Proof For convenience, we extend the sequence $(\eta _j)$ to all integers indices by defining $\eta _0=1$ and $\eta _j=0$ when $1\le j\le t-1$ or $j\le -1$ . Assume throughout that $\sigma>1$ . Then

(2.1) $$ \begin{align} \zeta(ts)^{-\eta_t}A(s) = \prod_{p} \bigg( \sum_{j\in\mathbb{Z}} \frac{\eta_j}{p^{js}} \bigg) \bigg( 1-\frac{1}{p^{ts}} \bigg)^{\eta_t}. \end{align} $$

The lemma is trivial if $\eta _t=0$ ; we consider two cases according to the sign of $\eta _t$ .

If $\eta _t>0$ , then equation (2.1) becomes

$$ \begin{align*} \zeta(ts)^{-\eta_t}A(s) &= \prod_{p} \bigg( \sum_{j\in\mathbb{Z}} \frac{\eta_j}{p^{js}} \!\bigg) \bigg( \sum_{m=0}^{\eta_t} \frac{(-1)^m \binom{\eta_t}{m}}{p^{m ts}} \bigg) = \prod_{p} \bigg( \sum_{k\in Z} \frac1{p^{ks}} \sum_{m=0}^{\eta_t} (-1)^m \binom{\eta_t}{m} \eta_{k-tm} \!\bigg). \end{align*} $$

It follows that

(2.2) $$ \begin{align} A(s) = \zeta(ts)^{\eta_t} \cdot \prod_{p} \bigg( 1 + \sum_{j\in\mathbb{Z}} \frac{\eta_j'}{p^{js}} \bigg) \quad\text{with}\quad \eta_j'= \sum_{m=0}^{\eta_t} (-1)^m \binom{\eta_t}{m} \eta_{j-tm}. \end{align} $$

In particular,

$$ \begin{align*} \eta_j'= \begin{cases} \eta_j=0, &\text{if } j\le -1 \text{ or } 1\leq j\leq t-1, \\ \eta_0=1, &\text{if } j=0, \\ \eta_t-\binom{\eta_t}1\eta_0 = 0, &\text{if } j=t, \\ \eta_j-\binom{\eta_t}1\eta_{j-t} = \eta_j, &\text{if } t+1\leq j \leq 2t-1, \\ \eta_{2t}-\binom{\eta_t}1\eta_t+\binom{\eta_t}{2}\eta_0 = \eta_{2t}-\frac12(\eta_t^2+\eta_t), &\text{if } j=2t. \end{cases} \end{align*} $$

From equation (2.2), we also deduce that

$$ \begin{align*}|\eta_j'|\leq \sum_{m=0}^{\eta_t} \binom{\eta_t}{m} |\eta_{j-tm}|\leq \max_{n\leq j} |\eta_n| \sum_{m=0}^{\eta_t} \binom{\eta_t}{m}\leq 2^{\eta_t} \max_{n\leq j} |\eta_n|, \end{align*} $$

which establishes the lemma in the $\eta _t>0$ case.

On the other hand, if $\eta _t<0$ , then equation (2.1) becomes

$$ \begin{align*} \zeta(ts)^{-\eta_t}A(s) &= \prod_{p} \bigg( \sum_{j\in\mathbb{Z}} \frac{\eta_j}{p^{js}} \bigg) \bigg( \sum_{m=0}^{\infty} \frac{\binom{m-\eta_t-1}{-\eta_t-1}}{p^{m ts}} \bigg) \\ &= \prod_{p} \bigg( \sum_{k\in Z} \frac1{p^{ks}} \sum_{m=0}^{\infty} \binom{m-\eta_t-1}{-\eta_t-1} \eta_{k-tm} \bigg). \end{align*} $$

and it follows that

(2.3) $$ \begin{align} A(s) = \zeta(ts)^{\eta_t} \cdot \prod_{p} \bigg( 1 + \sum_{j\in\mathbb{Z}} \frac{\eta_j'}{p^{js}} \bigg) \quad\text{with}\quad \eta_j'= \sum_{m=0}^{\infty} \binom{m-\eta_t-1}{-\eta_t-1} \eta_{j-tm} \end{align} $$

holds for all $j\geq 0$ . In particular, we have $\eta _0'=1$ , and

$$ \begin{align*} \eta_j'= \begin{cases} \eta_j=0, &\text{if } j\le -1 \text{ or } 1\leq j\leq t-1, \\ \eta_0=1, &\text{if } j=0, \\ \eta_t+\binom{-\eta_t}1\eta_0 = 0, &\text{if } j=t, \\ \eta_j+\binom{-\eta_t}1\eta_{j-t} = \eta_j, &\text{if } t+1\leq j \leq 2t-1, \\ \eta_{2t}+\binom{-\eta_t}1\eta_t+\binom{1-\eta_t}{2}\eta_0 = \eta_{2t}-\frac12(\eta_t^2+\eta_t), &\text{if } j=2t. \end{cases} \end{align*} $$

From equation (2.3), we also deduce that

$$ \begin{align*} |\eta_j'| \leq \sum_{m=0}^{\lfloor j/t \rfloor} \binom{m-\eta_t-1}{-\eta_t-1} |\eta_{j-tm}| &\leq \max_{n\leq j} |\eta_n| \sum_{m=0}^{\lfloor j/t \rfloor} \binom{m+|\eta_t|-1}{|\eta_t|-1} \\ &= \max_{n\leq j} |\eta_n| \cdot \frac1{|\eta_t|} \bigg\lfloor \frac{j}{t}+1 \bigg\rfloor \binom{\lfloor j/t \rfloor+|\eta_t|}{|\eta_t|-1} \\ &\leq \max_{n\leq j} |\eta_n| \bigg( \frac{j}{t}+1 \bigg) \bigg( \frac{j}{t}+|\eta_t|\bigg)^{|\eta_t|} \leq \max_{n\leq j} |\eta_n| (2j|\eta_t|)^{2|\eta_t|}, \end{align*} $$

which establishes the lemma in the $\eta _t<0$ case.

Proof To show absolute convergence, we must bound

$$ \begin{align*} \sum_{p} \sum_{j=n+1}^{\infty} \frac{|\eta_{j}|}{|p^{js}|} \le \sum_{p} \sum_{j=n+1}^{\infty} \frac{(Aj)^B}{p^{j\sigma}} &= A^B \sum_p \frac1{p^{(n+1)\sigma}} \sum_{i=0}^\infty \frac{(i+n+1)^B}{p^{i\sigma}} \\ &\le A^B \sum_p \frac1{p^{(n+1)\sigma}} \sum_{i=0}^\infty \frac{(n+1)^B(i+1)\cdots(i+B)}{p^{i\sigma}} \\ &= A^B \sum_p \frac{(n+1)^B}{p^{(n+1)\sigma}} B!(1-p^{-\sigma})^{-B-1} \ll_{A,B,n} \sum_p \frac1{p^{(n+1)\sigma}} \end{align*} $$

which converges by the assumption $\sigma>\frac 1{n+1}$ . Moreover, this convergence is locally uniform in s, which implies that the infinite product is indeed analytic.

Proof of Proposition 2.3

We begin by setting $\theta _j^{(k)}=\varepsilon _j$ for all j and writing

(2.6) $$ \begin{align} U_k(s) = \prod_p \bigg( 1 + \sum_{j=k}^\infty \frac{\theta_j^{(k)}}{p^{js}} \bigg) = D_f(s). \end{align} $$

We claim that for each $t=k,k+1,\dots ,2k+1$ , we can write

$$\begin{align*}D_f(s) = U_t(s) \prod_{j=k}^{t-1} \zeta(js)^{\varepsilon_j} \quad\text{with}\quad U_t(s) = \prod_p \bigg( 1 + \sum_{j=t}^\infty \frac{\theta_j^{(t)}}{p^{js}} \bigg), \end{align*}$$

where $\theta _j^{(t)} = \varepsilon _j$ for all $t\le j\le 2k$ and $|\theta _j^{(t)}| \le (2j)^{2(t-k)}$ for all $j\in \mathbb {N}$ . The base case $t=k$ is exactly equation (2.6), whereas deriving the case $t+1$ from the case t is a direct application of Lemma 2.1. (In the first step going from $t=k$ to $t=k+1$ , it is important to note that $\varepsilon _k=-1$ implies that $\varepsilon _{2k} - \frac 12(\varepsilon _k^2+\varepsilon _k)=\varepsilon _{2k}$ . The fact that $\varepsilon _k=-1$ also confirms that the critical index of f equals k.) At the end of this recursive process, the final case $t=2k+1$ is the statement of the proposition, with $\eta _j = \theta _j^{(2k+1)}$ .

