2.1 Notations

We employ some standard notation that will be used throughout the paper.

• • Throughout the paper, the expressions $f(X)=O(g(X))$ , $f(X) \ll g(X)$ , and $g(X) \gg f(X)$ are equivalent to the statement that $|f(X)| \leq C|g(X)|$ for all sufficiently large X, where $C>0$ is an absolute constant. A subscript of the form $\ll _{\alpha }$ means the implied constant may depend on the parameter $\alpha $ . Dependence on several parameters is indicated in an analogous manner, as in $\ll _{\alpha , \lambda }$ .

• • The notation $f = o(g)$ as $x\to a$ means that $\lim _{x\to a} f(x)/g(x) = 0$ and $f\sim g$ as ${x\to a}$ denotes $\lim _{x\to a} f(x)/g(x)= 1$ .

• • We denote the set of real numbers by $\mathbb {R}$ , the set of rational numbers by $\mathbb {Q}$ , the set of integers by $\mathbb {Z}$ , the set of natural numbers by $\mathbb {N}$ , and the set of complex numbers by  $\mathbb {C}$ .

• • For any set $\mathcal {A}$ , $\operatorname {card}(\mathcal {A})$ denotes the cardinality of the set $\mathcal {A}$ .

• • The notation $\operatorname {e}(x)$ stands for $\exp (2\pi i x)$ .

• • Dyadic sums are represented by $\sum _{n \sim N} f(n) = \sum _{N < n \leqslant 2N} f(n)$ .

• • The divisor function is denoted by $\tau (n)$ . The k-fold divisor function $\tau _k(n)$ is defined by the coefficients of the Dirichlet series

  $$\begin{align*}\zeta^k(s) = \sum_{n=1}^{\infty}\frac{\tau_k(n)}{n^s}, \quad \textrm{for } \operatorname{Re}(s)> 1. \end{align*}$$

• • The Möbius function is denoted by $\mu (n)$ , whereas the Popovici-Möbius function will be denoted by $\mu _k(n) = (\mu * \mu *\cdots * \mu )(n)$ , the convolution performed k times.

• • We denote by $\hat {\mu }$ the convolution of the Möbius function with the characteristic function of the squarefull numbers. More precisely, we write

  $$ \begin{align*} \hat{\mu}(n) := (\mu*f)(n), \end{align*} $$
  where f denotes the indicator function for the squarefull numbers.

• • The arithmetic function $\tilde \mu _k(n)$ is defined by the coefficients of the Dirichlet series

  $$\begin{align*}\sum_{n \geqslant 1} \frac{\tilde \mu_k(n)}{n^s} = \frac{1}{\zeta(ks)} \end{align*}$$
  for $\operatorname {Re}(s)>\frac {1}{k}$ .

2.2 Exponential sums and Dirichlet convolutions

We first state the following two lemmas, due to Koukoulopoulos, that provide bounds on exponential sums involving Dirichlet convolutions.

Lemma 2.1 (Type I estimate)

Let $f\colon \mathbb {N}\to \mathbb {C}$ be supported on $[1, y]$ , $v\geqslant 2$ , $X\geqslant 2$ , $\alpha \in \mathbb {R}$ and $a/q$ be a reduced fraction such that $|\alpha -a/q|\leqslant 1/q^2$ . Then,

$$ \begin{align*} \sum_{n\leqslant X}(f*\log^v)(n)\operatorname{e}(n\alpha)\ll \bigg(y + \dfrac{X}{q}+q\bigg)(\log X)^{v+1}\|f\|_{\infty}. \end{align*} $$

Lemma 2.2 (Type II estimate)

Let $f, g\colon \mathbb {N}\to \mathbb {C}$ be two arithmetic functions supported on $[1, y]$ and $[1, z]$ , respectively. Let $X\geqslant 2$ , $\alpha \in \mathbb {R}$ and consider a reduced fraction $a/q$ such that $|\alpha -a/q|\leqslant 1/q^2$ . Then,

$$ \begin{align*} \sum_{n\leqslant X}(f*g)(n)\operatorname{e}(n\alpha)\ll \bigg(q + y +z + \dfrac{yz}{q}\bigg)^{1/2}\sqrt{\log 2q}\cdot \|f\|_2\|g\|_2. \end{align*} $$

We would like to establish more general versions of Lemmas 2.1 and 2.2, in the case when $\alpha \in \mathbb {R}$ and there exists some reduced fraction $a/q$ such that $|\alpha -a/q|\leqslant \gamma /q^2$ , for some $\gamma \geqslant 1$ . In this regard, we first prove the following proposition.

Proposition 2.1 Let $\alpha \in \mathbb {R}$ and consider a reduced fraction $a/q$ such that $|\alpha -a/q|\leqslant \gamma /q^2$ for some $\gamma \geqslant 1$ . Then, for each $m_1 \in \{1,2,\dots ,q\}$ , there exists at most $14 \gamma $ choices of $m_2 \in \{1,2,\dots ,q \}$ such that

(2.1) $$ \begin{align} \bigg \lvert \left\lVert m_1\alpha\right\rVert- \left\lVert m_2\alpha\right\rVert \bigg \rvert < \frac{1}{q}, \end{align} $$

where $\left \lVert x\right \rVert $ denotes the smallest distance of x to an integer.

Proof Fix $m_1 \in \{1,2,\dots , q\}$ and suppose $m_2 \in \{1,2\dots ,q\}$ such that (2.1) is satisfied. Writing $\alpha =a/q+\beta $ with $\lvert \beta \rvert \leqslant \gamma /q^2$ , we have

$$\begin{align*}m_1 \alpha = \frac{m_1a}{q}+m_1\beta \quad \textrm{ and} \quad m_2 \alpha = \frac{m_2a}{q}+m_2\beta. \end{align*}$$

Therefore, we obtain

(2.2) $$ \begin{align} \bigg \lvert \left\lVert m_1\alpha\right\rVert- \left\lVert\frac{m_1 a}{q}\right\rVert \bigg \rvert \leqslant\lvert m_1\beta \rvert \leqslant\frac{\gamma}{q} \quad \mathrm{and} \quad \bigg \lvert \left\lVert m_2\alpha\right\rVert- \left\lVert\frac{m_2 a}{q}\right\rVert \bigg \rvert \leqslant\lvert m_2\beta \rvert \leqslant\frac{\gamma}{q}. \end{align} $$

From (2.1) and (2.2), it follows that

(2.3) $$ \begin{align} \bigg \lvert \left\lVert\frac{m_1 a}{q}\right\rVert- \left\lVert\frac{m_2 a}{q}\right\rVert \bigg \rvert \leqslant \frac{2\gamma+1}{q} \leqslant\frac{3\gamma}{q}. \end{align} $$

Next, we write

$$\begin{align*}\left\lVert\frac{m_1 a}{q}\right\rVert = \frac{\ell_1}{q} \quad \mathrm{and} \quad \left\lVert\frac{m_2 a}{q}\right\rVert = \frac{\ell_2}{q}, \end{align*}$$

where $\ell _1, \ell _2 \in \{0,1,\dots ,\lfloor q/2\rfloor \}$ . Note that when $m_1$ is fixed, $\ell _1$ is fixed. Also, for each choice of $\ell _2$ , there are at most two different choices of $m_2$ . Hence, in order to satisfy (2.3), there are at most $7\gamma $ choices of $\ell _2$ . This implies there are at most $14 \gamma $ choices of $m_2$ which satisfy (2.1).

With Proposition 2.1 in hand, we have the following two lemmas.

Lemma 2.3 Let $f\colon \mathbb {N}\to \mathbb {C}$ be supported on $[1, y]$ , $v\geqslant 2$ , $X\geqslant 2$ , $\alpha \in \mathbb {R}$ and $a/q$ be a reduced fraction such that $|\alpha -a/q|\leqslant \gamma /q^2$ for some $\gamma \geqslant 1$ . Then,

(2.4) $$ \begin{align} \sum_{n\leqslant X}(f*\log^v)(n)\operatorname{e}(n\alpha)\ll \gamma \bigg(y + \dfrac{X}{q}+q\bigg)(\log x)^{v+1}\|f\|_{\infty}, \end{align} $$

where the implied constant in (2.4) is absolute.

