May 31, 2026
In 1970, T. M. Apostol introduced the Möbius function \(\mu_{k}\) of order \(k\) for all positive integer \(k\), as a generalization of the Möbius function \(\mu = \mu_{1}\). For any integer \(k \ge 2\), he proved \(\sum_{n \le x} \mu_{k}(n) = A_{k} x + \relax_{k}(x^{1/k} \relax x)\) where \(A_{k}\) is a positive constant. In 2001, A. Bege conjectured both the conditional and unconditional estimates for the sum \(\sum_{n \le x, (n, q) = 1}\mu_{k}(n)\) for any positive integer \(q\). In this paper, we give affirmative solutions to the conditional version of Bege’s conjecture completely and the unconditional one partially. We also give a mean square estimate for the error term.
For any complex number \(s = \sigma + it\) where \(\sigma, t \in \mathbb{R}\), we let \(\zeta(s)\) denote the Riemann zeta-function. In 1970, T. M. Apostol [1] introduced the following generalization of the Möbius function \(\mu\). For each fixed positive integer \(k\), the Möbius function \(\mu_{k}\) of order \(k\) is defined by \[\mu_{k}(n) \coloneq \begin{cases*} 1 & if n = 1, \\ (-1)^{r} & if n = (p_{1} p_{2} \cdots p_{r})^{k} \prod_{i>r} p_{i}^{a_{i}} with 0\le a_{i}<k, \\ 0 & if p^{k + 1} \mid n for some prime p, \\ 1 & otherwise. \end{cases*} \label{eq:muk}\tag{1}\] Here \(p_{1}, p_{2}, p_{3}, \ldots\) are distinct primes. In other words, \(\mu_{k}(n)\) vanishes if \(n\) is divisible by \(p^{k + 1}\) for some prime \(p\); otherwise, \(\mu_{k}(n)\) is \(1\) unless the prime factorization of \(n\) contains the \(k\)-th powers of exactly \(r\) distinct primes, in which case \(\mu_{k}(n) = (-1)^{r}\). If \(k = 1\), \(\mu_{k}(n)\) is the usual Möbius function, that is \(\mu_{1}(n) = \mu(n)\). The asymptotic formula for the summatory function \(\sum_{n\leq x}\mu_{k}(n)\) was first considered by Apostol [1]. For any integer \(k \ge 2\) and any real number \(x \ge 2\), he proved that \[\sum_{n \le x}\mu_{k}(n) = A_{k}x + E_{k}(x) \label{eq:A}\tag{2}\] holds, where the error term \(E_{k}(x)\) is \(\relax_{k}(x^{1/k} \relax x)\), and the constant \(A_{k}\) is given by \[A_{k} = \prod_{p} \lr(1 - 2 p^{-k} + p^{-(k + 1)}). \label{eq:Ak}\tag{3}\] In 1977, under the Riemann Hypothesis, D. Suryanarayana [2] showed that \[E_{k}(x) = \relax(x^{4k/(4k^{2} + 1)}\relax(C_{0} \frac{\relax x}{\relax\relax x})), \label{eq:S}\tag{4}\] where \(C_{0}\) is an absolute positive constant.
We now consider a more general case by adding the condition that \(n\) is coprime to \(q\). Let \(x \ge 1\), we write for integers \(k \ge 2\) and \(q \ge 1\), \[\sum_{\substack{n \le x \\ (n, q) = 1}}\mu_{k}(n) = A_{k, q} x + E_{k, q}(x),\] where \(E_{k, q}(x)\) denotes the error term. In 2001, A. Bege [3] proposed the following conjecture on the size of \(E_{k, q}(x)\).
Conjecture 1. For any real number \(x \ge 3\) and any integers \(k \ge 2\) and \(q \ge 1\), we have \[E_{k, q}(x) \ll \theta(q) x^{1/k} \relax(-D \frac{(\relax x)^{3/5}}{(\relax\relax x)^{1/5}}) \label{eq:B-uncond}\qquad{(1)}\] where \[A_{k, q} \coloneq \frac{\varphi(q)}{q}\prod_{p \nmid q}\lr(1 - 2 p^{-k} + p^{-(k + 1)}), \label{eq:Akq}\qquad{(2)}\] \(\varphi(q)\) is Euler’s totient function, \(\theta(q)\) is the number of positive squarefree divisors of \(q\) and \(D\) is an absolute positive constant. In particular, when \(q = 1\) the conjecture is \[E_{k}(x) \ll x^{1/k} \relax(-D \frac{(\relax x)^{3/5}}{(\relax\relax x)^{1/5}}). \label{eq:B-uncond-1}\qquad{(3)}\]
If the Riemann Hypothesis is true, then the estimate is improved to \[E_{k, q}(x) \ll \theta(q) x^{2/(2 k + 1)} \relax(A \frac{\relax x}{\relax\relax x}) \label{eq:B-RH}\qquad{(4)}\] for any real number \(x\ge 3\) and any integer \(k\ge 2\) and \(q \ge 1\). Here \(A\) is an absolute positive constant. In particular, when \(q = 1\) the conjecture is \[E_{k}(x) \ll x^{2/(2 k + 1)}\relax(A \frac{\relax x}{\relax\relax x}). \label{eq:B-RH-1}\qquad{(5)}\]
In 2023, D. Banerjee, Y. Fujisawa, T. M. Minamide and Y. Tanigawa [4] improved Apostol’s result \(E_{k}(x) \ll_{k} x^{1/k} \relax x\) to \[E_{k}(x) \ll_{k} x^{1/k} \relax(-D_{0} k^{-13/5} \frac{(\relax x)^{3/5}}{(\relax\relax x)^{1/5}}) \label{eq:BFMT}\tag{5}\] unconditionally, where \(D_{0}\) is an absolute positive constant. This result solves the unconditional part of Bege’s conjecture when \(q = 1\) in a version which the \(O\)-constant and the constant \(D\) in ?? may depend on \(k\).
Recently, in 2025, under the Riemann Hypothesis, G. Martin and C. H. Yip [5] show that \[E_{k}(x) \ll_{k, \varepsilon} x^{1/(k + 1) + \varepsilon} \label{eq:MY}\tag{6}\] for any real number \(x \ge 1, \varepsilon > 0\) and any integer \(k \ge 2\). This result gives an affirmative solution to the conditional part of Bege’s conjecture with \(q = 1\) in a version where the \(O\)-constant in ?? may depend on \(k\). They also give the following oscillation result: \[E_{k}(x) = B_{k} x^{1/(k+1)} + \Omg_{\pm}(x^{1/(2 k)} \relax x), \quad x \to +\infty. \label{eq:MY-Omg}\tag{7}\] where \[B_{k} \coloneq \frac{\zeta(1/(k + 1))}{\zeta^{2}(k/(k + 1))}\prod_{p}\frac{\lr(1 - 2 p^{-k/(k + 1)} + p^{-1})\lr(1 - p^{-1})}{\lr(1 - p^{-k/(k + 1)})^{2}}. \label{eq:Bk}\tag{8}\]
The purpose of this paper is to generalize Banerjee, Fujisawa, Minamide and Tanigawa’s result 5 and Martin and Yip’s result 6 to the case \(q \ge 1\). we also estimate the mean square of the error term.
Theorem 1. For any real number \(x \ge 3\) and any integers \(k \ge 2\) and \(q \ge 1\), we have \[E_{k, q}(x) \ll_{k} \theta(q) x^{1/k} \relax(-D_{0} k^{-8/5} \frac{(\relax x)^{3/5}}{(\relax\relax x)^{1/5}}), \label{main}\qquad{(6)}\] where \(D_{0}\) is an absolute positive constant.
Remark 1. Theorem 1 answers a version of the unconditional part of Bege’s conjecture in which the \(O\)-constant and the constant \(D\) in ?? may depend on \(k\) in the affirmative. If \(q = 1\), it also slightly improves Banerjee, Fujisawa, Minamide and Tanigawa’s result 5 .
Theorem 2. Assume that the Riemann Hypothesis is true. For any real number \(x\ge 3\) and any integers \(k\ge 2\) and \(q \ge 1\), \[E_{k, q}(x) \ll \theta(q) x^{1/(k + 1)}\relax(C_{0} \frac{\relax x}{\relax\relax x}) \label{mainRH}\qquad{(7)}\] holds. Here \(C_0\) is an absolute positive constant.
Remark 2. Theorem 2 completely solves the conditional part of Bege’s conjecture ?? . If \(q = 1\), it also gives a little refinement to Martin and Yip’s result 6 .
