Note on sums involving the Euler function

TL;DR

This paper refines estimates for sums involving the Euler totient function and its ratios, improving understanding of their asymptotic behavior for large x, building on recent work by Bordellès et al. and Wu.

Contribution

It provides more precise asymptotic estimates for sums involving the Euler totient function and its ratios, extending recent research in the area.

Findings

Refined asymptotic estimates for sums involving vy totient function

Improved bounds on sums vy function ratios

Enhanced understanding of the sums' growth behavior

Abstract

In this note, we provide refined estimates of the following sums involving the Euler totient function: $$\sum_{n\le x} \phi\left(\left[\frac{x}{n}\right]\right) \qquad \text{and} \qquad \sum_{n\le x} \frac{\phi([x/n])}{[x/n]}$$ where $[x]$ denotes the integral part of real $x$. The above summations were recently considered by Bordell\`es et al. and Wu.

Equations

$$\sum _{n\le x}\varphi \left(\left[\frac{x}{n}\right]\right)\text{and}\sum _{n\le x}\frac{\varphi ([x/n])}{[x/n]}$$

$$S_f:=\sum _{n\le x}f\left(\left[\frac{x}{n}\right]\right).$$

$$\sum _{n\le x}\frac{\varphi ([x/n])}{[x/n]}=x\sum _{n\ge 1}\frac{\varphi (n)}{n^2(n+1)}+O(x^{1/2})$$

$$\left(\frac{2629}{4009}\cdot \frac{1}{\zeta (2)}+o(1)\right)x\log x$$

$$\le \sum _{n\le x}\varphi \left(\left[\frac{x}{n}\right]\right)\le \left(\frac{2629}{4009}\cdot \frac{1}{\zeta (2)}+\frac{1380}{4009}+o(1)\right)x\log x$$

$$\frac{2}{3}\cdot \frac{1}{\zeta (2)}x\log x+O(x)\le \sum _{n\le x}\varphi \left(\left[\frac{x}{n}\right]\right)\le \left(\frac{2}{3}\cdot \frac{1}{\zeta (2)}+\frac{1}{3}\right)x\log x+O(x).$$

$${\mathfrak{S}}_{\delta }(x,N):=\sum _{N<n\le 2N}\varphi (n)\psi \left(\frac{x}{n+\delta }\right)$$

$${\mathfrak{S}}_{\delta }^{\ast }(x,N):=\sum _{N<n\le 2N}\frac{\varphi (n)}{n}\psi \left(\frac{x}{n+\delta }\right)$$

$$\sum _{n\le x}\frac{\varphi ([x/n])}{[x/n]}=x\sum _{n\ge 1}\frac{\varphi (n)}{n^2(n+1)}+O(x^{131/416}{\left(\log x\right)}^{26947/8320}).$$

$$\frac{285}{416}\cdot \frac{1}{\zeta (2)}x\log x+O(x\log \log x)$$

$$\le \sum _{n\le x}\varphi \left(\left[\frac{x}{n}\right]\right)\le \left(\frac{285}{416}\cdot \frac{1}{\zeta (2)}+\frac{131}{416}\right)x\log x+O(x\log \log x).$$

$$\frac{285}{416}\cdot \frac{1}{\zeta (2)}\approx 0.41649$$

$$\frac{2}{3}\cdot \frac{1}{\zeta (2)}\approx 0.40528$$

$${\mathfrak{S}}_{\delta }^{\ast }(x,N):=\sum _{N<n\le 2N}\frac{\varphi (n)}{n}\psi \left(\frac{x}{n+\delta }\right).$$

$$\sum _{N<n\le 2N}\frac{\varphi (n)}{n}\psi \left(\frac{x}{n+\delta }\right)$$

$$=\sum _{k\le 2N}\frac{\mu (k)}{k}\sum _{N/k<\ell \le 2N/k}\psi \left(\frac{x}{k\ell +\delta }\right).$$

$$\frac{T}{M^j}\le |f^{(j)}(x)|\le c_j\frac{T}{M^j}.$$

$$\sum_{M<n\leq 2M}\psi\big{(}f(n)\big{)}\ll(MT)^{131/416}(\log MT)^{18627/8320}.$$

$$f^{(j)}(z)=(-1{)}^j\frac{j!k^{j}x}{(kz+\delta {)}^{j+1}}.$$

$$\sum _{N/k<\ell \le 2N/k}\psi \left(\frac{x}{k\ell +\delta }\right)=\sum _{[N/k]<\ell \le 2[N/k]}\psi \left(\frac{x}{k\ell +\delta }\right)+O(1).$$

