On Error Sum Functions for Approximations with Arithmetic Conditions

Abstract

Let $\mathcal{E}_{k,l}(\alpha ) =\sum _{q_{m}\equiv l\pmod k}\vert q_{m}\alpha - p_{m}\vert$ be error sum functions formed by convergents $p_{m}/q_{m}$ $(m \geq 0)$ of a real number $\alpha$ satisfying the arithmetical condition $q_{m} \equiv l\pmod k$ with $0 \leq l <k$. The functions $\mathcal{E}_{k,l}$ are Riemann-integrable on $[0,1]$, so that the integrals $\int _{0}^{1}\mathcal{E}_{k,l}(\alpha )\,d\alpha$ exist as the arithmetical means of the functions $\mathcal{E}_{k,l}$ on $[0,1]$. We express these integrals by multiple sums on rational terms and prove upper and lower bounds. In the case when $l$ vanishes (i.e. $k$ divides $q_{m}$) and when the smallest prime divisor $p_{1}$ of $k = p_{1}^{a_{1}}p_{2}^{a_{2}}\cdots p_{t}^{a_{t}}$ satisfies $p_{1}> k^{\varepsilon }$ for some positive real number $\varepsilon$, we have found an asymptotic expansion in terms of $k$, namely $\int _{0}^{1}\mathcal{E}_{k,0}(\alpha )\,d\alpha =\zeta (2)\big(2\zeta (3)k^{2}\big)^{-1} + \mathcal{O}\big(3^{t}k^{-2-\varepsilon }\big)$. This result includes all integers $k$ which are of the form $k = p^{a}$ for primes $p$ and integers $a \geq 1$.

Dedicated to the memory of Professor Wolfgang Schwarz

1 Introduction

There are many results in the literature concerned with rational approximations $p/q$ to irrational numbers, where $p$ and $q$ are restricted by additional arithmetical conditions. An important result in this direction is due to Uchiyama [11].

Theorem A

For every real irrational number $\alpha$ and integers $s> 1$ , $a \geq 0$ , $b \geq 0$ such that $a$ and $b$ are not simultaneously divisible by $s$ , there are infinitely many integers $p$ and $q\not =0$ satisfying

$$\displaystyle{\Big\vert \alpha -\frac{p} {q}\Big\vert \, <\, \frac{s^{2}} {4q^{2}}\,,\quad p \equiv a\pmod s\,,\quad q \equiv b\pmod s\,.}$$

In [3] the author proved that the constant 1/4 in Uchiyama’s paper cannot be improved. Let $\|\eta \|$ denote the distance of a real number $\eta$ from the nearest integer. Then we deduce the following corollary from Theorem A:

Corollary A1

Let $f: \mathbb{N} \rightarrow \mathbb{R}_{>0}$ be a function satisfying $f(q) = o(q)$ for positive integers $q$ tending to infinity. Then, for every integers $s> 0$ , $a \geq 0$ and every real irrational number $\alpha$ we have

$$\displaystyle{\liminf _{\begin{array}{c} q> 0 \\ q \equiv a\pmod s \end{array} }f(q)\|q\alpha \|\, =\, 0\,.}$$

In particular cases stronger results are possible, e.g., for the number $e =\exp (1)$ by Theorem 1.3 in [4].

Theorem B

Let $a$ and $s$ be arbitrary positive integers. Then

$$\displaystyle{\liminf _{\begin{array}{c} q> 0 \\ q \equiv a\pmod s\end{array} }q\|qe\|\, =\, 0\,.}$$

About 5 years later Komatsu [9, Theorem 4] showed that the result of Theorem B remains true for $e$ replaced by every number $e^{1/k}$ $(k \in \mathbb{N})$.

Recently, the author [5] studied the so-called error sum functions. Let

$$\displaystyle{\mathcal{E}(\alpha )\,:=\,\sum _{ m=0}^{\infty }\vert q_{ m}\alpha - p_{m}\vert \,,\qquad \mathcal{E}^{{\ast}}(\alpha )\,:=\,\sum _{ m=0}^{\infty }(q_{ m}\alpha - p_{m})\,,}$$

where for $m \geq 0$ the fraction $p_{m}/q_{m}$ is the $m$-th convergent of the real number $\alpha$. The numbers $p_{m}$ and $q_{m}$ can be computed recursively from the continued fraction expansion of $\alpha$. Various aspects of these functions have been investigated in [5–7], among them it is shown that $0 \leq \mathcal{E}(\alpha ) \leq (1 + \sqrt{5})/2$ and $0 \leq \mathcal{E}^{{\ast}}(\alpha ) \leq 1$ for all real numbers $\alpha$. Both, $\mathcal{E}(\alpha )$ and $\mathcal{E}^{{\ast}}(\alpha )$, measure the average of error terms for diophantine approximations of $\alpha$ by rationals. Moreover, $\mathcal{E}(\alpha ) \in \mathbb{Q}(\alpha )$ holds for real numbers of algebraic degree 1 and 2. For $e =\exp (1)$ we have the formula

$$\displaystyle{\mathcal{E}(e)\, =\, 2e\int _{0}^{1}\exp (-t^{2})\,dt - e\, =\, 1.3418751\ldots \,,}$$

which proves that $\mathcal{E}(e)\not\in \mathbb{Q}(e)$. The function $\mathcal{E}(\alpha )$ is continuous for every real irrational point $\alpha$, and discontinuous for all rational numbers $\alpha$ (see [6, Theorem 2]). Therefore, the function $\mathcal{E}$ is Riemann-integrable on $[0,1]$. It turns out [6, Theorem 5] that

$$\displaystyle{\int _{0}^{1}\mathcal{E}(\alpha )\,d\alpha \, =\, -\frac{5} {8} + \frac{3\zeta (2)\log 2} {2\zeta (3)} \, =\, 0.79778798\ldots \,,}$$

where $\zeta$ denotes the Riemann Zeta function. This integral represents the arithmetical mean of the function $\mathcal{E}$ on $[0,1]$. This result can be generalized. Let n = 1, 2, 3, … and

$$\displaystyle{I_{n}\,:=\,\int _{ 0}^{1}\sum _{ m=0}^{\infty }\big\vert q_{ m}\alpha - p_{m}\big\vert ^{n}\,d\alpha \,.}$$

It can be shown [6, Sec. 4] that

$$\displaystyle{I_{n}\, =\, \frac{1} {n + 1}\left (1 - \frac{1} {2^{n+1}} -\frac{2\zeta (n + 1,-1)} {\zeta (n + 2)} \right )\,,}$$

where

$$\displaystyle{\zeta (n + 1,-1)\,:=\,\sum _{m_{2}>m_{1}>0} \frac{(-1)^{m_{1}}} {m_{1}m_{2}^{n+1}}\, =\,\sum _{ m_{2}=1}^{\infty } \frac{1} {m_{2}^{n+1}}\sum _{m_{1}=1}^{m_{2}-1}\frac{(-1)^{m_{1}}} {m_{1}} }$$

is known as multiple Zeta function. Borwein et al. [2] expressed ζ(n + 1, −1) in terms of log2, ζ(2), ζ(3), …, ζ(n + 2). Thus we obtain the following results:

Theorem C

Let n ≥ 1 be an integer. Then we have

$$\displaystyle\begin{array}{rcl} & & \int _{0}^{1}\sum _{ m=0}^{\infty }\big\vert q_{ m}\alpha - p_{m}\big\vert ^{n}\,d\alpha = \frac{1} {n + 1}\left (1 - \frac{1} {2^{n+1}}\right ) - 1 + \frac{4\log (2)\,\zeta (n + 1)} {(n + 1)\zeta (n + 2)}\left (1 - \frac{1} {2^{n+1}}\right ) {}\\ & & \qquad +\, \frac{1} {(n + 1)\zeta (n + 2)}\sum _{k=1}^{n-1}\left (1 - \frac{1} {2^{k}}\right )\left (1 - \frac{1} {2^{n-k}}\right )\zeta (k + 1)\zeta (n + 1 - k)\,. {}\\ \end{array}$$