Proof of Theorem 2.6

We first show that Algorithm 1 terminates in finitely many steps; equivalently, we show that the critical index of the sequence $(\varepsilon _j)$ is finite. If there is a positive integer j such that $\varepsilon _{j}=-1$ , then the second if statement of the algorithm ensures that the critical index of $(\varepsilon _j)$ is at most j. Thus, we can assume that $\varepsilon _j \in \{0,1\}$ for all j. Let k be the initial index of f.

• • If $\varepsilon _j=0$ for all indices j that are not multiples of k, then let $k'$ be the smallest multiple of k such that $\varepsilon _{k'}\neq 1$ (such a $k'$ must exist by the exclusion of trivial fake $\mu $ ’s from the family ${\mathcal F}$ ). Then, the second if statement of the algorithm ensures that the critical index of $(\varepsilon _j)$ is at most $k'$ .

• • Otherwise, let $k'$ be the smallest integer with $\varepsilon _{j}=1$ that is not a multiple of k. Then, the first if statement of the algorithm sets $c_2=k'$ . Since $\operatorname {lcm}(k,k')$ has at least two representations from $\{k,k'\} = \{c_1,c_2\}$ , the second if statement of the algorithm ensures that the critical index of $(\varepsilon _j)$ is at most $\operatorname {lcm}(k,k')$ .

Next, we make the initial definitions $D_0(s) =D_f(s)$ and $\theta _j^{(0)}= \varepsilon _j$ for all j.

Main goal: We will show inductively, for each $1\le m \leq M$ , that $c_m$ is the smallest positive integer j such that $\theta _j^{(m-1)}$ is nonzero, and moreover that $\theta _{c_m}^{(m-1)}=1$ . These claims are trivial for $m=1$ , as $c_1$ is the initial index of f, and $\varepsilon _{c_1}=1$ since f is not of Möbius-type.

For the inductive step, fix $1\leq m \leq M-1$ , and assume both that $c_m$ is the smallest positive integer j such that $\theta _j^{(m-1)}$ is nonzero and that $\theta _{c_m}^{(m-1)}=1$ . We write ${\theta _0^{(m-1)}=1}$ , and for convenience, we adopt the convention that $\theta _j^{(m-1)}=0$ when $j\le -1$ . Set $D_m(s) = D_{m-1}(s)/\zeta (c_ms)$ , and let $\theta _j^{(m)}$ be defined by

$$ \begin{align*}D_m(s) = \prod_{p} \bigg( \sum_{j=0}^{\infty} \frac{\theta_j^{(m)}}{p^{js}} \bigg) \end{align*} $$

for $\sigma>1$ . By the definition of $D_m(s),$ we have $\theta _j^{(m)}= \theta _j^{(m-1)}-\theta _{j-c_m}^{(m-1)}$ , and then by induction, it is easy to show that

(2.9) $$ \begin{align} \theta_j^{(m)}= \sum_{I \subset \{1,\ldots,m\}} (-1)^{\#I} \theta^{(0)}_{j-\sum_{i \in I}c_i} = \sum_{I \subset \{1,\ldots,m\}} (-1)^{\#I} \varepsilon_{j-\sum_{i \in I}c_i}. \end{align} $$

We define three parameters $r_m,s_m,t_m$ as follows. Let $r_m$ be the smallest positive integer with $\varepsilon _{r_m}\neq 0$ such that $r_m$ has no representations from $\{c_1,c_2,\ldots , c_m\}$ . Let $s_m$ be the smallest positive integer with $\varepsilon _{s_m}\neq 1$ that has exactly one representation from $\{c_1,c_2,\ldots , c_m\}$ . Finally, let $t_m$ be the smallest positive integer that has at least two representations from $\{c_1,c_2,\ldots , c_m\}$ . If any of these numbers $r_m,s_m,t_m$ does not exist, we regard it as $+\infty $ .

Now set $c_{m+1}= \min \{r_m,s_m,t_m\}$ ; we claim that $c_{m+1}$ is finite. This is easy to see for $m \geq 2$ , since, in this case, $c_1c_2$ has at least two representations from $\{c_1,c_2,\ldots , c_m\}$ and thus $c_{m+1}\leq t_m\leq c_1c_2$ . It remains to consider the case $m=1$ , in which every nonnegative integer automatically has at most one representation from $\{c_1\}$ . But $r_1=s_1=+\infty $ would mean that $\varepsilon _j=1$ if j is a multiple of $c_1$ and $\varepsilon _j=0$ otherwise; however, this sequence results in a trivial fake $\mu $ which has been ruled out in the definition of the family ${\mathcal F}$ .

Subgoal: Next, we consider the range $1\leq j\leq c_{m+1}$ and determine the values $\theta _{j}^{(m)}$ in this range. We will show that $\theta _{j}^{(m)}=0$ when $j<c_{m+1}$ , and also that $\theta _{c_{m+1}}^{(m)}=\varepsilon _{c_{m+1}} - n_{c_{m+1}} \ne 0$ . We will need to consider three different cases depending on which of $r_m,s_m,t_m$ is smallest. Note that $c_{m+1}\le r_m$ and $c_{m+1}\le s_m$ and $c_{m+1}\le t_m$ by definition, so we will not need to consider values of j above any of these parameters.

1. (a) Assume that j has no representations from $\{c_1,c_2,\ldots , c_m\}$ . For each subset I of $\{1,\dots ,m\}$ , it follows that $j-\sum _{i \in I}c_i$ has no representations from $\{c_1,c_2,\ldots , c_m\}$ either and thus $\varepsilon _{j-\sum _{i \in I}c_i}=0$ by the definition of $r_m$ . Equation (2.9) then implies that $\theta _{j}^{(m)}= \varepsilon _{j}$ . By the definition of $r_m$ , if $j<r_m$ then $\theta _{j}^{(m)}= \varepsilon _{j}=0$ , while if $j=r_m$ then $j=c_{m+1}$ and $\theta _{c_{m+1}}^{(m)}= \varepsilon _{c_{m+1}} \in \{-1,1\}$ .

2. (b) Next, assume that $j= \sum _{i=1}^m \alpha _i c_i$ has exactly one representation from $\{c_1,c_2,\ldots , c_m\}$ . Let $X= \{1\leq i \leq m\colon \alpha _i>0\}$ . Then for each subset I of X, it follows that $j-\sum _{i \in J} c_i$ also has exactly one representation from $\{c_1,c_2,\ldots , c_m\}$ and thus $\varepsilon _{j-\sum _{i \in J} c_i}=1$ by the definition of $s_m$ . On the other hand, if I is a subset of $\{1,\dots ,m\}$ such that $I \not \subset X$ , then $j-\sum _{i \in I} c_i$ has no representations from $\{c_1,c_2,\ldots , c_m\}$ , and thus $\varepsilon _{j-\sum _{i \in I}c_i}=0$ by the definition of $r_m$ . Equation (2.9) and the binomial theorem then imply that

   $$ \begin{align*}\theta_{j}^{(m)}= \sum_{I \subset X} (-1)^{\#I} \varepsilon_{j-\sum_{i \in I}c_i}= \varepsilon_{j}+\sum_{b=1}^{\#X} (-1)^b \binom{\#X}{b}= \varepsilon_{j}-1. \end{align*} $$

   By the definition of $s_m$ , if $j<s_m$ then $\theta _{j}^{(m)}= \varepsilon _{j}-1=0$ , while if $j=s_m$ then $j=c_{m+1}$ and $\theta _{c_{m+1}}^{(m)}= \varepsilon _{c_{m+1}}-1 \in \{-2, -1\}$ .

3. (c) Finally, assume that j has $n_j\ge 2$ representations from $\{c_1,c_2,\ldots , c_m\}$ ; by the definition of $t_m$ , we must have $j=t_m=c_{m+1}$ . Write these representations as $j=\sum _{i=1}^m \alpha ^{(h)}_i c_i$ for $1\leq h \leq n_j$ . Let $X_h= \{1\leq i \leq m\colon \alpha _i^{(h)}>0\}$ ; we claim that $X_h$ are pairwise disjoint. Indeed, if we had $\alpha _i^{(h_1)}>0$ and $\alpha _i^{(h_2)}>0$ for some $1\le i\le m$ and $1\le h_1<h_2\le n_j$ , then $j-c_i$ would have at least two representations from $\{c_1,c_2,\ldots , c_m\}$ , violating the definition of $t_m$ .

