. Then on
replacing $\alpha_{-2}(m)$ with $\beta_{-2}(m)$ and re-extending the sums
(which by the same arguments incurs an error of $o(1)$) we have

$\mathcal{K}=\sum_{\begin{subarray}{c}m_{1}n_{1}n_{2}=m_{2}n_{3}n_{4}\\\
m_{j}\in
S(X)\end{subarray}}\frac{\beta_{-2}(m_{1})\beta_{-2}(m_{2})}{(m_{1}m_{2})^{1/2}n_{1}^{1/2+\alpha_{1}}n_{2}^{1/2+\alpha_{2}}n_{3}^{1/2-\alpha_{3}}n_{4}^{1/2-\alpha_{4}}}V\Big{(}\frac{n_{1}n_{2}n_{3}n_{4}}{t^{2}}\Big{)}+o(1).$

Unfolding the integral for $V(x)$ and pushing the sum through we find

(36) $\mathcal{K}=\frac{1}{2\pi
i}\int_{c-i\infty}^{c+i\infty}\mathcal{F}_{\alpha,X}(s)\frac{G(s)}{s}\bigg{(}\frac{t}{2\pi}\bigg{)}^{2s}ds+o(1)$

where

$\displaystyle\mathcal{F}_{\alpha,X}(s)=$
$\displaystyle\sum_{\begin{subarray}{c}m_{1}n_{1}n_{2}=m_{2}n_{3}n_{4}\\\
m_{j}\in
S(X)\end{subarray}}\frac{\beta_{-2}(m_{1})\beta_{-2}(m_{2})}{(m_{1}m_{2})^{1/2}n_{1}^{1/2+\alpha_{1}+s}n_{2}^{1/2+\alpha_{2}+s}n_{3}^{1/2-\alpha_{3}+s}n_{4}^{1/2-\alpha_{4}+s}}$
$\displaystyle=$
$\displaystyle\sum_{\begin{subarray}{c}m_{1}n_{1}=m_{2}n_{2}\\\ m_{j}\in
S(X)\end{subarray}}\frac{\beta_{-2}(m_{1})\beta_{-2}(m_{2})\sigma_{\alpha_{1},\alpha_{2}}(n_{1})\sigma_{-\alpha_{3},-\alpha_{4}}(n_{2})}{(m_{1}m_{2})^{1/2}(n_{1}n_{2})^{1/2+s}}$

with $\sigma_{u,v}(n)=\sum_{d_{1}d_{2}=n}d_{1}^{-u}d_{2}^{-v}$. Expressing
this as an Euler product we have

$\displaystyle\mathcal{F}_{\alpha,X}(s)=$
$\displaystyle\mathcal{A}_{\alpha}(s)\mathcal{G}_{\alpha,X}(s)$

where

$\mathcal{A}_{\alpha}(s)=\\\
\frac{\zeta(1+\alpha_{1}-\alpha_{3}+2s)\zeta(1+\alpha_{1}-\alpha_{4}+2s)\zeta(1+\alpha_{2}-\alpha_{3}+2s)\zeta(1+\alpha_{2}-\alpha_{4}+2s)}{\zeta(2+\alpha_{1}+\alpha_{2}-\alpha_{3}-\alpha_{4}+4s)}$

and

$\mathcal{G}_{\alpha,X}(s)=\prod_{p\leqslant
X}\sum_{\begin{subarray}{c}m_{1}+n_{1}\\\
=m_{2}+n_{2}\end{subarray}}\frac{\beta_{-2}(p^{m_{1}})\beta_{-2}(p^{m_{2}})\sigma_{\alpha_{1},\alpha_{2}}(p^{n_{1}})\sigma_{-\alpha_{3},-\alpha_{4}}(p^{n_{2}})}{p^{\tfrac{1}{2}(m_{1}+m_{2})+(\tfrac{1}{2}+s)(n_{1}+n_{2})}}\\\
/\sum_{n\geqslant
0}\frac{\sigma_{\alpha_{1},\alpha_{2}}(p^{n})\sigma_{-\alpha_{3},-\alpha_{4}}(p^{n})}{p^{n(1+2s)}}.$

Shifting the line of integration in (36) to $\Re(s)=-1/\log X$ we pick up a
simple pole only at $s=0$ (the poles of $A_{\alpha}(s)$ being cancelled by the
zeros of $G(s)$). Since $\beta_{-2}(n)\ll d(n)$ and
$\sigma_{\alpha_{i},\alpha_{j}}(p^{n})\ll p^{n/\log T}d(p^{n})$ we find that
on the new line of integration

$\mathcal{G}_{\alpha,X}(s)\ll(\log X)^{O(1)}.$

Therefore, on combining this with the bound for $G(s)$ given in (35) we see
the integral over the new line is bounded by

$\ll t^{-2/\log X}(\log T)^{O(1)}=o(1)$

since $t\asymp T$. Hence

$\mathcal{K}_{\alpha}=\mathcal{A}_{\alpha}(0)\mathcal{G}_{\alpha,X}(0)+o(1).$

We have satisfactorily computed the contribution from a single $Z$ term and
thus it remains to find the combinatorial sum of these which appears in
Theorem 4. Using the results of [13] we can express this sum as a multiple
contour integral. Precisely, Lemma 2.5.1 there gives

$\sum_{\tau\in\Xi}\Big{(}\frac{t}{2\pi}\Big{)}^{\sum_{j=1}^{4}\alpha_{\tau(j)}-\alpha_{j}}\mathcal{K}_{\tau(\alpha)}\\\
=\frac{1}{4(2\pi i)^{4}}\int_{\begin{subarray}{c}|z_{j}|=3^{j}/\log T\\\
1\leqslant j\leqslant
4\end{subarray}}\frac{\mathcal{A}_{z_{1},z_{2},z_{3},z_{4}}(0)\mathcal{G}_{z_{1},z_{2},z_{3},z_{4},X}(0)\Delta(z_{1},z_{2},z_{3},z_{4})^{2}}{\prod_{i,j=1}^{4}(z_{i}-\alpha_{j})}\\\
\times\Big{(}\frac{t}{2\pi}\Big{)}^{\sum_{j=1}^{2}(z_{j+2}-z_{j})/2}d\underline{z}+o(1)$

where $\Delta(\underline{z})$ denotes the vandermonde determinant. A short
calculation shows that

$\frac{\partial}{\partial
z_{j}}\mathcal{G}_{\underline{z},X}(0)\bigg{|}_{\underline{z}=\underline{0}}\ll\mathcal{G}_{\underline{0},X}(0)\sum_{p\leqslant
X}\frac{\log p}{p}\ll\mathcal{G}_{\underline{0},X}(0)\log X$

and hence we acquire the Taylor expansion

$\mathcal{G}_{z_{1},z_{2},z_{3},z_{4},X}(0)=\mathcal{G}_{\underline{0},X}(0)\Big{(}1+O\Big{(}\log
X\sum_{j=1}^{4}|z_{j}|\Big{)}\Big{)}$

whilst from the Laurent expansion of the zeta function we get

$\mathcal{A}_{z_{1},z_{2},z_{3},z_{4},X}(0)=\frac{1}{\zeta(2)}\prod_{i,j=1}^{2}\frac{1}{(z_{i}-z_{j+2})}\Big{(}1+O\Big{(}\sum_{j=1}^{4}|z_{j}|\Big{)}\Big{)}.$

On setting the shifts $\alpha_{j}$ equal to zero, substituting $z_{j}\mapsto
z_{j}/\tfrac{1}{2}\log(t/2\pi)$, and applying these expansions we find

$J_{2,\Phi}=\int_{\mathbb{R}}\Phi\Big{(}\frac{t}{T}\Big{)}\Big{(}c_{4}\log^{4}(t/2\pi)\cdot\frac{\mathcal{G}_{\underline{0},X}(0)}{\zeta(2)}\Big{(}1+O\Big{(}\frac{\log
X}{\log T}\Big{)}\Big{)}dt+o(T)$

where

$c_{4}=\frac{1}{4\cdot 2^{4}(2\pi
i)^{4}}\int_{\begin{subarray}{c}|z_{j}|=3^{j}\\\ 1\leqslant j\leqslant
4\end{subarray}}\frac{\Delta(z_{1},z_{2},z_{3},z_{4})^{2}}{\prod_{i,j=1}^{2}(z_{i}-z_{j+2})}e^{\sum_{j=1}^{2}z_{j+2}-z_{j}}\prod_{j=1}^{4}\frac{d{z_{j}}}{z_{j}^{4}}.$

From section 2.7 of [13] we know that $c_{4}=g(2)=1/12$ where $g(k)$ is given
by (2). Furthermore,

$\displaystyle\mathcal{G}_{\underline{0},X}(0)=\prod_{p\leqslant
X}\sum_{\begin{subarray}{c}m_{1}+n_{1}\\\
=m_{2}+n_{2}\end{subarray}}\frac{\beta_{-2}(p^{m_{1}})\beta_{-2}(p^{m_{2}})d(p^{n_{1}})d(p^{n_{2}})}{p^{\tfrac{1}{2}(m_{1}+m_{2}+n_{1}+n_{2})}}/\sum_{n\geqslant
0}\frac{d(p^{n})^{2}}{p^{n}}.$

The denominator here is

$\prod_{p\leqslant X}\bigg{(}\sum_{n\geqslant
0}\frac{d(p^{n})^{2}}{p^{n}}\bigg{)}^{-1}=\prod_{p\leqslant
X}\frac{(1-p^{-1})^{4}}{1-p^{-2}}\sim\frac{\zeta(2)}{(e^{\gamma}\log X)^{4}}$

whereas the numerator is

$\displaystyle\mathcal{G}_{\underline{0},X}(0)=$
$\displaystyle\prod_{p\leqslant
X}\bigg{(}1+O\Big{(}\sum_{\begin{subarray}{c}m_{1}+n_{1}\\\ =m_{2}+n_{2}\\\
p^{m_{1}}>X\end{subarray}}\frac{d(p^{m_{1}})d(p^{m_{2}})d(p^{n_{1}})d(p^{n_{2}})}{p^{\tfrac{1}{2}(m_{1}+m_{2}+n_{1}+n_{2})}}\Big{)}\bigg{)}$

since $\beta_{-2}(p^{m})=d_{-2}(p^{m})$ for $p^{m}\leqslant X$ and
$\beta_{-2}(p^{m})\ll d(p^{m})$ in general. Then since $n_{1}\geqslant 0$, the
sum in the error is $\ll\sum_{m:\,p^{m}>X}d_{4}(p^{m})^{2}/p^{m}$ after
forming the convolution. Therefore, we can apply the same argument which gave
(26) to find that the numerator is $1+O(X^{-1/2+\epsilon})$. Consequently, we
have

$J_{2,\Phi}\sim\frac{\hat{\Phi}(0)}{12}\cdot T\bigg{(}\frac{\log
T}{e^{\gamma}\log X}\bigg{)}^{4}=\frac{T}{12}\bigg{(}\frac{\log
T}{e^{\gamma}\log X}\bigg{)}^{4}+O(T^{1-\epsilon})$

and so the result follows by (34).

