A conditional explicit result for the prime number theorem in short intervals

Abstract

This paper gives an explicit bound for the prime number theorem in short intervals under the assumption of the Riemann hypothesis.

1 Introduction

The von Mangoldt function is defined as

$$\begin{aligned} \Lambda (n) = \left\{ \begin{array}{ll} \log p &{} : n=p^m, p\text {is prime}, m \in {\mathbb {N}}\\ 0 &{} : \text {otherwise,} \end{array} \right. \end{aligned}$$

and we will consider the sum $\psi (x) = \sum _{n \le x} \Lambda (n)$. The prime number theorem (PNT) is the statement $\psi (x) \sim x$ as $x \rightarrow \infty$. For the PNT in short intervals, it is known that

$$\begin{aligned} \psi (x+h) - \psi (h) \sim h \end{aligned}$$

(1)

provided that h grows suitably with respect to x. Heath-Brown [9] has shown that one can take $h = x^{\frac{7}{12} - \epsilon }$ provided that $\epsilon \rightarrow 0$ as $x \rightarrow \infty$. Assuming the Riemann hypothesis (RH), Selberg [14] showed that (1) is true for any $h=h(x)$ such that $h/(x^{1/2} \log x) \rightarrow \infty$ as $x \rightarrow \infty$. On the other hand, Maier [11] has shown that the statement is false for $h = (\log x)^{\lambda }$ for any $\lambda > 1$.

In this paper we prove the following explicit version of Selberg’s result.

Theorem 1

Assuming RH, for any h satisfying $\sqrt{x}\log x\le h \le x^\frac{3}{4}$ and all $x \ge e^{10}$ we have

$$\begin{aligned} |\psi (x+h) - \psi (x)-h|< \frac{1}{\pi } \sqrt{x} \log x \log \bigg ( \frac{h}{\sqrt{x}\log x} \bigg ) + 2\sqrt{x} \log x. \end{aligned}$$

(2)

Selberg’s result follows from Theorem 1 for any $h = f(x)\sqrt{x}\log x$ with unbounded $f(x)=o(x)$, in that we would have

$$\begin{aligned} |\psi (x+h) - \psi (x)-h| \ll \sqrt{x} \log x \log \left( f(x) \right) = o(h). \end{aligned}$$

For $h = c \sqrt{x} \log x$, Theorem 1 implies Cramér’s [6] result on primes in the interval $(x, x + h)$ for all sufficiently large x and c. In an earlier paper [7], the author showed that $c = 1 + \epsilon$ is suitable for any $\epsilon > 0$ and for all sufficiently large x. Carneiro, Milinovich and Soundararajan [4] have since shown that we can take $c=22/55$ for all $x\ge 4$. The same methods used in [7] are applied to reach Theorem 1. As such, it could be possible to sharpen Theorem 1 using the techniques in [4].

The closest result to Theorem 1 is the following from Schoenfeld [13].

Theorem 2

Assuming RH, for $x \ge 73.2$ we have

$$\begin{aligned} |\psi (x) - x|< \frac{1}{8 \pi } \sqrt{x} \log ^2 x. \end{aligned}$$

(3)

Schoenfeld’s result confirms Selberg’s theorem for the slightly stronger condition of $h/(\sqrt{x} \log ^2 x) \rightarrow \infty$. One also has from the above

$$\begin{aligned} |\psi (x+h) - \psi (x)-h | < \frac{1}{4 \pi } \sqrt{x+h} \log ^2 (x+h). \end{aligned}$$

When x is sufficiently large, Theorem 1 improves the leading constant in this bound for any choice of $h\le x^{0.735}$.