   Let I be a nonempty subset of $\{1,\dots ,m\}$ . Then, $j-\sum _{i \in I} c_i$ has at most one representation from $\{c_1,c_2,\ldots , c_m\}$ by the definition of $t_m$ . If $I \subset X_{h}$ for some $1\leq h \leq n_j$ , then $j-\sum _{i \in I} c_i>0$ does have a representation from $\{c_1,c_2,\ldots , c_m\}$ , and thus $\varepsilon _{j-\sum _{i \in I} c_i}=1$ by the definition of $s_m$ and the assumption that $t_m<s_m$ . On the other hand, if $I \not \subset X_{h}$ for every $1\leq h \leq n_j$ , then $j-\sum _{i \in I} c_i$ has no representations from $\{c_1,c_2,\ldots , c_m\}$ (for any such representation would induce an additional representation of j itself). Thus, it follows that $\varepsilon _{j-\sum _{i \in I} c_i}=0$ by the definition of $r_m$ and the assumption that $t_m<r_m$ . Equation (2.9) and the binomial theorem then imply that

   $$ \begin{align*}\theta_{t_m}^{(m)}= \varepsilon_{t_m}+\sum_{h=1}^{n_j} \sum_{\substack{I \subset X_{h}\\ I \neq \emptyset}} (-1)^{\#I} \varepsilon_{t_m-\sum_{i \in I}c_i}= \varepsilon_{t_m}+\sum_{h=1}^{n_j} \sum_{b=1}^{\#X_{h}} (-1)^b \binom{\#X_{h}}{b}= \varepsilon_{t_m}-n_j\leq -1. \end{align*} $$

We have therefore achieved our subgoal of showing that $\theta _{j}^{(m)}=0$ for $j<c_{m+1}$ and that $\theta ^{(m)}_{c_{m+1}}= \varepsilon _{c_{m+1}}-n_{c_{m+1}}\ne 0$ . We now consider the sign of $\theta ^{(m)}_{c_{m+1}}$ :

• • The only way that $\theta ^{(m)}_{c_{m+1}}>0$ is when $\varepsilon _{c_{m+1}}=1$ and $n_{c_{m+1}}=0$ , and thus $\theta ^{(m)}_{c_{m+1}}=1$ . Therefore, we have finished the proof for the induction step for $m+1$ . At this point, the first if statement of the algorithm appends $c_{m+1}$ to the list of principal indices, increases m by $1$ , and repeats the while loop.

• • On the other hand, $\theta ^{(m)}_{c_{m+1}}<0$ means that $\varepsilon _{c_{m+1}}<n_{c_{m+1}}$ . In this event, the second if statement sets $\ell =c_{m+1}$ and $m=M$ and terminates the algorithm.

These observations complete the verification of our main goal.

Note that for $\sigma>1$ ,

(2.10) $$ \begin{align} D_M(s) = \frac{D_f(s)}{\prod_{j=1}^{M} \zeta(c_j s)}= \prod_{p} \bigg( \sum_{j=0}^{\infty} \frac{\theta_j^{(M)}}{p^{js}} \bigg). \end{align} $$

We have shown that $\theta _j^{(M)}=0$ for $1\leq j<\ell $ and that $\theta _\ell ^{(M)}= \varepsilon _\ell -n_\ell <0.$ Also, note that equation (2.9) implies that $|\theta _j^{(M)}|\leq 2^M$ for all j. In particular,

$$ \begin{align*}\bigg| \theta_{2\ell}^{(M)}-\frac{(\theta_{\ell}^{(M)})^2+\theta_{\ell}^{(M)}}{2} \bigg|\leq 2^M+\frac{4^M+2^M}{2}< 2^{2M+1}. \end{align*} $$

Now, we can apply Lemma 2.1 inductively $\ell +1$ times to $D_M(s)$ to obtain

(2.11) $$ \begin{align} D_M(s) = \prod_{j= \ell}^{2\ell-1} \zeta(js)^{\theta_j^{(M)}} \cdot \zeta(2\ell s)^{\theta_{2\ell}^{(M)}-\frac12((\theta_{\ell}^{(M)})^2+\theta_{\ell}^{(M)})} U_{2\ell}(s), \end{align} $$

where

$$ \begin{align*}U_{2\ell}(s) = \prod_{p} \bigg(1+\sum_{j=2\ell+1}^{\infty} \frac{\eta_j}{p^{js}} \bigg) \end{align*} $$

with $|\eta _j|\leq 2^M (2j \cdot 2^{2M+1})^{4^{M+1} (\ell +1)}$ . (Note that this bound can be made to depend upon $\ell $ alone since $M<\ell $ .) Combining equations (2.9)–(2.11) establishes equation (2.7), which completes the proof of the theorem.

Proof We follow the notation used in Algorithm 1. We have $c_1=1$ and thus $n_j \geq 1$ for all j since $j=1+1+\cdots +1$ . Since $\varepsilon _2= \cdots = \varepsilon _{k-1}=1$ and $\varepsilon _k<1=n_k$ , we have $\ell =k$ and $M=1$ . Also, note that $\varepsilon _k \in \{-1,0\}$ implies that $-\frac 12((\varepsilon _k-1)^2+\varepsilon _k-1)= \varepsilon _k = -|\varepsilon _k|$ . The conclusion thus follows immediately from Theorem 2.6.

Proof Since $h(s)/s$ is also meromorphic with a pole of order $\xi $ at $\sigma _0$ , we can write

$$ \begin{align*} \frac{h(s)}s = \sum_{k=-\xi}^\infty c_k(s-\sigma_0)^k \quad\text{and}\quad x^s = x^{\sigma_0} e^{(s-\sigma_0)\log x} = x^{\sigma_0} \sum_{k=0}^\infty \frac{(s-\sigma_0)^k}{k!}(\log x)^k \end{align*} $$

for some constants $c_k$ with $c_{-\xi }\ne 0$ . When we multiply these series together, the coefficient of $(s-\sigma _0)^{-1}$ depends on the first $\xi $ terms in each series; more precisely, the residue we seek equals

$$\begin{align*}x^{\sigma_0} \sum_{j=0}^{\xi-1} \frac{(\log x)^j}{j!} c_{-1-j} = P(\log x)x^{\sigma_0} \quad\text{where}\quad P(t) = \sum_{j=0}^{\xi-1} \frac{c_{-1-j}}{j!} t^j. \end{align*}$$

Notation 3.3 Let $f\in {\mathcal F}$ be defined via the sequence $(\varepsilon _j)$ . Let $\ell $ be the critical index of f. By Proposition 2.3 and Theorem 2.6, there are integers $a_1, a_2, \ldots , a_{2\ell }$ and a function $U_{2\ell }(s)$ such that the following conditions hold:

• • We have $a_j\geq 0$ for $1\leq j\le \ell -1$ , while $a_\ell <0$ ;

• • For $\sigma>1$ , we can write

  (3.1) $$ \begin{align} D_f(s) =U_{2\ell}(s) \cdot \prod_{j=1}^{2\ell} \zeta(js)^{a_j} \quad{where }\quad U_{2\ell}(s) = \prod_{p} \bigg(1+\sum_{j=2\ell+1}^{\infty} \frac{\eta_j}{p^{js}} \bigg); \end{align} $$

• • There are positive constants $A,B$ , depending only on $\ell $ , such that $|\eta _j|\leq (Aj)^B$ for all j; in particular, $U_{2\ell }(s)$ is analytic for $\sigma>\frac 1{2\ell +1}$ by Lemma 2.2, and is nonzero provided each of its factors is nonzero.

• • Given equation (3.1), in the region $\sigma>\frac {1}{2\ell +1}$ , the only possible real poles of $D_f(s)$ are at $s=1, \frac {1}{2}, \ldots , \frac {1}{2\ell }$ . Let $\xi _j$ be the order of the pole of $D_f(s)$ at $s=\frac 1j$ , so that $\xi _j=0$ if $a_j\le 0$ and $0\le \xi _j\le a_j$ if $a_j\ge 1$ .

• • By Lemma 3.2, equation (1.4) becomes

  $$\begin{align*}G_{f}(x) = \sum_{j=1}^{2\ell} \operatorname{Res}\bigg(D_f(s) \frac{x^s}{s}, \frac{1}{j} \bigg) = \sum_{\substack{ 1\leq j \leq 2\ell \\ \xi_j \geq 1}} P_j(\log x) x^{1/j}, \end{align*}$$
  where each $P_j(t)$ is a polynomial of degree $\xi _j-1$ .

