# The error term in counting prime pairs

Leon Chou  Department of Mathematics, University of California, Berkeley, 970
Evans Hall, Berkeley, California 94720, United States of America
<EMAIL_ADDRESS>,  Summer Haag  Department of Mathematics, University
of Colorado Boulder, 2300 Colorado Avenue, Boulder, Colorado 80309, United
States of America<EMAIL_ADDRESS>,  Jake Huryn  Department of
Mathematics, The Ohio State University, 100 Math Tower, 231 West 18th Avenue,
Columbus, Ohio 43210, United States of America<EMAIL_ADDRESS>and  Andrew
Ledoan  Department of Mathematical Sciences, Northern Illinois University, 320
Watson Hall, 1425 West Lincoln Highway, DeKalb, Illinois 60115, United States
of America<EMAIL_ADDRESS>
###### Abstract.

We relate the size of the error term in the Hardy–Littlewood conjectured
formula for the number of prime pairs to the $L^{1}$ norm of an exponential
sum over the primes formed with the von Mangoldt function.

###### Key words and phrases:

Hardy–Littlewood prime $k$-tuple conjecture; primes; exponential sum; singular
series

###### 2020 Mathematics Subject Classification:

Primary 11N05; Secondary 11P32, 11N36

##  1\. Introduction

Let $S(\alpha)$ denote the exponential sum

$S(\alpha)=\sum_{n\leq N}\Lambda(n)e(n\alpha),$

where $e(u)=e^{2\pi iu}$ and $\Lambda(n)$ is von Mangoldt’s function defined
by

$\Lambda(n)=\left\\{\begin{array}[]{ll}\log p&\mbox{if $n=p^{m}$ for some
prime $p$ and integer $m\geq 1$,}\\\ 0&\mbox{otherwise.}\end{array}\right.$

The usual generating function for prime pairs is

$\lvert S(\alpha)\rvert^{2}=\sum_{\lvert k\rvert\leq
N}\psi_{2}(N,k)e(k\alpha),$ (1)

where

$\psi_{2}(N,k)=\sum_{\begin{subarray}{c}n,n^{\prime}\leq N\\\
n^{\prime}-n=k\end{subarray}}\Lambda(n)\Lambda(n^{\prime}).$ (2)

If $k$ is zero, then the prime number theorem gives

$\psi_{2}(N,0)=\sum_{n\leq N}\Lambda(n)^{2}=N\log N+O(N)\hskip
10.00002pt\textrm{as $N\to\infty$.}$ (3)

If $k$ is nonzero and even, then the Hardy–Littlewood conjecture [9] for prime
pairs states that

$\psi_{2}(N,k)\sim\mathfrak{S}(k)(N-\lvert k\rvert)\hskip 10.00002pt\textrm{as
$N\to\infty$,}$

where

$\mathfrak{S}(k)=\left\\{\begin{array}[]{ll}{\displaystyle
2C_{2}\prod_{\begin{subarray}{c}p\mid k\\\
p>2\end{subarray}}\left(\frac{p-1}{p-2}\right)}&\mbox{if $k$ is nonzero and
even,}\\\ 0&\mbox{if $k$ is odd,}\end{array}\right.$

and

$C_{2}=\prod_{p>2}\left(1-\frac{1}{(p-1)^{2}}\right)=0.66016\ldots.$

If $k$ is odd, then $\psi_{2}(N,k)$ has nonzero terms only when $n$ or
$n^{\prime}$ is a power of two. Hence,

$\psi_{2}(N,k)\ll(\log N)^{2}.$

From the recent works of Zhang [21], Maynard [13, 14], and Tao [18, 19] we now
know that there exist infinitely many integers $k$ and a constant $c_{1}$ for
which

$\psi_{2}(N,k)\gg\frac{N}{(\log N)^{c_{1}}},$

but the method used is unlikely to lead to asymptotic formulas or to prove
this result for all even $k$. In this paper, we are interested with the size
of the error in the Hardy–Littlewood conjectured formula for $\psi_{2}(N,k)$.
We show, in particular, that it cannot always be small. This problem has
previously been dealt with by Montgomery and Vaughan [16] for sums of primes.
We will see that their method is less successful here. In the first place, for
a given fixed even value of $k$, nothing that is nontrivial is currently
known. Heath-Brown [10] proposed the following problem for twin primes at an
Oberwolfach meeting in 1994:

> If $p_{n}$ denotes the $n$th twin prime, then  $p_{n+1}-p_{n}=\Omega((\log
> p_{n})^{2}).$ Improve on this result. Hence, or otherwise, improve on the
> trivial relation  $\pi_{2}(x)-c_{2}\int_{2}^{x}\frac{dt}{(\log
> t)^{2}}=\Omega(1),$ where $c_{2}$ is a positive constant.

This is equivalent to proving

$\psi_{2}(N,2)\neq\mathfrak{S}(2)N+O((\log N)^{2}),$

but there has been no progress on this problem. Thus, instead of individual
$k$, we will consider the total error, or variance, over all $k$ and define

$E(N)=\sum_{1\leq\lvert k\rvert\leq N}(\psi_{2}(N,k)-\mathfrak{S}(k)(N-\lvert
k\rvert))^{2}.$

$\begin{array}[]{c}\includegraphics[width=241.84842pt]{plot1.png}\end{array}$

Figure 1. $\psi_{2}(1000,k)$ (black dot) and $\mathfrak{S}(k)(1000-k)$ (gray
square) for $1\leq k\leq 100$.

$\begin{array}[]{c}\includegraphics[width=241.84842pt]{plot2.png}\end{array}$

Figure 2. $\psi_{2}(1000,k)$ and $\mathfrak{S}(k)(1000-k)$ for $900\leq k\leq
1000$.

Figures 1 and 2 give a clear indication of the close agreement between
$\psi_{2}(N,k)$ and $\mathfrak{S}(k)(N-\lvert k\rvert)$. Based on this and
more extensive numerical evidence supplied by Funkhouser et al. [2], we
propose the following conjecture concerning $E(N)$.

######  Prime Pair Error Conjecture.

There exists a positive constant $c_{3}$ such that

$E(N)\sim c_{3}N^{2}(\log N)^{2}\hskip 10.00002pt\textrm{as $N\to\infty$.}$

We show in Theorem 1 that the Prime Pair Error Conjecture implies the Riemann
Hypothesis. Obtaining upper bounds for $E(N)$ is a very hard problem. Assuming
the Generalized Riemann Hypothesis for Dirichlet’s $L$-functions, we have the
bound

$E(N)\ll N^{5/2}(\log N)^{c_{4}}$

for some positive constant $c_{4}$. This falls short of the bound in the Prime
Pair Error Conjecture by a factor of $N^{1/2}$.

Our main results connect $E(N)$ to the $L^{1}$ norm of $S(\alpha)$. The latter
was first studied by Vaughan [20]. He proved the lower bound

$\int_{0}^{1}\lvert S(\alpha)\rvert\,d\alpha\gg N^{1/2}.$ (4)

It is easy to obtain an upper bound not much larger than this. By the
Cauchy–Schwarz inequality and Parseval’s equation, together with (1) through
(3),

$\displaystyle\int_{0}^{1}\lvert S(\alpha)\rvert\,d\alpha$
$\displaystyle\leq\left(\int_{0}^{1}\lvert
S(\alpha)\rvert^{2}\,d\alpha\right)^{1/2}=\psi_{2}(N,0)^{1/2}$
$\displaystyle\leq(1+o(1))N^{1/2}(\log N)^{1/2}\hskip 10.00002pt\textrm{as
$N\to\infty$.}$

This upper bound has been improved slightly by Goldston [6], who proved, for
any positive $\epsilon$ and $N\geq N_{0}(\epsilon)$,

$\int_{0}^{1}\lvert
S(\alpha)\rvert\,d\alpha\leq\left[\left(\frac{1}{2}+\epsilon\right)N\log
N\right]^{1/2}.$ (5)

Equations (4) and (5) are respectively the current best upper and lower bounds
known for the $L^{1}$ norm of $S(\alpha)$. As we will see below, it appears to
be a difficult problem to improve on either of these bounds. “It seems quite
likely,” Vaughan [20] wrote, that there exists a positive constant $c_{5}$
such that the $L^{1}$ norm of $S(\alpha)$ is asymptotically $c_{5}(N\log
N)^{1/2}$ as $N\to\infty$, “but if true this must lie very deep.” This problem
is still unsolved. We will formulate it as the following conjecture.