In particular, for n = 2, 3, 4 we have the identities

$$\displaystyle\begin{array}{rcl} \int _{0}^{1}\sum _{ m=0}^{\infty }\big\vert q_{ m}\alpha - p_{m}\big\vert ^{2}\,d\alpha & =& \frac{7\log 2\,\zeta (3)} {6\zeta (4)} + \frac{\zeta ^{2}(2)} {12\zeta (4)} -\frac{17} {24}\, =\, 0,39813\ldots \,, {}\\ \int _{0}^{1}\sum _{ m=0}^{\infty }\big\vert q_{ m}\alpha - p_{m}\big\vert ^{3}\,d\alpha & =& \frac{15\log 2\,\zeta (4)} {16\zeta (5)} + \frac{3\zeta (2)\zeta (3)} {16\zeta (5)} -\frac{49} {64}\, =\, 0,27019\ldots \,, {}\\ \int _{0}^{1}\sum _{ m=0}^{\infty }\big\vert q_{ m}\alpha - p_{m}\big\vert ^{4}\,d\alpha & =& \frac{31\log 2\,\zeta (5)} {40\zeta (6)} + \frac{7\zeta (2)\zeta (4)} {40\zeta (6)} + \frac{9\zeta ^{2}(3)} {80\zeta (6)} -\frac{129} {160}\, =\, 0,20731\ldots \,. {}\\ \end{array}$$

Taking $\zeta (2s) \in \mathbb{Q}(\pi )$ into account, it follows that

$$\displaystyle\begin{array}{rcl} I_{1}& \in & \mathbb{Q}\big(\pi,\log (2),\zeta (3)\big)\,, {}\\ I_{2}& \in & \mathbb{Q}\big(\pi,\log (2),\zeta (3)\big)\,, {}\\ I_{3}& \in & \mathbb{Q}\big(\pi,\log (2),\zeta (3),\zeta (5)\big)\,, {}\\ I_{4}& \in & \mathbb{Q}\big(\pi,\log (2),\zeta (3),\zeta (5)\big)\,, {}\\ I_{5}& \in & \mathbb{Q}\big(\pi,\log (2),\zeta (3),\zeta (5),\zeta (7)\big)\,, {}\\ I_{6}& \in & \mathbb{Q}\big(\pi,\log (2),\zeta (3),\zeta (5),\zeta (7)\big)\,; {}\\ \end{array}$$

in particular we know $I_{1},\ldots,I_{6} \in \mathbb{Q}\big(\pi,\log (2),\zeta (3),\zeta (5),\zeta (7)\big)$. This proves that I ₁, …, I ₆ are algebraically dependent over $\mathbb{Q}$. But indeed a stronger result holds, which can be verified using a suitable computer algebra system.

Corollary C1

The numbers I ₁ ,I ₂ ,I ₃ ,I ₄ are algebraically dependent over $\mathbb{Q}$ . For x _i = I _i (i = 1,2,3,4) the algebraic equation

$$\displaystyle\begin{array}{rcl} 0& =& 10240x_{1}x_{3}x_{4} - 2976x_{1}^{2}x_{ 2} - 1488x_{1}^{2} - 5952x_{ 1}x_{2} + 5120x_{1}x_{3} + 7840x_{1}x_{4} {}\\ & & -2592x_{2}x_{3} +\, 6400x_{3}x_{4} + 944x_{1} - 4542x_{2} + 1904x_{3} + 4900x_{4} + 179 {}\\ \end{array}$$

holds.

The proof works by substituting the above expressions for I ₁, I ₂, I ₃, I ₄ into the equation given in the corollary, where additionally

$$\displaystyle{\zeta (2)\, =\, \frac{\pi ^{2}} {6}\,,\qquad \zeta (4)\, =\, \frac{\pi ^{4}} {90}\,,\qquad \zeta (6)\, =\, \frac{\pi ^{6}} {945}}$$

must be taken into account.

Note that

$$\displaystyle{I_{1},\ldots,I_{2n}\, \in \, \mathbb{Q}\big(\mathop{\underbrace{\pi,\log (2),\zeta (3),\zeta (5),\ldots,\zeta (2n + 1)}}\limits _{n+2}\big)\,.}$$

This proves

Corollary C2

For every integer n ≥ 3 any n + 3 numbers from the set {I ₁ ,I ₂ ,…I _2n } are algebraically dependent over $\mathbb{Q}$ .

In this paper we focus our interest on a generalized error sum function. Let $\alpha$ be a real number, and let $k \geq 1$ and $0 \leq l <k$ be integers. We define

$$\displaystyle{\mathcal{E}_{k,l}(\alpha )\,:=\,\sum _{ \begin{array}{c} m = 0 \\ q_{m} \equiv l\pmod k \end{array} }^{\infty }\vert q_{ m}\alpha -p_{m}\vert \,,}$$

in particular we set $\mathcal{E}_{k}(\alpha ) = \mathcal{E}_{k,0}(\alpha )$. It is clear that $\mathcal{E}_{1}(\alpha ) = \mathcal{E}(\alpha )$, and

$$\displaystyle{\sum _{l=0}^{k-1}\mathcal{E}_{ k,l}(\alpha )\, =\, \mathcal{E}(\alpha )\qquad (k \geq 1)\,.}$$

For $k> 1$ the error sum function $\mathcal{E}_{k}(\alpha )$ can be transformed into a more striking form. Since $k$ does not divide $q_{0} = 1$, the term for $m = 0$ in $\mathcal{E}_{k}(\alpha )$ does not occur. Moreover, for the convergents $p_{m}/q_{m}$ $(m \geq 1$) of $\alpha$ satisfying $q_{m} \equiv 0\pmod k$ we obtain the inequalities

$$\displaystyle{\vert q_{m}\alpha - p_{m}\vert \,\leq \, \frac{1} {q_{m}}\, \leq \, \frac{1} {k}\, \leq \, \frac{1} {2}\,.}$$

This proves

$$\displaystyle{ \mathcal{E}_{k}(\alpha )\, =\,\sum _{ \begin{array}{c} m = 1 \\ k\vert q_{m} \end{array} }^{\infty }\|q_{ m}\alpha \|\qquad (k> 1)\,. }$$

(1.1)

We continue to point out more basic properties of $\mathcal{E}_{k}(\alpha )$ for $k> 1$. Since $q_{m}$ and $q_{m+1}$ are coprime, at most every second term in $\mathcal{E}_{k}(\alpha )$ does not vanish. So we obtain the following upper bound for $\mathcal{E}_{k}(\alpha )$:

$$\displaystyle{\mathcal{E}_{k}(\alpha )\, =\,\sum _{ \begin{array}{c} m = 1 \\ k\vert q_{m} \end{array} }^{\infty }\vert q_{ m}\alpha -p_{m}\vert \,\leq \,\sum _{m=0}^{\infty }\vert q_{ 2m+1}\alpha -p_{2m+1}\vert \, =\, \frac{\mathcal{E}(\alpha ) -\mathcal{E}^{{\ast}}(\alpha )} {2} \qquad (k> 1)\,.}$$

The identities

$$\displaystyle{\mathcal{E}_{k}(\alpha )\, =\, \frac{\mathcal{E}(\alpha ) -\mathcal{E}^{{\ast}}(\alpha )} {2} \, =\,\alpha }$$

with $k> 1$ hold for all numbers $\alpha$ given by their continued fraction expansion

$$\displaystyle{\alpha \,=\, [0;k,1,k,1,k,1,\ldots ]\, =\, [0;\overline{k,1}]\, =\, \sqrt{\frac{1} {4} + \frac{1} {k}} -\frac{1} {2}\,,}$$

since for $m \geq 0$ we have the congruence relations $q_{2m+1} \equiv 0\pmod k$ and $q_{2m} \equiv 1\pmod k$. Moreover,

$$\displaystyle{\mathcal{E}_{k}(\alpha )\, =\,\sum _{ m=0}^{\infty }\vert q_{ 2m+1}\alpha - p_{2m+1}\vert \, =\,\sum _{ m=0}^{\infty }\left (\frac{k + 2} {2} -\sqrt{\Big(\frac{k + 2} {2} \Big)^{2} - 1}\,\right )^{m+1}\, =\,\alpha \,.}$$

There exist real irrational numbers $\alpha$ for which the series $\mathcal{E}_{k}(\alpha )$ consists of at most finitely many terms, contrary to the series $\mathcal{E}(\alpha )$. To prove the existence of such an irrational number, we define $\alpha$ recursively by its continued fraction expansion $\alpha = [0;a_{1},a_{2},\ldots ] = [0;2,1,1,2,2,4,6,\ldots ]$ as follows. We have

$$\displaystyle{q_{1} = 2\,,\quad q_{2} = 3\,,\quad q_{3} = 5\,,\quad q_{4} = 13\,,\quad q_{5} = 31\,,\quad q_{6} = 137\,,\quad q_{7} = 853\,.}$$

Now let us assume that for $m \geq 8$ the denominators $q_{m-1}$ and $q_{m-2}$ are primes. Then, by the Dirichlet prime number theorem, there are infinitely many positive integers $a$ such that $q_{m} = aq_{m-1} + q_{m-2} \in \mathbb{P}$. The number $a_{m}$ is uniquely defined by the smallest positive integer $a$ satisfying this condition. Then, for every integer $k> 1$, the series $\mathcal{E}_{k}(\alpha )$ consists of at most one term. Furthermore, there are many situations in which $\mathcal{E}_{k}(\alpha )$ vanishes.