Lemma 2.4 Let $f, g\colon \mathbb {N}\to \mathbb {C}$ be two arithmetic functions supported on $[1, y]$ and $[1, z]$ , respectively. Let $X\geqslant 2$ , $\alpha \in \mathbb {R}$ and consider a reduced fraction $a/q$ such that $|\alpha -a/q|\leqslant \gamma /q^2$ for some $\gamma \geqslant 1$ . Then,

(2.5) $$ \begin{align} \sum_{n\leqslant X}(f*g)(n)\operatorname{e}(n\alpha)\ll \gamma \bigg(q + y +z + \dfrac{yz}{q}\bigg)^{1/2}\sqrt{\log 2q}\cdot \|f\|_2\|g\|_2, \end{align} $$

where the implied constant in (2.5) is absolute.

We only prove Lemma 2.3. The proof of Lemma 2.4 is similar.

Proof of Lemma 2.3

Following the proof of Theorem 23.5 in [Reference Koukoulopoulos16], it suffices to bound the sum

$$\begin{align*}\sum_{k \leqslant y} \min \bigg \{\frac{X}{k},\frac{1}{\left\lVert k\alpha\right\rVert} \bigg \}. \end{align*}$$

To do so, we divide the interval $[1,y]$ into subintervals of length q and write

$$ \begin{align*}S_m:=\sum_{m q<k \leqslant(m+1) q} \min \bigg \{\frac{X}{k},\frac{1}{\left\lVert k\alpha\right\rVert} \bigg \}. \end{align*} $$

Note that

$$\begin{align*}\sum_{k \leqslant y} \min \bigg \{\frac{X}{k},\frac{1}{\left\lVert k\alpha\right\rVert} \bigg \} \leqslant\sum_{m=0}^{\lfloor y/q \rfloor} S_m. \end{align*}$$

We write $\alpha =a/q+\beta $ with $\lvert \beta \rvert \leqslant \gamma /q^2$ and consider two cases.

Case 1: $m \geqslant 1$ . Writing $k=mq+r$ where $1 \leqslant r \leqslant q$ , we have $ \left \lVert k\alpha \right \rVert = \left \lVert r\alpha +mq\beta \right \rVert $ . Define the intervals

$$\begin{align*}I _j = \bigg [\frac{j-1}{2q}, \frac{j}{2q}\bigg ), \quad j \in \{1,2,\dots, \lceil q \rceil \}. \end{align*}$$

Fix some $k_1=mq+r_1$ , such that $ \left \lVert k_1\alpha \right \rVert \in I _j$ for some j. Since $I _j$ is an interval of length $1/2q$ , it follows that $\left \lVert r_1\alpha \right \rVert $ is contained in the union of at most two intervals, say $T_1$ and $T_2$ , each of length at most $1/2q$ . Suppose for some $k_2=mq+r_2$ , $ \left \lVert k_2\alpha \right \rVert \in I _j$ as well. Then, $\left \lVert r_2\alpha \right \rVert $ is also contained in $T_1 \cup T_2$ . But using Proposition 2.1, each of $T_1$ or $T_2$ can contain at most $14 \gamma $ choices of $r_2$ . Hence, there can be at most $28 \gamma $ different $k_2$ such that both $\left \lVert k_1 \alpha \right \rVert $ and $\left \lVert k_2 \alpha \right \rVert $ lie in $I _\ell $ . Therefore, the contribution from those k’s for which $\left \lVert k \alpha \right \rVert \in I _j$ to the sum $S_m$ is

$$\begin{align*}\leqslant\frac{60 \gamma q}{j-1}, \end{align*}$$

when $j>1$ . For the specific case when $j=1$ , we trivially bound the contribution by $30\gamma /mq$ . Therefore, for $m \geqslant 1$ , we obtain

$$ \begin{align*}S_m \leqslant 30 \gamma \bigg (\frac{ X}{m q}+\sum_{2 \leqslant j \leqslant \lceil q \rceil } \frac{2 q}{j-1} \bigg) \ll \gamma \left (\frac{ X}{m q}+q \log q \right ). \end{align*} $$

Case 2: $m=0$ . For all $r \geqslant q/2\gamma $ , an argument similar to the previous case shows that the contribution from such terms is $\ll \gamma (X/q+q \log q)$ . Finally, we are left with the cases when $r<q/2\gamma $ . Then, $\lvert r \beta \rvert \leqslant 1/2q$ , which implies $\left \lVert r \alpha \right \rVert \geqslant 1/2q$ . Therefore, the total contribution from the cases when $r < q/2\gamma $ is

$$\begin{align*}\ll \gamma(2q+q \log q) \ll \gamma q \log q. \end{align*}$$

The rest of the proof follows along the lines of the proof of [Reference Koukoulopoulos16, Theorem 23.5], and noting that there are $\ll y/q$ choices of m.

2.3 Exponential sums and the divisor function

We record the following result achieved by Mikawa that provides bounds on exponential sums involving the three-fold divisor function.

Lemma 2.5 Let $\alpha \in \mathbb {R}$ and consider a reduced fraction $a/q$ such that $|\alpha -a/q|\leqslant 1/q^2$ . Then, for $0<M$ , $J\leqslant X $ , we have

$$\begin{align*}\sum_{m\sim M}\sum_{j\sim J}\tau_{3}(j)\min\bigg(\frac{X}{m^{2}j},\frac{1}{\Vert\alpha m^{2}j\Vert}\bigg) \ll MJ(\log X)^{3}+\dfrac{X^{3/4}}{M}\bigg(Xq^{-1}+XM^{-1}+q\bigg)^{1/4}(\log X)^{8},\end{align*}$$

where $\tau _3$ denotes the three-fold divisor function.

2.4 The Möbius function in arithmetic progressions

We collect below a result on the Möbius function in arithmetic progressions, due to Davenport, which is an application of the Siegel-Walfisz theorem with the standard zero-free region for Dirichlet L-functions.

Lemma 2.6 Let $X\geqslant 2$ and $B>0$ . Let $r, q\geqslant 1$ be two integers with $q\leqslant (\log X)^B$ . Then, there exists a constant $c_B>0$ depending only on B such that

$$ \begin{align*} \sum_{\substack{n \leqslant X \\ n \equiv r \operatorname{mod} q}} \mu(n) \ll_B X\exp(-c_B\sqrt{\log X}). \end{align*} $$

2.5 A divisor sum estimate

Let $k \in \mathbb {N}$ and let $\tau _k(n)$ be the k-fold divisor function. Norton established the following result.

Lemma 2.7 For $X \geqslant 1$ , and $k, r \geqslant 2$ , we have

$$ \begin{align*} \sum_{n \leqslant X } (\tau_k(n))^r \ll X(\log X)^{k^r-1}. \end{align*} $$

Proof See [Reference Norton24, Eq. 1.10].

2.6 Bounding $1/\zeta (s)$ in the critical strip.

In order to prove the explicit formula in Section 8, we will require a contour integration involving the term $1/\zeta ^2(s)$ . To prove our results unconditionally, we shall use the following result due to works of Ramachandra and Sankaranarayanan, and Inoue, which provide unconditional upper bounds for $1/\zeta (s)$ inside the critical strip along a specific vertical line segment.

Lemma 2.8 Let $T>0$ be a sufficiently large positive number and $H=T^{1/3}$ . Then, one has

$$ \begin{align*} \min_{T \leqslant t \leqslant T+H} \max_{\frac{1}{2} \leqslant \sigma \leqslant 2} |\zeta(\sigma+it)|^{-1} \leqslant \exp( C (\log \log T)^2) \end{align*} $$

with an absolute constant $C>0$ . In particular, there exists a real $T_* \in [T,T+T^{1/3}]$ such that

$$ \begin{align*} \frac{1}{\zeta(\sigma+iT_*)} \leqslant T_*^\varepsilon \quad \mathrm{where} \quad \frac{1}{2} \leqslant \sigma \leqslant 2, \end{align*} $$

for any $\varepsilon>0$ .

