# Negative moments of the Riemann zeta-function

Hung M. Bui and Alexandra Florea  Department of Mathematics, University of
Manchester, Manchester M13 9PL, UK<EMAIL_ADDRESS>UC Irvine,
Mathematics Department, Rowland Hall, Irvine 92697, USA<EMAIL_ADDRESS>
###### Abstract.

Assuming the Riemann Hypothesis we study negative moments of the Riemann zeta-
function and obtain asymptotic formulas in certain ranges of the shift in
$\zeta(s)$. For example, integrating $|\zeta(1/2+\alpha+it)|^{-2k}$ with
respect to $t$ from $T$ to $2T$, we obtain an asymptotic formula when the
shift $\alpha$ is roughly bigger than $\frac{1}{\log T}$ and $k<1/2$. We also
obtain non-trivial upper bounds for much smaller shifts, as long as
$\log\frac{1}{\alpha}\ll\log\log T$. This provides partial progress towards a
conjecture of Gonek on negative moments of the Riemann zeta-function, and
settles the conjecture in certain ranges. As an application, we also obtain an
upper bound for the average of the generalized Möbius function.

###### Key words and phrases:

Riemann zeta-function, moments, negative moments, Möbius function.

###### 2010 Mathematics Subject Classification:

11M06, 11M50

##  1\. Introduction

For $k>0$, the $2k^{\text{th}}$ moment of the Riemann zeta-function is given
by

$I_{k}(T)=\int_{0}^{T}|\zeta(\tfrac{1}{2}+it)|^{2k}\,dt.$

Hardy and Littlewood [11] computed the second moment, and Ingham [17] computed
the fourth moment. It is conjectured that

$I_{k}(T)\sim c_{k}T(\log T)^{k^{2}}$ (1.1)

for an explicit constant $c_{k}$, whose value was predicted by Keating and
Snaith [18] using analogies with random matrix theory. Their conjecture was
later refined by Conrey, Farmer, Keating, Rubinstein and Snaith [6] to include
lower order powers of $\log T$, for integer $k$. Under the Riemann Hypothesis
(RH), Soundararajan [29] established almost sharp upper bounds for all the
positive moments. This result was later improved by Harper [12], who obtained
upper bounds of the conjectural magnitude as in (1.1). There is a wealth of
literature on obtaining lower and upper bounds for positive moments of
$\zeta(s)$; for (an incomplete) list of results, we refer the reader to [27,
28, 15, 24, 26, 3, 13, 14].

In this paper, we are interested in studying negative moments of the Riemann
zeta-function. For $k>0$ and $\alpha>0$, let

$I_{-k}(\alpha,T)=\frac{1}{T}\int_{0}^{T}|\zeta(\tfrac{1}{2}+\alpha+it)|^{-2k}\,dt.$

A conjecture due to Gonek [10] states the following.

######  Conjecture 1.1 (Gonek).

Let $k>0$. Uniformly for $\frac{1}{\log T}\leq\alpha\leq 1$,

$I_{-k}(\alpha,T)\asymp\Big{(}\frac{1}{\alpha}\Big{)}^{k^{2}},$

and uniformly for $0<\alpha\leq\frac{1}{\log T}$,

$I_{-k}(\alpha,T)\asymp\begin{cases}(\log T)^{k^{2}}&\mbox{ if }k<1/2,\\\
\big{(}\log\frac{e}{\alpha\log T}\big{)}(\log T)^{k^{2}}&\mbox{ if }k=1/2.\\\
\end{cases}$

Gonek’s original conjecture predicted formulas for $k>1/2$ and
$\alpha\leq\frac{1}{\log T}$ as well, which seem to be contradicted however by
more recent evidence (see the next paragraph). Under RH, Gonek [10] also
proved lower bounds of the conjectured order of magnitude for all $k>0$ and
$\frac{1}{\log T}\leq\alpha\leq 1$, and for $k<1/2$ and
$0<\alpha\leq\frac{1}{\log T}.$

No other progress has been made towards Gonek’s conjecture so far, but more
recent random matrix theory computations due to Berry and Keating [2] and
Forrester and Keating [9] suggest that certain corrections to the above
conjecture are due in some ranges. Namely, when $\alpha\leq\frac{1}{\log T}$,
random matrix theory computations seem to contradict Gonek’s prediction for
the negative moments when $k\geq 3/2$. In particular, the work in [2, 9]
suggests certain transition regimes when $k=(2n+1)/2$, for $n$ a positive
integer (Gonek’s conjecture already captures the first transition at $k=1/2$
featuring a logarithmic correction, and in this case the two conjectures do
agree.)

Reinterpreting the random matrix theory computations in [2], one would expect
that for a shift $0<\alpha\leq\frac{1}{\log T}$ and $j$ a natural number such
that $(2j-1)/2<k<(2j+1)/2$,

$I_{-k}(\alpha,T)\asymp(\log T)^{k^{2}}(\alpha\log T)^{-j(2k-j)},$ (1.2)

while for $k=(2j-1)/2$ and $j$ a natural number, one would expect

$I_{-k}(\alpha,T)\asymp\log\Big{(}\frac{e}{\alpha\log T}\Big{)}(\log
T)^{k^{2}}(\alpha\log T)^{-j(2k-j)}.$

We note that the above prediction indeed agrees with Conjecture 1.1 for
$k=1/2$ and $\alpha\leq\frac{1}{\log T}$. We remark that one could also
predict (1.2) for integer $k$ using heuristic ideas similar as in [6].

In this paper, we study the negative moments of the Riemann zeta-function.
While obtaining lower bounds for the negative moments is a more tractable
problem (see the comment after Conjecture 1.1), no progress has been made so
far on obtaining asymptotic formulas or non-trivial upper bounds. When the
shift $\alpha$ is “big enough”, we obtain upper bounds which are almost sharp
according to Conjecture 1.1, up to some logarithmic factors. We also obtain
the first non-trivial upper bounds for the negative moments for a wide range
of much smaller shifts $\alpha$ (roughly $\alpha\gg(\log T)^{-O(1)}$);
however, the bounds in these cases are far from sharp.

More precisely, we prove the following.

######  Theorem 1.2.

Assume RH. Let $k\geq 1/2,\alpha>0$ and $\varepsilon,\delta>0$, such that
$u=\frac{\log\frac{1}{\alpha}}{\log\log T}\ll 1$. Then

$\frac{1}{T}\int_{T}^{2T}|\zeta(\tfrac{1}{2}+\alpha+it)|^{-2k}\,dt\ll$
$\displaystyle(\log\log T)^{k}(\log T)^{k^{2}}$ if $\alpha\gg\frac{(\log\log
T)^{\frac{4}{k}+\varepsilon}}{(\log T)^{\frac{1}{2k}}},$ (1.3)
$\displaystyle\exp\Big{(}\frac{(4+\varepsilon)\log T\log\log\log T}{\log\log
T}\Big{)}$ if $\frac{1}{(\log
T)^{\frac{1}{2k}}}\ll\alpha=o\big{(}\frac{(\log\log
T)^{\frac{4}{k}+\varepsilon}}{(\log T)^{\frac{1}{2k}}}\big{)},$ (1.4)
$\displaystyle T^{(1+\delta)(ku-\frac{1}{2}+k\varepsilon)}$ if
$\alpha\ll\frac{1}{(\log T)^{\frac{1}{2k}}}$. (1.5)

We also have the following bounds for $k<1/2$.

######  Theorem 1.3.

Assume RH. Let $k<1/2,\alpha>0$ and $\varepsilon,\delta>0$, such that
$u=\frac{\log\frac{1}{\alpha}}{\log\log T}\ll 1$. Then

$\frac{1}{T}\int_{T}^{2T}|\zeta(\tfrac{1}{2}+\alpha+it)|^{-2k}\,dt\ll$
$\displaystyle(\log\log T)^{k}\Big{(}\frac{\log(\alpha\log
T)}{\alpha}\Big{)}^{k^{2}}$ if $\alpha\gg\frac{\log\log T}{\log T},$ (1.6)
$\displaystyle\exp\Big{(}C_{1}(\log\log T)\Big{(}\log\frac{\log\log
T}{\alpha\log T}\Big{)}^{2}\Big{)}$ if $\frac{1}{\log
T}\ll\alpha=o\big{(}\frac{\log\log T}{\log T}\big{)},$ (1.7)
$\displaystyle\exp\Big{(}C_{2}\Big{(}\frac{1}{\alpha\log
T}\Big{)}^{\frac{2k}{1-2k-k\varepsilon}}\Big{)}$ if $\frac{1}{(\log
T)^{\frac{1}{2k}-\varepsilon}}\ll\alpha=o\big{(}\frac{1}{\log T}\big{)},$
(1.8) $\displaystyle
T^{(1+\delta)(ku-\frac{1}{2}+k\varepsilon)}\exp\Big{(}O\Big{(}\frac{\log
T\log\log\log T}{\log\log T}\Big{)}\Big{)}$ if $\alpha=o\big{(}\frac{1}{(\log
T)^{\frac{1}{2k}-\varepsilon}}\big{)}$, (1.9)

for some constants $C_{1},C_{2}>0$.

Remarks.

  1. (1)

We remark that results about negative moments of quadratic Dirichlet
$L$–functions were recently proven in the function field setting, first in [4]
for shifts big enough, and later improved in [8] to include small shifts.

  2. (2)

We note that the bound (1.3) in Theorem 1.2 and the bound (1.6) in Theorem 1.3
above are “almost sharp”. The former is off possibly by some powers of $\log$,
while the latter is off by some $\log\log$ factor.

We also further use the upper bounds to obtain an asymptotic formula for the
negative moments in the following ranges.

######  Theorem 1.4.

Assume RH. Let $k>0$, $C,\varepsilon>0$ and
$\alpha\geq\max\Big{\\{}C\frac{(\log\log T)^{\frac{4}{k}+\varepsilon}}{(\log
T)^{\frac{1}{2k}}},\frac{(1+\varepsilon)\log\log T}{2\log T}\Big{\\}}$. Then
we have

$\frac{1}{T}\int_{T}^{2T}|\zeta(\tfrac{1}{2}+\alpha+it)|^{-2k}\,dt=\big{(}1+o(1)\big{)}\zeta(1+2\alpha)^{k^{2}}\prod_{p}\Big{(}1-\frac{1}{p^{1+2\alpha}}\Big{)}^{k^{2}}\Big{(}1+\sum_{j=1}^{\infty}\frac{\mu_{k}(p^{j})^{2}}{p^{(1+2\alpha)j}}\Big{)},$

where $\mu_{k}(n)$ denotes the $n^{\text{th}}$ Dirichlet coefficient of
$\zeta(s)^{-k}$.

As an application, we study averages of the generalized Möbius function. Under
RH, Littlewood [20] proved that $\sum_{n\leq x}\mu(n)\ll
x^{\frac{1}{2}+\varepsilon}$. This was improved in several works [References,
References, References]. Soundararajan [30] showed that

$\sum_{n\leq x}\mu(n)\ll\sqrt{x}\exp\Big{(}\sqrt{\log x}(\log\log
x)^{14}\Big{)}$

on RH. A note of Balazard and de Roton [1] on Soundararajan’s paper [30]
improves the power of $\log\log x$ in the bound above, and it is stated in [1]
that the power $(\log\log x)^{\frac{5}{2}+\varepsilon}$ is the limitation of
the ideas in [30, 1].

Assuming RH and the conjectured order of magnitude of the negative second
moment of $\zeta^{\prime}(\rho)$ by Gonek [10] and Hejhal [16], Ng [23] also
showed that

$\sum_{n\leq x}\mu(n)\ll\sqrt{x}(\log x)^{\frac{3}{2}}.$

Using Theorem 1.2, we improve the result in [30]. As before, let $\mu_{k}(n)$
denote the $n^{\text{th}}$ Dirichlet coefficient of $\zeta(s)^{-k}$. We prove
the following.

######  Theorem 1.5.

Assume RH. We have

$\sum_{n\leq
x}\mu_{k}(n)\ll\begin{cases}\sqrt{x}\exp\big{(}\varepsilon\sqrt{\log
x}\big{)}&\emph{if }k<1,\\\ \sqrt{x}\exp\Big{(}(\log
x)^{\frac{k}{k+1}}(\log\log x)^{\frac{7}{k+1}+\varepsilon}\Big{)}&\emph{if
}k\geq 1.\end{cases}$

Also, if we assume Conjecture 1.1, then

$\sum_{n\leq x}\mu_{k}(n)\ll\begin{cases}\sqrt{x}(\log
x)^{\frac{k^{2}}{4}}&\emph{if }k\leq 2,\\\ \sqrt{x}\exp\Big{(}(\log
x)^{\frac{k-2}{k}}(\log\log x)^{-1+\varepsilon}\Big{)}&\emph{if
}k>2.\end{cases}$

Proving Theorem 1.5 requires good bounds for the negative moments of
$\zeta(s)$ when roughly $\alpha\gg\frac{1}{(\log T)^{\frac{1}{2k}}}$. We
remark that obtaining the bound (1.3) in Theorem 1.2 is the most delicate part
of the proof and requires a recursive argument which allows us to obtain
improved estimates at each step.

The ideas in the proof of Theorems 1.2 and 1.3 are sieve-theoretic inspired
ideas, as in the work of Soundararajan [29] and Harper [12]. However, unlike
in the work in [29] and [12], the contributions of zeros of $\zeta(s)$ play an
important part, and one needs to be careful about the choice of parameters in
the sieve-theoretic argument, in order to account for the big contributions
coming from the zeros.

The paper is organized as follows. In Section 2, we prove some lemmas
providing a lower bound for the logarithm of $\zeta(s)$, as well as a
pointwise lower bound for $|\zeta(s)|$. In Section 3, we introduce the setup
of the problem, and prove three main propositions. In Section 4, we consider
the case of a “big” shift $\alpha$, and do the first steps towards proving
Theorems 1.2 and 1.3 in this case. In Section 5, we obtain the bound (1.3) in
the smaller region $\alpha\gg\frac{1}{(\log T)^{\frac{1}{2k}-\varepsilon}}$,
and prove the bounds (1.6), (1.7) and (1.8) from Theorem 1.3 as well. In
Section 6, we obtain the bound (1.3) in the wider region stated in the theorem
by using a recursive argument. We consider the cases of “small” shifts in
Section 7 and prove the bounds (1.5) and (1.9). We then prove Theorem 1.4 in
Section 8, and Theorem 1.5 in Section 9.

Throughout the paper $\varepsilon$ denotes an arbitrarily small positive
number whose value may change from one line to the next.

Acknowledgments. The authors would like to thank Steve Gonek, Jon Keating and
Nathan Ng for helpful conversations and comments. The second author gratefully
acknowledges support from NSF grant DMS-2101769 while working on this paper.

##  2\. Preliminary lemmas

The first two lemmas concern the lower bounds for $|\zeta(s)|$.

######  Lemma 2.1.

Assume RH. Let $\alpha>0$. Then

$\displaystyle\log|\zeta(\tfrac{1}{2}+\alpha+it)|\geq\frac{\log\frac{t}{2\pi}}{2\pi\Delta}\log\big{(}1-e^{-2\pi\Delta\alpha}\big{)}+\Re\Big{(}\sum_{p\leq
e^{2\pi\Delta}}\frac{(\log p)a_{\alpha}(p;\Delta)}{p^{1/2+\alpha+it}}\Big{)}$
$\displaystyle\qquad\qquad-\frac{\log\log\log
T}{2}-\Big{(}\log\frac{1}{\Delta\alpha}\Big{)}^{\gamma(\Delta)}+O\Big{(}\frac{\Delta^{2}e^{\pi\Delta}}{1+\Delta
t}+\frac{\Delta\log(1+\Delta\sqrt{t})}{\sqrt{t}}+1\Big{)},$

where

$a_{\alpha}(n;\Delta)=\sum_{j=0}^{\infty}\Big{(}\frac{(j+1)}{\log n+2\pi
j\Delta}e^{-2\pi j\Delta\alpha}-\frac{(j+1)n^{2\alpha}}{2\pi(j+2)\Delta-\log
n}e^{-2\pi(j+2)\Delta\alpha}\Big{)}$ (2.1)

and

$\gamma(\Delta)=\begin{cases}1&\text{if }\Delta\alpha=o(1),\\\ 0&\text{if
}\Delta\alpha\gg 1.\end{cases}$

###### Proof.

We use the work in [5]. Let

$f_{\alpha}(x)=\log\Big{(}\frac{4+x^{2}}{\alpha^{2}+x^{2}}\Big{)},$

and let $m_{\Delta}(x)$ be an extremal majorant for $f_{\alpha}$, whose
Fourier transform $\widehat{m}_{\Delta}$ is supported in $[-\Delta,\Delta]$,
satisfying the properties from Lemma 8 in [5]. Equation $3.1$ in [5] gives

$\log|\zeta(\tfrac{1}{2}+\alpha+it)|\geq\Big{(}1-\frac{\alpha}{2}\Big{)}\log\frac{t}{2}-\frac{1}{2}\sum_{\gamma}m_{\Delta}(t-\gamma)+O(1),$
(2.2)

where the sum over $\gamma$ is over the ordinates of the zeros of $\zeta(s)$
on the critical line. Using the explicit formula, Equation $3.2$ in [5] leads
to