• • As always, we have $F_f(x) = \sum _{n\le x} f(n)$ and $E_f(x) = F_f(x) - G_f(x)$ .

Proof In the usual notation

$$ \begin{align*} N (T) &= \# \{\rho= \beta+i\gamma\colon \zeta(\rho) =0,\, 0< \gamma \leq T\} \\ N (\sigma, T) &= \# \{\rho= \beta+i\gamma\colon \zeta(\rho) =0,\, 0< \gamma \leq T, \, \beta\ge\sigma\}, \end{align*} $$

we know that $N(T) = \frac{T}{2\pi} \log T + O(T)$ (see, for example, [Reference Montgomery and Vaughan25, Corollary 14.2]), while well-known zero-density estimates (first proved by Bohr and Landau [Reference Bohr and Landau6]) imply that $N(\sigma , T)\ll _\sigma T$ for $\sigma>\frac {1}{2}$ . We use this latter estimate to bound how many of the $N(T)$ zeros up to height T are not included in $Z_\ell (T)$ . For the rest of this proof, all implicit constants may depend on $\ell $ .

The upper bound $\beta <\frac {\ell +1}{2\ell +1}$ excludes $N(\frac {\ell +1}{2\ell +1},T) \ll T$ zeros; also, by the symmetries of $\zeta (s)$ , the lower bound $\beta>\frac {\ell }{2\ell +1}$ excludes $N(1-\frac {\ell }{2\ell +1},T) \ll T$ zeros. Suppose that $\rho $ is a zero that has not been excluded so far. If $\ell +1\le j\le 2\ell $ , then $\Re (\frac {j}{\ell }\rho ) \ge \frac {\ell +1}\ell \frac \ell {2\ell +1} = \frac {\ell +1}{2\ell +1}> \frac 12$ , and so the number of these $\frac {j}{\ell }\rho $ that can be zeros of $\zeta (s)$ is at most $N(\frac {\ell +1}{2\ell +1},T) \ll T$ . Similarly, if $1\le j\le \ell -1$ , then $\Re (\frac {j}{\ell }\rho ) \le \frac {\ell -1}\ell \frac {\ell +1}{2\ell +1} < \frac 12$ , and so the number of these $\frac {j}{\ell }\rho $ that can be zeros of $\zeta (s)$ is at most $N(1-\frac {\ell -1}\ell \frac {\ell +1}{2\ell +1},T) \ll T$ . Similar observations hold for the $\zeta \big ( \frac {j(1-\rho )}{\ell } \big )$ factors. We conclude that $\#Z_\ell (T) =N(T)+O_\ell (T) = \frac{T}{2\pi} \log T+O_\ell (T)$ as desired.

Proof Landau’s formula [Reference Landau17] tells us that if $x>1$ ,

$$ \begin{align*}\sum_{0< \gamma \leq T} x^\rho=-\frac{T}{2\pi}\Lambda(x)+O(\log T), \end{align*} $$

where $\Lambda (x)$ is the von Mangoldt function when x is an integer and $\Lambda (x)=0$ otherwise. When $0<x<1$ , one can derive from Landau’s formula (see, for example, [Reference Gonek10, “Corollary”]) that

$$ \begin{align*}\sum_{0< \gamma \leq T} x^\rho=-\frac{Tx}{2\pi}\Lambda\bigg( \frac{1}{x} \bigg)+O(\log T). \end{align*} $$

In both cases, the right-hand side is $\ll _x T$ as required.

Notation 3.8 For each prime p, define

$$ \begin{align*}C_p(s) =1+\sum_{j=1}^{\infty} \frac{\varepsilon_j}{p^{js}} \quad\text{and}\quad \widetilde{C}_p(s) =1+\sum_{j=2\ell+1}^{\infty} \frac{\eta_{j}}{p^{js}}, \end{align*} $$

so that $D_f(s) = \prod _{p} C_p(s)$ for $\sigma>1$ and $U_{2\ell }(s) = \prod _{p} \widetilde {C}_p(s)$ for $\sigma>\frac {1}{2\ell +1}$ . Note that the bounds $|\varepsilon _j| \le 1$ and $|\eta _j| \le (Aj)^B$ imply that the series defining $C_p(s)$ and $\widetilde C_p(s)$ both converge for $\sigma>0$ .

Proof We certainly have

$$\begin{align*}\bigg| 1 - \widetilde{C}_p\bigg( \frac{\rho}{\ell} \bigg) \bigg| \le \sum_{j=2\ell+1}^{\infty} \frac{(Aj)^B}{p^{j\Re\rho/\ell}} < \sum_{j=2\ell+1}^{\infty} \frac{(Aj)^B}{p^{j/(2\ell+1)}}. \end{align*}$$

The right-hand side is a positive series that is decreasing in p and tends to $0$ termwise; by the monotone convergence theorem, the series itself tends to $0$ as $p \to \infty $ . In particular, there exists a positive integer P such that the series is less than $1$ when $p\ge P$ , and for these primes, we deduce that $\widetilde {C}_p(\rho /\ell ) \ne 0$ . Moreover, since A and B depend only on $\ell $ , the same is true of P.

As we observed in Notation 3.3, $U_{2\ell }(\rho /\ell ) =0$ only if $\widetilde {C}_p(\rho /\ell ) =0$ for some prime p, and such a prime must necessarily be less than P. On the other hand, observe that equation (3.1) implies that $\widetilde {C}_p(s) = C_p(s) \prod _{j=1}^{2\ell } (1-p^{-js})^{-a_j}$ for $\Re s>1$ , and so this identity remains true in the larger-half plane $\Re s>0$ where all terms converge. In particular,

$$ \begin{align*}\widetilde{C}_p\bigg( \frac{\rho}{\ell} \bigg) = C_p\bigg( \frac{\rho}{\ell} \bigg) \prod_{j=1}^{2\ell} (1-p^{-j\rho/\ell})^{-a_j}, \end{align*} $$

and thus $\widetilde {C}_p(\rho /\ell ) =0$ if and only if $C_p(\rho /\ell ) =0$ , which completes the proof of the lemma.

Proof of Proposition 3.4

For notational convenience, set $\varepsilon _0=1$ . Note that uniformly for all primes p, all positive integers n, and all zeros $\rho $ of $\zeta (s)$ with $\Re (\rho )\ge \frac {\ell }{2\ell +1}$ ,

$$ \begin{align*} C_p\bigg( \frac{\rho}{\ell} \bigg) &= \sum_{j=0}^n \frac{\varepsilon_j}{p^{{j\rho}/{\ell}}} + O\bigg( \sum_{j=n+1}^\infty \frac1{p^{{j\Re\rho}/{\ell}}} \bigg) \\ &= \sum_{j=0}^n \frac{\varepsilon_j}{p^{{j\rho}/{\ell}}} + O\bigg( \sum_{j=n+1}^\infty \frac1{p^{{j}/{(2\ell+1})}} \bigg) \\ &= \sum_{j=0}^n \frac{\varepsilon_j}{p^{{j\rho}/{\ell}}} + O\bigg( \frac{1}{p^{(n+1)/(2\ell+1)}} \bigg). \end{align*} $$

Let P be the positive integer (depending only on $\ell $ ) from Lemma 3.9, and set

$$ \begin{align*}Q_n = \bigg\{ \prod_{p \leq P} p^{\alpha(p)}\colon 0 \leq \alpha(p)\leq n \bigg\}. \end{align*} $$

Note that $|Q_n|\leq (n+1)^P$ . By expanding the product, we see that uniformly for all $n\in \mathbb {N}$ and for all zeros $\rho $ of $\zeta (s)$ with $\frac {\ell }{2\ell +1}<\Re (\rho )<\frac {\ell +1}{2\ell +1}$ ,

$$ \begin{align*} \prod_{p \leq P} & C_p\bigg( \frac{\rho}{\ell} \bigg) C_p\bigg( \frac{1-\rho}{\ell} \bigg) \\ &= \prod_{p \leq P} \bigg( \sum_{j=0}^n \frac{\varepsilon_j}{p^{{j\rho}/{\ell}}}+O\bigg( \frac{1}{p^{(n+1)/(2\ell+1)}} \bigg)\bigg) \bigg( \sum_{j=0}^n \frac{\varepsilon_j}{p^{{j(1-\rho)}/{\ell}}}+O\bigg( \frac{1}{p^{(n+1)/(2\ell+1)}} \bigg)\bigg) \\ &= \bigg( \sum_{q \in Q_{n}} \frac{f(q)}{q^{\rho/\ell}} \bigg) \bigg( \sum_{q \in Q_{n}} \frac{f(q)}{q^{(1-\rho)/\ell}} \bigg)+O\bigg( \frac{(n+1)^{2P}-1}{2^{(n+1)/(2\ell+1)}} \bigg). \end{align*} $$

