ON THE SECOND MOMENT FOR PRIMES

IN AN ARITHMETIC PROGRESSION

D. A. Goldston^{1} and C. Y. Yıldırım^{2}

Abstract. Assuming the Generalized Riemann Hypothesis, we obtain a lower bound within a constant factor of the conjectured asymptotic result for the second moment for primes in an individual arithmetic progression in short intervals. Previous results were averaged over all progression of a given modulus. The method uses a short divisor sum approximation for the von Mangoldt function, together with some new results for binary correlations of this divisor sum approximation in arithmetic progressions.

1. Introduction and Statement of results

In this paper we calculate a lower bound, of the same order of magnitude as conjectured, for the second moment of primes in an arithmetic progression. Specifically we examine

I(x, h, q, a) := ∫_{x}^{2x} (ψ(y + h; q, a) − ψ(y; q, a) − \frac{h}{ϕ(q)})^{2} dy  (1.1)

where

ψ(x; q, a) = ∑_{{n≤x \atop n≡a(mod q)}} Λ(n),  (1.2)

and Λ is the von Mangoldt function. We will take

(a, q) = 1, x ≥ 2, 1 ≤ q ≤ h ≤ x,  (1.3)

(other ranges not being interesting). We shall assume the truth of the Generalized Riemann Hypothesis (GRH), which implies, in particular,

E(x; q, a) := ψ(x; q, a) − \frac{x}{ϕ(q)} ≪ x^{\frac{1}{2}} log^{2} x , (q ≤ x).  (1.4)

The idea of our method originates from the work of Goldston [3] for the case of all primes, corresponding in the present formulation to q = 1. An improved and generalized version of this result appeared in [5] as

Typeset by AMS-TEX

1991 Mathematics Subject Classification. Primary:11M26.

1 Research at MSRI is supported in part by NSF grant DMS-9701755, also supported by an NSF grant

2 Research at MSRI is supported in part by NSF grant DMS-9701755, also supported by a TÜBITAK-NATO-B2 grant

Theorem A. Assume GRH. Then for any ϵ > 0 and 1 ≤ \frac{h}{q} ≪ \frac{x^{\frac{1}{3}}}{q^{ϵ} log^{3} x} we have

∑_{{a(mod q) \atop (a,q)=1}} I(x, h, q, a) ≥ \frac{1}{2}xh log ((\frac{q}{h})^{3}x) − O(xh(log log x)^{3}).  (1.5)

Moreover, for almost all q with h^{3/4} log^{5} x ≤ q ≤ h we have

I(x, h, q) ∼ xh log(\frac{xq}{h}).  (1.6)

For an individual arithmetic progression Özlük [11] proved unconditionally

Theorem B. For 1 ≤ q ≤ (log x)^{1−δ}, and h ≤ (log x)^{c} (δ and c are any fixed positive numbers) satisfying q ≤ h, we have

I(x, h, q, a) > (\frac{1}{2} − ϵ)\frac{xh}{ϕ(q)} log x  (1.7)

for any ϵ and x ≥ X(ϵ, c).

We shall see below that the GRH implies a result of the type in Theorem B for much wider ranges of q and h. An asymptotic estimate for I(x, h, q, a) in certain ranges was shown by Yıldırım [12] to be implied by GRH and a pair correlation conjecture for the zeros of Dirichlet’s L-functions.

Theorem C. Assume GRH. Let α_{1}, α_{2}, η be fixed and satisfying 0 < η < α_{1} ≤ α_{2} ≤ 1, and let δ = x^{−α} where α_{1} ≤ α ≤ α_{2}. Assume, as x → ∞, uniformly for

q ≤ min(x^{\frac{1}{2}} δ^{\frac{1}{2}} log^{A} x, δ^{−1}x^{−η})  (q: prime or 1)  (1.8)

and

\frac{x^{α_{1}}}{ϕ(q)} log^{−3} x ≤ T ≤ ϕ(q)x^{α_{2}} log^{3} x  (1.9)

that, for (a, q) = 1,

∑_{χ_{1},χ_{2}(mod q)} \overline{χ_{1}}(a)χ_{2}(a) ∑_{0<γ_{1},γ_{2}≤T ; {L( \frac{1}{2} +iγ_{1},χ_{1})=0 \atop L( \frac{1}{2} +iγ_{2},χ_{2})=0}} x^{i(γ_{1}−γ_{2})} \frac{4}{4 + (γ_{1} − γ_{2})^{2}} ∼ ϕ(q)\frac{T}{2π} log qT.