$\displaystyle\sum_{\gamma}m_{\Delta}(t-\gamma)$
$\displaystyle=m_{\Delta}\Big{(}t-\frac{1}{2i}\Big{)}+m_{\Delta}\Big{(}t+\frac{1}{2i}\Big{)}-\frac{\log\pi}{2\pi}\widehat{m}_{\Delta}(0)$
(2.3) $\displaystyle\
+\frac{1}{2\pi}\int_{-\infty}^{\infty}m_{\Delta}(x)\Re\bigg{(}\frac{\Gamma^{\prime}}{\Gamma}\Big{(}\frac{1}{4}+\frac{i(t-x)}{2}\Big{)}\bigg{)}\,dx-\frac{1}{\pi}\Re\bigg{(}\sum_{n=2}^{\infty}\frac{\Lambda(n)}{n^{1/2+it}}\widehat{m}_{\Delta}\Big{(}\frac{\log
n}{2\pi}\Big{)}\bigg{)},$

where

$\displaystyle\widehat{m}_{\Delta}(\xi)$
$\displaystyle=\sum_{j=0}^{\infty}\bigg{(}\frac{j+1}{\xi+j\Delta}\Big{(}e^{-2\pi(\xi+j\Delta)\alpha}-e^{-4\pi(\xi+j\Delta)}\Big{)}$
$\displaystyle\qquad\qquad\qquad\qquad-\frac{j+1}{(j+2)\Delta-\xi}\Big{(}e^{-2\pi((j+2)\Delta-\xi)\alpha}-e^{-4\pi((j+2)\Delta-\xi)}\Big{)}\bigg{)}.$

For $n\leq e^{2\pi\Delta}$ we note that

$\widehat{m}_{\Delta}\Big{(}\frac{\log n}{2\pi}\Big{)}=\frac{2\pi
a_{\alpha}(n;\Delta)}{n^{\alpha}}+O\Big{(}\frac{1}{n^{2}\log n}\Big{)}.$

Combining that with (2.2), (2.3) and Equations $(3.3),(3.4),(3.5)$ in [5] we
obtain

$\displaystyle\log|\zeta(\tfrac{1}{2}+\alpha+it)|$
$\displaystyle\geq\frac{\log\frac{t}{2\pi}}{2\pi\Delta}\log\big{(}1-e^{-2\pi\Delta\alpha}\big{)}+\Re\Big{(}\sum_{n\leq
e^{2\pi\Delta}}\frac{\Lambda(n)a_{\alpha}(n;\Delta)}{n^{1/2+\alpha+it}}\Big{)}$
(2.4)
$\displaystyle\qquad\qquad+O\Big{(}\frac{\Delta^{2}e^{\pi\Delta}}{1+\Delta
t}+\frac{\Delta\log(1+\Delta\sqrt{t})}{\sqrt{t}}+1\Big{)}.$

It is easy to see that $a_{\alpha}(n;\Delta)\geq 0$. Also, with $\log n\leq
2\pi\Delta$ we have

$\displaystyle a_{\alpha}(n;\Delta)$ $\displaystyle<\frac{1}{\log
n}+\sum_{j=1}^{\infty}(j+1)e^{-2\pi j\Delta\alpha}\Big{(}\frac{1}{\log n+2\pi
j\Delta}-\frac{1}{2\pi(j+2)\Delta-\log n}\Big{)}$
$\displaystyle\qquad\qquad+\sum_{j=1}^{\infty}\frac{(j+1)e^{-2\pi
j\Delta\alpha}}{2\pi(j+2)\Delta-\log
n}\big{(}1-n^{2\alpha}e^{-4\pi\Delta\alpha}\big{)}$
$\displaystyle\leq\frac{1}{\log n}+2(2\pi\Delta-\log
n)\sum_{j=1}^{\infty}\frac{(j+1)e^{-2\pi j\Delta\alpha}}{(\log n+2\pi
j\Delta)(2\pi(j+2)\Delta-\log n)}$ (2.5)
$\displaystyle\qquad\qquad+\frac{1-n^{2\alpha}e^{-4\pi\Delta\alpha}}{2\pi\Delta}\sum_{j=1}^{\infty}e^{-2\pi
j\Delta\alpha}$ $\displaystyle<\frac{1}{\log n}+\frac{2(2\pi\Delta-\log
n)}{(2\pi\Delta)^{2}}\sum_{j=1}^{\infty}\frac{e^{-2\pi
j\Delta\alpha}}{j}+\frac{1-n^{2\alpha}e^{-4\pi\Delta\alpha}}{2\pi\Delta(1-e^{-2\pi\Delta\alpha})}$
$\displaystyle=\frac{1}{\log n}-\frac{2(2\pi\Delta-\log
n)}{(2\pi\Delta)^{2}}\log\big{(}1-e^{-2\pi\Delta\alpha}\big{)}+O\Big{(}\frac{1}{\Delta}\Big{)}.$
(2.6)

Note that if $\Delta\alpha\gg 1$, then the second term above is
$O(\Delta^{-1})$, and if $\Delta\alpha=o(1),$ then it is

$\leq\frac{1}{\pi\Delta}\log\frac{1}{\Delta\alpha}.$ (2.7)

Now we use the expression (2.6) for $a_{\alpha}(n;\Delta)$ and evaluate the
contribution from the prime squares in (2.4). When dealing with the main term
of size $1/\log n$ above, we proceed similarly as in [12] or Lemma $2$ in
[30]. For the second term, we trivially bound the sum over primes and use the
bound (2.7), and it follows that the contribution from prime squares is

$\displaystyle\geq-\frac{\log\log\log
T}{2}-\Big{(}\log\frac{1}{\Delta\alpha}\Big{)}^{\gamma(\Delta)}-O(1).$

Dealing similarly with the sum over prime cubes and higher powers and
combining the equation above and (2.4) finishes the proof. ∎

######  Lemma 2.2.

Assume RH. If $0<\alpha=o(\frac{1}{\log\log t})$, then

$\displaystyle\log|\zeta(\tfrac{1}{2}+\alpha+it)|$
$\displaystyle\geq\frac{\log t}{2\log\log t}\log\big{(}1-(\log
t)^{-2\alpha}\big{)}$ $\displaystyle\qquad\qquad+O\Big{(}\frac{\log
t}{\log\log t}\Big{)}+O\Big{(}\frac{\log t}{(\log\log
t)^{2}}\log\frac{1}{1-(\log t)^{-2\alpha}}\Big{)},$ (2.8)

and if $1/2-\alpha\ll\frac{1}{\log\log t}$, then

$\displaystyle\log|\zeta(\tfrac{1}{2}+\alpha+it)|\geq-\log\log\log t+O(1),$

otherwise

$\displaystyle\log|\zeta(\tfrac{1}{2}+\alpha+it)|\geq-\Big{(}\frac{1}{2}+\frac{8\alpha}{1-4\alpha^{2}}\Big{)}\frac{(\log
t)^{1-2\alpha}}{\log\log t}-\log\log\log t+O\Big{(}\frac{(\log
t)^{1-2\alpha}}{(1-2\alpha)^{2}(\log\log t)^{2}}\Big{)}.$

###### Proof.

Carneiro and Chandee established the last two bounds in [5, Theorem 2] and in
the case $0<\alpha=o(\frac{1}{\log\log t})$ they obtained

$\log|\zeta(\tfrac{1}{2}+\alpha+it)|\geq\frac{\log t}{2\log\log
t}\log\big{(}1-(\log t)^{-2\alpha}\big{)}+O\Big{(}\frac{(\log
t)^{1-2\alpha}}{(\log\log t)^{2}(1-(\log t)^{-2\alpha})}\Big{)}.$

We will now prove the improved bound (2.2).

Let $\Delta=\frac{\log\log t}{\pi}$. From (2.4) and (2.6) we have

$\displaystyle\log|\zeta(\tfrac{1}{2}+\alpha+it)|\geq\frac{\log\frac{t}{2\pi}}{2\log\log
t}\log\big{(}1-(\log t)^{-2\alpha}\big{)}$
$\displaystyle\qquad-\sum_{n\leq(\log
t)^{2}}\frac{\Lambda(n)}{n^{1/2+\alpha}}\bigg{(}\frac{1}{\log
n}-\frac{2(2\log\log t-\log n)}{(2\log\log t)^{2}}\log\big{(}1-(\log
t)^{-2\alpha}\big{)}+O\Big{(}\frac{1}{\log\log t}\Big{)}\bigg{)}$
$\displaystyle\qquad+O(1).$

We now consider the sum over $n$ above, which expands into three terms. The
contribution of the first and the last term is

$\ll\frac{\log t}{\log\log t},$

while the second term is

$\displaystyle=-\frac{\log(1-(\log t)^{-2\alpha})}{\log\log
t}\Big{(}\frac{2(\log t)^{1-2\alpha}}{1-2\alpha}+O\big{(}(\log\log
t)^{3}\big{)}\Big{)}$ $\displaystyle\qquad\qquad+\frac{\log(1-(\log
t)^{-2\alpha})}{2(\log\log t)^{2}}\Big{(}\frac{4(\log t)^{1-2\alpha}\log\log
t}{1-2\alpha}+O(\log t)\Big{)}$ $\displaystyle=O\Big{(}\frac{\log t}{(\log\log
t)^{2}}\log\frac{1}{1-(\log t)^{-2\alpha}}\Big{)},$

by the prime number theorem and partial summation. This establishes (2.2). ∎

The next lemma is the classical mean value theorem (see, for instance, [22,
Theorem 6.1]).

######  Lemma 2.3.

For any complex numbers $a(n)$ we have

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

Now let $\ell$ be an integer, and for $z\in\mathbb{C}$, let

$E_{\ell}(z)=\sum_{s\leq\ell}\frac{z^{s}}{s!}.$

If $z\in\mathbb{R}$, then for $\ell$ even, $E_{\ell}(z)>0$.

We have the following elementary inequality.

######  Lemma 2.4.

Let $\ell$ be an even integer and let $z$ be a complex number such that
$|z|\leq\ell/e^{2}$. Then

$e^{\Re(z)}\leq\max\Big{\\{}1,|E_{\ell}(z)|\Big{(}1+\frac{1}{15e^{\ell}}\Big{)}\Big{\\}}.$

###### Proof.

We have

$\displaystyle e^{\Re(z)}=|e^{z}|\leq|e^{z}-E_{\ell}(z)|+|E_{\ell}(z)|.$

Now we have

$|e^{z}-E_{\ell}(z)|\leq\sum_{j=\ell+1}^{\infty}\frac{|z|^{j}}{j!},$

and we can proceed as in [25] (see Lemma $1$) to get that

$|e^{z}-E_{\ell}(z)|\leq\frac{1}{16e^{\ell}}.$

Hence

$e^{\Re(z)}\leq|E_{\ell}(z)|+\frac{1}{16e^{\ell}}.$

If $|E_{\ell}(z)|\leq\frac{15}{16}$, then we have $e^{\Re(z)}\leq 1$.
Otherwise, if $|E_{\ell}(z)|>\frac{15}{16}$, we get that

$e^{\Re(z)}\leq|E_{\ell}(z)|\Big{(}1+\frac{1}{15e^{\ell}}\Big{)},$

and this completes the proof. ∎

Let $\nu(n)$ be the multiplicative function given by

$\nu(p^{j})=\frac{1}{j!},$

for $p$ a prime. We will frequently use the following fact. For any interval
$I$, $s\in\mathbb{N}$ and $a(n)$ a completely multiplicative function, we have

$\Big{(}\sum_{p\in I}a(p)\Big{)}^{s}=s!\sum_{\begin{subarray}{c}p|n\Rightarrow
p\in I\\\ \Omega(n)=s\end{subarray}}a(n)\nu(n),$ (2.9)

where $\Omega(n)$ denotes the number of prime factors of $n$, counting
multiplicity.

##  3\. Setup of the proof and main propositions

We will first introduce some notation and state some key lemmas, before
proceeding to the proof of Theorems 1.2 and 1.3.

Let

$I_{0}=(1,T^{\beta_{0}}],\ I_{1}=(T^{\beta_{0}},T^{\beta_{1}}],\ \ldots,\
I_{K}=(T^{\beta_{K-1}},T^{\beta_{K}}]$

for a sequence $\beta_{0},\ldots,\beta_{K}$ to be chosen later such that
$\beta_{j+1}=r\beta_{j}$, for some $r>1$.

Also let $\ell_{j}$ be even parameters which we will also choose later on. Let
$s_{j}$ be integers. For now, we can think of $s_{j}\beta_{j}\asymp 1$, and
$\sum_{h=0}^{K}\ell_{h}\beta_{h}\ll 1$.

We let $T^{\beta_{j}}=e^{2\pi\Delta_{j}}$ for every $0\leq j\leq K$ and let

$P_{u,v}(t)=\sum_{p\in
I_{u}}\frac{b_{\alpha}(p;\Delta_{v})}{p^{1/2+\alpha+it}},$

where $b_{\alpha}(n;\Delta)$ is a completely multiplicative function in the
first variable and

$\displaystyle b_{\alpha}(p;\Delta)=-a_{\alpha}(p;\Delta)\log p.$

For $p\leq e^{2\pi\Delta}$, if $\Delta\alpha\gg 1$, we note from equation
(2.5) that

$\displaystyle|b_{\alpha}(p;\Delta)|$ $\displaystyle\leq
1+\frac{2\pi\Delta-\log p}{2\pi\Delta}\sum_{j=1}^{\infty}\frac{2\log p}{\log
p+2\pi j\Delta}e^{-2\pi j\Delta\alpha}+\frac{\log
p}{2\pi\Delta}\sum_{j=1}^{\infty}e^{-2\pi j\Delta\alpha}$ $\displaystyle\leq
1+\sum_{j=1}^{\infty}e^{-2\pi
j\Delta\alpha}=\frac{1}{1-e^{-2\pi\Delta\alpha}}.$

On the other hand, if $\Delta\alpha=o(1)$, then from (2.6) we get that

$\displaystyle|b_{\alpha}(p;\Delta)|\leq\frac{2(2\pi\Delta-\log p)\log
p}{(2\pi\Delta)^{2}}\sum_{j=1}^{\infty}\frac{e^{-2\pi
j\Delta\alpha}}{j}+O(1)\leq\frac{1}{2}\log\frac{1}{\Delta\alpha}+O(1).$

Thus we can rewrite the bound for $b_{\alpha}(p;\Delta)$ as a single bound,

$|b_{\alpha}(p;\Delta)|\leq
b(\Delta)\Big{(}\log\frac{1}{\Delta\alpha}\Big{)}^{\gamma(\Delta)},$ (3.1)

where $\gamma(\Delta)=0$ if $\Delta\alpha\gg 1$ and $\gamma(\Delta)=1$ if
$\Delta\alpha=o(1)$, and where

$b(\Delta)=\begin{cases}\frac{1}{1-e^{-2\pi\Delta\alpha}}&\mbox{ if
}\Delta\alpha\gg 1,\\\ \frac{1}{2}+\varepsilon&\mbox{ if
}\Delta\alpha=o(1).\end{cases}$ (3.2)

Let

$\mathcal{T}_{u}=\Big{\\{}T\leq t\leq 2T:\max_{u\leq v\leq
K}|P_{u,v}(t)|\leq\frac{\ell_{u}}{ke^{2}}\Big{\\}}.$

Denote the set of $t$ such that $t\in\mathcal{T}_{u}$ for all $u\leq K$ by
$\mathcal{T}^{\prime}$. For $0\leq j\leq K-1$, let $\mathcal{S}_{j}$ denote
the subset of $t\in[T,2T]$ such that $t\in\mathcal{T}_{h}$ for all $h\leq j$,
but $t\notin\mathcal{T}_{j+1}$.

We will prove the following lemma.

######  Lemma 3.1.

For $t\in[T,2T]$, we either have

$\max_{0\leq v\leq K}|P_{0,v}(t)|>\frac{\ell_{0}}{ke^{2}},$

or

$|\zeta(\tfrac{1}{2}+\alpha+it)|^{-2k}\leq S_{1}(t)+S_{2}(t),$

where

$\displaystyle S_{1}(t)=(\log\log
T)^{k}\Big{(}\frac{1}{1-T^{-\beta_{K}\alpha}}\Big{)}^{\frac{2k}{\beta_{K}}}\exp\Big{(}2k\Big{(}\log\frac{1}{\Delta_{K}\alpha}\Big{)}^{\gamma(\Delta_{K})}\Big{)}$
$\displaystyle\
\times\prod_{h=0}^{K}\max\Big{\\{}1,|E_{\ell_{h}}(kP_{h,K}(t))|^{2}\Big{(}1+\frac{1}{15e^{\ell_{h}}}\Big{)}^{2}\Big{\\}}\exp\Big{(}O\Big{(}\frac{\Delta_{K}^{2}e^{\pi\Delta_{K}}}{1+\Delta_{K}t}+\frac{\Delta_{K}\log(1+\Delta_{K}\sqrt{t})}{\sqrt{t}}+1\Big{)}\Big{)}$

and

$\displaystyle S_{2}(t)$ $\displaystyle=(\log\log
T)^{k}\sum_{j=0}^{K-1}\sum_{v=j+1}^{K}\Big{(}\frac{1}{1-T^{-\beta_{j}\alpha}}\Big{)}^{\frac{2k}{\beta_{j}}}\exp\Big{(}2k\Big{(}\log\frac{1}{\Delta_{j}\alpha}\Big{)}^{\gamma(\Delta_{j})}\Big{)}$
$\displaystyle\qquad\times\prod_{h=0}^{j}\max\Big{\\{}1,|E_{\ell_{h}}(kP_{h,j}(t))|^{2}\Big{(}1+\frac{1}{15e^{\ell_{h}}}\Big{)}^{2}\Big{\\}}\Big{(}\frac{ke^{2}}{\ell_{j+1}}|P_{j+1,v}(t)|\Big{)}^{2s_{j+1}}$
$\displaystyle\qquad\times\exp\Big{(}O\Big{(}\frac{\Delta_{j}^{2}e^{\pi\Delta_{j}}}{1+\Delta_{j}t}+\frac{\Delta_{j}\log(1+\Delta_{j}\sqrt{t})}{\sqrt{t}}+1\Big{)}\Big{)}.$

###### Proof.

For $T\leq t\leq 2T$, we have the following possibilities:

  1. (1)

$t\notin\mathcal{T}_{0}$;

  2. (2)

$t\in\mathcal{T}^{\prime}$;

  3. (3)

$t\in\mathcal{S}_{j}$ for some $0\leq j\leq K-1.$

If $t\notin\mathcal{T}_{0}$, then the first condition is automatically
satisfied. Now suppose that $t\in\mathcal{T}_{0}$.

