ON SUMS INVOLVING THE EULER TOTIENT FUNCTION

Abstract

Let $\gcd (n_{1},\ldots ,n_{k})$ denote the greatest common divisor of positive integers $n_{1},\ldots ,n_{k}$ and let $\phi $ be the Euler totient function. For any real number $x>3$ and any integer $k\geq 2$, we investigate the asymptotic behaviour of $\sum _{n_{1}\ldots n_{k}\leq x}\phi (\gcd (n_{1},\ldots ,n_{k})). $

1 Introduction and main results

Let $s=\sigma +it$ be the complex variable and let $\zeta (s)$ denote the Riemann zeta-function. For any positive integer $k\geq 2$ , let $\tau _{k}$ denote the k-factors divisor function defined by ${\textbf {1}*\textbf {1}*\cdots *\textbf {1}}_{}$ and $\tau =\tau _{2}$ . Here $*$ denotes the Dirichlet convolution of arithmetic functions and $\textbf {1}$ is given by $\textbf {1}(n)=1$ for any positive integer n. We define the error term $\Delta _{k}(x)$ in the generalised divisor problem by

(1.1) $$ \begin{align} \sum_{n\leq x}\tau_{k}(n) = Q_{k}(\log x) x + \Delta_{k}(x), \end{align} $$

where $Q_{k}(\log x)={\mathrm {Res}}_{s=1}\,\zeta ^{k}(s){x^{s-1}}/{s}$ is a polynomial in $\log x$ of degree $k-1$ . The order of magnitude of $\Delta _{k}(x)$ as $x\to \infty $ is an open problem called the Piltz divisor problem and it has attracted much interest in analytic number theory. It has been conjectured that

(1.2) $$ \begin{align} \Delta_{k}(x) =O(x^{{(k-1)}/{2k}+\varepsilon}) \end{align} $$

for any integer $k\geq 2$ and any $\varepsilon>0$ (see Ivić [7, Chapter 13] or Titchmarsh [14]). Let  $\mu $ denote the Möbius function defined by

$$ \begin{align*} \mu(n)=\left\{\begin{array}{ll} \displaystyle 1 &\ \ \ \mathrm{if} \ n=1, \ \ \mbox{} \\ \displaystyle (-1)^k &\ \ \ \mathrm{if} \ n\ \mathrm{is\ squarefree\ and}\ n=p_1p_2\ldots p_k, \ \ \mbox{} \\ \displaystyle 0 &\ \ \ \mathrm{if} \ n\ \mathrm{is\ not\ squarefree}, \\ \end{array} \right. \end{align*} $$

and let $\gcd (n_{1},\ldots ,n_{k})$ denote the greatest common divisor of the positive integers $n_{1},\ldots ,n_{k}$ for any integer $k\geq 2$ . For a real number $x>3$ , let $S_{f,k}(x)$ denote the summatory function

(1.3) $$ \begin{align} S_{f,k}(x) := \sum_{n_{1}\ldots n_{k}\leq x}f(\gcd(n_{1},\ldots,n_{k})), \end{align} $$

where f is any arithmetic function, by summing over the hyperbolic region $\{(n_{1},\ldots ,n_{k}) \in {\mathbb N}^{k}:\ n_{1}\ldots n_{k}\leq x \}$ . In 2012, Krätzel et al. [10] showed that

$$ \begin{align*} S_{f,k}(x) = \sum_{n\leq x}g_{f,k}(n), \end{align*} $$

where

(1.4) $$ \begin{align} g_{f,k}(n) = \sum_{n=m^{k}l}(\mu*f)(m)\tau_{k}(l) \end{align} $$

(see also Heyman and Tóth [3], Kiuchi and Saad Eddin [8]). If f is multiplicative, then (1.4) is multiplicative. We use (1.4) to get the formal Dirichlet series

(1.5) $$ \begin{align} \sum_{n=1}^{\infty}\frac{g_{f,k}(n) }{n^s} = \frac{\zeta^{k}(s)}{\zeta(ks)}\sum_{n=1}^{\infty}\frac{f(n)}{n^{ks}}, \end{align} $$

which converges absolutely in the half-plane $\sigma>\sigma _{0}$ , where $\sigma _{0}$ depends on f and k.