Next, we consider the mean square of the function \[\Xi_{k, q}(x) \coloneq E_{k, q}(x) - B_{k, q} x^{1/(k + 1)}\] where \[\begin{align} B_{k, q} &\coloneq \frac{\varphi(q) \J{1/(k + 1)}{q} q^{(k - 2)/(k + 1)} \zeta(1/(k + 1))}{\J^{2}{k/(k + 1)}{q} \zeta^{2}(k/(k + 1))}\\ &\phantom{{}\coloneq{}}\quad \times \prod_{p \nmid q} \frac{\lr(1 - 2 p^{-k/(k + 1)} + p^{-1})\lr(1 - p^{-1})}{\lr(1 - p^{-k/(k + 1)})^{2}} \end{align}\label{eq:Bkq}\tag{9}\] and \[\J{s}{q} \coloneq q^{s} \prod_{p \mid q} \lr(1 - p^{-s}). \label{eq:jordan}\tag{10}\] In other words, \(\Xi_{k, q}(x)\) measures the discrepancy of the size of the error term \(E_{k, q}(x)\) from its expected size.
Theorem 3. For any real number \(T \ge 3\) and any integers \(k \ge 2\) and \(q \ge 1\), we have \[\int^{T}_{1} \Xi_{k, q}^{2}(x) \mathop{{\operatorfont{d}}x} \ll C_{k, q}^{2} T^{1 + 1/k} \relax(G_{0}^{k} \lr(\relax T)^{G_{1} k/(G_{1} k + 1)}) \label{eq:sq}\qquad{(8)}\] where \[C_{k, q} \coloneq \prod_{p \mid q} \frac{\lr(1 + p^{-1/(2 k)}) \lr(1 + p^{-(k + 1)/(2 k)})}{\displaystyle \lr(1 - p^{-1/2})^{2} \lr(1 - p^{-1}) \lr(1 - p^{-(k + 1)/k})} \label{eq:Ckn}\qquad{(9)}\] and \(G_{0}\) and \(G_{1}\) are absolute positive constants.
Note 1. For any real number \(\varepsilon > 0\) and any integer \(k \ge 2\) and \(q \ge 1\), we have \[C_{k, q} \ll_{\varepsilon} \theta^{1 + \varepsilon}(q)\] and \[C_{k, q} \ll_{k, \varepsilon} \theta^{\varepsilon}(q).\]
Lemma 1. For any integer \(k \ge 2\), we decompose the function \(\mu_{k}\) as follows. \[\mu_{k} = f_{k} \ast c_{k} \label{eq:decomp1}\qquad{(10)}\] where \[f_{k}(n) \coloneq \sum_{d^{k} \mid n}\mmu(d). \label{eq:fk}\qquad{(11)}\] Then the function \(c_{k}\) satisfies \[\sum_{n = 1}^{\infty}\frac{c_{k}(n)}{n^{s}} = \prod_{p}\frac{1 - 2 p^{-k s} + p^{-(k + 1) s}}{\lr(1 - p^{-k s})^{2}} \label{eq:ck}\qquad{(12)}\] for \(\relax(s) > 1/(k + 1)\). The Dirichlet series and the Euler product in ?? converge absolutely for \(\relax(s) > 1/(k + 1)\).
Proof. See the proof of Lemma 2.5 in [4]. ◻
Lemma 2. For any integers \(k \ge 2\) and \(n \ge 1\), we have \[c_{k}(n) = \sum_{d^{k + 1} \mid n} b_{k}\lr(\frac{n}{d^{k + 1}}) \label{eq:ckbk}\qquad{(13)}\] where \[\sum_{n = 1}^{\infty} \frac{b_{k}(n)}{n^{s}} = \prod_{p} \frac{\lr(1 - 2 p^{-k s} + p^{-(k + 1) s})\lr(1 - p^{-(k + 1) s})}{\lr(1 - p^{-k s})^{2}} \label{eq:bk}\qquad{(14)}\] for \(\relax(s) > 1/(2 k)\). Both sides of Equation ?? converge absolutely for \(\relax(s) > 1/(2k)\). In addition, \[\sum_{n = 1}^{\infty}\frac{\left|b_{k}(n)\right|}{n^{1/(k + 1)}} \ll 1 \label{eq:absbk}\qquad{(15)}\] holds.
Proof. Since \[\begin{align} \frac{\lr(1 - 2 p^{-k s} + p^{-(k + 1) s})\lr(1 - p^{-(k + 1) s})}{\lr(1 - p^{-k s})^{2}} &= 1 - p^{-2 k s} \lr(\frac{1 - p^{-s}}{1 - p^{-k s}})^{2}, \end{align}\] we obtain \[\left|b_{k}(n)\right| \le b^{\ast}_{k}(n)\] where the function \(b^{\ast}_{k}\) is defined by the following equation: \[\sum_{n = 1}^{\infty} \frac{b^{\ast}_{k}(n)}{n^{s}} = \prod_{p} \lr(1 + p^{-2 k s} \lr(\frac{1 + p^{-s}}{1 - p^{-k s}})^{2}).\] For \(\sigma > 1/(2k)\), we see that \[\lr(\frac{1 + p^{-\sigma}}{1 - p^{-k \sigma}})^{2} \ll 1.\] We then immediately obtain Lemma 2. ◻
Lemma 3. For any integer \(k \ge 2\), we have \[\sum_{n = 1}^{\infty} \frac{b_{k}(n)}{n^{s}} = \frac{\zeta^{2}((2 k + 1) s)}{\zeta(2 k s)\zeta((2 k + 2) s)} D_{k}(s)\] for \(\relax(s) > 1/(2 k)\). Here \[\begin{align} D_{k}(s) &= \sum_{n = 1}^{\infty} \frac{d_{k}(n)}{n^{s}}\\ &= \prod_{p} \frac{\lr(1 - 2 p^{-k s} + p^{-(k + 1)s}) \lr(1 - p^{-(k + 1) s}) \lr(1 - p^{-(2 k + 1) s})^{2}}{\lr(1 - p^{-k s})^{2} \lr(1 - p^{-2 k s}) \lr(1 - p^{-(2 k + 2) s})} \end{align} \label{eq:dk}\qquad{(16)}\] for \(\relax(s) > 1/(3 k)\). The Dirichlet series and the Euler product in ?? converge absolutely for \(\relax(s) > 1/(3 k)\). Furthermore, \[\sum_{n = 1}^{\infty} \frac{\left|d_{k}(n)\right|}{n^{1/(2 k)}} \ll 1\] holds.