Suppose that $t\in\mathcal{T}^{\prime}$. We use Lemma 2.1 with
$\Delta=\Delta_{K}$ and the inequality in Lemma 2.4 with $z=k\Re P_{h,K}(t)$.

Now if $t\in\mathcal{S}_{j}$, then we use Lemma 2.1 with $\Delta=\Delta_{j}$,
the inequality in Lemma 2.4 with $z=k\Re P_{h,j}(t)$ and the fact that there
exists some $v\geq j+1$ such that $\frac{ke^{2}}{\ell_{j+1}}|P_{j+1,v}(t)|>1$.
∎

We will also need the following propositions.

######  Proposition 3.2.

For $0\leq v\leq K$ and $\beta_{0}s_{0}\leq 1$, we have

$\int_{T}^{2T}|P_{0,v}(t)|^{2s_{0}}\,dt\ll
Ts_{0}!b(\Delta_{0})^{2s_{0}}\Big{(}\log\frac{1}{\Delta_{0}\alpha}\Big{)}^{2s_{0}\gamma(\Delta_{0})}(\log\log
T^{\beta_{0}})^{s_{0}}.$

###### Proof.

Using (2.9), we have

$\displaystyle
P_{0,v}(t)^{s_{0}}=s_{0}!\sum_{\begin{subarray}{c}p|n\Rightarrow p\leq
T^{\beta_{0}}\\\
\Omega(n)=s_{0}\end{subarray}}\frac{b_{\alpha}(n;\Delta_{v})\nu(n)}{n^{1/2+\alpha+it}}.$

It then follows from Lemma 2.3 that

$\displaystyle\int_{T}^{2T}|P_{0,v}(t)|$
${}^{2s_{0}}\,dt=\big{(}T+O(T^{\beta_{0}s_{0}})\big{)}(s_{0}!)^{2}\sum_{\begin{subarray}{c}p|n\Rightarrow
p\leq T^{\beta_{0}}\\\
\Omega(n)=s_{0}\end{subarray}}\frac{b_{\alpha}(n;\Delta_{v})^{2}\nu(n)^{2}}{n^{1+2\alpha}}$
$\displaystyle\ll Ts_{0}!\bigg{(}\sum_{p\leq
T^{\beta_{0}}}\frac{b_{\alpha}(p;\Delta_{v})^{2}}{p^{1+2\alpha}}\bigg{)}^{s_{0}}\leq
Ts_{0}!b(\Delta_{v})^{2s_{0}}\Big{(}\log\frac{1}{\Delta_{v}\alpha}\Big{)}^{2s_{0}\gamma(\Delta_{v})}(\log\log
T^{\beta_{0}})^{s_{0}},$

where we have used the fact that $\nu(n)^{2}\leq\nu(n)$ and the bound (3.1).
Now since

$b(\Delta_{v})\Big{(}\log\frac{1}{\Delta_{v}\alpha}\Big{)}^{2s_{0}\gamma(\Delta_{v})}\leq
b(\Delta_{0})\Big{(}\log\frac{1}{\Delta_{0}\alpha}\Big{)}^{2s_{0}\gamma(\Delta_{0})},$

the conclusion follows. ∎

######  Proposition 3.3.

Let $0\leq j\leq K-1$. For
$\sum_{h=0}^{j}\ell_{h}\beta_{h}+s_{j+1}\beta_{j+1}\leq 1$ and for $j+1\leq
v\leq K$, we have

$\displaystyle\int_{T}^{2T}$
$\displaystyle\prod_{h=0}^{j}|E_{\ell_{h}}(kP_{h,j}(t))|^{2}|P_{j+1,v}(t)|^{2s_{j+1}}\,dt\ll
Ts_{j+1}!\big{(}\log
T^{\beta_{j}}\big{)}^{k^{2}b(\Delta_{j})^{2}(\log\frac{1}{\Delta_{j}\alpha})^{2\gamma(\Delta_{j})}}$
$\displaystyle\times
b(\Delta_{j+1})^{2s_{j+1}}\Big{(}\log\frac{1}{\Delta_{j+1}\alpha}\Big{)}^{2s_{j+1}\gamma(\Delta_{j+1})}(\log
r)^{s_{j+1}}.$

###### Proof.

Using (2.9), we have

$\displaystyle\int_{T}^{2T}\prod_{h=0}^{j}|E_{\ell_{h}}(kP_{h,j}(t))|^{2}|P_{j+1,v}(t)|^{2s_{j+1}}\,dt=(s_{j+1}!)^{2}\int_{T}^{2T}\Big{|}\sum_{n\leq
T^{\sum_{h=0}^{j}\ell_{h}\beta_{h}+s_{j+1}\beta_{j+1}}}\frac{c(n)\nu(n)}{n^{1/2+\alpha+it}}\Big{|}^{2}\,dt,$
(3.3)

where

$c(n)=\sum_{\begin{subarray}{c}n=n_{1}\ldots n_{j+1}\\\ p|n_{h}\Rightarrow
p\in I_{h}\\\ \Omega(n_{h})\leq\ell_{h},h=0,\ldots,j\\\
\Omega(n_{j+1})=s_{j+1}\end{subarray}}\Big{(}\prod_{h=0}^{j}b_{\alpha}(n_{h};\Delta_{j})k^{\Omega(n_{h})}\Big{)}b_{\alpha}(n_{j+1};\Delta_{v}).$

Using Lemma 2.3 in (3.3) and the fact that $\nu(m)^{2}\leq\nu(m)$ for any $m$,
we obtain that

$\displaystyle\eqref{to_bd}$
$\displaystyle\ll\Big{(}T+T^{\sum_{h=0}^{j}\ell_{h}\beta_{h}+s_{j+1}\beta_{j+1}}\Big{)}(s_{j+1}!)^{2}\Big{(}\prod_{h=0}^{j}\sum_{\begin{subarray}{c}p|n_{h}\Rightarrow
p\in I_{h}\\\
\Omega(n_{h})\leq\ell_{h}\end{subarray}}\frac{|b_{\alpha}(n_{h};\Delta_{j})|^{2}k^{2\Omega(n_{h})}\nu(n_{h})}{n_{h}^{1+2\alpha}}\Big{)}$
$\displaystyle\qquad\times\sum_{\begin{subarray}{c}p|n_{j+1}\Rightarrow p\in
I_{j+1}\\\
\Omega(n_{j+1})=s_{j+1}\end{subarray}}\frac{|b_{\alpha}(n_{j+1};\Delta_{v})|^{2}\nu(n_{j+1})}{n_{j+1}^{1+2\alpha}}.$

Now we use the assumption that
$\sum_{h=0}^{j}\ell_{h}\beta_{h}+s_{j+1}\beta_{j+1}\leq 1$. Bounding
$\nu(n_{h})\leq 1$, removing the condition on the number of primes of $n_{h}$,
and using (3.1), we get that

$\displaystyle\eqref{to_bd}$ $\displaystyle\ll Ts_{j+1}!\prod_{p\leq
T^{\beta_{j}}}\Big{(}1-\frac{k^{2}|b_{\alpha}(p;\Delta_{j})|^{2}}{p^{1+2\alpha}}\Big{)}^{-1}\Big{(}\sum_{p\in
I_{j+1}}\frac{|b_{\alpha}(p;\Delta_{v}|^{2}}{p^{1+2\alpha}}\Big{)}^{s_{j+1}}$
$\displaystyle\ll Ts_{j+1}!\big{(}\log
T^{\beta_{j}}\big{)}^{k^{2}b(\Delta_{j})^{2}(\log\frac{1}{\Delta_{j}\alpha})^{2\gamma(\Delta_{j})}}b(\Delta_{v})^{2s_{j+1}}\Big{(}\log\frac{1}{\Delta_{v}\alpha}\Big{)}^{2s_{j+1}\gamma(\Delta_{v})}\Big{(}\log\frac{\beta_{j+1}}{\beta_{j}}\Big{)}^{s_{j+1}}$
$\displaystyle\ll Ts_{j+1}!\big{(}\log
T^{\beta_{j}}\big{)}^{k^{2}b(\Delta_{j})^{2}(\log\frac{1}{\Delta_{j}\alpha})^{2\gamma(\Delta_{j})}}b(\Delta_{j+1})^{2s_{j+1}}\Big{(}\log\frac{1}{\Delta_{j+1}\alpha}\Big{)}^{2s_{j+1}\gamma(\Delta_{j+1})}(\log
r)^{s_{j+1}},$

which finishes the proof of the proposition. ∎

A minor modification of the proposition above (where we do not have the
contribution from the $j+1$ interval) yields the following.

######  Proposition 3.4.

For $\sum_{h=0}^{K}\ell_{h}\beta_{h}\leq 1$, we have

$\displaystyle\int_{T}^{2T}$
$\displaystyle\prod_{h=0}^{K}|E_{\ell_{h}}(kP_{h,K}(t))|^{2}\,dt\ll
T\big{(}\log
T^{\beta_{K}}\big{)}^{k^{2}b(\Delta_{K})^{2}(\log\frac{1}{\Delta_{K}\alpha})^{2\gamma(\Delta_{K})}}.$

##  4\. The case $\alpha\gg\frac{1}{(\log T)^{\frac{1}{2k}-\varepsilon}}$,
first steps

Here, we will consider the case $\alpha\gg\frac{1}{(\log
T)^{\frac{1}{2k}-\varepsilon}}$, which will be a starting point in the proof
of Theorems 1.2 and 1.3.

We choose

$\beta_{0}=\frac{a(2d-1)(\log\log T)^{2}}{(1+2\varepsilon)k(\log
T)\big{(}\log\frac{1}{1-(\log T)^{-2\alpha}}\big{)}},\quad
s_{0}=\Big{[}\frac{a}{\beta_{0}}\Big{]},\quad\ell_{0}=2\Big{\lceil}\frac{s_{0}^{d}}{2}\Big{\rceil}$
(4.1)

and

$\beta_{j}=r^{j}\beta_{0},\quad
s_{j}=\Big{[}\frac{a}{\beta_{j}}\Big{]},\quad\ell_{j}=2\Big{\lceil}\frac{s_{j}^{d}}{2}\Big{\rceil},\quad
1\leq j\leq K,$ (4.2)

where we can pick, for example,

$a=\frac{4-3k\varepsilon}{2(2-k\varepsilon)},\,r=\frac{2}{2-k\varepsilon},\,d=\frac{8-7k\varepsilon}{2(4-3k\varepsilon)},$

so that

$\frac{a(2d-1)}{r}=1-k\varepsilon.$ (4.3)

Here, $K$ is chosen such that it is the maximal integer for which

$\beta_{K}\leq\begin{cases}\frac{\log(\alpha\log T)}{\alpha\log T}&\mbox{ if
}\alpha\log T\to\infty,\\\ c&\mbox{ if }\alpha\ll\frac{1}{\log T},\end{cases}$
(4.4)

where $c>0$ is a small constant such that

$c^{1-d}\Big{(}\frac{a^{d}r^{1-d}}{r^{1-d}-1}+\frac{2r}{r-1}\Big{)}\leq 1-a.$
(4.5)

Note that the above ensures that the conditions in Propositions 3.3 and 3.4
are satisfied.

If $t\notin\mathcal{T}_{0}$, then there exists $0\leq v\leq K$ such that
$\frac{ke^{2}}{\ell_{0}}|P_{0,v}(t)|>1$. Then we have

$\displaystyle\int_{[T,2T]\setminus\mathcal{T}_{0}}|\zeta(\tfrac{1}{2}+\alpha+it)|^{-2k}\,dt$
$\displaystyle\leq\int_{T}^{2T}\Big{(}\frac{ke^{2}}{\ell_{0}}|P_{0,v}(t)|\Big{)}^{2s_{0}}|\zeta(\tfrac{1}{2}+\alpha+it)|^{-2k}\,dt$
$\displaystyle\leq\Big{(}\frac{ke^{2}}{\ell_{0}}\Big{)}^{2s_{0}}\Big{(}\frac{1}{1-(\log
T)^{-2\alpha}}\Big{)}^{\frac{(1+\varepsilon)k\log T}{\log\log
T}}\int_{T}^{2T}|P_{0,v}(t)|^{2s_{0}}\,dt,$

by using the pointwise bound in Lemma 2.2,

$|\zeta(\tfrac{1}{2}+\alpha+it)|^{-1}\ll\Big{(}\frac{1}{1-(\log
T)^{-2\alpha}}\Big{)}^{\frac{(1+\varepsilon)\log T}{2\log\log T}}.$ (4.6)

By Proposition 3.2 and Stirling’s formula, we get that

$\displaystyle\int_{[T,2T]\setminus\mathcal{T}_{0}}|\zeta(\tfrac{1}{2}+\alpha+it)|^{-2k}\,dt\ll
T\Big{(}\frac{1}{1-(\log T)^{-2\alpha}}\Big{)}^{\frac{(1+\varepsilon)k\log
T}{\log\log T}}$
$\displaystyle\qquad\times\sqrt{s_{0}}\exp\bigg{(}-(2d-1)s_{0}\log
s_{0}+2s_{0}\log\Big{(}ke^{3/2}b(\Delta_{0})\Big{(}\log\frac{1}{\Delta_{0}\alpha}\Big{)}^{\gamma(\Delta_{0})}\sqrt{\log\log
T^{\beta_{0}}}\Big{)}\bigg{)}$ $\displaystyle\quad=o(T),$ (4.7)

using the choice of $s_{0}$ in (4.1).

Now assume that $t\in\mathcal{T}_{0}$. Using Lemma 3.1, we have that

$\int_{\mathcal{T}_{0}}|\zeta(\tfrac{1}{2}+\alpha+it)|^{-2k}\,dt\leq\int_{T}^{2T}S_{1}(t)\,dt+\int_{T}^{2T}S_{2}(t)\,dt.$

With the choice (4.4) of $\beta_{K}$, we have that

$\Big{(}\frac{1}{1-T^{-\beta_{K}\alpha}}\Big{)}^{\frac{2k}{\beta_{K}}}\ll\\\
\begin{cases}1&\mbox{ if }\alpha\gg\frac{1}{\log T},\\\ (\log T)^{O(1)}&\mbox{
if }\alpha=o\big{(}\frac{1}{\log T}\big{)}\end{cases}$

and

$\exp\Big{(}O\Big{(}\frac{\Delta_{K}^{2}e^{\pi\Delta_{K}}}{1+\Delta_{K}t}+\frac{\Delta_{K}\log(1+\Delta_{K}\sqrt{t})}{\sqrt{t}}+1\Big{)}\Big{)}=O(1),$

so

$\displaystyle\int_{T}^{2T}S_{1}(t)\,dt$ $\displaystyle\ll(\log\log
T)^{k}\exp\Big{(}2k\Big{(}\log\frac{1}{\Delta_{K}\alpha}\Big{)}^{\gamma(\Delta_{K})}\Big{)}(\log
T)^{c_{1}\gamma(\Delta_{K})}$
$\displaystyle\qquad\times\int_{T}^{2T}\prod_{h=0}^{K}\max\Big{\\{}1,|E_{\ell_{h}}(kP_{h,K}(t))|^{2}\Big{(}1+\frac{1}{15e^{\ell_{h}}}\Big{)}^{2}\Big{\\}}\,dt.$

Note that in the inequality above, we can assume without loss of generality
that

$\max\Big{\\{}1,|E_{\ell_{h}}(kP_{h,K}(t))|^{2}\Big{(}1+\frac{1}{15e^{\ell_{h}}}\Big{)}^{2}\Big{\\}}=|E_{\ell_{h}}(kP_{h,K}(t))|^{2}\Big{(}1+\frac{1}{15e^{\ell_{h}}}\Big{)}^{2}.$

Using Proposition 3.4 and the observation that $\gamma(\Delta_{K})=0$ in the
first two cases, and $\gamma(\Delta_{K})=1$ in the third case, we get that

$\int_{T}^{2T}S_{1}(t)\,dt\ll\begin{cases}T(\log\log
T)^{k}\big{(}\frac{\log(\alpha\log T)}{\alpha}\big{)}^{k^{2}}&\mbox{ if
}\alpha\log T\to\infty,\\\ T(\log\log T)^{k}(\log
T)^{k^{2}(\frac{1}{1-T^{-c\alpha}})^{2}}&\mbox{ if }\alpha\asymp\frac{1}{\log
T},\\\ T(\log\log
T)^{k}\exp\Big{(}k^{2}\big{(}\frac{1}{4}+\varepsilon\big{)}(\log\log
T)\big{(}\log\frac{1}{\alpha\log T}\big{)}^{2}\Big{)}&\mbox{ if
}\alpha=o\big{(}\frac{1}{\log T}\big{)}.\end{cases}$ (4.8)

