The error term in the truncated Perron formula for the logarithm of an L-function

Abstract

We improve upon the traditional error term in the truncated Perron formula for the logarithm of an L-function. All our constants are explicit.

1 Introduction

The truncated Perron formula relates the summatory function of an arithmetic function to a contour integral that may be estimated using techniques from complex analysis. Let $F(s) = \sum _{n=1}^{\infty } f(n) n^{-s}$ be absolutely convergent on $\operatorname {Re} s> c_F$ ; examples include the Riemann zeta function, Dirichlet L-functions, the Dedekind zeta function associated with a number field, and Artin L-functions. The truncated Perron formula tells us that if $x>0$ is not an integer, $T\geq 1$ , and $c> c_F$ , then

(1.1) $$ \begin{align} \sum_{n \leq x} f(n) = \frac{1}{2\pi i }\int_{c-iT}^{c+iT} F(s) \frac{x^s}{s}\,ds + O^*\bigg( \sum_{n=1}^{\infty} \bigg( \frac{x}{n} \bigg)^{c} |f(n)| \min\left\{ 1, \frac{1}{T |\log \frac{x}{n}|}\right\}\bigg), \end{align} $$

in which $O^*(g(x)) = h(x)$ means $|h(x)|\leq g(x)$ (see [8, Chapter 7], [10, Section 5.1], [11, Example 4.4.15], and [15, Section II.2]). We let T depend on x, and let $c = c_F + 1/\log {x}$ , so that $x^c = ex^{c_F}$ . A variation of (1.1) improves the order of the error term by truncating the integral at $\pm T^*$ for an unknown $T^*\in [T,O(T)]$ [3], although this is inconvenient if one must avoid $T^*$ that correspond to the ordinates of nontrivial zeros of $F(s)$ . The authors of [3] have also informed us in a personal communication that their paper inherited an unfortunate typo from another paper, so the error term in their variation of the truncated Perron formula could be worse by a factor of $\log {x}$ ; this means that our main result (Theorem 1.1) will be comparable in strength and more straightforward to apply when compared against the outcome of their result.

For $\operatorname {Re}{s}> 1$ , the logarithm of the Riemann zeta function $\zeta (s) = \sum _{n=1}^{\infty } n^{-s}$ is $\log \zeta (s) = \sum _{n=1}^{\infty } \Lambda (n)(\log n)^{-1}n^{-s}$ , in which $\Lambda (n)$ is the von Mangoldt function. The logarithm of a typical L-function is of the form $\sum _{n=1}^{\infty } \Lambda (n)a_n(\log n)^{-1}n^{-s}$ , in which the $a_n$ are easily controlled. For example, $|a_n| \leq 1$ for Dirichlet L-functions and $|a_n|\leq d$ for Artin L-functions of degree d (see [7, Chapter 5]. In these cases, the error term in (1.1) with $c = 1/\log {x}$ is on the order of

(1.2) $$ \begin{align} \sum_{n=2}^{\infty} \Big( \frac{x}{n}\Big)^{\frac{1}{\log x}} \frac{ \Lambda(n)}{n \log n} \min\left\{ 1 , \frac{1}{T | \log \frac{x}{n} |} \right\} = O\bigg( \frac{\log x}{T} \bigg). \end{align} $$

Granville and Soundararajan used ( 1.2 ) with Dirichlet L-functions to study large character sums [6, equation (8.1)]. Cho and Kim applied it to Artin L-functions to obtain asymptotic bounds on Dedekind zeta residues [1, Proposition 3.1]. A bilinear relative of (1.2) appears in Selberg’s work on primes in short intervals [14, Lemma 4]. Analogous sums arise with the logarithmic derivative of an L-function in [4, p. 106] and [12, p. 44].

We improve upon (1.2) asymptotically and explicitly in the following result.

Theorem 1.1 If $x \geq 3.5$ is a half integer and $T \geq (\log \frac {3}{2})^{-1}> 2.46$ , then

(1.3) $$ \begin{align} \sum_{n=2}^{\infty} \Big( \frac{x}{n}\Big)^{\frac{1}{\log x}} \frac{ \Lambda(n)}{n \log n} \min\left\{ 1 , \frac{1}{T | \log \frac{x}{n} |} \right\} \leq \frac{R(x)}{T}, \end{align} $$

in which

(1.4) $$ \begin{align} R(x) = 40.23 \log \log x +58.12 +\frac{3.87}{\log x} +\frac{5.22 \log x}{\sqrt{x}} -\frac{1.84}{\sqrt{x}}. \end{align} $$

Our result has a wide and explicit range of applicability. For example, the following corollary employs (1.1) with $T=x$ and $c = 1/\log {x}$ . Since one can use analytic techniques to see the integral below is asymptotic to $\log {L(1,\chi )}$ , one can relate $\log {L(1,\chi )}$ to a short sum. We hope to do so explicitly in the future.