3.1 Proof of Theorem 1.1

We can assume $q \leqslant X$ ; otherwise, the conclusion follows trivially. Let $U, V \geqslant 2$ be parameters to be chosen later on. Applications of the Möbius inversion formula yield

(3.1) $$ \begin{align} \mu(n) &= \mu_{\leqslant U}(n) + \mu_{\leqslant V}(n)+ (\mu_{>U}*\mu_{>V}*1)(n) -(\mu_{\leqslant U}*\mu_{\leqslant V}*1)(n) \notag \\ & = \mu^{(1)} + \mu^{(2)}+\mu^{(3)}-\mu^{(4)}, \end{align} $$

say. Consider $\mu ^{(1)}$ and $\mu ^{(2)}$ first. Since our choice of parameters U and V will be small compared to X, bounding trivially, we arrive at

(3.2) $$ \begin{align} \sum_{n\leqslant X }\mu^{(1)}(n)\operatorname{e}(n\alpha)&\ll U \end{align} $$

(3.3) $$ \begin{align} \sum_{n\leqslant X }\mu^{(2)}(n)\operatorname{e}(n\alpha)& \ll V. \end{align} $$

For $\mu ^{(3)}(n)$ , we will use Lemma 2.2. However, applying the lemma directly does not suffice our purpose. We consider breaking our range into dyadic intervals. By definition, we have

$$ \begin{align*}\mu^{(3)}(n)=\sum_{\substack{k \ell=n \\ k>U, \ell >V}} \mu_{>U}(k) (\mu_{>V} * 1)(\ell). \end{align*} $$

To obtain better control of the support of the variables k and $\ell $ , we break the interval $(U, X/ V]$ into dyadic intervals $\left (2^{j-1}, 2^j\right ]$ , where $2^j \in (U, 2 X / V]$ . If $k \in \left (2^{j-1}, 2^j\right ]$ , then $\ell =n / k \leqslant X / 2^{j-1}$ . With this more accurate decomposition, we have

$$ \begin{align*}\mu^{(3)}(n)=\sum_{U<2^j \leqslant 2 X / V}\left(f_j * g_j\right)(n), \quad \text { for } n \leqslant X , \end{align*} $$

where $f_j(k)=\mu _{>U}(k) \cdot 1_{2^{j-1}<k \leqslant 2^j}$ and $g_j(\ell )=(\mu _{>V} * 1)(\ell ) \cdot 1_{V<\ell \leqslant X/ 2^{j-1}}$ . Then, applying Lemma 2.2, we obtain

(3.4) $$ \begin{align} \sum_{n\leqslant X }\mu^{(3)}(n)\operatorname{e}(n\alpha) &= \sum_{U<2^j \leqslant 2 X / V} \sum_{n\leqslant X }\left(f_j * g_j\right)(n)\operatorname{e}(n\alpha) \notag \\ &\ll \sum_{U<2^j \leqslant 2 X / V} \bigg(q^{1/2} + \dfrac{X^{1/2}}{U^{1/2}} +\dfrac{X^{1/2}}{V^{1/2}} + \dfrac{X^{1/2}}{q^{1/2}}\bigg)\sqrt{\log 2q}\cdot \|f_j\|_2\|g_j\|_2 \notag \\ &\ll \bigg(X^{1/2}q^{1/2} + \dfrac{X}{U^{1/2}} +\dfrac{X}{V^{1/2}}+ \dfrac{X}{q^{1/2}}\bigg)(\log X)^3. \end{align} $$

Note. In the rest of this section, we will always apply the above more accurate decomposition technique before applying Lemma 2.2.

Finally, we focus on $\mu ^{(4)}(n)$ . We apply Lemma 2.1 with $y=UV$ , $v=0$ , and ${f = \mu _{\leqslant U}*\mu _{\leqslant V}}$ . Since $\|f\|_{\infty } \ll _{\varepsilon } (UV)^{\varepsilon } $ for any fixed $\varepsilon>0$ , it follows that

(3.5) $$ \begin{align} \sum_{n\leqslant X }\mu^{(4)}(n)\operatorname{e}(n\alpha) \ll_{\varepsilon} (UV)^{\varepsilon}\bigg( UV+ \dfrac{X}{q} + q\bigg)(\log X). \end{align} $$

We choose $U=V=\min \{ X^{2/5}, q, X/q \}$ and combine the estimates (3.2), (3.3), (3.4), and (3.5) to obtain

$$ \begin{align*} \sum_{n\leqslant X }\mu(n)\operatorname{e}(n\alpha) \ll_{\varepsilon} X^{4/5+\varepsilon}+ \dfrac{X(\log X)^3}{q^{1/2}} + X^{1/2}q^{1/2}(\log X)^3. \end{align*} $$

This ends the proof.

3.2 Proof of Theorem 1.3

The proof follows along the lines of the proof of Theorem 1.1. Again, we can assume $q \leqslant X$ . Let $U, V \geqslant 2$ be parameters to be chosen later on. Employing the Möbius inversion formula, we obtain

$$ \begin{align*} \hat{\mu}(n)=\hat{\mu}_{ \leqslant U}(n) + (\hat{\mu}_{> U}*\mu_{>V}*1)(n) + (f*\mu_{\leqslant V})(n) - (\hat{\mu}_{ \leqslant U}*\mu_{\leqslant V}*1)(n). \end{align*} $$

Trivially, we have

(3.6) $$ \begin{align} \sum_{n\leqslant X }\hat{\mu}_{ \leqslant U}(n)\operatorname{e}(n\alpha) \ll U \log U. \end{align} $$

Next, by Lemma 2.2, similar to how we obtained (3.4), we have

(3.7) $$ \begin{align} \sum_{n\leqslant X } (\hat{\mu}_{> U}*\mu_{>V}*1)(n)\operatorname{e}(n\alpha) &\ll X^{1/2}\bigg(\dfrac{X}{U}+\dfrac{X}{V}+q + \dfrac{X}{q} \bigg)^{1/2} (\log X)^{5/2}\notag \\ &\ll \bigg(X^{1/2}q^{1/2}+\dfrac{X}{U^{1/2}}+\dfrac{X}{V^{1/2}} + \dfrac{X}{q^{1/2}}\bigg)(\log X)^{5/2}. \end{align} $$

Observing that $\|f \cdot 1_{n\leqslant X} \|_2 \ll X^{1/4}$ , again, an application of Lemma 2.2 shows that

(3.8) $$ \begin{align} \sum_{n\leqslant X }(f*\mu_{\leqslant V})(n)\operatorname{e}(n\alpha) &\ll \bigg(q + X + V + \dfrac{X}{q}\bigg)^{1/2}X^{1/4}V^{1/4}(\log X)^{3/2}\notag \\ &\ll \bigg(X^{1/4}V^{1/4}q^{1/2}+X^{3/4}V^{1/4}\bigg)(\log X)^{3/2}. \end{align} $$

Arguing as in the proof of (3.5) for the case of the Möbius function, we have

(3.9) $$ \begin{align} \sum_{n\leqslant X }(\hat{\mu}_{ \leqslant U}*\mu_{\leqslant V}*1)(n)\operatorname{e}(n\alpha) \ll_{\varepsilon} (UV)^{\varepsilon}\bigg( UV+ \dfrac{X}{q} + q\bigg)(\log X), \end{align} $$

for any fixed $\varepsilon>0$ . Finally, we choose $U=V=\min \{ X^{1/3}, q, X/q \}$ and combine (3.6), (3.7), (3.8), and (3.9) to obtain

$$ \begin{align*} \sum_{n\leqslant X }\hat{\mu}(n)\operatorname{e}(n\alpha) \ll \bigg(X^{5/6} + q^{1/2}X^{1/2} + \dfrac{X}{q^{1/2}}\bigg)(\log X)^{5/2}, \end{align*} $$

which is the desired result.

3.3 Proof of Theorem 1.4

Similar to the previous proofs, we can assume $q \leqslant X$ . Let $U, V \geqslant 2$ be parameters to be chosen later on. We start with the following identity:

$$ \begin{align*} \tilde{\mu}_2(n) &= \tilde{\mu}_2(n)\cdot 1_{n\leqslant U} + (\tilde{\mu}_2\cdot1_{n>U}*\mu_{>V}*1)(n) \notag \\ &\quad + (\mu^2*\mu_{\leqslant V})(n)-(\tilde{\mu}_2 \cdot 1_{n \leqslant U}*\mu_{\leqslant V}*1)(n). \end{align*} $$

Arguing as in the previous proofs, we can show that

(3.10) $$ \begin{align} \sum_{n\leqslant X }\tilde{\mu}_2(n)\cdot 1_{n \leqslant U} \operatorname{e}(n\alpha) &\ll U \log U, \end{align} $$

(3.11) $$ \begin{align} \sum_{n\leqslant X }(\tilde{\mu}_2\cdot1_{n>U}*\mu_{ >V}*1)(n)\operatorname{e}(n\alpha) &\ll \bigg(X^{1/2}q^{1/2} + \dfrac{X}{U^{1/2}}+\dfrac{X}{V^{1/2}} +\dfrac{X}{q^{1/2}}\bigg) (\log X)^3, \end{align} $$