2 Proof of Theorem 1

2.1 A smooth explicit formula

The Riemann–von Mangoldt explicit formula relates $\psi (x)$ to the zeros of the Riemann zeta-function $\zeta (s)$ (e.g. see Ingham [10]). Tor all non-integer $x>0$,

$$\begin{aligned} \psi (x) = x - \sum _{\rho } \frac{x^\rho }{\rho }-\log 2\pi - \frac{1}{2} \log (1-x^{-2}), \end{aligned}$$

(4)

where the sum is over all non-trivial zeroes $\rho = \beta +i\gamma$ of $\zeta (s)$. We define the weighted sum

$$\begin{aligned} \psi _1 (x) = \sum _{n \le x} (x-n) \Lambda (n) = \int _2^x \psi (t) dt \end{aligned}$$

(5)

and use the following explicit formula, proved in [7] (see also Thm. 28 of [10]).

Lemma 3

For non-integer $x>0$ we have

$$\begin{aligned} \psi _1(x) = \frac{x^2}{2} - \sum _{\rho } \frac{x^{\rho +1}}{\rho (\rho +1)} - x \log (2\pi ) + \epsilon (x) \end{aligned}$$

(6)

where

$$\begin{aligned} 1.545< \epsilon (x) < 2.069. \end{aligned}$$

The bound on $\epsilon (x)$ has been reduced from [7], as we can write

$$\begin{aligned} \epsilon (x)&= 2\log 2\pi - 2 + \sum _\rho \frac{2^{\rho +1}}{\rho (\rho +1)} - \frac{1}{2} \int _2^x \log (1-t^{-2}) dt \\&< 2\log 2\pi - 2 + 2^\frac{3}{2}(\gamma + 2 - \log 4\pi ) + \log \frac{3\sqrt{3}}{4} < 2.069 \end{aligned}$$

and

$$\begin{aligned} \epsilon (x)> 2\log 2\pi - 2 - 2^\frac{3}{2}(\gamma + 2 - \log 4\pi ) > 1.545. \end{aligned}$$

Using a linear combination of Eq. (5), we can examine the distribution of prime powers in the interval $(x, x+h)$. For $2\le \Delta < \sqrt{x}\log x \le h \le x$, let

$$\begin{aligned} w(n) = \left\{ \begin{array}{ll} (n-x+\Delta )/\Delta &{} : x-\Delta \le n \le x\\ 1 &{} : x \le n \le x+h\\ (x+h+\Delta -n)/\Delta &{} : x+h \le n \le x+h+\Delta \\ 0 &{} : \text {otherwise.} \end{array} \right. \end{aligned}$$

This leads to the identity

$$\begin{aligned} \sum _{n} \Lambda (n) w(n)= & {} \frac{1}{\Delta } (\psi _1(x+h+\Delta ) - \psi _1(x+h) - \psi _1 (x) + \psi _1(x-\Delta )), \end{aligned}$$

which can be verified by expanding both sides. Notice that over $x \le n \le x+h$, the sum on the LHS is equal to $\psi (x+h)-\psi (x)$. We thus aim to estimate this expression by bounding the RHS of (7). Using Lemma 3 in the above equation gives the following.

Lemma 4

Let $2 \le \Delta < h \le x$ with $x \notin {\mathbb {Z}}$. Then

$$\begin{aligned} \sum _{n} \Lambda (n) w(n) = h + \Delta - \frac{1}{\Delta } \sum _{\rho } S(\rho ) + \epsilon (\Delta ) \end{aligned}$$

where

$$\begin{aligned} S(\rho ) =\frac{ (x+h+\Delta )^{\rho +1} - (x+h)^{\rho +1} - x^{\rho +1} +(x-\Delta )^{\rho +1}}{\rho (\rho +1)} \end{aligned}$$

and

$$\begin{aligned} |\epsilon (\Delta )| < \frac{21}{20\Delta }. \end{aligned}$$

It remains to estimate the sum over zeros. We will split it into three sums,

$$\begin{aligned} \sum _{\rho } S(\rho ) = \bigg ( \sum _{| \gamma | \le \alpha x/h} +\sum _{\alpha x/h<| \gamma | < \beta x/\Delta } + \sum _{ | \gamma | \ge \beta x/\Delta } \bigg ) S(\rho ) \end{aligned}$$

(7)

where $\alpha > 0$ and $\beta >0$ are parameters we can later optimise over.