Now we will need to bound the contribution from $S_{2}(t)$. Using Proposition
3.3 and Stirling’s formula, we have

$\displaystyle\int_{T}^{2T}S_{2}(t)\,dt$ $\displaystyle\ll T(\log\log
T)^{k}\sum_{j=0}^{K-1}(K-j)\sqrt{s_{j+1}}\exp\bigg{(}\frac{2k}{\beta_{j}}\log\frac{1}{1-T^{-\beta_{j}\alpha}}-(2d-1)s_{j+1}\log
s_{j+1}$
$\displaystyle\qquad\qquad+2s_{j+1}\log\Big{(}ke^{3/2}b(\Delta_{j+1})\Big{(}\log\frac{1}{\Delta_{j+1}\alpha}\Big{)}^{\gamma(\Delta_{j+1})}\Big{)}+2k\Big{(}\log\frac{1}{\Delta_{j}\alpha}\Big{)}^{\gamma(\Delta_{j})}\bigg{)}$
$\displaystyle\qquad\times\big{(}\log
T^{\beta_{j}}\big{)}^{k^{2}b(\Delta_{j})^{2}(\log\frac{1}{\Delta_{j}\alpha})^{2\gamma(\Delta_{j})}}.$

In the equation above, note that if $\beta_{j}\alpha\log T\geq\varepsilon$,
then

$\frac{2k}{\beta_{j}}\log\frac{1}{1-T^{-\beta_{j}\alpha}}\leq\frac{2k}{\beta_{j}}\log\frac{1}{1-e^{-\varepsilon}}<(2d-1)s_{j+1}\log
s_{j+1},$

so the contribution in this case will be $o(T(\log\log T)^{k})$. Now if
$\beta_{j}\alpha\log T<\varepsilon$, then

$\log\frac{1}{1-T^{-\beta_{j}\alpha}}<\log\frac{1}{\alpha}-\log(\beta_{j}\log
T)+\log\frac{1}{1-\varepsilon/2},$

so we obtain that

$\displaystyle\int_{T}^{2T}S_{2}(t)\,dt\ll T(\log\log
T)^{k}\sum_{j=0}^{K-1}(K-j)\sqrt{\frac{1}{\beta_{j+1}}}\exp\bigg{(}\frac{\log\log
T}{\beta_{j}}\Big{(}2k\frac{\log\frac{1}{\alpha}}{\log\log
T}-\frac{a(2d-1)}{r}\Big{)}$ $\displaystyle\ \ +\frac{\log(\beta_{j}\log
T)}{\beta_{j}}\Big{(}\frac{a(2d-1)}{r}-2k\Big{)}+\frac{2a}{r\beta_{j}}\log\Big{(}ke^{3/2}b(\Delta_{j+1})\Big{(}\log\frac{1}{\Delta_{j+1}\alpha}\Big{)}^{\gamma(\Delta_{j+1})}\Big{)}$
$\displaystyle\ \
+\frac{2k}{\beta_{j}}\log\frac{1}{1-\varepsilon/2}+\frac{a(2d-1)}{r\beta_{j}}\log\frac{r}{a}+2k\Big{(}\log\frac{1}{\Delta_{j}\alpha}\Big{)}^{\gamma(\Delta_{j}}\bigg{)}\big{(}\log
T^{\beta_{j}}\big{)}^{k^{2}b(\Delta_{j})^{2}(\log\frac{1}{\Delta_{j}\alpha})^{2\gamma(\Delta_{j})}}.$
(4.9)

##  5\. Proof of Theorems 1.2 and 1.3 for “big” shifts $\alpha$

Here, we will prove the bound (1.3) in Theorem 1.2 when
$\alpha\gg\frac{1}{(\log T)^{\frac{1}{2k}-\varepsilon}}$, and the bounds
(1.6), (1.7), (1.8) in Theorem 1.3. Recall that

$u=\frac{\log\frac{1}{\alpha}}{\log\log T}.$

The contribution from $t\notin\mathcal{T}_{0}$ and from $S_{1}(t)$ has already
been bounded in Section 4 (see equations (4.7) and (4.8)). Now we focus on
bounding the contribution from $S_{2}(t)$.

First assume that $\alpha\gg\frac{1}{(\log T)^{\frac{1}{2k}-\varepsilon}}$ and
$k\geq 1/2$. We rewrite (4.9) as

$\displaystyle\int_{T}^{2T}S_{2}(t)\,dt$ $\displaystyle\ll T(\log\log
T)^{k}\sum_{j=0}^{K-1}(K-j)\sqrt{\frac{1}{\beta_{j+1}}}\exp\bigg{(}\frac{\log\log
T}{\beta_{j}}\Big{(}2ku-\frac{a(2d-1)}{r}\Big{)}$
$\displaystyle\qquad+\frac{\log(\beta_{j}\log
T)}{\beta_{j}}\Big{(}\frac{a(2d-1)}{r}-2k\Big{)}+2k\Big{(}\log\frac{1}{\Delta_{j}\alpha}\Big{)}^{\gamma(\Delta_{j})}+O\Big{(}\frac{\log\log\log
T}{\beta_{j}}\Big{)}\bigg{)}$ $\displaystyle\quad\times\big{(}\log
T^{\beta_{j}}\big{)}^{k^{2}b(\Delta_{j})^{2}(\log\frac{1}{\Delta_{j}\alpha})^{2\gamma(\Delta_{j})}}.$

Note that

$\frac{a(2d-1)}{r}-2k<0,$

and using (4.3) and the fact that $\alpha\gg\frac{1}{(\log
T)^{\frac{1}{2k}-\varepsilon}}$, it follows that

$2ku-\frac{a(2d-1)}{r}\leq-k\varepsilon.$

Hence we get that

$\displaystyle\int_{T}^{2T}S_{2}(t)\,dt$ $\displaystyle\ll T(\log\log
T)^{k}\sum_{j=0}^{K-1}(K-j)\sqrt{\frac{1}{\beta_{j+1}}}\exp\bigg{(}-\frac{k\varepsilon\log\log
T}{\beta_{j}}+2k\Big{(}\log\frac{1}{\Delta_{j}\alpha}\Big{)}^{\gamma(\Delta_{j})}$
$\displaystyle\qquad\qquad+O\Big{(}\frac{\log\log\log
T}{\beta_{j}}\Big{)}\bigg{)}\big{(}\log
T^{\beta_{j}}\big{)}^{k^{2}b(\Delta_{j})^{2}(\log\frac{1}{\Delta_{j}\alpha})^{2\gamma(\Delta_{j})}}.$
(5.1)

We first consider the contribution from those $j$ for which
$\gamma(\Delta_{j})=1$. Let $R_{1}$ denote this contribution. Using the fact
that $K\ll\log\log T$ and after a relabeling of the $\varepsilon$, we have
that

$\displaystyle R_{1}$ $\displaystyle\ll T(\log\log
T)^{k}\sum_{j}\exp\bigg{(}-\frac{k\varepsilon\log\log
T}{\beta_{j}}+2k\log\frac{1}{\beta_{j}\alpha\log T}+O\Big{(}\frac{\log\log\log
T}{\beta_{j}}\Big{)}$
$\displaystyle\qquad\qquad+\Big{(}\frac{1}{4}+\varepsilon\Big{)}k^{2}\Big{(}\log\frac{1}{\beta_{j}\alpha\log
T}\Big{)}^{2}\log(\beta_{j}\log T)\bigg{)},$

where the sum over $j$ is such that $\gamma(\Delta_{j})=1$. Since
$\alpha\gg\frac{1}{\log T}$, we have

$\Big{(}\log\frac{1}{\beta_{j}\alpha\log T}\Big{)}^{2}\log(\beta_{j}\log
T)\ll\Big{(}\log\frac{1}{\beta_{j}}\Big{)}^{2}\log\log T.$

As $\gamma(\Delta_{j})=1$, we have $\beta_{j}\to 0$ in the sum in $R_{1}$
above, and then it follows that

$R_{1}=o\big{(}T(\log\log T)^{k}\big{)}.$ (5.2)

Now we consider the contribution in (7.7) from those $j$ with
$\gamma(\Delta_{j})=0$. Let $R_{2}$ denote that term. We have that

$\displaystyle R_{2}\ll T(\log\log
T)^{k}\sum_{j}\exp\bigg{(}-\frac{k\varepsilon\log\log
T}{\beta_{j}}+O\Big{(}\frac{\log\log\log
T}{\beta_{j}}\Big{)}+k^{2}b(\Delta_{j})^{2}\log(\beta_{j}\log T)\bigg{)},$

where the sum is over $j$ such that $\gamma(\Delta_{j})=0$. Keeping in mind
the choices for $\beta_{j}$ (equations (4.2) and (4.4)), it then follows that

$R_{2}=\begin{cases}o(T(\log\log T)^{k})&\mbox{ if }\alpha\log T\to\infty,\\\
o\Big{(}T(\log\log T)^{k}(\log
T)^{k^{2}(\frac{1}{1-T^{-c\alpha}})^{2}}\Big{)}&\mbox{ if
}\alpha\asymp\frac{1}{\log T}.\end{cases}$ (5.3)

Combining the bounds (4.7), (4.8), (5.2) and (5.3), the bound (1.3) follows
when $\alpha\gg\frac{1}{(\log T)^{\frac{1}{2k}-\varepsilon}}$ and $k\geq 1/2$.

Now assume that $k<1/2$ and $\alpha\gg\frac{1}{\log T}$. If
$\frac{a(2d-1)}{r}-2k\leq 0$, then the same argument as before works. Hence we
assume we have $\frac{a(2d-1)}{r}-2k>0$. We rewrite the bound (4.9) for
$S_{2}(t)$ as

$\displaystyle\int_{T}^{2T}S_{2}(t)\,dt\ll T(\log\log
T)^{k}\sum_{j=0}^{K-1}(K-j)\sqrt{\frac{1}{\beta_{j+1}}}\exp\bigg{(}\frac{\log\log
T}{\beta_{j}}\Big{(}-\frac{2k\log(\alpha\log T)}{\beta_{j}}$
$\displaystyle\qquad+\frac{\log\beta_{j}}{\beta_{j}}\Big{(}\frac{a(2d-1)}{r}-2k\Big{)}+\frac{2}{\beta_{j}}\Big{(}\log\log\frac{1}{\Delta_{j+1}\alpha}\Big{)}^{\gamma(\Delta_{j+1})}+2k\Big{(}\log\frac{1}{\Delta_{j}\alpha}\Big{)}^{\gamma(\Delta_{j})}$
$\displaystyle\qquad+k^{2}b(\Delta_{j})^{2}\Big{(}\log\frac{1}{\Delta_{j}\alpha}\Big{)}^{2\gamma(\Delta_{j})}\log(\beta_{j}\log
T)+O\Big{(}\frac{1}{\beta_{j}}\Big{)}\bigg{)}.$ (5.4)

In the equation above, we first consider the contribution from those $j$ for
which $\gamma(\Delta_{j})=1$, i.e., those $j$ for which
$\beta_{j}=o(\frac{1}{\alpha\log T})$. As before, we denote this contribution
by $R_{1}$. We let

$\displaystyle f(x)=-$ $\displaystyle\frac{2k\log(\alpha\log T)}{x}+\frac{\log
x}{x}\Big{(}\frac{a(2d-1)}{r}-2k\Big{)}+\frac{2}{x}\log\log\frac{1}{x\alpha\log
T}+2k\log\frac{1}{x\alpha\log T}$
$\displaystyle+k^{2}\Big{(}\frac{1}{4}+\varepsilon\Big{)}\Big{(}\log\frac{1}{x\alpha\log
T}\Big{)}^{2}\log(x\log T).$

By taking the derivative, we see that when $\frac{1}{\log
T}\ll\alpha=o\big{(}\frac{\log\log T}{\log T}\big{)}$, the maximum of $f(x)$
(when $x=o(\frac{1}{\alpha\log T})$) is attained at some
$x_{0}\asymp\frac{1}{\log\log T}$, while if $\alpha\gg\frac{\log\log T}{\log
T}$, the function $f(x)$ is increasing on the interval under consideration.
Hence, we get that

$R_{1}=\begin{cases}o(T(\log\log T)^{k})&\mbox{ if }\alpha\gg\frac{\log\log
T}{\log T},\\\ O\Big{(}T\exp\Big{(}C_{1}(\log\log T)\big{(}\log\frac{\log\log
T}{\alpha\log T}\big{)}^{2}\Big{)}\Big{)}&\mbox{ if }\frac{1}{\log
T}\ll\alpha=o\big{(}\frac{\log\log T}{\log T}\big{)},\end{cases}$ (5.5)

for some $C_{1}>0$.

Now we bound the contribution in (5.4) from those $j$ for which
$\gamma(\Delta_{j})=0$. It is easy to see that in this case, the function in
(5.4) is decreasing in $j$, so

$R_{2}=o(T(\log\log T)^{k}).$

Combining the above with (4.7), (4.8) and (5.5), the bounds (1.6) and (1.7)
follow.

Now we assume that $\frac{1}{(\log
T)^{\frac{1}{2k}-\varepsilon}}\ll\alpha=o\big{(}\frac{1}{\log T}\big{)}$. We
rewrite the bound (4.9) for $S_{2}(t)$ as

$\displaystyle\int_{T}^{2T}S_{2}(t)\,dt\ll T(\log\log
T)^{k}\sum_{j=0}^{K-1}(K-j)\sqrt{\frac{1}{\beta_{j+1}}}\exp\bigg{(}-\frac{2k\log(\alpha\log
T)}{\beta_{j}}$ $\displaystyle\ \
+\frac{\log\beta_{j}}{\beta_{j}}\Big{(}\frac{a(2d-1)}{r}-2k\Big{)}+\frac{2}{\beta_{j}}\log\log\frac{1}{\Delta_{j+1}\alpha}+2k\log\frac{1}{\Delta_{j}\alpha}+O\Big{(}(\log\log
T)^{3}+\frac{1}{\beta_{j}}\Big{)}\bigg{)}.$

In the sum over $j$ above, the maximum is attained at $j_{0}$ such that
$\beta_{j_{0}}\asymp(\alpha\log T)^{\frac{2k}{\frac{a(2d-1)}{r}-2k}}$. It then
follows that

$\int_{T}^{2T}S_{2}(t)\,dt\ll T\exp\Big{(}C_{2}\Big{(}\frac{1}{\alpha\log
T}\Big{)}^{\frac{2k}{1-2k-k\varepsilon}}\Big{)},$

for some $C_{2}>0$ (and after a relabeling of the $\varepsilon$). Combining
the above and (4.7), (4.8), the bound (1.8) follows when $\frac{1}{(\log
T)^{\frac{1}{2k}-\varepsilon}}\ll\alpha=o(\frac{1}{\log T})$.

##  6\. Proof of Theorem 1.2, bound (1.3); some recursive estimates

Here, we will prove the bound (1.4). To do that, we will use an inductive
argument, which will be performed in Subsection 6.2. The first step of the
argument in carried out in the next subsection.

###  6.1. The range $\alpha\gg\frac{(\log\log
T)^{\frac{4}{k}+\varepsilon}}{(\log T)^{\frac{1}{2k}}}$, $k\geq 1/2$, the
first step

We previously obtained the bound (1.3) in the region $\alpha\gg\frac{1}{(\log
T)^{\frac{1}{2k}-\varepsilon}}$ and $k\geq 1/2$. Hence, we will assume that

$\frac{(\log\log T)^{\frac{4}{k}+\varepsilon}}{(\log
T)^{\frac{1}{2k}}}\ll\alpha=o\Big{(}\frac{1}{(\log
T)^{\frac{1}{2k}-\varepsilon}}\Big{)},$ (6.1)

when $k\geq 1/2$.