(3.12) $$ \begin{align} \textrm{and} \quad \sum_{n\leqslant X }(\mu_{ \leqslant V}*\tilde{\mu}_2 \cdot 1_{n \leqslant U}*1)(n) \operatorname{e}(n\alpha) &\ll_{\varepsilon} (UV)^{\varepsilon}\bigg( UV+ \dfrac{X}{q} + q\bigg)(\log X), \end{align} $$

for any fixed $\epsilon>0$ . It remains to estimate the sum

$$ \begin{align*} \sum_{n\leqslant X }(\mu^2*\mu_{\leqslant V})(n)\operatorname{e}(n\alpha). \end{align*} $$

A treatment like (3.8) is not enough in this case because $\|\mu ^2 \cdot 1_{n\leqslant X} \|_2 \ll X^{1/2}$ . Therefore, we proceed in a different way. Using the relation $\mu ^2(n)=\sum _{ab^2=n}\mu (b)$ , we can write

(3.13) $$ \begin{align} \sum_{n\leqslant X }(\mu^2*\mu_{\leqslant V})(n)\operatorname{e}(n\alpha) &=\sum_{\substack{mb^2\leqslant X \\ m\leqslant V}}\mu(b)\mu(m)\sum_{a\leqslant X /(mb^2)}\operatorname{e}(amb^2\alpha) \notag \\ &\ll \sum_{b\leqslant X ^{1/2}}\sum_{m\leqslant V}\min\bigg\{\dfrac{X}{b^2m}, \dfrac{1}{\|b^2m\alpha\|}\bigg\}. \end{align} $$

We need to break the range of b in (3.13) into two parts: for large b’s, we will use the bound from Lemma 2.5, and for small b’s, we will use the Cauchy-Schwarz inequality along with Lemma 2.1. Indeed, we write

$$ \begin{align*} \sum_{b\leqslant X ^{1/2}}\sum_{m\leqslant V}\min\bigg\{\dfrac{X}{b^2m}, \dfrac{1}{\|b^2m\alpha\|}\bigg\} = \bigg(\sum_{b\leqslant B}\sum_{m\leqslant V} + \sum_{B<b\leqslant X ^{1/2}}\sum_{m\leqslant V} \bigg) \min\bigg\{\dfrac{X}{b^2m}, \dfrac{1}{\|b^2m\alpha\|}\bigg\}. \end{align*} $$

Then, by applying the Cauchy-Schwarz inequality,

(3.14) $$ \begin{align} \sum_{b\leqslant B}\sum_{m\leqslant V}\min\bigg\{\dfrac{X}{b^2m}, \dfrac{1}{\|b^2m\alpha\|}\bigg\} &\leqslant\sum_{m\leqslant B^2V}\tau_3(m)\min\bigg\{\dfrac{X}{m}, \dfrac{1}{\|m\alpha\|}\bigg\} \notag \\ &\ll X^{1/2}\bigg(B^2V + q + \dfrac{X}{q}\bigg)^{1/2} (\log X)^6 \notag\\ &\ll \bigg(X^{1/2}BV^{1/2} + X^{1/2}q^{1/2} + \dfrac{X}{q^{1/2}}\bigg)(\log X)^6. \end{align} $$

Next, by Lemma 2.5 dyadically, we have

(3.15) $$ \begin{align} \!\sum_{B<b\leqslant X ^{1/2}}\sum_{m\leqslant V}\min\bigg\{\dfrac{X}{b^2m}, \dfrac{1}{\|b^2m\alpha\|}\bigg\} \kern1.3pt{\ll}\kern1.3pt \bigg(X^{1/2}V + \dfrac{X}{q^{1/4}B} + \dfrac{X}{B^{5/4}} + \dfrac{X^{3/4}q^{1/4}}{B}\bigg)(\log X)^{10}. \end{align} $$

Choose $B=(X/V)^{2/9}$ and combining (3.14) and (3.15), we obtain

(3.16) $$ \begin{align} \sum_{b\leqslant X ^{1/2}}\sum_{m\leqslant V}\min\bigg\{\dfrac{X}{b^2m}, \dfrac{1}{\|b^2m\alpha\|}\bigg\}\ll \bigg(X^{13/18}V^{5/18} + X^{1/2}V + \dfrac{X}{q^{1/4}} + X^{3/4}q^{1/4}\bigg)(\log X)^{10}. \end{align} $$

Finally, we choose $U=V=\min \{ X^{5/14}, q, X/q \}$ and combine (3.10), (3.11), (3.12), and (3.16) to obtain

$$ \begin{align*} \sum_{n\leqslant X }\tilde{\mu}_2(n)\operatorname{e}(n\alpha)\ll \bigg(X^{23/28} + \dfrac{X}{q^{1/4}} + X^{3/4}q^{1/4}\bigg)(\log X)^{10}, \end{align*} $$

as it was to be shown.

4.1 Initial step

Let

$$ \begin{align*} S_{\mu_2}(X, \alpha):=\sum_{n\leqslant X }(\mu*\mu)(n)\operatorname{e}(n\alpha). \end{align*} $$

Then, for some parameters $2\leqslant M , N\leqslant X $ to be chosen later, we can write

(4.2) $$ \begin{align} S_{\mu_2}(X, \alpha)=&\: \sum_{\substack{mn\leqslant X \\ m> M, n>N}}\mu(m)\mu(n)\operatorname{e}(mn\alpha) + \sum_{m\leqslant M }\mu(m)\sum_{n\leqslant X /m}\mu(n)\operatorname{e}(mn\alpha)\notag \\ \notag &+\sum_{n\leqslant N}\mu(n)\sum_{m\leqslant X /n}\mu(m)\operatorname{e}(mn\alpha)-\sum_{\substack{m\leqslant M\\ n\leqslant N }}\mu(m)\mu(n)\operatorname{e}(mn\alpha)\\ =&\: S_{\mu_2}^{(1)}(X, \alpha)+S_{\mu_2}^{(2)}(X, \alpha) + S_{\mu_2}^{(3)}(X, \alpha)-S_{\mu_2}^{(4)}(X, \alpha) , \end{align} $$

say. We now turn our attention to each of these terms.

4.2 Estimation of $S_{\mu _2}^{(1)}(X, \alpha )$ and $S_{\mu _2}^{(4)}(X, \alpha )$ .

Applying Lemma 2.2 after the dyadic decomposition following the proof of Theorem 1.1, we deduce that

(4.3) $$ \begin{align} S_{\mu_2}^{(1)}(X, \alpha)&\ll X^{1/2}\bigg(q+\dfrac{X}{M} + \dfrac{X}{N} + \dfrac{X}{q}\bigg)^{1/2}(\log X)^2\notag\\ &\ll \bigg(X^{1/2}q^{1/2}+\dfrac{X}{M^{1/2}} + \dfrac{X}{N^{1/2}} + \dfrac{X}{q^{1/2}}\bigg)(\log X)^2. \end{align} $$

Trivially, we have

(4.4) $$ \begin{align} S_{\mu_2}^{(4)}(X, \alpha) \ll MN. \end{align} $$

4.3 Estimation of $S_{\mu _2}^{(2)}(X, \alpha )$ and $S_{\mu _2}^{(3)}(X, \alpha )$ .

Fix $\varepsilon>0$ . Writing $\alpha = a/q + \beta $ , one has

$$\begin{align*}m\alpha = \dfrac{ma}{q} + m\beta = \dfrac{ma/(m, q)}{q/(m, q)} + m\beta \quad \text{and}\quad |m\beta|\leqslant\dfrac{m/(m, q)^2}{q^2/(m, q)^2} \leqslant\dfrac{m/(m, q)}{q^2/(m, q)^2}.\end{align*}$$

Therefore, using Lemma (4.1), we obtain

(4.5) $$ \begin{align} S_{\mu_2}^{(2)}(X, \alpha) & \ll_{\varepsilon} \sum_{m\leqslant M}\dfrac{m}{(m, q)}\bigg\{\bigg(\dfrac{X}{m}\bigg)^{4/5+\varepsilon} + \dfrac{X(m, q)^{1/2}(\log X)^3}{mq^{1/2}} + \bigg(\dfrac{X}{m}\bigg)^{1/2}\dfrac{q^{1/2}(\log X)^3}{(m, q)^{1/2}}\bigg \}\notag \\ &\ll_{\varepsilon} X^{4/5+\varepsilon}M^{6/5-\varepsilon} + \dfrac{XM(\log X)^3}{q^{1/2}} + X^{1/2}q^{1/2}M^{3/2}(\log X)^3. \end{align} $$

