Numerically Explicit Estimates for the Distribution of Rough Numbers

Kai (Steve) Fan Department of Mathematics, Dartmouth College, Hanover, NH 03755, USA [email protected]

Abstract.

For $x\geq y>1$ and $u\colonequals\log x/\log y$ , let $\Phi(x,y)$ denote the number of positive integers up to $x$ free of prime divisors less than or equal to $y$ . In 1950 de Bruijn [Br] studied the approximation of $\Phi(x,y)$ by the quantity

\mu_{y}(u)e^{\gamma}x\log y\prod_{p\leq y}\left(1-\frac{1}{p}\right),

where $\gamma=0.5772156...$ is Euler’s constant and

\mu_{y}(u)\colonequals\int_{1}^{u}y^{t-u}\omega(t)\,dt.

He showed that the asymptotic formula

\Phi(x,y)=\mu_{y}(u)e^{\gamma}x\log y\prod_{p\leq y}\left(1-\frac{1}{p}\right)+O\left(\frac{xR(y)}{\log y}\right)

holds uniformly for all $x\geq y\geq 2$ , where $R(y)$ is a positive decreasing function related to the error estimates in the Prime Number Theorem. In this paper we obtain numerically explicit versions of de Bruijn’s result.

Key words and phrases:

Rough numbers, Buchstab’s function, the Prime Number Theorem

2020 Mathematics Subject Classification:

11N25

1. Introduction

Let $x\geq y>1$ be positive real numbers. Throughout the paper, we shall always write $u\colonequals\log x/\log y$ , and the letters $p$ and $q$ will always denote primes. We say that a positive integer $n$ is $y$ -rough if all the prime divisors of $n$ are greater than $y$ . Let $\Phi(x,y)$ denote the number of $y$ -rough numbers up to $x$ . Explicitly, we have

\Phi(x,y)=\sum_{\begin{subarray}{c}n\leq x\\ P^{-}(n)>y\end{subarray}}1,

where $P^{-}(n)$ denotes the least prime divisor of $n$ , with the convention that $P^{-}(1)=\infty$ . When $1\leq u\leq 2$ , or equivalently when $\sqrt{x}\leq y\leq x$ , we simply have $\Phi(x,y)=\pi(x)-\pi(y)+1$ , where $\pi(\cdot)$ is the prime-counting function. The function $\Phi(x,y)$ is closely related to the sieve of Eratosthenes, one of the most ancient algorithms for finding primes, and it has been extensively studied by mathematicians. A simple application of the inclusion-exclusion principle enables us to write

\Phi(x,y)=\sum_{d\mid P(y)}\mu(d)\left\lfloor\frac{x}{d}\right\rfloor,

(1.1)

where $\lfloor a\rfloor$ is the integer part of $a$ for any $a\in\mathbb{R}$ , $\mu$ is the Möbius function, and $P(y)$ denotes the product of primes up to $y$ . If $y$ is relatively small in comparison with $x$ , say $y=x^{o(1)}$ , the above formula can be used to obtain $\Phi(x,y)\sim e^{-\gamma}x/\log y$ as $y\to\infty$ , where $\gamma=0.5772156...$ is Euler’s constant. However, it turns out that this nice asymptotic formula does not hold uniformly, as already exemplified by the base case $1\leq u\leq 2$ .

In 1937, Buchstab [Bu] showed that for any fixed $u>1$ , one has $\Phi(x,y)\sim\omega(u)x/\log y$ as $x\to\infty$ , where $\omega(u)$ is defined to be the unique continuous solution to the delay differential equation $(u\omega(u))^{\prime}=\omega(u-1)$ for $u\geq 2$ , subject to the initial value condition $\omega(u)=1/u$ for $u\in[1,2]$ . Comparing this result with the asymptotic formula obtained from (1.1), one would expect that $\omega(u)\to e^{-\gamma}$ as $u\to\infty$ . Indeed, it can be shown [T, Corollary III.6.5] that $\omega(u)=e^{-\gamma}+O(u^{-u/2})$ for $u\geq 1$ . Moreover, it is known that $\omega(u)$ oscillates above and below $e^{-\gamma}$ infinitely often. It is convenient to extend the definition of $\omega(u)$ by setting $\omega(u)=0$ for all $u<1$ , so that $\omega(u)$ satisfies the same delay differential equation on $\mathbb{R}\setminus\{1,2\}$ . In the sequel, we shall write $\omega^{\prime}(1)$ and $\omega^{\prime}(2)$ for the right derivatives of $\omega(u)$ at $u=1$ and $u=2$ , respectively. With this convention, we have $(u\omega(u))^{\prime}=\omega(u-1)$ for all $u\in\mathbb{R}$ .

Buchstab’s asymptotic formula can be proved easily based on the following identity [T, Theorem III.6.3] named after him:

\Phi(x,y)=\Phi(x,z)+\sum_{y<p\leq z}\sum_{v\geq 1}\Phi(x/p^{v},p)

(1.2)

for any $z\in[y,x]$ . The Buchstab function $\omega(u)$ then appears naturally in the iteration process, starting with $\Phi(x,y)\sim x/(u\log y)$ in the range $1<u\leq 2$ . Since $1/2\leq\omega(u)\leq 1$ for $u\in[1,\infty)$ , Buchstab’s asymptotic formula suggests that the relation $\Phi(x,y)\asymp x/\log y$ holds uniformly for $x\geq y>1$ . Thus, it is of interest to seek numerically explicit estimates for $\Phi(x,y)$ that are applicable in wide ranges. Confirming a conjecture of Ford, the author [F] showed that $\Phi(x,y)<x/\log y$ holds uniformly for $x\geq y>1$ , which is essentially best possible when $x^{1-\epsilon}\leq y\leq\epsilon x$ , where $\epsilon\in(0,1)$ is fixed. On the other hand, the values of $\omega(u)$ indicate that improvements should be expected in the narrower range $2\leq y\leq\sqrt{x}$ . In recent work jointly with Pomerance [FP], the author proved that $\Phi(x,y)<0.6x/\log y$ holds uniformly for $3\leq y\leq\sqrt{x}$ . This inequality provides a fairly good upper bound for $\Phi(x,y)$ , especially considering that the absolute maximum of $\omega(u)$ over $[2,\infty)$ is given by $M_{0}=0.5671432...$ , attained at the unique critical point $u=2.7632228...$ of the function $(\log(u-1)+1)u^{-1}$ on $[2,3]$ . With a bit more effort, one can show, using the Buchstab identity (1.2), that

\Phi(x,y)=\frac{x}{\log y}\left(\omega(u)+O\left(\frac{1}{\log y}\right)\right)

(1.3)

uniformly for $2\leq y\leq\sqrt{x}$ (see [T, Theorem III.6.4]). In Section 2, we shall derive a numerically explicit lower bound of this type that suits our needs. Our method can also be modified with ease to obtain a numerically explicit upper bound of the same type.

In [Br] de Bruijn provided a more precise approximation for $\Phi(x,y)$ than $\omega(u)x/\log y$ . Let us fix some $y_{0}\geq 2$ for the moment. Suppose that there exist a positive constant $C_{0}(y_{0})$ and a positive decreasing function $R(z)$ defined on $[y_{0},\infty)$ , such that $R(z)\gg\,z^{-1}$ , that $R(z)\downarrow 0$ as $z\to\infty$ and that for all $z\geq y_{0}$ we have

|\pi(z)-\operatorname{li}(z)|\leq\frac{z}{\log z}R(z)

(1.4)

and

\int_{z}^{\infty}\frac{|\pi(t)-\operatorname{li}(t)|}{t^{2}}\,dt\leq C_{0}(y_{0})R(z),

(1.5)

where $\operatorname{li}(z)$ is the logarithmic integral defined by

\operatorname{li}(z)\colonequals\int_{0}^{z}\frac{dt}{\log t}.

The classical version of the Prime Number Theorem allows us to take $R(z)=\exp(-c\sqrt{\log z})$ for some suitable constant $c>0$ . Using the zero-free region of Korobov and Vinogradov for the Riemann zeta-function, we obtain $R(z)=\exp(-c^{\prime}(\log z)^{3/5}(\log\log z)^{-1/5})$ for some absolute constant $c^{\prime}>0$ . If the Riemann Hypothesis holds, then one can take $R(z)=c^{\prime\prime}z^{-1/2}\log^{2}z$ , where $c^{\prime\prime}>0$ is an absolute constant.

To state de Bruijn’s result, we define

\mu_{y}(u)\colonequals\int_{1}^{u}y^{t-u}\omega(t)\,dt.

It is easy to see that $0\leq\mu_{y}(u)\log y\leq 1-y^{1-u}$ and that for every fixed $u\geq 1$ , we have $\mu_{y}(u)\log y\to\omega(u)$ as $y\to\infty$ . Precise expansions for $\mu_{y}(u)$ in terms of the powers of $\log y$ can be found in [T, Theorem III.6.18]. When $1\leq u\leq 2$ , the change of variable $t=\log v/\log y$ shows that

\mu_{y}(u)x=\int_{1}^{u}t^{-1}y^{t}\,dt=\int_{y}^{x}\frac{dv}{\log v}=\operatorname{li}(x)-\operatorname{li}(y).

Since $\Phi(x,y)=\pi(x)-\pi(y)+1$ when $1\leq u\leq 2$ , (1.4) clearly implies that

\Phi(x,y)=\mu_{y}(u)x+(\pi(x)-\operatorname{li}(x))-(\pi(y)-\operatorname{li}(y))+1=\mu_{y}(u)x+O\left(\frac{xR(y)}{\log y}\right).

It can be shown using (1.4) and (1.5) that

\prod_{p\leq y}\left(1-\frac{1}{p}\right)=\frac{e^{-\gamma}}{\log y}\left(1+O(R(y))\right).

Thus we have, equivalently,

\Phi(x,y)=\mu_{y}(u)e^{\gamma}x\log y\prod_{p\leq y}\left(1-\frac{1}{p}\right)+O\left(\frac{xR(y)}{\log y}\right).

(1.6)

Essentially, de Bruijn [Br] showed that this formula holds uniformly for $x\geq y\geq y_{0}$ . In Section 3 we shall derive an explicit version of (1.6), which will be applied in Section 4 to obtain numerically explicit estimates with suitable $y_{0}$ and $R(y)$ . Our main results are summarized in the following theorem.

Theorem 1.1.

Uniformly for $x\geq y\geq 2$ , we have

\left|\Phi(x,y)-\mu_{y}(u)e^{\gamma}x\log y\prod_{p\leq y}\left(1-\frac{1}{p}\right)\right|<4.403611\frac{x}{(\log y)^{3/4}}\exp\left(-\sqrt{\frac{\log y}{6.315}}\right).