When $f={\mathrm {id}}$ , it follows from (1.5) that

$$ \begin{align*}\sum_{n=1}^{\infty}\frac{g_{{\mathrm{id}},k}(n) }{n^s} = \frac{\zeta^{k}(s)\zeta(ks-1)}{\zeta(ks)} \quad\mbox{for Re } s>1. \end{align*} $$

Here the symbol $\mathrm {id}$ is given by $\mathrm {id}(n)=n$ for any positive integer n. For $k=2$ , Krätzel et al. [10] used the following three methods:

1. (1) the complex integration approach (see [7, 14]);

2. (2) a combination of fractional part sums and the theory of exponent pairs (see [2, 9]); and

3. (3) Huxley’s method (see [4–6]),

to prove

(1.6) $$ \begin{align} \sum_{ab\leq x}\gcd(a,b) &= P_{2}(\log x) x + O(x^{\theta}(\log x)^{\theta'}). \end{align} $$

Here $\theta $ satisfies $\tfrac 12 < \theta <1$ , $\theta '$ is some real number and $P_{2}$ is a certain quadratic polynomial with

$$ \begin{align*} P_{2}(\log{x})= \underset{s=1}{\mathrm{Res}}~\frac{\zeta^{2}(s)\zeta(2s-1)}{\zeta(2s)}\frac{x^{s-1}}{s}. \end{align*} $$

They showed that methods (1), (2) and (3) imply the results $\theta =\tfrac 23$ and $\theta '={16}/{9}$ , $\theta ={925}/{1392}$ and $\theta '=0$ , and $\theta ={547}/{832}$ and $\theta '={26947}/{8320}$ , respectively. Let $\phi $ denote the Euler totient function defined by $ \phi ={\mathrm {id}}*\mu. $ The Dirichlet series (1.5) with $f=\phi $ implies that

(1.7) $$ \begin{align} \sum_{n=1}^{\infty}\frac{g_{\phi,k}(n)}{n^{s}} = \frac{\zeta^{k}(s) \zeta(ks-1)}{\zeta^{2}(ks)} \quad\mbox{for Re } s>1. \end{align} $$

We consider some properties of the hyperbolic summation for the Euler totient function involving the gcd. The first purpose of this paper is to investigate the asymptotic behaviour of (1.3) with $f=\phi $ for $k=2$ . Applying fractional part sums and the theory of exponent pairs, we obtain the following result.

Theorem 1.1. For any real number $x>3$ ,

(1.8) $$ \begin{align} &\sum_{ab\leq x}\phi(\gcd(a,b)) = \frac{1}{4\zeta^{2}(2)} x \log^{2} x + \frac{1}{\zeta^{2}(2)}\bigg(2\gamma - \frac12 - 2\frac{\zeta'(2)}{\zeta(2)}\bigg) x \log x \nonumber\\ &\quad + \frac{1}{2\zeta^{2}(2)}\bigg({5}\gamma^{2}+6\gamma_{1}-4\gamma + 1 -4(4\gamma -1)\frac{\zeta'(2)}{\zeta(2)} - 4\frac{\zeta"(2)}{\zeta(2)} +12\bigg(\frac{\zeta'(2)}{\zeta(2)}\bigg)^{2}\bigg) x\nonumber \\ &\quad + O(x^{{55}/{84}+\varepsilon}), \end{align} $$

where $\gamma $ and $\gamma _{1}$ are the Euler constant and the first Stieltjes constant, respectively.

We note that the main term of (1.8) is given by (7.2) below.

Remark 1.2. Note that $\tfrac 12 < {55}/{84}=\tfrac 12+ {13}/{84}<{547}/{832}= \tfrac 12 +{131}/{832}$ .

The summation for the arithmetic functions $h(n)$ in the Dirichlet series

$$ \begin{align*} F_{h}(s)= \sum_{n=1}^{\infty}\frac{h_{}(n)}{n^s} = \zeta^{2}(s)\zeta(2s-1)\zeta^{M}(2s) \end{align*} $$

for $\mathrm {Re}~s>1$ and any fixed integer M was considered by Kühleitner and Nowak [11] in 2013. We use their results to show that the error term on the right-hand side of (1.8) is $\Omega (x^{1/2}{\log ^{2}x}/{\log \log x})$ as $x\to \infty $ . This suggests the following conjecture.

Conjecture 1.3. The order of magnitude of the error term on the right-hand side of (1.8) is $ O(x^{1/2}(\log x)^{A}) $ with $A>2$ .

When $k=3$ , Krätzel et al. [10] also derived the formula

$$ \begin{align*} \sum_{abc\leq x}\gcd(a,b,c) &= M_{3}(x) + O(x^{1/2}(\log x)^{5}), \end{align*} $$

where

$$ \begin{align*} M_{3}(x) & = \sum_{s_{0}=1,2/3} \underset{s=s_{0}}{\mathrm{Res}} \bigg(\frac{\zeta^{3}(s)\zeta(3s-1)}{\zeta(3s)}\frac{x^s}{s}\bigg) \\ &= x(0.6842 \ldots \log^{2}x - 0.6620\ldots \log x + 4.845\ldots) - 4.4569\ldots x^{2/3}. \end{align*} $$

For $k=3$ , we derive an asymptotic formula for (1.3) with $f=\phi $ by using the complex integration approach.