######  Vaughan’s Conjecture.

There exists a positive constant $c_{6}$ satisfying $0<c_{6}\leq 1/\sqrt{2}$
such that

$\int_{0}^{1}\lvert S(\alpha)\rvert\,d\alpha\sim c_{6}N^{1/2}(\log
N)^{1/2}\hskip 10.00002pt\textrm{as $N\to\infty$.}$

We emphasize that both the Prime Pair Error Conjecture and Vaughan’s
Conjecture are extremely difficult problems and there is little hope at
present for making progress on proving them. What seems like a much easier
problem is the following conjecture.

######  Prime Pair Error Lower Bound Conjecture.

We have

$E(N)\gg N^{2}(\log N)^{2}.$

$\begin{array}[]{c}\includegraphics[width=241.84842pt]{plot3.png}\end{array}$

Figure 3. $E(N)/(N^{2}(\log N)^{2})$ for $1\leq N\leq 1000$. $N$ | $E(N)/(N^{2}(\log N)^{2})$
---|---
$1\times 10^{3}$ | $0.09464$
$1\times 10^{4}$ | $0.12327$
$2\times 10^{4}$ | $0.13061$
$3\times 10^{4}$ | $0.14507$
$4\times 10^{4}$ | $0.15081$
$5\times 10^{4}$ | $0.15480$
$6\times 10^{4}$ | $0.16124$
$7\times 10^{4}$ | $0.17745$
$8\times 10^{4}$ | $0.15953$
$9\times 10^{4}$ | $0.16192$
$1\times 10^{5}$ | $0.16857$
Table 1. Values of $E(N)/(N^{2}(\log N)^{2})$ truncated to five decimal
places.

Figure 3 and Table 1 provide some initial numerical evidence in support of the
Prime Pair Error Lower Bound Conjecture. We have not been able to prove this
conjecture, but there does not appear to be any barrier to doing so. This is
somewhat surprising, because Montgomery and Vaughan [16] proved essentially
this conjecture for the corresponding problem for the error in the Goldbach
problem. Moreover, their method which we partly employ here almost works. Many
of our results hold when we replace Vaughan’s $L^{1}$ Conjecture with the
following lower bound conjecture.

######  Vaughan Lower Bound Conjecture.

We have

$\int_{0}^{1}\lvert S(\alpha)\rvert\,d\alpha\gg N^{1/2}(\log N)^{1/2}\hskip
10.00002pt\textrm{as $N\to\infty$.}$

In Section 2, we will state all of our results in detail. Our first theorem
collects together a number of simple results from Corollaries 1, 2, and 3.

######  Theorem 1.

(i) The Prime Pair Error Conjecture implies Vaughan’s Lower Bound Conjecture.
(ii) If Vaughan’s Conjecture holds with $c_{7}<1/\sqrt{2}$, then the Prime
Pair Error Lower Bound Conjecture is true. (iii) Moreover, if

$E(N)=o(N^{2}(\log N)^{2}\hskip 10.00002pt\textrm{as $N\to\infty$,}$

so that the Prime Pair Error Lower Bound Conjecture is false for all
sufficiently large $N$, then Vaughan’s Conjecture is true with
$c_{7}=1/\sqrt{2}$.

##  2\. Statement of results

It is not hard to relate $E(N)$ to the classical results on the error in the
prime number theorem. This follows Montgomery and Vaughan’s work [16], but
here it is particularly simple. We easily obtain the following result.

######  Theorem 2.

We have

$E(N)=\Omega(N^{2}(\log\log\log N)^{2}).$

If $\Theta$ is the supremum of the real parts of the zeros of Riemann’s zeta
function, then

$E(N)=\Omega(N^{1+2\Theta-\epsilon})$

for every positive $\epsilon$.

If the Riemann Hypothesis is false, then the latter estimate in Theorem 2 is
stronger than the former estimate. Moreover, we see that the Prime Pair Error
Conjecture implies that $\Theta=1/2$ and thus the Riemann Hypothesis. Theorem
2 uses the variability in the number of primes up to $N$ to directly generate
the growth in $E(N)$. However, we expect that the growth of $E(N)$ is more
strongly influenced by the additive generating function for prime pairs.

######  Theorem 3.

If, for sufficiently large $N$ and a computable large positive constant
$c_{8}$,

$\left(\int_{0}^{1}\lvert
S(\alpha)\rvert\,d\alpha\right)^{2}\leq\frac{1}{2}N\log N-c_{8}N,$ (6)

then

$E(N)\gg\left[\frac{1}{2}N\log N-\left(\int_{0}^{1}\lvert
S(\alpha)\rvert\,d\alpha\right)^{2}\right]^{2}.$ (7)

In particular, (6) implies

$E(N)\gg N^{2}.$

Unlike the lower bound in Theorem 2, Theorem 3 shows that, subject to (6),
$E(N)\gg N^{2}$ is never small. We immediately obtain the following corollary.

######  Corollary 1.

If, for sufficiently large $N$ and a fixed constant $\delta$ satisfying
$0<\delta<1/2$,

$\left(\int_{0}^{1}\lvert
S(\alpha)\rvert\,d\alpha\right)^{2}\leq\left(\frac{1}{2}-\delta\right)N\log
N,$

then

$E(N)\gg\delta^{2}N^{2}(\log N)^{2}.$

Hence, if the Vaughan $L^{1}$ Conjecture holds for any constant
$c_{5}<1/\sqrt{2}$, then the Prime Pair Error Lower Bound Conjecture holds.

######  Corollary 2.

If

$E(N)=o(N^{2}(\log N)^{2}),$

then

$\int_{0}^{1}\lvert S(\alpha)\rvert\,d\alpha\sim\frac{1}{\sqrt{2}}N^{1/2}(\log
N)^{1/2}\hskip 10.00002pt\textrm{as $N\to\infty$.}$

To see this, we first note that, by Theorem 3, this follows from (7) subject
to (6). On the other hand, if (6) is false for some set of values of $N$, then
by Theorem 5 below, for those values of $N$,

$\left(\frac{1}{2}N\log N-c_{9}N\right)^{1/2}\leq\int_{0}^{1}\lvert
S(\alpha)\rvert\,d\alpha\leq\left(\frac{1}{2}N\log
N\right)^{1/2}+c_{10}\left(\frac{N}{\log N}\right)^{1/2}\log\log N.$

The conclusion of Corollary 2 follows independent of the hypothesis that

$E(N)=o(N^{2}(\log N)^{2}.$

Our second main result shows that $E(N)$ and the $L^{1}$ norm of $S(\alpha)$
cannot both be too small.

######  Theorem 4.

We have

$(E(N)+N^{2}(\log N)^{2})\left(\int_{0}^{1}\lvert
S(\alpha)\rvert\,d\alpha+O(N^{1/2}(\log\log N)^{1/2})\right)^{2}\gg N^{3}(\log
N)^{3}.$

The following corollaries are immediate.