(1.10)

Then

∫_{x}^{2x} (ψ(u + uδ; q, a) − ψ(u; q, a) − \frac{uδ}{ϕ(q)})^{2} du ∼ \frac{3}{2} \frac{δx^{2}}{ϕ(q)} log \frac{q}{δ}  (1.11)

uniformly for x^{−α_{2}} ≤ δ ≤ x^{−α_{1}} and q as in (1.8).

It was also shown in [12] that the left-hand side of (1.10) is ∼ ϕ(q)\frac{T}{2π} log x for

1 ≤ q ≤ x^{\frac{1}{2}} log^{−3} x when \frac{x}{q} log x ≤ T ≤ e^{x^{\frac{1}{4}}} . These asymptotic values are what the diagonal terms (χ_{1} = χ_{2}) would contribute, so the assumption (1.10) is a way of expressing that the zeros of different Dirichlet L-functions are uncorrelated. Theorem C is a generalization of one half of a result of Goldston and Montgomery [4] for the case q = 1, where an equivalence between the pair correlation conjecture for ζ(s) and the second moment for primes was established. Since the argument in [4] works reversibly, a suitable converse to Theorem C is also provable. The restriction to prime q was made in order to avoid the presence of imprimitive characters. The formula (1.11) involving differences uδ which vary with u can be converted to a formula involving a fixed-difference h.

Our main result is the following theorem.

Theorem 1. Assume GRH. Then for any ϵ > 0 and

q ≤ h ≤ (xq)^{\frac{1}{3} −ϵ},  (1.12)

we have

I(x, h, q, a) ≥ \frac{1}{2} \frac{xh}{ϕ(q)} log(\frac{xq}{h^{3}} ) − O(\frac{xh}{ϕ(q)}(log log 3q)^{3}).  (1.13)

Notice that the conditions in (1.12) imply that both h and q are ≪ x^{\frac{1}{2} −ϵ}.

The proof of the theorem uses some new results on the function λ_{R}(n) used as an approximation for the von Mangoldt function in our earlier work. Propositions 2, 3, and 4 embody these results, and we expect they will have further applications to other problems.

2. Preliminaries

We shall need the following in our calculations. Let

f(n, x, h) = ∫_{[x,2x]∩[n−h,n)} 1 dy = \{ \begin{matrix}n − x, & for x ≤ n < x + h \\ h, & for x + h ≤ n ≤ 2x \\ 2x − n + h, & for 2x < n ≤ 2x + h \\ 0, & elsewhere. \end{matrix} (2.1)

Lemma 1. For real numbers a_{n} and b_{n} we have

∫_{x}^{2x} ( ∑_{y<n≤y+h} a_{n}) ( ∑_{y<m≤y+h} b_{m}) dy = ∑_{x<n≤2x+h} a_{n}b_{n}f(n, x, h)

+ ∑_{0<k≤h} ( ∑_{x<n≤2x+h−k}(a_{n}b_{n+k} + a_{n+k}b_{n})f(n, x, h − k)) .

Lemma 2. Let C(x) = ∑_{n≤x} c_{n}. Then we have

∑_{x<n≤2x+h} c_{n}f(n, x, h) = ∫_{2x}^{2x+h} C(u) du − ∫_{x}^{x+h} C(u) du.

Lemma 1 and Lemma 2 were proved in [5]. We take this opportunity to correct a minor error in Lemma 1 of [5]. In that lemma an extraneous term h(c_{x+h} − c_{2x}) was incorrectly included and should be removed. This term then contributed an unnecessary error term in equations (2.7),(2.14), and (2.15) of [5]. However these same error terms correctly occurred for a different reason in equation (2.9) so that starting with equation (2.16) these error terms were correctly included in the rest of [5].