Proposition 1.1

For every integer $k> 1$ there are uncountably many irrational numbers $\alpha$ such that $\mathcal{E}_{k}(\alpha ) = 0$ .

To prove this proposition, let $k> 1$ be any integer. We define an irrational number $\alpha$ depending on $k$ and on a sequence $(b_{n})_{n\geq 2}$ of positive integers by

$$\displaystyle{\alpha \,=\, \left [0;1,kb_{2},kb_{3},kb_{4},\ldots \right ]\, =\, \left [0;a_{1},a_{2},a_{3},\ldots \right ]\,.}$$

The denominators $q_{m}$ of the convergents $p_{m}/q_{m}$ of $\alpha$ satisfy the recurrence formula

$$\displaystyle{q_{0} = 1\,,\quad q_{1} = a_{1} = 1\,,\quad q_{m+2} = a_{m+2}q_{m+1} + q_{m}\quad (m = 0,1,2,\ldots )\,.}$$

Since $q_{m+2} \equiv q_{m}\pmod k$ for $m = 0,1,2,\ldots$ it follows recursively that $1 \equiv q_{0} \equiv q_{2} \equiv q_{4} \equiv \ldots \equiv q_{2m}\pmod k$ and, similarly, $1 \equiv q_{1} \equiv q_{3} \equiv q_{5} \equiv \ldots \equiv q_{2m+1}\pmod k$ for $m = 0,1,2,\ldots$. This proves that no denominator $q_{m}$ is divisible by $k$. Hence, $\mathcal{E}_{k}(\alpha ) = 0$. By Cantor’s counting principle we have found uncountably many real numbers $\alpha$ satisfying $\mathcal{E}_{k}(\alpha ) = 0$.

The main goal of this paper is to study the behaviour of the numbers $\int _{0}^{1}\mathcal{E}_{k,l}(\alpha )\,d\alpha$ depending on $k$ and $l$. For $l = 0$ and $k$ restricted to those numbers having no small prime divisors we prove the asymptotic behaviour of these integrals for $k$ tending to infinity (Theorem 2.1 and Corollaries 2.2–2.5). For integers $k$ having many small prime divisors the numbers $\int _{0}^{1}\mathcal{E}_{k}(\alpha )\,d\alpha$ tend more quickly to zero than in the case $k = p^{a}$ for fixed $a \geq 1$ and primes $p$ (Theorem 2.6 and Corollary 2.4). The integrals on the error sum functions $\mathcal{E}_{k,l}$ with $l> 0$ are treated in Theorem 2.7.

2 Statement of the Results

Let $\mu: \mathbb{N} \rightarrow \{-1,0,1\}$ be the Möbius function, and let $\zeta (s) =\sum _{ n=1}^{\infty }1/n^{s}$ for $s \geq 2$ be the Riemann Zeta function. By $J_{3}: \mathbb{N} \rightarrow \mathbb{N}$ we denote Jordan’s arithmetical function defined by $J_{3}(1) = 1$ and

$$\displaystyle{ J_{3}(n)\, =\, n^{3}\prod _{ p\vert n}\Big(1 - \frac{1} {p^{3}}\Big)\qquad (n> 1)\,, }$$

(2.1)

where $p$ runs through all prime divisors of $n$. Moreover, for any integer $n$ let $\mathcal{D}_{n}$ denote the set of all positive divisors of $n$. For every positive integer $r$ we define the number $T_{r}$ by

$$\displaystyle{ T_{r}\,:=\,\sum _{ n=r}^{\infty }\sum _{ 1\leq m\leq n/r}\frac{(-1)^{m+1}} {mn^{2}} \,. }$$

(2.2)

The identity from the following theorem can be considered as the main result of this paper, which contrasts with the property of the function $\mathcal{E}_{k}(\alpha )$ given by Proposition 1.1.

Theorem 2.1

For every integer $k> 1$ we have

$$\displaystyle{\int _{0}^{1}\mathcal{E}_{ k}(\alpha )\,d\alpha \, =\, \frac{1} {\zeta (3)}\sum _{r\in \mathcal{D}_{k}}\sum _{s\in \mathcal{D}_{r}}\frac{\mu (s)\mu (ks/r)T_{r}} {rJ_{3}(ks/r)} \,.}$$

Corollary 2.2

Let $k> 1$ be any integer having $t$ prime divisors, where $P$ denotes the smallest prime divisor of $k$ . Then we have

$$\displaystyle{\int _{0}^{1}\mathcal{E}_{ k}(\alpha )\,d\alpha \, =\, \frac{\zeta (2)} {2\zeta (3)k^{2}} + \mathcal{O}\Big( \frac{1} {k^{3}} + \frac{3^{t}} {k^{2}P}\Big)\,.}$$

Corollary 2.3

For all primes $p$ we have

$$\displaystyle{\int _{0}^{1}\mathcal{E}_{ p}(\alpha )\,d\alpha \, =\, \frac{1} {p^{3} - 1}\Big(\frac{p^{2}T_{p}} {\zeta (3)} -\frac{3\zeta (2)\log 2} {2\zeta (3)} + \frac{1} {4}\Big)\, =\, \frac{\zeta (2)} {2\zeta (3)p^{2}} + \mathcal{O}\Big( \frac{1} {p^{3}}\Big)}$$

and

$$\displaystyle{ \frac{2} {77p^{2}}\, <\,\int _{ 0}^{1}\mathcal{E}_{ p}(\alpha )\,d\alpha \, <\, \frac{97} {109p^{2}}\qquad (\,p \geq 3)\,.}$$

Corollary 2.4

Let $p$ be a prime and $a$ be a positive integer. Set $k:= p^{a}$ . Then we have

$$\displaystyle{\int _{0}^{1}\mathcal{E}_{ k}(\alpha )\,d\alpha \, =\, \frac{\zeta (2)} {2\zeta (3)k^{2}} + \mathcal{O}\Big( \frac{1} {k^{2+1/a}}\Big)\,.}$$

Corollary 2.5

Let $k> 1$ be an integer having at most $t$ prime divisors. The smallest prime divisor $P$ of $k$ satisfies $P> k^{\varepsilon }$ for any $0 <\varepsilon <1$ . Then we have

$$\displaystyle{\int _{0}^{1}\mathcal{E}_{ k}(\alpha )\,d\alpha \, =\, \frac{\zeta (2)} {2\zeta (3)k^{2}} + \mathcal{O}\Big( \frac{3^{t}} {k^{2+\varepsilon }}\Big)\,.}$$

To state the results in the subsequent theorems we need Euler’s totient $\varphi$.

Theorem 2.6

For every integer $k \geq 3$ we have

$$\displaystyle{ \frac{1} {k^{2}\log \log k}\, \ll \, \frac{\varphi (k)} {4k^{3}}\, <\,\int _{ 0}^{1}\mathcal{E}_{ k}(\alpha )\,d\alpha \, <\, \frac{\zeta (2)} {k^{2}} \,.}$$

For the numbers $k = p_{1}p_{2}\cdots p_{r}$ given by the product on the first $r \geq 2$ primes $p_{1} = 2$ , $p_{2} = 3$ , …we have

$$\displaystyle{\int _{0}^{1}\mathcal{E}_{ k}(\alpha )\,d\alpha \, \asymp \, \frac{1} {k^{2}\log \log k}\,.}$$

Theorem 2.6 shows that $\int _{0}^{1}\mathcal{E}_{k}(\alpha )\,d\alpha \asymp k^{-2}$ does not hold for $k \in \mathbb{N}$ tending to infinity. In the following theorem we estimate the integral on the error sum function $\mathcal{E}_{k,l}(\alpha )$ for $l> 0$, where the case $l = 1$ is treated separately. By $(a,b)$ we denote the greatest common divisor of two integers $a$ and $b$.