##  5\. Some useful tools

In this short section we describe some tools which will come in handy when
proving Proposition 3. The first relates to the exponential truncation of a
more general prime sum and will be used extensively throughout.

Given a general Dirichlet polynomial of the form $D(s)=\sum_{p\leqslant
Y}a(p)p^{-s}$, suppose $t\in[T,2T]$ is such that $|kD(s)|\leqslant Z$. Then by
(17) we have

(37) $\bigg{|}e^{kD(s)}-\sum_{0\leqslant j\leqslant
10Z}\frac{(kD(s))^{j}}{j!}\bigg{|}\leqslant e^{-10Z}.$

By the multinomial theorem the truncated exponential series can be written as

(38) $\sum_{0\leqslant j\leqslant 10Z}\frac{1}{j!}\bigg{(}k\sum_{p\leqslant
Y}\frac{a(p)}{p^{s}}\bigg{)}^{j}=\sum_{\begin{subarray}{c}\Omega(n)\leqslant
10Z\\\ p|n\implies p\leqslant
Y\end{subarray}}\frac{k^{\Omega(n)}a(n)\mathfrak{g}(n)}{n^{s}}$

where we recall $\mathfrak{g}$ is the multiplicative function such that
$\mathfrak{g}(p^{\alpha})=1/\alpha!$, and $a(n)$ is the completely
multiplicative extension of $a(p)$. Observe this is a Dirichlet polynomial of
length $Y^{10Z}$.

Our remaining observations relate to mean values of Dirichlet polynomials. We
first state the mean value theorem of Montgomery and Vaughan [36] which gives
for any complex coefficients $a(n)$,

(39) $\frac{1}{T}\int_{T}^{2T}\Big{|}\sum_{n\leqslant
N}\frac{a(n)}{n^{it}}\Big{|}^{2}dt=(1+O(N/T))\sum_{n\leqslant N}|a(n)|^{2}.$

Suppose we are given $R$ Dirichlet polynomials

$A_{j}(s)=\sum_{n\in{\mathcal{S}}_{j}}a_{j}(n)n^{-s},$

where the $\prod_{j=1}^{R}n_{j}\leqslant N=o(T)$ for all
$n_{j}\in{\mathcal{S}}_{j}$ i.e. the product of the $A_{j}(s)$ is short. Then
the Montgomery–Vaughan mean value theorem readily implies

$\displaystyle\frac{1}{T}\int_{T}^{2T}\prod_{j=1}^{R}\big{|}A_{j}(it)\big{|}^{2}dt\sim$
$\displaystyle\sum_{n\leqslant N}\Big{|}\sum_{\begin{subarray}{c}n=n_{1}\cdots
n_{R}\\\ n_{j}\in S_{j}\end{subarray}}a_{1}(n_{1})\cdots
a_{R}(n_{R})\Big{|}^{2}$ (40) $\displaystyle=$
$\displaystyle\sum_{\begin{subarray}{c}n_{1}\cdots n_{R}=n_{R+1}\cdots
n_{2R}\\\ n_{j}\in S_{j}\end{subarray}}a_{1}(n_{1})\cdots
a_{R}(n_{R})\overline{a_{1}(n_{R+1})\cdots a_{R}(n_{2R})}.$

Suppose in addition that for any $j_{1},j_{2}$ with $j_{1}\neq j_{2}$ the
elements of $S_{j_{1}}$ are all coprime to the elements of $S_{j_{2}}$. Then
there is at most one way to write $n=\prod_{j=1}^{R}n_{j}$ with
$n_{j}\in{\mathcal{S}}_{j}$ and thus several applications of the mean value
theorem imply

$\displaystyle\frac{1}{T}\int_{T}^{2T}\prod_{j=1}^{R}|A_{j}(it)|^{2}dt$
$\displaystyle=(1+O(NT^{-1}))\sum_{n\leq
N}\Big{|}\sum_{\begin{subarray}{c}n=n_{1}\cdots n_{R}\\\
n_{j}\in{\mathcal{S}}_{j}\end{subarray}}\prod_{j=1}^{R}a_{j}(n_{j})\Big{|}^{2}$
$\displaystyle=(1+O(NT^{-1}))\prod_{j=1}^{R}\Big{(}\sum_{n_{j}\in{\mathcal{S}}_{j}}|a_{j}(n_{j})|^{2}\Big{)}$
(41)
$\displaystyle=(1+O(NT^{-1}))^{1-R}\prod_{j=1}^{R}\Big{(}\frac{1}{T}\int_{T}^{2T}|A_{j}(it)|^{2}dt\Big{)}.$

We now move on to proving the upper bound of Proposition 3.

##  6\. The upper bound of Proposition 3

###  6.1. Initial cleaning

In this section we are required to show that on RH,

$\frac{1}{T}\int_{T}^{2T}|Z_{X}(\tfrac{1}{2}+it)|^{2k}dt\ll\bigg{(}\frac{\log
T}{\log X}\bigg{)}^{k^{2}}.$

On assuming RH, it is a simple task to replace $|P_{X}(1/2+it)|^{-2k}$ by
$|D(t,-k)|^{2}$ on the left hand side. Indeed; from Harper’s [24] conditional
bound $\int_{T}^{2T}|\zeta(1/2+it)|^{2k}\ll T(\log T)^{k^{2}}$, the bound for
the moments of $D(t,k)$ in (24), Lemmas 3, 4, 5, and the usual arguments
involving the decomposition $[T,2T]=\mathcal{S}\cup\mathcal{E}$ along with
Hölder’s inequality we have

(42)
$\frac{1}{T}\int_{T}^{2T}|Z_{X}(\tfrac{1}{2}+it)|^{2k}dt\sim\frac{1}{T}\int_{T}^{2T}|\zeta(\tfrac{1}{2}+it)|^{2k}|D(t,-k)|^{2}dt$

for all $k\geqslant 0$ on RH.

To bound the right hand side we use an upper bound for $\zeta(1/2+it)$ which
incorporates the recent developments of Harper [24] on moments of the zeta
function, although we present the result more in the style of
Radziwiłł–Soundararajan [38] (see the key inequality of section 3 there). Such
a treatment is similar to that of [35].

###  6.2. An upper bound for $\zeta(\tfrac{1}{2}+it)$

We start with a proposition of Soundararajan in a mildly adapted form of
Harper.

######  Lemma 6.

Assume RH. Let $t\in[T,2T]$ be large and suppose $4\leqslant x\leqslant
T^{2}$. Then

$\log|\zeta(\tfrac{1}{2}+it)|\leqslant\Re\sum_{p\leqslant
x}\frac{1}{p^{{1}/{2}+1/\log x+it}}\frac{\log(x/p)}{\log
x}+\Re\sum_{p\leqslant\min(\sqrt{x},\log T)}\frac{1}{2p^{1+2it}}+\frac{\log
T}{\log x}+O(1).$

###### Proof.

This is Proposition 1 of [24]. ∎

For the splitting of the prime sums we denote

$T_{-1}=A,\,\,\,\,\,\,T_{j}=T^{\frac{e^{j}}{(\log\log T)^{2}}}$

where $A>0$ is fixed and $j=0,\ldots,J$ with $J$ the maximal integer such that
$e^{J}/(\log\log T)^{2}\leqslant 1/10^{12}$ (so that $J\asymp\log\log\log T$).
In this section we take $A=1$ although later we need to take it sufficiently
large. Let

$\theta_{j}=\frac{e^{j}}{(\log\log
T)^{2}},\,\,\,\,\,\ell_{j}=\theta_{j}^{-3/4}$

so that $T_{j}=T^{\theta_{j}}$ for $0\leqslant j\leqslant J$. Now, write

$w_{j}(p)=\frac{1}{p^{1/\theta_{j}\log T}}\frac{\log(T_{j}/p)}{\log T_{j}}$

and

$\mathcal{P}_{i,j}(t):=\sum_{T_{i-1}<p\leqslant
T_{i}}\frac{w_{j}(p)}{p^{1/2+it}}.$

With this notation the first sum over primes in Lemma 6 with $x=T_{j}$ can be
written as

(43) $\sum_{p\leqslant
T_{j}}\frac{w_{j}(p)}{p^{{1}/{2}+it}}=\sum_{i=0}^{j}\mathcal{P}_{i,j}(t).$

Also, write

$\mathcal{N}_{i,j}(t,k)=\sum_{\begin{subarray}{c}\Omega(n)\leqslant
10\ell_{i}\\\ p|n\implies T_{i-1}<p\leqslant
T_{i}\end{subarray}}\frac{k^{\Omega(n)}w_{j}(n)\mathfrak{g}(n)}{n^{1/2+it}}$

and note that on the set of $t\in[T,2T]$ such that
$|k\mathcal{P}_{i,j}(t)|\leqslant\ell_{i}$ we have

(44)
$\exp(2k\Re\mathcal{P}_{i,j}(t))=(1+O(e^{-9\ell_{i}}))\big{|}\mathcal{N}_{i,j}(t,k)\big{|}^{2}$

by (37) and (38). Accordingly, if $t$ is such that
$|k\mathcal{P}_{i,j}(t)|\leqslant\ell_{i}$ for all $0\leqslant i\leqslant j$
then on applying (43) we have

(45) $\exp\Big{(}2k\Re\sum_{p\leqslant
T_{j}}\frac{w_{j}(p)}{p^{{1}/{2}+it}}\Big{)}\ll\prod_{i=0}^{j}\big{|}\mathcal{N}_{i,j}(t,k)\big{|}^{2}$

since $\sum_{i=0}^{j}e^{-9\ell_{i}}$ is a rapidly converging series. Note this
is a Dirichlet polynomial of length

(46)
$\leqslant\prod_{i=0}^{J}T_{i}^{10\ell_{i}}=T^{10\sum_{i=0}^{J}\theta_{j}^{1/4}}\leqslant
T^{20e^{J/4}/(\log\log T)^{1/2}}\leqslant T^{1/50}.$

We can now state an upper bound for the zeta function in terms of these short
Dirichlet polynomials.