######  Corollary 3.

If

$E(N)\ll N^{2}(\log N)^{c_{11}}$

for some positive constant $c_{11}$ satisfying $2\leq c_{11}<3$, then

$\int_{0}^{1}\lvert S(\alpha)\rvert\,d\alpha\gg N^{2}(\log N)^{(3-c_{11})/2}.$

In particular, if $c_{11}=2$ and if

$E(N)\ll N^{2}(\log N)^{2},$

then

$\int_{0}^{1}\lvert S(\alpha)\rvert\,d\alpha\gg N^{1/2}(\log N)^{1/2}.$

Consequently, the Prime Pair Error Conjecture implies Vaughan’s Lower Bound
Conjecture.

######  Corollary 4.

If

$\int_{0}^{1}\lvert S(\alpha)\rvert\,d\alpha\ll N^{1/2}(\log N)^{c_{12}}$

for some positive constant $c_{12}$ satisfying $0<c_{12}\leq 1/2$, then

$E(N)\gg N^{2}(\log N)^{3-2c_{12}}.$

######  Corollary 5.

If

$\int_{0}^{1}\lvert S(\alpha)\rvert\,d\alpha\ll N^{1/2}(\log\log N)^{1/2},$

then

$E(N)\gg\frac{N^{2}(\log N)^{3}}{\log\log N}.$

Many of our results depend on improving the upper bound in (5). Turning to
this question, we prove the following result.

######  Theorem 5.

We have

$\left(\int_{0}^{1}\lvert
S(\alpha)\rvert\,d\alpha\right)^{2}\leq\frac{1}{2}N\log N+\frac{3}{2}N\log\log
N+O(N).$

Alternatively, for some positive constant $c_{10}$ and $N$ sufficiently large,

$\int_{0}^{1}\lvert S(\alpha)\rvert\,d\alpha\leq\left(\frac{1}{2}N\log
N\right)^{1/2}+c_{10}\left(\frac{N}{\log N}\right)^{1/2}\log\log N.$

From Theorem 3, even small improvements in Theorem 5 would have important
consequences for $E(N)$.

##  3\. Proof of Theorem 2

For positive $N$, using

$\psi(N)=\sum_{n\leq N}\Lambda(n),$

we obtain

$\psi(N)^{2}=\sum_{n,n^{\prime}\leq
N}\Lambda(n)\Lambda(n^{\prime})=\sum_{\lvert k\rvert\leq
N}\sum_{\begin{subarray}{c}n,n^{\prime}\leq N\\\
n^{\prime}-n=k\end{subarray}}\Lambda(n)\Lambda(n^{\prime})=\sum_{\lvert
k\rvert\leq N}\psi_{2}(N,k).$

Let

$\psi(N)=N+R(N).$

The prime number theorem is equivalent to

$R(N)=o(N)\hskip 10.00002pt\textrm{as $N\to\infty$.}$

Since

$\psi(N)^{2}=N^{2}+2NR(N)+R(N)^{2}$

and

$\sum_{1\leq\lvert k\rvert\leq N}(N-\lvert
k\rvert)\mathfrak{S}(k)=N^{2}+O(N\log N)$

(see [3, 15]), we deduce that

$\sum_{1\leq\lvert k\rvert\leq N}(\psi_{2}(N,k)-(N-\lvert
k\rvert)\mathfrak{S}(k))=R(N)(2N+R(N))+O(N\log N).$

By the Cauchy–Schwarz inequality,

$\left\lvert\sum_{1\leq\lvert k\rvert\leq N}(\psi_{2}(N,k)-(N-\lvert
k\rvert)\mathfrak{S}(k))\right\rvert^{2}\leq 2NE(N).$

We conclude that

$E(N)\geq 2N\left[R(N)\left(1+\frac{R(N)}{2N}\right)+O(\log N)\right]^{2}.$

By Littlewood’s theorem,

$R(N)=\Omega_{\pm}(N^{1/2}\log\log\log N)$

(see [17]), and the first assertion of Theorem 2 follows. If the Riemann
Hypothesis is false, we use

$R(N)=\Omega_{\pm}(N^{\Theta-\epsilon})$

for every positive $\epsilon$ (see [12, 17]). In fact, Littlewood proved that

$\limsup_{N\to\infty}\frac{\pm R(N)}{N^{1/2}\log\log\log N}\geq\frac{1}{2},$

and so it follows that

$\limsup_{N\to\infty}\frac{E(N)}{N^{2}(\log\log\log N)^{2}}\geq\frac{1}{2}.$

##  4\. Proof of Theorem 3

Our main results depend on approximating $\lvert S(\alpha)\rvert^{2}$ by

$V_{y}(\alpha)=\sum_{q\leq
y}\frac{\mu(q)^{2}}{\phi(q)^{2}}\sum_{\begin{subarray}{c}1\leq a\leq q\\\
(a,q)=1\end{subarray}}\left\lvert
I\left(\alpha-\frac{a}{q}\right)\right\rvert^{2},$

where

$I(\alpha)=\sum_{n\leq N}e(n\alpha),$

where $\mu(q)$ is the Möbius function and $\phi(q)$ is the Euler totient
function. Here, $V_{y}(\alpha)$ is obtained by taking the diagonal terms from
squaring the absolute value of the major arc approximation $J_{Q}(\alpha)$ of
$S(\alpha)$ used by Goldston [6] and removing the restricted support for each
arc. (This is explained in Section 7. Our use of $V_{y}(\alpha)$ is
independent of these heuristics.) Obviously, $V_{y}(\alpha)$ is real-valued
and nonnegative. Using the relation

$\lvert I(\alpha)\rvert^{2}=\sum_{\lvert k\rvert\leq N}(N-\lvert
k\rvert)e(k\alpha),$

we obtain the Fourier series

$\displaystyle V_{y}(\alpha)$ $\displaystyle=\sum_{\lvert k\rvert\leq
N}(N-\lvert k\rvert)\left(\sum_{q\leq
y}\frac{\mu(q)^{2}}{\phi(q)^{2}}\sum_{\begin{subarray}{c}1\leq a\leq q\\\
(a,q)=1\end{subarray}}e\left(-\frac{ka}{q}\right)\right)e(k\alpha)$
$\displaystyle=\sum_{\lvert k\rvert\leq N}(N-\lvert
k\rvert)\mathfrak{S}_{y}(k)e(k\alpha),$

where

$\mathfrak{S}_{y}(k)=\sum_{q\leq y}\frac{\mu(q)^{2}}{\phi(q)^{2}}c_{q}(-k)$

and

$c_{q}(n)=\sum_{\begin{subarray}{c}1\leq a\leq q\\\
(a,q)=1\end{subarray}}e\left(\frac{an}{q}\right).$

The Ramanujan sum $c_{q}(n)$ is an even function in $n$ and a multiplicative
function of $q$ for a fixed value of $n$, and satisfies

$c_{q}(n)=\sum_{d\mid(q,n)}d\mu\left(\frac{q}{d}\right)=\mu\left(\frac{q}{(q,n)}\right)\frac{\phi(q)}{\displaystyle\phi\left(\frac{q}{(q,n)}\right)}$
(8)

(see [17, Theorem 4.1, (4.7)]).