Lemma 5

Let $2 \le \Delta < h \le x$ and assume RH. We have

$$\begin{aligned} \left| \sum _{| \gamma | \ge \beta x/\Delta } S(\rho ) \, \right| < \frac{4 \Delta (x+h + \Delta )^{3/2} }{\pi \beta x}\log (\beta x/\Delta ) \end{aligned}$$

provided that $\beta x / \Delta \ge \gamma _1 = 14.13\ldots$, the ordinate of the first zero of $\zeta (s)$.

Proof

On RH, one has

$$\begin{aligned} |S(\rho )| \le \frac{4 (x+h+\Delta )^{3/2}}{\gamma ^2}. \end{aligned}$$

The result follows from Lemma 1(ii) of Skewes [15], that for all $T \ge \gamma _1$,

$$\begin{aligned} \sum _{\gamma \ge T} \frac{1}{\gamma ^2} < \frac{1}{2\pi } \frac{\log T}{T}. \end{aligned}$$

$\square$

The following lemmas require estimates on the zero-counting function N(T), which counts the number of zeros of $\zeta (s)$ in the critical strip $0<\beta <1$ with $0<\gamma \le T$. Backlund [1] showed that $N(T) = P(T) + Q(T)$, where

$$\begin{aligned} P(T) := \frac{T}{2\pi }\log {\frac{T}{2\pi }} - \frac{T}{2\pi } + \frac{7}{8} \end{aligned}$$

and $Q(T) = O(\log T)$. Hasanalizade, Shen, and Wong [8, Cor. 1.2] have given the most recent explicit version of this, of

$$\begin{aligned} |Q(T)| \le R(T) = a_1\log {T} + a_2\log \log {T} + a_3 \end{aligned}$$

(8)

with $a_1= 0.1038$, $a_2=0.2573$, and $a_3 = 9.3675$, for all $T \ge e$.

Lemma 6

Let $2 \le \Delta < h \le x$ and assume RH. We have

$$\begin{aligned} \left| \sum _{| \gamma | \le \alpha x/h} S(\rho ) \, \right| < \frac{\alpha x (h+\Delta ) \Delta }{ \pi h \sqrt{x-\Delta }} \log (\alpha x/h). \end{aligned}$$

Proof

We can write

$$\begin{aligned} S(\rho ) = \int _{x+h}^{x+h+\Delta } \int _{u-h-\Delta }^u t^{\rho -1} dt du, \end{aligned}$$

so, under RH, one has

$$\begin{aligned} |S(\rho )| < \frac{ (h+\Delta ) \Delta }{ \sqrt{x-\Delta }}. \end{aligned}$$

With (8), we can use

$$\begin{aligned} N(T) < \frac{T \log T}{2 \pi }, \end{aligned}$$

from which the result immediately follows. $\square$

For the middle sum of (7), we will use the following lemma. It follows directly from Lemma 3 of [2], in whose notation we use $\phi (\gamma ) = \gamma ^{-1}$, and takes constants $A_0$ and $A_1$ from Trudgian [16, Thm. 2.2] and $A_2$ from [2, Lem. 2].

Lemma 7

For $2\pi \le T_1 \le T_2$ we have

$$\begin{aligned} \sum _{T_1< \gamma < T_2} \frac{1}{\gamma }&= \frac{1}{4 \pi }\log \frac{T_2}{T_1} \log \frac{T_2 T_1}{4\pi ^2} + \frac{Q(T_2)}{T_2} - \frac{Q(T_1)}{T_1} + E(T_1), \end{aligned}$$

(9)

where $|Q(T)|\le R(T)$, defined in (8), and

$$\begin{aligned} |E(T)| \le \frac{2A_1\log T + 2A_0 + A_1 + A_2}{T^2} \end{aligned}$$

with $A_0 = 2.067$, $A_1 = 0.059$, $A_2 = 1/150$.