Proof. Since \[\begin{align} &\frac{\lr(1 - 2 p^{-k s} + p^{-(k + 1)s}) \lr(1 - p^{-(k + 1) s}) \lr(1 - p^{-(2 k + 1) s})^{2}}{\lr(1 - p^{-k s})^{2} \lr(1 - p^{-2 k s}) \lr(1 - p^{-(2 k + 2) s})}\\ &\quad = 1 - p^{-3 k s} \lr(1 - p^{-k s})^{-2} \lr(1 - p^{-2 k s})^{-1} \lr(1 - p^{-(2 k + 2) s})^{-1}\\ &\quad \phantom{{} = {}} \quad \times \left(\vphantom{2 - 4 p^{-s} + 2 p^{-2 s} - p^{-k s} + 3 p^{-(k + 2) s} - 2 p^{-(k + 3) s} + p^{-(3 k + 2)} - 2 p^{-(3 k + 3) s} + p^{-(3 k + 4) s}} 2 - 4 p^{-s} + 2 p^{-2 s} - p^{-k s} + 3 p^{-(k + 2) s}\right.\\ &\quad \phantom{{} = {}} \quad \phantom{{} \times \left(\vphantom{2 - 4 p^{-s} + 2 p^{-2 s} - p^{-k s} + 3 p^{-(k + 2) s} - 2 p^{-(k + 3) s} + p^{-(3 k + 2)} - 2 p^{-(3 k + 3) s} + p^{-(3 k + 4) s}}\right.}\quad \left.{} - 2 p^{-(k + 3) s} + p^{-(3 k + 2)} - 2 p^{-(3 k + 3) s} + p^{-(3 k + 4) s} \vphantom{2 - 4 p^{-s} + 2 p^{-2 s} - p^{-k s} + 3 p^{-(k + 2) s} - 2 p^{-(k + 3) s} + p^{-(3 k + 2)} - 2 p^{-(3 k + 3) s} + p^{-(3 k + 4) s}}\right), \end{align}\] we find that \[\left|d_{k}(n)\right| \le d^{\ast}_{k}(n)\] where the function \(d^{\ast}_{k}\) is defined by the following equation: \[\begin{align} &\sum_{n = 1}^{\infty} \frac{d^{\ast}_{k}(n)}{n^{s}}\\ &\quad = \prod_{p} \left(\vphantom{1 + p^{-3 k s} \lr(1 - p^{-k s})^{-2} \lr(1 - p^{-2 k s})^{-1} \lr(1 - p^{-(2 k + 2) s})^{-1} \left(2 + 4 p^{-s} + 2 p^{-2 s} + p^{-k s} + 3 p^{-(k + 2) s} + 2 p^{-(k + 3) s} + p^{-(3 k + 2)} + 2 p^{-(3 k + 3) s} + p^{-(3 k + 4) s}\right)} 1 + p^{-3 k s} \lr(1 - p^{-k s})^{-2} \lr(1 - p^{-2 k s})^{-1} \lr(1 - p^{-(2 k + 2) s})^{-1}\right.\\ &\quad \phantom{{} = \prod_{p} \left(\vphantom{1 + p^{-3 k s} \lr(1 - p^{-k s})^{-2} \lr(1 - p^{-2 k s})^{-1} \lr(1 - p^{-(2 k + 2) s})^{-1} \left(2 + 4 p^{-s} + 2 p^{-2 s} + p^{-k s} + 3 p^{-(k + 2) s} + 2 p^{-(k + 3) s} + p^{-(3 k + 2)} + 2 p^{-(3 k + 3) s} + p^{-(3 k + 4) s}\right)}\right.} \quad \times \left(\vphantom{2 + 4 p^{-s} + 2 p^{-2 s} + p^{-k s} + 3 p^{-(k + 2) s} + 2 p^{-(k + 3) s} + p^{-(3 k + 2)} + 2 p^{-(3 k + 3) s} + p^{-(3 k + 4) s}} 2 + 4 p^{-s} + 2 p^{-2 s} + p^{-k s} + 3 p^{-(k + 2) s} + 2 p^{-(k + 3) s}\right.\\ &\quad \phantom{{} = \prod_{p} \left(\vphantom{1 + p^{-3 k s} \lr(1 - p^{-k s})^{-2} \lr(1 - p^{-2 k s})^{-1} \lr(1 - p^{-(2 k + 2) s})^{-1} \left(2 + 4 p^{-s} + 2 p^{-2 s} + p^{-k s} + 3 p^{-(k + 2) s} + 2 p^{-(k + 3) s} + p^{-(3 k + 2)} + 2 p^{-(3 k + 3) s} + p^{-(3 k + 4) s}\right)}\right.} \quad \phantom{{} \times \left(\vphantom{2 + 4 p^{-s} + 2 p^{-2 s} + p^{-k s} + 3 p^{-(k + 2) s} + 2 p^{-(k + 3) s} + p^{-(3 k + 2)} + 2 p^{-(3 k + 3) s} + p^{-(3 k + 4) s}}\right.} \quad \left.\left.{} + p^{-(3 k + 2)} + 2 p^{-(3 k + 3) s} + p^{-(3 k + 4) s} \vphantom{2 + 4 p^{-s} + 2 p^{-2 s} + p^{-k s} + 3 p^{-(k + 2) s} + 2 p^{-(k + 3) s} + p^{-(3 k + 2)} + 2 p^{-(3 k + 3) s} + p^{-(3 k + 4) s}}\right) \vphantom{1 + p^{-3 k s} \lr(1 - p^{-k s})^{-2} \lr(1 - p^{-2 k s})^{-1} \lr(1 - p^{-(2 k + 2) s})^{-1} \left(2 + 4 p^{-s} + 2 p^{-2 s} + p^{-k s} + 3 p^{-(k + 2) s} + 2 p^{-(k + 3) s} + p^{-(3 k + 2)} + 2 p^{-(3 k + 3) s} + p^{-(3 k + 4) s}\right)}\right). \end{align}\] For \(\sigma > 1/(3 k)\), we see that \[\begin{align} &\lr(1 - p^{-k \sigma})^{-2} \lr(1 - p^{-2 k \sigma})^{-1} \lr(1 - p^{-(2 k + 2) \sigma})^{-1}\\ &\quad \times \left(\vphantom{2 + 4 p^{-\sigma} + 2 p^{-2 \sigma} + p^{-k \sigma} + 3 p^{-(k + 2) \sigma} + 2 p^{-(k + 3) \sigma} + p^{-(3 k + 2)} + 2 p^{-(3 k + 3) \sigma} + p^{-(3 k + 4) \sigma}} 2 + 4 p^{-\sigma} + 2 p^{-2 \sigma} + p^{-k \sigma} + 3 p^{-(k + 2) \sigma} \right.\\ &\quad \phantom{{} \times \left(\vphantom{2 + 4 p^{-\sigma} + 2 p^{-2 \sigma} + p^{-k \sigma} + 3 p^{-(k + 2) \sigma} + 2 p^{-(k + 3) \sigma} + p^{-(3 k + 2)} + 2 p^{-(3 k + 3) \sigma} + p^{-(3 k + 4) \sigma}}\right.} \quad \left. {} + 2 p^{-(k + 3) \sigma} + p^{-(3 k + 2)} + 2 p^{-(3 k + 3) \sigma} + p^{-(3 k + 4) \sigma} \vphantom{2 + 4 p^{-\sigma} + 2 p^{-2 \sigma} + p^{-k \sigma} + 3 p^{-(k + 2) \sigma} + 2 p^{-(k + 3) \sigma} + p^{-(3 k + 2)} + 2 p^{-(3 k + 3) \sigma} + p^{-(3 k + 4) \sigma}}\right)\\ &\quad \ll 1. \end{align}\] The above immediately gives Lemma 3. ◻
Lemma 4 ([6]). For any real number \(x \ge 0\) and any integer \(q \ge 1\), we have \[\sum_{\substack{n \le x \\ (n, q) = 1}}1 = \frac{\varphi(q)}{q} x + \relax(\theta(q)). \label{eq:cohen}\qquad{(17)}\]
Lemma 5. For any real number \(x \ge 1\), we have \[\sum_{n \le x} \tau(n) \ll x \relax(1 + x).\]
Proof. This lemma follows from [7]. ◻
Lemma 6. For any real number \(x \ge 1\), any integer \(q \ge 1\) and any multiplicative function \(f\), we have \[\sum_{\substack{n \le x \\ (n, q) = 1}} f(n) = \sum_{\substack{d \le x \\ \rad(d) \mid n}} f^{\ast(-1)}(d) \sum_{l \le x/d} f(l) \label{eq:coprime}\qquad{(18)}\] where \(f^{\ast(-1)}\) is the Dirichlet inverse function of \(f\) and \[\rad(d) \coloneq \prod_{p \mid d} p. \label{eq:rad}\qquad{(19)}\]
Proof. For any integer \(n \ge 1\), we see that \[f(n) = f\lr(\lr(n, q^{n}) \frac{n}{\lr(n, q^{n})}) = f\lr(\lr(n, q^{n})) f\lr(\frac{n}{\lr(n, q^{n})}) = \sum_{\substack{a b = n \\ \rad(a) \mid q \\ (b, q) = 1}} f(a) f(b) \label{eq:coprime-1}\tag{11}\] and \[\sum_{\substack{d l = n \\ \rad(d) \mid q \\ \rad(l) \mid q}} f^{\ast(-1)}(d) f(l) = \sum_{\substack{d l = n \\ \rad(n) \mid q}} f^{\ast(-1)}(d) f(l) = \left\lfloor\frac{1}{n}\right\rfloor. \label{eq:coprime-2}\tag{12}\] Combinig 11 and 12 , we obtain \[\begin{align} \sum_{\substack{d l = n \\ \rad(d) \mid q}} f^{\ast(-1)}(d) f(l) &= \sum_{\substack{d a b = n \\ \rad(d) \mid q \\ \rad(a) \mid q \\ (b, q) = 1}} f^{\ast(-1)}(d) f(a) f(b)\\ &= \sum_{\substack{m b = n \\ (b, q) = 1}} \left\lfloor\frac{1}{m}\right\rfloor f(b)\\ &= f(n) \left\lfloor\frac{1}{(n, q)}\right\rfloor. \end{align}\label{eq:coprime-3}\tag{13}\] Summing up 13 with respect to all positive integer \(n \le x\) concludes the proof. ◻
Lemma 7. For any real number \(x \ge 1\), we have \[\sum_{n \le x}\mmu(n) \ll x \dlt(x) \label{eq:mmu}\qquad{(20)}\] where \[\dlt(x) \coloneq \relax(-\frac{D_{1} \lr(\relax(x + e^{e^{1/3}}))^{3/5}}{\displaystyle \lr(\relax\relax(x + e^{e^{1/3}}))^{1/5}})\] and \(D_{1}\) is an absolute constant.