Let

$\alpha=\frac{(\log\log T)^{b}}{(\log T)^{\frac{1}{2k}}},$ (6.2)

where $b\geq 4/k+\varepsilon$. From (6.1), we have that $b=o(\frac{\log\log
T}{\log\log\log T})$. We will show that for any $\delta>0$, we have

$\int_{T}^{2T}|\zeta(\tfrac{1}{2}+\alpha+it)|^{-2k}\,dt\ll
T\exp\bigg{(}\frac{(\log
T)^{\frac{3(1+\delta)}{kb-1}}}{\exp\big{(}\frac{2\log\log T\log\log\log\log
T}{(kb-1)\log\log\log T}\big{)}}\bigg{)}.$

We choose $\beta_{0},\ell_{0},s_{0}$ as in (4.1) and
$\beta_{j},\ell_{j},s_{j}$ are chosen as in (4.2). We choose $a,d,r$ such that

$\frac{a(2d-1)}{r}=1-\frac{n\log\log\log T}{\log\log T},$ (6.3)

where

$n=(2kb-2)\Big{(}1-\frac{10\delta}{12(1+\delta)}\Big{)}.$ (6.4)

For simplicity of notation, let $x=\frac{\log\log\log T}{\log\log T}$. We can
take

$\displaystyle a$
$\displaystyle=\frac{1-n(1-\frac{\delta}{24})x}{1-n(1-\frac{\delta}{12})x},\qquad
d=1-\frac{n}{2}\Big{(}1-\frac{\delta}{12}\Big{)}x,\qquad
r=\frac{1-n(1-\frac{\delta}{24})x}{1-nx}.$ (6.5)

We choose $\beta_{K}$ such that

$\beta_{K}^{1-d}\Big{(}\frac{a^{d}r^{1-d}}{r^{1-d}-1}+\frac{2r}{r-1}\Big{)}\leq
1-a.$

Again, the above inequality ensures that the conditions in Propositions 3.3
and 3.4 are satisfied. Note that the condition above can be re-expressed as

$\beta_{K}^{1-d}\leq c_{1}(1-a)(r-1)(1-d),$

for some constant $c_{1}>0$. We then choose $K$ such that $\beta_{K}$ is the
largest of the form in (4.2) such that

$\beta_{K}\leq\frac{\exp\big{(}\frac{6\log\log T\log\log\log\log
T}{n\log\log\log T}\big{)}}{(\log T)^{\frac{6}{n(1-\frac{\delta}{12})}}}.$
(6.6)

If $t\notin\mathcal{T}_{0}$, then we proceed as in Section 5 and similarly to
equation (4.7), we get that

$\displaystyle\int_{[T,2T]\setminus\mathcal{T}_{0}}|\zeta(\tfrac{1}{2}+\alpha+it)|^{-2k}\,dt\ll
T\Big{(}\frac{1}{1-(\log T)^{-2\alpha}}\Big{)}^{\frac{(1+\varepsilon)k\log
T}{\log\log T}}$
$\displaystyle\qquad\times\sqrt{s_{0}}\exp\bigg{(}-(2d-1)s_{0}\log
s_{0}+2s_{0}\log\Big{(}ke^{3/2}b(\Delta_{0})\Big{(}\log\frac{1}{\Delta_{0}\alpha}\Big{)}^{\gamma(\Delta_{0})}\sqrt{\log\log
T^{\beta_{0}}}\Big{)}\bigg{)}$ $\displaystyle\quad=o(T).$ (6.7)

Now we suppose that $t\in\mathcal{T}_{0}$. Similarly as in Section 5, using
Proposition 3.4 we get that

$\displaystyle\int_{T}^{2T}S_{1}(t)\,dt$ $\displaystyle\ll T(\log\log
T)^{k}\exp\Bigg{(}c_{1}\frac{(\log
T)^{\frac{6}{n(1-\frac{\delta}{12})}}\log\log T}{\exp\big{(}\frac{6\log\log
T\log\log\log\log T}{n(1-\frac{\delta}{12})\log\log\log
T}\big{)}}\Bigg{)}\exp\Big{(}k^{2}(\log\log T)^{3}\Big{)}$ $\displaystyle\ll
T\exp\Bigg{(}\frac{(\log
T)^{\frac{6}{n(1-\frac{\delta}{12})}}}{\exp\big{(}\frac{4\log\log
T\log\log\log\log T}{n\log\log\log T}\big{)}}\Bigg{)},$ (6.8)

for some $c_{1}>0$ (note that the constant $c_{1}$ can change from line to
line).

To bound the contribution from $S_{2}(t)$, we proceed as in equation (4.9) and
obtain that

$\displaystyle\int_{T}^{2T}S_{2}(t)\,dt\ll T(\log\log
T)^{k}\sum_{j=0}^{K-1}(K-j)\sqrt{\frac{1}{\beta_{j+1}}}\exp\bigg{(}\frac{\log\log
T}{\beta_{j}}\Big{(}2ku-\frac{a(2d-1)}{r}\Big{)}$ $\displaystyle\ \
+\frac{\log(\beta_{j}\log
T)}{\beta_{j}}\Big{(}\frac{a(2d-1)}{r}-2k\Big{)}+\frac{2a}{r\beta_{j}}\log\Big{(}ke^{3/2}b(\Delta_{j+1})\Big{(}\log\frac{1}{\Delta_{j+1}\alpha}\Big{)}^{\gamma(\Delta_{j+1})}\Big{)}$
$\displaystyle\ \
+\frac{2k}{\beta_{j}}\log\frac{1}{1-\varepsilon/2}+\frac{a(2d-1)}{r\beta_{j}}\log\frac{r}{a}+2k\Big{(}\log\frac{1}{\Delta_{j}\alpha}\Big{)}^{\gamma(\Delta_{j})}\bigg{)}\big{(}\log
T^{\beta_{j}}\big{)}^{k^{2}b(\Delta_{j})^{2}(\log\frac{1}{\Delta_{j}\alpha})^{2\gamma(\Delta_{j})}}.$
(6.9)

Since $\alpha=\frac{(\log\log T)^{b}}{(\log T)^{\frac{1}{2k}}}$ and given
(6.3), we get that

$\displaystyle\int_{T}^{2T}S_{2}(t)\,dt$ $\displaystyle\ll T(\log\log
T)^{k}\sum_{j=0}^{K-1}\exp\bigg{(}\frac{\log\log\log
T}{\beta_{j}}(n-2kb+2)+(\log\log
T)^{3}+O\Big{(}\frac{1}{\beta_{j}}\Big{)}\bigg{)},$

where we used the fact that $K\ll\log\log T$ and the fact that

$\log\log\frac{1}{\beta_{j}\alpha\log
T}\leq\log\log\frac{1}{\beta_{0}\alpha\log T}\leq\frac{\log\log\log T}{2k}.$

With the choice (6.4), we have that $n-2kb+2\leq-\delta$, and then

$\displaystyle\int_{T}^{2T}$ $\displaystyle S_{2}(t)\,dt\ll T(\log\log
T)^{k}\sum_{j=0}^{K-1}\exp\bigg{(}-\frac{\delta\log\log\log
T}{\beta_{j}}+(\log\log T)^{3}+O\Big{(}\frac{1}{\beta_{j}}\Big{)}\bigg{)}.$

In the sum over $j$ above, the maximum is attained at $j=K-1$, and given the
choice (6.6) for $\beta_{K}$, it follows that

$\int_{T}^{2T}S_{2}(t)\,dt=o(T).$ (6.10)

Combining equations (6.7), (6.8) and (6.10), it follows that

$\int_{T}^{2T}|\zeta(\tfrac{1}{2}+\alpha+it)|^{-2k}\,dt\ll
T\exp\Bigg{(}\frac{(\log
T)^{\frac{3(1+\delta)}{kb-1}}}{\exp\big{(}\frac{2\log\log T\log\log\log\log
T}{(kb-1)\log\log\log T}\big{)}}\Bigg{)}.$ (6.11)

###  6.2. The range $\alpha\gg\frac{(\log\log
T)^{\frac{4}{k}+\varepsilon}}{(\log T)^{\frac{1}{2k}}}$, $k\geq 1/2$, a
recursive bound

Here, we have the same setup as in the previous subsection. Namely, we assume
(6.1) and (6.2).

We will perform the same argument as before, but with a different choice of
parameters. We suppose that at step $m-1$, for any $\delta>0$, we have the
bound

$\int_{T}^{2T}|\zeta(\tfrac{1}{2}+\alpha+it)|^{-2k}\,dt\ll
T\exp\Bigg{(}\frac{(\log T)^{(1+\delta)(\frac{3}{kb-1})^{m-1}}\log\log
T}{\exp\big{(}\frac{2\cdot 3^{m-2}\log\log T\log\log\log\log
T}{(kb-1)^{m-1}\log\log\log
T}\big{)}}\Bigg{)}\exp\Big{(}(1+\delta)k^{2}(\log\log T)^{3}\Big{)}.$ (6.12)

Note that we proved the first step of the induction in Subsection 6.1 (see
equation (6.11)). Using (6.12), we will show that

$\int_{T}^{2T}|\zeta(\tfrac{1}{2}+\alpha+it)|^{-2k}\,dt\ll
T\exp\Bigg{(}\frac{(\log T)^{(1+\delta)(\frac{3}{kb-1})^{m}}\log\log
T}{\exp\big{(}\frac{2\cdot 3^{m-1}\log\log T\log\log\log\log
T}{(kb-1)^{m}\log\log\log
T}\big{)}}\Bigg{)}\exp\Big{(}(1+\delta)k^{2}(\log\log T)^{3}\Big{)}.$ (6.13)

Let

$\varepsilon^{\prime}=\frac{\delta(kb-1)\log\log T}{4kb(m-1)(\log\log
T-2kb\log\log\log T+\frac{\delta(kb-1)\log\log\log T}{2(m-1)})},$ (6.14)

and

$p=\frac{\log\log T}{\log\log T-2kb\varepsilon^{\prime}\log\log\log T},\qquad
q=\frac{\log\log T}{2kb\varepsilon^{\prime}\log\log\log T},$ (6.15)

so that $1/p+1/q=1$. Let

$f=b(1-\varepsilon^{\prime}).$ (6.16)

We will perform the inductive argument as long as

$\Big{(}\frac{kpf-1}{3}\Big{)}^{m-1}<\frac{\delta\log\log T}{9\log\log\log
T}.$ (6.17)

We choose

$\beta_{0}=\frac{3(2d-1)\big{(}1+\frac{\delta}{3}\big{)}(\frac{3}{kpf-1})^{m-1}\exp\big{(}\frac{2\cdot
3^{m-2}\log\log T\log\log\log\log T}{(kpf-1)^{m-1}\log\log\log
T}\big{)}}{4q(\log
T)^{(1+\frac{\delta}{3})(\frac{3}{kpf-1})^{m-1}}},\,s_{0}=\Big{[}\frac{1}{q\beta_{0}}\Big{]},\,\ell_{0}=2\Big{\lceil}\frac{s_{0}^{d}}{2}\Big{\rceil},$
(6.18)

where

$\displaystyle a$
$\displaystyle=\frac{1-n(1-\frac{\delta}{12})x}{1-n(1-\frac{\delta}{6})x},\qquad
d=1-\frac{n}{2}\Big{(}1-\frac{\delta}{6}\Big{)}x,\qquad
r=\frac{1-n\big{(}1-\frac{\delta}{12}\big{)}x}{1-nx}.$

As before, we have

$\frac{a(2d-1)}{r}=1-nx.$

Recall that $x=\frac{\log\log\log T}{\log\log T}$ and we choose

$n=\frac{2(kb-1)(kpf-1)^{m-1}\big{(}1-\frac{\delta}{24}\big{)}}{3^{m-1}\big{(}1+\frac{\delta}{3}\big{)}}.$
(6.19)

We also choose $\beta_{K}$ such that $K$ is the maximal integer for which

$\beta_{K}\leq\frac{\exp\big{(}\frac{6\log\log T\log\log\log\log
T}{n\log\log\log T}\big{)}}{(\log T)^{\frac{6}{n(1-\frac{\delta}{24})}}}.$
(6.20)

If $t\notin\mathcal{T}_{0}$, then there exists $0\leq v\leq K$ such that
$\frac{ke^{2}}{\ell_{0}}|P_{0,v}(t)|>1$. Using Hölder’s inequality, we have

$\displaystyle\int_{[T,2T]\setminus\mathcal{T}_{0}}$
$\displaystyle|\zeta(\tfrac{1}{2}+\alpha+it)|^{-2k}\,dt\leq\int_{T}^{2T}\Big{(}\frac{ke^{2}}{\ell_{0}}|P_{0,v}(t)|\Big{)}^{2s_{0}}|\zeta(\tfrac{1}{2}+\alpha+it)|^{-2k}\,dt$
$\displaystyle\leq\Big{(}\frac{ke^{2}}{\ell_{0}}\Big{)}^{2s_{0}}\Big{(}\int_{T}^{2T}|\zeta(\tfrac{1}{2}+\alpha+it)|^{-2kp}\,dt\Big{)}^{\frac{1}{p}}\Big{(}\int_{T}^{2T}|P_{0,v}(t)|^{2s_{0}q}\,dt\Big{)}^{\frac{1}{q}}.$

For the first integral above, we will use the bound (6.12) with
$\delta\mapsto\delta/3$. Note that

$\alpha=\frac{(\log\log T)^{f}}{(\log T)^{\frac{1}{2kp}}}.$

Then using the recursive bound (6.12), we obtain that

$\displaystyle\int_{T}^{2T}|\zeta(\tfrac{1}{2}+\alpha+it)|^{-2kp}$
$\displaystyle\ll T\exp\Bigg{(}\frac{(\log
T)^{(1+\frac{\delta}{3})(\frac{3}{kpf-1})^{m-1}}\log\log
T}{\exp\big{(}\frac{2\cdot 3^{m-2}\log\log T\log\log\log\log
T}{(kpf-1)^{m-1}\log\log\log T}\big{)}}\Bigg{)}$
$\displaystyle\qquad\qquad\times\exp\Big{(}\Big{(}1+\frac{\delta}{3}\Big{)}p^{2}k^{2}(\log\log
T)^{3}\Big{)}.$ (6.21)

Now using Proposition (3.2), we have that

$\displaystyle\int_{T}^{2T}|P_{0,v}(t)|^{2s_{0}q}\,dt\ll
T(s_{0}q)!b(\Delta_{0})^{2s_{0}q}\Big{(}\log\frac{1}{\Delta_{0}\alpha}\Big{)}^{2s_{0}q\gamma(\Delta_{0})}(\log\log
T^{\beta_{0}})^{s_{0}q}.$

Combining the bound above and (6.2) and using Stirling’s formula, we get that

$\displaystyle\int_{[T,2T]\setminus\mathcal{T}_{0}}$
$\displaystyle|\zeta(\tfrac{1}{2}+\alpha+it)|^{-2k}\,dt\ll
T\exp\Bigg{(}\frac{(\log
T)^{(1+\frac{\delta}{3})(\frac{3}{kpf-1})^{m-1}}\log\log
T}{\exp\big{(}\frac{2\cdot 3^{m-2}\log\log T\log\log\log\log
T}{(kpf-1)^{m-1}\log\log\log T}\big{)}}\Bigg{)}$
$\displaystyle\times\exp\Big{(}\Big{(}1+\frac{\delta}{3}\Big{)}pk^{2}(\log\log
T)^{3}\Big{)}s_{0}^{\frac{1}{2q}}\exp\bigg{(}-(2d-1)s_{0}\log s_{0}$
$\displaystyle\qquad\qquad+2s_{0}\log\Big{(}ke^{3/2}\sqrt{q}b(\Delta_{0})\Big{(}\log\frac{1}{\Delta_{0}\alpha}\Big{)}^{\gamma(\Delta_{0})}\sqrt{\log\log
T^{\beta_{0}}}\Big{)}\bigg{)}.$ (6.22)

Recall the choice (6.18) for $s_{0}$. Note that we have

$\log\log\frac{1}{\Delta_{0}\alpha}\leq\log\log\log
T,\qquad\log\log(\beta_{0}\log T)<\log\log\log T$

and

$\log q\leq\log\log\log T.$

Using the three bounds above in (6.22), it follows that

$\displaystyle\int_{[T,2T]\setminus\mathcal{T}_{0}}|\zeta(\tfrac{1}{2}+\alpha+it)|^{-2k}\,dt\ll
T\exp\Bigg{(}\frac{(\log
T)^{(1+\frac{\delta}{3})(\frac{3}{kpf-1})^{m-1}}\log\log
T}{\exp\big{(}\frac{2\cdot 3^{m-2}\log\log T\log\log\log\log
T}{(kpf-1)^{m-1}\log\log\log T}\big{)}}\Bigg{)}$
$\displaystyle\qquad\times\exp\Big{(}\Big{(}1+\frac{\delta}{3}\Big{)}pk^{2}(\log\log
T)^{3}\Big{)}\exp\Big{(}-(2d-1)s_{0}\log s_{0}+4s_{0}\log\log\log
T+O(s_{0})\Big{)}.$

By (6.17) we get

$\displaystyle\int_{[T,2T]\setminus\mathcal{T}_{0}}|\zeta(\tfrac{1}{2}+\alpha+it)|^{-2k}\,dt$
$\displaystyle\ll T\exp\Bigg{(}-\frac{(\log
T)^{(1+\frac{\delta}{3})(\frac{3}{kpf-1})^{m-1}}\log\log
T}{4\exp\big{(}\frac{2\cdot 3^{m-2}\log\log T\log\log\log\log
T}{(kpf-1)^{m-1}\log\log\log T}\big{)}}\Bigg{)}$
$\displaystyle\qquad\qquad\times\exp\Big{(}(1+\delta)k^{2}(\log\log
T)^{3}\Big{)}$ $\displaystyle\ll T\exp\Big{(}(1+\delta)k^{2}(\log\log
T)^{3}\Big{)}.$ (6.23)

Now suppose that $t\in\mathcal{T}_{0}$. Using Proposition 3.4 and proceeding
as before, we get that

$\displaystyle\int_{T}^{2T}S_{1}(t)\,dt\ll(\log\log
T)^{k}\exp\bigg{(}c_{1}(\log
T)^{\frac{3^{m}(1+\frac{\delta}{3})}{(kb-1)(kpf-1)^{m-1}(1-\frac{\delta}{24})^{2}}}\log\log
T$ $\displaystyle\qquad\times\exp\Big{(}-\frac{3^{m}\log\log T\log\log\log\log
T}{(kb-1)(kpf-1)^{m-1}\log\log\log
T}\Big{)}\bigg{)}\exp\Big{(}(1+\delta)k^{2}(\log\log T)^{3}\Big{)},$ (6.24)

for some $c_{1}>0$, and where we trivially bounded $\gamma(\Delta_{K})\leq 1$.
Now given the choice of parameters in (6.14), (6.15), (6.16), we have

$kpf-1=(kb-1)\Big{(}1-\frac{\delta}{4(m-1)}\Big{)}+O\Big{(}\frac{\log\log\log
T}{\log\log T}\Big{)}>(kb-1)\Big{(}1-\frac{7\delta}{24(m-1)}\Big{)}.$

Using this in (6.24) leads to

$\displaystyle\int_{T}^{2T}$ $\displaystyle S_{1}(t)\,dt\ll
T\exp\Bigg{(}\frac{(\log T)^{(1+\delta)(\frac{3}{kb-1})^{m}}\log\log
T}{\exp\big{(}\frac{2\cdot 3^{m-1}\log\log T\log\log\log\log
T}{(kb-1)^{m}\log\log\log
T}\big{)}}\Bigg{)}\exp\Big{(}(1+\delta)k^{2}(\log\log T)^{3}\Big{)}.$ (6.25)

To bound the contribution from $S_{2}(t)$, we proceed as in (6.9). We rewrite

$\displaystyle\int_{T}^{2T}S_{2}(t)\,dt$ $\displaystyle\ll T(\log\log
T)^{k}\sum_{j=0}^{K-1}\exp\bigg{(}\frac{\log\log\log
T}{\beta_{j}}(n-2kb+2)-\frac{n\log\log\log T\log(\beta_{j}\log
T)}{\beta_{j}\log\log T}$ $\displaystyle+\log\log T+k^{2}(\log\log
T)^{3}+O\Big{(}\frac{1}{\beta_{j}}\Big{)}\bigg{)},$ (6.26)

where we used the fact that $k\geq 1/2$. Note that the maximum in the sum over
$j$ is attained either at $j=0$ or $j=K-1$. Now given the choices (6.18) and
(6.19), the contribution from $j=0$ is

$\displaystyle\ll\exp\Big{(}\frac{\log\log\log
T}{\beta_{0}}\Big{(}-\frac{\delta(kb-1)}{4}+\frac{n\log\big{(}\frac{kpf-1}{3}\big{)}^{m-1}}{\log\log
T}\Big{)}+O\Big{(}\frac{1}{\beta_{0}}+(\log\log T)^{3}\Big{)}\Big{)},$

and again using the choices (6.19) and (6.17), it follows that the
contribution from $j=0$ is negligible.