In particular, we can choose n sufficiently large in terms of $\ell $ so that

(3.3) $$ \begin{align} \bigg| \prod_{p \leq P} C_p\bigg( \frac{\rho}{\ell} \bigg) C_p\bigg( \frac{1-\rho}{\ell} \bigg) \bigg| \ge \bigg| \bigg( \sum_{q \in Q_{n}} \frac{f(q)}{q^{\rho/\ell}} \bigg) \bigg( \sum_{q \in Q_{n}} \frac{f(q)}{q^{(1-\rho)/\ell}} \bigg) \bigg| - \frac12. \end{align} $$

For each nontrivial zero $\rho $ of $\zeta $ , we have

$$ \begin{align*} \bigg( \sum_{q \in Q_{n}} \frac{f(q)}{q^{\rho/\ell}} \bigg) \bigg( \sum_{q \in Q_{n}} \frac{f(q)}{q^{(1-\rho)/\ell}} \bigg) &= \sum_{q_1,q_2 \in Q_{n}} \frac{f(q_1)f(q_2)}{q_2^{1/\ell}} \bigg( \frac{q_2}{q_1} \bigg)^{\rho/\ell} \\ &= \sum_{q \in Q_{n}} \frac{f(q)^2}{q^{1/\ell}}+ \sum_{\substack{q_1,q_2 \in Q_{n}\\ q_1 \neq q_2}} \frac{f(q_1)f(q_2)}{q_2^{1/\ell}} \bigg( \frac{q_2}{q_1} \bigg)^{\rho/\ell}. \end{align*} $$

By Lemma 3.7, for each $q_1,q_2 \in Q_n$ with $q_1 \neq q_2$ ,

$$ \begin{align*}\sum_{0<\gamma \leq T} \bigg( \frac{q_2}{q_1} \bigg)^{\rho/\ell} \ll T; \end{align*} $$

while the implicit constant depends on $q_1$ and $q_2$ , both numbers come from the finite set $Q_n$ which depends only on $\ell $ . We conclude that

$$ \begin{align*} \sum_{0<\gamma \leq T} \bigg( \sum_{q \in Q_{n}} \frac{f(q)}{q^{\rho/\ell}} \bigg) \bigg( \sum_{q \in Q_{n}} \frac{f(q)}{q^{(1-\rho)/\ell}} \bigg) &= \bigg( \sum_{q \in Q_{n}} \frac{f(q)^2}{q^{1/\ell}} \bigg) N(T) + O_\ell(T) \\ &\ge 1\cdot N(T) + O_\ell(T) = \frac{T}{2\pi} \log T + O_\ell(T), \end{align*} $$

since $f(1)=1$ . Also, note that for each nontrivial zero $\rho $ of $\zeta (s)$ , the trivial bound $|f(q)|\le 1$ implies that

$$ \begin{align*}\bigg|\bigg( \sum_{q \in Q_{n}} \frac{f(q)}{q^{\rho/\ell}} \bigg) \bigg( \sum_{q \in Q_{n}} \frac{f(q)}{q^{(1-\rho)/\ell}} \bigg)\bigg|\leq |Q_n|^2 \leq (n+1)^{2P}. \end{align*} $$

Since the set $Z_\ell (T)$ defined in Lemma 3.6 excludes $\ll _\ell T$ zeros up to height T, it follows that

$$ \begin{align*} &\sum_{\rho \in Z_\ell(T)} \bigg( \sum_{q \in Q_{n}} \frac{f(q)}{q^{\rho/\ell}} \bigg) \bigg( \sum_{q \in Q_{n}} \frac{f(q)}{q^{(1-\rho)/\ell}} \bigg) \\ &\quad = \sum_{0<\gamma \leq T} \bigg( \sum_{q \in Q_{n}} \frac{f(q)}{q^{\rho/\ell}} \bigg) \bigg( \sum_{q \in Q_{n}} \frac{f(q)}{q^{(1-\rho)/\ell}} \bigg) + O_\ell(T (n+1)^{2P}) \\ &\quad \ge \frac{T}{2\pi} \log T + O_\ell(T). \end{align*} $$

Finally, equation (3.3) implies that

$$ \begin{align*}\sum_{\rho \in Z_\ell(T)} \bigg( \prod_{p \leq P} C_p\bigg( \frac{\rho}{\ell} \bigg) C_p\bigg( \frac{1-\rho}{\ell} \bigg)\bigg) &\ge \frac{T}{2\pi} \log T + O_\ell(T) - \frac12 \#Z_\ell(T) \\ &\ge \frac{T}{4\pi} \log T + O_\ell(T). \end{align*} $$

In particular, when T is sufficiently large in terms of $\ell $ , there exists $\rho \in Z_\ell (T)$ such that

$$ \begin{align*}\prod_{p \leq P} C_p\bigg( \frac{\rho}{\ell} \bigg) C_p\bigg( \frac{1-\rho}{\ell} \bigg)\neq 0, \end{align*} $$

which implies that

$$ \begin{align*}U_{2\ell}\bigg( \frac{\rho}{\ell} \bigg)U_{2\ell}\bigg( \frac{1-\rho}{\ell} \bigg) \neq 0 \end{align*} $$

by Lemma 3.9.

Since $\rho \in Z_\ell (T)$ implies $1-\rho \in Z_\ell (T)$ , we may assume that $\Re \rho \ge \frac 12$ . Then $\zeta (\rho )=0$ and

$$\begin{align*}\prod_{\substack{1\leq j \leq 2\ell\\j \neq \ell}}\zeta\bigg( \frac{j\rho}{\ell} \bigg)\neq 0. \end{align*}$$

by the definition of $Z_\ell (T)$ , and we have just shown that $U_{2\ell }\big ( \frac {\rho }{\ell } \big ) \ne 0$ , which confirms equation (3.2) and hence establishes the proposition.

Proof of Theorem 3.5

We show that $E_f(x) = \Omega _- \big (x^{1/2\ell }(\log x)^{|a_\ell |-1} \big )$ , as the proof of the corresponding $\Omega _+$ result is almost identical. For each $1\leq j \leq 2\ell $ with $\xi _j\ge 1$ , let the coefficients $b_{j,k}$ for $0\le k\le \xi _j-1$ be defined by $P_j(y) = \sum _{k=0}^{\xi _j-1} b_{j,k} y^{k}$ . Define

$$ \begin{align*}\widetilde{D}(s) =D_f(s)-\sum_{\substack{ 1\leq j \leq 2\ell \\ \xi_j \geq 1}} \sum_{k=0}^{\xi_j-1} \frac{ b_{j,k}s \cdot k!}{(s-\frac{1}{j})^{k+1}}. \end{align*} $$

Note that for each $k \geq 0$ ,

$$ \begin{align*}\frac{k!}{(s-\frac{1}{j})^{k+1}}= \int_{1}^{\infty} \frac{x^{1/j}(\log x)^k}{x^{s+1}} dx. \end{align*} $$

It follows that

$$ \begin{align*}\widetilde{D}(s) =s \int_{1}^{\infty} \frac{E_f(x)}{x^{s+1}}, \end{align*} $$

and thus $\widetilde {D}(s)$ has no real pole with $s\geq \frac {1}{2\ell }$ .

