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
$$ \begin{align} \sum_{n \leq x} f(n) = \frac{1}{2\pi i }\int_{c-iT}^{c+iT} F(s) \frac{x^s}{s}\,ds + O^*\bigg( \sum_{n=1}^{\infty} \bigg( \frac{x}{n} \bigg)^{c} |f(n)| \min\left\{ 1, \frac{1}{T |\log \frac{x}{n}|}\right\}\bigg), \end{align} $$
in which
$O^*(g(x)) = h(x)$
means
$|h(x)|\leq g(x)$
(see [Reference Koukoulopoulos8, Chapter 7], [Reference Montgomery and Vaughan10, Section 5.1], [Reference Murty11, Example 4.4.15], and [Reference Tenenbaum15, 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
$T^*\in [T,O(T)]$
[Reference Cully-Hugill and Johnston3], although this is inconvenient if one must avoid
$T^*$
that correspond to the ordinates of nontrivial zeros of
$F(s)$
. The authors of [Reference Cully-Hugill and Johnston3] 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 [Reference Iwaniec and Kowalski7, Chapter 5]. In these cases, the error term in (1.1) with
$c = 1/\log {x}$
is on the order of
$$ \begin{align} \sum_{n=2}^{\infty} \Big( \frac{x}{n}\Big)^{\frac{1}{\log x}} \frac{ \Lambda(n)}{n \log n} \min\left\{ 1 , \frac{1}{T | \log \frac{x}{n} |} \right\} = O\bigg( \frac{\log x}{T} \bigg). \end{align} $$
Granville and Soundararajan used ( 1.2 ) with Dirichlet L-functions to study large character sums [Reference Granville and Soundararajan6, equation (8.1)]. Cho and Kim applied it to Artin L-functions to obtain asymptotic bounds on Dedekind zeta residues [Reference Cho and Kim1, Proposition 3.1]. A bilinear relative of (1.2) appears in Selberg’s work on primes in short intervals [Reference Selberg14, Lemma 4]. Analogous sums arise with the logarithmic derivative of an L-function in [Reference Davenport4, p. 106] and [Reference Patterson12, 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
$$ \begin{align} \sum_{n=2}^{\infty} \Big( \frac{x}{n}\Big)^{\frac{1}{\log x}} \frac{ \Lambda(n)}{n \log n} \min\left\{ 1 , \frac{1}{T | \log \frac{x}{n} |} \right\} \leq \frac{R(x)}{T}, \end{align} $$
in which
$$ \begin{align} R(x) = 40.23 \log \log x +58.12 +\frac{3.87}{\log x} +\frac{5.22 \log x}{\sqrt{x}} -\frac{1.84}{\sqrt{x}}. \end{align} $$
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
$\zeta (s) \leq e^{\gamma (s-1)}/(s-1)$
[Reference Ramaré13, 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 [Reference Corless, Gonnet, Hare, Jeffrey and Knuth2], which decreases to
$-\infty $
as
$z\to 0^-$
(see Figure 1a). For
$n \geq 6>2e$
, we define the strictly increasing sequence
$$ \begin{align}\qquad y_n = \frac{-n}{2} W_{-1}\Big( \frac{-2}{n} \Big) \qquad \text{for }n \geq 6. \end{align} $$
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 [Reference Lóczi9, 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. 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)(see Figure 2a). The corresponding summand in ( 2.2 ) is zero if
$$ \begin{align} F_1(x) = \sum_{1 \leq n \leq 8} \frac{1}{ x-p_{-n} } \leq C \end{align} $$
$p_{-n}$
does not exist.