Calling Δψ = ψ(y + h; q, a) − ψ(y; q, a) for brevity, we have from (1.1)

I(x, h, q, a) = ∫_{x}^{2x}(Δψ)^{2} dy − \frac{2h}{ϕ(q)} ∫_{x}^{2x}(Δψ) dy + \frac{h^{2}x}{ϕ^{2}(q)}.

By the above lemmas and (1.4) we obtain

∫_{x}^{2x}(Δψ) dy = ∑_{{x<n≤2x+h \atop n≡a(mod q)}} Λ(n)f(n, x, h)

= \frac{xh}{ϕ(q)} + ∫_{2x}^{2x+h} E(y; q, a) dy − ∫_{x}^{x+h} E(y; q, a) dy

= \frac{xh}{ϕ(q)} + O(hx^{\frac{1}{2}} log^{2} x),

so that

I(x, h, q, a) = ∫_{x}^{2x}(Δψ)^{2} dy − \frac{xh^{2}}{ϕ^{2}(q)} + O(\frac{x^{\frac{1}{2}} h^{2} log^{2} x}{ϕ(q)}).  (2.2)

The integral ∫ (Δψ)^{2} leads to sums of the sort ∑ Λ(n)Λ(n + k) which are in the territory of the twin prime conjecture. In the uninteresting case 1 ≤ h ≤ q, only the sum ∑ Λ^{2}(n) is present, giving easily the evaluation

I(x, h, q, a) = \frac{xh}{ϕ(q)} log x − \frac{xh}{ϕ(q)} − \frac{xh^{2}}{ϕ^{2}(q)} + O(x^{\frac{1}{2}} h log^{3} x), (h ≤ q ≤ x). (2.3)

Now let λ_{R}(n) be any arithmetical function, and set

ψ_{R}(y; q, a) = ∑_{{n≤y \atop n≡a(mod q)}} λ_{R}(n) ;  Δψ_{R} = ψ_{R}(y + h; q, a) − ψ_{R}(y; q, a). (2.4)

PRIMES IN ARITHMETIC PROGRESSIONS 5

Trivially ∫ (Δψ − Δψ_{R})^{2} ≥ 0, so that

∫_{x}^{2x}(Δψ)^{2} dy ≥ 2 ∫_{x}^{2x}(Δψ) · (Δψ_{R}) dy − ∫_{x}^{2x}(Δψ_{R})^{2} dy.  (2.5)

We apply Lemma 1 and Lemma 2 to these integrals to obtain

∫_{x}^{2x}(Δψ)·(Δψ_{R}) dy = ∑_{{x<n≤2x+h \atop n≡a(mod q)}} Λ(n)λ_{R}(n)f(n, x, h)  (2.6 a)

+ ∑_{{0<k≤h \atop q|k}} ∑_{{x<n≤2x+h−k \atop n≡a(mod q)}} [Λ(n)λ_{R}(n + k) + Λ(n + k)λ_{R}(n)]f(n, x, h − k)

= (∫_{2x}^{2x+h} − ∫_{x}^{x+h}) ∑_{{n≤u \atop n≡a(mod q)}} Λ(n)λ_{R}(n) du

+ ∫_{2x}^{2x+h} ∑_{0<|j|≤ \frac{u−2x}{q}} ∑_{{N_{1}<n≤N_{2} \atop n≡a(mod q)}} λ_{R}(n)Λ(n + jq) du

− ∫_{x}^{x+h} ∑_{0<|j|≤ \frac{u−x}{q}} ∑_{{N_{1}<n≤N_{2} \atop n≡a(mod q)}} λ_{R}(n)Λ(n + jq) du ,  (2.6 b)

where N_{1} = max(0, −jq) and N_{2} = min(u, u − jq). Similarly

∫_{x}^{2x}(Δψ_{R})^{2} dy = ∑_{{x<n≤2x+h \atop n≡a(mod q)}} λ_{R}^{2}(n)f(n, x, h)