Theorem 1.4. For any real number $x>3$ ,

(1.9) $$ \begin{align} \sum_{abc\leq x} & \phi(\gcd(a,b,c)) \nonumber \\ &= \frac{\zeta(2)}{2\zeta^{2}(3)} x \log^{2} x + \frac{\zeta(2)}{\zeta^{2}(3)}\bigg(3\gamma - 1 + 3\frac{\zeta'(2)}{\zeta(2)} - 6\frac{\zeta'(3)}{\zeta(3)} \bigg) x \log x \nonumber \\ &\quad + \frac{\zeta(2)}{\zeta^{2}(3)}\bigg(3\gamma^{2}+3\gamma_{1}-3\gamma + 1 +3(3\gamma -1)\frac{\zeta'(2)}{\zeta(2)} -6(3\gamma -1) \frac{\zeta'(3)}{\zeta(3)} \bigg) x \nonumber \\ &\quad + \frac{\zeta(2)}{\zeta^{2}(3)}\bigg(27\bigg(\frac{\zeta'(3)}{\zeta(3)}\bigg)^{2} +\frac{9}{2}\frac{\zeta"(2)}{\zeta(2)} -9\frac{\zeta"(3)}{\zeta(3)}-18\frac{\zeta'(2)}{\zeta(2)}\frac{\zeta'(3)}{\zeta(3)} \bigg) x \nonumber \\ &\quad + \frac{\zeta^{3}(\frac{2}{3})}{2\zeta^{2}(2)}x^{{2}/{3}} + O(x^{{1}/{2}}\log^{5}x), \end{align} $$

where $\gamma $ and $\gamma _{1}$ are the Euler constant and the first Stieltjes constant, respectively.

We note that the main term of (1.9) is given by (7.3) below.

For $k=4$ , we use the complex integration approach to calculate the asymptotic formula for (1.3) with $f=\phi $ .

Theorem 1.5. For any real number $x>3$ ,

(1.10) $$ \begin{align} \sum_{abcd\leq x}\phi(\gcd(a,b,c,d)) &=x {P}_{\phi,4}(\log x) + O(x^{1/2} \log^{{17}/{3}} x ), \end{align} $$

where ${P}_{\phi ,4}(u)$ is a polynomial in u of degree three depending on $\phi $ .

For $k=5$ , from Ivić [7, Theorem 13.2], the error term $\Delta _{5}(x)$ is estimated by

(1.11) $$ \begin{align} \Delta_{5}(x) =O(x^{{11}/{20}+\varepsilon}) \end{align} $$

for any $\varepsilon>0$ . We use an elementary method and (1.1) to obtain the following result.

Theorem 1.6. For any real number $x>3$ ,

(1.12) $$ \begin{align} \sum_{abcde\leq x}\phi(\gcd(a,b,c,d,e)) &= x {P}_{\phi,5}(\log x) + \sum_{n\leq x^{1/5}}(\mu*\mu)(n)\sum_{m\leq x^{1/5}/n}m\Delta_{5}\bigg(\frac{x}{m^{5}n^{5}}\bigg), \end{align} $$

where ${P}_{\phi ,5}(u)$ is a polynomial in u of degree four depending on $\phi $ . In particular, it follows from (1.11) that the error term on the right-hand side of (1.12) is $O(x^{{11}/{20}+\varepsilon }).$

Remark 1.7. If we can use Conjecture (1.2) with $k=5$ , then the error term on the right-hand side of (1.12) becomes $O(x^{{2}/{5}+\varepsilon })$ .

Assuming Conjecture (1.2), it is easy to obtain an asymptotic formula for $S_{\phi ,k}(x)$ for any integer $k \geq 5$ .

Proposition 1.8. Assume Conjecture (1.2). With the previous notation,

$$ \begin{align*} \sum_{n_{1}n_{2}\ldots n_{k}\leq x}\phi(\gcd(n_{1},n_{2},\ldots,n_{k})) &= x {P}_{\phi,k}(\log x) + O(x^{{(k-1)}/{2k}+\varepsilon}) \end{align*} $$

for any real number $x>3$ , where ${P}_{\phi ,k}(u)\ (k \geq 5)$ is a polynomial in u of degree $k-1$ depending on $\phi $ .

Notation

We denote by $\varepsilon $ an arbitrary small positive number which may be different at each occurrence.

2 Auxiliary results

We will need the following lemma.

Lemma 2.1. For $t\geq t_{0}>0$ , uniformly in $\sigma $ ,

$$ \begin{align*} \zeta(\sigma+it) &\ll \left\{\begin{array}{ll} t^{(3-4\sigma)/6}\log t & \text{if } 0\leq \sigma \leq 1/2, \\ t^{(1-\sigma)/3} \log t & \text{if } 1/2 \leq \sigma \leq 1, \\ \log t & \text{if } 1\leq \sigma < 2, \\ 1 & \text{if } \sigma \geq 2. \end{array} \right. \end{align*} $$

Proof. The lemma follows from Tenenbaum [13, Theorem II.3.8]; see also Ivić [7] or Titchmarsh [14].