Theorem 2.7

• (i) For every integer $k \geq 2$ and $l = 1$ we have

  $$\displaystyle{\frac{5} {8}+ \frac{\varphi (k + 1)} {4(k + 1)^{3}} <\int _{ 0}^{1}\mathcal{E}_{ k,1}(\alpha )\,d\alpha =\sum _{ \begin{array}{c} a = 1 \\ a \equiv 1\pmod k \end{array} }^{\infty }\sum _{ \begin{array}{c} b = 0 \\ (a,b) = 1\end{array} }^{a-1} \frac{1} {a(a + b)^{2}}-\frac{3} {8} \leq \frac{5} {8}+\frac{\zeta (2)} {k^{2}} \,.}$$

• (ii) For integers $k \geq 3$ and $2 \leq l <k$ we have

  $$\displaystyle{ \frac{\varphi (l)} {4l^{3}} <\int _{ 0}^{1}\mathcal{E}_{ k,l}(\alpha )\,d\alpha =\sum _{ \begin{array}{c} a = 1 \\ a \equiv l\pmod k \end{array} }^{\infty }\sum _{ \begin{array}{c} b = 0 \\ (a,b) = 1\end{array} }^{a-1} \frac{1} {a(a + b)^{2}} \leq \frac{1} {l^{2}}+\frac{\zeta (2)} {k^{2}} \,.}$$

For two consecutive primes $p_{r-1}$ and $p_{r}$ $(r \geq 2)$ it follows from (ii) in Theorem 2.7 by Bertrand’s Postulate and Theorem 9 in [8] that

$$\displaystyle{\int _{0}^{1}\mathcal{E}_{ p_{r},p_{r-1}}(\alpha )\,d\alpha \, \asymp \, \frac{1} {p_{r}^{2}}\, \asymp \, \frac{1} {r^{2}\log ^{2}r}\,.}$$

3 Auxiliary Results

Lemma 3.1

Let $k> 1$ be an integer, and let $r$ be any positive divisor of $k$ . Then we have the identity

$$\displaystyle{\sum _{\begin{array}{c} d = 1 \\ (d,k) = k/r \end{array} }^{\infty }\frac{\mu (d)} {d^{3}} = \frac{1} {\zeta (3)}\sum _{s\in \mathcal{D}_{r}}\frac{\mu (s)\mu (ks/r)} {J_{3}(ks/r)} \,.}$$

Proof

We obtain

$$\displaystyle\begin{array}{rcl} S& \,:=\,& \sum _{\begin{array}{c} d = 1 \\ (d,k) = k/r \end{array} }^{\infty }\frac{\mu (d)} {d^{3}} \, =\,\sum _{ \begin{array}{c} m = 1 \\ (m,r) = 1 \end{array} }^{\infty } \frac{\mu (mk/r)} {(mk/r)^{3}}\, =\, \Big(\frac{r} {k}\Big)^{3}\sum _{ m=1}^{\infty }\sum _{ \begin{array}{c} s \geq 1 \\ s\vert (m,r) \end{array} }\frac{\mu (s)\mu (mk/r)} {m^{3}} {}\\ & \,=\,& \Big(\frac{r} {k}\Big)^{3}\sum _{ \begin{array}{c} s \geq 1 \\ s\vert r\end{array} }\mu (s)\sum _{\begin{array}{c} m = 1 \\ m \equiv 0\pmod s \end{array} }^{\infty }\frac{\mu (mk/r)} {m^{3}} \, =\, \Big(\frac{r} {k}\Big)^{3}\sum _{ s\in \mathcal{D}_{r}}\frac{\mu (s)} {s^{3}} \sum _{m=1}^{\infty }\frac{\mu (mks/r)} {m^{3}} \,. {}\\ \end{array}$$

For any positive integer $t$ we have

$$\displaystyle{\sum _{m=1}^{\infty }\frac{\mu (mt)} {m^{3}} \, =\,\mu (t)\sum _{\begin{array}{c} m = 1 \\ (m,t) = 1 \end{array} }^{\infty }\frac{\mu (m)} {m^{3}} \, =\, \frac{\mu (t)t^{3}} {\zeta (3)J_{3}(t)}\,,}$$

where the identity on the right-hand side can be obtained by using the method explained in [10]. Substituting the last expression into $S$ by setting $t = ks/r$, we complete the proof of the desired identity from the lemma. $\square$

Lemma 3.2

For every positive integer $r$ we have

$$\displaystyle\begin{array}{rcl} & T_{r}\, =\,\sum _{ n=1}^{\infty }\sum _{m=0}^{nr-1} \frac{1} {n(nr+m)^{2}} \,,\qquad T_{1}\, =\, \frac{3} {2}\zeta (2)\log 2 -\frac{1} {4}\zeta (3)\,, & {}\\ & T_{r}\, =\, \frac{\zeta (2)} {2r} + \mathcal{O}\Big( \frac{1} {r^{2}} \Big)\,,\qquad \mbox{ and}\qquad \frac{1} {2r}\, <\, T_{r}\, <\quad \left \{\begin{array}{lll} \frac{1} {r - 1} &\mbox{ if} \,\, &r \geq 2\,, \\ \\ 1 &\mbox{ if}&r \geq 1\,. \end{array} \right.& {}\\ \end{array}$$

In particular, we have $T_{r} <2/r$ for all $r \geq 1$ .

Proof

Let $r \geq 1$ be an integer. To prove the alternative expression of $T_{r}$, we first observe that

$$\displaystyle{ \sum _{n=1}^{\infty }\sum _{ m=0}^{nr-1} \frac{1} {n(nr + m)^{2}}\, =\,\sum _{ n=1}^{\infty }\frac{1} {n}\sum _{k=nr}^{2nr-1} \frac{1} {k^{2}}\, =\,\sum _{ k=r}^{\infty } \frac{1} {k^{2}}\sum _{n=\lfloor k/2r\rfloor +1}^{\lfloor k/r\rfloor }\frac{1} {n}\,, }$$

(3.1)

where the last identity follows by interchanging the order of summation, and where $\lfloor \eta \rfloor$ denotes the floor function, i.e. the greatest integer not exceeding $\eta$. Next, let $\beta \geq 1$ be a real number, and

$$\displaystyle{\delta \,:= \quad \left \{\begin{array}{rll} 1\,,&\mbox{ if} \,\, &\lfloor \beta \rfloor \equiv 1\pmod 2\,,\\ 0\,, &\mbox{ if} \,\, &\lfloor \beta \rfloor \equiv 0\pmod 2\,. \end{array} \right.}$$

Then we have

$$\displaystyle{\sum _{n=1}^{\lfloor \beta \rfloor }\frac{1} {n}\, =\,\sum _{ m=1}^{\lfloor \beta /2\rfloor +\delta } \frac{1} {2m - 1} +\sum _{ m=1}^{\lfloor \beta /2\rfloor } \frac{1} {2m}\,,}$$

which yields, equivalently,

$$\displaystyle\begin{array}{rcl} \sum _{n=\lfloor \beta /2\rfloor +1}^{\lfloor \beta \rfloor }\frac{1} {n}& \,=\,& \sum _{m=1}^{\lfloor \beta /2\rfloor +\delta } \frac{1} {2m - 1} + \frac{1} {2}\sum _{m=1}^{\lfloor \beta /2\rfloor } \frac{1} {m} -\sum _{n=1}^{\lfloor \beta /2\rfloor }\frac{1} {n} {}\\ & \,=\,& \sum _{m=1}^{\lfloor \beta /2\rfloor +\delta } \frac{1} {2m - 1} -\frac{1} {2}\sum _{m=1}^{\lfloor \beta /2\rfloor } \frac{1} {m} {}\\ & \,=\,& \sum _{m=1}^{\lfloor \beta \rfloor }\frac{(-1)^{m+1}} {m} \,. {}\\ \end{array}$$

With $\beta = k/r$ for $k \geq r$ we conclude from (3.1) and (2.2) that

$$\displaystyle{\sum _{n=1}^{\infty }\sum _{ m=0}^{nr-1} \frac{1} {n(nr + m)^{2}}\, =\,\sum _{ k=r}^{\infty } \frac{1} {k^{2}}\sum _{1\leq m\leq k/r}\frac{(-1)^{m+1}} {m} \, =\, T_{r}\,.}$$