Conditionally on the Riemann Hypothesis, we have

\left|\Phi(x,y)-\mu_{y}(u)e^{\gamma}x\log y\prod_{p\leq y}\left(1-\frac{1}{p}\right)\right|<0.449774\frac{x\log y}{\sqrt{y}}

uniformly for $x\geq y\geq 11$ .

The following consequence of Theorem 1.1 is sometimes more convenient to use.

Corollary 1.2.

Uniformly for $x\geq y\geq 2$ , we have

|\Phi(x,y)-\mu_{y}(u)x|<4.434084\frac{x}{(\log y)^{3/4}}\exp\left(-\sqrt{\frac{\log y}{6.315}}\right).

Conditionally on the Riemann Hypothesis, we have

|\Phi(x,y)-\mu_{y}(u)x|<0.460680\frac{x\log y}{\sqrt{y}}

uniformly for $x\geq y\geq 11$ .

2. Lower Bounds for $\Phi(x,y)$

Before moving on to the derivation of Theorem 1.1, we prove a clean lower bound for $\Phi(x,y)$ which is applicable in a wide range. This lower bound, which is interesting in itself, will be used in the proof of Theorem 1.1 and Corollary 1.2 in Section 4. We start by proving the following result, which provides a numerically explicit lower bound for the implicit constant in the error term in (1.3). As we already mentioned, our method can easily be adapted to yield a numerically explicit upper bound as well, though it will not be needed in the present paper.

Proposition 2.1.

Define $\Delta(x,y)$ by

\Phi(x,y)=\frac{x}{\log y}\left(\omega(u)+\frac{\Delta(x,y)}{\log y}\right)

for $2\leq y\leq\sqrt{x}$ . Let $y_{0}=602$ . For every positive integer $k\geq 3$ , we define

\Delta_{k}^{-}=\Delta_{k}^{-}(y_{0})\colonequals\inf\left\{\min(\Delta(x,y),0)\colon y\geq y_{0}\emph{~{}and~{}}2\leq u<k\right\}.

Then $\Delta_{3}^{-}>-0.563528$ , $\Delta_{4}^{-}>-0.887161$ , and $\Delta_{k}^{-}>-0.955421$ for all $k\geq 5$ .

Proof.

Let $y_{1}\colonequals 2{,}278{,383}$ . Suppose first that $y\geq y_{1}$ and set

G(v)\colonequals\sum_{x^{1/v}<p\leq\sqrt{x}}\frac{1}{p}

for $2\leq v\leq u$ . By [D, Theorem 5.6]¹¹1In [B] it is claimed that the proof of [D, Theorem 4.2] is incorrect due to the application of an incorrect zero density estimate of Ramaŕe [Rama, Theorem 1.1]. In a footnote on p. 2299 of the same paper, the authors state that the bounds asserted in [D] are likely affected for this reason. However, since they also give a correct proof of [D, Theorem 4.2] (see [B, Corollary 11.2]), one verifies easily that the proof of [D, Theorem 5.6], which relies only on [D, Theorem 4.2], partial summation, and numerical computation, remains valid., we have

\left|G(v)-\log\frac{v}{2}\right|\leq\frac{c_{1}}{\log^{2}y}

(2.1)

for all $y\geq y_{1}$ , where $c_{1}=0.4/\log y_{1}$ . We shall also make use of the following inequality [D, Corollary 5.2]²²2For the same reason mentioned above, it is reasonable to suspect that the bounds given in [D, Corollary 5.2] are also affected. However, one can verify these bounds without much difficulty. Indeed, (5.2) of [D, Corollary 5.2] is superseded by [RS, Corollary 1], while (5.3) and (5.4) of [D, Corollary 5.2] follow from [BD, Lemmas 3.2–3.4] and direct calculations.:

\frac{z}{\log z}\left(1+\frac{c_{3}}{\log z}\right)\leq\pi(z)\leq\frac{z}{\log z}\left(1+\frac{c_{2}}{\log z}\right),

(2.2)

where $c_{2}=1+2.53816/\log y_{1}$ and $c_{3}=1+2/\log y_{1}$ . We start with the range $2\leq u\leq 3$ . In this range, we have

	$\displaystyle\Phi(x,y)$	$\displaystyle=\#\{n\leq x\colon P^{-}(n)>y\text{~{}and~{}}\Omega(n)\leq 2\}$
		$\displaystyle=\pi(x)-\pi(y)+1+\sum_{y<p\leq\sqrt{x}}\sum_{p\leq q\leq x/p}1$
		$\displaystyle=\pi(x)-\pi(y)+1+\sum_{y<p\leq\sqrt{x}}(\pi(x/p)-\pi(p)+1),$

where $\Omega(n)$ denotes the total number of prime factors of $n$ , with multiplicity counted. Since

\sum_{y<p\leq\sqrt{x}}(\pi(p)-1)=\sum_{\pi(y)<j\leq\pi(\sqrt{x})}(j-1)=\frac{\pi(\sqrt{x})(\pi(\sqrt{x})-1)}{2}-\frac{\pi(y)(\pi(y)-1)}{2},

we see that

\pi(x)-\pi(y)+1-\sum_{y<p\leq\sqrt{x}}(\pi(p)-1)>\pi(x)-\frac{\pi(\sqrt{x})^{2}}{2}+\frac{\pi(\sqrt{x})}{2}.

It follows from (2.2) that

\Phi(x,y)>\frac{x}{\log x}\left(1+\frac{c_{3}}{\log x}\right)-\frac{x}{2\log^{2}\sqrt{x}}\left(1+\frac{c_{2}}{\log\sqrt{x}}\right)^{2}+\frac{\sqrt{x}}{2\log\sqrt{x}}+\sum_{y<p\leq\sqrt{x}}\pi(x/p).

(2.3)

To handle the sum in (2.3), we appeal to (2.2) again to arrive at

\sum_{y<p\leq\sqrt{x}}\pi(x/p)\geq\sum_{y<p\leq\sqrt{x}}\left(\frac{x}{p\log(x/p)}+\frac{c_{3}x}{p\log^{2}(x/p)}\right).

By partial summation we see that

\sum_{y<p\leq\sqrt{x}}\frac{1}{p\log(x/p)}=\frac{1}{\log x}\int_{2^{-}}^{u}\frac{v}{v-1}\,dG(v)=\frac{1}{\log y}\left(\frac{G(u)}{u-1}+\frac{1}{u}\int_{2^{-}}^{u}\frac{G(v)}{(v-1)^{2}}\,dv\right)

From (2.1) it follows that

\frac{G(u)}{u-1}\geq\frac{1}{u-1}\left(\log\frac{u}{2}-\frac{c_{1}}{\log^{2}y}\right),

and

	$\displaystyle\int_{2^{-}}^{u}\frac{G(v)}{(v-1)^{2}}\,dv$	$\displaystyle\geq\int_{2}^{u}\frac{1}{(v-1)^{2}}\left(\log\frac{v}{2}-\frac{c_{1}}{\log^{2}y}\right)\,dv$
		$\displaystyle=-\frac{1}{u-1}\log\frac{u}{2}+\int_{2}^{u}\frac{1}{v(v-1)}\,dv-\frac{c_{1}}{\log^{2}y}\left(1-\frac{1}{u-1}\right)$
		$\displaystyle=-\frac{u}{u-1}\log\frac{u}{2}+\log(u-1)-\frac{c_{1}}{\log^{2}y}\left(1-\frac{1}{u-1}\right).$

Hence

	$\displaystyle\sum_{y<p\leq\sqrt{x}}\frac{x}{p\log(x/p)}$	$\displaystyle\geq\frac{x}{\log y}\left(\frac{\log(u-1)}{u}-\frac{2c_{1}}{u\log^{2}y}\right)$
		$\displaystyle=\frac{x}{\log y}\left(\omega(u)-\frac{2c_{1}}{u\log^{2}y}\right)-\frac{x}{\log x}.$		(2.4)

Similarly, we have

\sum_{y<p\leq\sqrt{x}}\frac{1}{p\log^{2}(x/p)}=\frac{1}{\log^{2}x}\int_{2^{-}}^{u}\left(\frac{v}{v-1}\right)^{2}\,dG(v)=\frac{1}{\log^{2}x}\left(\frac{G(u)u^{2}}{(u-1)^{2}}+2\int_{2^{-}}^{u}\frac{vG(v)}{(v-1)^{3}}\,dv\right).

By (2.1) we have

\frac{G(u)u^{2}}{(u-1)^{2}}\geq\frac{u^{2}}{(u-1)^{2}}\left(\log\frac{u}{2}-\frac{c_{1}}{\log^{2}y}\right),

and

\int_{2^{-}}^{u}\frac{vG(v)}{(v-1)^{3}}\,dv\geq\int_{2}^{u}\frac{v}{(v-1)^{3}}\left(\log\frac{v}{2}-\frac{c_{1}}{\log^{2}y}\right)\,dv.

Since

	$\displaystyle\int_{2}^{u}\frac{v}{(v-1)^{3}}\log\frac{v}{2}\,dv$	$\displaystyle=-\left(\frac{1}{u-1}+\frac{1}{2(u-1)^{2}}\right)\log\frac{u}{2}+\int_{2}^{u}\left(\frac{1}{v-1}+\frac{1}{2(v-1)^{2}}\right)\frac{dv}{v}$
		$\displaystyle=-\frac{2u-1}{2(u-1)^{2}}\log\frac{u}{2}+\frac{1}{2}\int_{2}^{u}\left(\frac{1}{(v-1)^{2}}+\frac{1}{v(v-1)}\right)\,dv$
		$\displaystyle=-\frac{u^{2}}{2(u-1)^{2}}\log\frac{u}{2}+\frac{1}{2}\left(\log(u-1)+1-\frac{1}{u-1}\right)$

and

\int_{2}^{u}\frac{v}{(v-1)^{3}}\,dv=-\frac{2u-1}{2(u-1)^{2}}+\frac{3}{2},

we have

\sum_{y<p\leq\sqrt{x}}\frac{x}{p\log^{2}(x/p)}\geq\frac{x}{\log^{2}x}\left(\log(u-1)+\frac{u-2}{u-1}-\frac{4c_{1}}{\log^{2}y}\right).

(2.5)

Inserting (2) and (2.5) into (2.3) yields

\Delta(x,y)\geq g(u)-\frac{2c_{1}}{u\log y}+\frac{\log y}{uy^{3/2}}-\frac{1}{u^{2}}\left(2-c_{3}+\frac{4c_{1}c_{3}}{\log^{2}y}+\frac{8c_{2}}{u\log y}+\frac{8c_{2}^{2}}{u^{2}\log^{2}y}\right),

where

g(u)\colonequals\frac{c_{3}}{u^{2}}\left(\log(u-1)+\frac{u-2}{u-1}\right).