Proof. See [5]. ◻
Note 2. \(\delta(x)\) is decreasing function. The inequality \[\dlt(xy) \ge \dlt(x) \dlt(y)\] holds for any real numbers \(x \ge 0\) and \(y \ge 0\).
Lemma 8. For any real number \(x \ge 1\) and any integer \(q \ge 1\), we have \[\sum_{\substack{n \le x \\ (n, q) = 1}} \mmu(n) \ll \theta(q) x \dlt(x).\]
Proof. By Lemma 6 and Lemma 7, we obtain \[\begin{align} \sum_{\substack{n \le x \\ (n, q) = 1}} \mmu(n) &= \sum_{\substack{d \le x \\ \rad(d) \mid q}} \tau(d) \sum_{\substack{l \le x/d}} \mmu(l)\\ &\ll \sum_{\substack{d \le x \\ \rad(d) \mid q}} \tau(d) \frac{x}{d} \dlt(\frac{x}{d})\\ &\le x \dlt(x) \sum_{\rad(d) \mid q} \frac{\tau(d)}{d \dlt(d)}\\ &\ll x \dlt(x) \sum_{\rad(d) \mid q} \frac{\tau(d)}{d^{1/2}}\\ &= \frac{q}{\J^{2}{1/2}{q}} x \dlt(x)\\ &\ll \theta(q) x \dlt(x). \end{align}\] ◻
Lemma 9. For any real number \(x \ge 1\) and any integers \(k \ge 2\) and \(q \ge 1\), we have \[\sum_{\substack{n \le x \\ (n, q) = 1}} \frac{\mmu(n)}{n^{k}} = \frac{q^{k}}{\zeta^{2}(k) \J^{2}{k}{q}} + \relax(\theta(q) x^{-k + 1} \dlt(x)).\]
Proof. Applying Abel summation formula and Lemma 8, we obtain \[\begin{align} \sum_{\substack{n \le x \\ (n, q) = 1}} \frac{\mmu(n)}{n^{k}} &= \frac{q^{k}}{\zeta^{2}(k) \J^{2}{k}{q}} - \sum_{\substack{n > x \\ (n, q) = 1}} \frac{\mmu(n)}{n^{k}}\\ &= \frac{q^{k}}{\zeta^{2}(k) \J^{2}{k}{q}} + x^{-k} \sum_{\substack{n \le x \\ (n, q) = 1}} \mmu(n)\\ &\phantom{{}={}} \quad - k \int^{\infty}_{x} t^{-k - 1} \sum_{\substack{n \le t \\ (n, q) = 1}} \mmu(n) \mathop{{\operatorfont{d}}t}\\ &= \frac{q^{k}}{\zeta^{2}(k) \J^{2}{k}{q}} + \relax(\theta(q) x^{-k + 1} \dlt(x)). \end{align}\] ◻
Lemma 10. For any real numbers \(x \ge 1\) and \(0 < \sigma < 1\), we have \[\sum_{n \le x} n^{-\sigma} \ll \frac{x^{1 - \sigma}}{1 - \sigma}. \label{eq:zetapoly}\qquad{(21)}\]
Proof. This lemma follows from [7]. ◻
Lemma 11. For any real number \(x \ge 1\) and any integer \(k \ge 2\), we have \[\sum_{n \le x} \left|c_{k}(n)\right| \ll x^{1/(k +1)}.\]
Proof. By ?? , we see that \[\left|c_{k}(n)\right| \le \sum_{d^{k + 1} \mid n} \left|b_{k}\lr(\frac{n}{d^{k + 1}})\right|.\] Hence by ?? , we obtain \[\begin{align} \sum_{n \le x} \left|c_{k}(n)\right| &\le \sum_{l \le x} \left|b_{k}(l)\right| \sum_{d \le (x/l)^{1/(k + 1)}} 1\\ &\le x^{1/(k + 1)} \sum_{l \le x} \frac{\left|b_{k}(l)\right|}{l^{1/(k + 1)}}\\ &\ll x^{1/(k + 1)}. \end{align}\] ◻
Lemma 12. For any real number \(x \ge 1\) and any integers \(k \ge 2\) and \(q \ge 1\), we have \[\sum_{\substack{n \le x \\ (n, q) = 1}} \frac{c_{k}(n)}{n} = \frac{\zeta^{2}(k) \J^{2}{k}{q} A_{k, q}}{\varphi(q) q^{2 k - 1}} + \relax(x^{-1 + 1/(k + 1)}). \label{eq:ck1}\qquad{(22)}\]
Proof. Applying Abel summation formula and Lemma 11 easily completes the proof. ◻
Lemma 13. For any real number \(x \ge 1\) and any integer \(k \ge 2\), we have \[\sum_{n \le x} \frac{\left|c_{k}(n)\right|}{n^{1/(k + 1)}} \ll \relax(1 + x^{1/k}).\]
Proof. We again apply Abel summation formula and Lemma 11. ◻
Lemma 14 ([8]). For any arithmetic function \(\lr(a_{n})_{n \ge 1}\) and any real numbers \(\sigma_{0} > \relax\{0, \sigma_{\mathrm{a}}\}\), \(x > 0\) and \(T > 0\), we have \[\begin{align} \sideset{}{^\prime}{\sum}_{n \le x} a_{n} &= \frac{1}{2 \pi i} \int^{\sigma_{0} + iT}_{\sigma_{0} - iT} \sum_{n = 1}^{\infty} \frac{a_{n}}{n^{s}} \frac{x^{s}}{s} \mathop{{\operatorfont{d}}s}\\ &\phantom{{}={}} \quad + \relax(\sum_{\substack{x/2 < n < 2 x \\ n \ne x}} \left|a_{n}\right| \relax\{1, \frac{x}{T \left|x - n\right|}\}) \end{align}\] where \(\sigma_{\mathrm{a}}\) is the abscissa of absolute convergence of the Dirichlet series of \(\lr(a_{n})_{n \ge 1}\) and \[\sideset{}{^\prime}{\sum}_{n \le x} = \frac{1}{2} \lr(\sum_{n < x} + \sum_{n \le x}).\]
Lemma 15. For any real numbers \(c \ge 0\), \(\theta > 0\) and \(x \ge 0\), we define \[\begin{align} \omg{c}[\theta]{x} &\coloneq \relax(\frac{c \relax(x + e^{e^{2}})}{\displaystyle \relax\relax(x + e^{e^{2}})} \relax\{1, \relax\frac{e}{\displaystyle \theta \relax\relax(x + e^{e^{2}})}\}), \\ \omg{c}{x} &\coloneq \relax(\frac{c \relax(x + e^{e^{2}})}{\displaystyle \relax\relax(x + e^{e^{2}})}). \end{align}\] Then the following holds.
The function \(\omg{c}[\theta]{x}\) is increasing in \(c\) and \(x\) and decreasing in \(\theta\).
For any real numbers \(c \ge 0\), \(\theta > 0\) and \(x \ge 0\), we have \[\omg{c}[\theta]{x}=\omg{c}[\relax\{\theta, \frac{1}{\displaystyle \relax\relax(x + e^{e^{2}})}\}]{x}.\] In particular, we have \[\omg{c}[\theta]{x} = \omg{c}{x}\] if \(\theta \ge 1/\relax\relax(x + e^{e^{2}})\).
For any real numbers \(c \ge 0\), \(\theta > 0\), \(x \ge 0\) and \(y \ge 0\), we have \[\begin{align} \omg{c}[\theta]{x + y} &\le \omg{c}[\theta]{x}\omg{c}[\theta]{y}, \\ \omg{c}[\theta]{x y} &\le \omg{c}[\theta]{x}\omg{c}[\theta]{y}. \end{align}\]
For any real numbers \(c \ge 0\), \(\theta > 0\), \(x \ge 0\) and \(a \ge 1\), we have \[\omg{c}[\frac{\theta}{a}]{x} \le \omg{c(1+\relax a)}[\theta]{x}.\]
For any real numbers \(x \ge 0\) and \(\theta > 0\), we have \[\relax(1+\frac{x}{\theta}) \ll \omg{1}[\theta]{x}.\]
All the results in Lemma 15 can be proven easily, hence we omit the proof.
Lemma 16. For any integer \(n \ge 1\), we have \[\tau(n) \ll \omg{1}{n}. \label{eq:div-max}\qquad{(23)}\]
Proof. This lemma follows from [9]. ◻
Lemma 17. Assume that the Riemann Hypothesis is true. For \(\sigma > 1/2\), \[\left|\zeta^{-1}(s)\right| \le \omg{C_{1}}[\sigma - \frac{1}{2}]{\left|t\right|}\] holds. Here \(C_{1}\) is an absolute positive constant.