Let $r>0$ be a constant to be chosen later, and define

$$ \begin{align*}H(s) = \widetilde{D}(s)+\frac{rs (|a_\ell|-1)!}{(s-\frac{1}{2\ell})^{-a_\ell}}. \end{align*} $$

Then for $\sigma>1$ ,

(3.4) $$ \begin{align} H(s) =s \int_{1}^{\infty} \frac{E_f(x)+rx^{1/2\ell}(\log x)^{|a_\ell|-1}}{x^{s+1}}\,dx. \end{align} $$

Suppose that $E_f(x)+rx^{1/2\ell }(\log x)^{|a_\ell |-1}$ is positive when x is sufficiently large. Note that $\widetilde {D}(s)$ has no real singularity with $s\geq \frac {1}{2\ell }$ , and the smallest real singularity of $(s-\frac {1}{2\ell })^{a_\ell }$ is at $s= \frac {1}{2\ell }$ . Thus, Landau’s theorem (see, for example, [Reference Montgomery and Vaughan25, Lemma 15.1]) implies that $H(s)$ is analytic for $\sigma>\frac {1}{2\ell }$ and equation (3.4) holds for $\sigma>\frac {1}{2\ell }$ . On the other hand, by Proposition 3.4, there is a nontrivial zero $\rho $ of $\zeta $ such that $\Re (\rho )\geq \frac {1}{2}$ and

(3.5) $$ \begin{align} U_{2\ell}\bigg( \frac{\rho}{\ell} \bigg) \cdot \prod_{\substack{1\leq j \leq 2\ell\\j \neq \ell}}\zeta\bigg( \frac{j\rho}{\ell} \bigg)\neq 0. \end{align} $$

In particular, since $\zeta (\ell s)$ has a zero at $\rho /\ell $ , $D_f(s)$ does indeed have a pole at $\rho /\ell $ by equation (3.1), and thus so does $H(s)$ . Since $H(s)$ is analytic for $\sigma>\frac {1}{2\ell }$ , we deduce that $\Re (\rho ) = \frac {1}{2}$ . Set $\rho = \frac {1}{2}+i\gamma $ and $\gamma '= \gamma /\ell $ . Equation (3.4) implies that for $\sigma>\frac {1}{2\ell }$ ,

$$ \begin{align*} |H(\sigma+i \gamma')| & \leq |\sigma+i \gamma'| \int_{1}^{\infty} \frac{\big|E_f(x)+rx^{1/2\ell}(\log x)^{|a_\ell|-1} \big|}{x^{\sigma+1}}\,dx \\ &= \frac{|\sigma+i \gamma'|}{\sigma} H(\sigma) < 2\ell|\rho| H(\sigma). \end{align*} $$

Let m be the order of $\rho $ as a zero of $\zeta (s)$ , and set $m'=|a_\ell | m$ . The above inequality implies that

(3.6) $$ \begin{align} \lim_{\sigma \to \frac{1}{2\ell}^+} \bigg( \sigma-\frac{1}{2\ell} \bigg)^{m'} |H(\sigma+i \gamma')| \leq 2\ell|\rho|\lim_{\sigma \to \frac{1}{2\ell}^+} \bigg( \sigma-\frac{1}{2\ell} \bigg)^{m'} H(\sigma). \end{align} $$

Since $\widetilde {D}(s)$ is analytic at $\sigma = \frac {1}{2\ell }$ , the right-hand side of inequality (3.6) is equal to

$$ \begin{align*}2\ell|\rho|\lim_{\sigma \to \frac{1}{2\ell}^+} \bigg( \sigma-\frac{1}{2\ell} \bigg)^{m'} \bigg|\frac{r\sigma(|a_\ell|-1)!}{(\sigma-\frac{1}{2\ell})^{-a_\ell}} \bigg|= \begin{cases} r|\rho|(|a_\ell|-1)!, &\text{if } m=1,\\ 0, &\text{if } m>1. \end{cases} \end{align*} $$

Since $\rho $ is a zero of $\zeta (s)$ of order m, the left-hand side of inequality (3.6) is equal to

$$ \begin{align*} & \lim_{\sigma \to \frac{1}{2\ell}^+} \bigg( \sigma-\frac{1}{2\ell} \bigg)^{m'} \bigg|D_f(\sigma+i \gamma')-\sum_{\substack{ 1\leq j \leq 2\ell \\ \xi_j \geq 1}} \sum_{k=0}^{\xi_j-1} \frac{ b_{j,k}s \cdot k!}{(\sigma+i \gamma'-\frac{1}{j})^{k+1}}+\frac{r(\sigma+i \gamma')(|a_\ell|-1)!}{(\sigma+i \gamma'-\frac{1}{2\ell})^{-a_\ell}} \bigg|\\ &\quad = \lim_{\sigma \to \frac{1}{2\ell}^+} \bigg( \sigma-\frac{1}{2\ell} \bigg)^{m'} \big|D_f(\sigma+i \gamma')\big|= \bigg( \frac{m!}{\ell^m |\zeta^{(m)}(\rho)|} \bigg)^{-a_\ell} \bigg|U_{2\ell}\bigg( \frac{\rho}{\ell} \bigg)\bigg| \prod_{\substack{1\leq j \leq 2\ell \\ j \neq \ell}} \bigg|\zeta\bigg( \frac{j\rho}{\ell} \bigg)\bigg|^{a_j}. \end{align*} $$

Now inequality (3.5) implies that $m=1$ and thus

(3.7) $$ \begin{align} \frac{|U_{2\ell}(\frac{\rho}{\ell})|}{(\ell |\zeta'(\rho)|)^{-a_\ell}} \cdot \prod_{\substack{1\leq j \leq 2\ell \\ j \neq \ell}} \bigg|\zeta\bigg( \frac{j\rho}{\ell} \bigg)\bigg|^{a_j} \leq r|\rho|(|a_\ell|-1)!. \end{align} $$

Note that inequality (3.5) implies that the left-right side of inequality (3.7) is nonzero, and thus inequality (3.7) can only hold when r is sufficiently large. In other words, for smaller positive values of r, the difference $E_f(x)+rx^{1/2\ell }(\log x)^{|a_\ell |-1}$ cannot be always positive for sufficiently large x, which shows that $E_f(x) = \Omega _- \big (x^{1/2\ell }(\log x)^{|a_\ell |-1} \big )$ as required.

Proof of Theorem 1.10

For each $1\leq j \leq 2k$ , Proposition 2.3 tells us that $D_f(s)$ has at most a simple pole at $s= \frac {1}{j}$ ; therefore, the residue of $D_f(s) \cdot \frac {x^s}{s}$ at $s= \frac {1}{j}$ is ${\mathfrak a}_f(j)x^{1/j}$ , where ${\mathfrak a}_f(j)$ is the constant from Definition 1.9. The theorem then follows from Theorem 3.5 by applying the partial zeta-factorization of $D_f(s)$ in Proposition 2.3.

Proof of Theorem 1.14

Let $1 \leq j \leq 2k$ . By Proposition 2.8, $D_f(s)$ has a pole at $s= \frac {1}{j}$ with order at most $2$ . By Definition 1.9, the principal part of $D_f(s)$ is

$$\begin{align*}\frac{{\mathfrak b}_f(j)}{j^2(s-1/j)^2} + \frac{{\mathfrak a}_f(j)}{j(s-1/j)}. \end{align*}$$

Thus, the residue of $D_f(s) \cdot \frac {x^s}{s}$ at $s= \frac {1}{j}$ is given by

$$ \begin{align*}\lim_{s \to \frac{1}{j}} \frac{d}{ds} D_f(s) \cdot \frac{x^s}{s} &= \lim_{s \to \frac{1}{j}} \frac{d}{ds} \bigg( \frac{{\mathfrak b}_f(j)}{j^2(s-1/j)^2} + \frac{{\mathfrak a}_f(j)}{j(s-1/j)} \bigg) \cdot \frac{x^s}{s} \\ &= {\mathfrak a}_f(j)x^{1/j}+{\mathfrak b}_f(j) x^{1/j} \bigg( \frac{\log x}{j}-1\bigg). \end{align*} $$

The theorem now follows from Proposition 2.8 and Theorem 3.10.

Proof Since both $h_1$ and $\tau $ are multiplicative, it suffices to observe that $|h_1(p^j)| \le j+1 = \tau (p^j)$ for all prime powers $p^j$ .

Proof Note that $h_k(n)=h_1(m)$ if $n=m^k$ is a perfect kth power and otherwise $h_k(n)=0$ . It follows that $\sum _{n \leq x}|h_k(n)|=\sum _{n \leq x^{1/k}} |h_1(n)|$ ; thus it suffices to prove the lemma for $k=1$ . But since $|h_1(n)| \le \tau (n)$ for all $n\ge 1$ , the estimate $\sum _{n \leq x} |h_1(n)|\ll x\log x$ follows immediately from the classical evaluation of $\sum _{n\le x} \tau (n)$ (see, for example, [Reference Montgomery and Vaughan25, Theorem 2.3]).

Proof We have $H_k(x) = H_1(x^{1/k})$ by the same reasoning as in the proof of Lemma 4.3, and so it suffices to prove the lemma for $k=1$ . We leverage a known result for the Mertens function $M(x) = \sum _{n\le x} \mu (n)$ , which uses contour integration to show that

(4.1) $$ \begin{align} M(x) \ll x\exp\bigg(-c\frac{(\log x)^{3/5}}{(\log \log x)^{1/5}} \bigg) \end{align} $$

for some absolute positive constant c (see, for example, [Reference Ivić16, Theorem 12.7]). It is possible to slightly modify that proof to deal with $H_1(x)$ ; instead, however, we give an alternative derivation that uses the result (4.1) directly rather than modifying its proof.