Using Mathematica we find that $\Delta_{3}^{-}>-0.301223$ when $y\geq y_{1}$ .

Now we proceed to bound $\Delta_{k}^{-}$ for $k\geq 4$ recursively when $y\geq y_{1}$ . Let $k\geq 3$ be arbitrary. It is easily seen that the following variant of Buchstab’s identity (1.2) holds for any $z\in[y,x]$ :

\Phi(x,y)=\Phi(x,z)+\sum_{y<p\leq z}\Phi(x/p,p^{-}),

(2.6)

where $p^{-}<p$ is any real number sufficiently close to $p$ . For $3\leq k\leq u<k+1$ and $y\geq y_{1}$ , we obtain by taking $z=x^{1/3}$ that

\Phi(x,y)=\Phi\left(x,x^{1/3}\right)+\sum_{y<p\leq x^{1/3}}\Phi(x/p,p^{-}).

(2.7)

We have already shown that

\Phi\left(x,x^{1/3}\right)\geq\frac{x}{\log x^{1/3}}\left(\omega\left(\frac{\log x}{\log x^{1/3}}\right)+\frac{\Delta_{3}^{-}}{\log x^{1/3}}\right)=\frac{3x}{\log y}\left(\frac{\omega(3)}{u}+\frac{3\Delta_{3}^{-}}{u^{2}\log y}\right).

(2.8)

Note that $2<\log(x/p)/\log(p^{-})<k$ . Thus, we have

\Phi(x/p,p^{-})\geq\frac{x}{p\log(p^{-})}\left(\omega\left(\frac{\log(x/p)}{\log(p^{-})}\right)+\frac{\Delta_{k}^{-}}{\log(p^{-})}\right).

Since $\omega(u)$ is continuous on $[1,\infty)$ , it follows from (2.7) and (2.8) that

\Phi(x,y)\geq\frac{3x}{\log y}\left(\frac{\omega(3)}{u}+\frac{3\Delta_{3}^{-}}{u^{2}\log y}\right)+\sum_{y<p\leq x^{1/3}}\frac{x}{p\log p}\left(\omega\left(\frac{\log x}{\log p}-1\right)+\frac{\Delta_{k}^{-}}{\log p}\right).

(2.9)

By partial summation we see that

\sum_{y<p\leq x^{1/3}}\frac{1}{p\log^{2}p}<\int_{y}^{\infty}\frac{1}{t\log^{2}t}\,d\pi(t)=-\frac{\pi(y)}{y\log^{2}y}+\int_{y}^{\infty}\frac{\log t+2}{t^{2}\log^{3}t}\pi(t)\,dt,

which, by (2.2), is

	$\displaystyle<-\frac{1}{\log^{3}y}\left(1+\frac{c_{3}}{\log y}\right)+\int_{y}^{\infty}\frac{\log t+2}{t\log^{4}t}\left(1+\frac{c_{2}}{\log t}\right)\,dt$
	$\displaystyle=-\frac{1}{\log^{3}y}\left(1+\frac{c_{3}}{\log y}\right)+\frac{1}{2\log^{2}y}+\frac{c_{2}+2}{3\log^{3}y}+\frac{c_{2}}{2\log^{4}y}$
	$\displaystyle=\frac{1}{\log^{2}y}\left(\frac{1}{2}+\left(\frac{c_{2}}{3}-1\right)\frac{1}{\log y}+\left(\frac{c_{2}}{2}-c_{3}\right)\frac{1}{\log^{2}y}\right)$
	$\displaystyle<\frac{1}{\log^{2}y}\left(\frac{1}{2}+\left(\frac{c_{2}}{3}-1\right)\frac{1}{\log y}\right).$

Hence

\sum_{y<p\leq x^{1/3}}\frac{\Delta_{k}^{-}x}{p\log^{2}p}\geq\frac{\Delta_{k}^{-}x}{\log^{2}y}\left(\frac{1}{2}+\left(\frac{c_{2}}{3}-1\right)\frac{1}{\log y}\right).

(2.10)

On the other hand, we have

	$\displaystyle\sum_{y<p\leq x^{1/3}}\frac{1}{p\log p}\omega\left(\frac{\log x}{\log p}-1\right)$	$\displaystyle=\frac{1}{\log x}\int_{3^{-}}^{u}v\omega(v-1)\,dG(v)$
		$\displaystyle=\frac{1}{\log x}\left(\int_{3}^{u}\omega(v-1)\,dv+\int_{3^{-}}^{u}v\omega(v-1)\,d\left(G(v)-\log\frac{v}{2}\right)\right).$

Observe that

\int_{3}^{u}\omega(v-1)\,dv=u\omega(u)-3\omega(3)

and that

	$\displaystyle\int_{3^{-}}^{u}v\omega(v-1)\,d\left(G(v)-\log\frac{v}{2}\right)$	$\displaystyle=u\omega(u-1)\left(G(v)-\log\frac{v}{2}\right)-3\omega(2)\left(G(3)-\log\frac{3}{2}\right)$
		$\displaystyle\hskip 8.53581pt-\int_{3^{-}}^{u}\left(G(v)-\log\frac{v}{2}\right)\,d(v\omega(v-1)).$

By [T, (6.23), p. 562] and [T, Theorems III.5.7 & III.6.6], we have, for all $v\geq 3$ , that

	$\displaystyle\frac{d}{dv}(v\omega(v-1))$	$\displaystyle=\omega(v-2)+\omega^{\prime}(v-1)\geq\frac{1}{2}-\rho(v-1)\geq\frac{1}{2}-\rho(2)=\log 2-\frac{1}{2},$
	$\displaystyle\frac{d}{dv}(v\omega(v-1))$	$\displaystyle\leq 1+\rho(v-1)\leq 1+\rho(2)=2-\log 2,$

where $\rho$ is the Dickman-de Bruijn function defined to be the unique continuous solution to the delay differential equation $t\rho^{\prime}(t)+\rho(t-1)=0$ for $t\geq 1$ , subject to the initial value condition $\rho(t)=1$ for $0\leq t\leq 1$ . Moreover, we have

\lim\limits_{v\to 3^{-}}\frac{d}{dv}(v\omega(v-1))=\lim\limits_{v\to 3^{-}}\left(\omega(v-2)+\omega^{\prime}(v-1)\right)=-\frac{1}{4}.

It follows by (2.1) that

\int_{3^{-}}^{u}\left(G(v)-\log\frac{v}{2}\right)\,d(v\omega(v-1))\leq\frac{c_{1}}{\log^{2}y}\left(u\omega(u-1)-3\omega(2)\right).

Thus we have

\int_{3^{-}}^{u}v\omega(v-1)\,d\left(G(v)-\log\frac{v}{2}\right)\geq-\frac{2c_{1}u\omega(u-1)}{\log^{2}y}\geq-\frac{2c_{1}M_{0}u}{\log^{2}y},

where $M_{0}=0.5671432...$ . Hence we have shown that

\sum_{y<p\leq x^{1/3}}\frac{x}{p\log p}\omega\left(\frac{\log x}{\log p}-1\right)\geq\frac{x}{\log y}\left(\omega(u)-\frac{3\omega(3)}{u}-\frac{2c_{1}M_{0}}{\log^{2}y}\right).

(2.11)

Combining (2.9), (2.10) and (2.11), we deduce that

\Delta(x,y)\geq\frac{9\Delta_{3}^{-}}{u^{2}}+\frac{\Delta_{k}^{-}}{2}-\frac{1}{\log y}\left(2c_{1}M_{0}-\left(\frac{c_{2}}{3}-1\right)\Delta_{k}^{-}\right)

for $k\leq u<k+1$ . Therefore, $\Delta_{k+1}^{-}\geq\min(\Delta_{k}^{-},a_{k}^{-})$ for all $k\geq 3$ , where

a_{k}^{-}\colonequals\frac{9\Delta_{3}^{-}}{k^{2}}+\frac{\Delta_{k}^{-}}{2}-\frac{1}{\log y_{1}}\cdot\max\left(2c_{1}M_{0}-\left(\frac{c_{2}}{3}-1\right)\Delta_{k}^{-},0\right).

Consequently, we have $\Delta_{4}^{-}>-0.451835$ and $\Delta_{k}^{-}>-0.480075$ for all $k\geq 5$ .

Suppose now that $602\leq y\leq y_{1}$ . By [RS, Theorem 20] we can replace (2.1) with

\left|G(v)-\log\frac{v}{2}\right|\leq\frac{d_{1}}{\sqrt{y}\log y},

where $d_{1}=2$ . Moreover, (2.2) remains true if we replace $c_{2}$ and $c_{3}$ by $d_{2}=1.2762$ and $d_{3}=1$ , respectively, according to [D, Corollary 5.2]. With these changes, we run the same argument used to handle the case $y\geq y_{1}$ and get

\Delta(x,y)>g(u)-\frac{2d_{1}}{u\sqrt{y}}+\frac{\log y}{uy^{3/2}}-\frac{1}{u^{2}}\left(2-d_{3}+\frac{4d_{1}d_{3}}{\sqrt{y}\log y}+\frac{8d_{2}}{u\log y}+\frac{8d_{2}^{2}}{u^{2}\log^{2}y}\right).

when $2\leq u\leq 3$ and

\Delta(x,y)\geq\frac{9\Delta_{3}^{-}}{u^{2}}+\frac{\Delta_{k}^{-}}{2}-\frac{1}{\log y}\left(\frac{2d_{1}M_{0}\log y}{\sqrt{y}}-\left(\frac{d_{2}}{3}-1\right)\Delta_{k}^{-}\right)

when $3\leq k\leq u<k+1$ , so that we can take

a_{k}^{-}=\frac{9\Delta_{3}^{-}}{k^{2}}+\frac{\Delta_{k}^{-}}{2}-\frac{1}{\log y_{0}}\cdot\max\left(\frac{2d_{1}M_{0}\log y_{0}}{\sqrt{y_{0}}}-\left(\frac{d_{2}}{3}-1\right)\Delta_{k}^{-},0\right).

As a consequence, we have $\Delta_{3}^{-}>-0.563528$ , $\Delta_{4}^{-}>-0.887161$ and $\Delta_{k}^{-}>-0.955421$ for all $k\geq 5$ . This completes the proof of the proposition. ∎

The next result provides a numerical lower bound for $\omega(u)$ on $[3,\infty)$ .

Lemma 2.2.