Corollary 1.2 Let $L(s, \chi )$ be an entire Artin L-function of degree d such that

$$ \begin{align*} L(s,\chi) = \prod_{p}\prod_{i=1}^{d} \left(1 - \frac{\alpha_i(p)}{p^s}\right)^{-1} \quad \text{for }\operatorname{Re} s> 1 \end{align*} $$

with $a(p^k) = \alpha _1(p)^k + \cdots + \alpha _d(p)^k$ for prime p. Then, with $R(x)$ as in ( 1.4 ), we have

$$ \begin{align*} \sum_{1<n < x} \frac{\Lambda(n) a(n)}{n \log{n}} = \frac{1}{2\pi i }\int_{\tfrac{1}{\log{x}}-ix}^{\tfrac{1}{\log{x}}+ix} \frac{x^s}{s} \log{L(1+s,\chi)}\,ds + O^*\!\left( \frac{d\,R(x)}{x} \right). \end{align*} $$

2 Preliminaries

Here, we establish several lemmas needed for the proof of Theorem 1.1.

Lemma 2.1 If $\sigma>0$ , then $\log \zeta (1+\sigma ) \leq - \log \sigma + \gamma \sigma $ .

Proof For $s> 1$ , we have $\zeta (s) \leq e^{\gamma (s-1)}/(s-1)$ [13, Lemma 5.4]. Let $s = 1+\sigma $ and take logarithms to obtain the desired result.

For real $z,w$ , the equation $z = we^w$ can be solved for w if and only if $z \geq -e^{-1}$ . There are two branches for $-e^{-1} \leq z < 0$ . The lower branch defines the Lambert $W_{-1}(z)$ function [2], which decreases to $-\infty $ as $z\to 0^-$ (see Figure 1a). For $n \geq 6>2e$ , we define the strictly increasing sequence

(2.1) $$ \begin{align}\qquad y_n = \frac{-n}{2} W_{-1}\Big( \frac{-2}{n} \Big) \qquad \text{for }n \geq 6. \end{align} $$

Figure 1 Graphs relevant to the construction of the sequence $y_n$ .

Lemma 2.2 For $n \geq 8$ , we have $\frac {2y_n}{\log y_n} = n$ and $ y_n \geq \frac {n}{2} \log n$ .

Proof For $n \geq 6$ , the definition of $W_{-1}$ and (2.1) confirm that $\frac {2y_n}{\log y_n} = n$ . Thus, the desired inequality is equivalent to $W_{-1}(\frac {-2}{n} ) \leq -\log n$ . Since $f(w) = we^w$ decreases on $(-\infty ,-1]$ (Figure 1b) and $-\frac {1}{e} < -\frac {2}{n} <0$ , the desired inequality is equivalent to

$$ \begin{align*} -\frac{2}{n} \geq f(-\log n) = (-\log n)e^{-\log n} = -\frac{\log n}{n}, \end{align*} $$

which holds whenever $\log n \geq 2$ . This occurs for $n \geq e^2 \approx 7.38906$ .

Remark 2.3 For all $-e^{-1} \leq x < 0$ , the bound $W_{-1}(x) \leq \log (-x) - \log (-\log (-x))$ is valid (see [9, equations (8) and (39)]). It follows from this observation and (2.1) that

$$ \begin{align*} y_n \geq \frac{n}{2} \left(\log\left(\frac{1}{2}\log\frac{n}{2}\right) + \log{n} \right), \end{align*} $$

which also implies Lemma 2.2 for $n\geq 15$ .

The next lemma is needed later to handle a few exceptional primes.

Lemma 2.4 Let $x>1$ be a half integer, and let $C = \frac {1284699552}{444215525}= 2.89206\ldots $ .

1. 1. Let $p_{-8} < p_{-7} < \cdots < p_{-1} < x$ denote the largest eight primes (if they exist) in the interval $(\frac {x}{2},x)$ . We have the sharp bound

   (2.2) $$ \begin{align} F_1(x) = \sum_{1 \leq n \leq 8} \frac{1}{ x-p_{-n} } \leq C \end{align} $$
   (see Figure 2a). The corresponding summand in ( 2.2 ) is zero if $p_{-n}$ does not exist.

2. 2. Let $x<p_1 <p_2<\cdots < p_8$ denote the smallest eight primes (if they exist) in the interval $(x,\frac {3x}{2})$ . We have the sharp bound

   (2.3) $$ \begin{align} F_2(x) = \sum_{1 \leq n \leq 8} \frac{1}{p_n -x} \leq C \end{align} $$
   (see Figure 2a). The corresponding summand in ( 2.3 ) is zero if $p_{n}$ does not exist.

Figure 2 The functions $F_1(x)$ and $F_2(x)$ behave erratically.