3 Proof of Theorem 1.1

Our main work is to evaluate the sum $ A(x)=\sum _{mnl^{2}\leq x,\, m,n,l>0}l $ . We utilise [10, Section 3.2] to derive the formula

$$ \begin{align*}A(x) = M_{1}(x) + \Delta(x), \end{align*} $$

where the error term $\Delta (x)$ is estimated by $O(x^{1/4+(\alpha+\beta)/2})$ . Here $(\alpha ,\beta )$ is an exponent pair (see [2, 7]) and $M_{1}(x)$ is the main term given by

$$ \begin{align*}M_{1}(x)=\underset{s=1}{\mathrm{Res}}~\zeta^{2}(s)\zeta(2s-1)\frac{x^{s}}{s}. \end{align*} $$

From (1.7), this gives

(3.1) $$ \begin{align} \sum_{ab\leq x}\phi(\gcd(a,b)) = \sum_{l\leq \sqrt{x}}(\mu*\mu)(l)A\bigg(\frac{x}{l^2}\bigg) = M_{2}(x) + O(x^{1/4+(\alpha+\beta)/2+\varepsilon}), \end{align} $$

where the main term $M_{2}(x)$ is given by

$$ \begin{align*} M_{2}(x) & =\underset{s=1}{\mathrm{Res}}~\frac{\zeta^{2}(s)\zeta(2s-1)}{\zeta^{2}(2s)}\frac{x^{s}}{s} \\ &= \frac{1}{4\zeta^{2}(2)} x \log^{2} x + \frac{1}{\zeta^{2}(2)}\bigg(2\gamma - \frac12 - 2\frac{\zeta'(2)}{\zeta(2)}\bigg) x \log x \\ & \quad + \frac{1}{2\zeta^{2}(2)}\bigg({5}\gamma^{2}+6\gamma_{1}-4\gamma + 1 -4(4\gamma -1)\frac{\zeta'(2)}{\zeta(2)} - 4\frac{\zeta"(2)}{\zeta(2)} +12\bigg(\frac{\zeta'(2)}{\zeta(2)}\bigg)^{2}\bigg) x \end{align*} $$

by (7.2) below. Choosing, in particular, the exponent pair

$$ \begin{align*}(\alpha,\beta)=(\tfrac{13}{84}+\varepsilon,\tfrac{55}{84}+\varepsilon), \end{align*} $$

discovered by Bourgain [1, Theorem 6], we obtain the order of magnitude $ O(x^{{55}/{84}+\varepsilon }) $ of the error term on the right-hand side of (3.1). This completes the proof of Theorem 1.1.

4 Preparations for the proof of Theorems 1.4 and 1.5

In order to derive the formulas (1.9) and (1.10), we use the following notation. Let k be any integer such that $k \geq 3$ and let $ \sigma _{0} =1+{1}/{k}+\varepsilon. $ Consider the estimation of the error terms of Perron’s formula (see [12, Theorem 5.2 and Corollary 5.3]) for (1.4) with $f=\phi $ . The estimation of $g_{\phi ,k}(n)$ is given by

$$ \begin{align*} g_{\phi,k}(n) = \sum_{n=m^{k}l}({\mathrm{id}} *\mu*\mu)(m) \tau_{k}(l) & = \sum_{n=d^{k}m^{k}l}d (\mu*\mu)(m) \tau_{k}(l) \\ & \ll n^{1/k} \sum_{n=d^{k}m^{k}l} \tau_{}(m)\tau_{k}(l) \ll n^{{1}/{k}+\varepsilon}. \end{align*} $$

In Perron’s formula,

$$ \begin{align*} R \ll x^{{1}/{k} +\varepsilon} \bigg(1+\frac{x}{T}\sum_{1\leq k \leq x}\frac{1}{k}\bigg) + \frac{(4x)^{\sigma_0}}{T} \bigg|\frac{\zeta^{k}(\sigma_{0})\zeta(k\sigma_{0}-1)}{\zeta^{2}(k\sigma_{0})}\bigg| \ll \frac{x^{\sigma_0}}{T} \end{align*} $$

for $T\leq x$ . Hence, from Perron’s formula and (1.7),

$$ \begin{align*}S_{\phi,k}(x) = \frac{1}{2\pi i}\int_{\sigma_0-iT}^{\sigma_0+iT}\frac{\zeta^{k}(s)\zeta(ks-1)}{\zeta^{2}(ks)} \frac{x^s}{s} \,ds + O\bigg(\frac{x^{\sigma_0}}{T}\bigg) \end{align*} $$

for any real number $x>3$ . When $k=3$ , we move the line of integration to $\mathrm {Re}~s=\tfrac 12$ and consider the rectangular contour formed by the line segments joining the points $c_{0}-iT$ , $c_{0}+iT$ , $\tfrac 12+iT$ , $\tfrac 12-iT$ and $c_{0}-iT$ in the anticlockwise sense. We observe that the integrand has a triple pole at $s=1$ and a simple pole at $s=\tfrac 23$ . Thus, we obtain the main term from the sum of the residues coming from the poles at $s=1$ and $\tfrac 23$ . Hence, using the Cauchy residue theorem,