We have $\omega(u)>0.549307$ for all $u\geq 3$ .

Proof.

Consider first the case $u\in[3,4]$ . Since $(t\omega(t))^{\prime}=\omega(t-1)$ for $t\geq 2$ and $\omega(t)=(\log(t-1)+1)/t$ for $t\in[2,3]$ , we have

\omega(u)=\frac{1}{u}\left(\log 2+1+\int_{3}^{u}\frac{\log(t-2)+1}{t-1}\,dt\right)

for $u\in[3,4]$ . Note that $u\omega^{\prime}(u)=\omega(u-1)-\omega(u)=S(u)/u$ , where

S(u)\colonequals\frac{u(\log(u-2)+1)}{u-1}-\log 2-1-\int_{3}^{u}\frac{\log(t-2)+1}{t-1}\,dt.

Since

	$\displaystyle S^{\prime}(u)$	$\displaystyle=\frac{1}{u-1}\left(\log(u-2)+1+\frac{u}{u-2}-\frac{u(\log(u-2)+1)}{u-1}-(\log(u-2)+1)\right)$
		$\displaystyle=\frac{u(1-(u-2)\log(u-2))}{(u-2)(u-1)^{2}},$

we know that $S(u)$ is strictly increasing on $[3,u_{1}]$ and strictly decreasing on $[u_{1},4]$ , where $u_{1}=3.7632228...$ is the unique solution to the equation $(u-2)\log(u-2)=1$ . But $S(3)=1/2-\log 2<0$ and

S(4)=\frac{\log 2+1}{3}-\int_{3}^{4}\frac{\log(t-2)+1}{t-1}\,dt>0.

It follows that $S(u)$ has a unique zero $u_{2}\in[3,4]$ . The numerical value of $u_{2}$ is given by $u_{2}=3.4697488...$ , according to Mathematica. Hence $S(u)<0$ for $u\in[3,u_{2})$ and $S(u)>0$ for $u\in(u_{2},4]$ . The same is true for $\omega^{\prime}(u)$ , which implies that $\omega(u)$ is strictly decreasing on $[3,u_{2}]$ and strictly increasing on $[u_{2},4]$ . Thus, $\omega(u)\geq\omega(u_{2})=0.5608228...$ for $u\in[3,4]$ .

Consider now the case $u\in[4,\infty)$ . It is known [JR] that $\omega(t)$ satisfies

|\omega(t)-e^{-\gamma}|\leq\frac{\rho(t-1)}{t}

for all $t\geq 1$ . Since $\rho(t)$ is strictly decreasing on $[4,\infty)$ , we have $\omega(u)\geq e^{-\gamma}-\rho(3)/4$ for all $u\geq 4$ . To find the value of $\rho(3)$ , we use $t\rho^{\prime}(t)+\rho(t-1)=0$ for $t\geq 1$ and $\rho(t)=1-\log t$ for $t\in[1,2]$ to obtain

\rho(u)=1-\log 2-\int_{2}^{u}\frac{1-\log(t-1)}{t}\,dt

for $u\in[2,3]$ . It follows that

\omega(u)\geq e^{-\gamma}-\frac{1}{4}\left(1-\log 2-\int_{2}^{3}\frac{1-\log(t-1)}{t}\,dt\right)=0.5493073...

for all $u\geq 4$ . We have therefore shown that $\omega(u)>0.549307$ for all $u\geq 3$ . ∎

We are now ready to prove the following clean lower bound for $\Phi(x,y)$ that we alluded to.

Theorem 2.3.

We have $\Phi(x,y)>0.4x/\log y$ uniformly for all $7\leq y\leq x^{2/3}$ .

Proof.

In the range $\max(7,x^{2/5})\leq y\leq x^{2/3}$ , we have trivially $\Phi(x,y)\geq\pi(x)-\pi(y)+1$ . By [D, Corollary 5.2] we have

	$\displaystyle\pi(x)-\pi(y)$	$\displaystyle\geq\frac{x}{\log x}\left(1+\frac{1}{\log x}\right)-\frac{y}{\log y}\left(1+\frac{1.2762}{\log y}\right)$
		$\displaystyle=\left(\frac{1}{u}\left(1+\frac{1}{\log x}\right)-\frac{y}{x}\left(1+\frac{1.2762u}{\log x}\right)\right)\frac{x}{\log y}$
		$\displaystyle>\left(\frac{2}{5}\left(1+\frac{1}{\log x}\right)-\frac{1}{x^{1/3}}\left(1+\frac{3.1905}{\log x}\right)\right)\frac{x}{\log y}>0.4\frac{x}{\log y}$

whenever $x\geq 41{,}217$ . Furthermore, we have verified $\Phi(x,y)>0.4x/\log y$ for $\max(7,x^{2/5})\leq y\leq x^{2/3}$ with $x\leq 41{,}217$ using Mathematica. Hence, $\Phi(x,y)>0.4x/\log y$ holds in the range $\max(7,x^{2/5})\leq y\leq x^{2/3}$ .

Consider now the case $\max(x^{1/3},7)\leq y\leq x^{2/5}$ . Following the proof of Proposition 2.1, we have

	$\displaystyle\Phi(x,y)$	$\displaystyle=\pi(x)-\pi(y)+1+\sum_{y<p\leq\sqrt{x}}(\pi(x/p)-\pi(p)+1)$
		$\displaystyle=\pi(x)-M(x,y)+\sum_{y<p\leq x^{1/2}}\pi(x/p),$		(2.12)

where

M(x,y)\colonequals\frac{1}{2}\pi\left(\sqrt{x}\right)^{2}-\frac{1}{2}\pi\left(\sqrt{x}\right)-\frac{1}{2}\pi(y)^{2}+\frac{3}{2}\pi(y)-1.

To handle the sum in (2), we appeal to Theorem 5 and its corollary from [RS] to arrive at

G(v)-\log\frac{v}{2}>-\frac{1}{2\log^{2}\sqrt{x}}-\frac{1}{\log^{2}y}\geq-\frac{33}{25\log^{2}y}

in the range $\max(x^{1/3},7)\leq y\leq x^{2/5}$ . By [RS, Corollary 1] we have

\sum_{y<p\leq x^{1/2}}\pi(x/p)>x\sum_{y<p\leq x^{1/2}}\frac{1}{p\log(x/p)}=\frac{x}{\log x}\int_{2^{-}}^{u}\frac{v}{v-1}\,dG(v),

provided that $x\geq 289$ . The right-hand side of the above can be estimated in the same way as in the proof of Proposition 2.1, so we obtain

\sum_{y<p\leq\sqrt{x}}\pi(x/p)>\frac{x}{\log y}\left(\omega(u)-\frac{66}{25u\log^{2}y}\right)-\frac{x}{\log x}.

On the other hand, we see by [D, Corollary 5.2] and [RS, Corollary 2] that

\pi(x)-M(x,y)>\pi(x)-\frac{1}{2}\pi\left(\sqrt{x}\right)^{2}\geq\frac{x}{\log x}\left(1+\frac{1}{\log x}\right)-\frac{25x}{8\log^{2}x}=\frac{x}{\log x}-\frac{17x}{8\log^{2}x}.

for $x\geq 114^{2}$ . Collecting the estimates above and using the inequality $\omega(u)\geq\omega(5/2)=2(\ln(3/2)+1)/5$ for $u\in[5/2,3]$ , we find that

\Phi(x,y)>\frac{\omega(5/2)x}{\log y}-\frac{17x}{8\log^{2}x}-\frac{66x}{25u\log^{3}y}\geq\frac{\omega(5/2)x}{\log y}-\frac{17x}{50\log^{2}y}-\frac{132x}{125\log^{3}y}>0.4\frac{x}{\log y}

for all $\max(46,x^{1/3})\leq y\leq x^{2/5}$ . For $x^{1/3}\leq y\leq x^{2/5}$ with $7\leq y\leq 46$ , we have verified the inequality $\Phi(x,y)>0.4x/\log y$ directly through numerical computation.

Next, we consider the range $7\leq y<x^{1/3}$ . By Proposition 2.1 and Lemma 2.2 we have

\Phi(x,y)>\frac{x}{\log y}\left(0.549307-\frac{0.955421}{\log y}\right)>0.4\frac{x}{\log y},

provided that $y\geq 602$ . To deal with the range $7\leq y\leq\min(x^{1/3},602)$ , we follow the inclusion-exclusion technique used in [FP, Section 3]. For any integer $n\geq 1$ , let $\nu(n)$ denote the number of distinct prime factors of $n$ . We start by “pre-sieving” with the primes 2, 3, and 5: for any $x\geq 1$ the number of integers $n\leq x$ with $\gcd(n,30)=1$ is $(4/15)x+r_{x}$ , where $|r_{x}|\leq 14/15$ . Let $P_{5}(y)$ be the product of the primes in $(5,y]$ . Then we have by the Bonferroni inequalities that

\Phi(x,y)\geq\sum_{\begin{subarray}{c}d\mid P_{5}(y)\\ \nu(d)\leq 3\end{subarray}}\mu(d)\left(\frac{4}{15}\cdot\frac{x}{d}+r_{x/d}\right)\geq a(y)x-b(y),

where

	$\displaystyle a(y)$	$\displaystyle\colonequals\frac{4}{15}\sum_{\begin{subarray}{c}d\mid P_{5}(y)\\ \nu(d)\leq 3\end{subarray}}\frac{\mu(d)}{d}=\frac{4}{15}\sum_{j=0}^{3}(-1)^{j}\sum_{\begin{subarray}{c}d\mid P_{5}(y)\\ \nu(d)=j\end{subarray}}\frac{1}{d},$
	$\displaystyle b(y)$	$\displaystyle\colonequals\frac{14}{15}\sum_{j=0}^{3}\binom{\pi(y)-3}{j}.$

By Newton’s identities, the inner sum in the definition of $a(y)$ can be represented in terms of the power sums of $1/p$ over all primes $5<p\leq y$ . Thus, we have $\Phi(x,y)>0.4x/\log y$ whenever $a(y)>0.4/\log y$ and $x>b(y)/(a(y)-0.4/\log y)$ . Using Mathematica, we find that the inequality $\Phi(x,y)>0.4x/\log y$ holds for $7\leq y\leq 602$ and $x\geq 13{,}160{,}748$ . Finally, we have verified the inequality $\Phi(x,y)>0.4x/\log y$ directly for $7\leq y\leq x^{1/3}$ with $x\leq 13{,}160{,}748$ by numerical calculations, completing the proof of our theorem. ∎

Remark 2.1.