######  Lemma 7.

Assume RH. Then either

$|k\mathcal{P}_{0,j}(t)|>\ell_{0}$

for some $0\leqslant j\leqslant J$ or

$|\zeta(\tfrac{1}{2}+it)|^{2k}\ll\bigg{(}\prod_{i=0}^{J}\big{|}\mathcal{N}_{i,J}(t,k)\big{|}^{2}+\sum_{\begin{subarray}{c}0\leqslant
j\leqslant J-1\\\ j+1\leqslant l\leqslant
J\end{subarray}}\exp\Big{(}\frac{2k}{\theta_{j}}\Big{)}\bigg{(}\frac{|k\mathcal{P}_{j+1,l}(t)|}{\ell_{j+1}}\bigg{)}^{2s_{j}}\prod_{i=0}^{j}\big{|}\mathcal{N}_{i,j}(t,k)\big{|}^{2}\bigg{)}\\\
\times|\mathcal{M}(t,k)|^{2}$

for any positive integers $s_{j}$ where

$\mathcal{M}(t,k)=\sum_{\begin{subarray}{c}\Omega(n)\leqslant
10k(\log_{2}T)^{2}\\\ p|n\implies p\leqslant\log
T\end{subarray}}\frac{(k/2)^{\Omega(n)}\mathfrak{g}(n)}{n^{1+2it}}.$

###### Proof.

Suppose $|k\mathcal{P}_{0,j}(t)|<\ell_{0}$. For $0\leqslant j\leqslant J-1$
let

$S(j)=\left\\{t\in[T,2T]:\begin{array}[]{lr}&|k\mathcal{P}_{i,l}(t)|\leqslant\ell_{i}\,\,\,\,\,\,\,\,\,\,\,\forall
1\leqslant i\leqslant j,\,\,\forall j\leqslant l\leqslant J;\\\
&|k\mathcal{P}_{j+1,l}(t)|>\ell_{j+1}\text{ for some }j+1\leqslant l\leqslant
J\end{array}\right\\}$

and

$S(J)=\bigg{\\{}t\in[T,2T]:|k\mathcal{P}_{i,J}(t)|\leqslant\ell_{i}\,\,\,\,\,\,\,\forall
1\leqslant i\leqslant J\bigg{\\}}.$

Then since $[T,2T]=\cup_{j=0}^{J}S(j)$, for $t\in[T,2T]$ we have

(47) $|\zeta(\tfrac{1}{2}+it)|^{2k}\leqslant\mathds{1}_{t\in
S(J)}\cdot|\zeta(\tfrac{1}{2}+it)|^{2k}+\sum_{\begin{subarray}{c}0\leqslant
j\leqslant J-1\\\ j+1\leqslant l\leqslant J\end{subarray}}\mathds{1}_{t\in
S_{l}(j)}\cdot|\zeta(\tfrac{1}{2}+it)|^{2k}$

where

$S_{l}(j)=\left\\{t\in[T,2T]:\begin{array}[]{lr}|k\mathcal{P}_{i,l}(t)|\leqslant\ell_{i}&\,\,\,\,\,\,\,\,\,\,\,\forall
1\leqslant i\leqslant j,\,\,\forall j\leqslant l\leqslant J;\\\
|k\mathcal{P}_{j+1,l}(t)|>\ell_{j+1}&\end{array}\right\\}.$

We apply Lemma 6 to each zeta function on the right hand side of (47). If
$t\in S_{l}(j)$ then we take $x=T_{j}$ to give

$|\zeta(\tfrac{1}{2}+it)|^{2k}\ll\exp\bigg{(}2k\Re\sum_{p\leqslant
T_{j}}\frac{w_{j}(p)}{p^{1/2+it}}+2k\Re\sum_{p\leqslant\log
T}\frac{1}{2p^{1+2it}}+\frac{2k}{\theta_{j}}\bigg{)}.$

For the first sum over primes in the exponential we apply (45). For the second
sum we note that, since $|\sum_{p\leqslant\log T}\frac{1}{p^{1+2it}}|\leqslant
2\log_{3}T\leqslant(\log\log T)^{2}$, we have

$\begin{split}\exp\Big{(}2k\Re\sum_{p\leqslant\log
T}\frac{1}{2p^{1+2it}}\Big{)}=&\big{(}1+O(e^{-9k(\log_{2}T)^{2}})\big{)}\bigg{|}\sum_{\begin{subarray}{c}\Omega(n)\leqslant
10k(\log\log T)^{2}\\\ p|n\implies p\leqslant\log
T\end{subarray}}\frac{(k/2)^{\Omega(n)}\mathfrak{g}(n)}{n^{1+2it}}\bigg{|}^{2}\end{split}$

by (37) and (38). This is $\ll|\mathcal{M}(t,k)|^{2}$. Finally, to capture the
small size of the set, we multiply by

$\bigg{(}\frac{|k\mathcal{P}_{j+1,l}(t)|}{\ell_{j+1}}\bigg{)}^{2s_{j}}>1.$

If $t\in S(J)$ then we omit this last step since of course there is no such
$P_{J+1,l}(t)$. ∎

###  6.3. Proof of the upper bound in Proposition 3

From (42) we are required to show that

$\frac{1}{T}\int_{T}^{2T}|\zeta(\tfrac{1}{2}+it)|^{2k}|D(t,-k)|^{2}dt\ll\bigg{(}\frac{\log
T}{\log X}\bigg{)}^{k^{2}}.$

To apply Lemma 7 we must consider two cases; that where
$|k\mathcal{P}_{0,j}(t)|>\ell_{0}$ and otherwise. We consider the former case
first since this is simpler.

So, let $E\subset[T,2T]$ be the subset on which
$|k\mathcal{P}_{0,j}(t)|>\ell_{0}$, that is, when

$\bigg{|}\sum_{p\leqslant T^{1/(\log\log
T)^{2}}}\frac{w_{j}(p)}{p^{1/2+it}}\bigg{|}>\frac{(\log\log T)^{3/2}}{k}.$

By Chebyshev’s inequality and Lemma 2 we have

$\displaystyle\mu(E)\leqslant$ $\displaystyle\bigg{(}\frac{k^{2}}{(\log\log
T)^{3}}\bigg{)}^{m}\int_{T}^{2T}\bigg{|}\sum_{p\leqslant T^{1/(\log\log
T)^{2}}}\frac{w_{j}(p)}{p^{1/2+it}}\bigg{|}^{2m}dt$ $\displaystyle\ll$
$\displaystyle Tm!\bigg{(}\frac{k^{2}}{(\log\log
T)^{3}}\bigg{)}^{m}\bigg{(}\sum_{p\leqslant T^{1/(\log\log
T)^{2}}}\frac{1}{p}\bigg{)}^{m}$

provided $m\leqslant(1-o(1))(\log\log T)^{2}$ where in the last line we have
used $|w_{j}(p)|\leqslant 1$ for all $j$. By Stirling’s formula and Mertens’
theorem this is

(48) $\mu(E)\ll Tm^{1/2}\bigg{(}\frac{k^{2}m}{e(\log\log
T)^{2}}\bigg{)}^{m}\leqslant Te^{-c(\log\log T)^{2}}$

for some $c>1$ on choosing $m=\lfloor\min(1,\tfrac{1}{k^{2}})(\log\log
T)^{2}\rfloor$. Therefore by Hölder’s inequality, Harper’s bound for the
moments of zeta and (24) we have

$\frac{1}{T}\int_{E}|\zeta(\tfrac{1}{2}+it)|^{2k}|D(t,-k)|^{2}dt\ll
e^{-C(\log\log T)^{2}}(\log T)^{O(1)}=o(1).$

We may now consider the second case where $|k\mathcal{P}_{0,j}(t)|>\ell_{0}$
and accordingly concentrate on the integral

$\frac{1}{T}\int_{[T,2T]\backslash
E}|\zeta(\tfrac{1}{2}+it)|^{2k}|D(t,-k)|^{2}dt.$

By the second part of Lemma 7 this is

$\displaystyle\ll$
$\displaystyle\frac{1}{T}\int_{T}^{2T}\bigg{(}\prod_{i=0}^{J}\big{|}\mathcal{N}_{i,J}(t,k)\big{|}^{2}+\sum_{\begin{subarray}{c}0\leqslant
j\leqslant J-1\\\ j+1\leqslant l\leqslant
J\end{subarray}}\exp\Big{(}\frac{2k}{\theta_{j}}\Big{)}\bigg{(}\frac{|k\mathcal{P}_{j+1,l}(t)|}{\ell_{j+1}}\bigg{)}^{2s_{j}}\prod_{i=0}^{j}\big{|}\mathcal{N}_{i,j}(t,k)\big{|}^{2}\bigg{)}$
(49) $\displaystyle\qquad\qquad\times|\mathcal{M}(t,k)|^{2}|D(t,-k)|^{2}dt.$

To compute the resultant integrals we apply the following lemma.

######  Lemma 8.

For $0\leqslant s_{j}\leqslant 1/(10\theta_{j})$ we have

$\frac{1}{T}\int_{T}^{2T}|D(t,-k)|^{2}|\mathcal{M}(t,k)|^{2}|\mathcal{P}_{j+1,l}(t)|^{2s_{j}}\prod_{i=0}^{j}\big{|}\mathcal{N}_{i,j}(t,k)\big{|}^{2}dt\ll
s_{j}!P_{j+1}^{s_{j}}\bigg{(}\frac{\log T_{j}}{\log X}\bigg{)}^{k^{2}}$

where

$P_{j+1}=\sum_{T_{j}<p\leqslant T_{j+1}}\frac{1}{p}.$

###### Proof.

We write the integrand as a multiple sum. First off, by the multinomial
theorem we have

$\mathcal{P}_{j+1,l}(t)^{s_{j}}=\bigg{(}\sum_{T_{j}<p\leqslant
T_{j+1}}\frac{w_{l}(p)}{p^{1/2+it}}\bigg{)}^{s_{j}}=s_{j}!\sum_{\begin{subarray}{c}\Omega(n)=s_{j}\\\
p|n\implies T_{j}<p\leqslant
T_{j+1}\end{subarray}}\frac{w_{l}(n)\mathfrak{g}(n)}{n^{1/2+it}}.$

Therefore

$\displaystyle
D(t,-k)\mathcal{M}(t,k)\mathcal{P}_{j+1,l}(t)^{s_{j}}\prod_{i=0}^{j}\mathcal{N}_{i,j}(t,k)$
$\displaystyle=$ $\displaystyle s_{j}!\sum_{n\in
S(X)}\frac{\alpha_{-k}(n)}{n^{1/2+it}}\sum_{\begin{subarray}{c}\Omega(n)\leqslant
10k(\log_{2}T)^{2}\\\ p|n\implies p\leqslant\log
T\end{subarray}}\frac{(k/2)^{\Omega(n)}\mathfrak{g}(n)}{n^{1+2it}}\sum_{\begin{subarray}{c}\Omega(n)=s_{j}\\\
p|n\implies T_{j}<p\leqslant
T_{j+1}\end{subarray}}\frac{w_{l}(n)\mathfrak{g}(n)}{n^{1/2+it}}$
$\displaystyle\times\prod_{i=0}^{j}\bigg{(}\sum_{\begin{subarray}{c}\Omega(n)\leqslant
10\ell_{i}\\\ p|n\implies T_{i-1}<p\leqslant
T_{i}\end{subarray}}\frac{k^{\Omega(n)}w_{j}(n)\mathfrak{g}(n)}{n^{1/2+it}}\bigg{)}.$

Since $X\leqslant T_{0}$ we may group together all the sums over
$T_{0}$-smooth numbers as a single sum and write the above as