Similarly, we can show that

(4.6) $$ \begin{align} S_{\mu_2}^{(3)}(X, \alpha)\ll_{\varepsilon} X^{4/5+\varepsilon}N^{6/5-\varepsilon} + \dfrac{XN(\log X)^3}{q^{1/2}} + X^{1/2}q^{1/2}N^{3/2}(\log X)^3. \end{align} $$

4.4 Parameter optimization

Substituting the estimates (4.3), (4.5), (4.6), and (4.4) in (4.2) and choosing $M=N$ , we deduce that

(4.7) $$ \begin{align} S_{\mu_2}(X, \alpha) \ll_{\varepsilon} X^{4/5+\varepsilon}M^{6/5-\varepsilon}+\bigg(\dfrac{X}{M^{1/2}} + X^{1/2}q^{1/2} + \dfrac{XM}{q^{1/2}} + X^{1/2}q^{1/2}M^{3/2}\bigg)(\log X)^3. \end{align} $$

We are now left with the crucial step: to choose our parameter M optimally. Since we will choose $M \geqslant 1$ , we may omit the term $X^{1/2}q^{1/2}$ in (4.7). Let $M=y^5$ , $A= \sqrt {q}$ , $B = X^{1/5}$ , and $C = \sqrt {{X}/{q}}$ . To optimally choose M, it suffices to minimize the expression

(4.8) $$ \begin{align} H(y):= \max \bigg \{ \dfrac{1}{y^{5/2}}, \dfrac{y^5}{A}, \dfrac{y^6}{B}, \dfrac{y^{15/2}}{C} \bigg \}. \end{align} $$

Note that as a function of y, only $1/y^{5/2}$ is decreasing. Hence, the minimum value in (4.8) is attained when $1/y^{5/2}$ is equal to one of the other three functions. We consider the following cases.

Case 1. Suppose we have

(4.9) $$ \begin{align} \dfrac{y^5}{A} \geqslant \max \bigg \{ \dfrac{y^6}{B}, \dfrac{y^{15/2}}{C} \bigg \}. \end{align} $$

Then the minimum value in (4.8) is attained when

$$\begin{align*}\frac{1}{y^{5/2}} = \frac{y^5}{A}, \end{align*}$$

which implies that $y = A^{2/15}$ . With this particular choice of y, we see that

$$\begin{align*}\dfrac{y^5}{A} \geqslant \dfrac{y^6}{B} \end{align*}$$

is equivalent to $A^{-1/3} \geqslant A^{4/5}B^{-1}$ which occurs when $1 \leqslant q \leqslant X^{6/17}$ . However, the inequality

$$\begin{align*}\dfrac{y^5}{A} \geqslant \dfrac{y^{15/2}}{C} \end{align*}$$

is equivalent to $A^{-1/3} \geqslant AC^{-1}$ which occurs when $1 \leqslant q \leqslant X^{6/14}$ . Combining the two ranges, we see that (4.9) holds when $1 \leqslant q \leqslant X^{6/17}$ . In this range, we have ${H(y) =A^{-1/3} = q^{-1/6}}$ , and thus,

(4.10) $$ \begin{align} S_{\mu_2}(X, \alpha) \ll \dfrac{X(\log X)^3}{q^{1/6}}. \end{align} $$

Case 2. Now let’s assume the scenario

(4.11) $$ \begin{align} \dfrac{y^6}{B} \geqslant \max \bigg \{ \dfrac{y^5}{A} , \dfrac{y^{15/2}}{C} \bigg \}. \end{align} $$

Then the minimum value in (4.8) is attained when

$$\begin{align*}\frac{1}{y^{5/2}} = \frac{y^6}{B}, \end{align*}$$

which implies that $y = B^{2/17}$ . Therefore, the inequality

$$\begin{align*}\dfrac{y^6}{B} \geqslant \dfrac{y^5}{A} \end{align*}$$

holds with $y=B^{2/17}$ when $A \geqslant B^{15/17}$ , which implies that $q \geqslant X^{6/17}$ . However, the inequality

$$\begin{align*}\dfrac{y^6}{B} \geqslant \dfrac{y^{15/2}}{C} \end{align*}$$

holds with $y=B^{2/17}$ when $B \geqslant C^{17/20}$ , which implies that $q \leqslant X^{9/17}$ . Putting together the two ranges, we see that (4.11) holds when $X^{6/17}\leqslant q \leqslant X^{9/17}$ . In this range, ${H(y)=B^{-5/17} =X^{-1/17}}$ , and we have

(4.12) $$ \begin{align} S_{\mu_2}(X, \alpha) \ll_{\varepsilon} X^{16/17+\varepsilon}. \end{align} $$

Case 3. Arguments similar to Cases 1 and 2 show that when $X^{9/17}\leqslant q \leqslant X$ , then the minimum value in (4.8) is $H(y) = C^{-1/4}=q^{1/8}X^{-1/8} $ . Therefore in this range, we obtain

(4.13) $$ \begin{align} S_{\mu_2}(X, \alpha) \ll X^{7/8}q^{1/8}(\log X)^3. \end{align} $$

Combining the estimates (4.10), (4.12), and (4.13), we conclude that

$$\begin{align*}\sum_{n\leqslant X }(\mu*\mu)(n)\operatorname{e}(n\alpha) \ll_{\varepsilon} X^{16/17+\varepsilon}+\bigg(\dfrac{X}{q^{1/6}} + X^{7/8}q^{1/8}\bigg)(\log X)^{3}. \end{align*}$$

This completes the proof for the case when $k=2$ .

4.5 Induction step

Fix $\varepsilon>0$ . Assume that for $X \geqslant 2$ ,

$$ \begin{align*} \sum_{n\leqslant X }\mu_k(n)\operatorname{e}(n\alpha) \ll_{\varepsilon} X^{a_k+\varepsilon}+ \dfrac{X(\log X)^{k^2}}{q^{b_k}} + X^{c_k}q^{1-c_k}(\log X)^{k^2}, \end{align*} $$

for some $k=\ell \geqslant 2$ , and we want to prove our desired result when $k=\ell +1$ . Similar to Lemma 4.1, one can show that if $\alpha \in \mathbb {R}$ such that $|\alpha -a/q|\leqslant \gamma /q^2$ for some $(a, q)=1$ and $\gamma \geqslant 1$ , then

(4.14) $$ \begin{align} \sum_{n\leqslant X }\mu_\ell(n)\operatorname{e}(n\alpha) \ll_{\varepsilon} \gamma \bigg(X^{a_{\ell}+\varepsilon}+ \dfrac{X(\log X)^{\ell^2}}{q^{b_{\ell}}} + X^{c_{\ell}}q^{1-c_{\ell}}(\log X)^{\ell^2} \bigg), \end{align} $$

where the implied constant in (4.14) depends at most on $\varepsilon $ . Equipped with (4.14), for some parameters $2\leqslant M, N\leqslant X$ to be chosen later, we can write

(4.15) $$ \begin{align} S_{\mu_{\ell+1}}(X, \alpha)=&\: \sum_{\substack{mn\leqslant X \\ m> M, n>N}}\mu(m)\mu_\ell (n)\operatorname{e}(mn\alpha) + \sum_{m\leqslant M}\mu(m)\sum_{n\leqslant X/m}\mu_\ell (n)\operatorname{e}(mn\alpha)\notag \\ \notag &+\sum_{n\leqslant N}\mu_\ell (n)\sum_{m\leqslant X/n}\mu(m)\operatorname{e}(mn\alpha)-\sum_{\substack{m\leqslant M\\ n\leqslant N}}\mu(m)\mu _\ell (n)\operatorname{e}(mn\alpha)\\ =&\: S_{\mu _{\ell+1} }^{(1)}(X, \alpha)+S_{\mu _{\ell+1} }^{(2)}(X, \alpha) + S_{\mu _{\ell+1} }^{(3)}(X, \alpha)-S_{\mu _{\ell+1} }^{(4)}(X, \alpha). \end{align} $$

Again, applying Lemma 2.2, we obtain

(4.16) $$ \begin{align} S_{\mu _{\ell+1} }^{(1)}(X, \alpha) \ll \bigg(X^{1/2}q^{1/2}+\dfrac{X}{M^{1/2}} + \dfrac{X}{N^{1/2}} + \dfrac{X}{q^{1/2}}\bigg)(\log X)^{(\ell+1)^2}. \end{align} $$