Proof (a) If $x \geq 10.5$ , then $2,3,5 \notin (\frac {x}{2},x)$ . Computation confirms that

$$\begin{align*}F_1(x) \leq F_1(3.5) = \frac{8}{3} =2.66\ldots\end{align*}$$

for $x \leq 9.5$ . If $x \geq 10.5$ , then any prime in $(\frac {x}{2},x)$ is congruent to one of $1, 7, 11, 13, 17, 19, 23, 29 \pmod {30}$ . There are finitely many patterns modulo $30$ that the $p_{-8},p_{-7},\ldots ,p_{-1}$ may assume. Among these, computation confirms that $F_1(x)$ is maximized if

$$ \begin{align*} p_{-1} &= \lfloor x \rfloor \equiv 19\pmod{30}, & p_{-5} &= \lfloor x \rfloor-12 \equiv 7\pmod{30},\\ p_{-2} &= \lfloor x \rfloor-2 \equiv 17\pmod{30}, & p_{-6} &= \lfloor x \rfloor-18 \equiv 1\pmod{30},\\ p_{-3} &= \lfloor x \rfloor-6 \equiv 13\pmod{30}, & p_{-7} &= \lfloor x \rfloor-20 \equiv 29\pmod{30},\\ p_{-4} &= \lfloor x \rfloor-8 \equiv 11\pmod{30}, & p_{-8} &= \lfloor x \rfloor-26 \equiv 23\pmod{30}, \end{align*} $$

which yields the desired upper bound C. This prime pattern first occurs for $x = 88,819.5$ (see https://oeis.org/A022013).

(b) If $x \geq 5.5$ , then $2,3,5 \notin (x, \frac {3x}{2})$ . Observe that $F_2(x) \leq 2$ for $x \leq 4.5$ (attained at $x=1.5, 2.5, 4.5$ ). If $x \geq 5.5$ , then (as in (a)), any prime in $(x,\frac {3x}{2})$ is congruent to one of $1, 7, 11, 13, 17, 19, 23, 29 \pmod {30}$ . It follows that $F_2(x)$ is maximized if

$$ \begin{align*} p_{1} &= \lceil x \rceil \equiv 11\pmod{30}, & p_{5} &= \lceil x \rceil+12 \equiv 23\pmod{30},\\ p_{2} &= \lceil x \rceil+2 \equiv 13\pmod{30}, & p_{6} &= \lceil x \rceil+18 \equiv 29\pmod{30},\\ p_{3} &= \lceil x \rceil+6 \equiv 17\pmod{30}, & p_{7} &= \lceil x \rceil+20 \equiv 1\pmod{30}, \\ p_{4} &= \lceil x \rceil+8 \equiv 19\pmod{30}, & p_8 &= \lceil x \rceil + 26 \equiv 7 \pmod{30}, \end{align*} $$

which yields the desired upper bound C. This prime pattern occurs for $x=10.5$ , but not all eight primes lie in $(x,\frac {3}{2}x)$ . Therefore, the first admissible value is $x = 15,760,090.5$ (see https://oeis.org/A022011).

We also need an elementary estimate on kth powers in intervals.

Lemma 2.5 Let $X> 1$ be a noninteger, $h>1$ , and $k\geq 2$ .

1. (1) There are at most $N_k + 1$ perfect kth powers in $[X,X+h)$ , in which $N_k \leq \frac {h}{k \sqrt {X}}$ .

2. (2) The shortest gap between kth powers in $[X,X+h)$ (if they exist) is $G_k \geq k \sqrt {X}$ .

Proof We may assume that X is so large that $N_k \geq 1$ . Let $m = \lceil X^{\frac {1}{k}} \rceil $ so that $m^k$ is the first kth power larger than X. Consider the gaps $g_1,g_2,\ldots ,g_{N_k}$ between the $N_k$ consecutive kth powers in $[X,X+h)$ (see Figure 3). Then

$$ \begin{align*} G_k = \min\{ g_1,g_2,\ldots,g_{N_k}\} = g_1 = (m+1)^k - m^k \geq km^{k-1} \geq k X^{\frac{k-1}{k}} \geq k \sqrt{X}. \end{align*} $$

The desired inequality follows since $N_k G_k \leq g_1 + g_2+ \cdots + g_{N_k} \leq h$ .

Figure 3 Proof of Lemma 2.5.

Finally, we need an estimate on the nth harmonic number $H_n = \sum _{j=1}^n \frac {1}{j}$ :

(2.4) $$ \begin{align}\qquad \frac{1}{2n+ \frac{2}{5}} < H_n - \log n - \gamma < \frac{1}{2n + \frac{1}{3}} \leq \frac{3}{7} \quad\text{for}\quad n \geq 1, \end{align} $$

in which $\gamma $ is the Euler–Mascheroni constant [16]. We require the upper bound

(2.5) $$ \begin{align} \sum_{\ell=1}^n \frac{1}{2\ell - 1} &= \left(1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{2n}\right) - \left( \frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{2n} \right) = H_{2n} - \frac{1}{2} H_n \nonumber \\ &\leq \bigg(\log 2n +\gamma + \frac{1}{4n + \frac{1}{3}} \bigg) - \frac{1}{2} \bigg( \log n + \gamma + \frac{1}{2n+\frac{2}{5}} \bigg) \nonumber \\ &\leq \frac{1}{2}\log n + \frac{1}{2}\gamma + \log 2 + \frac{1}{4n + \frac{1}{3}}- \frac{1}{4n+\frac{4}{5}} \nonumber \\ &= \frac{1}{2}\log n + \frac{1}{2}\gamma + \log 2 + \frac{7}{240 n^2+68 n+4} \nonumber\\ &\leq \frac{1}{2}\log n + \frac{1}{2}\gamma + \log 2 + \frac{7}{312} \quad \text{for}\quad n \geq 1. \end{align} $$

3 Proof of Theorem 1.1

In what follows, $x\in \mathbb {N}+ \frac {1}{2} = \{ \frac {3}{2}, \frac {5}{2}, \frac {7}{2},\ldots \}$ and $c= \frac {1}{\log x}$ . Minor improvements below are possible; these were eschewed in favor of a final estimate of simple shape.