$s_{j}!\sum_{n}\frac{\gamma(n)}{n^{1/2+it}}\sum_{\begin{subarray}{c}\Omega(n)=s_{j}\\\
p|n\implies T_{j}<p\leqslant
T_{j+1}\end{subarray}}\frac{w_{l}(n)\mathfrak{g}(n)}{n^{1/2+it}}\prod_{i=1}^{j}\bigg{(}\sum_{\begin{subarray}{c}\Omega(n)\leqslant
10\ell_{i}\\\ p|n\implies T_{i-1}<p\leqslant
T_{i}\end{subarray}}\frac{k^{\Omega(n)}w_{j}(n)\mathfrak{g}(n)}{n^{1/2+it}}\bigg{)}$

for some coefficients $\gamma(n)$ where the product is empty if $j=0$. Since
this is a short Dirichlet polynomial we find by (5) that

(50)
$\frac{1}{T}\int_{T}^{2T}|D(t,-k)|^{2}|\mathcal{M}(t,k)|^{2}|\mathcal{P}_{j+1,l}(t)|^{2s_{j}}\prod_{i=0}^{j}\big{|}\mathcal{N}_{i,j}(t,k)\big{|}^{2}dt\\\
\ll{(s_{j}!)^{2}}\sum_{n}\frac{\gamma(n)^{2}}{n}\sum_{\begin{subarray}{c}\Omega(n)=s_{j}\\\
p|n\implies T_{j}\leqslant
p<T_{j+1}\end{subarray}}\frac{w_{l}(n)^{2}\mathfrak{g}(n)^{2}}{n}\\\
\prod_{i=1}^{j}\bigg{(}\sum_{\begin{subarray}{c}\Omega(n)\leqslant
10\ell_{i}\\\ p|n\implies T_{i-1}<p\leqslant
T_{i}\end{subarray}}\frac{k^{\Omega(n)}w_{j}(n)^{2}\mathfrak{g}(n)^{2}}{n}\bigg{)}.$

Now, by (5) we explicitly have

$\sum_{n}\frac{\gamma(n)^{2}}{n}=\sideset{}{{}^{\prime}}{\sum}_{\begin{subarray}{c}n_{1}n_{2}n_{3}^{2}=\\\
n_{4}n_{5}n_{6}^{2}\end{subarray}}\frac{\alpha_{-k}(n_{1})\alpha_{-k}(n_{4})k^{\Omega(n_{2}n_{5})}(k/2)^{\Omega(n_{3}n_{6})}W(\underline{n})}{(n_{1}n_{2}n_{4}n_{5})^{1/2}n_{3}n_{6}}$

where

$W(\underline{n})=w_{j}(n_{2})w_{j}(n_{5})\mathfrak{g}(n_{2})\mathfrak{g}(n_{3})\mathfrak{g}(n_{5})\mathfrak{g}(n_{6})$

and the $\prime$ in the sum denotes that $n_{1},n_{3}\in S(X)$ and

$\begin{array}[]{cc}\Omega(n_{2}),\Omega(n_{5})\leqslant
10\ell_{0}&\qquad\Omega(n_{3}),\Omega(n_{6})\leqslant 10k(\log_{2}T)^{2}\\\
p|n_{2},n_{5}\implies p<T_{0}&\qquad p|n_{3},n_{6}\implies p\leqslant\log
T.\end{array}$

We first estimate the terms with $\Omega(n_{1}),\Omega(n_{4})>W_{0}$. By the
usual arguments, for $1<r<2$ these terms are bounded by,

$\displaystyle\ll$ $\displaystyle
r^{-W_{0}}\sum_{\begin{subarray}{c}n_{1}n_{2}n_{3}^{2}=\\\
n_{4}n_{5}n_{6}^{2}\\\ p|n_{j}\implies p\leqslant
T\end{subarray}}\frac{r^{\Omega(n_{1})}d_{k}(n_{1})d_{k}(n_{4})k^{\Omega(n_{2}n_{3}n_{5}n_{6})}W(\underline{n})}{(n_{1}n_{2}n_{4}n_{5})^{1/2}n_{3}n_{6}}$
$\displaystyle=$ $\displaystyle r^{-W_{0}}\prod_{p\leqslant
T}\bigg{(}1+\frac{(2r+2)k^{2}}{p}+O(p^{-2})\bigg{)}\ll e^{-W_{0}}(\log
T)^{6k^{2}}=o(1).$

We then replace $\alpha_{-k}(n)$ with $\beta_{-k}(n)$ in the remaining sum.
The usual arguments also allow us to remove the restrictions on all the
$\Omega(n_{j})$ at the cost of an error of size $o(1)$. Expressing the
resultant sum as an Euler product we get

$\displaystyle\sum_{n}\frac{\gamma(n)^{2}}{n}=$
$\displaystyle\prod_{p\leqslant
X}\bigg{(}1+O(p^{-2})\bigg{)}\prod_{X<p\leqslant
T_{0}}\bigg{(}1+\frac{k^{2}}{p}+O(p^{-2})\bigg{)}+o(1)$ $\displaystyle\asymp$
$\displaystyle\bigg{(}\frac{\log T_{0}}{\log X}\bigg{)}^{k^{2}}$

since $\beta_{-k}(n)$ is supported on $X$-smooth numbers where it satisfies
$\beta_{-k}(p)=-k$ and $\beta_{-k}(p^{m})\ll d_{k}(p^{m})$ for $m\geqslant 2$.

The second sum in (50) is

$(s_{j}!)^{2}\sum_{\begin{subarray}{c}\Omega(n)=s_{j}\\\ p|n\implies
T_{j}\leqslant
p<T_{j+1}\end{subarray}}\frac{w_{l}(n)^{2}\mathfrak{g}(n)^{2}}{n}\leqslant
s_{j}!\sum_{\begin{subarray}{c}\Omega(n)=s_{j}\\\ p|n\implies T_{j}\leqslant
p<T_{j+1}\end{subarray}}\frac{s_{j}!\mathfrak{g}(n)}{n}=s_{j}!\bigg{(}\sum_{T_{j}\leqslant
p<T_{j+1}}\frac{1}{p}\bigg{)}^{s_{j}},$

since $\mathfrak{g}(n)\leqslant 1$, whilst the third sum is

$\displaystyle\prod_{i=1}^{j}\sum_{\begin{subarray}{c}\Omega(n)\leqslant
10\ell_{i}\\\ p|n\implies T_{i-1}<p\leqslant
T_{i}\end{subarray}}\frac{k^{2\Omega(n)}w_{j}(n)^{2}\mathfrak{g}(n)^{2}}{n}\ll$
$\displaystyle\prod_{i=1}^{j}\prod_{T_{i-1}\leqslant
p<T_{i}}\bigg{(}1+\frac{k^{2}}{p}+O(p^{-2})\bigg{)}$ $\displaystyle\asymp$
$\displaystyle\bigg{(}\frac{\log T_{j}}{\log T_{0}}\bigg{)}^{k^{2}}.$

Combining these estimates gives the result. ∎

###### Completion of proof of upper bound in Proposition 3.

Applying Lemma 8 in (6.3) gives an upper bound of the form

$\displaystyle\bigg{(}\frac{\log T_{J}}{\log
X}\bigg{)}^{k^{2}}+\sum_{\begin{subarray}{c}0\leqslant j\leqslant J-1\\\
j+1\leqslant l\leqslant
J\end{subarray}}\exp\Big{(}\frac{2k}{\theta_{j}}\Big{)}s_{j}!\bigg{(}\frac{k{P}_{j+1}}{\ell_{j+1}^{2}}\bigg{)}^{s_{j}}\bigg{(}\frac{\log
T_{j}}{\log X}\bigg{)}^{k^{2}}.$

On noting that

$J-j\ll\log(1/\theta_{j}),\qquad P_{j+1}=\log\bigg{(}\frac{\log T_{j+1}}{\log
T_{j}}\bigg{)}+o(1)\leqslant 2,\qquad\ell_{j+1}^{2}=\theta_{j}^{-3/2}$

and setting $s_{j}=1/(10\theta_{j})$, then by Stirling’s formula this is

$\displaystyle\ll$ $\displaystyle\bigg{(}\frac{\log T}{\log
X}\bigg{)}^{k^{2}}\bigg{(}1+\sum_{0\leqslant j\leqslant
J-1}\log(\tfrac{1}{\theta_{j}})\theta_{j}^{-{1/10\theta_{j}}}c^{1/\theta_{j}}\theta_{j}^{{3}/{20\theta_{j}}}\bigg{)}$
$\displaystyle\ll$ $\displaystyle\bigg{(}\frac{\log T}{\log
X}\bigg{)}^{k^{2}}\bigg{(}1+\sum_{0\leqslant j\leqslant
J-1}e^{-{C}\theta_{j}^{-1}\log(1/\theta_{j})}\bigg{)}$

for some constants $c,C>0$. Since this last series is bounded the result
follows. ∎

##  7\. Lower bound for $0\leqslant k\leqslant 1$

We keep a similar setup to the previous section but with a few minor changes.
There is no longer any need for the weights $w_{j}(p)$ so we can simplify our
notation and let

$\mathcal{N}_{i}(t,k)=\sum_{\begin{subarray}{c}\Omega(n)\leqslant
10\ell_{i}\\\ p|n\implies T_{i-1}<p\leqslant
T_{i}\end{subarray}}\frac{k^{\Omega(n)}\mathfrak{g}(n)}{n^{1/2+it}}$

where the $T_{i}$ and $\ell_{i}$ are as before for $0\leqslant i\leqslant J$.
As remarked earlier, in this section we take $T_{-1}=A$ with $A$ sufficiently
large to be chosen later. We then form the product

(51) $\mathcal{N}(t,k):=\prod_{i=0}^{J}\mathcal{N}_{i}(t,k)=\sum_{n\leqslant
Y}\frac{\gamma_{k}(n)}{n^{1/2+it}}$

for some coefficients $\gamma_{k}(n)$ where from (46) we have $Y=T^{1/50}$. We
can think of this as an approximation to $\zeta(1/2+it)^{k}$; It possesses
several nice features akin to an Euler product whilst also being a short
Dirichlet polynomial.

We acquire our lower bound by applying Hölder’s inequality, the form of which
will depend on whether $0<k\leqslant 1$ or $k\geqslant 1$. The latter case is
somewhat simpler so we give details for the case $0<k\leqslant 1$ first. By
Hölder’s inequality, we have

$\displaystyle\Big{|}\frac{1}{T}\int_{\mathcal{S}}\zeta(\tfrac{1}{2}+it)\mathcal{N}(t,k-1)\overline{\mathcal{N}(t,k)}|D(t,-k)|^{2}dt\Big{|}$
$\displaystyle\ll\Big{(}\frac{1}{T}\int_{\mathcal{S}}|\zeta(\tfrac{1}{2}+it)|^{2k}|D(t,-k)|^{2}dt\Big{)}^{\frac{1}{2}}\Big{(}\frac{1}{T}\int_{T}^{2T}|\zeta(\tfrac{1}{2}+it){\mathcal{N}}(t,k-1)|^{2}|D(t,-k)|^{2}dt\Big{)}^{\frac{1-k}{2}}$
$\displaystyle\hskip
72.26999pt\times\Big{(}\frac{1}{T}\int_{T}^{2T}|{\mathcal{N}}(t,k)|^{\frac{2}{k}}|{\mathcal{N}}(t,k-1)|^{2}|D(t,-k)|^{2}dt\Big{)}^{\frac{k}{2}}.$

Since $D(t,-k)\sim P_{X}(1/2+it)^{-k}$ for $t\in\mathcal{S}$ the first
integral on the right hand side is
$\ll\frac{1}{T}\int_{T}^{2T}|Z_{X}(1/2+it)|^{2k}dt$.