It is now easy to verify that $\mathfrak{S}(k)$ in the Hardy–Littlewood
conjecture for prime pairs has the series representation

$\mathfrak{S}(k)=\sum_{q=1}^{\infty}\frac{\mu(q)^{2}}{\phi(q)^{2}}c_{q}(-k)$

for $k\neq 0$. While we do not define $\mathfrak{S}(k)$ for $k=0$, we have
from [17, 2.1.1, Ex. 17] that

$\mathfrak{S}_{y}(0)=\sum_{q\leq y}\frac{\mu(q)^{2}}{\phi(q)}=\log y+O(1).$
(9)

Let

$a_{0}(N,y)=\sum_{n\leq N}\Lambda(n)^{2}-N\mathfrak{S}_{y}(0).$ (10)

By (3) and (9), for $1\leq y\leq N$,

$a_{0}(N,y)=N\log\left(\frac{N}{y}\right)+O(N).$ (11)

Let

$V(\alpha)=\sum_{1\leq\lvert k\rvert\leq N}(N-\lvert
k\rvert)\mathfrak{S}(k)e(k\alpha).$

By Parseval’s identity,

$\displaystyle E(N)$ $\displaystyle=\int_{0}^{1}\left\lvert\lvert
S(\alpha)\rvert^{2}-V(\alpha)-\sum_{n\leq
N}\Lambda(n)^{2}\right\rvert^{2}d\alpha$
$\displaystyle=\int_{0}^{1}\lvert(\lvert
S(\alpha)\rvert^{2}-V_{y}(\alpha)-a_{0}(N,y))-(V(\alpha)-V_{y}(\alpha)+N\mathfrak{S}_{y}(0))\rvert^{2}\,d\alpha.$

Let now

$J=J(N,y)=\int_{0}^{1}\lvert\lvert
S(\alpha)\rvert^{2}-V_{y}(\alpha)-a_{0}(N,y)\rvert^{2}\,d\alpha$

and

$W=W(N,y)=\int_{0}^{1}\lvert
V(\alpha)-V_{y}(\alpha)+N\mathfrak{S}_{y}(0)\rvert^{2}\,d\alpha.$

By the Cauchy–Schwarz inequality,

$E(N)\geq J-2(JW)^{1/2}+W=(J^{1/2}-W^{1/2})^{2}.$ (12)

We obtain a lower bound for $E(N)$ from a lower bound for $J$ and an
asymptotic formula for $W$. From Parseval’s identity and [7, Theorem 2], for
$1\leq y\leq N^{1/2}$,

$\begin{split}W&=\sum_{1\leq\lvert k\rvert\leq N}(N-\lvert
k\rvert)^{2}(\mathfrak{S}(k)-\mathfrak{S}_{y}(k))^{2}\\\
&=\frac{1}{3}N^{3}H(y)-\frac{1}{4}N^{2}\left[\log\left(\frac{N}{y^{2}}\right)^{2}\right]+MN^{2}\log\left(\frac{N}{y^{2}}\right)\\\
&\hskip
10.00002pt+O(N^{2})+O\left(\frac{N^{2}}{y^{1/2}}\log(2N)\right),\end{split}$
(13)

where

$H(y)=\frac{L}{y^{2}}(1+o(1)),\hskip
10.00002ptL=\prod_{p}\left(1+\frac{2p-1}{p(p-1)^{2}}\right),$

and

$M=\frac{3}{4}-\frac{1}{2}\log(2\pi)+\frac{1}{2}\sum_{p}\frac{(p-2)\log
p}{p(p-1)^{2}}.$

Hence, for $1\leq y\leq N^{1/2}$,

$W\asymp\frac{N^{3}}{y^{2}}.$ (14)

The implied upper and lower bound constants are computable since, as mentioned
in [7], the term $O(N^{2})$ in (13) can be replaced by $c_{13}N^{2}+o(N^{2})$
for a computable constant $c_{13}$.

Montgomery and Vaughan [16] obtained a lower bound for what corresponds to $J$
for the Goldbach problem. Their argument is that, by the Cauchy–Schwarz
inequality,

$J\geq T^{2},$ (15)

where

$T=T(N,y)=\int_{0}^{1}\lvert\lvert
S(\alpha)\rvert^{2}-V_{y}(\alpha)-a_{0}(N,y)\rvert\,d\alpha.$

But, by (10),

$\displaystyle T$ $\displaystyle\geq\int_{0}^{1}(\lvert
S(\alpha)\rvert^{2}-V_{y}(\alpha)-a_{0}(N,y))\,d\alpha$
$\displaystyle=\int_{0}^{1}\lvert
S(\alpha)\rvert^{2}\,d\alpha-\int_{0}^{1}V_{y}(\alpha)\,d\alpha-a_{0}(N,y)$
$\displaystyle=\sum_{n\leq
N}\Lambda(n)^{2}-N\mathfrak{S}_{y}(0)-a_{0}(N,y)\equiv 0,$

and this argument fails. In the Goldbach problem, the term $a_{0}(N,y)=0$,
which is why it succeeds there. We now modify this argument to obtain a lower
bound for $J$ in terms of the $L^{1}$ norm of $S(\alpha)$.

######  Lemma 1.

If, for $y_{0}\leq y\leq N^{1/2}$ for some $y_{0}$ satisfying $N^{1/2}/\log
N\leq y_{0}\leq N^{1/2}$ and a computable positive constant $c_{14}$,

$\left(\int_{0}^{1}\lvert S(\alpha)\rvert\,d\alpha\right)^{2}\leq
a_{0}(N,y)-c_{14}\frac{N^{3/2}}{y},$

then

$E(N)\geq\frac{1}{4}\left[a_{0}(N,y)-\left(\int_{0}^{1}\lvert
S(\alpha)\rvert\,d\alpha\right)^{2}\right]^{2}\gg\frac{N^{3}}{y^{2}}.$

Theorem 3 is the special case of Lemma 1 where we take $y_{0}=N^{1/2}$.

###### Proof.

We utilize the inequality, for any two real numbers $a$ and $b$,

$\lvert a^{2}-b^{2}\rvert\geq(\lvert a\rvert-\lvert b\rvert)^{2}.$

This inequality holds true because of the identity

$\lvert a^{2}-b^{2}\rvert=(\lvert a\rvert-\lvert b\rvert)^{2}+2\min(\lvert
a\rvert,\lvert b\rvert)\lvert\lvert a\rvert-\lvert b\rvert\rvert,$

which in turn holds true since we can assume without loss of generality that
$a\geq b\geq 0$, in which case it is obvious. Since $V_{y}(\alpha)\geq 0$ and
$a_{0}(N,y)>0$ for $1\leq y\leq N^{1/2}$, we have

$\displaystyle T$ $\displaystyle\geq\int_{0}^{1}\lvert\lvert
S(\alpha)\rvert^{2}-a_{0}(N,y)\rvert\,d\alpha-\int_{0}^{1}V_{y}(\alpha)\,d\alpha$
$\displaystyle\geq\int_{0}^{1}\lvert\lvert S(\alpha)\rvert-
a_{0}(N,y)^{\frac{1}{2}}\rvert^{2}\,d\alpha-N\mathfrak{S}_{y}(0)$
$\displaystyle=\int_{0}^{1}\lvert
S(\alpha)\rvert^{2}\,d\alpha-2a_{0}(N,y)^{1/2}\int_{0}^{1}\lvert
S(\alpha)\rvert\,d\alpha+a_{0}(N,y)-N\mathfrak{S}_{y}(0)$
$\displaystyle=2a_{0}(N,y)^{1/2}\left(a_{0}(N,y)^{1/2}-\int_{0}^{1}\lvert
S(\alpha)\rvert\,d\alpha\right).$

Provided

$\int_{0}^{1}\lvert S(\alpha)\rvert\,d\alpha\leq a_{0}(N,y)^{1/2},$

substitution into (15) yields

$\displaystyle J$ $\displaystyle\geq T^{2}\geq
4a_{0}(N,y)\left(a_{0}(N,y)^{1/2}-\int_{0}^{1}\lvert
S(\alpha)\rvert\,d\alpha\right)^{2}$
$\displaystyle=4a_{0}(N,y)\frac{\displaystyle\left[a_{0}(N,y)-\left(\int_{0}^{1}\lvert
S(\alpha)\rvert\,d\alpha\right)^{2}\right]^{2}}{\displaystyle\left(a_{0}(N,y)^{1/2}+\int_{0}^{1}\lvert
S(\alpha)\rvert\,d\alpha\right)^{2}}$ $\displaystyle\geq
4a_{0}(N,y)\frac{\displaystyle\left[a_{0}(N,y)-\left(\int_{0}^{1}\lvert
S(\alpha)\rvert\,d\alpha\right)^{2}\right]^{2}}{[2a_{0}(N,y)^{1/2}]^{2}}$
$\displaystyle=\left[a_{0}(N,y)-\left(\int_{0}^{1}\lvert
S(\alpha)\rvert\,d\alpha\right)^{2}\right]^{2}.$

Observe that the lower bound for $J$ we have just obtained is $\leq
a_{0}(N,y)^{2}\ll N^{2}(\log N)^{2}$. In order for (12) to give a lower bound
for $E(N)$, we need $J$ to be larger than $W$ which, by (14), requires
$N^{2}(\log N)^{2}\gg N^{3}/y^{2}$ and, therefore, $y\gg N^{1/2}/\log N$.