Proof. This lemma follows from [8]. ◻
Lemma 18. Assume that the Riemann Hypothesis is true. Then \[\left|\zeta(s)\right| \le \omg{C_{2}}{\left|t\right|} \lr(1 + \frac{1}{\left|s\right|})\] holds for any complex number \(s\) with \(\relax(s) \ge 1/2\) and \(s \ne 1\).
Proof. This lemma follows from [8]. ◻
Lemma 19. Assume that the Riemann Hypothesis is true. We have \[\zeta(s) \ll \lr(1 + \left|t\right|)^{1/2 - \sigma} \omg{C_{2}}{\left|t\right|}\] for \(0 \le \sigma \le 1/2\).
Proof. This lemma follows from [8]. ◻
Lemma 20. Assume that the Riemann Hypothesis is true. For any real number \(Y \ge 1\) and any integer \(q \ge 1\), we have \[\begin{align} g_{Y, q}(s) &\coloneq \zeta^{-2}(s) \prod_{p \mid q} \lr(1 - p^{-s})^{-2} - \sum_{\substack{n \le Y \\ (n, q) = 1}} \frac{\mmu(n)}{n^{s}}\\ &\ll 2^{\sigma} \kappa(q) Y^{1/2 - \sigma} \omg{C_{3}}[\sigma - \frac{1}{2}]{Y^{\sigma}} \omg{C_{3}}[\sigma - \frac{1}{2}]{\left|t\right|} \end{align}\] for \(\sigma > 1/2\). Here \[\kappa(q) \coloneq \prod_{p \mid q} \lr(1 - p^{-1/2})^{-2}\] and \(C_{3}\) is an absolute positive constant.
Proof. By Lemma 14 and Lemma 16, we see that \[\sum_{\substack{n \le Y \\ (n, q) = 1}} \frac{\mmu(n)}{n^{s}} = \frac{1}{2 \pi i} \int^{\alpha + i T}_{\alpha - i T} \zeta^{-2}(s + w)\prod_{p \mid q} \lr(1 - p^{-(s + w)})^{-2} \frac{Y^{w}}{w} \mathop{{\operatorfont{d}}w} + R\] where \[\begin{align} \alpha &= \frac{1}{2} + \frac{1}{\relax(1 + Y)}\\ T &= Y^{\sigma} \end{align}\] and \[\begin{align} R &\ll 2^{\sigma} Y^{-\sigma} \omg{1}{2 Y} \lr(1 + \frac{Y \relax(1 + Y)}{T}) + \frac{Y^{1/2} \lr(\relax(1 + Y))^{2}}{T}\\ &\ll 2^{\sigma} Y^{1/2 - \sigma} \omg{3}{Y^{\sigma}}. \end{align}\] We rewrite the above integral as \[\begin{align} &\frac{1}{2 \pi i} \int^{\alpha + i T}_{\alpha - i T} \zeta^{-2}(s + w)\prod_{p \mid q} \lr(1 - p^{-(s + w)})^{-2} \frac{Y^{w}}{w} \mathop{{\operatorfont{d}}w}\\ &\quad = \frac{1}{2 \pi i} \left(\int^{\alpha + i T}_{\beta + i T} + \int^{\beta + i T}_{\beta - i T} + \int^{\beta - i T}_{\alpha - i T}\right) \zeta^{-2}(s + w)\prod_{p \mid q} \lr(1 - p^{-(s + w)})^{-2} \frac{Y^{w}}{w} \mathop{{\operatorfont{d}}w}\\ &\quad \phantom{{}={}} \quad + \zeta^{-2}(s) \prod_{p \mid q} \lr(1 - p^{-s})^{-2}\\ &\quad \eqcolon I_{1} + I_{2} + I_{3} + \zeta^{-2}(s) \prod_{p \mid q} \lr(1 - p^{-s})^{-2} \end{align}\] where \[\beta = \frac{1}{2} - \sigma + \relax\{\frac{1}{\displaystyle \relax\relax(\left|t\right| + T + e^{e^{2}})}, \frac{1}{2} \lr(\sigma - \frac{1}{2})\}.\] By Lemma 17, we have \[\begin{align} I_{1} &\ll \omg{2 C_{1}}[\sigma + \beta - \frac{1}{2}]{\left|t + T\right|} \kappa(q) \frac{Y^{\alpha}}{T}\\ &\ll \kappa(q) Y^{1/2 - \sigma} \omg{4C}[\sigma - \frac{1}{2}]{\left|t\right| + Y^{\sigma}}, \\ I_{2} &\ll \omg{2 C_{1}}[\sigma + \beta - \frac{1}{2}]{\left|t\right| + T} \kappa(q) Y^{\beta} \int^{T}_{-T}\frac{\mathop{{\operatorfont{d}}v}}{\left|\beta\right| + \left|v\right|}\\ &\ll \kappa(q) \omg{4 C_{1}}[\sigma - \frac{1}{2}]{\left|t\right| + Y^{\sigma}}\\ &\phantom{{}\ll {}}\quad \times Y^{1/2 - \sigma} \relax(\frac{\relax Y}{\displaystyle \relax\relax(\left|t\right| + T + e^{e^{2}})}) \relax(1 + \frac{T}{\left|b\right|})\\ &\ll \kappa(q) Y^{1/2 - \sigma} \omg{4 C_{1} + 4}[\sigma - \frac{1}{2}]{\left|t\right| + Y^{\sigma}} \intertext{and} I_{3} &\ll \omg{4 C_{1}}[\sigma + \beta - \frac{1}{2}]{\left|t - T\right|} \kappa(q) \frac{Y^{\alpha}}{T}\\ &\ll \kappa(q) Y^{1/2 - \sigma} \omg{4 C_{1}}[\sigma - \frac{1}{2}]{\left|t\right| + Y^{\sigma}}. \end{align}\] Combining the above estimates completes the proof. ◻
We begin by estimating the sum \(\sum_{n \le x, (n, q) = 1} f_{k}(n)\). Let \(z = x^{1/k}\) and \(0 < \rho \le 1\) and split the sum as follows: \[\begin{align} \sum_{\substack{n \le x \\ (n, q) = 1}} f_{k}(n) &= \sum_{\substack{d^{k} l \le x \\ (d, q) = (l, q) = 1}} \mmu(d)\\ &= \lr(\sum_{\substack{d^{k} l \le x \\ d \le \rho z \\ (d, q) = (l, q) = 1}} + \sum_{\substack{d^{k} l \le x \\ l \le \rho^{-k} \\ (d, q) = (l, q) = 1}} - \sum_{\substack{d \le \rho z \\ l \le \rho^{-k} \\ (d, q) = (l, q) = 1}}) \mmu(d)\\ &\eqcolon S_{1} + S_{2} - S_{3}. \end{align}\] By Lemma 4, Lemma 9 and Lemma 5, we see that \[\begin{align} S_{1} &= \sum_{\substack{d \le \rho z \\ (d, q) = 1}} \mmu(d) \sum_{\substack{l \le x/d^{k} \\ (l, q) = 1}} 1\\ &= \frac{\varphi(q)}{q} x \sum_{\substack{d \le \rho z \\ (d, q) = 1}} \frac{\mmu(d)}{d^{k}} + \relax(\theta(q) \sum_{\substack{d \le \rho z \\ (d, q) = 1}} \left|\mmu(d)\right|)\\ &= \frac{\varphi(n) n^{2 k - 1}}{\zeta^{2}(k) \J^{2}{k}{n}} x + \relax(\theta(q) \rho^{1 - k} z \dlt(\rho z)) +\relax(\theta(q) \rho z \relax(1 + \rho z)). \end{align}\label{eq:s1}\tag{14}\] Applying Lemma 8 and Lemma 10, we obtain \[\begin{align} S_{2} &= \sum_{\substack{l \le \rho^{-k} \\ (l, q) = 1}} \sum_{\substack{d \le (x/l)^{1/k} \\ (d, q) = 1}} \mmu(d)\\ &\ll \theta(q) x^{1/k} \sum_{\substack{l \le \rho^{-k} \\ (l, q) = 1}} l^{-1/k} \dlt(\lr(\frac{x}{l})^{1/k})\\ &\ll \theta(q) \rho^{1 - k} z \dlt(\rho z). \end{align}\label{eq:s2}\tag{15}\] Lemma 8 also gives \[S_{3} = \sum_{\substack{d \le \rho z \\ (d, q) = 1}} \mmu(d) \sum_{\substack{l \le \rho^{-k} \\ (l, q) = 1}} 1 \ll \theta(q) \rho^{1 - k} z \dlt(\rho z). \label{eq:s3}\tag{16}\] Putting \(\rho = \dlt^{1/k}(z \dlt^{1/k}(z))\), we have \[\begin{align} \rho^{1 - k} z \dlt(\rho z) &= \rho z \dlt^{-1}(z \dlt^{1/k}(z)) \dlt(z \dlt^{1/k}(z \dlt^{1/k}(z)))\\ &\le \rho z \dlt^{-1}(z \dlt^{1/k}(z)) \dlt(z \dlt^{1/k}(z))\\ &= \rho z\\ &= z \dlt^{1/k}(z \dlt^{1/k}(z))\\ &\le z \dlt^{1/k}(z^{1/2})\\ &\le z \dlt^{1/(2k)}(z). \end{align}\label{eq:rhoz}\tag{17}\] Combining 14 , 15 , 16 and 17 , we find that \[\sum_{\substack{n \le x \\ (n, q) = 1}} f_{k}(n) = \frac{\varphi(q) q^{2 k - 1}}{\zeta^{2}(k) \J^{2}{k}{n}} x + \relax(\theta(q) x^{1/k}\dlt^{1/(2k)}(x^{1/k})\relax(1+x^{1/k})).