######  Remark.

Note also that in this argument we can modify the definition of $D$ to be

$D(t,k)=\sum_{p|n\implies A<p\leqslant X}\frac{\alpha_{k}(n)}{n^{1/2+it}}$

since the removal of the $A$-smooth numbers from the sum in

$\exp\Big{(}k\sum_{n\leqslant X}\frac{\Lambda(n)}{n^{1/2+it}\log n}\Big{)}$

leads to a bounded multiplicative factor which can be absorbed into the $\ll$
sign. Also, to save space in the future we may absorb the condition
$p|n\implies A<p\leqslant X$ into the coefficients $\alpha_{k}(n)$.

The lower bound in the case $0<k\leqslant 1$ now follows from the subsequent
Propositions.

######  Proposition 4.

Suppose $X\leqslant\eta_{k}(\log T)^{2}(\log_{2}T)^{2}$. Then

$\frac{1}{T}\int_{\mathcal{S}}\zeta(\tfrac{1}{2}+it)\mathcal{N}(t,k-1)\overline{\mathcal{N}(t,k)}|D(t,-k)|^{2}dt\geqslant
C\bigg{(}\frac{\log T}{\log X}\bigg{)}^{k^{2}}$

for some $C>0$. Assuming RH we may take $X\leqslant(\log
T)^{\theta_{k}-\epsilon}$.

######  Proposition 5.

For $X\leqslant T^{1/(\log_{2}T)^{2}}$ we have

$\frac{1}{T}\int_{T}^{2T}|\zeta(\tfrac{1}{2}+it){\mathcal{N}}(t,k-1)|^{2}|D(t,-k)|^{2}dt\ll\bigg{(}\frac{\log
T}{\log X}\bigg{)}^{k^{2}}.$

######  Proposition 6.

For $X\leqslant T^{1/(\log_{2}T)^{2}}$ we have

$\frac{1}{T}\int_{T}^{2T}|{\mathcal{N}}(t,k)|^{\frac{2}{k}}|{\mathcal{N}}(t,k-1)|^{2}|D(t,-k)|^{2}dt\ll\bigg{(}\frac{\log
T}{\log X}\bigg{)}^{k^{2}}.$

Note that our argument works unconditionally provided $X\leqslant\eta_{k}(\log
T)^{2}(\log_{2}T)^{2}$ as claimed in the introduction.

###  7.1. Proof of Proposition 4

Our first job is to extend the range of integration to the full set $[T,2T]$.
By the usual argument involving Hölder’s inequality the integral over
$\mathcal{E}$ is $o(1)$. Indeed, from the conditions on $X$ and Lemma 3 we
have $T^{-1}\mu(\mathcal{E})\ll e^{-2|k|V_{0}}$ which is enough to kill any
power of $\log T$. We also have the second moment bound for the zeta function,
and by (24) the $m$th moment of $D(t,-k)$ is also $(\log T)^{O(1)}$. The only
new ingredient required is a moment bound for $\mathcal{N}(t,k)$ but by the
Montgomery–Vaughan mean value theorem this is, for $m\leqslant 50$,

$\displaystyle\frac{1}{T}\int_{T}^{2T}|\mathcal{N}(t,k)|^{2m}dt\ll$
$\displaystyle\sum_{\begin{subarray}{c}n_{1}\cdots n_{m}=n_{m+1}\cdots
n_{2m}\\\ n_{j}\leqslant
T^{1/50}\end{subarray}}\frac{\gamma_{k}(n_{1})\cdots\gamma_{k}(n_{2m})}{(n_{1}\cdots
n_{2m})^{1/2}}$ $\displaystyle=$ $\displaystyle\prod_{p\leqslant
T}\bigg{(}1+\frac{m^{2}k^{2}}{p}+O(p^{-2})\bigg{)}\ll(\log T)^{m^{2}k^{2}}.$

Therefore,

$\frac{1}{T}\int_{\mathcal{S}}\zeta(\tfrac{1}{2}+it)\mathcal{N}(t,k-1)\overline{\mathcal{N}(t,k)}|D(t,-k)|^{2}dt=I_{3}+o(1)$

where

$I_{3}=\frac{1}{T}\int_{T}^{2T}\zeta(\tfrac{1}{2}+it)\mathcal{N}(t,k-1)\overline{\mathcal{N}(t,k)}|D(t,-k)|^{2}dt.$

To lower bound $I_{3}$ we apply the approximation

$\zeta(\tfrac{1}{2}+it)=\sum_{n\leqslant
T}\frac{1}{n^{1/2+it}}+O\Big{(}\frac{1}{T^{1/2}}\Big{)}.$

The other terms of the integrand satisfy pointwise bounds of the form

$\mathcal{N}(t,k)\ll Y^{1/2+\epsilon}\ll
T^{1/100+\epsilon},\qquad\sum_{n}\frac{\alpha_{-k}(n)}{n^{1/2+it}}\ll
X^{W_{0}(1/2+\epsilon)}\ll T^{1/100};$

since $k^{\Omega(n)}$ has average order $(\log n)^{k-1}$ and
$\alpha_{-k}(n)\ll d_{k}(n)\ll n^{\epsilon}$. We then see that the error term
in the approximation for zeta leads to an error of size $o(1)$.

Therefore

$I_{3}=\frac{1}{T}\int_{T}^{2T}\sum_{\begin{subarray}{c}n_{1}\leqslant
T,\,\,n_{2},n_{3}\leqslant
Y\end{subarray}}\frac{\gamma_{k-1}(n_{2})\gamma_{k}(n_{3})\alpha_{-k}(n_{4})\alpha_{-k}(n_{5})}{(n_{1}n_{2}n_{3}n_{4}n_{5})^{1/2}}\Big{(}\frac{n_{3}n_{5}}{n_{1}n_{2}n_{4}}\Big{)}^{it}dt+o(1).$

By direct integration, the off-diagonal terms for which $n_{1}n_{2}n_{4}\neq
n_{3}n_{5}$ lead to an error of size

$\frac{YX^{W_{0}}}{T}\sum_{n_{1}\leqslant T,\,\,n_{2},n_{3}\leqslant
Y}\frac{|\gamma_{k-1}(n)|\gamma_{k}(n)}{(n_{1}n_{2}n_{3})^{1/2}}\bigg{(}\sum_{n}\frac{|\alpha_{-k}(n)|}{n^{1/2}}\bigg{)}^{2}\ll\frac{Y^{2+\epsilon}X^{W_{0}(2+\epsilon)}}{T^{1/2}}=o(1)$

since $|\log(n_{3}n_{5}/n_{1}n_{2}n_{4})|\gg 1/(YX^{W_{0}})$. Accordingly,

$I_{3}=\sum_{\begin{subarray}{c}n_{1}n_{2}n_{4}=n_{3}n_{5}\\\ n_{1}\leqslant
T,n_{2},n_{3}\leqslant
Y\end{subarray}}\frac{\gamma_{k-1}(n_{2})\gamma_{k}(n_{3})\alpha_{-k}(n_{4})\alpha_{-k}(n_{5})}{(n_{1}n_{2}n_{3}n_{4}n_{5})^{1/2}}+o(1).$

Since $n_{3}n_{5}\leqslant YX^{W_{0}}\leqslant T$ we may remove the condition
$n_{1}\leqslant T$.

Now, since $\alpha_{-k}(n)$ is supported on prime powers $p^{m}$ with
$A<p\leqslant X$ we may write our sum as

(52)
$\displaystyle\sum_{n_{1}n_{2}n_{4}=n_{3}n_{5}}=\sum_{\begin{subarray}{c}n_{1}n_{2}n_{4}=n_{3}n_{5}\\\
p|n_{j}\implies A<p\leqslant
X\end{subarray}}\cdot\sum_{\begin{subarray}{c}n_{1}n_{2}=n_{3}\\\
p|n_{j}\implies X<p\leqslant
T_{0}\end{subarray}}\cdot\prod_{i=1}^{J}\sum_{\begin{subarray}{c}n_{1}n_{2}=n_{3}\\\
p|n_{j}\implies T_{i-1}<p\leqslant T_{i}\end{subarray}}$

and note that we have essentially taken $n_{4}=n_{5}=1$ in the second two sums
on the right. Unfolding the coefficients using (51) gives the first sum as

(53) $\sum_{\begin{subarray}{c}n_{1}n_{2}n_{4}=n_{3}n_{5}\\\ p|n_{j}\implies
A<p\leqslant X\\\ \Omega(n_{2}),\Omega(n_{3})\leqslant
10\ell_{0}\end{subarray}}\frac{(k-1)^{\Omega(n_{2})}k^{\Omega(n_{3})}\mathfrak{g}(n_{2})\mathfrak{g}(n_{3})\alpha_{-k}(n_{4})\alpha_{-k}(n_{5})}{(n_{1}n_{2}n_{3}n_{4}n_{5})^{1/2}}$

The usual arguments now allow us to replace $\alpha_{-k}(n_{j})$ with
$\beta_{-k}(n_{j})$ and remove the restrictions on
$\Omega(n_{2}),\Omega(n_{3})$ at the cost of a term of size $o(1)$. We then
express the resultant sum as an Euler product. Since $\beta_{-k}(p)=-k$ we
find that the leading terms cancel and that (53) is

(54) $\prod_{A<p\leqslant X}\Big{(}1+O_{k}(p^{-2})\Big{)}+o(1).$

On taking $A$ sufficiently large we can guarantee that the term
$O_{k}(p^{-2})$ is always $<1$ in modulus and hence the above product is
$\geqslant c$ for some constant $c>0$.