Assuming that $J\geq 4W$, we have by (12) that

$E(N)\geq(J^{1/2}-W^{1/2})^{2}\geq\frac{1}{4}J,$

and thus we conclude

$E(N)\geq\frac{1}{4}\left[a_{0}(N,y)-\left(\int_{0}^{1}\lvert
S(\alpha)\rvert\,d\alpha\right)^{2}\right]^{2},$

provided that

$\left(\int_{0}^{1}\lvert S(\alpha)\rvert\,d\alpha\right)^{2}\leq a_{0}(N,y).$

By (13) and the comment following it, we can find a constant $c_{15}$ such
that $W\leq c_{15}N^{3}/y^{2}$, and thus the condition $J\geq 4W$ is satisfied
when

$\left[a_{0}(N,y)-\left(\int_{0}^{1}\lvert
S(\alpha)\rvert\,d\alpha\right)^{2}\right]^{2}\geq
4c_{15}\frac{N^{3}}{y^{2}},$

which is equivalent to

$\left(\int_{0}^{1}\lvert S(\alpha)\rvert\,d\alpha\right)^{2}\leq
a_{0}(N,y)-2c_{15}^{1/2}\frac{N^{3}}{y^{2}}.$

This proves Lemma 1 with $c_{14}=2c_{15}^{1/2}$. ∎

##  5\. Proof of Theorem 4

If $1<p<\infty$ and $1/p+1/q=1$, and if $f\in L^{p}$ and $g\in L^{q}$, then
$fg\in L^{1}$ and Hölder’s inequality is that

$\int\lvert fg\rvert=\|fg\|_{1}\leq\|f\|_{p}\,\|g\|_{q}.$

We apply this inequality with $f=\lvert h\rvert^{1/3}$ and $g=\lvert
h\rvert^{2/3}$, and thus

$\|h\|_{1}\leq\|\lvert h\rvert^{1/3}\|_{p}\,\|\lvert h\rvert^{2/3}\|_{q}.$

If $p=3/2$, then $q=3$, and we have

$\left(\int_{0}^{1}\lvert
h(\alpha)\rvert\,d\alpha\right)^{3}\leq\left(\int_{0}^{1}\lvert
h(\alpha)\rvert^{1/2}\,d\alpha\right)^{2}\int_{0}^{1}\lvert
h(\alpha)\rvert^{2}\,d\alpha.$

We take

$h(\alpha)=\lvert S(\alpha)\rvert^{2}-V_{y}(\alpha)$

and have

$\left(\int_{0}^{1}\lvert\lvert
S(\alpha)\rvert^{2}-V_{y}(\alpha)\rvert\,d\alpha\right)^{3}\leq\left(\int_{0}^{1}\lvert\lvert
S(\alpha)\rvert^{2}-V_{y}(\alpha)\rvert^{1/2}\,d\alpha\right)^{2}\int_{0}^{1}\lvert\lvert
S(\alpha)\rvert^{2}-V_{y}(\alpha)\rvert^{2}\,d\alpha.$

We now set $y=N^{1/2}$. Using $V_{y}(\alpha)\geq 0$, and equations (10) and
(11), we have

$\displaystyle\int_{0}^{1}\lvert\lvert
S(\alpha)\rvert^{2}-V_{y}(\alpha)\rvert\,d\alpha$
$\displaystyle\geq\int_{0}^{1}\lvert
S(\alpha)\rvert^{2}\,d\alpha-\int_{0}^{1}V_{y}(\alpha)\,d\alpha$
$\displaystyle=a_{0}(N,y)\sim\frac{1}{2}N\log N\hskip 10.00002pt\textrm{as
$N\to\infty$,}$

and therefore

$\left(\int_{0}^{1}\lvert\lvert
S(\alpha)\rvert^{2}-V_{y}(\alpha)\rvert\,d\alpha\right)^{3}\gg N^{3}(\log
N)^{3}.$

By Parseval’s identity, and making use of $W\ll N^{2}$ from (14), we have

$\displaystyle\int_{0}^{1}\lvert\lvert
S(\alpha)\rvert^{2}-V_{y}(\alpha)\rvert^{2}\,d\alpha$
$\displaystyle=\sum_{1\leq\lvert k\rvert\leq
N}(\psi_{2}(N,k)-\mathfrak{S}_{y}(k)(N-\lvert k\rvert))^{2}+a_{0}(N,y)^{2}$
$\displaystyle\ll E(N)+W+\frac{1}{4}N^{2}(\log N)^{2}(1+o(1))$
$\displaystyle\leq E(N)+N^{2}(\log N)^{2}.$

The proof will be completed once we prove

$\int_{0}^{1}\lvert\lvert
S(\alpha)\rvert^{2}-V_{y}(\alpha)\rvert^{1/2}\,d\alpha\ll\int_{0}^{1}\lvert
S(\alpha)\rvert\,d\alpha+O(N^{1/2}(\log\log N)^{1/2}).$

Using the inequality

$\lvert\lvert a\rvert\pm\lvert b\rvert\rvert^{1/2}\leq\lvert
a\rvert^{1/2}+\lvert b\rvert^{1/2},$

we obtain

$\int_{0}^{1}\lvert\lvert
S(\alpha)\rvert^{2}-V_{y}(\alpha)\rvert^{1/2}\,d\alpha\leq\int_{0}^{1}\lvert
S(\alpha)\rvert\,d\alpha+\int_{0}^{1}\lvert
V_{y}(\alpha)\rvert^{1/2}\,d\alpha,$

with $y=N^{1/2}$. It remains only to prove that

$\int_{0}^{1}\lvert V_{y}(\alpha)\rvert^{1/2}\,d\alpha\ll N^{1/2}(\log\log
N)^{1/2}.$

Let now

$w(q)=\sum_{\begin{subarray}{c}1\leq a\leq q\\\
(a,q)=1\end{subarray}}\frac{\mu(q)^{2}}{\phi(q)^{2}}\left\lvert
I\left(\alpha-\frac{a}{q}\right)\right\rvert^{2}.$

We have

$\int_{0}^{1}\lvert
V_{y}(\alpha)\rvert^{1/2}\,d\alpha=\int_{0}^{1}\left\lvert\sum_{1\leq q\leq
y_{0}}w(q)+\sum_{y_{0}<q\leq y}w(q)\right\rvert^{1/2}d\alpha\leq I_{1}+I_{2},$

where

$I_{1}=\int_{0}^{1}\left\lvert\sum_{1\leq q\leq
y_{0}}w(q)\right\rvert^{1/2}d\alpha$

and

$I_{2}=\int_{0}^{1}\left\lvert\sum_{y_{0}<q\leq
y}w(q)\right\rvert^{1/2}d\alpha.$

The integral $I_{1}$ is treated using the inequality

$\left(\sum_{n=1}^{\infty}\lvert
a_{n}\rvert\right)^{1/2}\leq\sum_{n=1}^{\infty}\lvert a_{n}\rvert^{1/2},$

which holds true since, on squaring, the left-hand side is just the diagonal
terms of the right-hand side. We obtain