(4.1) $$ \begin{align} S_{\phi,3}(x) &= J_{3}(x,T) + I_{3,1}(x,T) + I_{3,2}(x,T) - I_{3,3}(x,T) + O\bigg(\frac{x^{{4}/{3}+\varepsilon}}{T}\bigg), \end{align} $$

where

(4.2) $$ \begin{align} J_{3}(x,T) &= \Big(\underset{s=1}{\mathrm{Res}} + \underset{s=\frac23}{\mathrm{Res}}\Big)~\frac{\zeta^{3}(s)\zeta(3s-1)}{\zeta^{2}(3s)}\frac{x^{s}}{s}. \end{align} $$

Here the integrals are given by

(4.3) $$ \begin{align} I_{3,1}(x,T) &= \frac{1}{2\pi i}\int_{{1}/{2}+iT}^{4/3+\varepsilon +iT} \frac{\zeta^{3}(s)\zeta(3s-1)}{\zeta^{2}(3s)}\frac{x^{s}}{s} \,ds, \end{align} $$

(4.4) $$ \begin{align} I_{3,2}(x,T) &= \frac{1}{2\pi i}\int_{{1}/{2}-iT}^{{1}/{2}+iT} \frac{\zeta^{3}(s)\zeta(3s-1)}{\zeta^{2}(3s)}\frac{x^{s}}{s}\,ds, \end{align} $$

$$ \begin{align*} I_{3,3}(x,T) &= \frac{1}{2\pi i}\int_{{1}/{2}-iT}^{4/3+\varepsilon -iT} \frac{\zeta^{3}(s)\zeta(3s-1)}{\zeta^{2}(3s)}\frac{x^{s}}{s} \,ds. \end{align*} $$

Similarly,

$$ \begin{align*} S_{\phi,4}(x) &= J_{4}(x,T) + I_{4,1}(x,T) + I_{4,2}(x,T) - I_{4,3}(x,T) + O\bigg(\frac{x^{{5}/{4}+\varepsilon}}{T}\bigg), \end{align*} $$

where

(4.5) $$ \begin{align} J_{4}(x,T) &= \underset{s=1}{\mathrm{Res}}~\frac{\zeta^{4}(s)\zeta(4s-1)}{\zeta^{2}(4s)}\frac{x^{s}}{s}, \end{align} $$

(4.6) $$ \begin{align} I_{4,1}(x,T) &= \frac{1}{2\pi i}\int_{{1}/{2}+a+iT}^{5/4+\varepsilon +iT} \frac{\zeta^{4}(s)\zeta(4s-1)}{\zeta^{2}(4s)}\frac{x^{s}}{s} \,ds, \end{align} $$

$$ \begin{align*} I_{4,2}(x,T) &= \frac{1}{2\pi i} \int_{{1}/{2}+a-iT}^{{1}/{2}+a+iT} \frac{\zeta^{4}(s)\zeta(4s-1)}{\zeta^{2}(4s)}\frac{x^{s}}{s}\,ds, \end{align*} $$

$$ \begin{align*} I_{4,3}(x,T) &= \frac{1}{2\pi i}\int_{{1}/{2}+a-iT}^{5/4+\varepsilon -iT} \frac{\zeta^{4}(s)\zeta(4s-1)}{\zeta^{2}(4s)}\frac{x^{s}}{s} \,ds \end{align*} $$

with $ a={1}/{\log T}$ for any large number $T(>5)$ .

5 Proofs of Theorems 1.4 and 1.5

5.1 Proof of the formula (1.9)

Consider the estimate $I_{3,1}(x,T)$ . We use Lemma 2.1 and (4.3) to deduce the estimation

$$ \begin{align*} I_{3,1}(x;T) &= \frac{1}{2\pi i} \int_{1/2}^{4/3+\varepsilon} \frac{\zeta^{}(\sigma+iT)^3 \zeta(3\sigma-1+3iT)}{\zeta^{}(3\sigma+3iT)^2 (\sigma+iT)} x^{\sigma+iT} \,d\sigma \\ &\ll \frac{1}{T} \bigg(\int_{1/2}^{2/3}+\int_{2/3}^{1} + \int_{1}^{4/3 +\varepsilon}\bigg) |\zeta^{}(\sigma+iT)|^3 |\zeta(3\sigma-1+3iT)| x^{\sigma} \,d\sigma \\ &\ll T^{2/3} \log^{4}T \int_{1/2}^{2/3} \bigg(\frac{x}{T^{2}}\bigg)^{\sigma}\,d\sigma + \log^{4}T \int_{2/3}^{1} \bigg(\frac{x}{T^{}}\bigg)^{\sigma} \,d\sigma + \frac{\log^{4}T}{T} \int_{1}^{4/3+\varepsilon} {x}^{\sigma}\,d\sigma \\ &\ll \frac{x^{{4}/{3}+\varepsilon}}{T^{}}\log^{4}T. \end{align*} $$