With a similar computation the second sum in (52) is

(55) $\begin{split}&\sum_{\begin{subarray}{c}n_{1}n_{2}=n_{3}\\\
p|n_{2},n_{3}\implies X<p\leqslant T_{0}\\\
\Omega(n_{2}),\Omega(n_{3})\leqslant
10\ell_{0}\end{subarray}}\frac{(k-1)^{\Omega(n_{2})}k^{\Omega(n_{3})}\mathfrak{g}(n_{2})\mathfrak{g}(n_{3})}{(n_{1}n_{2}n_{3})^{1/2}}\\\
=&\sum_{\begin{subarray}{c}n_{1}n_{2}=n_{3}\\\ p|n_{2},n_{3}\implies
X<p\leqslant
T_{0}\end{subarray}}\frac{(k-1)^{\Omega(n_{2})}k^{\Omega(n_{3})}\mathfrak{g}(n_{2})\mathfrak{g}(n_{3})}{(n_{1}n_{2}n_{3})^{1/2}}+O(e^{-10\ell_{0}}(\tfrac{\log
T_{0}}{\log X})^{O(1)})\\\ =&\prod_{X<p\leqslant
T_{0}}\bigg{(}1+\frac{k^{2}}{p}+O(p^{-2})\bigg{)}+o(1)\geqslant
c\bigg{(}\frac{\log T_{0}}{\log X}\bigg{)}^{k^{2}}.\end{split}$

For the sums inside the product in (52), we must be a little more careful. The
sums in question are given by

(56) $\sum_{\begin{subarray}{c}n_{1}n_{2}=n_{3}\\\ p|n_{2},n_{3}\implies
T_{i-1}<p\leqslant T_{i}\\\ \Omega(n_{2}),\Omega(n_{3})\leqslant
10\ell_{i}\end{subarray}}\frac{(k-1)^{\Omega(n_{2})}k^{\Omega(n_{3})}\mathfrak{g}(n_{2})\mathfrak{g}(n_{3})}{(n_{1}n_{2}n_{3})^{1/2}}.$

Since $0<k\leqslant 1$ and $d(n)\leqslant 2^{\Omega(n)}\leqslant
e^{\Omega(n)}$, the error incurred from dropping the condition on
$\Omega(n_{2})$ is, in absolute value,

$\displaystyle\leqslant$ $\displaystyle
e^{-10\ell_{i}}\sum_{\begin{subarray}{c}n_{1}n_{2}=n_{3}\\\
p|n_{2},n_{3}\implies T_{i-1}<p\leqslant
T_{i}\end{subarray}}\frac{e^{\Omega(n_{2})}\mathfrak{g}(n_{2})\mathfrak{g}(n_{3})}{(n_{1}n_{2}n_{3})^{1/2}}\leqslant
e^{-10\ell_{i}}\sum_{\begin{subarray}{c}p|n\implies T_{i-1}<p\leqslant
T_{i}\end{subarray}}\frac{e^{2\Omega(n)}\mathfrak{g}(n)}{n}$ $\displaystyle=$
$\displaystyle\exp\Big{(}-10\ell_{i}+e^{2}\sum_{T_{i-1}<p\leqslant
T_{i}}\frac{1}{p}\Big{)}\leqslant\exp\Big{(}-10\ell_{i}+e^{2}\log(\tfrac{\log
T_{i}}{\log T_{i-1}})+o(1)\Big{)}\leqslant e^{-9\ell_{i}}$

since $\log(\tfrac{\log T_{i}}{\log T_{i-1}})\leqslant 2$ and
$\ell_{i}\geqslant 10^{9}$. Doing the same for $n_{3}$ gives an error of the
same size and hence the sum in (56) is

(57) $\begin{split}\geqslant&\prod_{T_{i-1}<p\leqslant
T_{i}}\bigg{(}1+\frac{k^{2}}{p}+O(p^{-2})\bigg{)}-e^{-8\ell_{i}}\\\
\geqslant&\big{(}1-e^{-7\ell_{i}}\big{)}\prod_{T_{i-1}<p\leqslant
T_{i}}\bigg{(}1+\frac{k^{2}}{p}+O(p^{-2})\bigg{)}.\end{split}$

Hence, combining (54), (LABEL:second_sum) and (57) we find that

$I_{3}\geqslant c\bigg{(}\frac{\log T_{0}}{\log
X}\bigg{)}^{k^{2}}\prod_{i=1}^{J}\big{(}1-e^{-7\ell_{i}}\big{)}\prod_{T_{i-1}<p\leqslant
T_{i}}\bigg{(}1+\frac{k^{2}}{p}+O(p^{-2})\bigg{)}\geqslant C\bigg{(}\frac{\log
T}{\log X}\bigg{)}^{k^{2}}$

since, again, $\sum_{i=1}^{J}e^{-7\ell_{i}}$ is a rapidly converging series.
This completes the proof of Proposition 4.

###  7.2. Proof of Proposition 5

We are required to show

$I_{4}:=\frac{1}{T}\int_{T}^{2T}|\zeta(\tfrac{1}{2}+it){\mathcal{N}}(t,k-1)|^{2}|D(t,-k)|^{2}dt\ll\bigg{(}\frac{\log
T}{\log X}\bigg{)}^{k^{2}}.$

From the conditions on $X$ given in the statement of the proposition,
$D(t,-k)$ is a Dirichlet polynomial of length $X^{W_{0}}=T^{o(1)}$ which is
still short. Thus, we can apply the results of [5, 7] after combining the two
Dirichlet polynomials into a single sum. This gives

$I_{2}=\sum_{m,n}\frac{h_{k}(m)h_{k}(n)(m,n)}{mn}\log\Big{(}\frac{BT(m,n)^{2}}{mn}\Big{)}+o(1)$

for some constant $B$ where

$h_{k}(n)=\sum_{n_{1}n_{2}=n}\gamma_{k-1}(n_{1})\alpha_{-k}(n_{2}).$

As in [27], we write

$\log\Big{(}\frac{BT(m,n)^{2}}{mn}\Big{)}=\frac{1}{2\pi i}\int_{|z|=1/\log
T}\Big{(}\frac{BT(m,n)^{2}}{mn}\Big{)}^{z}\frac{dz}{z^{2}},$

so that the main term in $I_{2}$ becomes

$\frac{1}{2\pi i}\int_{|z|=1/\log
T}\sum_{m,n}\frac{h_{k}(m)h_{k}(n)(m,n)}{mn}\Big{(}\frac{BT(m,n)^{2}}{mn}\Big{)}^{z}\frac{dz}{z^{2}}.$

Then after a trivial estimate we get

(58) $I_{4}\ll(\log T)\max_{|z|=1/\log
T}\bigg{|}\sum_{m,n}\frac{h_{k}(m)h_{k}(n)(m,n)^{1+2z}}{(mn)^{1+z}}\bigg{|}$

and thus we are required to show this last sum is $\ll\tfrac{1}{\log X}(\log
T/\log X)^{k^{2}-1}$.

Unfolding the coefficients, the above sum becomes

$\displaystyle\sum_{m_{1},m_{2},n_{1},n_{2}}\frac{\gamma_{k-1}(m_{1})\gamma_{k-1}(n_{1})\alpha_{-k}(m_{2})\alpha_{-k}(n_{2})(m_{1}m_{2},n_{1}n_{2})^{1+2z}}{(m_{1}m_{2}n_{1}n_{2})^{1+z}}.$

Estimating the terms with $\Omega(m_{2}),\Omega(n_{2})\geqslant W_{0}$ in the
usual way we may replace $\alpha_{-k}(n)$ by $\beta_{-k}(n)$ and then re-
extend the sums at the cost of $o(1)$. Then by multiplicativity we can express
the resultant sum as

$\sum_{\begin{subarray}{c}m_{1},m_{2},n_{1},n_{2}\\\ p|m_{j}n_{j}\implies
A<p\leqslant X\end{subarray}}\cdot\sum_{\begin{subarray}{c}m_{1},n_{1}\\\
p|m_{1}n_{1}\implies X<p\leqslant
T_{0}\end{subarray}}\cdot\prod_{i=1}^{J}\sum_{\begin{subarray}{c}m_{1},n_{1}\\\
p|m_{1}n_{1}\implies T_{i-1}<p\leqslant T_{i}\end{subarray}}$

where, again, we have taken $m_{2}=n_{2}=1$ in the second two sums since the
functions $\beta_{k}(n)$ are supported on $X$-smooth numbers. As usual, we
drop the conditions on $\Omega(m_{j}),\Omega(n_{j})$ in these sums.
Considering, for the moment, the first sum without these conditions we get

$\displaystyle\prod_{A<p\leqslant
X}\bigg{(}1+\sum_{\begin{subarray}{c}0\leqslant m_{j},n_{j}\leqslant 1\\\
m_{1}+m_{2}+n_{1}+n_{2}\neq
0\end{subarray}}\frac{(k-1)^{m_{1}+n_{1}}(-k)^{m_{2}+n_{2}}(p^{m_{1}+m_{2}},p^{n_{1}+n_{2}})^{1+2z}}{p^{(m_{1}+m_{2}+n_{1}+n_{2})(1+z)}}+O(p^{-2})\bigg{)}$
$\displaystyle=$ $\displaystyle\prod_{A<p\leqslant
X}\bigg{(}1+\frac{2(k-1)-2k}{p^{1+z}}+\frac{(k-1)^{2}+k^{2}-2k(k-1)}{p}+O(p^{-2})\bigg{)}$
$\displaystyle=$ $\displaystyle\prod_{A<p\leqslant
X}\bigg{(}1-\frac{1}{p}+O(p^{-2})+O\Big{(}\frac{1}{\log T}\frac{\log
p}{p}\Big{)}\bigg{)}\ll\frac{1}{\log X}$

where we have used $\beta_{-k}(n)\ll d_{k}(n)$ and $|z|\leqslant 1/\log T$
along with the bound $\frac{1}{\log T}\sum_{p\leqslant X}(\log p)/p\ll 1$. As
usual, for the error term we apply Rankin’s trick along with similar Euler
product computations to give an error of size

$\ll e^{-10\ell_{0}+O(\log\log X)}$

which is $o(1/\log X)$.

For the second sum we get, with a similar argument,

$\displaystyle\sum_{\begin{subarray}{c}m_{1},n_{1}\\\ p|m_{1}n_{1}\implies
X<p\leqslant T_{0}\end{subarray}}$
$\displaystyle\frac{\gamma_{k-1}(m_{1})\gamma_{k-1}(n_{1})(m_{1},n_{1})^{1+2z}}{(m_{1}n_{1})^{1+z}}$
$\displaystyle=$ $\displaystyle\prod_{X<p\leqslant
T_{0}}\bigg{(}1+\frac{k^{2}-1}{p}+O(p^{-2})\bigg{)}+O(e^{-10\ell_{0}+O(\log(\log
T_{0}/\log X))})$

which is $\ll(\frac{\log T_{0}}{\log X})^{k^{2}-1}$. Finally, for the product
of sums we have

$\displaystyle\prod_{i=1}^{J}$
$\displaystyle\sum_{\begin{subarray}{c}m_{1},n_{1}\\\ p|m_{1}n_{1}\implies
T_{i-1}<p\leqslant
T_{i}\end{subarray}}\frac{\gamma_{k-1}(m_{1})\gamma_{k-1}(n_{1})(m_{1},n_{1})^{1+2z}}{(m_{1}n_{1})^{1+z}}$
$\displaystyle=$
$\displaystyle\prod_{i=1}^{J}\bigg{[}\prod_{T_{i-1}<p\leqslant
T_{i}}\bigg{(}1+\frac{k^{2}-1}{p}+O(p^{-2})\bigg{)}+O(e^{-10\ell_{i}+O(\log(\log
T_{i}/\log T_{i-1}))})\bigg{]}$ $\displaystyle=$
$\displaystyle\prod_{i=1}^{J}\bigg{[}\big{(}1+O(e^{-10\ell_{i}})\big{)}\prod_{T_{i-1}<p\leqslant
T_{i}}\bigg{(}1+\frac{k^{2}-1}{p}+O(p^{-2})\bigg{)}\bigg{]}$

since $\log T_{i}/\log T_{i-1}\leqslant 2$. This last term is $\ll(\frac{\log
T}{\log T_{0}})^{k^{2}-1}$ and so combining these bounds in (58) we get

$I_{4}\ll\log T\cdot\frac{1}{\log X}\cdot\bigg{(}\frac{\log T_{0}}{\log
X}\bigg{)}^{k^{2}-1}\cdot\bigg{(}\frac{\log T}{\log
T_{0}}\bigg{)}^{k^{2}-1}\ll\bigg{(}\frac{\log T}{\log X}\bigg{)}^{k^{2}}$

which completes the proof of Proposition 5.