For the asymptotic expansion of $T_{r}$ we apply Euler’s summation formula to the function $f(x) = 1/(x + nr - 1)^{2}$: Let $B(\eta ) =\eta -[\eta ] - 1/2$. Then,

$$\displaystyle\begin{array}{rcl} \sum _{m=1}^{nr} \frac{1} {(x + nr - 1)^{2}}& \,=\,& \int _{1}^{nr}f(x)\,dx +\int _{ 1}^{nr}B(x)f'(x)\,dx - B(1)f(1) - B(nr)f(nr) {}\\ & \,=\,& \frac{1} {nr} - \frac{1} {2nr - 1} + \mathcal{O}\Big( \frac{1} {n^{2}r^{2}}\Big)\,, {}\\ & & {}\\ \end{array}$$

which yields

$$\displaystyle\begin{array}{rcl} T_{r}& \,=\,& \sum _{n=1}^{\infty }\frac{1} {n}\sum _{m=1}^{nr} \frac{1} {(x + nr - 1)^{2}} {}\\ & \,=\,& \sum _{n=1}^{\infty } \frac{1} {n^{2}r} -\sum _{n=1}^{\infty } \frac{1} {n(2nr - 1)} + \mathcal{O}\Big(\sum _{n=1}^{\infty } \frac{1} {n^{3}r^{2}}\Big) {}\\ & \,=\,& \frac{\zeta (2)} {r} -\sum _{n=1}^{\infty }\Big( \frac{1} {2n^{2}r} + \frac{1} {2n^{2}r(2nr - 1)}\Big) + \mathcal{O}\Big( \frac{1} {r^{2}}\Big) {}\\ & \,=\,& \frac{\zeta (2)} {r} -\frac{\zeta (2)} {2r} + \mathcal{O}\Big(\sum _{n=1}^{\infty } \frac{1} {n^{2}r(2nr - 1)}\Big) + \mathcal{O}\Big( \frac{1} {r^{2}}\Big) {}\\ & \,=\,& \frac{\zeta (2)} {2r} + \mathcal{O}\Big( \frac{1} {r^{2}}\Big)\,. {}\\ \end{array}$$

$T_{1}$ is a special case of the multivariate zeta function $\zeta (m,n)$, see [1, Sect. 2.6]:

$$\displaystyle\begin{array}{rcl} T_{1}& \,=\,& \sum _{n=1}^{\infty }\sum _{ m=1}^{n}\frac{(-1)^{m+1}} {mn^{2}} \, =\,\sum _{ n=1}^{\infty }\sum _{ m=1}^{n-1}\frac{(-1)^{m+1}} {mn^{2}} +\sum _{ n=1}^{\infty }\frac{(-1)^{n+1}} {n^{3}} {}\\ & \,=\,& -\zeta (2,-1) + \frac{3} {4}\zeta (3)\, =\, \frac{3} {2}\zeta (2)\log 2 -\frac{1} {4}\zeta (3)\,. {}\\ \end{array}$$

The bounds for $T_{r}$ stated in the lemma follow from (2.2) by using the inequalities

$$\displaystyle\begin{array}{rcl} & \frac{1} {2}\, \leq \,\sum _{1\leq m\leq n/r}\frac{(-1)^{m+1}} {m} \, \leq \, 1\qquad (n \geq r)\,, & {}\\ & \sum _{m=r}^{\infty } \frac{1} {m^{2}} <\int _{ r-1}^{\infty }\frac{dt} {t^{2}} = \frac{1} {r-1} \leq \frac{2} {r}\quad (r \geq 2)\,,\qquad T_{1} = 1.409757\ldots <2\,,& {}\\ \end{array}$$

and

$$\displaystyle{\sum _{m=r}^{\infty } \frac{1} {m^{2}}>\int _{ r}^{\infty }\frac{dt} {t^{2}} = \frac{1} {r}\quad (r \geq 1)\,.}$$

4 Proof of Theorem 2.1

Let $\chi _{k,l}: \mathbb{N} \rightarrow \{ 0,1\}$ be defined by

$$\displaystyle{\chi _{k,l}(n)\,:= \quad \left \{\begin{array}{ll} 1\,&\mbox{ if $n \equiv l\pmod k$}\,,\\ 0\, &\mbox{ otherwise}\,. \end{array} \right.}$$

Note that $\chi _{k,l}(1) = 1$ holds if and only if $l = 1$. At the beginning of the proof of Theorem 2.1 we follow the lines in Sect. 4 in [6] and modify the arguments. Let $m$ and $a_{1},\ldots,a_{m}$ be positive integers. We define the rational numbers $\xi _{1}$, $\xi _{2}$ by their continued fraction expansion:

$$\displaystyle{\xi _{1}\, =\, \left [0;a_{1},\ldots,a_{m-1},a_{m}\right ]\qquad \mbox{ and}\qquad \xi _{2}\, =\, \left [0;a_{1},\ldots,a_{m-1},a_{m} + 1\right ]\,.}$$

We have $\xi _{1} <\xi _{2}$ for even $m$ and $\xi _{2} <\xi _{1}$ for odd $m$. We define the interval $I_{m}$ by $I_{m} = (\xi _{1},\xi _{2})$ for even $m$ and $I_{m} = (\xi _{2},\xi _{1})$ otherwise. It is well known that the intervals $I_{m}$ are disjoint for different positive integers $a_{1},\ldots,a_{m}$, and that for any fixed $m$ the union of all closed intervals $\overline{I}_{m}$ gives the interval $[0,1]$. For this decomposition of $[0,1]$ we express the integral as follows:

$$\displaystyle\begin{array}{rcl} \int _{0}^{1}\mathcal{E}_{ k,l}(\alpha )\,d\alpha & \,=\,& \int _{0}^{1}\sum _{ m=0}^{\infty }(-1)^{m}\chi _{ k,l}(q_{m})(q_{m}\alpha - p_{m})\,d\alpha {}\\ & \,=\,& \frac{\chi _{k,l}(1)} {2} +\sum _{ m=1}^{\infty }(-1)^{m}\int _{ 0}^{1}\chi _{ k,l}(q_{m})(q_{m}\alpha - p_{m})\,d\alpha {}\\ & \,=\,& \frac{\chi _{k,l}(1)} {2} +\sum _{ m=1}^{\infty }(-1)^{m}\sum _{ a_{1}=1}^{\infty }\ldots \sum _{ a_{m}=1}^{\infty }\int _{ I_{m}}\chi _{k,l}(q_{m})(q_{m}\alpha - p_{m})\,d\alpha {}\\ & \,=\,& \frac{\chi _{k,l}(1)} {2} +\sum _{ m=1}^{\infty }\sum _{ a_{1}=1}^{\infty }\ldots \sum _{ a_{m}=1}^{\infty }\chi _{ k,l}(q_{m})\int _{\xi _{1}}^{\xi _{2} }(q_{m}\alpha - p_{m})\,d\alpha \,. {}\\ \end{array}$$

Note that $p_{m}$ and $q_{m}$ depend on $a_{1},\ldots,a_{m}$. The continued fraction expansion of every point $\alpha \in I_{m}$ has the form $\alpha = [0;a_{1},\ldots,a_{m-1},a_{m},\ldots ]$. Hence the convergents $p_{\nu }/q_{\nu }$ for $\nu \leq m$ depend on $I_{m}$, but not on $\alpha \in I_{m}$. Therefore we compute the above integral on $[\xi _{1},\xi _{2}]$ by

$$\displaystyle{\int _{\xi _{1}}^{\xi _{2} }(q_{m}\alpha - p_{m})\,d\alpha = (\xi _{2} -\xi _{1})\,\frac{(\xi _{2} +\xi _{1})q_{m} - 2p_{m}} {2} \,.}$$

Using

$$\displaystyle{\xi _{1} = \frac{p_{m}} {q_{m}}\quad \mbox{ and}\quad \xi _{2} = \frac{(a_{m} + 1)p_{m-1} + p_{m-2}} {(a_{m} + 1)q_{m-1} + q_{m-2}} }$$

we compute the expressions

$$\displaystyle{\xi _{2} -\xi _{1} = \frac{(-1)^{m}} {(q_{m} + q_{m-1})q_{m}}}$$

and

$$\displaystyle{\xi _{2} +\xi _{1} = \frac{p_{m-1}q_{m} + q_{m-1}p_{m} + 2p_{m}q_{m}} {(q_{m} + q_{m-1})q_{m}} \,,}$$

which give

$$\displaystyle{\int _{\xi _{1}}^{\xi _{2} }(q_{m}\alpha - p_{m})\,d\alpha = \frac{1} {2q_{m}(q_{m} + q_{m-1})^{2}}\,,}$$

and consequently

$$\displaystyle{ \int _{0}^{1}\mathcal{E}_{ k,l}(\alpha )\,d\alpha \, =\, \frac{\chi _{k,l}(1)} {2} +\sum _{ m=1}^{\infty }\sum _{ a_{1}=1}^{\infty }\ldots \sum _{ a_{m}=1}^{\infty } \frac{\chi _{k,l}(q_{m})} {2q_{m}(q_{m} + q_{m-1})^{2}}\,. }$$

(4.1)

For the denominators of two subsequent convergents of the continued fraction of α = 〈0; a ₁, …, a _m , …〉 it is well known that (q _m , q _m−1) = 1. For fixed q _m  = a we count the solutions of q _m−1 = b with (a, b) = 1 and 0 ≤ b ≤ a − 1 in the multiple sum on the left-hand side of (4.1). It is necessary to distinguish the cases m ≥ 2 and m = 1.