\] Therefore, applying Lemma 1, Lemma 12 and Lemma 13, we obtain \[\begin{align} \sum_{\substack{n \le x \\ (n, q) = 1}} \mu_{k}(n) &= \sum_{\substack{d \le x \\ (d, n) = 1}} c_{k}(d) \sum_{\substack{l \le x/d \\ (l, n) = 1}} f_{k}(l)\\ &= \frac{\varphi(q) q^{2 k - 1}}{\zeta^{2}(k) \J^{2}{k}{n}} x \sum_{\substack{d \le x \\ (d, q) = 1}} \frac{c_{k}(d)}{d}\\ &\phantom{{}={}} + O\left(\theta(q) x^{1/k} \dlt^{1/(2k)}(x^{1/k}) \relax(1 + x^{1/k}) \vphantom{\sum_{\substack{d \le x \\ (d, q) = 1}} \frac{\left|c_{k}(d)\right|}{d^{1/k} \dlt^{1/(2k)}(d^{1/k})}}\right.\\ &\phantom{\phantom{{}={}} + O\left(\vphantom{\sum_{\substack{d \le x \\ (d, q) = 1}} \frac{\left|c_{k}(d)\right|}{d^{1/k} \dlt^{1/(2k)}(d^{1/k})}}\right.} \quad \left. {}\times \sum_{\substack{d \le x \\ (d, q) = 1}} \frac{\left|c_{k}(d)\right|}{d^{1/k} \dlt^{1/(2k)}(d^{1/k})}\right)\\ &= A_{k, q} x + \relax(\theta(q) x^{1/k} \dlt^{1/(2k)}(x^{1/k}) \lr(\relax(1 + x^{1/k}))^{2})\\ &= A_{k, q} x + \relax_{k}(\theta(q) x^{1/k} \dlt^{1/(3 k)}(x^{1/k})). \end{align}\] Finally, if \(x \ge 3\), we see that \[\begin{align} \dlt^{1/(3 k)}(x^{1/k})&= \relax(-\frac{D_{1}}{3 k} \frac{\lr(\relax(x^{1/k} + e^{e^{1/3}}))^{3/5}}{\displaystyle \lr(\relax\relax(x^{1/k} + e^{e^{1/3}}))^{1/5}})\\ &\le \relax(-\frac{D_{1}}{3 k} \frac{\lr(k^{-1} \relax(x + e^{e^{1/3}}))^{3/5}}{\displaystyle \lr(\relax\relax(x + e^{e^{1/3}}))^{1/5}})\\ &\le \relax(-\frac{D_{1}}{4} k^{-8/5} \frac{\lr(\relax x)^{3/5}}{\lr(\relax\relax x)^{1/5}}), \end{align}\] and we are done.
We first split the sum \(\sum_{n \le x, (n, q) = 1} f_{k}(n)\) as follows: \[\begin{align} \sum_{\substack{n \le x \\ (n, q) = 1}} f_{k}(n) &= \sum_{\substack{d^{k} l \le x \\ (d, q) = (l, q) = 1}} \mmu(d)\\ &= \lr(\sum_{\substack{d \le Y \\ (d, q) = 1}} + \sum_{\substack{Y < d \le x^{1/k} \\ (d, q) = 1}}) \mmu(d) \sum_{\substack{l \le x/d^{k} \\ (l, q) = 1}} 1\\ &\eqcolon S_{1} + S_{2} \end{align}\] where \(Y \ge 1\). By Lemma 4 and Lemma 5, we see that \[\begin{align} S_{1} &= \sum_{\substack{d \le Y \\ (d, q) = 1}} \mmu(d) \sum_{\substack{l \le x/d^{k} \\ (l, q) = 1}} 1\\ &= \sum_{\substack{d \le Y \\ (d, q) = 1}} \mmu(d) \lr(\frac{\varphi(q)}{q} \frac{x}{d^{k}} + \relax(\theta(q)))\\ &= \frac{\varphi(q)}{q} x \sum_{\substack{d \le Y \\ (d, q) = 1}} \frac{\mmu(d)}{d^{k}} + \relax(\theta(q) Y \lr(1 + \relax Y)). \end{align}\] Since \[\left|\sum_{\substack{d^{k} \mid n \\ d > Y}} \mmu(d)\right| \le \sum_{d^{k} \mid n} \tau(d) \le \sum_{d \mid n} \tau(d) \le \sum_{d \mid n} \tau(n) = \tau^{2}(n),\] by Lemma 14 and Lemma 16, we obtain \[\begin{align} S_{2} &= \sum_{\substack{Y < d \le x^{1/k} \\ (d, q) = 1}} \mmu(d) \sum_{\substack{l \le x/d^{k} \\ (l, q) = 1}} 1 \\ &= \frac{1}{2 \pi i} \int^{\alpha + i T}_{\alpha - i T} \zeta(s) \prod_{p \mid q} \lr(1 - p^{-s}) g_{Y, q}(ks) \frac{x^{s}}{s} \mathop{{\operatorfont{d}}s} + R \end{align} \label{eq:s2-RH}\tag{18}\] where \[\begin{align} \alpha &= 1 + \frac{1}{\relax(1 + x)}, \\ T &= x \end{align}\] and \[\begin{align} R &\ll \omg{2}{2 x} \lr(1 + \frac{x \relax(1 + x)}{T}) + \frac{x \lr(\relax(1 + x))^{3}}{T}\\ &\ll \omg{3}{x}. \end{align}\] We move the line segment \([\alpha - i T, \alpha + i T]\) which is the contour for the first term on the right hand side of 18 to the contour consisting of the line segments \([\alpha - i T, 1/2 - i T]\), \([1/2 - i T, 1/2 + i T]\) and \([1/2 + i T, \alpha +i T]\). Then we have \[\begin{align} &\frac{1}{2 \pi i} \int^{\alpha + i T}_{\alpha - i T} \zeta(s) \prod_{p \mid q} \lr(1 - p^{-s}) g_{Y, q}(ks) \frac{x^{s}}{s} \mathop{{\operatorfont{d}}s}\\ &\quad = \frac{\varphi(q)}{q} x \lr(\frac{q^{2 k}}{\zeta^{2}(k) \J^{2}{k}{q}} - \sum_{\substack{n \le Y \\ (n, q) = 1}} \frac{\mmu(n)}{n^{k}}) + I_{1} + I_{2} + I_{3} \end{align}\] where \(I_{1}\) and \(I_{3}\) denote the integrals over the horizontal line segments and \(I_{2}\) is the integral over the vertical line segment. Applying Lemma 18 and Lemma 20, we have \[\begin{align} I_{1}, I_{3} &\ll \omg{C_{2}}{T} \eta(q) (2 e)^{k} \kappa(q) Y^{(1 - k)/2} \omg{3 C_{3}}{Y^{k}} \omg{C_{3}}{k} \omg{C_{3}}{T}\\ &\ll \theta(q) \relax(\lr(2 + C_{3}) k) Y^{(1 - k)/2} \omg{3 C_{3}}{Y^{k}} \omg{C_{2} + C_{3}}{x} \end{align}\] and \[\begin{align} I_{2} &\ll \omg{C_{2}}{T} \eta(q) 2^{k} \kappa(q) Y^{(1 - k)/2} \omg{C_{3}}{Y^{k}} \omg{C_{3}}{k} \omg{C_{3}}{T} x^{1/2} \relax(1 + T)\\ &\ll \theta(q) \relax(\lr(1 + C_{3}) k) x^{1/2} \omg{C_{2} + C_{3} + 1}{x} Y^{(1 - k)/2} \omg{C_{3}}{Y^{k}} \end{align}\] where \[\eta(q) \coloneq \prod_{p \mid q} \lr(1 + p^{-1/2}).\] Putting \(Y = \relax(2 C_{3} + 4) x^{1/(k + 1)}\), we obtain \[\sum_{\substack{n \le x \\ (n, q) = 1}} f_{k}(n) = \frac{\varphi(q) q^{2 k - 1}}{\zeta^{2}(k) \J^{2}{k}{q}} x + \relax(\theta(q) x^{1/(k + 1)} \omg{C_{2} + 4 C_{3} + 3}{x}).\] Here we note that \(S_{2} = 0\) if \(Y \ge x^{1/k}\). Finally, by Lemma 1, Lemma 12 and Lemma 13, we obtain \[\begin{align} \sum_{\substack{n \le x \\ (n, q) = 1}} \mu_{k}(n) &= \sum_{\substack{d \le x \\ (d, q) = 1}} c_{k}(d) \sum_{\substack{l \le x/d \\ (l, q) = 1}} f_{k}(l)\\ &= \frac{\varphi(q) q^{2 k - 1}}{\zeta^{2}(k) \J^{2}{k}{q}} x \sum_{\substack{d \le x \\ (d, q) = 1}} \frac{c_{k}(d)}{d}\\ &\phantom{{}={}} \quad + \relax(\theta(q) x^{1/(k + 1)} \omg{C_{2} + 4 C_{3} + 3}{x} \sum_{d \le x} \frac{\left|c_{k}(d)\right|}{d^{1/(k + 1)}})\\ &= A_{k, q} x + \relax(\theta(q) x^{1/(k + 1)} \omg{C_{2} + 4 C_{3} + 4}{x}). \end{align}\] Since \[\frac{\relax x}{\relax\relax x} \ll \frac{\relax(x + e^{e^{2}})}{\displaystyle \relax\relax(x + e^{e^{2}})}\] for \(x \ge 3\), this completes the proof.