###  7.3. Proof of Proposition 6

We begin with the following lemma from [27].

######  Lemma 9.

Let

$\mathcal{P}_{j}(t):=\sum_{T_{j-1}<p\leqslant T_{j}}\frac{1}{p^{1/2+it}}.$

Then for $0\leqslant j\leqslant J$

$|{\mathcal{N}}_{j}(t,k-1){\mathcal{N}}_{j}(t,k)^{\frac{1}{k}}|^{2}\leqslant|{\mathcal{N}}_{j}(t,k)|^{2}(1+O(e^{-9\ell_{j}}))+O\big{(}{\mathcal{Q}}_{j}(t)\big{)},$

where the implied constants are absolute, and

${\mathcal{Q}}_{j}(t)=\Big{(}\frac{e|{\mathcal{P}}_{j}(t)|}{10\ell_{j}}\Big{)}^{20\ell_{j}}\sum_{r=0}^{10\ell_{j}/k}\Big{(}\frac{2e|{\mathcal{P}}_{j}(t)|}{r+1}\Big{)}^{2r}.$

###### Proof.

This is essentially Lemma 1 of [27]. Our sequence $T_{j}$ is defined slightly
differently but one can check that this makes no difference to the end result.
∎

######  Lemma 10.

With the above notation

$\frac{1}{T}\int_{T}^{2T}{\mathcal{Q}}_{0}(t)|D(t,-k)|^{2}dt\ll
e^{-10\ell_{0}}(\log X)^{2k^{2}}$

and for $1\leqslant j\leqslant J$

$\frac{1}{T}\int_{T}^{2T}{\mathcal{Q}}_{j}(t)dt\ll e^{-10\ell_{j}}.$

###### Proof.

We prove the first bound since this is new, the second bound follows similarly
(and is essentially Lemma 2 of [27]). Let $L=10\ell_{0}$. From the definition
of $\mathcal{Q}_{0}(t)$ we have

$\displaystyle\frac{1}{T}\int_{T}^{2T}$
$\displaystyle{\mathcal{Q}}_{0}(t)|D(t,-k)|^{2}dt$ $\displaystyle=$
$\displaystyle\Big{(}\frac{12}{L}\Big{)}^{2L}\sum_{r=0}^{L/k}\Big{(}\frac{2e}{r+1}\Big{)}^{2r}\cdot\frac{1}{T}\int_{T}^{2T}|{\mathcal{P}}_{0}(t)|^{2L+2r}|D(t,-k)|^{2}dt$

By the Cauchy–Schwarz inequality and (3) the integral is

$\ll(\log
X)^{2k^{2}}\bigg{(}\frac{1}{T}\int_{T}^{2T}|{\mathcal{P}}_{0}(t)^{2L+2r}|^{2}dt\bigg{)}^{1/2}.$

Since

${\mathcal{P}}_{0}(t)^{2L+2r}=(2L+2r)!\sum_{\begin{subarray}{c}\Omega(n)=2L+2r\\\
p|n\implies p\leqslant
T_{0}\end{subarray}}\frac{\mathfrak{g}(n)}{n^{1/2+it}},$

the Montgomery–Vaughan mean value theorem gives that our original integral is

(59) $\begin{split}\ll&(\log
X)^{2k^{2}}\Big{(}\frac{12}{L}\Big{)}^{2L}\sum_{r=0}^{L/k}\Big{(}\frac{2e}{r+1}\Big{)}^{2r}\cdot\bigg{(}(2L+2r)!^{2}\sum_{\begin{subarray}{c}\Omega(n)=2L+2r\\\
p|n\implies p\leqslant
T_{0}\end{subarray}}\frac{\mathfrak{g}(n)^{2}}{n}\bigg{)}^{1/2}\\\
\leqslant&(\log
X)^{2k^{2}}\Big{(}\frac{12}{L}\Big{)}^{2L}\sum_{r=0}^{L/k}\Big{(}\frac{2e}{r+1}\Big{)}^{2r}\cdot(2L+2r)!^{1/2}\bigg{(}\sum_{p\leqslant
T_{0}}\frac{1}{p}\bigg{)}^{L+r}\end{split}$

since $\mathfrak{g}(n)^{2}\leqslant\mathfrak{g}(n)$. Letting
$P=\sum_{p\leqslant T_{0}}p^{-1}$ we find by Stirling’s formula that the
summand is

$\ll(2/e)^{L}(8e)^{r}(r+1)^{-2r}({L+r})^{L+r+{1}/{4}}P^{L+r}$

which is maximised at the solution of $r^{2}=8P(L+r)(1+O(1/r))$. Since
$P\leqslant 2\log\log T=o(L)$ this solution $r=r_{0}$ satisfies
$2\sqrt{2}(PL)^{1/2}\leqslant r_{0}\leqslant 3(PL)^{1/2}$. Therefore, (59) is

$\ll(\log
X)^{2k^{2}}\Big{(}\frac{c}{L}\Big{)}^{2L}\cdot\frac{L}{k}\cdot(3L)!^{1/2}P^{L+r_{0}}r_{0}^{-2r_{0}}$

since $2r_{0}\leqslant L$. This is then

$\ll(\log X)^{2k^{2}}L^{-L/2+o(1)}\ll e^{-10\ell_{0}}$

and the result follows. ∎

By Lemma 9 and (5) we find that

$\displaystyle I_{5}:=$
$\displaystyle\frac{1}{T}\int_{T}^{2T}|{\mathcal{N}}(t,k)|^{\frac{2}{k}}|{\mathcal{N}}(t,k-1)|^{2}|D(t,-k)|^{2}dt$
$\displaystyle\ll$
$\displaystyle\frac{1}{T}\int_{T}^{2T}\Big{(}|{\mathcal{N}}_{0}(t,k)|^{2}(1+O(e^{-10\ell_{0}}))+O\big{(}{\mathcal{Q}}_{0}(t)\big{)}\Big{)}|D(t,-k)|^{2}dt$
$\displaystyle\times\prod_{j=1}^{J+1}\frac{1}{T}\int_{T}^{2T}\Big{(}|{\mathcal{N}}_{j}(t,k)|^{2}(1+O(e^{-10\ell_{j}}))+O\big{(}{\mathcal{Q}}_{j}(t)\big{)}\Big{)}dt$

since $J\asymp\log\log\log T$. By Lemma 10 we have

(60)
$\begin{split}&\prod_{j=1}^{J}\frac{1}{T}\int_{T}^{2T}\Big{(}|{\mathcal{N}}_{j}(t,k)|^{2}(1+O(e^{-10\ell_{j}}))+O\big{(}{\mathcal{Q}}_{j}(t)\big{)}\Big{)}dt\\\
=&\prod_{j=1}^{J}\bigg{(}(1+O(e^{-10\ell_{j}}))\sum_{\begin{subarray}{c}\Omega(n)\leqslant
10\ell_{j}\\\ p|n\implies T_{i-1}<p\leqslant
T_{i}\end{subarray}}\frac{k^{2\Omega(n)}\mathfrak{g}(n)^{2}}{n}+O(e^{-10\ell_{j}})\bigg{)}\\\
\ll&\prod_{j=1}^{J}\prod_{T_{i-1}<p\leqslant
T_{i}}\bigg{(}1+\frac{k^{2}}{p}+O(p^{-2})\bigg{)}\ll\bigg{(}\frac{\log T}{\log
T_{0}}\bigg{)}^{k^{2}}.\end{split}$

By Lemma 10 again we find

$\displaystyle\frac{1}{T}\int_{T}^{2T}\Big{(}|{\mathcal{N}}_{0}(t,k)|^{2}(1+O(e^{-10\ell_{0}}))+O\big{(}{\mathcal{Q}}_{0}(t)\big{)}\Big{)}|D(t,-k)|^{2}dt$
$\displaystyle=$
$\displaystyle(1+O(e^{-10\ell_{0}}))\frac{1}{T}\int_{T}^{2T}|{\mathcal{N}}_{0}(t,k)|^{2}\bigg{|}\sum_{n}\frac{\alpha_{-k}(n)}{n^{{1}/{2}+it}}\bigg{|}^{2}dt+o(1).$

This last integral is

$\sum_{\begin{subarray}{c}m_{1}n_{1}=m_{2}n_{2}\\\ \Omega(m_{j})\leqslant
10\ell_{0}\\\ p|m_{j}\implies A<p\leqslant
T_{0}\end{subarray}}\frac{k^{\Omega(m_{1})+\Omega(m_{2})}\mathfrak{g}(m_{1})\mathfrak{g}(m_{2})\alpha_{-k}(n_{1})\alpha_{-k}(n_{2})}{(m_{1}m_{2}n_{1}n_{2})^{1/2}}+o(1).$

The usual arguments allow us to remove the conditions on $\Omega(m_{j})$ and
replace $\alpha_{-k}(n)$ by $\beta_{-k}(n)$ at the cost of $o(1)$. We then
find that the resultant sum is

$\prod_{A<p\leqslant X}\Big{(}1+O(p^{-2})\Big{)}\prod_{X<p\leqslant
T_{0}}\Big{(}1+\frac{k^{2}}{p}+O(p^{-2})\Big{)}\ll\bigg{(}\frac{\log
T_{0}}{\log X}\bigg{)}^{k^{2}}$

since the leading terms cancel over $A<p\leqslant X$. Combining this with (60)
gives

$I_{5}\ll\bigg{(}\frac{\log T}{\log X}\bigg{)}^{k^{2}}$

thus completing the proof of Proposition 6.

##  8\. The lower bound of Proposition 3 for $k\geqslant 1$

The lower bound for $k\geqslant 1$ is similar to the case $0\leqslant
k\leqslant 1$, if not a little simpler. In this case we take

$T_{-1}=Bk^{2}$

for some $B>0$ to be chosen and we alter the definition of $J$ slightly so
that it is the maximal integer such that $e^{J}/(\log\log T)^{2}\leqslant
1/(10^{12}k^{4})$. This implies that $\ell_{j}=\theta_{j}^{-3/4}\geqslant
10^{9}k^{3}$ for all $j\leqslant J$.