-
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)(see Figure 2a). The corresponding summand in ( 2.3 ) is zero if
$$ \begin{align} F_2(x) = \sum_{1 \leq n \leq 8} \frac{1}{p_n -x} \leq C \end{align} $$
$p_{n}$
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) 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) 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
$H_n = \sum _{j=1}^n \frac {1}{j}$
:
$$ \begin{align}\qquad \frac{1}{2n+ \frac{2}{5}} < H_n - \log n - \gamma < \frac{1}{2n + \frac{1}{3}} \leq \frac{3}{7} \quad\text{for}\quad n \geq 1, \end{align} $$
in which
$\gamma $
is the Euler–Mascheroni constant [Reference Tóth and Mare16]. We require the upper bound
$$ \begin{align} \sum_{\ell=1}^n \frac{1}{2\ell - 1} &= \left(1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{2n}\right) - \left( \frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{2n} \right) = H_{2n} - \frac{1}{2} H_n \nonumber \\ &\leq \bigg(\log 2n +\gamma + \frac{1}{4n + \frac{1}{3}} \bigg) - \frac{1}{2} \bigg( \log n + \gamma + \frac{1}{2n+\frac{2}{5}} \bigg) \nonumber \\ &\leq \frac{1}{2}\log n + \frac{1}{2}\gamma + \log 2 + \frac{1}{4n + \frac{1}{3}}- \frac{1}{4n+\frac{4}{5}} \nonumber \\ &= \frac{1}{2}\log n + \frac{1}{2}\gamma + \log 2 + \frac{7}{240 n^2+68 n+4} \nonumber\\ &\leq \frac{1}{2}\log n + \frac{1}{2}\gamma + \log 2 + \frac{7}{312} \quad \text{for}\quad n \geq 1. \end{align} $$
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
$c = \frac {1}{\log x}$
)
$$ \begin{align} \sum_{\substack{n \leq \frac{x}{2}\, \text{or}\\ n \geq \frac{3x}{2}}} \Big( \frac{x}{n}\Big)^c \frac{ \Lambda(n) }{n \log n} \min\Big\{ 1 , \frac{1}{T | \log \frac{x}{n} |} \Big\} &\leq x^c \sum_{\substack{n \leq \frac{x}{2}\, \text{or}\\ n \geq \frac{3x}{2} }} \frac{1}{n^{1+c}}\bigg( \frac{\Lambda(n)}{\log n}\bigg) \bigg(\frac{1}{T \log \frac{3}{2} } \bigg) \nonumber \\ &\leq \frac{ x^c}{T\log \frac{3}{2}} \log \zeta(1+c) \nonumber \\ &\leq \frac{ x^c}{T \log \frac{3}{2}} ( - \log c + \gamma c) \nonumber \\ &= \frac{e}{T \log \frac{3}{2}}\bigg(\log \log x + \frac{\gamma}{\log x} \bigg). \end{align} $$
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
$$ \begin{align} | \log(1-z) |> \bigg(2 \log \frac{3}{2} \bigg) |z| \quad\text{for}\quad |z| < \frac{1}{2}, \end{align} $$
whose validity is illustrated in Figure 4b. Therefore,
$$ \begin{align} \bigg| \log \frac{x}{n} \bigg|> \bigg( 2 \log \frac{3}{2}\bigg) \bigg|1 - \frac{n}{x}\bigg| \quad\text{for}\quad \frac{x}{2} < n < \frac{3x}{2}, \end{align} $$
Figure 4 Graphs relevant to the derivation of (3.2).
and hence