Lemma 8

Let $2 \le \Delta < h \le x$ and assume RH. For $\alpha x/h\ge 15$ we have

$$\begin{aligned} \left| \sum _{\alpha x/h<| \gamma |< \beta x/\Delta } S(\rho ) \, \right| < \Delta (x+h+\Delta )^{1/2} \bigg ( \frac{1}{\pi }\log \bigg ( \frac{\beta h}{\alpha \Delta } \bigg ) \log \bigg ( \frac{\alpha \beta x^2}{4\pi ^2 h \Delta } \bigg ) + 5.4\bigg ). \end{aligned}$$

Proof

We can write

$$\begin{aligned} S(\rho ) = \frac{1}{\rho } \bigg ( \int _{x+h}^{x+h+\Delta } t^{\rho } dt - \int _{x-\Delta }^x t^{\rho } dt \bigg ), \end{aligned}$$

and so bounding trivially gives

$$\begin{aligned} | S(\rho )| \le \frac{2 (x+h+\Delta )^{1/2} \Delta }{| \gamma |}. \end{aligned}$$

It follows that

$$\begin{aligned} \left| \sum _{\alpha x/h<| \gamma |< \beta x/\Delta } S(\rho ) \, \right| \le 4 (x+h+\Delta )^{1/2} \Delta \sum _{\alpha x/h< \gamma < \beta x /\Delta } \frac{1}{\gamma }, \end{aligned}$$

on which we apply Lemma 7, and bound the smaller order terms with the assumption of $T_1\ge 15$ to obtain the result. Note that the bound on $T_1$ is to reduce the constant 5.4, but not restrict $\alpha$ too much. $\square$

2.2 Bounding the PNT in intervals

From Lemma 4 we can write

$$\begin{aligned}&\bigg | \psi (x+h) - \psi (x) - h \bigg |< \ \frac{1}{\Delta } \Bigg | \sum _{\rho } S(\rho ) \Bigg | + \Delta + \frac{21}{20 \Delta } + \sum _{x-\Delta< n \le x} w(n) \Lambda (n) \\&+ \sum _{x+h < n \le x+h+\Delta } w(n) \Lambda (n) \end{aligned}$$

As the smooth weight has $|w(n)| \le 1$, the above bound is no greater than

$$\begin{aligned} \frac{1}{\Delta } \Bigg | \sum _{\rho } S(\rho ) \Bigg | + \Delta + \frac{21}{20 \Delta } + 2\sum _{\begin{array}{c} x+h < p^k \le x+h+\Delta \\ k\ge 1 \end{array}} \log p. \end{aligned}$$

(10)

The largest term in this bound comes from the sum over $\rho$, in particular, the section estimated in Lemma 8. Larger $\Delta$ results in a smaller main-term constant, so we will set $\Delta = C\sqrt{x}\log x$ and later choose an optimal value of $C\in (0,1)$. The reason for not taking larger $\Delta$ is two-fold: to keep $\Delta < h$ and ensure the smaller terms in (10) are $O(\sqrt{x}\log x)$.

To bound the sum over prime powers we can use Montgomery and Vaughan’s version of the Brun–Titchmarsh theorem for primes in intervals [12, Eq. 1.12]. Defining $\theta (x) = \sum _{p\le x}\log p$, Eq. (1.12) of [12] implies

$$\begin{aligned} \theta (x+h) - \theta (x) = \sum _{x<p\le x+h} \log p \le \frac{2h\log (x+h)}{\log h}. \end{aligned}$$