$\displaystyle I_{1}$ $\displaystyle\leq\int_{0}^{1}\sum_{1\leq q\leq
y_{0}}\sum_{\begin{subarray}{c}1\leq a\leq q\\\
(a,q)=1\end{subarray}}\left\lvert\frac{\mu(q)^{2}}{\phi(q)^{2}}\left\lvert
I\left(\alpha-\frac{a}{q}\right)\right\rvert^{2}\right\rvert^{1/2}d\alpha$
$\displaystyle=\int_{0}^{1}\sum_{1\leq q\leq
y_{0}}\sum_{\begin{subarray}{c}1\leq a\leq q\\\
(a,q)=1\end{subarray}}\frac{\mu(q)^{2}}{\phi(q)}\left\lvert
I\left(\alpha-\frac{a}{q}\right)\right\rvert d\alpha$
$\displaystyle=\sum_{1\leq q\leq y_{0}}\mu(q)^{2}\int_{0}^{1}\lvert
I(\alpha)\rvert\,d\alpha\ll y_{0}\log N,$

by the $L^{1}$ estimate for the Dirichlet kernel. Thus, if $y_{0}=N^{1/2}/\log
N$,

$I_{1}\ll N^{1/2}.$ (17)

The integral $I_{2}$ is treated using the Cauchy–Schwarz inequality and
Parseval’s identity. We have, using (9) in the last step,

$\displaystyle I_{2}$ $\displaystyle\leq\left(\int_{0}^{1}\sum_{y_{0}<q\leq
y}\sum_{\begin{subarray}{c}1\leq a\leq q\\\
(a,q)=1\end{subarray}}\frac{\mu(q)^{2}}{\phi(q)^{2}}\left\lvert
I\left(\alpha-\frac{a}{q}\right)\right\rvert^{2}d\alpha\right)^{1/2}$
$\displaystyle=\left(\sum_{y_{0}<q\leq y}\sum_{\begin{subarray}{c}1\leq a\leq
q\\\
(a,q)=1\end{subarray}}\frac{\mu(q)^{2}}{\phi(q)^{2}}\int_{0}^{1}\left\lvert
I\left(\alpha-\frac{a}{q}\right)\right\rvert^{2}d\alpha\right)^{1/2}$
$\displaystyle=\left(N\sum_{y_{0}<q\leq
y}\frac{\mu(q)^{2}}{\phi(q)}\right)^{1/2}=\left[N\left(\log\left(\frac{y}{y_{0}}\right)+O(1)\right)\right]^{1/2}$
$\displaystyle\ll N^{1/2}(\log\log N)^{1/2}.$ (18)

Substitution of (17) and (18) in (5) yields Theorem 4.

##  6\. First Proof of Theorem 5

If

$n=\prod_{k=1}^{r}p_{k}^{a_{k}},$

then

$\log n=\sum_{k=1}^{r}a_{k}\log p_{k}=\sum_{d\mid n}\Lambda(d),$

and so by Möbius inversion

$\Lambda(n)=\sum_{d\mid n}\mu(d)\log(n/d).$

Following [4], we approximate $\Lambda(n)$ by

$\Lambda_{R}(n)=\sum_{\begin{subarray}{c}d\mid n\\\ d\leq
R\end{subarray}}\mu(d)\log\left(\frac{R}{d}\right),$

which truncates the number of divisors $d$ of $n$ and smooths the sum. If
$1<n\leq R$, then the condition $d\leq R$ can be dropped, so that
$\Lambda_{R}(n)=\Lambda(n)$. Therefore, we can interpret $\Lambda_{R}(n)$ as
an approximation of $\Lambda(n)$ which gets better as $R$ increases up to $n$
where it becomes $\Lambda(n)$.

We thus introduce the approximation for $S(\alpha)$,

$S_{R}(\alpha)=\sum_{n\leq N}\Lambda_{R}(n)e(n\alpha).$

We make use of the formula, for $1\leq R\leq N$,

$\sum_{n\leq N}\Lambda_{R}(n)^{2}=N\log R+O(N)$

due to Graham [8]. In the range $1\leq R\leq N^{1/2}/\log N$, this can be
obtained by an elementary calculation. By the prime number theorem, we also
obtain, for $1\leq R\leq N$ and any positive constant $c_{13}$,

$\sum_{n\leq N}\Lambda_{R}(n)\Lambda(n)=N\log R+O(R)+O(N(\log N)^{-c_{13}})$

(see [4]). By Parseval’s identity, we have by (3) and the above estimates, for
$1\leq R\leq N$,

$\displaystyle\int_{0}^{1}\lvert S(\alpha)-S_{R}(\alpha)\rvert^{2}\,d\alpha$
$\displaystyle=\int_{0}^{1}\left\lvert\sum_{n\leq
N}(\Lambda(n)-\Lambda_{R}(n))e(n\alpha)\right\rvert^{2}\,d\alpha$
$\displaystyle=\sum_{n\leq N}(\Lambda(n)-\Lambda_{R}(n))^{2}$
$\displaystyle=\sum_{n\leq N}\Lambda(n)^{2}-2\sum_{n\leq
N}\Lambda(n)\Lambda_{R}(n)+\sum_{n\leq N}\Lambda_{R}(n)^{2}$
$\displaystyle=N\log N-2N\log R+N\log R+O(N)$
$\displaystyle=N\log\left(\frac{N}{R}\right)+O(N).$

Hence, by the Cauchy–Schwarz inequality, for $1\leq R\leq N$,

$\displaystyle\int_{0}^{1}\lvert S(\alpha)-S_{R}(\alpha)\rvert\,d\alpha$
$\displaystyle\leq\left(\int_{0}^{1}\lvert
S(\alpha)-S_{R}(\alpha)\rvert^{2}\,d\alpha\right)^{1/2}$
$\displaystyle\leq\left(N\log\left(\frac{N}{R}\right)+O(N)\right)^{1/2}.$

By the reverse triangle inequality,

$\displaystyle\left\lvert\int_{0}^{1}\lvert
S(\alpha)\rvert\,d\alpha-\int_{0}^{1}\lvert
S_{R}(\alpha)\rvert\,d\alpha\right\rvert$ $\displaystyle\leq\int_{0}^{1}\lvert
S(\alpha)-S_{R}(\alpha)\rvert\,d\alpha$
$\displaystyle\leq\left(N\log\left(\frac{N}{R}\right)+O(N)\right)^{1/2}.$

We now make use of the following result that we will prove below.

######  Lemma 2.

We have

$\int_{0}^{1}\lvert S_{R}(\alpha)\rvert\,d\alpha\ll R\log N.$

Thus, we conclude

$\int_{0}^{1}\lvert
S(\alpha)\rvert\,d\alpha\leq\left(N\log\left(\frac{N}{R}\right)+O(N)\right)^{1/2}+O(R\log
N).$

We take $R=N^{1/2}/(\log N)^{3/2}$ and on squaring obtain

$\left(\int_{0}^{1}\lvert
S(\alpha)\rvert\,d\alpha\right)^{2}\leq\frac{1}{2}N\log N+\frac{3}{2}N\log\log
N+O(N),$

from which

$\int_{0}^{1}\lvert S(\alpha)\rvert\,d\alpha\leq\left(\frac{1}{2}N\log
N\right)^{1/2}+c_{10}\left(\frac{N}{\log N}\right)^{1/2}\log\log N$

for a positive constant $c_{10}$. This proves Theorem 5.