$$ \begin{align} & \sum_{\frac{x}{2} < n < \frac{3x}{2}} \Big( \frac{x}{n}\Big)^c \frac{ \Lambda(n) }{n \log n} \min\left\{ 1 , \frac{1}{T | \log \frac{x}{n} |} \right\} \nonumber \\&\qquad\leq \frac{x^c}{T} \sum_{ \frac{x}{2} < n < \frac{3x}{2}} \bigg( \frac{ \Lambda(n) }{n^{1+c} \log n}\bigg) \bigg( \frac{1}{ | \log \frac{x}{n} |} \bigg) \nonumber \\&\qquad\leq \frac{x^c}{T} \sum_{\frac{x}{2} < n < \frac{3x}{2}} \bigg( \frac{ \Lambda(n) }{n^{1+c} \log n}\bigg) \frac{1}{ (2\log \frac{3}{2}) | 1 - \frac{n}{x} |} && (\text{by (3.3)}) \nonumber \\ &\qquad\leq \frac{ x^c}{T (2\log\frac{3}{2}) } \sum_{\frac{x}{2} < n < \frac{3x}{2}} \frac{2}{x} \bigg( \frac{ \Lambda(n) }{n^c \log n}\bigg)\frac{1}{ | 1 - \frac{n}{x} |} && (\text{since }\tfrac{x}{2} < n) \nonumber \\ &\qquad\leq \frac{x^c }{T \log \frac{3}{2}} \sum_{\frac{x}{2} < n < \frac{3x}{2}} \bigg( \frac{ \Lambda(n) }{n^c \log n}\bigg)\frac{1}{ | x-n |} \nonumber \\ &\qquad\leq \frac{x^c }{T \log\frac{3}{2}} \sum_{\frac{x}{2} < p^k < \frac{3 x}{2}} \bigg( \frac{ \log p }{(p^k)^c k\log p}\bigg)\frac{1}{ | x-p^k |} && (\text{def. of }\Lambda)\nonumber \\ &\qquad\leq \frac{e }{T \log\frac{3}{2}} \sum_{\frac{x}{2} < p^k < \frac{3x}{2}} \frac{1}{k | x-p^k |} &&(\text{since }c=\tfrac{1}{\log x}), \end{align} $$
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 [Reference Goldston5, Lemma 2] to estimate
$$ \begin{align} \sum_{\frac{x}{2} < p^k < \frac{3x}{2}} \frac{1}{ k | x-p^k |} = \underbrace{\sum_{\frac{x}{2} < p < \frac{3x}{2}} \frac{1}{ | x-p |} }_{S_{\mathrm{prime}}(x)} + \underbrace{ \sum_{ \substack{\frac{x}{2} < p^k < \frac{3x}{2}\\ k\geq 2}} \frac{1}{ k | x-p^k |} }_{S_{\mathrm{power}}(x)}. \end{align} $$
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 [Reference Montgomery and Vaughan10, Corollary 2]):
$$ \begin{align} \pi(X+Y) - \pi(X) \leq \frac{2Y}{\log{Y}},\quad\text{where }\pi(x) = \sum_{p\leq x}1, X> 0\text{, and }Y > 1. \end{align} $$
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
$$ \begin{align} \frac{1}{x-p_{-(n+1)}} \leq \frac{1}{y_n} \qquad \text{for }6 \leq n \leq k-1. \end{align} $$
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
$$ \begin{align} \frac{1}{p_{n+1} - x} \leq \frac{1}{y_n} \qquad \text{for}\qquad 6 \leq n \leq k. \end{align} $$
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
$$ \begin{align} S_{\mathrm{prime}}(x) = \sum_{ \frac{x}{2} < p < \frac{3x}{2}} \frac{1}{ | x-p |} \leq 2C +4 \log\log x. \end{align} $$
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
$$ \begin{align} S_{\mathrm{sqf}}(x) = \sum_{ \substack{ \frac{x}{2} < n^k <\frac{3x}{2} \\ k\geq 2 \\ n \geq 2 \text{sq.~free} }} \frac{1}{ k | x-n^k |}, \end{align} $$
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
$n \geq 2$
)
$$ \begin{align} k \leq \frac{\log \frac{3x}{2} }{\log 2} \leq \lfloor 2.4 \log x \rfloor \qquad \text{for}\qquad x \geq 3.5. \end{align} $$
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
$$ \begin{align} \mathcal{N}_k \subseteq \big\{ \lfloor x^{\frac{1}{k}} \rfloor^k , \lceil x^{\frac{1}{k}} \rceil^k \big\} \end{align} $$