Similarly, the estimation of $ I_{3,3}(x,T) $ is of the same order. Hence, taking $T=x^{}$ in the estimations of $I_{3,1}(x,T)$ and $I_{3,3}(x,T)$ , we find that the total contribution of the horizontal lines in absolute value is

(5.1) $$ \begin{align} \ll x^{1/3 +\varepsilon}. \end{align} $$

Now we estimate $I_{3,2}(x,T)$ . We use (4.4), the estimate $ \zeta (\tfrac 32+it) \asymp 1 $ for $t\geq 1$ , the well-known estimate

(5.2) $$ \begin{align} \int_{1}^{T}\frac{|\zeta(\frac12 +iu)|^{4}}{u}\,du \ll \log^{5}T \end{align} $$

for any large T and the Hölder inequality to obtain the estimate

(5.3) $$ \begin{align} I_{3,2}(x,T) & = \frac{1}{2\pi}\int_{-T}^{T}\frac{\zeta^{3}(\frac{1}{2}+it)\zeta(\frac{1}{2}+3it)}{\zeta^{2}(\frac32+3it)} \frac{x^{1/2 +it}}{\frac12 +it} \,dt \nonumber \\ &\ll x^{1/2} + x^{1/2} \int_{1}^{T} \frac{|\zeta(\frac12 +it)|^3}{|\zeta(\frac32 +3it)|^2}\cdot \frac{|\zeta(\frac12+3it)|}{t}\,dt \nonumber \\ &\ll x^{1/2} \bigg(\int_{1}^{T} \frac{|\zeta(\frac12 +it)|^4}{t} \,dt\bigg)^{3/4} \bigg(\int_{1}^{T} \frac{|\zeta(\frac12 +3it)|^4}{3t} \,dt\bigg)^{1/4} \nonumber \\ &\ll x^{1/2} \log^{5}T. \end{align} $$

Taking $T=x^{}$ in (5.1), (5.3) and (4.1) with $k=3$ , and substituting the above and the residue (4.2) into (4.1) with $k=3$ , we obtain the formula (1.9).

5.2 Proof of the formula (1.10)

Let $a={1}/{\log T}\ (T\geq 5)$ . From (4.5) with $k=4$ ,

(5.4) $$ \begin{align} J_{4}(x,T) = x {P}_{\phi,4}(\log x), \end{align} $$

since $s=1$ is a pole of $\zeta ^{4}(s)$ of order four, where $ {P}_{\phi ,4}(u)$ is a polynomial in u of degree three depending on $\phi $ . Consider the estimate $I_{4,1}(x,T)$ . From (4.6) and Lemma 2.1,

$$ \begin{align*} I_{4,1}(x,T) &\ll \frac{1}{T} \bigg(\int_{1/2+a}^{1}+\int_{1}^{5/4 +\varepsilon}\bigg) |\zeta^{}(\sigma+iT)|^4 |\zeta(4\sigma-1+4iT)| x^{\sigma} \,d\sigma \\ &\ll T^{1/3}\log^{5}T \int_{1/2+a}^{1} \bigg(\frac{x}{T^{4/3}}\bigg)^{\sigma}\,d\sigma + \frac{\log^{5}T}{T} \int_{1}^{5/4 +\varepsilon} {x}^{\sigma} \,d\sigma \\ &\ll x^{1/2+a}\frac{\log^{5}T}{T^{1/3}} + x^{5/4 +\varepsilon} \frac{\log^{5}T}{T}. \end{align*} $$

Similarly, the estimation of $ I_{4,3}(x,T) $ is of the same order. Hence, taking $T=x^{}$ in the estimations $I_{4,1}(x,T)$ and $I_{4,3}(x,T)$ , we find that the total contribution of the horizontal lines in absolute value is

(5.5) $$ \begin{align} \ll x^{1/4+\varepsilon}. \end{align} $$

We use (5.2) and the estimation $ \zeta (1+it) \ll {\log ^{2/3} t} $ for $t\geq t_{0}$ (see [7, Theorem 6.3]) to obtain the estimation

(5.6) $$ \begin{align} I_{4,2}(x,T) \ll x^{1/2+a} + x^{1/2+a}\log^{2/3}T \int_{1}^{T} \frac{|\zeta(\frac12 +it)|^4}{t} \,dt \ll x^{1/2} \log^{{17}/{3}}T. \end{align} $$

We take $T=x^{}$ in (5.4), (5.5), (5.6) and (4.1) with $k=4$ to complete the proof of the formula (1.10).