Since $h_1$ is the Dirichlet convolution of $\mu $ with itself, the hyperbola method implies that

(4.2) $$ \begin{align} H_1(x) &= \sum_{n \leq \sqrt{x}} \mu(n) M\bigg(\frac{x}{n}\bigg) + \sum_{m \leq \sqrt{x}} \mu(m) M\bigg(\frac{x}{m}\bigg) - M(\sqrt{x})M(\sqrt{x}) \notag \\ &\ll \sum_{n \leq \sqrt{x}} \bigg|M\bigg(\frac{x}{n}\bigg)\bigg| + \big| M(\sqrt{x}) \big|^2. \end{align} $$

Inequalities (4.1) and (4.2) imply that there are absolute constants $0<c"<c'<c$ such that

$$ \begin{align*} H_1(x) &\ll\sum_{n \leq \sqrt{x}} \frac{x}{n}\exp\bigg(-c\frac{(\log (x/n))^{3/5}}{(\log \log (x/n))^{1/5}} \bigg) + x \exp\bigg(-2c\frac{(\log \sqrt{x})^{3/5}}{(\log \log \sqrt{x})^{1/5}} \bigg)\\ &\ll \sum_{n \leq \sqrt{x}} \frac{x}{n}\exp\bigg(-c'\frac{(\log x)^{3/5}}{(\log \log x)^{1/5}} \bigg) \\ &\ll x\log x\cdot \exp\bigg(-c' \frac{(\log x)^{3/5}}{(\log \log x)^{1/5}} \bigg) \ll x\exp\bigg(-c"\frac{(\log x)^{3/5}}{(\log \log x)^{1/5}} \bigg), \end{align*} $$

as desired.

Proof Since $D_{g_k}(s) = \zeta (s)\zeta (ks)^{-2}$ , we have $G_{g_k}(x)={x}/{\zeta (k)^2}$ . Since $g_k$ is the Dirichlet convolution of $h_k$ and $1$ , we can apply the hyperbola method, with a parameter $z \in (\sqrt {x},x)$ to be determined and with $y=\frac xz$ :

$$ \begin{align*} F_{g_k}(x) = \sum_{mn \leq x} h_k(n) &= \sum_{n \leq z} h_k(n) \bigg\lfloor \frac{x}{n} \bigg\rfloor + \sum_{m\leq y} H_k\bigg(\frac{x}{m}\bigg) - \lfloor y \rfloor H_k(z) \\ &= x \sum_{n\le z} \frac{h_k(n)}n + O \bigg( \sum_{n\le z} |h_k(n)| + \sum_{m\le y} \bigg| H_k\bigg(\frac{x}{m}\bigg) \bigg| + y |H_k(z)| \bigg). \end{align*} $$

Since $\sum _{n=1}^{\infty } h_k(n)/n = 1/\zeta (k)^2$ , we use Lemma 4.3 to conclude that

(4.3) $$ \begin{align} E_{g_k}(x) = F_{g_k}(x) - G_{g_k}(x) \ll x \bigg| \sum_{n>z} \frac{h_k(n)}n \bigg| + z^{1/k}\log z + \sum_{m\le y} \bigg| H_k\bigg(\frac{x}{m}\bigg) \bigg| + y |H_k(z)|. \end{align} $$

By Lemma 4.4 and partial summation,

(4.4) $$ \begin{align} \frac{1}{\zeta(k)^2}-\sum_{n\leq z} \frac{h_k(n)}{n} &\ll \frac{|H_k(z)|}{z}+\int_{z}^\infty \frac{|H_k(t)|}{t^2} \,dt \nonumber\\ &\ll z^{1/k-1} \exp\bigg(-c\frac{(\log z)^{3/5}}{(\log \log z)^{1/5}} \bigg). \end{align} $$

Since $y \leq \sqrt {x}$ , we have $x/m \geq \sqrt {x}$ for each $m \leq y$ . Thus, by Lemma 4.4, there is a constant $c'\in (0,c)$ such that

(4.5) $$ \begin{align} \sum_{m\leq y} \bigg(H_k\bigg(\frac{x}{m}\bigg)-H_k(z)\bigg) &\ll \sum_{m \leq y} \bigg(\frac{x}{m}\bigg)^{1/k} \exp\bigg(-c\frac{(\log (x/m))^{3/5}}{(\log \log (x/m))^{1/5}} \bigg)\notag\\ &\ll x^{1/k}\sum_{m \leq y} m^{-1/k} \exp\bigg(-c'\frac{(\log x)^{3/5}}{(\log \log x)^{1/5}} \bigg)\notag\\ &\ll \frac{x}{z^{1-1/k}}\exp\bigg(-c'\frac{(\log x)^{3/5}}{(\log \log x)^{1/5}} \bigg). \end{align} $$

Combining inequalities (4.3)–(4.5), we conclude that

(4.6) $$ \begin{align} E_g(x) &\ll \frac{x}{z^{1-1/k}}\exp\bigg(-c'\frac{(\log x)^{3/5}}{(\log \log x)^{1/5}} \bigg)+\sum_{n \leq z}|h_k(n)|\notag\\ &\ll \frac{x}{z^{1-1/k}}\exp\bigg(-c'\frac{(\log x)^{3/5}}{(\log \log x)^{1/5}} \bigg)+z^{1/k}\log x. \end{align} $$

If we set

$$ \begin{align*}z=x \exp\bigg(-c'\frac{(\log x)^{3/5}}{(\log \log x)^{1/5}} \bigg), \end{align*} $$

then we conclude that there exists $c" \in (0,c')$ such that $E_g(x)\ll x^{1/k}\exp \big (-c"\frac {(\log x)^{3/5}}{(\log \log x)^{1/5}} \big )$ .

Proof Recall that $\sum _{n=1}^{\infty } h_1(n)n^{-s}=\zeta (s)^{-2}$ for $\sigma>1$ . Assume that $\frac {1}{2}+\varepsilon \leq \sigma \leq \sigma _0$ . By Lemma 4.2, we have $|h_1(n)n^{-s}|<\tau (n)n^{-1/2}\ll n^{-1/2+\varepsilon }$ . Thus, a truncated Perron’s formula [Reference Montgomery and Vaughan25, Corollary 5.3] implies that

(4.7) $$ \begin{align} \sum_{n\leq x} h_1(n)n^{-s}=\frac{1}{2\pi i}\int_{2-iT}^{2+iT} \frac{1}{\zeta(s+w)^2}\frac{x^w}{w} \,dw +R(x), \end{align} $$

where

(4.8) $$ \begin{align} R(x) \ll\sum_{\substack{x/2<n<2x \\n \neq x}} |h_1(n)n^{-s}| \min \bigg(1, \frac{x}{T|x-n|} \bigg)+ \frac{4^{2}+x^{2}}{T} \sum_{n=1}^{\infty} \frac{|h_1(n)n^{-s}|}{n^{2}}\ll \frac{x^2}{T} \end{align} $$

(we may assume that x is an integer). Let $\varepsilon '=\varepsilon /(\sigma _0+\frac 52)$ . We shift the contour integral in equation (4.7) leftwards from $\Re (w)=2$ to $\Re (w)=\frac {1}{2}-\sigma +\varepsilon '$ , noting that the only pole of ${x^w}/{w}{\zeta (s+w)^2}$ inside the contour is the simple pole at $w=0$ . Thus,

(4.9) $$ \begin{align} &\frac{1}{2\pi i}\int_{2-iT}^{2+iT} \frac{1}{\zeta(s+w)^2}\frac{x^w}{w} \,dw \nonumber\\ &=\frac{1}{\zeta(s)^2}+\frac{1}{2\pi i}\bigg(\int_{\frac{1}{2}-\sigma+\varepsilon'+iT}^{2+iT} + \int_{\frac{1}{2}-\sigma+\varepsilon'-iT}^{\frac{1}{2}-\sigma+\varepsilon'+iT} + \int_{2-iT}^{\frac{1}{2}-\sigma+\varepsilon'-iT}\bigg) \frac{1}{\zeta(s+w)^2}\frac{x^w}{w} \,dw. \end{align} $$