+ 2 ∑_{{0<k≤h \atop q|k}} ∑_{{x<n≤2x+h−k \atop n≡a(mod q)}} λ_{R}(n)λ_{R}(n + k)f(n, x, h − k)

= (∫_{2x}^{2x+h} − ∫_{x}^{x+h}) ∑_{{n≤u \atop n≡a(mod q)}} λ_{R}^{2}(n) du

+ 2 ∫_{2x}^{2x+h} ∑_{0<j≤ \frac{u−2x}{q}} ∑_{{n≤u−jq \atop n≡a(mod q)}} λ_{R}(n)λ_{R}(n + jq) du

− 2 ∫_{x}^{x+h} ∑_{0<j≤ \frac{u−x}{q}} ∑_{{n≤u−jq \atop n≡a(mod q)}} λ_{R}(n)λ_{R}(n + jq) du. (2.7)

3. The choice of λ_{R}(n) and some number-theoretic sums

As the auxiliary function we use

λ_{R}(n) := ∑_{r≤R} \frac{μ^{2}(r)}{ϕ(r)} ∑_{d|(r,n)} dμ(d).  (3.1)

This function is known ([3], [8]) to exhibit behavior similar to Λ(n) when considered on average in arithmetic progressions, and it has been employed in related problems ([1], [2], [5], [6]). An upper bound for λ_{R}(n) is

|λ_{R}(n)| ≤ ∑_{d|n} d ∑_{{r≤R \atop d|r}} \frac{1}{ϕ(r)} ≤ max_{r≤R}(\frac{r}{ϕ(r)}) ∑_{d|n} d ∑_{{r≤R \atop d|r}} \frac{1}{r} ≪ d(n) log R log log R. (3.2)

To evaluate the sums which arise when (3.1) is used in (2.6) and (2.7) we shall need some lemmas. In the following p will denote a prime number.

Lemma 3. (Hildebrand [9]) We have for each positive integer k, uniformly in R ≥ 1,

L_{k}(R) := ∑_{{n≤R \atop (n,k)=1}} \frac{μ^{2}(n)}{ϕ(n)} = \frac{ϕ(k)}{k}(log R + c + v(k)) + O(\frac{w(k)}{\sqrt{R}}),  (3.3)

where

c :=γ + ∑_{p} \frac{log p}{p(p − 1)}; v(k) := ∑_{p|k} \frac{log p}{p};

w(k) := ∑_{d|k} \frac{μ^{2}(d)}{\sqrt{d}} = ∏_{p|k}(1 + \frac{1}{\sqrt{p}}); v(1) = 0, w(1) = 1.  (3.4)

Lemma 4. We have

v(k) ≪ log log 3k,  (3.5)
∑_{p|k} \frac{1}{\sqrt{p}} ≪ \frac{\sqrt{log k}}{log log 3k} ,  (3.6)

and

g(k) := ∏_{p|k}(1 + \frac{p}{p − 1}) ≪ 2^{ν(k)}(log log 3k)  (3.7)

Proof. We show (3.7); the other inequalities can be proved similarly. Let ν(k) be the

number of distinct prime factors of k, which satisfies the bound ν(k) ≪ \frac{log k}{log log k}.

PRIMES IN ARITHMETIC PROGRESSIONS 7

We have

log g(k) = ∑_{p|k} log(2 + \frac{1}{p − 1})

< ν(k) log 2 + ∑_{p|k} \frac{1}{p}

< ν(k) log 2 + ∑_{p≤2 log 2k} \frac{1}{p}

= ν(k) log 2 + log log log 21k + O(1),

where the prime number theorem and Mertens’ theorem have been employed. Exponentiating both sides we obtain (3.7). Lemma 5. We have

∑_{ℓ|k} \frac{μ^{2}(ℓ)}{ϕ(ℓ)} log ℓ = \frac{k}{ϕ(k)}v(k).  (3.8)

Proof.