By Lemma 14 and Lemma 3, we see that \[\begin{align} \sideset{}{^\prime}{\sum}_{\substack{n \le x \\ (n, q) = 1}} \mu_{k}(n) &= \frac{1}{2 \pi i} \int^{\alpha + i V}_{\alpha - i V} \frac{\zeta(s) \zeta((k + 1) s)}{\displaystyle \zeta^{2}(k s) \zeta(2 k s) \zeta((2 k + 2) s)}\\ &\hphantom{{} = \frac{1}{2 \pi i}\int} \times \prod_{p \mid q} \frac{\lr(1 - p^{-s}) \lr(1 - p^{-(k + 1) s})}{\displaystyle \lr(1 - p^{-k s})^{2} \lr(1 - p^{-2 k s}) \lr(1 - p^{-(2 k + 2) s})}\\ &\hphantom{{} = \frac{1}{2 \pi i}\int} \times \lr(\sum_{\substack{n = 1 \\ (n, q) = 1}}^{\infty} \frac{1}{n^{(2 k + 1) s}})^{2} D_{k, q}(s) \frac{x^{s}}{s} \mathop{{\operatorfont{d}}s}\\ &\hphantom{{} = {}} \quad + \relax(\frac{x \relax(1 + x)}{V \left\|x\right\|^{\prime}}) \end{align} \label{eq:s-sq}\tag{19}\] where \[\begin{align} \alpha &= 1 + \frac{1}{\relax(1 + x)}, \\ V &\ge 1, \\ D_{k, q}(s) &\coloneq \sum_{\substack{n = 1 \\ (n, q) = 1}}^{\infty} \frac{d_{k}(n)}{n^{s}}\\ &= \prod_{p \nmid q} \frac{\lr(1 - 2 p^{-k s} + p^{-(k + 1)s}) \lr(1 - p^{-(k + 1) s}) \lr(1 - p^{-(2 k + 1) s})^{2}}{\lr(1 - p^{-k s})^{2} \lr(1 - p^{-2 k s}) \lr(1 - p^{-(2 k + 2) s})} \end{align}\] for \(\relax(s) > 1/(3 k)\), \[\left\|x\right\|^{\prime} \coloneq \relax_{\substack{n \in \mathbb{Z}\\ n \ne x}}\left|x - n\right|\] and \[\sideset{}{^\prime}{\sum}_{\substack{n \le x \\ (n, q) = 1}} \coloneq \frac{1}{2} \lr(\sum_{\substack{n \le x \\ (n, q) = 1}} + \sum_{\substack{n < x \\ (n, q) = 1}}).\] We move the line segment \([\alpha - i V, \alpha + i V]\) which is the contour for the first term on the right hand side of 19 to the contour consisting of the line segments \([\alpha - i V, \beta - i V]\), \([\beta - i V, \beta + i V]\) and \([\beta + i V, \alpha + i V]\) with \(1/(2 k) < \beta \le 3/(5 k)\). We denote the integrals over the horizontal line segments by \(I_{1}\) and \(I_{3}\), and the integral over the vertical line segment by \(I_{2}\). Then we have \[\begin{align} &\frac{1}{2 \pi i} \int^{\alpha + i V}_{\alpha - i V} \frac{\zeta(s) \zeta((k + 1) s)}{\displaystyle \zeta^{2}(k s) \zeta(2 k s) \zeta((2 k + 2) s)} \\ &\hphantom{\frac{1}{2 \pi i} \int^{\alpha + i V}_{\alpha - i V}} \quad \times \prod_{p \mid q} \frac{\lr(1 - p^{-s}) \lr(1 - p^{-(k + 1) s})}{\displaystyle \lr(1 - p^{-k s})^{2} \lr(1 - p^{-2 k s}) \lr(1 - p^{-(2 k + 2) s})}\\ &\hphantom{\frac{1}{2 \pi i} \int^{\alpha + i V}_{\alpha - i V}} \quad \times \lr(\sum_{\substack{n = 1 \\ (n, q) = 1}}^{\infty} \frac{1}{n^{(2 k + 1) s}})^{2} D_{k, q}(s) \frac{x^{s}}{s} \mathop{{\operatorfont{d}}s}\\ &\quad = A_{k, q} x + B_{k, q} x^{1/(k + 1)} + I_{1} + I_{2} + I_{3}. \end{align}\] Applying Lemma 18, Lemma 19, Lemma 17 and Lemma 3, we get \[\begin{align} I_{1}, I_{3} &\ll V^{1/2} \omg{C_{2}}{V} \omg{C_{2}}{k V} \omg{2 C_{1}}[k \beta - \frac{1}{2}]{k V} \omg{2 C_{1}}{k V} C_{k, q} k^{2} \frac{x}{V}\\ &\longrightarrow 0 \end{align}\] as \(V \to +\infty\), where \[C_{k, q} \coloneq \prod_{p \mid q} \frac{\lr(1 + p^{-1/(2 k)}) \lr(1 + p^{-(k + 1)/(2 k)})}{\displaystyle \lr(1 - p^{-1/2})^{2} \lr(1 - p^{-1}) \lr(1 - p^{-(k + 1)/k})}.\] Hence, we obtain \[\begin{align} \widehat{\Xi}_{k, q}(x) &= \frac{1}{2 \pi i} \int^{\beta + i \infty}_{\beta - i \infty} \frac{\zeta(s) \zeta((k + 1) s)}{\displaystyle \zeta^{2}(k s) \zeta(2 k s) \zeta((2 k + 2) s)} \\ &\hphantom{{} = \frac{1}{2 \pi i} \int^{\beta + i \infty}_{\beta - i \infty}} \quad \times \prod_{p \mid q} \frac{\lr(1 - p^{-s}) \lr(1 - p^{-(k + 1) s})}{\displaystyle \lr(1 - p^{-k s})^{2} \lr(1 - p^{-2 k s}) \lr(1 - p^{-(2 k + 2) s})} \\ &\hphantom{{} = \frac{1}{2 \pi i} \int^{\beta + i \infty}_{\beta - i \infty}} \quad \times \lr(\sum_{\substack{n = 1 \\ (n, q) = 1}}^{\infty} \frac{1}{n^{(2 k + 1) s}})^{2} D_{k, q}(s) \frac{x^{s}}{s} \mathop{{\operatorfont{d}}s} \end{align}\] where \[\widehat{\Xi}_{k, q}(x) \coloneq \sideset{}{^\prime}{\sum}_{\substack{n \le x \\ (n, q) = 1}} \mu_{k}(n) - A_{k, q} x - B_{k, q} x^{1/(k + 1)}.