Trivially, we have

(4.17) $$ \begin{align} S_{\mu _{\ell+1}}^{(4)}(X, \alpha) \ll MN (\log N)^{\ell-1}. \end{align} $$

Choosing $\gamma = m/(m,q)$ and applying (4.14), we arrive at

(4.18) $$ \begin{align} S_{\mu _{\ell+1} }^{(2)}(X, \alpha) \ll_{\varepsilon} X^{a_\ell+\varepsilon}M^{2-a_\ell-\varepsilon}+ \dfrac{XM(\log X)^{(\ell+1)^2}}{q^{b_\ell}} + X^{c _\ell}q^{1-c _\ell}M^{2-c_\ell}(\log X)^{(\ell+1)^2}. \end{align} $$

Note that Lemma 2.7 shows that

$$\begin{align*}\sum_{n \leqslant N } \tau_\ell^2(n) \ll N(\log N)^{\ell^2-1}.\end{align*}$$

Hence, applying Lemma (4.1) and Cauchy-Schwartz inequality in conjunction, we can write

(4.19) $$ \begin{align} S_{\mu _{\ell+1} }^{(3)}(X, \alpha) &\ll_{\varepsilon} \sum_{n \leqslant N } \tau_\ell(n) \cdot \dfrac{n}{(n, q)}\bigg\{\bigg(\dfrac{X}{n}\bigg)^{4/5+\varepsilon} + \dfrac{X(n, q)^{1/2}(\log X)^3}{nq^{1/2}} + \bigg(\dfrac{X}{n}\bigg)^{1/2}\dfrac{q^{1/2}(\log X)^3}{(n, q)^{1/2}}\bigg\} \notag \\ &\ll_{\varepsilon}X^{4/5+\varepsilon}N^{6/5-\varepsilon}+ \dfrac{XN(\log X)^{(\ell+1)^2}}{q^{1/2}} + X^{1/2}q^{1/2}N^{3/2}(\log X)^{(\ell+1)^2}. \end{align} $$

Substituting the estimates (4.16), (4.17), (4.18), and (4.19) in (4.15) and choosing ${M=N}$ , we have

$$ \begin{align*} S_{\mu _{\ell+1} }(X, \alpha) \ll_{\varepsilon} X^{a _\ell+\varepsilon}M^{2-a _\ell-\varepsilon}+ \bigg(\dfrac{X}{M^{1/2}} +\dfrac{XM}{q^{b _\ell}} + X^{c _\ell}q^{1-c _\ell}M^{2-c _\ell} + X^{1/2}q^{1/2}M^{3/2}\bigg)(\log X)^{(\ell+1)^2}. \end{align*} $$

An optimization procedure similar to that for the case $k=2$ then shows that

$$ \begin{align*} S_{\mu _{\ell+1} }(X, \alpha) \ll_{\varepsilon} X^{a _{\ell+1}+\varepsilon}+\dfrac{X(\log X)^{(\ell+1)^2}}{q^{b _{\ell+1}}} + X^{c _{\ell+1}}q^{1-c _{\ell+1}}(\log X)^{(\ell+1)^2}, \end{align*} $$

thereby completing the proof.

Initial setup

Recall that

(8.1) $$ \begin{align} \Phi_{\mu * \mu}(\rho \operatorname{e}(\theta)) = \sum_{j=1}^\infty \sum_{n=1}^\infty \frac{(\mu * \mu)(n)}{j} \exp\bigg( -jn \bigg(\frac{1}{X}-2\pi i \theta\bigg)\bigg), \end{align} $$

where $\rho =\exp (-1/X)$ . We first apply the Cahen-Mellin formula and rewrite (8.1) to obtain

$$ \begin{align*} \Phi_{\mu * \mu}(\rho \operatorname{e}(\theta)) = \sum_{j=1}^\infty \sum_{n=1}^\infty \frac{(\mu * \mu)(n)}{j} \frac{1}{2 \pi i} \int_{(c)} \Gamma(s) j^{-s} n^{-s} \bigg(\frac{X}{1-2\pi i X \theta}\bigg)^s ds, \end{align*} $$

for $c>0$ . Interchanging the summations and the integration, for any real $c>1$ , we have

(8.2) $$ \begin{align} \Phi_{\mu * \mu}(\rho \operatorname{e}(\theta)) &= \frac{1}{2 \pi i} \int_{(c)} \Gamma(s) \bigg(\sum_{j=1}^\infty \frac{1}{j^{s+1}}\bigg) \bigg(\sum_{n=1}^\infty \frac{(\mu * \mu)(n)}{n^s} \bigg) \bigg(\frac{X}{1-2\pi i X \theta}\bigg)^s ds \nonumber \\ &= \frac{1}{2 \pi i} \int_{(c)} \Gamma(s) \zeta(s+1) \frac{1}{\zeta^2(s)} \bigg(\frac{X}{1-2\pi i X \theta}\bigg)^s ds. \end{align} $$

Contour shifting.

The path of integration in 8.2 is a vertical line with $\operatorname {Re}(s)=c>1$ . We move the portion $|t| \leqslant T$ of this path to the left, replacing it by a rectangular path joining the points $c \pm i T$ and $-u \pm i T$ , where $u=2N+1$ , for some large $N \in \mathbb {N}$ . Thus, the new path of integration is of the form $L=\bigcup _{i=1}^5 L_i$ , with $L_1=(c-i \infty , c-i T], L_2=[c-i T, -u-i T], L_3=[-u-i T, -u+i T], L_4=[-u+i T, c+i T]$ and $ L_5=[c+i T, c+i \infty )$ . With this change of the path of integration, an application of the Cauchy residue theorem shows that

$$\begin{align*}\Phi_{\mu * \mu}(\rho \operatorname{e}(\theta)) = M+\frac{1}{2 \pi i} \sum_{j=1}^5 I_j, \end{align*}$$

where M denotes the contribution of the residues at singularities of the integrand in the region enclosed by the two paths and $I_j$ denotes the integral over the path $L_j$ . Our method will require an analysis of the $I_j$ ’s. Equipped with Lemma 2.8, we will establish our results unconditionally. For similar techniques, the reader is referred to the works of Bartz [Reference Bartz1], Ramachandra and Sankaranarayanan [Reference Ramachandra and Sankaranarayanan25], and Inoue [Reference Inoue14].

Residue computations

Set

$$\begin{align*}y:=\frac{X}{1-2\pi i X\theta}. \end{align*}$$

A singularity analysis regarding the integrand in (8.2) shows the following. We have a double pole at $s=0$ coming from $\Gamma (s)\zeta (s+1)$ for which

$$ \begin{align*} \mathop{\operatorname{res}}\limits_{s=0} \Gamma(s) \zeta(s+1) \frac{1}{\zeta^2(s)} y^s = 4 \log y - 8 \log(2 \pi). \end{align*} $$

The simple pole at $s=-1$ yields

$$ \begin{align*} \mathop{\operatorname{res}}\limits_{s=-1} \Gamma(s) \zeta(s+1) \frac{1}{\zeta^2(s)} y^s = - \frac{\zeta(0)}{\zeta^2(-1)y} = \frac{72}{y}. \end{align*} $$

Unlike previous cases in the literature [Reference Berndt, Malik and Zaharescu2, Reference Berndt, Robles, Zaharescu and Zeindler3, Reference Das, Robles, Zaharescu and Zeindler8, Reference Dunn and Robles10, Reference Gafni11, Reference Gafni12], we now capture residues from the trivial and nontrivial zeros of the Riemann zeta-function. We start with the nontrivial zeros. Those take place at $s = \varpi $ and yield double poles for which

$$ \begin{align*} \mathop{\operatorname{res}}\limits_{s=\varpi} \Gamma(s) \zeta(s+1) \frac{1}{\zeta^2(s)} y^s &= y^{\varpi} \Gamma(\varpi) \frac{\zeta(1+\varpi)}{\zeta'(\varpi)^2}\bigg(\log y + \psi(\varpi) - \frac{\zeta"}{\zeta'}(\varpi) + \frac{\zeta'}{\zeta}(1+\varpi) \bigg), \end{align*} $$

where we have used the simplicity of the zeros to alleviate the notation. For the trivial zeros, we have triple poles at $s=-2n$ , for $n=1,2,3,\dots $ , coming from $\Gamma (s)/\zeta ^2(s)$ . These residues can be written as

$$ \begin{align*} \mathop{\operatorname{res}}\limits_{s=-2n} \Gamma(s) \zeta(s+1) \frac{1}{\zeta^2(s)} y^s = y^{-2 n}(\mathfrak{c}_1(n)(\log y)^2 + \mathfrak{c}_2(n)\log y + \mathfrak{c}_3(n)), \end{align*} $$

where the coefficients are given by

(8.3) $$ \begin{align} \mathfrak{c}_1(n) &:= -\frac{B_{2 n}}{8 n^2 \Gamma (2 n) \zeta '(-2 n)^2},\end{align} $$