Case 1: :

m ≥ 2. First let a ₁ = 1. Then,

$$\displaystyle{\frac{q_{m-1}} {q_{m}} =\langle 0;a_{m},\ldots,a_{2},1\rangle =\langle 0;a_{m},\ldots,a_{2} + 1\rangle \,.}$$

For a ₁ ≥ 2 we have

$$\displaystyle{\frac{q_{m-1}} {q_{m}} =\langle 0;a_{m},\ldots,a_{2},a_{1}\rangle =\langle 0;a_{m},\ldots,a_{2},a_{1} - 1,1\rangle \,.}$$

Case 2: :

m = 1. For a ₁ = 1 we have a unique representation of the fraction

$$\displaystyle{\frac{q_{m-1}} {q_{m}} = \frac{q_{0}} {q_{1}} = \frac{1} {a_{1}} = \frac{1} {1} =\langle 0;1\rangle \,,}$$

since the integer part a ₀ = 0 must not be changed. For a ₁ ≥ 2 there are again two representations:

$$\displaystyle{\frac{q_{m-1}} {q_{m}} = \frac{q_{0}} {q_{1}} = \frac{1} {a_{1}} =\langle 0;a_{1}\rangle =\langle 0;a_{1} - 1,1\rangle \,.}$$

Therefore it becomes clear that for any fixed $q_{m} = a$ every coprime integer $b$ with $0 \leq b \leq a - 1$ occurs exactly two times in the multiple sum on the right-hand side of (4.1), except for $m = 1$ and $a_{1} = 1$. For this exceptional case we separate the term

$$\displaystyle{ \frac{\chi _{k,l}(q_{1})} {2q_{1}(q_{1} + q_{0})^{2}}\, =\, \frac{\chi _{k,l}(1)} {8} }$$

from the multiple sum. Therefore we obtain

$$\displaystyle\begin{array}{rcl} & & \int _{0}^{1}\mathcal{E}_{ k,l}(\alpha )\,d\alpha \\ & \,=\,& \frac{\chi _{k,l}(1)} {2} +\sum _{ m=2}^{\infty }\sum _{ a_{1}=1}^{\infty }\ldots \sum _{ a_{m}=1}^{\infty } \frac{\chi _{k,l}(q_{m})} {2q_{m}(q_{m} + q_{m-1})^{2}} +\sum _{ a_{1}=2}^{\infty } \frac{\chi _{k,l}(q_{1})} {2q_{1}(q_{1} + 1)^{2}} + \frac{\chi _{k,l}(1)} {8} \\ & \,=\,& \frac{5\chi _{k,l}(1)} {8} +\sum _{ a=1}^{\infty }\sum _{ \begin{array}{c} b = 1 \\ (a,b) = 1 \end{array} }^{a-1} \frac{\chi _{k,l}(a)} {a(a + b)^{2}} \\ & \,=\,& -\frac{3\chi _{k,l}(1)} {8} +\sum _{ \begin{array}{c} a = 1 \\ a \equiv l\pmod k \end{array} }^{\infty }\sum _{ \begin{array}{c} b = 0 \\ (a,b) = 1 \end{array} }^{a-1} \frac{1} {a(a + b)^{2}}\,.{}\end{array}$$

(4.2)

Note that for $b = 0$ the condition $(a,0) = 1$ holds for $a = 1$ only. For the proof of Theorem 2.1 we now assume that $l = 0$, so that $\chi _{k,l}(1)$ vanishes. Then (4.2) simplifies to

$$\displaystyle{ \int _{0}^{1}\mathcal{E}_{ k}(\alpha )\,d\alpha \, =\,\sum _{ \begin{array}{c} a = 1 \\ k\vert a \end{array} }^{\infty }\sum _{ \begin{array}{c} b = 0 \\ (a,b) = 1 \end{array} }^{a-1} \frac{1} {a(a + b)^{2}}\,. }$$

(4.3)

Next, we express the arithmetic condition $(a,b) = 1$ on $a$ and $b$ from the inner sum by the Möbius function. Then we proceed by interchanging the order of the resulting triple sum. Here, $[d,k]$ denotes the least common multiple of $d$ and $k$.

$$\displaystyle\begin{array}{rcl} \int _{0}^{1}\mathcal{E}_{ k}(\alpha )\,d\alpha & \,=\,& \sum _{\begin{array}{c} a = 1\\ k\vert a \end{array} }^{\infty }\sum _{ b=0}^{a-1}\sum _{ \begin{array}{c} d> 0 \\ d\vert (a,b) \end{array} } \frac{\mu (d)} {a(a + b)^{2}}\, =\,\sum _{ d=1}^{\infty }\sum _{ \begin{array}{c} a = 1 \\ \left [d,k\right ]\vert a \end{array} }^{\infty }\sum _{ \begin{array}{c} b = 0 \\ d\vert b \end{array} }^{a-1} \frac{\mu (d)} {a(a + b)^{2}} {}\\ & \,=\,& \sum _{d=1}^{\infty }\sum _{ n=1}^{\infty }\sum _{ m=0}^{(\left [d,k\right ]n-1)/d} \frac{\mu (d)} {[d,k]n\big([d,k]n + dm\big)^{2}} {}\\ & \,=\,& \sum _{r\in \mathcal{D}_{k}}\sum _{\begin{array}{c} d = 1 \\ (d,k) = k/r \end{array} }^{\infty }\sum _{ n=1}^{\infty }\sum _{ m=0}^{\left [d,k\right ]n/d-1} \frac{\mu (d)} {[d,k]n\big([d,k]n + dm\big)^{2}}. {}\\ \end{array}$$

The condition $(d,k) = k/r$ implies that

$$\displaystyle{[d,k]\, =\, \frac{dk} {(d,k)}\, =\, dr\,.}$$

Hence the above multiple sum takes the form

$$\displaystyle\begin{array}{rcl} \int _{0}^{1}\mathcal{E}_{ k}(\alpha )\,d\alpha & \,=\,& \sum _{r\in \mathcal{D}_{k}}\sum _{\begin{array}{c} d = 1 \\ (d,k) = k/r \end{array} }^{\infty }\sum _{ n=1}^{\infty }\sum _{ m=0}^{nr-1} \frac{\mu (d)} {nrd(nrd + md)^{2}} {}\\ & \,=\,& \sum _{r\in \mathcal{D}_{k}}\frac{1} {r}\Big(\sum _{\begin{array}{c} d = 1 \\ (d,k) = k/r \end{array} }^{\infty }\frac{\mu (d)} {d^{3}} \Big)\Big(\sum _{n=1}^{\infty }\sum _{ m=0}^{nr-1} \frac{1} {n(nr + m)^{2}}\Big)\,.{}\\ \end{array}$$

Finally, we express the two terms in brackets by the identities given in Lemma 3.1 and Lemma 3.2, respectively. This completes the proof of the theorem. $\square$