Note that for $y\in[5,7)$ we have

\Phi(x,y)\geq\frac{4}{15}x-\frac{14}{15}>0.4\frac{x}{\log 5}\geq 0.4\frac{x}{\log y},

provided that $x\geq 52$ . Combined with Theorem 2.3 and numerical examination of the case $11\leq x\leq 52$ , this implies that the inequality $\Phi(x,y)>0.4x/\log y$ holds in the slightly larger range $5\leq y\leq x^{2/3}$ if one assumes $x\geq 41$ .

3. An Explicit Version of de Bruijn’s Estimate

To prove Theorem 1.1, we shall first develop an explicit version of (1.6) with a general $R(y)$ , following [Br], where $R(y)$ is a positive decreasing function satisfying the same conditions described in the introduction. Suppose that $y_{0}\geq 3$ . For each $z\geq 2$ , put

Q(z)\colonequals\prod_{p\leq z}\left(1-\frac{1}{p}\right).

We start by estimating $Q(y)$ for $y\geq y_{0}$ . Using a Stieltjes integral, we may write

\log\frac{Q(z)}{Q(y)}=\int_{y}^{z}\log\left(1-t^{-1}\right)\,d\operatorname{li}(y)+\int_{y}^{z}\log\left(1-t^{-1}\right)\,d(\pi(y)-\operatorname{li}(t)),

(3.1)

where $z\geq y\geq y_{0}$ . The first integral on the right-hand side of the above is equal to

\displaystyle\int_{y}^{z}\log\left(1-t^{-1}\right)\frac{dt}{\log t}=-\log\frac{\log z}{\log y}+\int_{y}^{z}\left(t^{-1}+\log\left(1-t^{-1}\right)\right)\frac{dt}{\log t}.

Since

-\frac{1}{2t(t-1)}<t^{-1}+\log\left(1-t^{-1}\right)<0

for all $t\geq y_{0}$ , we have

-\frac{1}{2}\int_{y}^{\infty}\frac{dt}{t(t-1)\log t}<\int_{y}^{z}\left(t^{-1}+\log\left(1-t^{-1}\right)\right)\frac{dt}{\log t}<0.

But a change of variable shows that

\int_{y}^{\infty}\frac{dt}{t(t-1)\log t}=\int_{1}^{\infty}\frac{dt}{t(y^{t}-1)}\leq\frac{1}{y-1}\int_{1}^{\infty}\frac{dt}{t^{2}}=\frac{1}{y-1},

where we have used the inequality $y^{t}-1\geq(y-1)t$ for $t\geq 1$ and $y\geq y_{0}$ . It follows that

-\frac{1}{2(y-1)}\leq\int_{y}^{z}\log\left(1-t^{-1}\right)\,d\operatorname{li}(y)+\log\frac{\log z}{\log y}<0.

(3.2)

Now we estimate the second integral on the right-hand side of (3.1). By (1.4) and partial integration we have

	$\displaystyle\left\|\int_{y}^{z}\log\left(1-t^{-1}\right)\,d(\pi(y)-\operatorname{li}(t))\right\|$	$\displaystyle\leq\log\left(1-y^{-1}\right)^{-1}\frac{y}{\log y}R(y)+\log\left(1-z^{-1}\right)^{-1}\frac{z}{\log z}R(z)$
		$\displaystyle\hskip 5.69054pt+\int_{y}^{z}\frac{\|\pi(t)-\operatorname{li}(t)\|}{t(t-1)}\,dt.$

Using (1.5) we see that

\int_{y}^{z}\frac{|\pi(t)-\operatorname{li}(t)|}{t(t-1)}\,dt\leq\frac{C_{0}(y_{0})y_{0}}{y_{0}-1}R(y).

It is clear that the function

\log\left(1-t^{-1}\right)^{-1}\frac{t}{\log t}=\frac{1}{\log t}\sum_{n=0}^{\infty}\frac{t^{-n}}{n+1}

is strictly decreasing for $t\in(1,\infty)$ . Since $R(t)$ is decreasing on $[y_{0},\infty)$ , we find that

\left|\int_{y}^{z}\log\left(1-t^{-1}\right)\,d(\pi(y)-\operatorname{li}(t))\right|\leq\left(2\log\left(1-y_{0}^{-1}\right)^{-1}\frac{y_{0}}{\log y_{0}}+\frac{C_{0}(y_{0})y_{0}}{y_{0}-1}\right)R(y).

Combining this inequality with (3.1) and (3.2) yields

-C_{2}(y_{0})R(y)\leq\log\frac{Q(z)}{Q(y)}+\log\frac{\log z}{\log y}\leq C_{1}(y_{0})R(y)

(3.3)

for $z\geq y\geq y_{0}$ , where

	$\displaystyle C_{1}(y_{0})$	$\displaystyle=2\log\left(1-y_{0}^{-1}\right)^{-1}\frac{y_{0}}{\log y_{0}}+\frac{C_{0}(y_{0})y_{0}}{y_{0}-1},$
	$\displaystyle C_{2}(y_{0})$	$\displaystyle=C_{1}(y_{0})+\sup_{t\geq y_{0}}\frac{1}{2(t-1)R(t)}.$

Exponentiating (3.3) we obtain

-C_{4}(y_{0})R(y)\leq\frac{Q(z)\log z}{Q(y)\log y}-1\leq C_{3}(y_{0})R(y)

(3.4)

for $z\geq y\geq y_{0}$ , where

	$\displaystyle C_{3}(y_{0})$	$\displaystyle=\sup_{t\geq y_{0}}\frac{\exp(C_{1}(y_{0})R(t))-1}{R(t)}=\frac{\exp(C_{1}(y_{0})R(y_{0}))-1}{R(y_{0})},$
	$\displaystyle C_{4}(y_{0})$	$\displaystyle=\sup_{t\geq y_{0}}\frac{1-\exp(-C_{2}(y_{0})R(t))}{R(t)}=C_{2}(y_{0}).$

As a consequence, we have by letting $z\to\infty$ in (3.4) and using the fact that $Q(z)\log z\to e^{-\gamma}$ as $z\to\infty$ , that

e^{\gamma}\log y(1-C_{4}(y_{0})R(y))\leq\frac{1}{Q(y)}\leq e^{\gamma}\log y(1+C_{3}(y_{0})R(y)).

(3.5)

Similarly, we derive from (3.3) that

\frac{e^{-\gamma}}{\log y}(1-C_{6}(y_{0})R(y))\leq Q(y)\leq\frac{e^{-\gamma}}{\log y}(1+C_{5}(y_{0})R(y))

(3.6)

for $y\geq y_{0}$ , where

	$\displaystyle C_{5}(y_{0})$	$\displaystyle=\sup_{t\geq y_{0}}\frac{\exp(C_{2}(y_{0})R(t))-1}{R(t)}=\frac{\exp(C_{2}(y_{0})R(y_{0}))-1}{R(y_{0})},$
	$\displaystyle C_{6}(y_{0})$	$\displaystyle=\sup_{t\geq y_{0}}\frac{1-\exp(-C_{1}(y_{0})R(t))}{R(t)}=C_{1}(y_{0}).$

For $x\geq y\geq 2$ , we define

\psi(x,y)\colonequals\frac{\Phi(x,y)}{xQ(y)}.

We then need to estimate $\eta(x,y)=\psi(x,y)-\lambda(x,y)$ , where $\lambda(x,y)\colonequals e^{\gamma}\mu_{y}(u)\log y$ . For $1\leq u\leq 2$ this can be done straightforward. Indeed, we have $\Phi(x,y)=\pi(x)-\pi(y)+1$ and $\omega(u)=1/u$ when $1\leq u\leq 2$ , so that

\eta(x,y)=\frac{\pi(x)-\pi(y)+1}{xQ(y)}-e^{\gamma}\log y\int_{1}^{u}t^{-1}y^{t-u}\,dt.

Note that

\left|\pi(x)-\pi(y)-x\int_{1}^{u}t^{-1}y^{t-u}\,dt\right|=\left|\pi(x)-\pi(y)-\int_{y}^{x}\frac{dt}{\log t}\right|\leq\left(\frac{x}{\log x}+\frac{y}{\log y}\right)R(y).

From (3.5) it follows that $|\eta(x,y)|\leq e^{\gamma}\alpha_{y}(u)R(y)$ for $y\geq y_{0}$ and $u\in[1,2]$ , where

\alpha_{y}(u)\colonequals\frac{\log y}{y^{u}R(y)}+C_{3}(y_{0})\left(\frac{\log y}{y^{u}}+\log y\int_{1}^{u}t^{-1}y^{t-u}\,dt\right)+(1+C_{3}(y_{0})R(y))\left(\frac{1}{u}+y^{1-u}\right).

Integration by parts enables us to write

\log y\int_{1}^{u}t^{-1}y^{t-u}\,dt=\frac{1}{u}-y^{1-u}+\int_{1}^{u}t^{-2}y^{t-u}\,dt

for $y\geq y_{0}$ . Hence $|\eta(x,y)|\leq e^{\gamma}\eta_{1}(y)R(y)$ for $y\geq y_{0}$ and $u\in[1,2]$ , where

\eta_{1}(y)\colonequals\sup_{t\geq y}\frac{\log t}{tR(t)}+\max_{u\in[1,2]}\left(C_{3}(y_{0})I_{y}(u)+(1+C_{3}(y_{0})R(y))\left(\frac{1}{u}+y^{1-u}\right)\right)

(3.7)

with

I_{y}(u)\colonequals\frac{1}{u}+\int_{1}^{u}t^{-2}y^{t-u}\,dt.

We remark that $I_{y}(u)$ is strictly decreasing on $[1,2]$ and hence satisfies $I_{y}(u)<1$ for $u\in(1,2]$ , since its derivative is

I_{y}^{\prime}(u)=-\int_{1}^{u}t^{-2}y^{t-u}\log y\,dt<0.

Thus, (3.7) simplifies to

\eta_{1}(y)=\sup_{t\geq y}\frac{\log t}{tR(t)}+C_{3}(y_{0})+2(1+C_{3}(y_{0})R(y)).

(3.8)

Suppose now that $y\geq y_{0}$ and $u\geq 2$ . From (2.6) it follows that

\psi(x,y)=\psi(x,z)\frac{Q(z)}{Q(y)}+\sum_{y<p\leq z}\psi(x/p,p^{-})\cdot\frac{1}{p}\prod_{y<q<p}\left(1-\frac{1}{q}\right),

(3.9)

where $z\geq y\geq y_{0}$ . Put $h\colonequals\log z/\log y\geq 1$ and

H_{y}(v)\colonequals\sum_{y<p\leq y^{v}}\frac{1}{p}\prod_{y<q<p}\left(1-\frac{1}{q}\right)

(3.10)

for $v\geq 1$ . Then we have $H_{y}(v)=1-Q(y^{v})/Q(y)$ . By partial summation, we see that (3.9) becomes

\psi(x,y)=\psi(y^{u},y^{h})(1-H_{y}(h))+\int_{1}^{h}\psi(y^{u-v},(y^{v})^{-})\,dH_{y}(v).