(8.4) $$ \begin{align} \mathfrak{c}_2(n) &:= \frac{B_{2 n} \zeta "(-2 n)+\zeta '(-2 n) \left(2 n \zeta '(1-2 n)-B_{2 n} \psi (2n+1)\right)}{2 n (2 n)! \zeta '(-2 n)^3}, \end{align} $$

and

(8.5) $$ \begin{align} \mathfrak{c}_3(n) &:= \frac{1}{48 n^2 \Gamma (2 n) \zeta '(-2 n)^4} \nonumber \\&\quad \times \big[ B_{2 n} \big\{-9 \zeta "(-2 n)^2-2 \left(3 \psi(2 n+1)^2-3 \psi'(2 n+1)+\pi ^2\right) \zeta '(-2 n)^2 \nonumber \\ &\quad \quad +4 \zeta '(-2 n) (\zeta "'(-2 n)+3 \psi(2 n+1) \zeta "(-2 n))\big\} \nonumber \\ &\quad \quad +12 n \zeta '(-2 n) \left(\zeta '(-2 n) \zeta "(1-2 n)+2 \zeta '(1-2 n) \left(\psi(2 n+1) \zeta '(-2 n)-\zeta "(-2 n)\right)\right) \kern-1pt\big].\nonumber\\ \end{align} $$

Here, $\psi (x) = \Gamma '/\Gamma (x)$ denotes the digamma function and $B_n$ denotes the Bernoulli numbers.

Estimating the vertical integral $I_3$

Using Stirling’s formula, for $\operatorname {Re}(s) \leqslant -3/2$ ,

(8.6) $$ \begin{align} \Gamma(s)\zeta(s+1)\zeta(s) \ll (2 \pi)^{2\operatorname{Re}(s)}\lvert s \rvert^{-\operatorname{Re}(s)-1/2} e^{-\pi \lvert t \rvert/2}. \end{align} $$

Therefore, we have

(8.7) $$ \begin{align} I_3 = \int_{-u-iT}^{-u+iT} \Gamma(s)\zeta(s+1)\frac{1}{\zeta^2(s)}y^s \,ds &\ll \int_{-u-iT}^{-u+iT} \frac{(2 \pi)^{-2u}\lvert s \rvert^{u-1/2} e^{-\pi \lvert t \rvert/2}}{\zeta^3(s)}y^s \,ds \notag \\ &\ll \frac{1}{(2\pi)^{2u}} \int_{-u-iT}^{-u+iT} \frac{(u^2+\lvert t \rvert^2 )^{u/2-1/4} e^{-\pi \lvert t \rvert/2}}{\zeta^3(s)}y^s \,ds. \end{align} $$

We apply the functional equation for $\zeta (s)$ in (8.7) alongside the observation that $\zeta ^{-1}(1-s) \ll 1$ . Also, following [Reference Vaughan33] (See Lemma 2.8),

$$\begin{align*}\lvert y \rvert^s \leqslant(X\Delta)^{-u}e^{-\lvert t \rvert(\Delta-\pi/2)}.\end{align*}$$

Hence, we obtain

(8.8) $$ \begin{align} I_3 &\ll \frac{2^{3u} \pi^{3(u-1)}}{(2\pi)^{2u} (X \Delta)^{u}} \int_{-u-iT}^{-u+iT} \frac{(u^2+\lvert t \rvert^2 )^{u/2-1/4} e^{-\lvert t \rvert \Delta}}{\sin^3(\pi s/2)\Gamma^3(1-s)} \,ds. \end{align} $$

Next, we note that in our range of integration,

$$ \begin{align*} \frac{1}{\Gamma(1-s)} \ll e^{(u+1) - (u+\frac{1}{2})\log (u+1) + \frac{1}{2}\pi |t|} \quad \mathrm{and} \quad \frac{1}{\sin(\pi s/2)} \ll e^{- \frac{1}{2}\pi |t|}. \end{align*} $$

Inserting these two bounds in (8.8), we arrive at

(8.9) $$ \begin{align} I_3 &\ll \frac{(2\pi)^{u}e^{3(u+1) - 3(u+\frac{1}{2})\log (u+1)}}{ (X \Delta)^{u}} \int_{-T}^{T} (u^2+\lvert t \rvert^2 )^{u/2-1/4} e^{-\lvert t \rvert \Delta} \,dt, \notag \\ &\ll \frac{(2\pi)^{u}e^{3(u+1) - 3(u+\frac{1}{2})\log (u+1)}}{ (X \Delta)^{u}} \int_{0}^{T} (u^2+t^2 )^{u/2-1/4} e^{-t \Delta} \,dt. \end{align} $$

We consider the inside integral in (8.9). We divide this into two cases. When $t \leqslant u$ , we have

(8.10) $$ \begin{align} \int_{0}^{T} (u^2+t^2 )^{u/2-1/4} e^{-t \Delta} \,dt \ll 2^{u/2}u^{u-1/2} \int_{0}^{\infty} e^{-t \Delta} \,dt \ll 2^{u/2}u^{u-1/2} \Delta^{-1}. \end{align} $$

However, when $t>u$ ,

(8.11) $$ \begin{align} \int_{0}^{T} (u^2+t^2 )^{u/2-1/4} e^{-t \Delta} \,dt \ll 2^{u/2} \int_{0}^{\infty} t^{u-1/2} e^{-t \Delta} \,dt \ll 2^{u/2}\Gamma(u+1/2)\Delta^{-u-1/2}. \end{align} $$

Therefore, applying (8.10) and (8.11) in (8.9) and using Stirling’s approximation for $\Gamma (u+1/2)$ , we obtain

(8.12) $$ \begin{align} I_3 &\ll \frac{(2\sqrt{2}\pi)^{u}e^{3(u+1) - 3(u+\frac{1}{2})\log (u+1)}}{ (X \Delta)^{u}} (u^{u-1/2} \Delta^{-1}+ \Gamma(u+1/2)\Delta^{-u-1/2})\notag \\ &\ll \bigg (\frac{2\sqrt{2}\pi e^2}{X\Delta^2} \bigg)^{u}( u^{-2u-3/2} \Delta^{-1/2}). \end{align} $$

Estimating the horizontal integrals $I_2$ and $I_4$ .

We consider $I_4$ only. The treatment for $I_2$ is similar. Dividing the contour into two parts, we write

$$\begin{align*}I_4 = \int_{-u+iT}^{-3/2+iT}+ \int_{-3/2+iT}^{c+iT} = I_{4,1}+I_{4,2}, \end{align*}$$

say. We work with $I_{4,1}$ first. Applying (8.6), we have

(8.13) $$ \begin{align} I_{4,1} = \int_{-u+iT}^{-3/2+iT} \Gamma(s) \zeta(s+1) \frac{1}{\zeta^2(s)} y^s \,ds &\ll \int_{-u+iT}^{-3/2+iT} \frac{(2 \pi)^{2\sigma}\lvert s \rvert^{-\sigma-1/2} e^{-\pi T/2}}{\zeta^3(s)}y^s \,ds \notag \\ &\ll e^{-\pi T/2} \int_{-u+iT}^{-3/2+iT} \frac{(2\pi)^{2\sigma}(\sigma^2+T^2 )^{-\sigma/2-1/4}}{ \zeta^3(s)}y^s \,ds. \end{align} $$

Proceeding as in the case of $I_3$ , we apply the functional equation for $\zeta (s)$ in (8.13) to obtain

$$ \begin{align*} I_{4,1} &\ll e^{-T\Delta} \int_{-u+iT}^{-3/2+iT} \frac{(2\pi)^{2\sigma} (X\Delta)^\sigma (\sigma^2+T^2 )^{-\sigma/2-1/4}}{2^{3\sigma} \pi^{3(\sigma-1)}\sin^3(\pi s/2)\Gamma^3(1-s)} \,ds. \end{align*} $$