∑_{ℓ|k} \frac{μ^{2}(ℓ)}{ϕ(ℓ)} log ℓ = log ∏_{ℓ|k} ℓ\frac{μ^{2}(ℓ)}{ϕ(ℓ)} = log ∏_{p|k} p ^{∑_{ℓ|k,p|ℓ} \frac{μ^{2}(ℓ)}{ϕ(ℓ)}}

= log ∏_{p|k} p^{\frac{1}{ϕ(p)} ∑_{ℓ′| \frac{k}{p} ,(ℓ′,p)=1} \frac{μ^{2}(ℓ′)}{ϕ(ℓ′)}} = log ∏_{p|k} p\frac{1}{ϕ(p)} \frac{1}{1 + \frac{1}{ϕ(p)}} ∏_{p′|k}(1 + \frac{1}{ϕ(p′)})

= log ∏_{p|k} p ^{\frac{1}{p} \frac{k}{ϕ(k)}} = \frac{k}{ϕ(k)} ∑_{p|k} \frac{log p}{p}.

Lemma 6. (Goldston [3]) We have

∑_{r≤R} \frac{μ^{2}(r)}{ϕ(r)} ∑_{{d|r \atop (d,k)=1}} \frac{dμ(d)}{ϕ(d)} = 𝔖(k) + O(\frac{k d(k)}{Rϕ(k)}),  (3.9)

where

𝔖(k) = \{ \begin{matrix}2C ∏_{{p|k \atop p>2}} (\frac{p − 1}{p − 2}) , & if k is even, k ≠ 0; \\ 0, & if k is odd; \end{matrix}  (3.10)

8 GOLDSTON AND YILDIRIM

with

C = ∏_{p>2} (1 − \frac{1}{(p − 1)^{2}} ) .  (3.11)

Proof. The proof can be found in [3]; we just note that

∑_{{d|r \atop (d,k)=1}} \frac{dμ(d)}{ϕ(d)} = \frac{μ(r)}{ϕ(r)}μ((r, k))ϕ((r, k)) ,  (3.12)

so the left-hand side of (3.9) may be expressed as

∑_{r=1}^{∞} \frac{μ(r)μ((r, k))ϕ((r, k))}{ϕ^{2}(r)} + O(∑_{r>R} \frac{μ^{2}(r)μ^{2}((r, k))ϕ((r, k))}{ϕ^{2}(r)}).  (3.13)

Here the first sum is 𝔖(k) and the error term is ≪ the O-term in (3.9). Lemma 7. (Goldston and Friedlander [1])We have

∑_{0<j≤ \frac{h}{q}}(h − jq)𝔖(jq) = \frac{h^{2}}{2ϕ(q)} − \frac{h}{2} log \frac{h}{q} + O(h(log log 3q)^{3}).  (3.14)

Lemma 8. (Hooley [10]) Assuming GRH, we have

∑_{{a(mod q) \atop (a,q)=1}} max_{u≤x} |E(u; q, a)|^{2} ≪ x log^{4} x; for q ≤ x.  (3.15)

Lemma 9. We have

∑_{r≤R} \frac{μ^{2}(r)σ(r)}{ϕ(r)} ≪ R  (3.16)
∑_{r≤R} \frac{μ^{2}(r)σ_{\frac{1}{2}} (r)}{ϕ(r)} ≪ \sqrt{R}  (3.17)
∑_{0<r≤R} \frac{rd(r)}{ϕ(r)} ≪ R log 2R.  (3.18)

Proof. To prove (3.16), note

∑_{r≤R} \frac{μ^{2}(r)σ(r)}{ϕ(r)} = ∑_{r≤R} μ^{2}(r) ∏_{p|r}(1 + \frac{2}{p − 1}) = ∑_{r≤R} μ^{2}(r) ∑_{d|r} \frac{2^{ν(d)}}{ϕ(d)}

= ∑_{d≤R} \frac{μ^{2}(d)2^{ν(d)}}{ϕ(d)} ∑_{ℓ≤ \frac{R}{d}} μ^{2}(ℓ) ≤ R ∑_{d≤R} \frac{μ^{2}(d)2^{ν(d)}}{dϕ(d)}

≤ R ∏_{p}(1 + \frac{2}{p(p − 1)}) ≪ R.