$$\sum _{N/k<\ell \le 2N/k}\psi \left(\frac{x}{k\ell +\delta }\right)\ll {\left(\frac{x}{k}\right)}^{131/416}{\left(\log \frac{Cx}{k}\right)}^{18627/8320}.$$

$$\sum _{N/k<\ell \le 2N/k}\psi \left(\frac{x}{k\ell +\delta }\right)\ll 1.$$

$$\sum _{N<n\le 2N}\frac{\varphi (n)}{n}\psi \left(\frac{x}{n+\delta }\right)$$

$$\ll \sum _{k<N}\frac{1}{k}{\left(\frac{x}{k}\right)}^{131/416}{\left(\log x\right)}^{18627/8320}$$

$$\ll x^{131/416}{\left(\log x\right)}^{18627/8320}.$$

$$\sum _{N<n\le 2N}\frac{\varphi (n)}{n}\psi \left(\frac{x}{n+\delta }\right)\ll x^{131/416}{\left(\log x\right)}^{18627/8320}$$

$$\sum _{n\le x}\frac{\varphi ([x/n])}{[x/n]}=\sum _{1/C<n\le x}\frac{\varphi ([x/n])}{[x/n]}+O(1).$$

$$\sum _{1/C<n\le x}\frac{\varphi ([x/n])}{[x/n]}$$

$$\displaystyle=\sum_{d\leq Cx}\frac{\phi(d)}{d}\Bigg{(}\frac{x}{d(d+1)}+\psi\left(\frac{x}{d+1}\right)-\psi\left(\frac{x}{d}\right)\Bigg{)}$$

$$\displaystyle=x\sum_{d\geq 1}\frac{\phi(d)}{d^{2}(d+1)}+O(1)+\sum_{d\leq Cx}\frac{\phi(d)}{d}\Bigg{(}\psi\left(\frac{x}{d+1}\right)-\psi\left(\frac{x}{d}\right)\Bigg{)}.$$

$$\sum _{d\le Cx}\frac{\varphi (d)}{d}\psi \left(\frac{x}{d+\delta }\right)\ll x^{131/416}{\left(\log x\right)}^{26947/8320}.$$

$$\sum _{n\le x}\varphi \left(\left[\frac{x}{n}\right]\right)=\sum _{n\le D}\varphi \left(\left[\frac{x}{n}\right]\right)+\sum _{D<n\le x}\varphi \left(\left[\frac{x}{n}\right]\right),$$

$$\sum _{D<n\le x}\varphi \left(\left[\frac{x}{n}\right]\right)$$

$$\displaystyle=x\sum_{d\leq x/D}\frac{\phi(d)}{d(d+1)}+\sum_{d\leq x/D}\phi(d)\Bigg{(}\psi\left(\frac{x}{d+1}\right)-\psi\left(\frac{x}{d}\right)\Bigg{)}$$

$$\displaystyle=\frac{1}{\zeta(2)}x\log\frac{x}{D}+O(x)+\sum_{d\leq x/D}\phi(d)\Bigg{(}\psi\left(\frac{x}{d+1}\right)-\psi\left(\frac{x}{d}\right)\Bigg{)}.$$

Full text

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where $[x]$ denotes the integral part of real $x$. The above summations were recently considered by Bordellès et al. and Wu.

Keywords. Euler totient function, integral part, asymptotic behavior.

2010MSC. 11A25, 11L07.

1. Introduction

Let $[x]$ denote the integral part of a real number $x$. In a recent paper of Bordellès, Dai, Heyman, Pan and Shparlinski [3], the asymptotic behavior of the following function was studied:

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In particular, if $f(n)$ is set to be $\phi(n)$ and $\phi(n)/n$ where $\phi(n)$ is the Euler totient function, Bordellès et al. obtained the following estimates.

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and

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for $x\to\infty$.

Subsequently, Wu respectively improved in [7] the upper and lower bounds in (1.2) and in [8] the error term in (1.1). More precisely, Wu showed that the error term in (1.1) can be sharpened to $O(x^{1/3}\log x)$, while the bounds in (1.2) can be refined as

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To bound $\displaystyle\sum_{n\leq x}\phi\left(\left[\frac{x}{n}\right]\right)$, the main idea in Bordellès et al. [3] and Wu [7] relies on an estimate of the following summation

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for $x\geq 2$ and $1\leq N\leq x$ where $\psi(x)=x-[x]-\frac{1}{2}$ and $\delta\in\{0,1\}$. Such an estimate is built on Vaaler’s expansion formula of $\psi(x)$ (cf. [6] or Theorem 6.1 in [2]) and the theory of exponential pairs (cf. Section 6.6.3 in [2]). Further, as Wu has shown in [8], the estimate of a similar summation

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will be useful to deduce the error term in (1.1).