5 Proofs of Corollaries 2.2–2.5

Proof of Corollary 2.2

From the multiple sum in Theorem 2.1 we separate the term for $r = k$ and $s = 1$:

$$\displaystyle\begin{array}{rcl} \int _{0}^{1}\mathcal{E}_{ k}(\alpha )\,d\alpha & \,=\,& \frac{T_{k}} {\zeta (3)k} + \frac{1} {\zeta (3)}\mathop{\sum \quad \sum }\limits _{\begin{array}{c} r \in \mathcal{D}_{k}\,\,s \in \mathcal{D}_{r} \\ r\not =k \vee s\not =1 \end{array} }\frac{\mu (s)\mu (ks/r)T_{r}} {rJ_{3}(ks/r)} \\ & \,=\,& \frac{T_{k}} {\zeta (3)k} + \mathcal{O}\Big(\mathop{\sum \quad \sum }\limits _{\begin{array}{c} r \in \mathcal{D}_{k}\,\,s \in \mathcal{D}_{r} \\ r\not =k \vee s\not =1 \end{array} }\frac{\vert \mu (s)\mu (ks/r)\vert T_{r}} {rJ_{3}(ks/r)} \Big) \\ & \,=\,& \frac{T_{k}} {\zeta (3)k} + \mathcal{O}\Big(\mathop{\sum \quad \sum }\limits _{\begin{array}{c} r \in \mathcal{D}_{k}\,\,s \in \mathcal{D}_{r} \\ r\not =k \vee s\not =1 \end{array} }\vert \mu (s)\mu (ks/r)\vert \, \frac{r} {k^{3}s^{3}}\Big)\,.{}\end{array}$$

(5.1)

Here we have applied the inequalities $T_{r} \ll 1/r$ (Lemma 3.2) and

$$\displaystyle\begin{array}{rcl} J_{3}(n)\, =\, n^{3}\prod _{ p\vert n}\Big(1 - \frac{1} {p^{3}}\Big)\,>\, n^{3}\prod _{ p\in \mathbb{P}}\Big(1 - \frac{1} {p^{3}}\Big)\, =\, \frac{n^{3}} {\zeta (3)}\qquad (n \geq 1)& &{}\end{array}$$

(5.2)

(see [8, Theorem 280]). In order to estimate $k^{3}s^{3}/r$ we discuss the following two cases. Recall that $r\vert k$ and $s\vert r$, and that the number $P$ is the smallest prime divisor of $k$.

Case 1: :

$1 \leq r <k$ and $s = 1$.

$$\displaystyle{\frac{k^{3}s^{3}} {r} \, =\, \frac{k^{3}} {r} \, \geq \, \frac{k^{3}} {k/P}\, =\, Pk^{2}\,.}$$

Case 2: :

$1 \leq r \leq k$ and $s \geq P$.

$$\displaystyle{\frac{k^{3}s^{3}} {r} \, \geq \, \frac{k^{3}s^{3}} {k} \, \geq \, P^{3}k^{2}\, \geq \, Pk^{2}\,.}$$

Using these bounds we estimate the error term in (5.1). This gives

$$\displaystyle\begin{array}{rcl} \int _{0}^{1}\mathcal{E}_{ k}(\alpha )\,d\alpha & \,=\,& \frac{T_{k}} {\zeta (3)k} + \mathcal{O}\Big(\mathop{\sum \quad \sum }\limits _{\begin{array}{c} r \in \mathcal{D}_{k}\,\,s \in \mathcal{D}_{r} \\ r\not =k \vee s\not =1 \end{array} }\vert \mu (s)\mu (ks/r)\vert \, \frac{1} {Pk^{2}}\Big) {}\\ & \,=\,& \frac{T_{k}} {\zeta (3)k} + \mathcal{O}\Big( \frac{1} {Pk^{2}}\sum _{r\in \mathcal{D}_{k}}\sum _{s\in \mathcal{D}_{r}}\vert \mu (s)\mu (ks/r)\vert \Big) {}\\ & \,=\,& \frac{\zeta (2)} {2\zeta (3)k^{2}} + \mathcal{O}\Big( \frac{1} {k^{3}}\Big) + \mathcal{O}\Big( \frac{1} {Pk^{2}}\sum _{r\in \mathcal{D}_{k}}\sum _{s\in \mathcal{D}_{r}}\vert \mu (s)\mu (ks/r)\vert \Big)\,, {}\\ \end{array}$$

where we have applied the asymptotic formula for $T_{k}$ from Lemma 3.2. To complete the proof of the corollary we finally prove the identity

$$\displaystyle{q_{k}:=\sum _{r\in \mathcal{D}_{k}}\sum _{s\in \mathcal{D}_{r}}\vert \mu (s)\mu (ks/r)\vert \, =\, 3^{t}\qquad (k = p_{ 1}^{a_{1} }\cdots p_{t}^{a_{t} })}$$

by induction with respect to $t$. For $t = 1$ let $k = p^{a}$. We count three pairs $[r,s] \in \mathcal{D}_{k} \times \mathcal{D}_{r}$ such that $\vert \mu (s)\mu (ks/r)\vert = 1$ given by $[p^{a},1]$, $[p^{a},p]$, and $[p^{a-1},1]$. Now we assume that q _k′ = 3^t holds for all integers $k'$ having $t$ prime divisors. Let $k = p_{1}^{a_{1}}\cdots p_{t}^{a_{t}}p_{t+1}^{a_{t+1}}$ and $k' = p_{1}^{a_{1}}\cdots p_{t}^{a_{t}}$. While $[r',s'] \in \mathcal{D}_{k'} \times \mathcal{D}_{r'}$ runs through all $3^{t}$ pairs which are counted for $q_{k'}$, we obtain $q_{k}$ by counting the $3 \cdot 3^{t}$ pairs $[r,s] \in \mathcal{D}_{k} \times \mathcal{D}_{r}$ given by $[r'p_{t+1}^{a_{t+1}},s']$, $[r'p_{t+1}^{a_{t+1}},s'p_{t+1}]$, and $[r'p_{t+1}^{a_{t+1}-1},s']$. This completes the proof of Corollary 2.2. $\square$

Proof of Corollary 2.3

For $k = p \in \mathbb{P}$ the multiple sum in Theorem 2.1 consists of three terms corresponding to $[r = 1,s = 1]$, $[r = p,s = 1]$, and $[r = p,s = p]$. Therefore, we obtain

$$\displaystyle{\int _{0}^{1}\mathcal{E}_{ p}(\alpha )\,d\alpha \, =\, \frac{1} {\zeta (3)}\Big(- \frac{T_{1}} {J_{3}(\,p)} + \frac{T_{p}} {p} + \frac{T_{p}} {pJ_{3}(\,p)}\Big)\,.}$$

With $J_{3}(\,p) = p^{3} - 1$ and

$$\displaystyle{T_{1}\, =\, \frac{3} {2}\zeta (2)\log 2 -\frac{1} {4}\zeta (3)}$$

given in Lemma 3.2 we prove the identity stated in the corollary. Now, for $p \geq 3$ we have with Lemma 3.2:

$$\displaystyle{\int _{0}^{1}\mathcal{E}_{ p}(\alpha )\,d\alpha \, <\, \frac{1} {p^{3} - 1}\Big( \frac{p^{2}} {(\,p - 1)\zeta (3)} -\frac{3\zeta (2)\log 2} {2\zeta (3)} + \frac{1} {4}\Big)\, <\, \frac{97} {109p^{2}}\,,}$$

and

$$\displaystyle{\int _{0}^{1}\mathcal{E}_{ p}(\alpha )\,d\alpha \,>\, \frac{1} {p^{3} - 1}\Big( \frac{p} {2\zeta (3)} -\frac{3\zeta (2)\log 2} {2\zeta (3)} + \frac{1} {4}\Big)\,>\, \frac{2} {77p^{2}}\,.}$$

This proves the inequalities stated in Corollary 2.3. $\square$

Proofs of Corollaries 2.4 and 2.5

We apply Corollary 2.2 with $k = p^{a}$, $t = 1$, and $P = p$. Then the error term in Corollary 2.2 takes the form

$$\displaystyle{\mathcal{O}\Big( \frac{1} {k^{3}} + \frac{3^{t}} {k^{2}P}\Big)\, =\, \mathcal{O}\Big( \frac{1} {k^{3}} + \frac{1} {k^{2}p}\Big)\, =\, \mathcal{O}\Big( \frac{1} {k^{3}} + \frac{1} {k^{2}k^{1/a}}\Big)\, =\, \mathcal{O}\Big( \frac{1} {k^{2+1/a}}\Big)\,.}$$

Thus Corollary 2.4 is proven. Also Corollary 2.5 follows immediately from Corollary 2.2. $\square$