The proof of (3.17) is similar, and (3.18) was shown in [3].

4. The proof of the Theorem

In this section we calculate the right-hand sides of (2.6) and (2.7), and so obtain our result.

Proposition 1. Assuming GRH, we have

∑_{{n≤N \atop n≡a(mod q)}} Λ(n)λ_{R}(n) = \frac{N log R}{ϕ(q)}+\frac{cN}{ϕ(q)}+O(\frac{N}{ϕ(q)\sqrt{R}})+O(N^{\frac{1}{2}} log^{3} N)+O(R log N).

(4.1)

Proof. Starting from the definition (3.1) and recalling L_{k}(R) from (3.3), we have

∑_{{n≤N \atop n≡a(mod q)}} Λ(n)λ_{R}(n) = ∑_{r≤R} \frac{μ^{2}(r)}{ϕ(r)} ∑_{d|r} dμ(d) ∑_{n≤N ; {n≡a(mod q) \atop d|n}} Λ(n)

= L_{1}(R)ψ(N; q, a) − ∑_{r≤R} \frac{μ^{2}(r)}{ϕ(r)} ∑_{p|r} p log p ∑_{k≥1 ; {p^{k}≤N \atop p^{k}≡a(mod q)}} 1.

Here the sum over k is trivially of size O(\frac{log N}{log p}), so that by Lemma 3

∑_{r≤R} \frac{μ^{2}(r)}{ϕ(r)} ∑_{p|r} p log p ∑_{k≥1 ; {p^{k}≤N \atop p^{k}≡a(mod q)}} 1 ≪ log N ∑_{r≤R} \frac{μ^{2}(r)}{ϕ(r)} ∑_{p|r} p

= log N ∑_{p≤R} \frac{p}{ϕ(p)} ∑_{{m≤ \frac{R}{p} \atop (m,p)=1}} \frac{μ^{2}(m)}{ϕ(m)}

≪ log N ∑_{p≤R} log \frac{R}{p} ≪ R log N.

By the prime number theorem we obtain (4.1). In order for the main term to dominate the error terms in (4.1) we will require that

qR ≤ \frac{N}{log N} , q ≤ \frac{N^{\frac{1}{2}}}{log^{3} N} .  (4.2)

Hence the relevant contribution to (2.6 b) will be

(∫_{2x}^{2x+h} − ∫_{x}^{x+h}) ∑_{{n≤u \atop n≡a(mod q)}} Λ(n)λ_{R}(n) du  (4.3)

=\frac{xh}{ϕ(q)}(log R + c) + O(\frac{xh}{ϕ(q)\sqrt{R}}) + O(x^{\frac{1}{2}} h log^{3} x) + O(Rh log x).

Proposition 2. Assuming GRH, we have for

1 ≤ q ≤ h ≤ x, qR ≤ x,  (4.4)

that

∑_{{0<k≤h \atop q|k}} ∑_{{x<n≤2x+h−k \atop n≡a(mod q)}} [Λ(n)λ_{R}(n + k) + Λ(n + k)λ_{R}(n)]f(n, x, h − k)

=\frac{xh^{2}}{ϕ^{2}(q)} − \frac{xh}{ϕ(q)} log \frac{h}{q} + O(\frac{xh}{ϕ(q)}(log log 3q)^{3}) + O(\frac{xh^{2}d(q)}{ϕ^{2}(q)R} log \frac{2h}{q})
+ O(\frac{x^{\frac{1}{2}} h^{\frac{3}{2}} R log^{2} x}{q^{\frac{1}{2}}}) + O(\frac{x^{\frac{1}{2}} h^{2}R^{\frac{1}{2}} log^{2} x}{q}).  (4.5)

Proof. We have

∑_{{N_{1}<n≤N_{2} \atop n≡a(mod q)}} λ_{R}(n)Λ(n + jq) = ∑_{r≤R} \frac{μ^{2}(r)}{ϕ(r)} ∑_{d|r} dμ(d) ∑_{N_{1}<n≤N_{2} ; {n≡a(mod q) \atop d|n}} Λ(n + jq).