Since $\zeta (z)^{-1}\ll _{\varepsilon '} T^{\varepsilon '}$ holds uniformly for $\frac {1}{2}+\varepsilon ' \leq \Re (z) \leq 2$ and $|\Im (z)|\leq T$ (see [Reference Montgomery and Vaughan25, Theorem 13.23]), it follows that

$$ \begin{align*} \int_{\frac{1}{2}-\sigma+\varepsilon'-iT}^{\frac{1}{2}-\sigma+\varepsilon'+iT} \frac{1}{\zeta(s+w)^2}\frac{x^w}{w} \,dw &\ll_{\varepsilon'} x^{1/2-\sigma+\varepsilon}T^{\varepsilon'} \\ \int_{\frac{1}{2}-\sigma+\varepsilon' \pm iT}^{2 \pm iT} \frac{1}{\zeta(s+w)^2}\frac{x^w}{w} \,dw &\ll_{\varepsilon'} \sigma T^{\varepsilon'} \cdot \frac{x^2}{T} \ll_{\varepsilon', \sigma_0} \frac{x^{2}}{T^{1-\varepsilon'}}. \end{align*} $$

Combining these estimates with equations (4.7)–(4.9), and setting $T=x^{3/2+\sigma }$ , we conclude that

$$ \begin{align*}\frac{1}{\zeta(s)^2}-\sum_{n<x} h_1(n)n^{-s} &\ll_{\varepsilon',\sigma_0} x^{1/2-\sigma+\varepsilon'} T^{\varepsilon'}+\frac{x^{2}}{T^{1-\varepsilon'}}+\frac{x^2}{T} \\ &\ll_{\varepsilon',\sigma_0} x^{1/2-\sigma+{} \varepsilon'(\sigma+5/2)}{}\leq x^{1/2-\sigma+ \varepsilon} \end{align*} $$

as required.

Proof We adapt the proof that Montgomery and Vaughan [Reference Montgomery and Vaughan24] used for the indicator function of k-free numbers. Recall that $h_k(n)=h_1(m)$ if $n=m^k$ is a perfect kth power and $h_k(n)=0$ otherwise. Let $y=x^{1/(k+1)}$ , and define

$$ \begin{align*}A(s)=\frac{1}{\zeta(s)^2}-\sum_{n \leq y} h_1(n)n^{-s}. \end{align*} $$

If we set $\widetilde {h}(n)=h_k(n)$ for $n>y^k$ and $\widetilde {h}(n)=0$ otherwise, we see that the Dirichlet series for $\widetilde {h}$ is precisely $A(ks)$ . Define $\widetilde {g}$ to be the Dirichlet convolution of $\widetilde {h}$ and $1$ , so that the Dirichlet series for $\widetilde {g}$ is $\zeta (s)A(ks)$ . Also note that $|\widetilde {g}(n)|\leq \sum _{d\mid n}|\widetilde {h}(d)|\leq \sum _{d\mid n}\tau (d)\ll _{\varepsilon } n^\varepsilon $ for each $\varepsilon>0$ by Lemma 4.2. Using the truncated Perron’s formula [Reference Montgomery and Vaughan25, Corollary 5.3] with the choice $T=x$ ,

(4.10) $$ \begin{align} \sum_{n \leq x} \widetilde{g}(n)=\frac{1}{2\pi i}\int_{c-ix}^{c+ix} \zeta(s)A(ks) \frac{x^s}{s} \,ds +R(x), \end{align} $$

where $c=1+1/(k+1)$ and

(4.11) $$ \begin{align} R(x) \ll\sum_{\substack{x/2<n<2x \\n \neq x}} |\widetilde{g}(n)| \min \bigg(1, \frac{x}{x|x-n|} \bigg)+ \frac{4^{c}+x^{c}}{x} \sum_{n=1}^{\infty} \frac{|\widetilde{g}(n)|}{n^{c}}\ll x^\varepsilon + x^{1/(k+1)}. \end{align} $$

We shift the contour leftwards from $\sigma =c$ to $\sigma =\frac {1}{2}$ and notice that the only pole of $\zeta (s)A(ks)$ inside the contour is the simple pole at $s=1$ . Thus,

(4.12) $$ \begin{align} &\frac{1}{2\pi i}\int_{c-ix}^{c+ix} \zeta(s)A(ks) \frac{x^s}{s} \nonumber \\ &\quad =xA(k)+\frac{1}{2\pi i}\bigg(\int_{\frac{1}{2}-ix}^{\frac{1}{2}+ix}+\int_{\frac{1}{2}+ix}^{c+ix}+\int_{c-ix}^{\frac{1}{2}-ix}\bigg) \zeta(s)A(ks) \frac{x^s}{s} \,ds. \end{align} $$

By [Reference Montgomery and Vaughan25, Theorem 13.18], $\zeta (s)\ll _{\varepsilon } x^\varepsilon $ uniformly for $\sigma \geq \frac {1}{2}$ and $1\leq |t|\leq x$ . By Lemma 4.6, $A(s)\ll _{\varepsilon } y^{1/2-\sigma +\varepsilon }$ holds uniformly for $\frac {1}{2}+\varepsilon \leq \sigma \leq k+2$ , and thus $A(ks)\ll _{\varepsilon } y^{1/2-k\sigma +\varepsilon }$ holds uniformly for $\frac {1}{2}+\varepsilon \leq \sigma \leq c$ . Therefore

$$ \begin{align*} \int_{\frac{1}{2}-ix}^{\frac{1}{2}+ix} \zeta(s)A(ks) \frac{x^s}{s} &\ll_{\varepsilon} x^{1/2+\varepsilon} y^{(1-k)/2+\varepsilon}, \\ \int_{\frac{1}{2}\pm ix}^{c\pm ix} \zeta(s)A(ks) \frac{x^s}{s} &\ll_{\varepsilon} x^{c-1+\varepsilon} y^{(1-k)/2+\varepsilon}. \end{align*} $$

Combining these estimates with equations (4.10)–(4.12), we conclude that

(4.13) $$ \begin{align} \sum_{n \leq x} \widetilde{g}(n)=xA(k)+O_{\varepsilon}\big(x^{1/(k+1)}+x^{1/2+\varepsilon} y^{(1-k)/2+\varepsilon}\big). \end{align} $$

On the other hand,

$$ \begin{align*} \sum_{mn \leq x} \big( h_k(n)-\widetilde{h}(n) \big) &=\sum_{\substack{mn \leq x\\n\leq y^k}} h_k(n) =\sum_{\substack{mn^k \leq x\\n \leq y}} h_1(n) =\sum_{n \leq y} h_1(n) \bigg\lfloor \frac{x}{n^k} \bigg\rfloor \\ &= x \sum_{n \leq y} \frac{h_1(n)}{n^k}+O \bigg( \sum_{n\le y} |h_1(n)| \bigg) \\ &=x \bigg( \frac{1}{\zeta(k)^2} -A(k)\bigg) + O(y\log y) \end{align*} $$

by Lemmas 4.6 and 4.3. From this estimate and equation (4.13), it follows that

$$ \begin{align*} F_{g_k}(x) &= \sum_{n \leq x} g_k(n)=\sum_{mn \leq x} h_k(n)\\ &=\sum_{mn \leq x} \big(h_k(n)-\widetilde{h}(n)\big)+\sum_{mn \leq x} \widetilde{h}(n) \\ &=\sum_{mn \leq x} \big(h_k(n)-\widetilde{h}(n)\big)+\sum_{n \leq x}\widetilde{g}(n) \\ &=\frac{x}{\zeta(k)^2}+O_{\varepsilon}\big( y\log y+ x^{1/(k+1)}+ x^{1/2+\varepsilon} y^{(1-k)/2+\varepsilon}\big). \end{align*} $$

Since $y=x^{1/(k+1)}$ , this error term is $O_{\varepsilon }\big ( x^{1/(k+1)+\varepsilon } \big )$ and therefore

$$ \begin{align*}E_{g_k}(x) = F_{g_k}(x) - G_{g_k}(x) \ll_{\varepsilon} x^{1/(k+1)+\varepsilon} \end{align*} $$

as required.

Proof of Lemma 4.8

Let $1\leq j \leq k$ and assume that $D_f(s)$ has a pole at $\frac 1j$ . Since $Z(s)$ converges absolutely for $\sigma>\frac {1}{k+1}$ and $D_f(s) =Y(s)Z(s)$ , it follows that $\frac 1j$ is a pole of $Y(s)$ , which must be simple since the residues in equation (4.14) have no logarithmic terms. We deduce that