3.1 When n is very far from x

Suppose that $n \leq \frac {x}{2}$ or $\frac {3x}{2} \geq n$ . Then $\log \frac {x}{n} \leq -\log \frac {3}{2}$ or $\log \frac {3}{2} <\log 2 \leq \log \frac {x}{n}$ , so $|\log \frac {x}{n}| \geq \log \frac {3}{2}$ . If $T \geq (\log \frac {3}{2})^{-1}> 2.46$ , then

$$ \begin{align*} \min\left\{ 1 , \frac{1}{T | \log \frac{x}{n}| } \right\} \leq \frac{1}{T |\log \frac{x}{n}|} \leq \frac{1}{T \log \frac{3}{2}}. \end{align*} $$

For such T, the previous inequality and Lemma 2.1 imply (recall that $c = \frac {1}{\log x}$ )

(3.1) $$ \begin{align} \sum_{\substack{n \leq \frac{x}{2}\, \text{or}\\ n \geq \frac{3x}{2}}} \Big( \frac{x}{n}\Big)^c \frac{ \Lambda(n) }{n \log n} \min\Big\{ 1 , \frac{1}{T | \log \frac{x}{n} |} \Big\} &\leq x^c \sum_{\substack{n \leq \frac{x}{2}\, \text{or}\\ n \geq \frac{3x}{2} }} \frac{1}{n^{1+c}}\bigg( \frac{\Lambda(n)}{\log n}\bigg) \bigg(\frac{1}{T \log \frac{3}{2} } \bigg) \nonumber \\ &\leq \frac{ x^c}{T\log \frac{3}{2}} \log \zeta(1+c) \nonumber \\ &\leq \frac{ x^c}{T \log \frac{3}{2}} ( - \log c + \gamma c) \nonumber \\ &= \frac{e}{T \log \frac{3}{2}}\bigg(\log \log x + \frac{\gamma}{\log x} \bigg). \end{align} $$

3.2 Reduction to a sum over prime powers

Suppose that $\frac {x}{2} < n < \frac {3x}{2}$ . Let $z = 1-\frac {n}{x}$ and observe that $|z| < \frac {1}{2}$ . Then

$$ \begin{align*} \log \frac{x}{n} = - \log(1-z) = z \left(- \frac{\log(1-z)}{z} \right), \end{align*} $$

in which the function in parentheses is positive and achieves its minimum value $2 \log \frac {3}{2} = 0.81093\ldots $ on $|z| < \frac {1}{2}$ at its left endpoint $-\frac {1}{2}$ (see Figure 4a). Then

(3.2) $$ \begin{align} | \log(1-z) |> \bigg(2 \log \frac{3}{2} \bigg) |z| \quad\text{for}\quad |z| < \frac{1}{2}, \end{align} $$

whose validity is illustrated in Figure 4b. Therefore,

(3.3) $$ \begin{align} \bigg| \log \frac{x}{n} \bigg|> \bigg( 2 \log \frac{3}{2}\bigg) \bigg|1 - \frac{n}{x}\bigg| \quad\text{for}\quad \frac{x}{2} < n < \frac{3x}{2}, \end{align} $$

Figure 4 Graphs relevant to the derivation of (3.2).

and hence

(3.4) $$ \begin{align} & \sum_{\frac{x}{2} < n < \frac{3x}{2}} \Big( \frac{x}{n}\Big)^c \frac{ \Lambda(n) }{n \log n} \min\left\{ 1 , \frac{1}{T | \log \frac{x}{n} |} \right\} \nonumber \\&\qquad\leq \frac{x^c}{T} \sum_{ \frac{x}{2} < n < \frac{3x}{2}} \bigg( \frac{ \Lambda(n) }{n^{1+c} \log n}\bigg) \bigg( \frac{1}{ | \log \frac{x}{n} |} \bigg) \nonumber \\&\qquad\leq \frac{x^c}{T} \sum_{\frac{x}{2} < n < \frac{3x}{2}} \bigg( \frac{ \Lambda(n) }{n^{1+c} \log n}\bigg) \frac{1}{ (2\log \frac{3}{2}) | 1 - \frac{n}{x} |} && (\text{by (3.3)}) \nonumber \\ &\qquad\leq \frac{ x^c}{T (2\log\frac{3}{2}) } \sum_{\frac{x}{2} < n < \frac{3x}{2}} \frac{2}{x} \bigg( \frac{ \Lambda(n) }{n^c \log n}\bigg)\frac{1}{ | 1 - \frac{n}{x} |} && (\text{since }\tfrac{x}{2} < n) \nonumber \\ &\qquad\leq \frac{x^c }{T \log \frac{3}{2}} \sum_{\frac{x}{2} < n < \frac{3x}{2}} \bigg( \frac{ \Lambda(n) }{n^c \log n}\bigg)\frac{1}{ | x-n |} \nonumber \\ &\qquad\leq \frac{x^c }{T \log\frac{3}{2}} \sum_{\frac{x}{2} < p^k < \frac{3 x}{2}} \bigg( \frac{ \log p }{(p^k)^c k\log p}\bigg)\frac{1}{ | x-p^k |} && (\text{def. of }\Lambda)\nonumber \\ &\qquad\leq \frac{e }{T \log\frac{3}{2}} \sum_{\frac{x}{2} < p^k < \frac{3x}{2}} \frac{1}{k | x-p^k |} &&(\text{since }c=\tfrac{1}{\log x}), \end{align} $$