\] By (A.3) and (A.5) in [10], we see that \[\begin{align} &\frac{1}{2 \pi} \int^{+\infty}_{-\infty} \frac{\left|\zeta(\beta + i t)\right|^{2} \left|\zeta((k + 1)(\beta + i t))\right|^{2}}{\left|\zeta(k (\beta + i t))\right|^{4} \left|\zeta(2 k (\beta + i t))\right|^{2} \left|\zeta((2 k + 2)(\beta + i t))\right|^{2} \left|\beta + i t\right|^{2}} \\ &\hphantom{\frac{1}{2 \pi} \int^{+\infty}_{-\infty}} \quad \times \prod_{p \mid q} \frac{\left|1 - p^{-(\beta + i t)}\right|^{2} \left|1 - p^{-(k + 1)(\beta + i t)}\right|^{2}}{\left|1 - p^{-k (\beta + i t)}\right|^{4} \left|1 - p^{-2 k (\beta + i t)}\right|^{2} \left|1 - p^{-(2 k + 2)(\beta + i t)}\right|^{2}} \\ &\hphantom{\frac{1}{2 \pi} \int^{+\infty}_{-\infty}} \quad \times \left|\sum_{\substack{n = 1 \\ (n, q) = 1}}^{\infty} \frac{1}{n^{(2 k + 1)(\beta + i t)}}\right|^{4} \mathop{{\operatorfont{d}}t} \\ &\quad = \int^{+\infty}_{0} \widehat{\Xi}_{k, q}^{2}(x) x^{-1 - 2 \beta} \mathop{{\operatorfont{d}}x} \\ &\quad = \int^{+\infty}_{0} \Xi_{k, q}^{2}(x) x^{-1 - 2 \beta} \mathop{{\operatorfont{d}}x}. \end{align} \label{eq:f-sq}\tag{20}\] Using Lemma 19, Lemma 18 and Lemma 17, we have \[\begin{align} &\frac{\left|\zeta(\beta + i t)\right|^{2} \left|\zeta((k + 1)(\beta + i t))\right|^{2}}{\left|\zeta(k (\beta + i t))\right|^{4} \left|\zeta(2 k (\beta + i t))\right|^{2} \left|\zeta((2 k + 2)(\beta + i t))\right|^{2} \left|\beta + i t\right|^{2}}\\ &\quad \ll \lr(1 + \left|t\right|)^{1 - 2 \beta} \omg{2 C_{2}}{\left|t\right|} \omg{2 C_{2}}{k \left|t\right|} \omg{4 C_{1}}[k \beta - \frac{1}{2}]{k \left|t\right|} \omg{2 C_{1}}{k \left|t\right|} \\ &\hphantom{\quad \ll {}} \quad \times \omg{2 C_{1}}{k \left|t\right|} \frac{k^{2}}{\lr(1 + \left|t\right|)^{2}} \\ &\quad \ll k^{2} \lr(1 + \left|t\right|)^{-1 - 2 \beta} \omg{8 C_{1} + 4 C_{2}}[k \beta - \frac{1}{2}]{k \left|t\right|}. \end{align}\] Therefore, we find that \[\begin{align} \int^{+\infty}_{0} \Xi_{k, q}^{2}(x) x^{-1 - 2 \beta} \mathop{{\operatorfont{d}}x} &\ll k^{6} C_{k, q}^{2} \int_{0}^{+\infty} \lr(t + e^{e^{2}})^{-1 - 2 \beta} \omg{C}[k \beta - \frac{1}{2}]{k t} \mathop{{\operatorfont{d}}t} \\ &\ll k^{7} C_{k, q}^{2} \int_{0}^{+\infty} \lr(k t + e^{e^{2}})^{-1 - 2 \beta} \omg{C}[k \beta - \frac{1}{2}]{k t} \mathop{{\operatorfont{d}}t} \\ &= k^{6} C_{k, q}^{2} \int_{0}^{+\infty} U^{-1 - 2 \beta} \omg{C}[k \beta - \frac{1}{2}]{u} \mathop{{\operatorfont{d}}u} \end{align}\] where \(C = 8 C_{1} + 4 C_{2}\) and \(U = u + e^{e^{2}}\). Letting \[y = \relax(\relax(2 C k \relax\frac{e}{2 k \beta - 1})),\] we see that \[\frac{C}{\relax\relax Y} \relax\{1, \relax\frac{e}{\lr(k \beta - 1/2) \relax\relax Y}\} \le \beta\] where \(Y = y + e^{e^{2}}\). Thus, we obtain \[\begin{align} \int^{y}_{0} U^{-1 - 2 \beta} \omg{C}[k \beta - \frac{1}{2}]{u} \mathop{{\operatorfont{d}}u} &\ll k \omg{C}[k \beta - \frac{1}{2}]{y}\\ &\ll k y^{1/k} \end{align}\] and \[\begin{align} &\int^{+\infty}_{y} U^{-1 - 2 \beta} \omg{C}[k \beta - \frac{1}{2}]{u} \mathop{{\operatorfont{d}}u} \\ &\quad \ll \int^{+\infty}_{y} \relax\left(\vphantom{\relax U \left(-1 - 2 \beta + \frac{C}{\relax\relax Y} \times \relax\{1, \relax\frac{e}{\lr(k \beta - 1/2) \relax\relax Y}\}\right)} \relax U \left(\vphantom{-1 - 2 \beta + \frac{C}{\relax\relax Y} \times \relax\{1, \relax\frac{e}{\lr(k \beta - 1/2) \relax\relax Y}\}} -1 - 2 \beta + \frac{C}{\relax\relax Y} \right.\right. \\ &\hphantom{\quad \ll \int^{+\infty}_{y} \relax\left(\vphantom{\relax U \left(-1 - 2 \beta + \frac{C}{\relax\relax Y} \times \relax\{1, \relax\frac{e}{\lr(k \beta - 1/2) \relax\relax Y}\}\right)}\right.} \quad \left.\left. {} \times \relax\{1, \relax\frac{e}{\lr(k \beta - 1/2) \relax\relax Y}\} \vphantom{-1 - 2 \beta + \frac{C}{\relax\relax Y} \times \relax\{1, \relax\frac{e}{\lr(k \beta - 1/2) \relax\relax Y}\}}\right) \vphantom{\relax U \left(-1 - 2 \beta + \frac{C}{\relax\relax Y} \times \relax\{1, \relax\frac{e}{\lr(k \beta - 1/2) \relax\relax Y}\}\right)}\right) \mathop{{\operatorfont{d}}u}\\ &\quad \ll k Y^{-1 - 2 \beta} \omg{C}[k \beta - \frac{1}{2}]{y} \\ &\quad \ll k y^{1/k}. \end{align}\] Then we have \[\int^{+\infty}_{0} \Xi_{k, q}^{2}(x) x^{-1 - 2 \beta} \mathop{{\operatorfont{d}}x} \ll k^{7} C_{k, q}^{2} \relax(\frac{1}{k}\relax(2 C k \relax\frac{e}{2 k \beta - 1}))\] immediately. Hence, we obtain \[\begin{align} \int^{T}_{1} \Xi_{k, q}^{2}(x) \mathop{{\operatorfont{d}}x} &= \int^{T}_{1} \Xi_{k, q}^{2}(x) x^{-1 - 2 \beta} x^{1 + 2 \beta} \mathop{{\operatorfont{d}}x} \\ &\le T^{1 + 2 \beta} \int^{T}_{1} \Xi_{k, q}^{2}(x) x^{-1 - 2 \beta} \mathop{{\operatorfont{d}}x} \\ &\ll T^{1 + 2 \beta} k^{7} C_{k, q}^{2} \relax(\frac{1}{k}\relax(2 C k \relax\frac{e}{2 k \beta - 1})). \end{align}\] Setting \[\beta \coloneq \frac{1}{2 k} + \frac{1}{10 k} \lr(\relax T)^{-1/(2 C k + 1)},\] we see that \[\begin{align} &T^{1 + 2 \beta} \relax(\frac{1}{k}\relax(2 C k \relax\frac{e}{2 k \beta - 1})) \\ &\quad = T^{1 + 1/k} \relax(\frac{1}{k} \lr(\frac{1}{5} + (5 e)^{2 C k}) \lr(\relax T)^{2 C k/(2 C k + 1)}), \end{align}\] and the proof is done.
The author thanks Professor Isao Kiuchi for his kind guidance and valuable comments.