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
$$ \begin{align} S_{\text{near}}(x) \,\,= \!\!\!\! \sum_{ \substack{ \frac{x}{2} < n^k < \frac{3x}{2} \\ k \geq 2 \\ n \geq 2\text{ sq.~free} \\ n^k \in \mathcal{N}_k}} \frac{1}{ k | x-n^k |} \quad \text{and} \quad S_{\text{far}}(x) \,\,= \!\!\!\! \sum_{ \substack{ \frac{x}{2} < n^k < \frac{3x}{2} \\ k \geq 2 \\ n \geq 2\text{ sq.~free} \\ n^k \notin \mathcal{N}_k}} \frac{1}{ k | x-n^k |}. \end{align} $$
3.4.3 Near Sum
For
$x\geq 3.5$
, a nearest-neighbor overestimate provides
$$ \begin{align} S_{\text{near}}(x) &= \sum_{ \substack{ \frac{x}{2} < n^k < \frac{3x}{2} \\ k \geq 2 \\ n \geq 2\text{ sq.~free} \\ n^k \in \mathcal{N}_k}} \frac{1}{ k | x-n^k |} && (\text{by 3.13}) \nonumber\\ &= \sum_{k=2}^{\lfloor 2.4\log x \rfloor } \sum_{ m \in \mathcal{N}_k } \frac{1}{ k | x-m |} && (\text{by 3.11}) \nonumber\\ &\leq \frac{1}{2} \sum_{k=2}^{\lfloor 2.4\log x \rfloor } \sum_{ m \in \mathcal{N}_k } \frac{1}{ | x-m |} && ( \text{since }k \geq 2 ) \nonumber\\ &\leq \frac{1}{2} \sum_{j=0}^{\lfloor 2.4\log x \rfloor-2 } \left( \frac{1}{ x - (\lfloor x \rfloor-j)} + \frac{1}{ (\lceil x \rceil +j) - x} \right) && (\text{see below}) \nonumber\\ &\leq \frac{1}{2} \sum_{\ell=1}^{\lfloor 2.4\log x \rfloor-1 } \frac{2}{\ell - \frac{1}{2}} \quad<\quad 2 \!\!\!\!\sum_{\ell=1}^{\lfloor 2.4\log x \rfloor } \frac{1}{2\ell - 1} \nonumber\\ &\leq \log( \lfloor 2.4\log x \rfloor) + \gamma + 2\log 2 + \frac{14}{312}&& ( \text{by 2.5}) \nonumber\\ &< \log \log x + \gamma + 2\log 2 + \log 2.4 + \frac{7}{156}. \end{align} $$
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
$$ \begin{align} S_{\text{far}}^-(x) = \sum_{k\geq 2} \sum_{ \substack{ \frac{x}{2} < n^k < x \\ n \geq 2, n^k \notin \mathcal{N}_k}} \frac{1}{ k | x-n^k |} \quad \text{and} \quad S_{\text{far}}^+(x) = \sum_{k\geq 2} \sum_{ \substack{ x < n^k < \frac{3x}{2} \\ n \geq 2, n^k \notin \mathcal{N}_k}} \frac{1}{ k | x-n^k |}. \end{align} $$
For
$k \geq 2$
, Lemma 2.5 with
$X =h= \frac {x}{2}$
, then with
$X = x$
and
$h = \frac {x}{2}$
, implies that
$$ \begin{align} G_k^- \geq \frac{k \sqrt{x}}{\sqrt{2}}, \quad N_k^- \leq \frac{\sqrt{x}}{2 \sqrt{2}}, \qquad \text{and} \qquad G_k^+ \geq k \sqrt{x}, \quad N_k^+ \leq \frac{ \sqrt{x}}{4} \end{align} $$
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
$$ \begin{align} \sum_{ \substack{\frac{x}{2} < n^k < x \\ n \geq 2, n^k \notin \mathcal{N}_k}} \frac{1}{ k | x-n^k |} = \sum_{j=1}^{N^-_k} \frac{1}{k |x-(m_- - j)^k|} \leq \frac{\sqrt{2}H_{N^-_k} }{k^2 \sqrt{x}} \end{align} $$
and
$$ \begin{align} \sum_{\substack{x < n^k < \tfrac{3x}{2} \\ n \geq 2, n^k \notin \mathcal{N}_k}} \frac{1}{ k | x-n^k |} = \sum_{j=1}^{N^+_k} \frac{1}{k |x-(m_+ + j)^k|} \leq \frac{H_{N^+_k}}{k^2 \sqrt{x}}. \end{align} $$