The contribution from higher prime powers is relatively small, and can be bounded with explicit estimates on the difference between the Chebyshev functions $\psi (x)$ and $\theta (x)$. Costa Pereira [5, Thm. 2,4,5] gives lower bounds for different ranges of x. These can be combined into

$$\begin{aligned} \psi (x)-\theta (x) > 0.999x^\frac{1}{2} + \frac{2}{3} x^\frac{1}{3} \end{aligned}$$

(11)

for all $x\ge 2187$. Broadbent et al. [3, Cor. 5.1] give

$$\begin{aligned} \psi (x) - \theta (x) < \alpha _1 x^\frac{1}{2} + \alpha _2 x^\frac{1}{3} \end{aligned}$$

(12)

with $\alpha _1= 1+ 1.93378 \cdot 10^{-8}$ and $\alpha _2 = 2.69$ for all $x\ge e^{10}$. Thus, we have

$$\begin{aligned} \psi (x+h+\Delta ) - \psi (x+h)&\le \theta (x+h+\Delta ) - \theta (x+h) + E_1(x) \\&\le \frac{2\Delta \log (x+h+\Delta )}{\log \Delta } + E_1(x) \end{aligned}$$

where $E_1(x) = \alpha _1 (x+h+\Delta )^\frac{1}{2} + \alpha _2 (x+h+\Delta )^\frac{1}{3} - 0.999(x+h)^\frac{1}{2} - \frac{2}{3} (x+h)^\frac{1}{3}$, and is bounded by $E_1(x) \le \beta _1 x^{\frac{1}{2}} + \beta _2 x^{\frac{1}{3}}$ with

$$\begin{aligned} \beta _1 = \sqrt{3}\alpha _1 - 0.999 \quad \text {and} \quad \beta _2 = 3^{\frac{1}{3}}\alpha _2 - \frac{2}{3}. \end{aligned}$$

Here and hereafter, let $x_0=e^{10}$. For $x\ge x_0$ we can bound the smaller order terms in (10),

$$\begin{aligned} \Delta + \frac{21}{20 \Delta } + 2\sum _{\begin{array}{c} x+h< p^k \le x+h+\Delta \\ k\ge 1 \end{array}} \log p < K_1 \sqrt{x}\log x \end{aligned}$$

where, for $h\le x^t$ with $t<1$,

$$\begin{aligned} K_1&= C + \frac{4C \log (x_0+2x_0^t)}{\log (C \sqrt{x_0}\log x_0)} + \frac{2\beta _1}{\log x_0} + \frac{2\beta _2}{x_0^{\frac{1}{6}}\log x_0} + \frac{21}{20C x_0\log ^2 x_0}. \end{aligned}$$

This, along with Lemmas 5 and 6, allow us to bound

$$\begin{aligned} \bigg | \psi (x+h) - \psi (x) - h \bigg |< \frac{1}{\Delta } \Bigg | \sum _{\alpha x/h< |\gamma | < \beta x / \Delta } S(\rho ) \Bigg | + E(x,h,\Delta ) \end{aligned}$$

(13)

where

$$\begin{aligned} E(x,h,\Delta )&= K_1 \sqrt{x} + \frac{\alpha x (h + \Delta )}{\pi h \sqrt{x-\Delta } } \log \left( \frac{\alpha x}{h}\right) + \frac{4 (x+h+\Delta )^{3/2}}{\pi \beta x} \log \left( \frac{\beta x}{\Delta }\right) . \end{aligned}$$

For $\sqrt{x}\log x\le h\le x^t$ we have

$$\begin{aligned}&E(x,h,\Delta ) \le \ K_1 \sqrt{x} + \frac{2\alpha x}{\pi \sqrt{x-C\sqrt{x}\log x}} \log \left( \frac{\alpha \sqrt{x}}{\log x}\right) \\&+ \frac{4 (x+x^t+C\sqrt{x}\log x)^{3/2}}{\pi \beta x} \log \left( \frac{\beta \sqrt{x}}{C\log x}\right) \le K_2 \sqrt{x}\log x, \end{aligned}$$

where, for $x\ge x_0\ge e^{\beta /C}$ and $0<\alpha \le 5$, we can take

$$\begin{aligned} K_2 = \&\frac{K_1}{\log x_0} + \frac{\alpha }{\pi } + \frac{2 (x_0+x_0^t + C\sqrt{x_0}\log x_0)^{3/2}}{\pi \beta x_0^{3/2}}. \end{aligned}$$