6 Proofs of Theorems 2.6 and 2.7

Proof of Theorem 2.6

For the upper bound we estimate the right-hand side of the identity in (4.3). Let $k \geq 2$. Then

$$\displaystyle{\int _{0}^{1}\mathcal{E}_{ k}(\alpha )\,d\alpha \, =\,\sum _{ m=1}^{\infty }\sum _{ \begin{array}{c} b = 1 \\ (km,b) = 1 \end{array} }^{km-1} \frac{1} {km(km + b)^{2}}\, <\,\sum _{ m=1}^{\infty }\sum _{ b=1}^{km-1} \frac{1} {k^{3}m^{3}}\, <\,\sum _{ m=1}^{\infty } \frac{1} {k^{2}m^{2}}\, =\, \frac{\zeta (2)} {k^{2}} \,.}$$

On the other hand, the lower bound for the integral in Theorem 2.6 follows from the identity (4.3), too. Here we assume that $k \geq 3$.

$$\displaystyle{ \int _{0}^{1}\mathcal{E}_{ k}(\alpha )\,d\alpha \, \geq \,\sum _{\begin{array}{c} b = 1 \\ (k,b) = 1 \end{array} }^{k-1} \frac{1} {k(k + b)^{2}}\, \geq \, \frac{\varphi (k)} {k(2k - 1)^{2}}\,>\, \frac{\varphi (k)} {4k^{3}}\, \gg \, \frac{1} {k^{2}\log \log k}\,. }$$

(6.1)

The inequality on the right-hand side involving a lower bound of Euler’s totient follows from Theorem 328 in [8].

Next, let $k = p_{1}p_{2}\cdots p_{r}$ for some positive integer $r \geq 2$. Then we have

$$\displaystyle{\log k\, =\,\sum _{\begin{array}{c} p \leq p_{r} \\ p \in \mathbb{P} \end{array} }\log p\, =\,\vartheta (\,p_{r})\, \ll \, p_{r}}$$

by Theorem 414 in [8]. Applying additionally Theorem 429 in [8], we obtain

$$\displaystyle{ \prod _{\begin{array}{c} p\vert k \\ p \in \mathbb{P}\end{array} }\Big(1-\frac{1} {p}\Big)\, =\,\prod _{\begin{array}{c} p \leq p_{r} \\ p \in \mathbb{P} \end{array} }\Big(1-\frac{1} {p}\Big)\, \ll \, \frac{1} {\log p_{r}}\, \ll \, \frac{1} {\log \log k}\,. }$$

(6.2)

Moreover,

$$\displaystyle\begin{array}{rcl} \int _{0}^{1}\mathcal{E}_{ k}(\alpha )\,d\alpha & \,\leq \,& \sum _{m=1}^{\infty }\sum _{ \begin{array}{c} b = 1 \\ (km,b) = 1 \end{array} }^{km-1} \frac{1} {k^{3}m^{3}}\, =\, \frac{1} {k^{3}}\sum _{m=1}^{\infty }\frac{\varphi (km)} {m^{3}} {}\\ & \,=\,& \frac{1} {k^{2}}\sum _{m=1}^{\infty } \frac{1} {m^{2}}\prod _{\begin{array}{c} p\vert km \\ p \in \mathbb{P}\end{array} }\Big(1 -\frac{1} {p}\Big)\, \leq \, \frac{\zeta (2)} {k^{2}} \prod _{\begin{array}{c} p\vert k \\ p \in \mathbb{P} \end{array} }\Big(1 -\frac{1} {p}\Big) {}\\ & \,\ll \,& \frac{1} {k^{2}\log \log k}\qquad \mbox{ (by (6.2))}\,. {}\\ \end{array}$$

Together with the lower bound (6.1) we complete the proof of the theorem. $\square$

Proof of Theorem 2.7

By (4.2) we have already shown the identities in the theorem. So it remains to prove the inequalities. First, we prove the upper bounds.

$$\displaystyle\begin{array}{rcl} & & \sum _{\begin{array}{c} a = 1 \\ a \equiv l\pmod k \end{array} }^{\infty }\sum _{ \begin{array}{c} b = 0 \\ (a,b) = 1 \end{array} }^{a-1} \frac{1} {a(a + b)^{2}} -\frac{3\chi _{k,l}(1)} {8} {}\\ & \,=\,& \sum _{m=0}^{\infty }\sum _{ \begin{array}{c} b = 0 \\ (km + l,b) = 1 \end{array} }^{km+l-1} \frac{1} {(km + l)(km + l + b)^{2}} -\frac{3\chi _{k,l}(1)} {8} {}\\ & \,\leq \,& \sum _{m=0}^{\infty }\sum _{ b=0}^{km+l-1} \frac{1} {(km + l)^{3}} -\frac{3\chi _{k,l}(1)} {8} \, =\, \frac{1} {l^{2}} +\sum _{ m=1}^{\infty } \frac{1} {(km + l)^{2}} -\frac{3\chi _{k,l}(1)} {8}{}\\ && {}\\ & \,\leq \,& \frac{1} {l^{2}} +\sum _{ m=1}^{\infty } \frac{1} {k^{2}m^{2}} -\frac{3\chi _{k,l}(1)} {8} \, =\, \frac{1} {l^{2}} + \frac{\zeta (2)} {k^{2}} -\frac{3\chi _{k,l}(1)} {8} {}\\ & \,=\,& \quad \left \{\begin{array}{lcl} \frac{5} {8} + \frac{\zeta (2)} {k^{2}} \, & \mbox{ if} \,\,l = 1\,,\\ \\ \frac{1} {l^{2}} + \frac{\zeta (2)} {k^{2}} \, & \mbox{ if} \,\, l> 1\,.\end{array} \right. {}\\ \end{array}$$

For the lower bounds we treat the cases $l = 1$ and $l> 1$ separately. First, let $l = 1$. Then

$$\displaystyle\begin{array}{rcl} & & \sum _{\begin{array}{c} a = 1 \\ a \equiv 1\pmod k \end{array} }^{\infty }\sum _{ \begin{array}{c} b = 0 \\ (a,b) = 1 \end{array} }^{a-1} \frac{1} {a(a + b)^{2}} -\frac{3\chi _{k,1}(1)} {8} {}\\ & =& \sum _{m=0}^{\infty }\sum _{ \begin{array}{c} b = 0 \\ (km + 1,b) = 1 \end{array} }^{km} \frac{1} {(km + 1)(km + 1 + b)^{2}} -\frac{3} {8} {}\\ & =& \frac{5} {8} +\sum _{ m=1}^{\infty }\sum _{ \begin{array}{c} b = 0 \\ (km + 1,b) = 1 \end{array} }^{km} \frac{1} {(km + 1)(km + 1 + b)^{2}} {}\\ & \geq & \frac{5} {8} +\sum _{ \begin{array}{c} b = 1 \\ (k + 1,b) = 1 \end{array} }^{k} \frac{1} {(k + 1)(k + 1 + b)^{2}} {}\\ &>& \frac{5} {8} +\sum _{ \begin{array}{c} b = 1 \\ (k + 1,b) = 1 \end{array} }^{k+1} \frac{1} {4(k + 1)^{3}}\, =\, \frac{5} {8} + \frac{\varphi (k + 1)} {4(k + 1)^{3}}\,. {}\\ \end{array}$$

Next, let $l> 1$. Then

$$\displaystyle\begin{array}{rcl} & & \sum _{\begin{array}{c} a = 1 \\ a \equiv l\pmod k \end{array} }^{\infty }\sum _{ \begin{array}{c} b = 0 \\ (a,b) = 1 \end{array} }^{a-1} \frac{1} {a(a + b)^{2}} -\frac{3\chi _{k,1}(1)} {8} {}\\ & =& \sum _{m=0}^{\infty }\sum _{ \begin{array}{c} b = 1 \\ (km + l,b) = 1 \end{array} }^{km+l-1} \frac{1} {(km + l)(km + l + b)^{2}} {}\\ & \geq & \sum _{\begin{array}{c} b = 1 \\ (l,b) = 1 \end{array} }^{l-1} \frac{1} {l(l + b)^{2}}\,>\,\sum _{ \begin{array}{c} b = 1 \\ (l,b) = 1 \end{array} }^{l} \frac{1} {4l^{3}}\, =\, \frac{\varphi (l)} {4l^{3}}\,.{}\\ \end{array}$$

This completes the proof of the theorem. $\square$