We perform Hölder’s inequality in the form

(61)
$\Big{|}\frac{1}{T}\int_{\mathcal{S}}\zeta(\tfrac{1}{2}+it)\mathcal{N}(t,k-1)\overline{\mathcal{N}(t,k)}|D(t,-k)|^{2}dt\Big{|}\\\
\ll\Big{(}\frac{1}{T}\int_{T}^{2T}|Z_{X}(\tfrac{1}{2}+it)|^{2k}dt\Big{)}^{\frac{1}{2k}}\Big{(}\frac{1}{T}\int_{T}^{2T}|{\mathcal{N}}(t,k-1){\mathcal{N}}(t,k)|^{\frac{2k}{2k-1}}|D(t,-k)|^{2}dt\Big{)}^{\frac{2k-1}{2k}}$

where again we have used Lemma 5. The integral on the left can be dealt with
rather similarly to Proposition 4 although the change in parameters requires
some modifications. We detail these alterations first before dealing with the
second integral on the right.

###  8.1. Modifying the proof of Proposition 4

We see that we can arrive at (52) in exactly the same way. When dealing with
(53), the errors incurred from dropping the conditions on $\Omega(n_{2})$ etc.
are now

$\ll e^{-10\ell_{0}}(\log X)^{Ck^{2}}$

which of course is still $o(1)$ since $k$ is fixed. Note that on writing (53)
as a single sum $\sum_{n}f(n)/n$, the coefficients are supported on $X$-smooth
numbers and satisfy the bounds $|f(n)|\leqslant
k^{2\Omega(n)}d_{3}(n)^{2}\leqslant(3k)^{2\Omega(n)}$. Then we find that the
equivalent of (53) is

$\prod_{Bk^{2}<p\leqslant
X}\bigg{(}1+O\Big{(}\frac{k^{4}}{p^{2}}\Big{)}\bigg{)}+o(1)\geqslant C$

for some $C$ on taking $B$ large enough. In a similar way we find that the
equivalent of (LABEL:second_sum) is

$\prod_{X<p\leqslant
T_{0}}\bigg{(}1+\frac{k^{2}}{p}+O\Big{(}\frac{k^{4}}{p^{2}}\Big{)}\bigg{)}+o(1)\geqslant
C\bigg{(}\frac{\log T_{0}}{\log X}\bigg{)}^{k^{2}}.$

It remains to deal with the term

(62) $\prod_{i=1}^{J}\sum_{\begin{subarray}{c}n_{1}n_{2}=n_{3}\\\
p|n_{2},n_{3}\implies T_{i-1}<p\leqslant T_{i}\\\
\Omega(n_{2}),\Omega(n_{3})\leqslant
10\ell_{i}\end{subarray}}\frac{(k-1)^{\Omega(n_{2})}k^{\Omega(n_{3})}\mathfrak{g}(n_{2})\mathfrak{g}(n_{3})}{(n_{1}n_{2}n_{3})^{1/2}}$

in the current context. The error from removing the condition on
$\Omega(n_{2})$ in the sums is

$\displaystyle\leqslant$ $\displaystyle
e^{-10\ell_{i}}\sum_{\begin{subarray}{c}n_{1}n_{2}=n_{3}\\\
p|n_{2},n_{3}\implies T_{i-1}<p\leqslant
T_{i}\end{subarray}}\frac{(ek)^{\Omega(n_{2})}k^{\Omega(n_{3})}\mathfrak{g}(n_{2})\mathfrak{g}(n_{3})}{(n_{1}n_{2}n_{3})^{1/2}}$
$\displaystyle\leqslant$ $\displaystyle
e^{-10\ell_{i}}\sum_{\begin{subarray}{c}p|n\implies T_{i-1}<p\leqslant
T_{i}\end{subarray}}\frac{(ek)^{2\Omega(n)}\mathfrak{g}(n)}{n}=\exp\Big{(}-10\ell_{i}+e^{2}k^{2}\sum_{T_{i-1}<p\leqslant
T_{i}}\frac{1}{p}\Big{)}$ $\displaystyle\leqslant$
$\displaystyle\exp\Big{(}-10\ell_{i}+e^{2}k^{2}\log(\tfrac{\log T_{i}}{\log
T_{i-1}})+o(1)\Big{)}\leqslant e^{-9\ell_{i}}$

where we have used $d(n)\leqslant 2^{\Omega(n)}\leqslant e^{\Omega(n)}$. Then
(62) is

$\begin{split}\geqslant&\prod_{i=1}^{J}\bigg{(}\prod_{T_{i-1}<p\leqslant
T_{i}}\bigg{(}1+\frac{k^{2}}{p}+O\Big{(}\frac{k^{4}}{p^{2}}\Big{)}\bigg{)}-e^{-8\ell_{i}}\bigg{)}\\\
\geqslant&\prod_{i=1}^{J}\big{(}1-e^{-7\ell_{i}}\big{)}\prod_{T_{i-1}<p\leqslant
T_{i}}\bigg{(}1+\frac{k^{2}}{p}+O\Big{(}\frac{k^{4}}{p^{2}}\Big{)}\bigg{)}\geqslant
C\bigg{(}\frac{\log T}{\log T_{0}}\bigg{)}^{k^{2}}.\end{split}$

Combining these we get the desired bound

(63)
$\frac{1}{T}\int_{\mathcal{S}}\zeta(\tfrac{1}{2}+it)\mathcal{N}(t,k-1)\overline{\mathcal{N}(t,k)}|D(t,-k)|^{2}dt\geqslant
C\bigg{(}\frac{\log T}{\log X}\bigg{)}^{k^{2}}.$

###  8.2. The remaining integral

We let

$I_{6}=\frac{1}{T}\int_{T}^{2T}|{\mathcal{N}}(t,k-1){\mathcal{N}}(t,k)|^{\frac{2k}{2k-1}}|D(t,-k)|^{2}dt.$

Then Proposition 3 in the case $k\geqslant 1$ will follow from Hölder’s
inequality (61) and (63) if we can show that

$I_{6}\ll\bigg{(}\frac{\log T}{\log X}\bigg{)}^{k^{2}}.$

######  Lemma 11.

For $0\leqslant i\leqslant J$ we have

$|{\mathcal{N}_{i}}(t,k-1){\mathcal{N}_{i}}(t,k)|^{\frac{2k}{2k-1}}\leqslant(1+O(e^{-9\ell_{i}}))|\mathcal{N}_{i}(t,k)|^{2}+\Big{(}\frac{ek|\mathcal{P}_{i}(t)|}{10\ell_{i}}\Big{)}^{40\ell_{i}}.$

###### Proof.

If $k|\mathcal{P}_{i}(t)|\leqslant 10\ell_{i}$ then by (37) we have

$\displaystyle|{\mathcal{N}_{i}}(t,k-1){\mathcal{N}_{i}}(t,k)|^{\frac{2k}{2k-1}}=$
$\displaystyle(1+O(e^{-9\ell_{i}}))\big{|}\exp\big{(}(k-1)\mathcal{P}_{i}(t)+k\mathcal{P}_{i}(t)\big{)}\big{|}^{\frac{2k}{2k-1}}$
$\displaystyle=$
$\displaystyle(1+O(e^{-9\ell_{i}}))\big{|}\exp\big{(}k\mathcal{P}_{i}(t)\big{)}\big{|}^{2}$
$\displaystyle=$
$\displaystyle(1+O(e^{-9\ell_{i}}))|\mathcal{N}_{i}(t,k)|^{2}.$

If $k|\mathcal{P}_{i}(t)|>10\ell_{i}$ then

$\displaystyle|{\mathcal{N}_{i}}(t,k-1)|\leqslant$
$\displaystyle\sum_{r=0}^{10\ell_{i}}\frac{((k-1)|\mathcal{P}_{i}(t)|)^{r}}{r!}\leqslant(k|\mathcal{P}_{i}(t)|)^{10\ell_{i}}\sum_{r=0}^{10\ell_{i}}(10\ell_{i})^{r-10\ell_{i}}\frac{1}{r!}$
$\displaystyle\leqslant$
$\displaystyle\Big{(}\frac{ek|\mathcal{P}_{i}(t)|}{10\ell_{i}}\Big{)}^{10\ell_{i}}.$

The same bound holds for $|\mathcal{N}_{i}(t,k)|$ and hence the result follows
since $2k/(2k-1)\leqslant 2$. ∎

######  Lemma 12.

We have

$\frac{1}{T}\int_{T}^{2T}\Big{(}\frac{ek|\mathcal{P}_{0}(t)|}{10\ell_{0}}\Big{)}^{40\ell_{0}}|D(t,-k)|^{2}dt\ll
e^{-10\ell_{0}}(\log X)^{2k^{2}}$

and for $1\leqslant i\leqslant J$.

$\frac{1}{T}\int_{T}^{2T}\Big{(}\frac{ek|\mathcal{P}_{i}(t)|}{10\ell_{i}}\Big{)}^{40\ell_{i}}dt\ll
e^{-10\ell_{i}}.$

###### Proof.

We prove the first formula since the second follows similarly. By the
Cauchy–Schwarz inequality and (3) we have

$\frac{1}{T}\int_{T}^{2T}|\mathcal{P}_{0}(t)|^{40\ell_{0}}|D(t,-k)|^{2}dt\ll(\log
X)^{2k^{2}}\bigg{(}\frac{1}{T}\int_{T}^{2T}|\mathcal{P}_{0}(t)^{40\ell_{0}}|^{2}dt\bigg{)}^{1/2}.$

Letting $L=10\ell_{0}=10(\log\log T)^{3/2}$ the last integral is

$\ll(4L)!^{2}\sum_{\begin{subarray}{c}\Omega(n)=4L\\\ p|n\implies
T_{-1}<p\leqslant
T_{0}\end{subarray}}\frac{\mathfrak{g}(n)^{2}}{n}\leqslant(4L)!\bigg{(}\sum_{T_{-1}<p\leqslant
T_{0}}\frac{1}{p}\bigg{)}^{4L}\leqslant(4L)!(\log\log T)^{4L}$

using $\mathfrak{g}(n)^{2}\leqslant\mathfrak{g}(n)$. Thus, the original
integral is

$\ll(\log
X)^{2k^{2}}\bigg{(}\frac{ek}{L}\bigg{)}^{4L}(4L)!^{1/2}\Big{(}\frac{L}{10}\Big{)}^{4L/3}\ll(\log
X)^{2k^{2}}L^{-2L/3}c^{L}$

by Stirling’s formula. The result follows. ∎

As in subsection 7.3 we now combine Lemmas 11 and 12 along with (5) and the
usual calculations to give

$I_{6}\ll\bigg{(}\frac{\log T}{\log X}\bigg{)}^{k^{2}}.$

This completes the proof of Proposition 3 in the case $k\geqslant 1$.

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