It remains to prove Lemma 2. We use

$\sum_{n\leq N}e(n\alpha)\ll\min\left(N,\frac{1}{\|\alpha\|}\right)$

to obtain

$\displaystyle S_{R}(\alpha)$ $\displaystyle=\sum_{n\leq
N}\sum_{\begin{subarray}{c}d\mid n\\\ d\leq
R\end{subarray}}\mu(d)\log\left(\frac{R}{d}\right)e(n\alpha)$
$\displaystyle=\sum_{d\leq R}\mu(d)\log\left(\frac{R}{d}\right)\sum_{m\leq
N/d}e(dm\alpha)$ $\displaystyle\ll\sum_{d\leq
R}\log\left(\frac{R}{d}\right)\min\left(\frac{N}{d},\frac{1}{\|d\alpha\|}\right).$

Thus,

$\int_{0}^{1}\lvert S_{R}(\alpha)\rvert\,d\alpha\ll\sum_{d\leq
R}\log\left(\frac{R}{d}\right)\int_{0}^{1}\min\left(\frac{N}{d},\frac{1}{\|d\alpha\|}\right)\,d\alpha.$

As a function of $\alpha$, the integrand on the right-hand side is a function
with period $1/d$. Thus,

$\displaystyle\int_{0}^{1}\min\left(\frac{N}{d},\frac{1}{\|d\alpha\|}\right)d\alpha$
$\displaystyle\ll
d\int_{0}^{1/d}\min\left(\frac{N}{d},\frac{1}{\|d\alpha\|}\right)d\alpha$
$\displaystyle\ll
d\left(\int_{0}^{1/N}\frac{N}{d}\,d\alpha+\int_{1/N}^{1/(2d)}\frac{1}{d\alpha}\,d\alpha\right)$
$\displaystyle\ll 1+\log\left(\frac{N}{2d}\right)\ll\log N.$

Hence,

$\int_{0}^{1}\lvert S_{R}(\alpha)\rvert\,d\alpha\ll\sum_{d\leq
R}\log\left(\frac{R}{d}\right)\log N\ll R\log N.$

This completes the proof of Lemma 2.

The idea here is that $S_{R}(\alpha)$ contributes $N\log R$ in the $L^{2}$
norm but nothing in the $L^{1}$ norm until $R$ is about size $N^{1/2}$. Thus,
for example, with $R=N^{1/2}/\log N$, we see $S_{R}(\alpha)$ removes half of
the $L^{2}$ norm of $S(\alpha)$, but leaves the $L^{1}$ norm of $S(\alpha)$
unchanged. Hence, the Cauchy–Schwarz inequality gives an improved upper bound
for the $L^{1}$ norm of $S(\alpha)$.

##  7\. Second Proof of Theorem 5

We now give another proof of Theorem 5 in which we use part of the circle
method to analyze $S(\alpha)$ and directly pull out the major arc
contribution. This provides a slightly better approximation for $S(\alpha)$
than $S_{R}(\alpha)$, although the result obtained is asymptotically the same.

Let $\alpha=a/q+\beta$, where $a/q$ is a fraction with $1\leq a\leq q$ and
$(a,q)=1$. Suppose $\beta$ is small. Then

$\displaystyle S\left(\frac{a}{q}+\beta\right)$ $\displaystyle=\sum_{n\leq
N}\Lambda(n)e\left(\frac{na}{q}\right)e(n\beta)$ $\displaystyle=\sum_{1\leq
b\leq q}e\left(\frac{ba}{q}\right)\sum_{\begin{subarray}{c}n\leq N\\\ n\equiv
b{\@displayfalse\pmod{q}}\end{subarray}}\Lambda(n)e(n\beta).$

If $\beta=0$, by the prime number theorem for arithmetic progressions [1,
Chapter 20, equation (11)], the inner sum is

$\sum_{\begin{subarray}{c}n\leq N\\\ n\equiv
b{\@displayfalse\pmod{q}}\end{subarray}}\Lambda(n)=\psi(N;q,b)=\operatorname{\boldsymbol{1}}_{(b,q)=1}\frac{N}{\phi(q)}+O[N\exp(-c_{11}(\log
N)^{1/2})].$

When $\beta$ is small, by partial summation,

$\sum_{\begin{subarray}{c}n\leq N\\\ n\equiv
b{\@displayfalse\pmod{q}}\end{subarray}}\Lambda(n)e(n\beta)=\operatorname{\boldsymbol{1}}_{(b,q)=1}\frac{I(\beta)}{\phi(q)}+\textrm{error}.$

Thus, we expect

$S\left(\frac{a}{q}+\beta\right)=c_{q}(a)\frac{I(\beta)}{\phi(q)}+\textrm{error}=\frac{\mu(q)}{\phi(q)}I(\beta)+\textrm{error},$

since $c_{q}(a)=\mu(q)$ when $(a,q)=1$. Hence, we obtain a spike function in
each interval which is largest when $q$ is small. In the circle method we
consider the set of all fractions $a/q$ of order $Q$ with $1\leq q\leq Q$,
$1\leq a\leq q$, and $(a,q)=1$ and divide the unit interval $[0,1]$ into
nonoverlapping intervals around each fraction $a/q$. Thus, with
$\beta=\alpha-a/q$ in [6],

$J_{Q}(\alpha)=\sum_{q\leq Q}\sum_{\begin{subarray}{c}1\leq a\leq q\\\
(a,q)=1\end{subarray}}\frac{\mu(q)}{\phi(q)}I\left(\alpha-\frac{a}{q}\right)\chi_{Q}(\alpha;q,a),$

where

$\chi_{Q}(\alpha;q,a)=\left\\{\begin{array}[]{ll}1&\mbox{if
$\displaystyle\alpha\in\left(\frac{a}{q}-\frac{1}{q(Q+u)},\frac{a}{q}+\frac{1}{q(Q+v)}\right],\
\ 1\leq u,v\leq q$,}\\\ 0&\mbox{otherwise.}\end{array}\right.$

We find $J_{Q}(\alpha)$ becomes a better and better approximation for
$S(\alpha)$ in $L^{2}$ as $Q$ increases up to size $N^{1/2}$ but then becomes
a worse approximation as the requirement that the intervals
$\chi_{Q}(\alpha;q,a)$ not overlap cuts off the full contribution of each
spike function approximation.

We thus define a new approximation by dropping $\chi_{Q}(\alpha;q,a)$ and
letting each spike $I(\alpha-a/q)$ cut off its contribution just by being
small when $\beta$ is not small. Thus, we define, replacing $Q$ by $R$,

$\mathscr{S}_{R}(\alpha)=\sum_{q\leq R}\sum_{\begin{subarray}{c}1\leq a\leq
q\\\
(a,q)=1\end{subarray}}\frac{\mu(q)}{\phi(q)}I\left(\alpha-\frac{a}{q}\right).$

This is our major arc approximation to $S(\alpha)$, although it does not have
arcs. Unlike $J_{Q}(\alpha)$, here we believe that $\mathscr{S}_{R}(\alpha)\to
S(\alpha)$ as $R\to N$, but as soon as $R>N^{1/2}$ we do not know how to prove
this. Thus, we say that $\mathscr{S}_{R}(\alpha)$ is the major arc
approximation to $S(\alpha)$ if $R\leq N^{1/2}$; but if $R>N^{1/2}$, then
$\mathscr{S}_{R}(\alpha)-\mathscr{S}_{N^{1/2}}(\alpha)$ is the minor arc
approximation to $S(\alpha)$. We have

$\displaystyle\mathscr{S}_{R}(\alpha)$ $\displaystyle=\sum_{q\leq
R}\sum_{\begin{subarray}{c}1\leq a\leq q\\\
(a,q)=1\end{subarray}}\frac{\mu(q)}{\phi(q)}\sum_{n\leq
N}e(n\alpha)e\left(-\frac{na}{q}\right)$ $\displaystyle=\sum_{n\leq
N}\sum_{q\leq R}\frac{\mu(q)}{\phi(q)}\sum_{\begin{subarray}{c}1\leq a\leq
q\\\ (a,q)=1\end{subarray}}e\left(-\frac{na}{q}\right)e(n\alpha)$
$\displaystyle=\sum_{n\leq N}\lambda_{R}(n)e(n\alpha),$ (19)

where

$\lambda_{R}(n)=\sum_{q\leq R}\frac{\mu(q)}{\phi(q)}c_{q}(-n).$ (20)