in which the final two sums run over all prime powers $p^k$ in the stated interval.

The remainder of the proof uses ideas from [5, Lemma 2] to estimate

(3.5) $$ \begin{align} \sum_{\frac{x}{2} < p^k < \frac{3x}{2}} \frac{1}{ k | x-p^k |} = \underbrace{\sum_{\frac{x}{2} < p < \frac{3x}{2}} \frac{1}{ | x-p |} }_{S_{\mathrm{prime}}(x)} + \underbrace{ \sum_{ \substack{\frac{x}{2} < p^k < \frac{3x}{2}\\ k\geq 2}} \frac{1}{ k | x-p^k |} }_{S_{\mathrm{power}}(x)}. \end{align} $$

3.3 The sum over primes

First observe that

$$ \begin{align*} S_{\mathrm{prime}}(x) \leq \underbrace{\sum_{\frac{x}{2} < p < x} \frac{1}{ x-p } }_{S_{\mathrm{prime}}^-(x)} + \underbrace{\sum_{x < p < \frac{3x}{2}} \frac{1}{ p-x} }_{S_{\mathrm{prime}}^+(x)}. \end{align*} $$

We require the Brun–Titchmarsh theorem (see [10, Corollary 2]):

(3.6) $$ \begin{align} \pi(X+Y) - \pi(X) \leq \frac{2Y}{\log{Y}},\quad\text{where }\pi(x) = \sum_{p\leq x}1, X> 0\text{, and }Y > 1. \end{align} $$

3.3.1 The lower sum over primes

Let $p_{-k} < p_{-(k-1)} < \cdots < p_{-2} < p_{-1}$ be the primes in $(\frac {x}{2}, x)$ ; note that $k \leq \frac {x}{2}$ . Apply (3.6) with $X = x-y_n$ and $Y = y_n$ to get

$$ \begin{align*}\qquad 0 \leq \pi(x) - \pi(x-y_n) \leq \frac{2y_n}{\log y_n} = n \qquad \text{for }6 \leq n \leq k \end{align*} $$

by Lemma 2.2, so $(x-y_n,x]$ contains at most n primes. Thus, $p_{-(n+1)} \leq x - y_n$ and

(3.7) $$ \begin{align} \frac{1}{x-p_{-(n+1)}} \leq \frac{1}{y_n} \qquad \text{for }6 \leq n \leq k-1. \end{align} $$

Then Lemma 2.2, which requires $k \geq 8$ , and the integral test provide

$$ \begin{align*} \sum_{ \frac{x}{2} < p < x} \frac{1}{ x-p } &=\sum_{ 1 \leq n\leq 8} \frac{1}{ x-p_{-n} } + \sum_{9 \leq n \leq k} \frac{1}{ x-p_{-n} } \\&\leq F_1(x) + \sum_{8\leq n \leq k-1} \frac{1}{y_n} && (\text{by (2.2) and (3.7)})\\&\leq C + 2 \sum_{7 < n \leq \frac{x}{2}} \frac{1}{n \log n} && (\text{by Lemma 2.2})\\&\leq C + 2 \log \log x, \end{align*} $$

which is valid for $k \leq 7$ since Lemma 2.4a shows that the sum is majorized by C.

3.3.2 The upper sum over primes

Let $p_1<p_2< \cdots < p_k$ denote the primes in $(x, \frac {3x}{2})$ and note that $k \leq \frac {x}{2}$ . Then (3.6) with $X = x$ and $Y = y_n$ ensures that

$$ \begin{align*} 0 \leq \pi(x+y_n) - \pi(x) \leq \frac{2y_n}{\log y_n} = n \qquad \text{for}\qquad 6 \leq n \leq k \end{align*} $$

by Lemma 2.2, so $(x, x+y_n]$ contains at most n primes. Thus, $p_{n+1} \geq x + y_n$ and

(3.8) $$ \begin{align} \frac{1}{p_{n+1} - x} \leq \frac{1}{y_n} \qquad \text{for}\qquad 6 \leq n \leq k. \end{align} $$

An argument similar to that above reveals that

$$ \begin{align*} \sum_{x < p < \frac{3x}{2} } \frac{1}{p-x} \leq \sum_{1\leq n \leq 8} \frac{1}{p_n -x} + \sum_{9\leq n \leq k} \frac{1}{p_n-x } \leq C +2 \log\log x. \end{align*} $$