Therefore,
$$ \begin{align} S_{\text{far}}^-(x) &=\sum_{k\geq 2} \sum_{ \substack{ \frac{x}{2} < n^k < x \\ n \geq 2, n^k \notin \mathcal{N}_k}} \frac{1}{ k | x-n^k |} \leq \frac{\sqrt{2}}{\sqrt{x}} \sum_{k\geq 2}\frac{H_{N^-_k} }{k^2 } && (\text{by (3.15) and (3.17)}) \nonumber\\ &\leq \frac{\sqrt{2}}{\sqrt{x}} \bigg(\frac{1}{2}\log{x} - \frac{3}{2}\log{2} + \gamma + \frac{3}{7} \bigg) \sum_{k\geq 2}\frac{1}{k^2} && (\text{by (2.4) and (3.16)}) \nonumber \\ &=\frac{\pi^2-6}{3\sqrt{2 x}} \bigg(\frac{1}{2}\log{x} - \frac{3}{2}\log{2} + \gamma + \frac{3}{7} \bigg) &&(\text{since }\zeta(2)-1 = \tfrac{\pi^2-6}{6}) \end{align} $$
and
$$ \begin{align} S_{\text{far}}^+(x) &=\sum_{k\geq 2} \sum_{ \substack{ x < n^k < \frac{3x}{2} \\ n \geq 2, n^k \notin \mathcal{N}_k}} \frac{1}{ k | x-n^k |} \leq \frac{1}{ \sqrt{x}} \sum_{k \geq 2} \frac{H_{N^+_k}}{k^2} && (\text{by (3.15) and (3.18)}) \nonumber\\ &\leq \frac{1}{ \sqrt{x}} \bigg(\frac{1}{2}\log{x} - 2\log{2} + \gamma + \frac{3}{7} \bigg) \sum_{k \geq 2} \frac{1}{k^2} && (\text{by (2.4) and (3.16)}) \nonumber \\ &= \frac{\pi^2-6}{6\sqrt{x}} \bigg(\frac{1}{2}\log{x} - 2\log{2} + \gamma + \frac{3}{7} \bigg) &&(\text{since }\zeta(2)-1 = \tfrac{\pi^2-6}{6}). \end{align} $$
4 Conclusion
For
$x \geq 3.5$
, with
$T \geq (\log \frac {3}{2})^{-1}$
, the sum (1.3) is bounded by
$$ \begin{align*} &\underbrace{\frac{e}{T \log \frac{3}{2}}\bigg(\log \log x + \frac{\gamma}{\log x} \bigg)}_{\text{by (3.1)}} + \underbrace{\frac{e }{T \log\frac{3}{2}} \sum_{\frac{x}{2} < p^k < \frac{3x}{2}} \frac{1}{k | x-p^k |} }_{\text{by (3.4)}} \\ &\quad \leq \frac{e}{T \log \frac{3}{2}} \bigg(\log \log x + \frac{\gamma}{\log x} \bigg) + \frac{e}{T \log \frac{3}{2}} \underbrace{ \big(S_{\text{prime}}(x) + S_{\text{sqf}}(x) \big)}_{\text{by (3.5) and (3.10)}} \\ &\quad \leq \frac{e}{T \log \frac{3}{2}} \bigg(\log \log x + \frac{\gamma}{\log x} \bigg) + \frac{e}{T \log \frac{3}{2}}\bigg[ \underbrace{2C +4 \log\log x}_{S_{\text{prime}}(x)\text{ bounded by (3.9)}} \\ &\qquad + \underbrace{ \bigg(\log \log x + \gamma + 2\log 2 + \log 2.4 +\frac{7}{156} \bigg) }_{S_{\text{near}}(x)\text{ bounded by (3.14)}} + \underbrace{\frac{\pi^2-6}{3\sqrt{2 x}} \bigg(\frac{1}{2}\log{x} - \frac{3}{2}\log{2} + \gamma + \frac{3}{7} \bigg)}_{S_{\text{far}}^-\text{ bounded by (3.19)}} \\ &\qquad + \underbrace{\frac{\pi^2-6}{6\sqrt{x}} \bigg(\frac{1}{2}\log{x} - 2\log{2} + \gamma + \frac{3}{7} \bigg)}_{S_{\text{far}}^+(x)\text{ bounded by (3.20)}} \bigg] \\ &\quad < \frac{1}{T}\bigg( 40.22465 \log \log x +58.11106 +\frac{3.86972}{\log x} +\frac{5.21918 \log x}{\sqrt{x}} -\frac{1.85268}{\sqrt{x}} \bigg). \end{align*} $$
Acknowledgment
E.S.L. thanks the Heilbronn Institute for Mathematical Research for their support. We also thank the referee for their suggestions.