In comparison to $S(\alpha)$, we expect that $\lambda_{R}(n)$ is an
approximation of $\Lambda(n)$. As before, we have

$\displaystyle\int_{0}^{1}\lvert
S(\alpha)-\mathscr{S}_{R}(\alpha)\rvert^{2}\,d\alpha$
$\displaystyle=\sum_{n\leq N}(\Lambda(n)-\lambda_{R}(n))^{2}$
$\displaystyle=\sum_{n\leq N}\Lambda(n)^{2}-2\sum_{n\leq
N}\lambda_{R}(n)\Lambda(n)+\sum_{n\leq N}\lambda_{R}(n)^{2}.$

Let

$L(R)=\sum_{q\leq R}\frac{\mu(q)^{2}}{\phi(q)}.$

By (9),

$L(R)=\log R+O(1).$

From Lemma 1 of [5, equation (2.5)] for $1\leq R\leq N$, and the prime number
theorem with remainder we have, for any positive constant $c_{16}$,

$\displaystyle\sum_{n\leq N}\lambda_{R}(n)\Lambda(n)$
$\displaystyle=\psi(N)L(R)+O(R\log N)$
$\displaystyle=NL(R)+O\left(\frac{N}{(\log N)^{c_{16}}}\right)+O(R\log N).$

From Lemma 2 of [5, equation (2.7)], for $1\leq R\leq N^{1/2}$,

$\sum_{n\leq N}\lambda_{R}(n)^{2}=NL(R)+O(R^{2}).$

Thus, for $R\leq N^{1/2}$, by (9) and (10),

$\displaystyle\int_{0}^{1}\lvert
S(\alpha)-\mathscr{S}_{R}(\alpha)\rvert^{2}\,d\alpha$
$\displaystyle=\sum_{n\leq
N}\Lambda(n)^{2}-NL(R)+O(R^{2})+O\left(\frac{N}{(\log N)^{c_{16}}}\right)$
$\displaystyle=a_{0}(N,R)+O(R^{2})+O\left(\frac{N}{(\log N)^{c_{16}}}\right).$

We now replace (9) by a more precise formula. Hildebrand [11, Hilfssatz 2]
proved that, for $x\geq 1$,

$\sum_{\begin{subarray}{c}n\leq x\\\
(n,k)=1\end{subarray}}\frac{\mu(n)^{2}}{\phi(n)}=\frac{\phi(k)}{k}(\log
x+D+g(k))+O\left(\frac{h(k)}{x^{1/2}}\right),$ (21)

where

$g(k)=\sum_{p\mid k}\frac{\log p}{p},\hskip 10.00002pth(k)=\sum_{d\mid
k}\frac{\mu(d)^{2}}{d^{1/2}}=\prod_{p\mid k}\left(1+\frac{1}{p^{1/2}}\right).$

and

$D=\gamma+\sum_{p}\frac{\log p}{p(p-1)}\hskip 10.00002pt\textrm{($\gamma=$
Euler constant).}$

Hence, we obtain a slightly improved estimate over the first proof of

$\displaystyle\int_{0}^{1}\lvert
S(\alpha)-\mathscr{S}_{R}(\alpha)\rvert^{2}\,d\alpha$
$\displaystyle=N\log\left(\frac{N}{R}\right)-N(1+D)+O(R^{2})+O\left(\frac{N}{R^{1/2}}\right)+O\left(\frac{N}{(\log
N)^{A}}\right)$ $\displaystyle\leq N\log\left(\frac{N}{R}\right)+O(R^{2}),$

if $N^{1/2}/(\log N)^{2}\leq R\leq N^{1/2}$.

We complete the proof of Theorem 5 as before by proving the following
estimate.

######  Lemma 3.

We have

$\int_{0}^{1}\lvert\mathscr{S}_{R}(\alpha)\rvert\,d\alpha\ll R\log N.$

To prove Lemma 3, substituting the first expression for $c_{q}(-n)$ of (8)
into (20) yields

$\lambda_{R}(n)=\sum_{q\leq
R}\frac{\mu(q)}{\phi(q)}\sum_{d\mid(q,n)}d\mu\left(\frac{d}{q}\right).$

If $d\mid q$, then $q=dm$. We have $\mu(q)=\mu(dm)=\mu(d)\mu(m)$ if $(d,m)=1$,
and $\mu(q)=0$ if $(d,m)>1$. Thus, in the nested double sum above, all the
nonzero terms when $\mu(q)\neq 0$ are when $(d,m)=1$, so that
$\mu(q/d)=\mu(m)=\mu(q)\mu(d)$. Hence,

$\lambda_{R}(n)=\sum_{q\leq
R}\frac{\mu(q)^{2}}{\phi(q)}\sum_{d\mid(q,n)}d\mu(d)=\sum_{\begin{subarray}{c}d\leq
R\\\ d\mid n\end{subarray}}d\mu(d)\sum_{\begin{subarray}{c}q\leq R\\\ d\mid
q\end{subarray}}\frac{\mu(q)^{2}}{\phi(q)}.$

Since

$\sum_{\begin{subarray}{c}q\leq R\\\ d\mid
q\end{subarray}}\frac{\mu(q)^{2}}{\phi(q)}=\sum_{dm\leq
R}\frac{\mu(dm)^{2}}{\phi(dm)}=\frac{\mu(d)^{2}}{\phi(d)}\sum_{\begin{subarray}{c}m\leq
R/d\\\ (m,d)=1\end{subarray}}\frac{\mu(m)^{2}}{\phi(m)},$

we have

$\lambda_{R}(n)=\sum_{\begin{subarray}{c}d\leq R\\\ d\mid
n\end{subarray}}\frac{d\mu(d)}{\phi(d)}\sum_{\begin{subarray}{c}m\leq R/d\\\
(n,d)=1\end{subarray}}\frac{\mu(n)^{2}}{\phi(n)}.$ (22)

Thus, from (21),

$\displaystyle\lambda_{R}(n)$ $\displaystyle=\sum_{\begin{subarray}{c}d\leq
R\\\ d\mid n\end{subarray}}\mu(d)\log\left(\frac{R}{d}\right)+\textrm{smaller
terms}$ $\displaystyle=\Lambda_{R}(n)+\textrm{smaller terms}.$

This shows that $\Lambda_{R}(n)$ is the first term in the more refined
approximation $\lambda_{R}(n)$.

As in the proof for $S_{R}(\alpha)$ in Lemma 2, by (19) and (22),

$\mathscr{S}_{R}(\alpha)=\sum_{n\leq N}\sum_{\begin{subarray}{c}d\mid n\\\
d\leq R\end{subarray}}a(d,R)e(n\alpha)=\sum_{d\leq R}a(d,R)\sum_{m\leq
N/d}e(md\alpha),$

where

$a(d,R)=\frac{d\mu(d)}{\phi(d)}\sum_{\begin{subarray}{c}m\leq R/d\\\
(n,d)=1\end{subarray}}\frac{\mu(m)^{2}}{\phi(m)}.$

Thus, as before,

$\displaystyle\int_{0}^{1}\lvert\mathscr{S}_{R}(\alpha)\rvert\,d\alpha$
$\displaystyle\leq\sum_{d\leq R}\lvert
a(d,R)\rvert\int_{0}^{1}\min\left(\frac{N}{d},\frac{1}{\|\alpha
d\|}\right)d\alpha$ $\displaystyle\ll\sum_{d\leq R}\lvert
a(d,R)\rvert\left[1+\log\left(\frac{N}{2d}\right)\right]$
$\displaystyle\ll\log N\sum_{d\leq R}\log\left(\frac{R}{d}\right)\ll R\log N.$

This completes the proof of Lemma 3.

######  Acknowledgments.

A. Ledoan wishes to express his thanks and appreciation to D.A. Goldston at
San José State University for enlightening discussions and comments on the
paper. Ledoan’s research is supported by the National Science Foundation under
Grant DMS-1852288.

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