In our range of integration,

$$ \begin{align*} \frac{1}{\Gamma(1-s)} \ll e^{1-\sigma - (\frac{1}{2}-\sigma)\log (1-\sigma) + \frac{1}{2}\pi T} \quad \mathrm{as\ well\ as} \quad \frac{1}{\sin(\pi s/2)} \ll e^{- \frac{1}{2}\pi T}. \end{align*} $$

Therefore, we arrive at

(8.14) $$ \begin{align} I_{4,1} &\ll e^{-T\Delta} \int_{-u}^{-3/2} (2\pi)^{-\sigma}(X\Delta)^\sigma (\sigma^2+T^2 )^{-\sigma/2-1/4}e^{3(1-\sigma) - 3(\frac{1}{2}-\sigma)\log (1-\sigma)} \,d\sigma \notag \\ &\ll e^{-T\Delta} \int_{3/2}^{u} \bigg (\frac{2\pi e^3}{X\Delta} \bigg)^{\sigma} \frac{(\sigma^2+T^2 )^{\sigma/2-1/4}}{ (\sigma+1)^{3(\sigma+\frac{1}{2})}} \,d\sigma. \end{align} $$

We write (8.14) as

$$\begin{align*}e^{-T\Delta}\int_{3/2}^{u} \bigg (\frac{2\pi e^3}{X\Delta} \bigg)^{\sigma} \frac{(\sigma^2+T^2 )^{\sigma/2-1/4}}{ (\sigma+1)^{3(\sigma+\frac{1}{2})}} \,d\sigma = e^{-T\Delta}\int_{3/2}^{T}+ e^{-T\Delta}\int_{T}^{u} = J_1 +J_2, \end{align*}$$

say. Consider $J_1$ first. We have

$$\begin{align*}J_1 \ll e^{-T\Delta}\int_{3/2}^{T} \bigg (\frac{2\sqrt{2}\pi e^3 T}{\sigma^3} \bigg)^{\sigma} \,d\sigma. \end{align*}$$

Let f be defined as

$$\begin{align*}f(\sigma) = \bigg (\frac{2\sqrt{2}\pi e^3 T}{\sigma^3} \bigg)^{\sigma}, \end{align*}$$

and suppose $g(\sigma ) = \log f(\sigma )$ . The maximum of $g(\sigma )$ is attained at $\sigma _{0} = (2\sqrt {2}\pi T )^{1/3}$ . Hence, the maximum value attained by f in our range of integration is $\exp (3(2\sqrt {2}\pi T )^{1/3})$ . Therefore, we have

(8.15) $$ \begin{align} J_1 \ll e^{-T\Delta} e^{3(2\sqrt{2}\pi T )^{1/3}} T \ll e^{-T\Delta/2}. \end{align} $$

For $J_2$ , we obtain

(8.16) $$ \begin{align} e^{-T\Delta}\int_{T}^{u} \bigg (\frac{2\pi e^3}{X\Delta} \bigg)^{\sigma} \frac{(\sigma^2+T^2 )^{\sigma/2-1/4}}{ (\sigma+1)^{3(\sigma+\frac{1}{2})}} \,d\sigma \ll e^{-T\Delta}\int_{T}^{u} \bigg (\frac{2\sqrt{2}\pi e^3}{\sigma^2} \bigg)^{\sigma} \,d\sigma \ll e^{-T\Delta/2}. \end{align} $$

Now, we consider $I_{4,2}$ , which is given by

$$ \begin{align*} I_{4,2} = \int_{-3/2+iT}^{c+iT} \Gamma(s) \zeta(s+1) \frac{1}{\zeta^2(s)} y^s \,ds. \end{align*} $$

In our range of integration,

(8.17) $$ \begin{align} \Gamma(s) \ll T^{\sigma-1/2}e^{-\pi T/2}, \quad \textrm{and} \quad \zeta(s+1) \ll T^\eta, \end{align} $$

for some fixed constant $\eta>0$ . We write

$$\begin{align*}I_{4,2} = \bigg(\int_{-3/2+iT}^{-1+iT} +\int_{-1+iT}^{c+iT}\bigg) \Gamma(s) \zeta(s+1) \frac{1}{\zeta^2(s)} y^s \,ds. \end{align*}$$

For the first part, we apply the functional equation for $\zeta (s)$ to obtain

(8.18) $$ \begin{align} \int_{-3/2+iT}^{-1+iT} \Gamma(s) \zeta(s+1) \frac{1}{\zeta^2(s)} y^s \,ds &\ll \frac{T^{\eta-1/2}}{e^{T\Delta}}\int_{-3/2}^{-1} \bigg(\frac{X\Delta T}{2\pi e}\bigg)^{\sigma}\frac{1}{(1-\sigma)^{1/2-\sigma}} \,d\sigma \notag\\ &\ll \frac{T^{\eta-1/2}}{e^{ T\Delta}}\int_{1}^{3/2} \bigg(\frac{2\pi e}{X\Delta T}\bigg)^{\sigma}\frac{1}{(1+\sigma)^{1/2+\sigma}} \,d\sigma \ll e^{- T\Delta /2}. \end{align} $$

For the second part, using Lemma 2.8, for $-1 \leqslant \operatorname {Re}(s)=\sigma \leqslant 2$ and any $\epsilon>0$ , for sufficiently large $V>0$ , there exists some $V \leqslant T_{*} \leqslant 2V$ such that

(8.19) $$ \begin{align} \frac{1}{\zeta^2(\sigma+iT_{*})}\ll T_{*}^{\epsilon}. \end{align} $$

Therefore, using the bounds (8.17) and (8.19) and choosing $T=T_{*}$ , we arrive at

(8.20) $$ \begin{align} \int_{-1+iT}^{c+iT} \Gamma(s) \zeta(s+1) \frac{1}{\zeta^2(s)} y^s \,ds &\ll \frac{T^{-1/2+\eta+\epsilon}}{e^{T\Delta}}\int_{-1}^{c} (X\Delta T)^{\sigma} \,d\sigma \ll e^{- T\Delta /2}. \end{align} $$

Finally, combining the estimates (8.15), (8.16), (8.18), and (8.20), we conclude that ${I_4 \ll e^{- T\Delta /2}.}$

Estimating the vertical integrals $I_1$ and $I_5$ .

Writing

$$\begin{align*}I_1 = \int_{c+iT}^{c+i\infty} \Gamma(s) \zeta(s+1) \frac{1}{\zeta^2(s)} y^s \,ds, \end{align*}$$

we note that in our range of integration,

$$ \begin{align*} \Gamma(s) &\ll e^{-\pi T/2}, \quad \zeta(s+1) \ll 1, \quad \zeta^{-2}(s) \ll 1, \quad \textrm{and} \quad \lvert y \rvert^s \leqslant(X\Delta)^{\sigma}e^{-T(\Delta-\pi/2)}. \end{align*} $$

Combining these estimates, we obtain $I_1 \ll e^{- T\Delta /2}.$ A similar argument holds true for  $I_5$ .

Final steps

Combining the estimates from the integrals and letting $N \to \infty $ , we obtain

(8.21) $$ \begin{align} \Phi_{\mu * \mu}(\rho \operatorname{e}(\theta)) &= 4 \log \frac{X}{1-2\pi i X\theta} - 8 \log(2 \pi) + 72 \frac{1-2\pi i X\theta}{X}\nonumber \\ & \quad +\sum_{|\operatorname{Im}(\varpi)| < T_\nu} f(X,\theta,\varpi) +\sum_{n=1}^{\infty} g(X,\theta,n) +O(e^{- T_\nu\Delta /2}), \end{align} $$

where $\nu \leqslant T_\nu \leqslant 2\nu$ , $\nu \in \mathbb {N}$ , the sum over $\varpi $ runs along all the nontrivial zeros of $\zeta $ , and $f,g$ are as defined in the statement of the theorem. Finally, we let $ \nu \to \infty $ and choose our sequence of $T_{\nu }$ appropriately such that (8.19) is satisfied. We conclude that

$$ \begin{align*} \Phi_{\mu * \mu}(\rho \operatorname{e}(\theta)) &= 4 \log \frac{X}{1-2\pi i X\theta} - 8 \log(2 \pi) + 72 \frac{1-2\pi i X\theta}{X}\nonumber \\ & \quad + \lim_{\nu \to \infty} \sum_{|\operatorname{Im}(\varpi)| < T_\nu} f(X,\theta,\varpi) +\sum_{n=1}^{\infty} g(X,\theta,n), \end{align*} $$

thereby finishing the last step of the proof.