The first term in (13) can be estimated with Lemma 8, so that

$$\begin{aligned} \frac{1}{\Delta } \Bigg | \sum _{\alpha x/h<| \gamma |< \beta x/\Delta } S(\rho ) \Bigg |&< (x+h+\Delta )^{1/2} \bigg ( \frac{1}{\pi }\log \bigg ( \frac{\beta h}{\alpha \Delta } \bigg ) \log \bigg ( \frac{\alpha \beta x^2}{4\pi ^2 h \Delta } \bigg ) + 5.4\bigg ) \\&< \frac{\sqrt{x}}{\pi }\log x\log \left( \frac{h}{\sqrt{x}\log x}\right) + K_3 \sqrt{x} \log x, \end{aligned}$$

in which, assuming $100 e^{-10} \le \frac{\alpha \beta }{4\pi ^2 C}\le 100$, we can take

$$\begin{aligned} K_3&= \frac{1}{\pi } \log \left( \frac{\beta }{\alpha C} \right) \log \left( \frac{\alpha \beta x_0}{4\pi ^2 C\log ^2 x_0} \right) \frac{1}{\log x_0} \\&\quad + \frac{x_0^{t/2-1/2}}{\pi \log x_0}\log \bigg ( \frac{\beta x_0^{t-1/2}}{\alpha C\log x_0} \bigg ) \log \bigg ( \frac{\alpha \beta x_0}{4\pi ^2 C\log ^2 x_0} \bigg ) \\&\quad + \frac{\sqrt{C}}{\pi x_0^{1/4} \sqrt{\log x_0}} \log \bigg ( \frac{\beta x_0^{t-1/2}}{\alpha C\log x_0} \bigg ) \log \bigg ( \frac{\alpha \beta x_0}{4\pi ^2 C\log ^2 x_0} \bigg ) \\&\quad + \frac{5.4}{\log x_0} \left( 1 + x_0^{t-1} + \frac{C\log x_0}{\sqrt{x_0}} \right) ^{1/2}. \end{aligned}$$

Note that the assumption for $\alpha$ and $\beta$ is to ensure certain terms are bounded for all $x\ge x_0$. Combining estimates, we have

$$\begin{aligned} \bigg | \psi (x+h) - \psi (x) - h \bigg | <&\frac{\sqrt{x}}{\pi }\log x\log \left( \frac{h}{\sqrt{x}\log x}\right) + K_4 \sqrt{x}\log x, \end{aligned}$$

(14)

where $K_4 = K_3 + K_2$. It remains to optimise over the parameters. Before deciding these values, recall that we have made the assumptions $\beta \le 10C$,

$$\begin{aligned} \frac{15 h}{x} \le \alpha \le 5,\quad \beta \ge \gamma _1 \frac{C\log x}{\sqrt{x}},\quad C \alpha < \beta \le \alpha , \quad \text {and}\quad \frac{100}{e^{10}}\le \frac{\alpha \beta }{4\pi ^2 C}\le 100. \end{aligned}$$

The restriction on $\alpha$ will be satisfied for all $\sqrt{x}\log x\le h\le x^\frac{3}{4}$ if we take $\alpha \ge 15 x_0^{-\frac{1}{4}}$. Optimising over C, $\alpha$, and $\beta$ to minimise $K_4$, we find that choosing $C= 0.25$ and $\alpha = \beta = 1.35$ allows us to take $K_4 = 2$ for all $x\ge x_0$.