(3.11)

By (3.4) we have

|H_{y}(v)-1+v^{-1}|\leq C_{7}(y_{0})R(y),

where $C_{7}(y_{0})\colonequals\max(C_{3}(y_{0}),C_{4}(y_{0}))$ . Thus, one can think of $1-v^{-1}$ as a smooth approximation to $H_{y}(v)$ . Since we also expect $\lambda(x,y)$ to be a smooth approximation to $\psi(x,y)$ , in view of (3.11) it is reasonable to expect

E_{1}(h;y,u)\colonequals\lambda(y^{u},y)-\lambda(y^{u},y^{h})h^{-1}-\int_{1}^{h}\lambda(y^{u-v},y^{v})v^{-2}\,dv

to be small in size as a function of $y$ . This can be easily verified when $1\leq h\leq u/2$ . Following de Bruijn [Br], we have

\frac{\partial}{\partial h}E_{1}(h;y,u)=-h^{-1}\cdot\frac{\partial}{\partial h}\lambda(y^{u},y^{h})+h^{-2}\lambda(y^{u},y)-h^{-2}\lambda(y^{u-h},y^{h}).

(3.12)

Since

\frac{\lambda(y^{u},y^{h})}{e^{\gamma}\log y}=h\int_{1}^{u/h}y^{ht-u}\omega(t)\,dt,

we find

\frac{\partial}{\partial h}\left(\frac{\lambda(y^{u},y^{h})}{e^{\gamma}\log y}\right)=\int_{1}^{u/h}y^{ht-u}\omega(t)\,dt+h\left(\log y\int_{1}^{u/h}y^{ht-u}(t\omega(t))\,dt-uh^{-2}\omega(uh^{-1})\right).

Recall that $(t\omega(t))^{\prime}=\omega(t-1)$ for $t\in\mathbb{R}$ with the obvious extension $\omega(t)=0$ for $t<1$ . It follows that

	$\displaystyle\log y\int_{1}^{u/h}y^{ht-u}(t\omega(t))\,dt$	$\displaystyle=h^{-1}y^{ht-u}(t\omega(t))\bigg{\|}_{1}^{u/h}-h^{-1}\int_{1}^{u/h}y^{ht-u}\omega(t-1)\,dt$
		$\displaystyle=uh^{-2}\omega(uh^{-1})-h^{-1}y^{h-u}-h^{-1}y^{h}\int_{1}^{u/h-1}y^{ht-u}\omega(t)\,dt$
		$\displaystyle=uh^{-2}\omega(uh^{-1})-h^{-1}y^{h-u}-\left(h^{2}e^{\gamma}\log y\right)^{-1}\lambda(y^{u-h},y^{h}).$

Hence we have

	$\displaystyle\frac{\partial}{\partial h}\lambda(y^{u},y^{h})$	$\displaystyle=e^{\gamma}\log y\left(\int_{1}^{u/h}y^{ht-u}\omega(t)\,dt-y^{h-u}\right)-h^{-1}\lambda(y^{u-h},y^{h})$
		$\displaystyle=h^{-1}\lambda(y^{u},y^{h})-e^{\gamma}y^{h-u}\log y-h^{-1}\lambda(y^{u-h},y^{h}).$

Inserting this in (3.12) yields

\frac{\partial}{\partial h}E_{1}(h;y,u)=h^{-1}e^{\gamma}y^{h-u}\log y.

Integrating both sides with respect to $h$ and using the initial value condition $E_{1}(1;y,u)=0$ , we obtain

E_{1}(h;y,u)=e^{\gamma}\log y\int_{1}^{h}t^{-1}y^{t-u}\,dt<e^{\gamma}y^{h-u}.

(3.13)

In what follows, we shall always suppose that $1\leq h\leq u/2$ . Following de Bruijn [Br], we proceed to show that

E_{3}(h;y,u)\colonequals\lambda(y^{u},y)-\lambda(y^{u},y^{h})(1-H(h))-\int_{1}^{h}\lambda(y^{u-v},y^{v})\,dH(h)

is small in size as a function of $y$ . This is intuitive, because

\lambda(y^{u},y^{h})h^{-1}-\int_{1}^{h}\lambda(y^{u-v},y^{v})v^{-2}\,dv,

which is a good approximation to $\lambda(y^{u},y)$ as we have already demonstrated, can be thought of as a smooth approximation to

\lambda(y^{u},y^{h})(1-H(h))-\int_{1}^{h}\lambda(y^{u-v},y^{v})\,dH(h).

Moreover, we have by (3.11) that

\eta(x,y)=\eta(y^{u},y^{h})(1-H_{y}(h))+\int_{1}^{h}\eta(y^{u-v},(y^{v})^{-})\,dH_{y}(v)-E_{3}(h;y,u),

(3.14)

which will later be used to estimate $\eta(x,y)$ . To estimate $E_{3}(h;y,u)$ , let us write $E_{3}(h;y,u)=E_{1}(h;y,u)+E_{2}(h;y,u)$ , where

E_{2}(h;y,u)\colonequals-\int_{1}^{h}\lambda(y^{u-v},y^{v})\,d\left(H(v)-1+v^{-1}\right)+(H(h)-1+h^{-1})\lambda(y^{u},y^{h}).

Then we expect $E_{2}(h;y,u)$ to be small in size as a function of $y$ . Using (3.10) and the observation that $H(1)=0$ , we have

|E_{2}(h;y,u)|\leq\left(\left|\lambda(y^{u},y^{h})-\lambda(y^{u-h},y^{h})\right|+\int_{1}^{h}\left|\frac{\partial}{\partial v}\lambda(y^{u-v},y^{v})\right|\,dv\right)C_{7}(y_{0})R(y).

(3.15)

Note that

	$\displaystyle\frac{\lambda(y^{u},y^{h})-\lambda(y^{u-h},y^{h})}{he^{\gamma}\log y}$	$\displaystyle=\int_{1}^{u/h}y^{ht-u}\omega(t)\,dt-\int_{2}^{u/h}y^{ht-u}\omega(t-1)\,dt$
		$\displaystyle=\int_{1}^{2}y^{ht-u}\omega(t)\,dt+\int_{2}^{u/h}y^{ht-u}(\omega(t)-\omega(t-1))\,dt$
		$\displaystyle=\int_{1}^{2}t^{-1}y^{ht-u}\,dt-\int_{2}^{u/h}y^{ht-u}t\omega^{\prime}(t)\,dt.$

By Theorems III.5.7 and III.6.6 in [T] we have

|\omega^{\prime}(t)|\leq\rho(t)\leq\frac{1}{\Gamma(t+1)}

(3.16)

for all $t\geq 1$ . It follows that

\left|\lambda(y^{u},y^{h})-\lambda(y^{u-h},y^{h})\right|\leq he^{\gamma}\log y\left(\int_{1}^{2}t^{-1}y^{ht-u}\,dt+\int_{2}^{u/h}y^{ht-u}t\rho(t)\,dt\right).

(3.17)

This inequality will later be used in conjunction with the formulas

h\log y\int_{1}^{2}t^{-1}y^{ht-u}\,dt=\frac{y^{2h-u}}{2}-y^{h-u}+\int_{1}^{2}t^{-2}y^{ht-u}\,dt

(3.18)

and

	$\displaystyle h\log y\int_{2}^{u/h}y^{ht-u}t\rho(t)\,dt$	$\displaystyle=uh^{-1}\rho(uh^{-1})-2\rho(2)y^{2h-u}-\int_{2}^{u/h}y^{ht-u}(t\rho(t))^{\prime}\,dt$
		$\displaystyle\leq uh^{-1}\rho(uh^{-1})-2\rho(2)y^{2h-u}+\int_{2}^{u/h}y^{ht-u}\rho(t-1)\,dt.$		(3.19)

On the other hand, we have

\frac{\lambda(y^{u-v},y^{v})}{e^{\gamma}\log y}=v\int_{2}^{u/v}y^{vt-u}\omega(t-1)\,dt,

which implies that

\frac{\partial}{\partial v}\left(\frac{\lambda(y^{u-v},y^{v})}{e^{\gamma}\log y}\right)=\int_{2}^{u/v}y^{vt-u}(1+tv\log y)\omega(t-1)\,dt-uv^{-1}\omega(uv^{-1}-1).

By partial integration, the right side of the above is easily seen to be

-2y^{2v-u}-\int_{2}^{u/v}y^{vt-u}t\omega^{\prime}(t-1)\,dt.

Hence, we arrive at

\int_{1}^{h}\left|\frac{\partial}{\partial v}\lambda(y^{u-v},y^{v})\right|\,dv\leq e^{\gamma}\log y\left(2\int_{1}^{h}y^{2v-u}\,dv+\int_{1}^{h}\int_{2}^{u/v}y^{vt-u}t|\omega^{\prime}(t-1)|\,dtdv\right).

Furthermore, we have by Fubini’s theorem that

\int_{1}^{h}\int_{2}^{u/v}y^{vt-u}t|\omega^{\prime}(t-1)|\,dtdv=\int_{2}^{u/h}\int_{1}^{h}y^{vt-u}t|\omega^{\prime}(t-1)|\,dv\,dt+\int_{u/h}^{u}\int_{1}^{u/t}y^{vt-u}t|\omega^{\prime}(t-1)|\,dv\,dt,

the right side of which is easily seen to be

\frac{1}{\log y}\left(\int_{2}^{u/h}y^{ht-u}|\omega^{\prime}(t-1)|\,dt+\int_{u/h}^{u}|\omega^{\prime}(t-1)|\,dt-\int_{2}^{u}y^{t-u}|\omega^{\prime}(t-1)|\,dt\right).

It follows that

\int_{1}^{h}\left|\frac{\partial}{\partial v}\lambda(y^{u-v},y^{v})\right|\,dv<e^{\gamma}\left(y^{2h-u}+\int_{2}^{u/h}y^{ht-u}|\omega^{\prime}(t-1)|\,dt+\int_{u/h}^{u}|\omega^{\prime}(t-1)|\,dt\right).

(3.20)

This estimate together with (3.17) will lead us to a good estimate for $E_{2}(h;y,u)$ .