6 Proof of Theorem 1.6

We use (1.4) with $k=5$ to deduce that

(6.1) $$ \begin{align} \sum_{abcde\leq x}\phi(\gcd(a,b,c,d,e)) =\sum_{lm^{5}n^{5}\leq x}(\mu*\mu)(n)m\tau_{5}(l) =\sum_{n\leq x^{1/5}}(\mu*\mu)(n)B\bigg(\frac{x}{n^5}\bigg), \end{align} $$

where $ B(x):=\sum _{lm^{5}\leq x}m\tau _{5}(l). $ From (1.1),

(6.2) $$ \begin{align} B(x) &= \sum_{m\leq x^{1/5}}m\sum_{n\leq {x}/{m^5}} \tau_{5}(n) \nonumber \\ &= \sum_{m\leq x^{1/5}}m\bigg(A_{1}\frac{x}{m^5}\log^{4}\frac{x}{m^5} + \cdots + A_{5}\frac{x}{m^5} + \Delta_{5}\bigg(\frac{x}{m^5}\bigg) \bigg) \nonumber \\ &= \widetilde{Q_{4}}(\log x) x + \sum_{m\leq x^{1/5}}m\Delta_{5}\bigg(\frac{x}{m^5}\bigg), \end{align} $$

where $\widetilde {{Q}_{4}}(u)$ is a polynomial in u of degree four and $A_{1},A_{2},\ldots , A_{5}$ are computable constants. Inserting (6.2) into (6.1),

$$ \begin{align*} \sum_{abcde\leq x}\phi(\gcd(a,b,c,d,e)) = x {P}_{\phi,5}(\log x) + \sum_{n\leq x^{1/5}}(\mu*\mu)(n)\sum_{m\leq x^{1/5}/n}m\Delta_{5}\bigg(\frac{x}{m^{5}n^{5}}\bigg). \end{align*} $$

Hence, we obtain the formula (1.12).□

7 Appendix

To calculate the main terms of Theorems 1.1 and 1.4, we use the Laurent expansion of the Riemann zeta-function at $s=1$ : that is,

(7.1) $$ \begin{align} \zeta(s) = \frac{1}{s-1} + \gamma + \gamma_{1}(s-1) +\gamma_{2}(s-1)^2 + \gamma_{3}(s-1)^3 + \cdots \end{align} $$

as $s\to \infty $ , where $\gamma $ is the Euler constant and $\gamma _{k}\ (k=1,2,3,\ldots )$ are the Stieltjes constants,

$$ \begin{align*}\gamma_{k}:=\frac{(-1)^k}{k!}\lim_{N\to \infty}\bigg(\sum_{m\leq N}\frac{\log^{k}m}{m}-\frac{\log^{k+1}N}{k+1}\bigg). \end{align*} $$

We need the following residues.

(7.2) $$ \begin{align} M_{2}(x) &:=\underset{s=1}{\mathrm{Res}}~\frac{\zeta^{2}(s)\zeta(2s-1)}{\zeta^{2}(2s)}\frac{x^{s}}{s} \nonumber\\ &= \frac{1}{4\zeta^{2}(2)} x \log^{2} x + \frac{1}{\zeta^{2}(2)}\bigg(2\gamma - \frac12 - 2\frac{\zeta'(2)}{\zeta(2)}\bigg) x \log x \nonumber\\ &\quad + \frac{1}{2\zeta^{2}(2)}\bigg({5}\gamma^{2}+6\gamma_{1}-4\gamma + 1 -4(4\gamma -1)\frac{\zeta'(2)}{\zeta(2)} - 4\frac{\zeta"(2)}{\zeta(2)} +12\bigg(\frac{\zeta'(2)}{\zeta(2)}\bigg)^{2}\bigg) x, \end{align} $$