3.3.3 Final bound over primes

For $x \in \mathbb {N} + \frac {1}{2}$ , the previous inequalities yield

(3.9) $$ \begin{align} S_{\mathrm{prime}}(x) = \sum_{ \frac{x}{2} < p < \frac{3x}{2}} \frac{1}{ | x-p |} \leq 2C +4 \log\log x. \end{align} $$

3.4 The sum over prime powers

We now majorize

$$ \begin{align*} S_{\mathrm{power}}(x) = \sum_{ \substack{ \frac{x}{2} < p^k <\frac{3x}{2} \\ k\geq 2}} \frac{1}{ k | x-p^k |}. \end{align*} $$

3.4.1 Initial reduction

To bound $S_{\mathrm {power}}(x)$ it suffices to majorize

(3.10) $$ \begin{align} S_{\mathrm{sqf}}(x) = \sum_{ \substack{ \frac{x}{2} < n^k <\frac{3x}{2} \\ k\geq 2 \\ n \geq 2 \text{sq.~free} }} \frac{1}{ k | x-n^k |}, \end{align} $$

in which the prime powers $p^k$ are replaced with the powers $n^k$ of square free $n \geq 2$ . The square-free restriction ensures that powers such as $2^6 = (2^2)^3 = (2^3)^2$ are not counted multiple times in (3.10). If $\frac {x}{2} < n^k < \frac {3x}{2}$ and $k \geq 2$ , then (since $n \geq 2$ )

(3.11) $$ \begin{align} k \leq \frac{\log \frac{3x}{2} }{\log 2} \leq \lfloor 2.4 \log x \rfloor \qquad \text{for}\qquad x \geq 3.5. \end{align} $$

3.4.2 Nearest-power sets

The largest contributions to $S_{\mathrm {sqf}}(x)$ come from the powers closest to x. We handle those summands separately and split the sum (3.10) accordingly. For each $k \geq 2$ , the inequalities $\lfloor x^{\frac {1}{k}} \rfloor ^k < x < \lceil x^{\frac {1}{k}} \rceil ^k$ exhibit the two kth powers nearest to x. Define

(3.12) $$ \begin{align} \mathcal{N}_k \subseteq \big\{ \lfloor x^{\frac{1}{k}} \rfloor^k , \lceil x^{\frac{1}{k}} \rceil^k \big\} \end{align} $$

according to the following rules:

• • $\mathcal {N}_k$ contains $\lfloor x^{\frac {1}{k}} \rfloor ^k$ if it is square free and belongs to $( \frac {x}{2}, \frac {3x}{2} )$ .

• • $\mathcal {N}_k$ contains $\lceil x^{\frac {1}{k}} \rceil ^k$ if it is square free and belongs to $( \frac {x}{2},\frac {3x}{2} )$ .

Consequently, $\mathcal {N}_k$ , if nonempty, contains only powers that satisfy the restrictions in (3.10). The square-free condition ensures that $\mathcal {N}_j \cap \mathcal {N}_k = \varnothing $ for $j \neq k$ .

Write $S_{\mathrm {sqf}}(x) = S_{\text {near}}(x) + S_{\text {far}}(x)$ , in which

(3.13) $$ \begin{align} S_{\text{near}}(x) \,\,= \!\!\!\! \sum_{ \substack{ \frac{x}{2} < n^k < \frac{3x}{2} \\ k \geq 2 \\ n \geq 2\text{ sq.~free} \\ n^k \in \mathcal{N}_k}} \frac{1}{ k | x-n^k |} \quad \text{and} \quad S_{\text{far}}(x) \,\,= \!\!\!\! \sum_{ \substack{ \frac{x}{2} < n^k < \frac{3x}{2} \\ k \geq 2 \\ n \geq 2\text{ sq.~free} \\ n^k \notin \mathcal{N}_k}} \frac{1}{ k | x-n^k |}. \end{align} $$

3.4.3 Near Sum

For $x\geq 3.5$ , a nearest-neighbor overestimate provides

(3.14) $$ \begin{align} S_{\text{near}}(x) &= \sum_{ \substack{ \frac{x}{2} < n^k < \frac{3x}{2} \\ k \geq 2 \\ n \geq 2\text{ sq.~free} \\ n^k \in \mathcal{N}_k}} \frac{1}{ k | x-n^k |} && (\text{by 3.13}) \nonumber\\ &= \sum_{k=2}^{\lfloor 2.4\log x \rfloor } \sum_{ m \in \mathcal{N}_k } \frac{1}{ k | x-m |} && (\text{by 3.11}) \nonumber\\ &\leq \frac{1}{2} \sum_{k=2}^{\lfloor 2.4\log x \rfloor } \sum_{ m \in \mathcal{N}_k } \frac{1}{ | x-m |} && ( \text{since }k \geq 2 ) \nonumber\\ &\leq \frac{1}{2} \sum_{j=0}^{\lfloor 2.4\log x \rfloor-2 } \left( \frac{1}{ x - (\lfloor x \rfloor-j)} + \frac{1}{ (\lceil x \rceil +j) - x} \right) && (\text{see below}) \nonumber\\ &\leq \frac{1}{2} \sum_{\ell=1}^{\lfloor 2.4\log x \rfloor-1 } \frac{2}{\ell - \frac{1}{2}} \quad<\quad 2 \!\!\!\!\sum_{\ell=1}^{\lfloor 2.4\log x \rfloor } \frac{1}{2\ell - 1} \nonumber\\ &\leq \log( \lfloor 2.4\log x \rfloor) + \gamma + 2\log 2 + \frac{14}{312}&& ( \text{by 2.5}) \nonumber\\ &< \log \log x + \gamma + 2\log 2 + \log 2.4 + \frac{7}{156}. \end{align} $$