Now we derive estimates for $E_{3}(h;y,u)$ that suit our needs. Suppose that $k\leq u<k+1$ and take $h=h_{k}=u/k$ , where $k\geq 2$ is a positive integer. We first consider the case $k=2$ . In view of (3.18), we see that (3.17) simplifies to

\left|\lambda\left(y^{u},y^{h_{2}}\right)-\lambda\left(y^{u-h_{2}},y^{h_{2}}\right)\right|<e^{\gamma}\left(\frac{1}{2}+\int_{1}^{2}t^{-2}y_{0}^{t-2}\,dt\right)=e^{\gamma}I_{y_{0}}(2)

for $y\geq y_{0}$ . By (3.20) we have

\int_{1}^{h_{2}}\left|\frac{\partial}{\partial v}\lambda(y^{u-v},y^{v})\right|\,dv\leq e^{\gamma}\left(1+\int_{2}^{3}|\omega^{\prime}(t-1)|\,dt\right)=\frac{3e^{\gamma}}{2},

since $\omega^{\prime}(t)=-1/t^{2}$ for $t\in[1,2)$ . Combining these estimates with (3.13) and (3.15), we obtain $E_{3}(h_{2};y,u)\leq e^{\gamma}\xi_{2}(y_{0})R(y)$ for $y\geq y_{0}$ and $2\leq u<3$ , where

\xi_{2}(y_{0})\colonequals\max_{t\geq y_{0}}\frac{1}{tR(t)}+C_{7}(y_{0})\left(I_{y_{0}}(2)+\frac{3}{2}\right).

Now we handle the case $k\geq 3$ . From (3.16)–(3.19) it follows that

\left|\lambda\left(y^{u},y^{h_{k}}\right)-\lambda\left(y^{u-h_{k}},y^{h_{k}}\right)\right|<e^{\gamma}\left(\frac{1}{\Gamma(k)}+\left(2\log 2-\frac{3}{2}\right)y^{2-k}+\int_{1}^{2}t^{-2}y^{t-k}\,dt+\int_{2}^{3}y^{t-k}(1-\log(t-1))\,dt+\int_{3}^{k}y^{t-k}\frac{dt}{\Gamma(t)}\right),

where we have used the fact that $\rho(t)=1-\log t$ for $t\in[1,2]$ . By (3.16) and (3.20) we have

\int_{1}^{h_{k}}\left|\frac{\partial}{\partial v}\lambda(y^{u-v},y^{v})\right|\,dv\leq e^{\gamma}\left(y^{2-k}+\int_{2}^{3}y^{t-k}\frac{dt}{(t-1)^{2}}+\int_{3}^{k}y^{t-k}\frac{dt}{\Gamma(t)}+\int_{k}^{k+1}\frac{dt}{\Gamma(t)}\right).

Together with (3.13) and (3.15), these inequalities imply that $E_{3}(h_{k};y,u)\leq e^{\gamma}\xi_{k}(y_{0})R(y)$ for $y\geq y_{0}$ and $3\leq k\leq u<k+1$ , where

\xi_{k}(y_{0})\colonequals\left(\max_{t\geq y_{0}}\frac{1}{tR(t)}\right)y_{0}^{2-k}+C_{7}(y_{0})\left(\frac{1}{(k-1)!}+\int_{k}^{k+1}\frac{dt}{\Gamma(t)}+\left(2\log 2-\frac{1}{2}\right)y_{0}^{2-k}+\int_{1}^{2}t^{-2}y_{0}^{t-k}\,dt+\int_{2}^{3}y_{0}^{t-k}\left(1-\log(t-1)+\frac{1}{(t-1)^{2}}\right)\,dt+2\int_{3}^{k}y_{0}^{t-k}\frac{dt}{\Gamma(t)}\right).

As a direct corollary, we obtain

\sum_{k=2}^{\infty}\xi_{k}(y_{0})=\frac{y_{0}}{y_{0}-1}\max_{t\geq y_{0}}\frac{1}{tR(t)}+C_{7}(y_{0})\left(e-\frac{1}{2}+\int_{3}^{\infty}\frac{dt}{\Gamma(t)}+\frac{1}{y_{0}-1}\left(2\log 2-1+y_{0}I_{y_{0}}(2)+\int_{2}^{3}y_{0}^{t-2}\left(1-\log(t-1)+\frac{1}{(t-1)^{2}}\right)\,dt+2\int_{3}^{\infty}y_{0}^{\{t\}}\frac{dt}{\Gamma(t)}\right)\right),

where we have applied partial summation to derive

	$\displaystyle\sum_{k=3}^{\infty}\int_{3}^{k}y_{0}^{t-k}\frac{dt}{\Gamma(t)}$	$\displaystyle=\left(\sum_{k=3}^{\infty}y_{0}^{-k}\right)\int_{3}^{\infty}y_{0}^{t}\frac{dt}{\Gamma(t)}-\int_{3}^{\infty}\left(\sum_{3\leq k\leq t}y_{0}^{-k}\right)y_{0}^{t}\frac{dt}{\Gamma(t)}$
		$\displaystyle=\frac{y_{0}^{-3}}{1-y_{0}^{-1}}\int_{3}^{\infty}y_{0}^{t}\frac{dt}{\Gamma(t)}-\int_{3}^{\infty}\frac{y_{0}^{t-3}(1-y_{0}^{-\lfloor t\rfloor+2})}{1-y_{0}^{-1}}\cdot\frac{dt}{\Gamma(t)}$
		$\displaystyle=\frac{1}{y_{0}-1}\int_{3}^{\infty}y_{0}^{\{t\}}\frac{dt}{\Gamma(t)}.$

For computational purposes, we can transform the last integral above by observing that

\int_{3}^{\infty}y_{0}^{\{t\}}\frac{dt}{\Gamma(t)}=\int_{0}^{1}\left(\sum_{n=0}^{\infty}\frac{1}{(t+2)\cdots(t+2+n)}\right)y_{0}^{t}\frac{dt}{\Gamma(t+2)}.

Let

\gamma(s,z)\colonequals\int_{0}^{z}v^{s-1}e^{-v}\,dv

be the lower incomplete gamma function, where $s\in\mathbb{C}$ with $\Re(s)>0$ and $z\geq 0$ . It is well known that

\gamma(s,z)=z^{s}e^{-z}\sum_{n=0}^{\infty}\frac{z^{n}}{s(s+1)\cdots(s+n)},

from which it follows that

\sum_{n=0}^{\infty}\frac{1}{(t+2)\cdots(t+2+n)}=\gamma(t+2,1)e.

Thus we obtain

\sum_{k=2}^{\infty}\xi_{k}(y_{0})=\frac{y_{0}}{y_{0}-1}\max_{t\geq y_{0}}\frac{1}{tR(t)}+C_{7}(y_{0})\left(e-\frac{1}{2}+\int_{3}^{\infty}\frac{dt}{\Gamma(t)}+\frac{1}{y_{0}-1}\left(2\log 2-1+y_{0}I_{y_{0}}(2)+\int_{2}^{3}y_{0}^{t-2}\left(1-\log(t-1)+\frac{1}{(t-1)^{2}}\right)\,dt+2e\int_{0}^{1}y_{0}^{t}\frac{\gamma(t+2,1)}{\Gamma(t+2)}\,dt\right)\right).

(3.21)

In Mathematica, the function $\gamma(t+2,1)$ can be evaluated by “Gamma[t+2,0,1]”.

Finally, we are ready to estimate $\eta(x,y)$ . Let

\eta_{k}(y)\colonequals\frac{1}{e^{\gamma}R(y)}\sup_{\begin{subarray}{c}u\in[k,k+1)\\ t\geq y\end{subarray}}|\eta(t^{u},t)|

for $k\geq 1$ and $y\geq y_{0}$ , where the value of $\eta_{1}(y)$ is provided by (3.8). Using (3.14) and the estimates for $E_{3}(h_{k};y,u)$ with $y\geq y_{0}$ and $2\leq k\leq u<k+1$ , we find

\eta_{k}(y)\leq\eta_{k-1}(y)+\xi_{k}(y_{0})

for all $k\geq 2$ and $y\geq y_{0}$ , from which we derive

\eta_{k}(y)\leq\eta_{1}(y)+\sum_{\ell=2}^{k}\xi_{\ell}(y_{0})

for all $k\geq 1$ and $y\geq y_{0}$ . Since $\eta_{1}(y)$ is decreasing on $[y_{0},\infty)$ , we have therefore shown that

|\eta(x,y)|\leq e^{\gamma}\left(\eta_{1}(y_{0})+\sum_{k=2}^{\infty}\xi_{k}(y_{0})\right)R(y)

(3.22)

for all $y\geq y_{0}$ , where the infinite sum can be evaluated using (3.21). To derive an explicit version of de Bruijn’s result (1.6), we observe that (3.6), (3.22) and [RS, Theorem 23] imply that $Q(y)|\eta(x,y)|\leq C_{8}(y_{0})R(y)/\log y$ for all $y\geq y_{0}$ , where

C_{8}(y_{0})\colonequals\beta(y_{0})\left(\eta_{1}(y_{0})+\sum_{k=2}^{\infty}\xi_{k}(y_{0})\right)

with

\beta(y_{0})\colonequals\begin{cases}~{}1,&\text{\hskip 5.69054ptif~{}}3\leq y_{0}<10^{8},\\ ~{}\exp(C_{2}(y_{0})R(y_{0})),&\text{\hskip 5.69054ptif~{}}y_{0}\geq 10^{8}.\end{cases}

Hence, it follows that

\left|\Phi(x,y)-\mu_{y}(u)e^{\gamma}x\log y\prod_{p\leq y}\left(1-\frac{1}{p}\right)\right|<\frac{C_{8}(y_{0})xR(y)}{\log y}

(3.23)

for all $y\geq y_{0}$ .