For the contribution from $j=K-1$, proceeding similarly as in the bound for
$S_{1}(t)$, it follows that

$\displaystyle\int_{T}^{2T}$ $\displaystyle S_{2}(t)\,dt\ll
T\exp\Bigg{(}\frac{(\log T)^{(1+\delta)(\frac{3}{kb-1})^{m}}\log\log
T}{\exp\big{(}\frac{2\cdot 3^{m-1}\log\log T\log\log\log\log
T}{(kb-1)^{m}\log\log\log
T}\big{)}}\Bigg{)}\exp\Big{(}(1+\delta)k^{2}(\log\log T)^{3}\Big{)}.$

Combining the above, (6.2) and (6.25), the induction conclusion (6.13)
follows.

Now taking $m$ maximal as in (6.17), we get that

$\int_{T}^{2T}|\zeta(\tfrac{1}{2}+\alpha+it)|^{-2k}\,dt\ll
T\exp\Big{(}(1+\delta)k^{2}(\log\log T)^{3}\Big{)}.$ (6.27)

###  6.3. The range $\alpha\gg\frac{(\log\log
T)^{\frac{4}{k}+\varepsilon}}{(\log T)^{\frac{1}{2k}}}$, $k\geq 1/2$, once
more

Here, we use the same setup as in Subsection 6.2. Once again, we assume (6.1)
and (6.2).

We will improve the bound (6.27). The proof is similar to the proof in the
previous cases, so we will skip some of the details.

As before, let

$p=\frac{\log\log T}{\log\log T-2kb\delta\log\log\log T}$

and

$q=\frac{\log\log T}{2kb\delta\log\log\log T},$ (6.28)

so that $1/p+1/q=1$. We choose

$\beta_{0}=\frac{(6d-5)\log\log\log T}{(1+2\delta)pqk^{2}(\log\log
T)^{3}},\quad
s_{0}=\Big{[}\frac{1}{q\beta_{0}}\Big{]},\quad\ell_{0}=2\Big{\lceil}\frac{s_{0}^{d}}{2}\Big{\rceil}.$
(6.29)

We choose $\beta_{j},s_{j},\ell_{j}$ as in (4.2), and $a,d,r$ are such that

$\frac{a(2d-1)}{r}=1-\delta.$

We also choose $K$ maximal such that

$\beta_{K}\leq c,$ (6.30)

where $c$ is a small constant as in (4.5). Note that the conditions in
Propositions 3.3 and 3.4 are satisfied.

We now proceed as before. If $t\notin\mathcal{T}_{0}$, then as in Subsection
(6.2), we have that

$\displaystyle\int_{[T,2T]\setminus\mathcal{T}_{0}}$
$\displaystyle|\zeta(\tfrac{1}{2}+\alpha+it)|^{-2k}\,dt\leq\Big{(}\frac{ke^{2}}{\ell_{0}}\Big{)}^{2s_{0}}\Big{(}\int_{T}^{2T}|\zeta(\tfrac{1}{2}+\alpha+it)|^{-2kp}\,dt\Big{)}^{\frac{1}{p}}\Big{(}\int_{T}^{2T}|P_{0,v}(t)|^{2s_{0}q}\,dt\Big{)}^{\frac{1}{q}}$
$\displaystyle\ll T\exp\Big{(}(1+\delta)pk^{2}(\log\log
T)^{3}\Big{)}\exp\bigg{(}-(2d-1)s_{0}\log s_{0}$
$\displaystyle\qquad\qquad\qquad+2s_{0}\log\Big{(}ke^{3/2}\sqrt{q}b(\Delta_{0})\Big{(}\log\frac{1}{\Delta_{0}\alpha}\Big{)}^{\gamma(\Delta_{0})}\sqrt{\log\log
T^{\beta_{0}}}\Big{)}\bigg{)}.$

Note that $\gamma(\Delta_{0})=0$ with the choice of parameters (6.29). We
deduce that

$\displaystyle\int_{[T,2T]\setminus\mathcal{T}_{0}}|\zeta(\tfrac{1}{2}+\alpha+it)|^{-2k}\,dt$
$\displaystyle\ll T\exp\Big{(}(1+\delta)pk^{2}(\log\log T)^{3}\Big{)}$
$\displaystyle\qquad\times\exp\Big{(}-(2d-1)s_{0}\log s_{0}+2s_{0}\log\log\log
T+O(s_{0})\Big{)}$ $\displaystyle=o(T),$ (6.31)

by (6.28) and the choice of parameters in (6.29). Using the choice of
$\beta_{K}$ in (6.30), we also get that

$\int_{T}^{2T}S_{1}(t)\,dt\ll T(\log\log T)^{k}(\log T)^{k^{2}}.$ (6.32)

To bound the contribution from $S_{2}(t)$, we rewrite equation (6.9) using the
fact that $\gamma(\Delta_{j})=0$ for all $j$ and that $K\ll\log\log T$. We
have

$\displaystyle\int_{T}^{2T}S_{2}(t)\,dt$ $\displaystyle\ll T(\log\log
T)^{k+1}\sum_{j=0}^{K-1}\exp\bigg{(}\frac{\log\log
T}{\beta_{j}}\Big{(}1-2k-\frac{2kb\log\log\log T}{\log\log T}\Big{)}$
$\displaystyle\qquad\qquad+\frac{4\log\log\log
T}{\beta_{j}}\Big{(}2k-\frac{a(2d-1)}{r}\Big{)}+k^{2}\log\log
T+O\Big{(}\frac{1}{\beta_{j}}\Big{)}\bigg{)}.$

In the sum over $j$ above, note that if $k=1/2$, then the exponential term is
bounded by

$\exp\Big{(}-\frac{c_{1}\log\log\log T}{\beta_{j}}+k^{2}\log\log T\Big{)},$

for some $c_{1}>0$. If $k>1/2$, then the exponential term is bounded by

$\exp\Big{(}-\frac{c_{1}\log\log T}{\beta_{j}}+k^{2}\log\log T\Big{)},$

for some $c_{1}>0$. In both cases, we obtain that

$\int_{T}^{2T}S_{2}(t)\,dt=o\Big{(}T(\log\log T)^{k}(\log T)^{k^{2}}\Big{)}.$

Combining the above, (6.32) and (6.31), the bound (1.3) follows.

###  6.4. The range $\frac{1}{(\log
T)^{\frac{1}{2k}}}\ll\alpha=o\big{(}\frac{(\log\log
T)^{\frac{4}{k}+\varepsilon}}{(\log T)^{\frac{1}{2k}}}\big{)}$

Here, we will prove the bound (1.4). We will only sketch the proof, since it
is similar to the proof in the previous cases. We choose the parameters
$\beta_{j},\ell_{j},s_{j}$ as in (4.1), (4.2), and $a,d,r$ as in (6.3) and
(6.5), where

$n=\frac{6}{1-\delta}.$ (6.33)

We also choose $\beta_{K}$ as in (6.6). We now proceed as in Subsection 6.1.

As in (6.7), we have

$\displaystyle\int_{[T,2T]\setminus\mathcal{T}_{0}}|\zeta(\tfrac{1}{2}+\alpha+it)|^{-2k}\,dt=o(T).$
(6.34)

Also, as in (6.8), and keeping in mind the choice (6.33) for $n$, we get that

$\displaystyle\int_{T}^{2T}S_{1}(t)\,dt\ll T\exp\Bigg{(}\frac{\log T\log\log
T}{\exp(\frac{2\log\log T\log\log\log\log T}{3\log\log\log T})}\Bigg{)}.$
(6.35)

To bound the contribution from $S_{2}(t)$, we proceed as in equation (6.9).
Since $\alpha\gg\frac{1}{(\log T)^{\frac{1}{2k}}}$, we have

$\displaystyle\int_{T}^{2T}S_{2}(t)\,dt$ $\displaystyle\ll T(\log\log
T)^{k}\sum_{j=0}^{K-1}\exp\bigg{(}\frac{\log\log\log
T}{\beta_{j}}(n+2)+(\log\log
T)^{3}+O\Big{(}\frac{1}{\beta_{j}}\Big{)}\bigg{)}.$

The maximum in the sum above is attained when $j=0$. So, keeping in mind the
choice of $\beta_{0}$ in (4.1), we obtain that

$\displaystyle\int_{T}^{2T}S_{2}(t)\,dt\ll T\exp\Big{(}\frac{(4+\delta)\log
T\log\log\log T}{\log\log T}\Big{)},$ (6.36)

after a relabeling of the $\delta$. Combining (6.34), (6.35) and (6.36), the
conclusion follows in this case.

##  7\. Proof of Theorems 1.2 and 1.3 for small shifts $\alpha$

Here, we will prove the bounds (1.5) and (1.9) . The proof will be similar to
the one in the previous subsection, but we choose the parameters differently.
Recall that

$u=\frac{\log\frac{1}{\alpha}}{\log\log T}.$

Let $\delta>0$. We choose

$\beta_{0}=\frac{(2ku+2d-1-\frac{a(2d-1)}{r})\log\log T}{(1+\delta)ku\log
T},\quad
s_{0}=\Big{[}\frac{1}{\beta_{0}}\Big{]},\quad\ell_{0}=2\Big{\lceil}\frac{s_{0}^{d}}{2}\Big{\rceil},$
(7.1)

where we pick

$a=\frac{1-3k\varepsilon}{1-2k\varepsilon},\quad
r=\frac{1}{1-2k\varepsilon},\quad
d=\frac{2-7k\varepsilon}{2(1-3k\varepsilon)},$

so that

$\frac{a(2d-1)}{r}=1-4k\varepsilon.$ (7.2)

We further pick $\beta_{j},\ell_{j},s_{j}$ as in (4.2). We choose $K$ to be
maximal such that

$\beta_{K}\leq c,$ (7.3)

for $c$ a small constant such that

$c^{1-d}\Big{(}\frac{ra^{d}}{r^{1-d}-1}+\frac{2r}{r-1}\Big{)}\leq
1-a-\beta_{0}^{1-d}.$ (7.4)

Note that the above ensures that the conditions in Propositions 3.2, 3.3 and
3.4 are satisfied.

If $t\notin\mathcal{T}_{0}$, then we proceed as before, and similarly to
equation (4.7) we get that

$\displaystyle\int_{[T,2T]\setminus\mathcal{T}_{0}}|\zeta(\tfrac{1}{2}+\alpha+it)|^{-2k}\,dt\ll
T^{1+(1+\delta)ku}$ $\displaystyle\qquad\times\exp\bigg{(}-(2d-1)s_{0}\log
s_{0}+2s_{0}\log\Big{(}ke^{3/2}b(\Delta_{0})\Big{(}\log\frac{1}{\Delta_{0}\alpha}\Big{)}^{\gamma(\Delta_{0})}\sqrt{\log\log
T^{\beta_{0}}}\Big{)}\bigg{)}.$

Keeping in mind the choice of parameters (7.1), we obtain that

$\int_{[T,2T]\setminus\mathcal{T}_{0}}|\zeta(\tfrac{1}{2}+\alpha+it)|^{-2k}\,dt\ll
T^{1+(1+\delta)\frac{2ku-\frac{a(2d-1)}{r}}{2ku-\frac{a(2d-1)}{r}+2d-1}}\exp\Big{(}O\Big{(}\frac{\log
T\log\log\log T}{\log\log T}\Big{)}\Big{)}.$ (7.5)

Now if $t\in\mathcal{T}_{0}$, then since $\gamma(\Delta_{K})=1$, we use
Proposition 3.4 and the expression (7.3) for $\beta_{K}$ as before to get that

$\int_{T}^{2T}S_{1}(t)\,dt\ll T(\log
T)^{O(1)}\exp\bigg{(}\Big{(}\frac{1}{4}+\varepsilon\Big{)}k^{2}\Big{(}\log\frac{1}{\alpha}\Big{)}^{2}\log\log
T\bigg{)}.$ (7.6)

Similarly as before (see equation (4.9)), we have

$\displaystyle\int_{T}^{2T}S_{2}(t)\,dt$ $\displaystyle\ll T(\log\log
T)^{k}\sum_{j=0}^{K-1}(K-j)\sqrt{\frac{1}{\beta_{j+1}}}\exp\bigg{(}\frac{\log\log
T}{\beta_{j}}\Big{(}2ku-\frac{a(2d-1)}{r}\Big{)}$ $\displaystyle\
+\frac{\log(\beta_{j}\log
T)}{\beta_{j}}\Big{(}\frac{a(2d-1)}{r}-2k\Big{)}+\frac{2a}{r\beta_{j}}\log\Big{(}e^{3/2}b(\Delta_{j+1})\Big{(}\log\frac{1}{\Delta_{j+1}\alpha}\Big{)}^{\gamma(\Delta_{j+1})}\Big{)}$
$\displaystyle\
+\frac{2k}{\beta_{j}}\log\frac{1}{1-\varepsilon/2}\bigg{)}\big{(}\log
T^{\beta_{j}}\big{)}^{k^{2}b(\Delta_{j})^{2}(\log\frac{1}{\Delta_{j}\alpha})^{2\gamma(\Delta_{j})}}.$

As $K\ll\log\log T$, we further write the above as

$\displaystyle\int_{T}^{2T}$ $\displaystyle S_{2}(t)\,dt\ll
T\sum_{j=0}^{K-1}\exp\bigg{(}\frac{\log\log
T}{\beta_{j}}\Big{(}2ku-\frac{a(2d-1)}{r}\Big{)}+O\Big{(}\frac{\log
T\log\log\log T}{\log\log T}\Big{)}\bigg{)}$

Since $\alpha=o\big{(}\frac{1}{(\log T)^{\frac{1}{2k}-\varepsilon}}\big{)},$
we have that

$2ku-\frac{a(2d-1)}{r}\geq 2k\varepsilon.$

Hence the sum over $j$ above achieves its maximum when $j=0$. Using (7.1), we
obtain that

$\displaystyle\int_{T}^{2T}S_{2}(t)\,dt\ll
T^{1+(1+\delta)\frac{2ku-\frac{a(2d-1)}{r}}{2ku-\frac{a(2d-1)}{r}+2d-1}}\exp\Big{(}O\Big{(}\frac{\log
T\log\log\log T}{\log\log T}\Big{)}\Big{)}.$ (7.7)

Combining equations (7.2), (7.5), (7.6) and (7.7) and after a relabeling of
the $\varepsilon$, the bounds (1.5) and (1.9) follow.