We may write the innermost sum as

∑_{N_{1}+jq<m≤N_{2}+jq ; {m≡a(mod q) \atop m≡jq(mod d)}} Λ(m).

Here m−jq = ℓd for some integer ℓ, and so a ≡ ℓd( mod q). Since (a, q) = 1, we can include only those d’s such that (d, q) = 1. Then there is a unique b, 0 < b < qd, such that m ≡ b(mod qd). We know (m, q) = 1, so that (m, d) = 1 if and only if (j, d) = 1. Hence the innermost sum is equal to

ψ(N_{2} + jq; qd, b)−ψ(N_{1} + jq; qd, b)

= \frac{N_{2} − N_{1}}{ϕ(qd)} E_{qd,b} + E(N_{2} + jq; qd, b) − E(N_{1} + jq; qd, b),

where E_{qd,b} = 1 if (qd, b) = 1, and E_{qd,b} = 0 if (qd, b) > 1. Thus

∑_{{N_{1}<n≤N_{2} \atop n≡a(mod q)}} λ_{R}(n)Λ(n + jq) = \frac{u − |j|q}{ϕ(q)} ∑_{r≤R} \frac{μ^{2}(r)}{ϕ(r)} ∑_{{d|r \atop (d,jq)=1}} \frac{dμ(d)}{ϕ(d)}  (4.6)

+ ∑_{r≤R} \frac{μ^{2}(r)}{ϕ(r)} ∑_{{d|r \atop (d,q)=1}} dμ(d)[E(N_{2} + jq; qd, b) − E(N_{1} + jq; qd, b)],

where the first term on the right-hand side is the main term, its value settled by Lemma 6, and its contribution to (2.6 b) will be

∫_{2x}^{2x+h} ∑_{0<|j|≤ \frac{u−2x}{q}} \frac{u − |j|q}{ϕ(q)} [𝔖(jq) + O(\frac{jqd(jq)}{Rϕ(jq)})] du

− ∫_{x}^{x+h} ∑_{0<|j|≤ \frac{u−x}{q}} \frac{u − |j|q}{ϕ(q)} [𝔖(jq) + O(\frac{jq d(jq)}{Rϕ(jq)})] du

=\frac{2x}{ϕ(q)} ∑_{0<j≤ \frac{h}{q}}(h − jq)𝔖(jq) + O(\frac{xhqd(q)}{Rϕ^{2}(q)} ∑_{0<j≤ \frac{h}{q}} \frac{jd(j)}{ϕ(j)})

=\frac{xh^{2}}{ϕ^{2}(q)} − \frac{xh}{ϕ(q)} log \frac{h}{q} + O(\frac{xh}{ϕ(q)}(log log 3q)^{3}) + O(\frac{xh^{2}d(q)}{ϕ^{2}(q)R} log \frac{2h}{q}), (4.7)

by Lemma 7 and (3.18). For the second term in the right-hand side of (4.6), if we use (1.4) directly, we will get the upper bound Rx^{\frac{1}{2}} log^{2} x, by (3.16). This will lead

to a contribution of O(\frac{x^{\frac{1}{2}} h^{2}R}{q} log^{2} x) in (2.6 b). Instead, in view of the averaging

over j in (2.6 b), we will use Hooley’s estimate quoted as Lemma 8 above. To do this note that some of the d’s may not be coprime to b, but we can discard them (from the j- and n-summations) with an error

≪ ∑_{{0<|j|≤ \frac{h}{q} \atop (j,d)>1}} ψ(3x; qd, b) ≪ ∑_{0<|j|≤ \frac{h}{q}} ∑_{p|d} ∑_{{n≤3x \atop p|n}} Λ(n) ≪ \frac{h}{q} log^{2} x ,

and this leads to an error of O(\frac{h^{2}R}{q} log^{2} x) in (2.6 b). Hence the contribution to