We observe that, with the aid of an elaborate result due to Huxley (cf. [5] or Theorem 6.40 in [2]), the estimate of $\mathfrak{S}_{\delta}^{*}(x,D)$ in [8] can be further sharpened. In fact, Huxley’s result is strong enough in the sense that the best known error term up to now for the Dirichlet divisor problem can be deduced from it.

In this note, we shall prove the following results.

Theorem 1.1**.**

We have, as $x\to\infty$,

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Theorem 1.2**.**

We have, as $x\to\infty$,

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We have two remarks to make.

Let $\tau(n)$ denote the number of divisors of $n$. It is known that the main term of $\displaystyle\sum_{n\leq x}\tau(n)$ is $x(\log x+2\gamma-1)$ where $\gamma$ is the Euler constant. The error term, denoted by $\Delta(x)$, can be trivially bounded to be $O(x^{1/2})$. Hardy [4] also showed that $\Delta(x)$ cannot be $o(x^{1/4})$. The best known bound up to now for $\Delta(x)$ is $O(x^{131/416}\left(\log x\right)^{26947/8320})$ which is due to Huxley as we have mentioned above. We can see that the error term in (1.4) can also reach this size. 2. 2.

Numerically, we have

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2. An auxiliary estimate

Let $\delta\in\{0,1\}$. We will focus on the following auxiliary function defined in the Introduction:

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One has

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Now we will apply the following result due to Huxley [5].

Lemma 2.1** (Huxley, cf. Theorem 6.40 in [2]).**

Let $r\geq 5$, $M\geq 1$ be integers and $f\in\mathcal{C}^{r}[M,2M]$ such that there exist real numbers $T\geq 1$ and $1\leq c_{0}\leq\cdots\leq c_{r}$ such that, for all $x\in[M,2M]$ and all $j\in\{0,\ldots,r\}$, we have

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Then we have

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Under the setting of Lemma 2.1, let us put $\displaystyle f(z)=\frac{x}{kz+\delta}$. It can be easily computed that for $j\geq 1$

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Notice that we have, trivially, $\displaystyle\left[\frac{N}{k}\right]\asymp\frac{N}{k}$ when $k<N$. It can be shown with almost no effort that $T$ can be chosen to be $\displaystyle\asymp\frac{x}{N}$. In fact, $\displaystyle T=C\frac{x}{N}$ is admissible where $\displaystyle C=\frac{6!}{3\cdot 6^{6}}$. Now we assume that $N\leq Cx$.

For $k<N$, we notice that

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It follows that, for $k<N\leq Cx$,

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Further, for $N\leq k\leq 2N$,

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Hence, by (2.1), we conclude that

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To summarize, we have

Proposition 2.2**.**

Let $\delta\in\{0,1\}$. Then

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uniformly for $\displaystyle 1\leq N\leq\frac{6!}{3\cdot 6^{6}}x$.

3. Proof of Theorem 1.1

Again, let $\displaystyle C=\frac{6!}{3\cdot 6^{6}}$. Following the argument in [8], we have

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Notice that if $d=[x/n]$, then $x/(d+1)<n\leq x/d$. Hence,

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Using a dyadical split together with Proposition 2.2, we see that for $\delta\in\{0,1\}$,

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We therefore arrive at Theorem 1.1.

4. Proof of Theorem 1.2

We first split the sum $\displaystyle\sum_{n\leq x}\phi\left(\left[\frac{x}{n}\right]\right)$ into two parts:

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where $D\geq 1/C$ is to be determined later.

Using a similar argument to that in the previous section, we have

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Here in the last identity we use the following standard result (cf. Exercise 3.6 in [1])

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so that

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Applying Abel’s summation formula to the last part in (4.2) yields

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Notice that for $t\in[1,x/D]$ and $\delta\in\{0,1\}$, by a dyadical split, it follows from Proposition 2.2 that

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It turns out that by (4.3)

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Let us choose

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It follows from (4.2) and (4.4) that

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We can also trivially bound

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Theorem 1.2 is a direct combination of (4.5) and (4.6).

Acknowledgements

I want to thank Jie Wu for sharing the manuscript of [8].