##  8\. The asymptotic formula

In this section we shall prove Theorem 1.4. We have

$\displaystyle\frac{1}{2\pi
i}\int_{1-iT/2}^{1+iT/2}e^{z^{2}/2}X^{z}\frac{1}{\zeta(\tfrac{1}{2}+\alpha+it+z)^{k}\zeta(\tfrac{1}{2}+\alpha-
it+z)^{k}}\frac{dz}{z}$ (8.1)
$\displaystyle\qquad\qquad=\sum_{m,n=1}^{\infty}\frac{\mu_{k}(m)\mu_{k}(n)}{(mn)^{1/2+\alpha}}\Big{(}\frac{m}{n}\Big{)}^{-it}W\Big{(}\frac{mn}{X}\Big{)},$

where

$W(x)=\frac{1}{2\pi i}\int_{1-iT/2}^{1+iT/2}e^{z^{2}/2}x^{-z}\frac{dz}{z},$
(8.2)

by writing the zeta-functions in (8.1) as Dirichlet series and integrating
term-by-term. On the other hand, by deforming the line of integration, the
left hand side of (8.1) is equal to

$\displaystyle|\zeta(\tfrac{1}{2}+\alpha+it)|^{-2k}+\frac{1}{2\pi
i}\int_{-(1-\varepsilon)\alpha-
iT/2}^{-(1-\varepsilon)\alpha+iT/2}e^{z^{2}/2}X^{z}\frac{1}{\zeta(\tfrac{1}{2}+\alpha+it+z)^{k}\zeta(\tfrac{1}{2}+\alpha-
it+z)^{k}}\frac{dz}{z}$
$\displaystyle\qquad\qquad+O\Big{(}\frac{e^{-T^{2}/8}X}{T}\max_{\varepsilon\alpha\leq\sigma\leq
1+\alpha}\frac{1}{|\zeta(\tfrac{1}{2}+\sigma+i(t\pm
T/2))|^{k}|\zeta(\tfrac{1}{2}+\sigma-i(t\mp T/2))|^{k}}\Big{)}.$

For $t\in[T,2T]$, the $O$-term is

$\displaystyle\ll\frac{e^{-T^{2}/8}X}{T}T^{\frac{k\log\frac{1}{\varepsilon\alpha}}{\log\log
T}}\ll e^{-T^{2}/8}XT^{O_{k}(1)},$

by Lemma 2.2. Hence, integrating over $t\in[T,2T]$ we obtain that

$\displaystyle\int_{T}^{2T}|\zeta(\tfrac{1}{2}+\alpha+it)|^{-2k}dt=\sum_{m,n=1}^{\infty}\frac{\mu_{k}(m)\mu_{k}(n)}{(mn)^{1/2+\alpha}}W\Big{(}\frac{mn}{X}\Big{)}\int_{T}^{2T}\Big{(}\frac{m}{n}\Big{)}^{-it}dt$
$\displaystyle\qquad\qquad-\frac{1}{2\pi i}\int_{-(1-\varepsilon)\alpha-
iT/2}^{-(1-\varepsilon)\alpha+iT/2}e^{z^{2}/2}X^{z}\int_{T}^{2T}\frac{1}{\zeta(\tfrac{1}{2}+\alpha+it+z)^{k}\zeta(\tfrac{1}{2}+\alpha-
it+z)^{k}}dt\frac{dz}{z}$
$\displaystyle\qquad\qquad+O\big{(}e^{-T^{2}/8}XT^{O_{k}(1)}\big{)}.$

Furthermore, by the Cauchy-Schwarz inequality the second term above is

$\displaystyle\ll_{\varepsilon}\frac{X^{-(1-\varepsilon)\alpha}}{\alpha}\int_{-T/2}^{T/2}e^{-z^{2}/2}\bigg{(}\int_{T}^{2T}|\zeta(\tfrac{1}{2}+\varepsilon\alpha+i(t+z))|^{-2k}dt\bigg{)}^{1/2}$
$\displaystyle\qquad\qquad\qquad\qquad\times\bigg{(}\int_{T}^{2T}|\zeta(\tfrac{1}{2}+\varepsilon\alpha-i(t-z))|^{-2k}dt\bigg{)}^{1/2}dz$
$\displaystyle\ll_{\varepsilon}\frac{X^{-(1-\varepsilon)\alpha}}{\alpha}T(\log\log
T)^{k}(\log T)^{k^{2}}\ll_{\varepsilon}TX^{-(1-\varepsilon)\alpha}(\log
T)^{k^{2}+1},$

in view of (1.3) and (1.6) in Theorems 1.2 and 1.3, and the fact that
$\frac{1}{\alpha}\ll\min\Big{\\{}\frac{(\log T)^{\frac{1}{2k}}}{(\log\log
T)^{\frac{4}{k}+\varepsilon}},\frac{\log T}{\log\log T}\Big{\\}}$. Thus,

$\displaystyle\int_{T}^{2T}|\zeta(\tfrac{1}{2}+\alpha+it)|^{-2k}dt$
$\displaystyle=\sum_{m,n=1}^{\infty}\frac{\mu_{k}(m)\mu_{k}(n)}{(mn)^{1/2+\alpha}}W\Big{(}\frac{mn}{X}\Big{)}\int_{T}^{2T}\Big{(}\frac{m}{n}\Big{)}^{-it}dt$
(8.3) $\displaystyle\qquad\qquad+O\big{(}TX^{-(1-\varepsilon)\alpha}(\log
T)^{k^{2}+1}\big{)}+O\big{(}e^{-T^{2}/8}XT^{O_{k}(1)}\big{)}.$

We next consider the contribution of off-diagonal terms $m\neq n$ on the right
hand side of (8.3), which is

$\displaystyle\ll\sum_{m\neq n}\frac{d_{k}(m)d_{k}(n)}{(mn)^{1/2+\alpha}|\log
m/n|}\Big{|}W\Big{(}\frac{mn}{X}\Big{)}\Big{|}.$

We note from (8.2) that $W(x)\ll x^{-1}$ trivially and

$W(x)=1+O(x)+O\Big{(}\frac{e^{-T^{2}/8}}{xT}\Big{)}$

if $x\leq 1$, by moving the contour to the $-1$-line, and so this contribution
is bounded by

$\displaystyle\ll\sum_{\begin{subarray}{c}m\neq n\\\ mn\leq
X\end{subarray}}\frac{d_{k}(m)d_{k}(n)}{(mn)^{1/2+\alpha}|\log
m/n|}+\frac{e^{-T^{2}/8}X}{T}\sum_{\begin{subarray}{c}m\neq n\\\ mn\leq
X\end{subarray}}\frac{d_{k}(m)d_{k}(n)}{(mn)^{3/2+\alpha}|\log m/n|}$
$\displaystyle\qquad\qquad+X\sum_{\begin{subarray}{c}m\neq n\\\
mn>X\end{subarray}}\frac{d_{k}(m)d_{k}(n)}{(mn)^{3/2+\alpha}|\log m/n|}$
$\displaystyle=E_{1}+E_{2},$

say, where $E_{1}$ and $E_{2}$ denote the sums with $\frac{m}{n}\notin[1/2,2]$
and $\frac{m}{n}\in[1/2,2]$, respectively. With $E_{1}$, $|\log m/n|\gg 1$ and
we get

$\displaystyle E_{1}$ $\displaystyle\ll\sum_{\begin{subarray}{c}mn\leq
X\end{subarray}}\frac{d_{k}(m)d_{k}(n)}{(mn)^{1/2+\alpha}}+\frac{e^{-T^{2}/8}X}{T}\sum_{\begin{subarray}{c}mn\leq
X\end{subarray}}\frac{d_{k}(m)d_{k}(n)}{(mn)^{3/2+\alpha}}+X\sum_{\begin{subarray}{c}mn>X\end{subarray}}\frac{d_{k}(m)d_{k}(n)}{(mn)^{3/2+\alpha}}$
$\displaystyle\ll X^{1/2-\alpha}(\log X)^{2k-1}+\frac{e^{-T^{2}/8}X}{T}.$
(8.4)

For $E_{2}$, we use the fact that

$\frac{d_{k}(m)d_{k}(n)}{(mn)^{\sigma}}\ll\frac{d_{k}(m)^{2}}{m^{2\sigma}}+\frac{d_{k}(n)^{2}}{n^{2\sigma}},$
(8.5)

and we have

$\displaystyle E_{2}$
$\displaystyle\ll\sum_{m\leq\sqrt{2X}}\frac{d_{k}(m)^{2}}{m^{1+2\alpha}}\sum_{\begin{subarray}{c}n\neq
m\\\ m/2\leq n\leq 2m\end{subarray}}\frac{1}{|\log
m/n|}+\frac{e^{-T^{2}/8}X}{T}\sum_{m\leq\sqrt{2X}}\frac{d_{k}(m)^{2}}{m^{3+2\alpha}}\sum_{\begin{subarray}{c}n\neq
m\\\ m/2\leq n\leq 2m\end{subarray}}\frac{1}{|\log m/n|}$
$\displaystyle\qquad\qquad+X\sum_{m>\sqrt{X/2}}\frac{d_{k}(m)^{2}}{m^{3+2\alpha}}\sum_{\begin{subarray}{c}n\neq
m\\\ m/2\leq n\leq 2m\end{subarray}}\frac{1}{|\log m/n|}$
$\displaystyle\ll\log
X\sum_{m\leq\sqrt{2X}}\frac{d_{k}(m)^{2}}{m^{2\alpha}}+\frac{e^{-T^{2}/8}X\log
X}{T}\sum_{m\leq\sqrt{2X}}\frac{d_{k}(m)^{2}}{m^{2+2\alpha}}+X\sum_{m>\sqrt{X/2}}\frac{d_{k}(m)^{2}\log
m}{m^{2+2\alpha}}$ $\displaystyle\ll X^{1/2-\alpha}(\log
X)^{k^{2}}+\frac{e^{-T^{2}/8}X\log X}{T}.$ (8.6)

We are left with the contribution of the diagonal terms $m=n$ on the right
hand side of (8.3). By (8.2) this is

$\displaystyle
T\sum_{n=1}^{\infty}\frac{\mu_{k}(n)^{2}}{n^{1+2\alpha}}W\Big{(}\frac{n^{2}}{X}\Big{)}=\frac{T}{2\pi
i}\int_{1-iT/2}^{1+iT/2}e^{z^{2}/2}X^{z}\sum_{n=1}^{\infty}\frac{\mu_{k}(n)^{2}}{n^{1+2\alpha+2z}}\frac{dz}{z}$
$\displaystyle\qquad=\frac{T}{2\pi
i}\int_{1-iT/2}^{1+iT/2}e^{z^{2}/2}X^{z}\zeta(1+2\alpha+2z)^{k^{2}}\prod_{p}\bigg{(}1-\frac{1}{p^{1+2\alpha+2z}}\bigg{)}^{k^{2}}\bigg{(}1+\sum_{j=1}^{\infty}\frac{\mu_{k}(p^{j})^{2}}{p^{(1+2\alpha+2z)j}}\bigg{)}\frac{dz}{z}.$

We move the contour to the $-(1-\varepsilon)\alpha$-line, crossing a simple
pole at $z=0$. In doing so, we get that this is equal to

$\displaystyle
T\zeta(1+2\alpha)^{k^{2}}\prod_{p}\bigg{(}1-\frac{1}{p^{1+2\alpha}}\bigg{)}^{k^{2}}\bigg{(}1+\sum_{j=1}^{\infty}\frac{\mu_{k}(p^{j})^{2}}{p^{(1+2\alpha)j}}\bigg{)}$
$\displaystyle\qquad\qquad+O\big{(}TX^{-(1-\varepsilon)\alpha}\alpha^{-(k^{2}+1)}\big{)}+O\big{(}e^{-T^{2}/8}XT^{O_{k}(1)}\big{)}.$

Thus,

$\displaystyle\int_{T}^{2T}|\zeta(\tfrac{1}{2}+\alpha+it)|^{-2k}dt=T\zeta(1+2\alpha)^{k^{2}}\prod_{p}\bigg{(}1-\frac{1}{p^{1+2\alpha}}\bigg{)}^{k^{2}}\bigg{(}1+\sum_{j=1}^{\infty}\frac{\mu_{k}(p^{j})^{2}}{p^{(1+2\alpha)j}}\bigg{)}$
$\displaystyle\qquad\qquad+O\big{(}TX^{-(1-\varepsilon)\alpha}(\log
T)^{k^{2}+1}\big{)}+O\big{(}e^{-T^{2}/8}XT^{O_{k}(1)}\log
X\big{)}+O\big{(}X^{1/2-\alpha}(\log X)^{k^{2}}\big{)},$

by combining the above with (8.3), (8) and (8). We choose $X=T^{2}$, then the
error terms above become $T^{1-2(1-\varepsilon)\alpha}(\log T)^{k^{2}+1}$, and
the conclusion follows after a relabeling of the $\varepsilon$.

##  9\. Proof of Theorem 1.5

###  9.1. Assuming RH: $k\geq 1$

Following the arguments in [7, Chapter 17], it is standard from Perron’s
formula that

$\sum_{n\leq x}\mu_{k}(n)=\frac{1}{2\pi
i}\int_{c-i[x]}^{c+i[x]}\frac{x^{s}}{\zeta(s)^{k}}\frac{ds}{s}+O\big{(}(\log
x)^{k}\big{)}$ (9.1)

with $c=1+\frac{1}{\log x}$. We deform the contour by replacing the line
segment $c+it$, $|t|\leq[x]$, with a piecewise linear path comprising of a
number of horizontal and vertical line segments,

$\bigcup_{j=1}^{J}\big{(}V_{j}\cup H_{j}\big{)}\bigcup V_{0},$

where

$\displaystyle V_{0}:\quad s=\frac{1}{2}+\frac{(\log\log
x_{0})^{\frac{8}{k}+\varepsilon}}{(\log x_{0})^{\frac{1}{k}}}+it,\quad|t|\leq
x_{0},$ $\displaystyle V_{j}:\quad
s=\frac{1}{2}+\frac{(\log\log(2^{j-1}x_{0}))^{\frac{8}{k}+\varepsilon}}{(\log(2^{j-1}x_{0}))^{\frac{1}{k}}}+it,\quad
2^{j-1}x_{0}\leq|t|\leq\min\\{2^{j}x_{0},[x]\\},\quad 1\leq j\leq J,$
$\displaystyle H_{j}:\quad s=\sigma\pm
i2^{j}x_{0},\quad\frac{1}{2}+\frac{(\log\log(2^{j}x_{0}))^{\frac{8}{k}+\varepsilon}}{(\log(2^{j}x_{0}))^{\frac{1}{k}}}\leq\sigma\leq\frac{1}{2}+\frac{(\log\log(2^{j-1}x_{0}))^{\frac{8}{k}+\varepsilon}}{(\log(2^{j-1}x_{0}))^{\frac{1}{k}}},$
$\displaystyle\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad
1\leq j\leq J-1,$ $\displaystyle H_{J}:\quad s=\sigma\pm
i[x],\quad\frac{1}{2}+\frac{(\log\log(2^{J-1}x_{0}))^{\frac{8}{k}+\varepsilon}}{(\log(2^{J-1}x_{0}))^{\frac{1}{k}}}\leq\sigma\leq
1+\frac{1}{\log x}$

and $x_{0}=\exp\big{(}(\log x)^{\frac{k}{k+1}}(\log\log
x)^{\frac{k+8}{k+1}}\big{)}$, $J=\lceil\log_{2}\frac{[x]}{x_{0}}\rceil\ll\log
x$. We encounter no pole in doing so.

We first consider the integral along $V_{0}$. Let

$\alpha_{0}:=\frac{(\log\log x_{0})^{\frac{8}{k}+\varepsilon}}{(\log
x_{0})^{\frac{1}{k}}}\qquad\text{and}\qquad T_{0}:=\exp\Big{(}\frac{\log
x_{0}}{(\log\log x_{0})^{8}}\Big{)}.$

For $T_{0}\leq T\leq x_{0}$ we have $\frac{1}{(\log
T)^{\frac{1}{k}}}\ll\alpha_{0}\ll\frac{(\log\log
T)^{\frac{8}{k}+\varepsilon}}{(\log T)^{\frac{1}{k}}}$ and so

$\displaystyle\frac{1}{2\pi
i}\int_{1/2+\alpha_{0}+iT/2}^{1/2+\alpha_{0}+iT}\frac{x^{s}}{\zeta(s)^{k}}\frac{ds}{s}$
$\displaystyle\ll\frac{x^{1/2+\alpha_{0}}}{T}\int_{1/2+\alpha_{0}+iT/2}^{1/2+\alpha_{0}+iT}\Big{|}\frac{ds}{\zeta(s)^{k}}\Big{|}$
$\displaystyle\ll x^{1/2+\alpha_{0}}\exp\Big{(}\log T(\log\log
T)^{-1+\varepsilon}\Big{)},$

by (1.4). We hence get

$\displaystyle\frac{1}{2\pi
i}\int_{1/2+\alpha_{0}+iT_{0}}^{1/2+\alpha_{0}+ix_{0}}\frac{x^{s}}{\zeta(s)^{k}}\frac{ds}{s}$
$\displaystyle\ll x^{1/2+\alpha_{0}}\exp\Big{(}\log x_{0}(\log\log
x_{0})^{-1+\varepsilon}\Big{)}$ $\displaystyle=\sqrt{x}\exp\Big{(}\frac{\log
x(\log\log x_{0})^{\frac{8}{k}+\varepsilon}}{(\log x_{0})^{\frac{1}{k}}}+\log
x_{0}(\log\log x_{0})^{-1+\varepsilon}\Big{)}$
$\displaystyle\ll\sqrt{x}\exp\Big{(}(\log x)^{\frac{k}{k+1}}(\log\log
x)^{\frac{7}{k+1}+\varepsilon}\Big{)},$ (9.2)

by diving the segment of integration into dyadic intervals. The same bound
holds for the integral along
$\int_{1/2+\alpha_{0}-ix_{0}}^{1/2+\alpha_{0}-iT_{0}}$. Furthermore, we note
from Lemma 2.2 that

$|\zeta(s)|^{-1}\ll(|t|+2)^{(\frac{1}{2k}+\varepsilon)\frac{\log\log
x_{0}}{\log\log(|t|+2)}}$ (9.3)

for $s\in V_{0}$. So

$\displaystyle\frac{1}{2\pi
i}\int_{1/2+\alpha_{0}-iT_{0}}^{1/2+\alpha_{0}+iT_{0}}\frac{x^{s}}{\zeta(s)^{k}}\frac{ds}{s}$
$\displaystyle\ll\sqrt{x}\exp\Big{(}\frac{\log x(\log\log
x_{0})^{\frac{8}{k}+\varepsilon}}{(\log x_{0})^{\frac{1}{k}}}+\frac{\log\log
x_{0}\log T_{0}}{\log\log T_{0}}\Big{)}$
$\displaystyle\ll\sqrt{x}\exp\Big{(}(\log x)^{\frac{k}{k+1}}(\log\log
x)^{\frac{7}{k+1}+\varepsilon}\Big{)}.$ (9.4)