Let us elaborate on a crucial step above. Consider the at most $\lfloor 2 \log x \rfloor - 1$ pairs of values $|x-m|$ that arise as m ranges over each $\mathcal {N}_k$ with $2 \leq k \leq \lfloor 2 \log x \rfloor $ (since $\mathcal {N}_j(x) \cap \mathcal {N}_k = \varnothing $ for $j \neq k$ , no m appears more than once). Replace these values with the absolute deviations of x from its $2 \times (\lfloor 2 \log x \rfloor - 1)$ nearest neighbors $\lfloor x \rfloor -j$ (to the left) and $\lceil x \rceil +j$ (to the right), in which $0 \leq j \leq \lfloor 2 \log x \rfloor -2$ . Since $x \in \mathbb {N}+\frac {1}{2}$ , these deviations are of the form $\ell - \frac {1}{2}$ for $1 \leq \ell \leq \lfloor 2\log x \rfloor -1$ .

3.4.4 Splitting the second sum

From (3.13), the second sum in question is

$$ \begin{align*} S_{\text{far}}(x) = \!\!\!\! \sum_{ \substack{ \frac{x}{2} < n^k < \frac{3x}{2} \\ k \geq 2 \\ n \geq 2\text{ sq.~free} \\ n^k \notin \mathcal{N}_k}} \frac{1}{ k | x-n^k |} \leq \!\!\!\! \sum_{ \substack{ \frac{x}{2} < n^k < \frac{3x}{2} \\ n,k \geq 2 \\ n^k \notin \mathcal{N}_k}} \frac{1}{ k | x-n^k |} = S_{\text{far}}^-(x) + S_{\text{far}}^+(x), \end{align*} $$

in which

(3.15) $$ \begin{align} S_{\text{far}}^-(x) = \sum_{k\geq 2} \sum_{ \substack{ \frac{x}{2} < n^k < x \\ n \geq 2, n^k \notin \mathcal{N}_k}} \frac{1}{ k | x-n^k |} \quad \text{and} \quad S_{\text{far}}^+(x) = \sum_{k\geq 2} \sum_{ \substack{ x < n^k < \frac{3x}{2} \\ n \geq 2, n^k \notin \mathcal{N}_k}} \frac{1}{ k | x-n^k |}. \end{align} $$

For $k \geq 2$ , Lemma 2.5 with $X =h= \frac {x}{2}$ , then with $X = x$ and $h = \frac {x}{2}$ , implies that

(3.16) $$ \begin{align} G_k^- \geq \frac{k \sqrt{x}}{\sqrt{2}}, \quad N_k^- \leq \frac{\sqrt{x}}{2 \sqrt{2}}, \qquad \text{and} \qquad G_k^+ \geq k \sqrt{x}, \quad N_k^+ \leq \frac{ \sqrt{x}}{4} \end{align} $$

are admissible in Figure 5. For $1\leq j\leq N^-_k$ and $1\leq j\leq N^+_k$ , respectively,

Figure 5 Analysis of kth powers in $[\tfrac {x}{2}, \tfrac {3x}{2}]$ , in which $m_- = \lfloor x^{1/k} \rfloor $ and $m_+ = \lceil x^{1/k} \rceil $ are excluded from consideration. There are at most $N_k^-$ admissible kth powers in $[\frac {x}{2},x)$ , with minimal gap size $G_k^-$ , and at most $N_k^+$ admissible kth powers in $[x,\frac {3x}{2})$ , with minimal gap size $G_k^+$ .

$$ \begin{align*} |x - (m_- - j)^k| \geq \frac{j k \sqrt{x}}{\sqrt{2}} \qquad\text{and}\qquad |x - (m_+ + j)^k| \geq j k \sqrt{x}. \end{align*} $$

Let $N_k^{\pm } \geq 1$ , since otherwise the corresponding sum estimated below is zero. Then

(3.17) $$ \begin{align} \sum_{ \substack{\frac{x}{2} < n^k < x \\ n \geq 2, n^k \notin \mathcal{N}_k}} \frac{1}{ k | x-n^k |} = \sum_{j=1}^{N^-_k} \frac{1}{k |x-(m_- - j)^k|} \leq \frac{\sqrt{2}H_{N^-_k} }{k^2 \sqrt{x}} \end{align} $$

and

(3.18) $$ \begin{align} \sum_{\substack{x < n^k < \tfrac{3x}{2} \\ n \geq 2, n^k \notin \mathcal{N}_k}} \frac{1}{ k | x-n^k |} = \sum_{j=1}^{N^+_k} \frac{1}{k |x-(m_+ + j)^k|} \leq \frac{H_{N^+_k}}{k^2 \sqrt{x}}. \end{align} $$