4. Deduction of Theorem 1.1 and Corollary 1.2

Now we apply (3.23) to obtain explicit estimates for $\Phi(x,y)$ with specific choices of $R(y)$ . Unconditionally, it has been shown [MT, Corollary 2] that

|\pi(z)-\operatorname{li}(z)|\leq 0.2593\frac{z}{(\log z)^{3/4}}\exp\left(-\sqrt{\frac{\log z}{6.315}}\right)

for all $z\geq 229$ . With $y_{0}\geq 229$ , the function

R(z)=0.2593(\log z)^{1/4}\exp\left(-\sqrt{\frac{\log z}{6.315}}\right)

is strictly decreasing on $[y_{0},\infty)$ and satisfies (1.4) and (1.5) with

C_{0}(y_{0})=2\sqrt{\frac{6.315}{\log y_{0}}},

since

	$\displaystyle\int_{z}^{\infty}\frac{1}{t(\log t)^{3/4}}\exp\left(-\sqrt{\frac{\log t}{6.315}}\right)\,dt$	$\displaystyle=2\int_{\sqrt{\log z}}^{\infty}\frac{1}{\sqrt{t}}\exp\left(-\frac{t}{\sqrt{6.315}}\right)\,dt$
		$\displaystyle<\frac{2}{(\log z)^{1/4}}\int_{\sqrt{\log z}}^{\infty}\exp\left(-\frac{t}{\sqrt{6.315}}\right)\,dt$
		$\displaystyle=\frac{2\sqrt{6.315}}{(\log z)^{1/4}}\exp\left(-\sqrt{\frac{\log z}{6.315}}\right)$

for $z\geq y_{0}$ . Numerical computation using Mathematica allows us to conclude that

\left|\Phi(x,y)-\mu_{y}(u)e^{\gamma}x\log y\prod_{p\leq y}\left(1-\frac{1}{p}\right)\right|<4.403611\frac{x}{(\log y)^{3/4}}\exp\left(-\sqrt{\frac{\log y}{6.315}}\right)

(4.1)

for all $x\geq y\geq 229$ . Suppose now that $2\leq y<229$ . Using the inequalities $\Phi(x,y)<x/\log y$ [F, Theorem], $\prod_{p\leq y}(1-1/p)<e^{-\gamma}/\log y$ [RS, Theorem 23] and $0\leq\mu_{y}(u)<1/\log y$ , we have

\left|\Phi(x,y)-\mu_{y}(u)e^{\gamma}x\log y\prod_{p\leq y}\left(1-\frac{1}{p}\right)\right|<\frac{2x}{\log y}<4.403611\frac{x}{(\log y)^{3/4}}\exp\left(-\sqrt{\frac{\log y}{6.315}}\right)

for all $2\leq y<229$ . Combining this with (4.1) proves the first half of Theorem 1.1.

Under the assumption of the Riemann Hypothesis, it is known [S, Corollary 1] that

|\pi(z)-\operatorname{li}(z)|<\frac{1}{8\pi}\sqrt{z}\log z

for all $z\geq 2{,}657$ . With $y_{0}=2{,}657$ and

R(z)=\frac{\log^{2}z}{8\pi\sqrt{z}},

we have

\int_{z}^{\infty}\frac{|\pi(t)-\operatorname{li}(t)|}{t^{2}}\,dt\leq\frac{1}{8\pi}\int_{z}^{\infty}\frac{\log t}{t^{3/2}}\,dt=\frac{\log z+2}{4\pi\sqrt{z}}\leq C_{0}(y_{0})R(z)

for $z\geq y_{0}$ , where

C_{0}(y_{0})=\frac{2(\log y_{0}+2)}{\log^{2}y_{0}}.

Therefore, we conclude by (3.23) and numerical calculations that

\left|\Phi(x,y)-\mu_{y}(u)e^{\gamma}x\log y\prod_{p\leq y}\left(1-\frac{1}{p}\right)\right|<0.184563\frac{x\log y}{\sqrt{y}}

(4.2)

for all $x\geq y\geq 2{,}657$ . The values of relevant constants are recorded in the table below.

Table: Numerical Constants

constants	unconditional estimates		conditional estimates
$y_{0}$	229	$10^{8}$	2,657	$10^{8}$
$R(y_{0})$	.156576	.097363	.047992	.001351
$C_{0}(y_{0})$	2.156096	1.171019	.317985	.120362
$C_{1}(y_{0})$	2.534430	1.279593	.571800	.228936
$C_{2}(y_{0})$	2.548436	1.279593	.575723	.228940
$C_{3}(y_{0})$	3.110976	1.362717	.579718	.228971
$C_{4}(y_{0})$	2.548436	1.279593	.575723	.228940
$C_{5}(y_{0})$	3.131827	1.362717	.583750	.228975
$C_{6}(y_{0})$	2.534430	1.279593	.571800	.228936
$C_{7}(y_{0})$	3.110976	1.362717	.579718	.228971
$C_{8}(y_{0})$	16.982691	9.079975	4.638553	2.967998
$\eta_{1}(y_{0})$	6.236726	3.628074	2.697198	2.229726
$\sum_{k=2}^{\infty}\xi_{k}(y_{0})$	10.745960	4.388310	1.941356	.737355

To complete the proof of the second half of Theorem 1.1, it remains to deal with the case $11\leq y\leq 2{,}657$ . For simplicity of notation we set

D(x,y)\colonequals\Phi(x,y)-\mu_{y}(u)e^{\gamma}x\log y\prod_{p\leq y}\left(1-\frac{1}{p}\right).

Using Mathematica we find that

	$\displaystyle M$	$\displaystyle\colonequals\max\limits_{11\leq z\leq 2{,}657}\frac{\operatorname{li}(z)-\pi(z)}{\sqrt{z}\log z}<0.259141,$
	$\displaystyle m$	$\displaystyle\colonequals\min\limits_{11\leq z\leq 2{,}657}e^{\gamma}\log z\prod_{p\leq z}\left(1-\frac{1}{p}\right)>0.876248.$

If $\sqrt{x}\leq y<x$ , then

\Phi(x,y)=\mu_{y}(u)x+(\pi(x)-\operatorname{li}(x))-(\pi(y)-\operatorname{li}(y))+1.

Note that $x\leq y^{2}<10^{8}$ . Since $\pi(z)<\operatorname{li}(z)$ for $2\leq z\leq 10^{8}$ by [RS, Theorem 16] and

\prod_{p\leq z}\left(1-\frac{1}{p}\right)<\frac{e^{-\gamma}}{\log z}

for $0<z\leq 10^{8}$ by [RS, Theorem 23], we have

	$\displaystyle\|D(x,y)\|$	$\displaystyle<(1-m)\left(1-y^{-1}\right)\frac{x}{\log y}+M\sqrt{x}\log x+1$
		$\displaystyle\leq\left((1-m)\left(1-y^{-1}\right)+M\frac{\log^{2}y}{\sqrt{y}}+\frac{\log y}{y}\right)\frac{x}{\log y},$		(4.3)

where we have used the fact that $\log x/\sqrt{x}$ is strictly decreasing on $[e^{2},\infty)$ . Numerical computation shows that the right side of (4.3) is $<0.449774x\log y/\sqrt{y}$ for $11\leq y\leq 2{,}657$ . Suppose now that $11\leq y\leq\sqrt{x}$ . By [FP, Theorem 1], Theorem 2.3 and [RS, Theorem 23] we have, for $11\leq y\leq 2{,}657$ ,

	$\displaystyle D(x,y)$	$\displaystyle\leq\left(0.6-\frac{m}{2}\left(1-y^{-1}\right)\right)\frac{x}{\log y}<0.449774\frac{x\log y}{\sqrt{y}},$
	$\displaystyle D(x,y)$	$\displaystyle>(0.4-M_{0})\frac{x}{\log y}>-0.449774\frac{x\log y}{\sqrt{y}}.$

This settles the case $11\leq y\leq 2{,}657$ and completes the proof of Theorem 1.1.

The proof of Corollary 1.2 is similar, and we shall only sketch it. When $y\geq y_{0}$ , where $y_{0}=229$ for the unconditional estimate and $y_{0}=2{,}657$ for the conditional estimate, we have by the triangle inequality that

|\Phi(x,y)-\mu_{y}(u)x|<|D(x,y)|+\left|1-e^{\gamma}\log y\prod_{p\leq y}\left(1-\frac{1}{p}\right)\right|\frac{x}{\log y}.

Then we bound $|D(x,y)|$ using the values of $C_{8}(y_{0})$ listed in the table above. To estimate the second term, we use (3.6) when $y\geq 10^{8}$ and the inequality

m(y)<e^{\gamma}\log y\prod_{p\leq y}\left(1-\frac{1}{p}\right)<1

when $y_{0}\leq y\leq 10^{8}$ , where $m(y)$ is given by

m(y)\colonequals\begin{cases}~{}0.983296,&\text{\hskip 5.69054ptif~{}}229\leq y\leq 2{,}657,\\ ~{}0.996426,&\text{\hskip 5.69054ptif~{}}2{,}657\leq y<210{,}000,\\ ~{}0.999643,&\text{\hskip 5.69054ptif~{}}210{,}000\leq y\leq 10^{8},\end{cases}

according to [RS, Theorem 23] and Mathematica. This leads to the asserted bounds for $y\geq y_{0}$ . Suppose now that $y\leq y_{0}$ . In this case, the proof of the unconditional bound is exactly the same as that of the unconditional bound in Theorem 1.1. As for the conditional bound, we argue in the same way as in the proof of Theorem 1.1 to get

|\Phi(x,y)-\mu_{y}(u)x|\leq\left(M\frac{\log^{2}y}{\sqrt{y}}+\frac{\log y}{y}\right)\frac{x}{\log y}

when $\sqrt{x}\leq y<x$ and

	$\displaystyle\|\Phi(x,y)-\mu_{y}(u)x\|$	$\displaystyle\leq\left(0.6-\frac{1}{2}\left(1-y^{-1}\right)\right)\frac{x}{\log y},$
	$\displaystyle\|\Phi(x,y)-\mu_{y}(u)x\|$	$\displaystyle>(0.4-M_{0})\frac{x}{\log y},$

when $11\leq y\leq\sqrt{x}$ . Together, these inequalities yield the asserted conditional bound.

Remark 4.1.

The bounds in Theorem 1.1 and its corrollary may be improved. For example, the numerical values of the sum $\sum_{k=2}^{\infty}\xi_{k}(y_{0})$ may be reduced by keeping $\rho$ (or even $|\omega^{\prime}|$ ) in all of the relevant integrals, but of course the computational complexity is expected to increase as a cost. In addition, our method would allow an extension of the range $x\geq y\geq 11$ in the second half of Theorem 1.1 to the entire range $x\geq y\geq 2$ if we argue with $y_{0}=2{,}657$ replaced by some smaller value and enlarge the constant 0.449774.

Acknowledgment. The author would like to thank his advisor C. Pomerance for his helpful comments and suggestions.

Numerically Explicit Estimates for the Distribution of Rough Numbers

Abstract.

Key words and phrases:

2020 Mathematics Subject Classification:

1. Introduction

Theorem 1.1.

Corollary 1.2.

2. Lower Bounds for Φ​(x,y)\Phi(x,y)

Proposition 2.1.

Proof.

Lemma 2.2.

Proof.

Theorem 2.3.

Proof.

Remark 2.1.

3. An Explicit Version of de Bruijn’s Estimate

4. Deduction of Theorem 1.1 and Corollary 1.2

Remark 4.1.

References

2. Lower Bounds for $\Phi(x,y)$