Combining (9.1) and (9.1) we obtain that

$\frac{1}{2\pi
i}\int_{V_{0}}\frac{x^{s}}{\zeta(s)^{k}}\frac{ds}{s}\ll\sqrt{x}\exp\Big{(}(\log
x)^{\frac{k}{k+1}}(\log\log x)^{\frac{7}{k+1}+\varepsilon}\Big{)}.$

For the contribution from the vertical segments $\cup_{j=1}^{J}V_{j}$, we
deduce from (1.3) that it is bounded by

$\displaystyle\ll\sqrt{x}(\log
x)^{\frac{k^{2}}{4}+\varepsilon}\sum_{j=0}^{J-1}\exp\Big{(}\frac{\log
x(\log\log(2^{j}x_{0}))^{\frac{8}{k}+\varepsilon}}{(\log(2^{j}x_{0}))^{\frac{1}{k}}}\Big{)}$
$\displaystyle\ll\sqrt{x}\exp\Big{(}\frac{\log x(\log\log
x)^{\frac{8}{k}+\varepsilon}}{(\log
x_{0})^{\frac{1}{k}}}\Big{)}\ll\sqrt{x}\exp\Big{(}(\log
x)^{\frac{k}{k+1}}(\log\log x)^{\frac{7}{k+1}+\varepsilon}\Big{)}.$ (9.5)

For $s\in H_{j}$, $1\leq j\leq J-1$, again like (9.3) we have

$|\zeta(s)|^{-1}\ll(2^{j}x_{0})^{\frac{1}{2k}+\varepsilon}.$

So the contribution to (9.1) from the horizontal segments
$\cup_{j=1}^{J-1}H_{j}$ is

$\displaystyle\ll\sqrt{x}\sum_{j=0}^{J-2}(2^{j}x_{0})^{-1/2+\varepsilon}\exp\Big{(}\frac{\log
x(\log\log(2^{j}x_{0}))^{\frac{8}{k}+\varepsilon}}{(\log(2^{j}x_{0}))^{\frac{1}{k}}}\Big{)}$
$\displaystyle\ll\sqrt{x}x_{0}^{-1/2+\varepsilon}\exp\Big{(}\frac{\log
x(\log\log x)^{\frac{8}{k}+\varepsilon}}{(\log
x_{0})^{\frac{1}{k}}}\Big{)}\ll\sqrt{x}.$ (9.6)

We are left with the integral along $H_{J}$, which is bounded by

$\max_{\frac{1}{2}+\frac{(\log\log x)^{\frac{8}{k}+\varepsilon}}{(\log
x)^{\frac{1}{k}}}\leq\sigma\leq 1+\frac{1}{\log x}}\exp\Big{(}(\sigma-1)\log
x-k\log|\zeta(\sigma\pm ix)|\Big{)}.$ (9.7)

For $\frac{1}{2}+\frac{(\log\log x)^{\frac{8}{k}+\varepsilon}}{(\log
x)^{\frac{1}{k}}}\leq\sigma\leq\frac{1}{2}+o(\frac{1}{\log\log x})$, Lemma 2.2
implies that this is

$\displaystyle\ll\exp\bigg{(}(\sigma-1)\log x+\frac{k\log x}{2\log\log
x}\log\frac{1}{1-(\log x)^{1-2\sigma}}$
$\displaystyle\qquad\qquad\qquad\qquad\qquad+O_{k}\Big{(}\frac{\log
x}{\log\log x}\Big{)}+O_{k}\Big{(}\frac{\log x}{(\log\log
x)^{2}}\log\frac{1}{1-(\log x)^{1-2\sigma}}\Big{)}\bigg{)}$
$\displaystyle\ll\exp\bigg{(}(\sigma-1)\log x+\frac{k\log x}{2\log\log
x}\log\frac{1}{(2\sigma-1)\log\log x}+O_{k}\big{(}(2\sigma-1)\log x\big{)}$
$\displaystyle\qquad\qquad\qquad\qquad\qquad+O_{k}\Big{(}\frac{\log
x}{\log\log x}\Big{)}+O_{k}\Big{(}\frac{\log x}{(\log\log
x)^{2}}\log\frac{1}{2\sigma-1}\Big{)}\bigg{)},$ (9.8)

which is decreasing with repsect to $\sigma$, and, hence,

$\ll\exp\bigg{(}-\frac{(8+k)\log x\log\log\log x}{2\log\log
x}+O_{k}\Big{(}\frac{\log x}{\log\log x}\Big{)}\bigg{)}\ll 1.$ (9.9)

For $1-\frac{1}{\log\log x}\leq\sigma\leq 1+\frac{1}{\log x}$, the second
estimate in Lemma 2.2 leads to a bound of size

$\displaystyle\ll\exp\Big{(}(\sigma-1)\log x+k\log\log\log
x+O_{k}(1)\Big{)}\ll(\log\log x)^{k}.$ (9.10)

Finally, for $\frac{1}{2}+O(\frac{1}{\log\log x})\leq\sigma\leq
1-\frac{1}{\log\log x}$, we use the last estimate in Lemma 2.2 to get the
bound

$\displaystyle\ll\exp\bigg{(}(\sigma-1)\log x+\frac{(\log
x)^{2-2\sigma}}{(1-\sigma)\log\log x}+\varepsilon\log
x+O_{k}\Big{(}\frac{(\log x)^{2-2\sigma}}{(1-\sigma)^{2}(\log\log
x)^{2}}\Big{)}\bigg{)}\ll x^{\varepsilon}.$ (9.11)

Combining the estimates we obtain the first part of the theorem for $k\geq 1$.

###  9.2. Assuming RH: $k<1$

The arguments are similar to the previous subsection. We replace the contour
in (9.1) with

$\bigcup_{j=1}^{J}\big{(}V_{j}\cup H_{j}\big{)}\bigcup V_{0},$

where

$\displaystyle V_{0}:\quad s=\frac{1}{2}+\varepsilon\frac{\log\log x_{1}}{\log
x_{1}}+it,\quad|t|\leq x_{1},$ $\displaystyle V_{j}:\quad
s=\frac{1}{2}+\varepsilon\frac{\log\log(2^{j-1}x_{1})}{\log(2^{j-1}x_{1})}+it,\quad
2^{j-1}x_{1}\leq|t|\leq\min\\{2^{j}x_{1},[x]\\},\quad 1\leq j\leq J,$
$\displaystyle H_{j}:\quad s=\sigma\pm
i2^{j}x_{1},\quad\frac{1}{2}+\varepsilon\frac{\log\log(2^{j}x_{1})}{\log(2^{j}x_{1})}\leq\sigma\leq\frac{1}{2}+\varepsilon\frac{\log\log(2^{j-1}x_{1})}{\log(2^{j-1}x_{1})},\quad
1\leq j\leq J-1,$ $\displaystyle H_{J}:\quad s=\sigma\pm
i[x],\quad\frac{1}{2}+\varepsilon\frac{\log\log(2^{J-1}x_{1})}{\log(2^{J-1}x_{1})}\leq\sigma\leq
1+\frac{1}{\log x}$

and $x_{1}=\exp(\sqrt{\varepsilon\log x}\log\log x)$,
$J=\lceil\log_{2}\frac{[x]}{x_{1}}\rceil\ll\log x$. Again we encounter no pole
in doing so.

For the integral along $V_{0}$, let

$\alpha_{1}:=\varepsilon\frac{\log\log x_{1}}{\log
x_{1}}\qquad\text{and}\qquad T_{1}:=\exp\Big{(}\frac{\log x_{1}}{\log\log
x_{1}}\Big{)}.$

If $T_{1}\leq T\leq x_{1}$, then $\frac{1}{\log
T}\ll\alpha_{1}\ll\frac{\log\log T}{\log T}$, and hence

$\displaystyle\frac{1}{2\pi
i}\int_{1/2+\alpha_{1}+iT/2}^{1/2+\alpha_{1}+iT}\frac{x^{s}}{\zeta(s)^{k}}\frac{ds}{s}$
$\displaystyle\ll\frac{x^{1/2+\alpha_{1}}}{T}\int_{1/2+\alpha_{1}+iT/2}^{1/2+\alpha_{1}+iT}\Big{|}\frac{ds}{\zeta(s)^{k}}\Big{|}\ll
x^{1/2+\alpha_{1}}\exp\Big{(}(\log\log T)^{1+\varepsilon}\Big{)},$

by (1.7). It follows that

$\displaystyle\frac{1}{2\pi
i}\int_{1/2+\alpha_{1}+iT_{1}}^{1/2+\alpha_{1}+ix_{1}}\frac{x^{s}}{\zeta(s)^{k}}\frac{ds}{s}$
$\displaystyle\ll x^{1/2+\alpha_{1}}\exp\Big{(}(\log\log
x_{1})^{1+\varepsilon}\Big{)}$
$\displaystyle=\sqrt{x}\exp\Big{(}\varepsilon\frac{\log x\log\log x_{1}}{\log
x_{1}}+(\log\log x_{1})^{1+\varepsilon}\Big{)}$
$\displaystyle\ll\sqrt{x}\exp\big{(}\varepsilon\sqrt{\log x}\big{)},$ (9.12)

by relabelling $\varepsilon$. The same bound holds for the integral along
$\int_{1/2+\alpha_{1}-ix_{1}}^{1/2+\alpha_{1}-iT_{1}}$. Also, by (9.3) we have

$\displaystyle\frac{1}{2\pi
i}\int_{1/2+\alpha_{1}-iT_{1}}^{1/2+\alpha_{1}+iT_{1}}\frac{x^{s}}{\zeta(s)^{k}}\frac{ds}{s}$
$\displaystyle\ll\sqrt{x}\exp\Big{(}\varepsilon\frac{\log x\log\log
x_{1}}{\log x_{1}}+\Big{(}\frac{k}{2}+\varepsilon\Big{)}\frac{\log\log
x_{1}\log T_{1}}{\log\log T_{1}}\Big{)}$
$\displaystyle\ll\sqrt{x}\exp\big{(}\varepsilon\sqrt{\log x}\big{)},$ (9.13)

by another relabelling of $\varepsilon$. Combining (9.2) and (9.2) we obtain
that

$\frac{1}{2\pi
i}\int_{V_{0}}\frac{x^{s}}{\zeta(s)^{k}}\frac{ds}{s}\ll\sqrt{x}\exp\big{(}\varepsilon\sqrt{\log
x}\big{)}.$ (9.14)

The same bound holds for the other integrals. Indeed, similar to (9.1) and
(9.1) we have

$\displaystyle\sum_{j=1}^{J}\frac{1}{2\pi
i}\int_{V_{j}}\frac{x^{s}}{\zeta(s)^{k}}\frac{ds}{s}$
$\displaystyle\ll\sqrt{x}(\log
x)^{\frac{k^{2}}{4}+\varepsilon}\sum_{j=0}^{J-1}\exp\Big{(}\varepsilon\frac{\log
x\log\log(2^{j}x_{1})}{\log(2^{j}x_{1})}\Big{)}$
$\displaystyle\ll\sqrt{x}(\log
x)^{\frac{k^{2}}{4}+1+\varepsilon}\exp\Big{(}\varepsilon\frac{\log x\log\log
x_{1}}{\log x_{1}}\Big{)}\ll\sqrt{x}\exp\big{(}\varepsilon\sqrt{\log
x}\big{)}$ (9.15)

and

$\displaystyle\sum_{j=1}^{J-1}\frac{1}{2\pi
i}\int_{H_{j}}\frac{x^{s}}{\zeta(s)^{k}}\frac{ds}{s}$
$\displaystyle\ll\sqrt{x}\sum_{j=0}^{J-2}(2^{j}x_{1})^{\frac{k}{2}-1+\varepsilon}\exp\Big{(}\varepsilon\frac{\log
x\log\log(2^{j}x_{1})}{\log(2^{j}x_{1})}\Big{)}$
$\displaystyle\ll\sqrt{x}x_{1}^{-1/2+\varepsilon}(\log
x)\exp\Big{(}\varepsilon\frac{\log x\log\log x_{1}}{\log
x_{1}}\Big{)}\ll\sqrt{x}\exp\big{(}\varepsilon\sqrt{\log x}\big{)}.$ (9.16)

And similar to (9.7), (9.1), (9.9), (9.10) and (9.11) we get

$\displaystyle\frac{1}{2\pi
i}\int_{H_{J}}\frac{x^{s}}{\zeta(s)^{k}}\frac{ds}{s}\ll\max_{\frac{1}{2}+\varepsilon\frac{\log\log
x}{\log x}\leq\sigma\leq 1+\frac{1}{\log x}}\exp\Big{(}(\sigma-1)\log
x-k\log|\zeta(\sigma\pm ix)|\Big{)}$ $\displaystyle\ \
\ll\begin{cases}\exp\Big{(}\frac{(k-1)\log x}{2}-\frac{k\log x\log\log\log
x}{\log\log x}+O_{k}\big{(}\frac{\log x}{\log\log x}\big{)}\Big{)}&\text{if
}\frac{1}{2}+\varepsilon\frac{\log\log x}{\log
x}\leq\sigma\leq\frac{1}{2}+o(\frac{1}{\log\log x}),\\\ (\log\log
x)^{k}&\text{if }1-\frac{1}{\log\log x}\leq\sigma\leq 1+\frac{1}{\log x},\\\
x^{\varepsilon}&\text{if }\frac{1}{2}+O(\frac{1}{\log\log x})\leq\sigma\leq
1-\frac{1}{\log\log x},\end{cases}$ $\displaystyle\ \ \ll x^{\varepsilon},$

by Lemma 2.2. This, together with (9.14), (9.2) and (9.2), establishes the
first part of the theorem for $k<1$.

###  9.3. Assuming Conjecture 1.1

We replace the contour in (9.1) by $H_{1}\cup V_{0}$, where

$\displaystyle V_{0}:\quad s=\frac{1}{2}+\alpha_{2}+it,\quad|t|\leq[x],$
$\displaystyle H_{1}:\quad s=\sigma\pm
i[x],\quad\frac{1}{2}+\alpha_{2}\leq\sigma\leq 1+\frac{1}{\log x}$

for some $\frac{1}{\log x}\leq\alpha_{2}\leq 1$ to be chosen later. We
encounter no pole in doing so.

On one hand, by Conjecture 1.1 we get

$\displaystyle\frac{1}{2\pi
i}\int_{V_{0}}\frac{x^{s}}{\zeta(s)^{k}}\frac{ds}{s}$
$\displaystyle\ll\sqrt{x}x^{\alpha_{2}}\Big{(}\frac{1}{\alpha_{2}}\Big{)}^{\frac{k^{2}}{4}}.$
(9.17)

On the other hand, we have

$\frac{1}{2\pi
i}\int_{H_{1}}\frac{x^{s}}{\zeta(s)^{k}}\frac{ds}{s}\ll\max_{\frac{1}{2}+\alpha_{2}\leq\sigma\leq
1+\frac{1}{\log x}}\exp\Big{(}(\sigma-1)\log x-k\log|\zeta(\sigma\pm
i[x])|\Big{)}.$

Like in (9.1) and (9.9), this is

$\displaystyle\ll\exp\bigg{(}(\sigma-1)\log x+\frac{k\log x}{2\log\log
x}\log\frac{1}{(2\sigma-1)\log\log x}+O_{k}\big{(}(2\sigma-1)\log x\big{)}$
$\displaystyle\qquad\qquad\qquad\qquad\qquad\qquad+O_{k}\Big{(}\frac{\log
x}{\log\log x}\Big{)}+O_{k}\Big{(}\frac{\log x}{(\log\log
x)^{2}}\log\frac{1}{2\sigma-1}\Big{)}\bigg{)}$
$\displaystyle\ll\exp\bigg{(}-\frac{\log x}{2}+\frac{k\log
x\log\frac{1}{\alpha_{2}}}{2\log\log x}-\frac{k\log x\log\log\log x}{2\log\log
x}$
$\displaystyle\qquad\qquad\qquad\qquad\qquad\qquad+O_{k}\big{(}\alpha_{2}\log
x\big{)}+O_{k}\Big{(}\frac{\log x}{\log\log x}\Big{)}+O_{k}\Big{(}\frac{\log
x\log\frac{1}{\alpha_{2}}}{(\log\log x)^{2}}\Big{)}\bigg{)}$ (9.18)

for $\frac{1}{2}+\alpha_{2}\leq\sigma\leq\frac{1}{2}+o(\frac{1}{\log\log x})$,
and as in (9.10) and (9.11), it is

$\ll(\log\log x)^{k}+x^{\varepsilon}$

for $\frac{1}{2}+O(\frac{1}{\log\log x})\leq\sigma\leq 1+\frac{1}{\log x}$. In
view of (9.3), we choose

$\alpha_{2}=\begin{cases}\frac{1}{\log x}&\text{if }k\leq 2,\\\ \frac{1}{(\log
x)^{\frac{2}{k}}(\log\log x)^{1-\varepsilon}}&\text{if }k>2,\end{cases}$

and then

$\frac{1}{2\pi
i}\int_{H_{1}}\frac{x^{s}}{\zeta(s)^{k}}\frac{ds}{s}\ll\sqrt{x}.$

Combining with (9.17) we obtain the second part of the theorem.

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