Therefore,

(3.19) $$ \begin{align} S_{\text{far}}^-(x) &=\sum_{k\geq 2} \sum_{ \substack{ \frac{x}{2} < n^k < x \\ n \geq 2, n^k \notin \mathcal{N}_k}} \frac{1}{ k | x-n^k |} \leq \frac{\sqrt{2}}{\sqrt{x}} \sum_{k\geq 2}\frac{H_{N^-_k} }{k^2 } && (\text{by (3.15) and (3.17)}) \nonumber\\ &\leq \frac{\sqrt{2}}{\sqrt{x}} \bigg(\frac{1}{2}\log{x} - \frac{3}{2}\log{2} + \gamma + \frac{3}{7} \bigg) \sum_{k\geq 2}\frac{1}{k^2} && (\text{by (2.4) and (3.16)}) \nonumber \\ &=\frac{\pi^2-6}{3\sqrt{2 x}} \bigg(\frac{1}{2}\log{x} - \frac{3}{2}\log{2} + \gamma + \frac{3}{7} \bigg) &&(\text{since }\zeta(2)-1 = \tfrac{\pi^2-6}{6}) \end{align} $$

and

(3.20) $$ \begin{align} S_{\text{far}}^+(x) &=\sum_{k\geq 2} \sum_{ \substack{ x < n^k < \frac{3x}{2} \\ n \geq 2, n^k \notin \mathcal{N}_k}} \frac{1}{ k | x-n^k |} \leq \frac{1}{ \sqrt{x}} \sum_{k \geq 2} \frac{H_{N^+_k}}{k^2} && (\text{by (3.15) and (3.18)}) \nonumber\\ &\leq \frac{1}{ \sqrt{x}} \bigg(\frac{1}{2}\log{x} - 2\log{2} + \gamma + \frac{3}{7} \bigg) \sum_{k \geq 2} \frac{1}{k^2} && (\text{by (2.4) and (3.16)}) \nonumber \\ &= \frac{\pi^2-6}{6\sqrt{x}} \bigg(\frac{1}{2}\log{x} - 2\log{2} + \gamma + \frac{3}{7} \bigg) &&(\text{since }\zeta(2)-1 = \tfrac{\pi^2-6}{6}). \end{align} $$

3.4.5 Final prime-power estimate

Using (3.14), (3.19), and (3.20), we can bound

$$ \begin{align*} S_{\mathrm{power}}(x) \leq S_{\mathrm{sqf}} (x) = S_{\mathrm{near}}(x) + S_{\mathrm{far}}^-(x) + S_{\mathrm{far}}^+(x). \end{align*} $$

We postpone doing this explicitly until the finale below.

4 Conclusion

For $x \geq 3.5$ , with $T \geq (\log \frac {3}{2})^{-1}$ , the sum (1.3) is bounded by

$$ \begin{align*} &\underbrace{\frac{e}{T \log \frac{3}{2}}\bigg(\log \log x + \frac{\gamma}{\log x} \bigg)}_{\text{by (3.1)}} + \underbrace{\frac{e }{T \log\frac{3}{2}} \sum_{\frac{x}{2} < p^k < \frac{3x}{2}} \frac{1}{k | x-p^k |} }_{\text{by (3.4)}} \\ &\quad \leq \frac{e}{T \log \frac{3}{2}} \bigg(\log \log x + \frac{\gamma}{\log x} \bigg) + \frac{e}{T \log \frac{3}{2}} \underbrace{ \big(S_{\text{prime}}(x) + S_{\text{sqf}}(x) \big)}_{\text{by (3.5) and (3.10)}} \\ &\quad \leq \frac{e}{T \log \frac{3}{2}} \bigg(\log \log x + \frac{\gamma}{\log x} \bigg) + \frac{e}{T \log \frac{3}{2}}\bigg[ \underbrace{2C +4 \log\log x}_{S_{\text{prime}}(x)\text{ bounded by (3.9)}} \\ &\qquad + \underbrace{ \bigg(\log \log x + \gamma + 2\log 2 + \log 2.4 +\frac{7}{156} \bigg) }_{S_{\text{near}}(x)\text{ bounded by (3.14)}} + \underbrace{\frac{\pi^2-6}{3\sqrt{2 x}} \bigg(\frac{1}{2}\log{x} - \frac{3}{2}\log{2} + \gamma + \frac{3}{7} \bigg)}_{S_{\text{far}}^-\text{ bounded by (3.19)}} \\ &\qquad + \underbrace{\frac{\pi^2-6}{6\sqrt{x}} \bigg(\frac{1}{2}\log{x} - 2\log{2} + \gamma + \frac{3}{7} \bigg)}_{S_{\text{far}}^+(x)\text{ bounded by (3.20)}} \bigg] \\ &\quad < \frac{1}{T}\bigg( 40.22465 \log \log x +58.11106 +\frac{3.86972}{\log x} +\frac{5.21918 \log x}{\sqrt{x}} -\frac{1.85268}{\sqrt{x}} \bigg). \end{align*} $$