and

(7.3) $$ \begin{align} J_{3}(x,T) &:= \Big(\underset{s=1}{\mathrm{Res}} + \underset{s=2/3}{\mathrm{Res}}\Big)~\frac{\zeta^{3}(s)\zeta(3s-1)}{\zeta^{2}(3s)}\frac{x^{s}}{s} \nonumber\\ & = \frac{\zeta(2)}{2\zeta^{2}(3)} x \log^{2} x + \frac{\zeta(2)}{\zeta^{2}(3)}\bigg(3\gamma - 1 + 3\frac{\zeta'(2)}{\zeta(2)} - 6\frac{\zeta'(3)}{\zeta(3)} \bigg) x \log x \nonumber \\ &\quad+ \frac{\zeta(2)}{\zeta^{2}(3)}\bigg(3\gamma^{2}+3\gamma_{1}-3\gamma + 1 +3(3\gamma -1)\frac{\zeta'(2)}{\zeta(2)} -6(3\gamma -1) \frac{\zeta'(3)}{\zeta(3)} \bigg) x \nonumber \\ &\quad+ \frac{\zeta(2)}{\zeta^{2}(3)}\bigg(27\bigg(\frac{\zeta'(3)}{\zeta(3)}\bigg)^{2} +\frac{9}{2}\frac{\zeta"(2)}{\zeta(2)} -9\frac{\zeta"(3)}{\zeta(3)}-18\frac{\zeta'(2)}{\zeta(2)}\frac{\zeta'(3)}{\zeta(3)} \bigg) x +\frac{\zeta(\frac{2}{3})^3}{2\zeta^{2}(2)}x^{{2}/{3}}. \end{align} $$

Proof. Suppose that $g(s)$ is regular in the neighbourhood of $s=1$ and $f(s)$ has only a triple pole at $s=1$ . Then the Laurent expansion of $f(s)$ implies that

$$ \begin{align*}f(s) := \frac{a}{(s-1)^3} + \frac{b}{(s-1)^2} + \frac{c}{s-1} + h(s), \end{align*} $$

where $h(s)$ is regular in the neighbourhood of $s=1$ and $a,b,c$ are computable constants. We use the residue calculation to deduce that

$$ \begin{align*} &\underset{s=1}{\mathrm{Res}} f(s)g(s) =\frac{a}{2} g"(1) + b g'(1) + c g(1). \end{align*} $$

To prove (7.3), we use (7.1) to deduce that

$$ \begin{align*} \zeta^{3}(s)&= \frac{1}{(s-1)^3} + \frac{3\gamma}{(s-1)^2} + \frac{3\gamma^{2}+3\gamma_{1}}{s-1} + O(1) \quad\mbox{as } s\to 1. \end{align*} $$

Setting

$$ \begin{align*}g(s):=\frac{\zeta(3s-1)}{\zeta^{2}(3s)}\cdot \frac{x^s}{s}, \end{align*} $$

we have

$$ \begin{align*}g(1) =\frac{\zeta(2)}{\zeta^{2}(3)}x, \quad g'(1) =\frac{\zeta(2)}{\zeta^{2}(3)}\bigg(\log x + 3\frac{\zeta'(2)}{\zeta(2)} -6\frac{\zeta'(3)}{\zeta(3)}-1\bigg)x, \end{align*} $$

$$ \begin{align*} g"(1)&= \frac{\zeta(2)}{\zeta^{2}(3)}x \log^{2} x + \frac{2\zeta(2)}{\zeta^{2}(3)}\bigg(3\frac{\zeta'(2)}{\zeta(2)} - 6\frac{\zeta'(3)}{\zeta(3)}-1 \bigg)x\log x \\ &\quad+ \frac{2\zeta(2)}{\zeta^{2}(3)}\bigg(1+6\frac{\zeta'(3)}{\zeta(3)} - 3\frac{\zeta'(2)}{\zeta(2)} + 27\bigg(\frac{\zeta'(3)}{\zeta(3)}\bigg)^2 \bigg)x \\ &\quad+ \frac{2\zeta(2)}{\zeta^{2}(3)}\bigg(\frac{9}{2}\frac{\zeta"(2)}{\zeta(2)} -9\frac{\zeta"(3)}{\zeta(3)} - 18\frac{\zeta'(2)}{\zeta(2)} \frac{\zeta'(3)}{\zeta(3)} \bigg)x. \end{align*} $$

Hence,

$$ \begin{align*} &\Big(\underset{s=1}{\mathrm{Res}} + \underset{s=2/3}{\mathrm{Res}}\Big)~\frac{\zeta^{3}(s)\zeta(3s-1)}{\zeta^{2}(3s)}\frac{x^{s}}{s} \\ &\quad=\frac{1}{2}g"(1) +3\gamma g'(1) + 3(\gamma_{1}+\gamma^2)g(1) + \frac{\zeta^{3}(\frac{2}{3})}{2\zeta^{2}(2)}x^{2/3}. \end{align*} $$

Hence, we obtain the stated identity. To prove (7.2), we use

$$ \begin{align*} \zeta^{2}(s)\zeta(2s-1)&= \frac{\frac12}{(s-1)^3} + \frac{2\gamma}{(s-1)^2} + \frac{\frac52 \gamma^{2}+3\gamma_{1}}{s-1} + O(1) \quad\mbox{as } s\to 1. \end{align*} $$

The proof of (7.2) is similar to that of (7.3).