(2.6 b) from the second term on the right-hand side of (4.6) is

≪\frac{h^{2}R}{q} log^{2} x + (∫_{2x}^{2x+h}− ∫_{x}^{x+h}) ∑_{r≤R} \frac{μ^{2}(r)}{ϕ(r)} ∑_{{d|r \atop (d,q)=1}}d ∑_{{0<|j|≤ \frac{h}{q} \atop (j,d)=1}} max_{u≤2x+h} |E(u; qd, b)| du

≪\frac{h^{2}R}{q} log^{2} x + h ∑_{r≤R} \frac{μ^{2}(r)}{ϕ(r)} ∑_{{d|r \atop (d,q)=1}} d (\frac{h}{q})^{\frac{1}{2}} ( ∑_{{0<|j|≤ \frac{h}{q} \atop (j,d)=1}} max_{u≤2x+h} |E(u; qd, b)|^{2})^{\frac{1}{2}}

≪\frac{h^{2}R}{q} log^{2} x + \frac{h^{\frac{3}{2}}}{q^{\frac{1}{2}}} ∑_{r≤R} \frac{μ^{2}(r)}{ϕ(r)}∑_{d|r} d(1 + (\frac{h}{qd}))^{\frac{1}{2}} ( ∑_{{j(mod d) \atop (j,d)=1}}max_{u≤3x} |E(u; qd, b)|^{2})^{\frac{1}{2}} . (4.8)

In the last sum as j runs through the reduced residues modulo d, b runs through those elements of the set {a, a + q, …, a + (d − 1)q} which are relatively prime to d (note that a ≡ ℓd(mod q) and (a, q) = 1 implies (a, d) = 1), and this correspondence is one-to-one. This is because m ≡ a( mod q) and m ≡ jq( mod d) if and only if m ≡ n_{1}da + n_{2}jq^{2}(mod qd) where n_{i} satisfy dn_{1} ≡ 1(mod q), qn_{2} ≡ 1(mod d), and we have n_{1}da + n_{2}jq^{2} ≡ a + tq(mod qd) if and only if j − t ≡ an_{2}(mod d). Hence we may replace the j-sum in (4.8) by

∑_{{t(mod d) \atop (a+tq,d)=1}} max_{1≤u≤3x} |E(u; qd, a + tq)|^{2}.  (4.9)

Although the last sum is over only \frac{1}{ϕ(q)} of the reduced residue classes modulo qd, we

shall use Hooley’s estimate as is. One would want to get a Hooley-type estimate for (4.9) itself, thereby saving a factor of ϕ(q), but this seems to require some estimates for certain integrals involving pairs of L-functions. We do not follow this path now. Recall that Theorem C, which gives an asymptotic estimate for our integral already rests upon such an assumption, (1.10), about L-functions. By Lemma 8, we take x log^{4} x as upper bound for (4.9) on the condition that qR ≤ x, and on applying Lemma 9 we obtain that the expression in (4.8) is

≪ \frac{x^{\frac{1}{2}} h^{\frac{3}{2}} R}{q^{\frac{1}{2}}} log^{2} x + \frac{x^{\frac{1}{2}} h^{2}R^{\frac{1}{2}}}{q} log^{2} x.  (4.10)

This completes the proof of Proposition 2.

Proposition 3. For (a, q) = 1 we have

∑_{{n≤N \atop n≡a(mod q)}} λ_{R}^{2}(n) = \frac{N}{ϕ(q)}[log R + c + O(v(q)) + O(R^{− \frac{1}{2} +ϵ})] + O(R^{2}). (4.11)

Proposition 4. For (a, q) = 1 and j ≠ 0 we have

∑_{{n≤N \atop n≡a(mod q)}} λ_{R}(n)λ_{R}(n + jq) = \frac{N}{ϕ(q)}𝔖(jq) + O(\frac{Ng(q)}{ϕ(q)R} \frac{jd(j)}{ϕ(j)}) + O(R^{2}). (4.12)

Proof. The beginning of the proof of Proposition 3 may be incorporated into that of Proposition 4 upon a notational stipulation for the case j = 0. When the positive integer t satisfies t | j, if j = 0 we will understand that t can be any positive integer; and we will take (t, 0) = t.

By definition (3.1),

∑_{{n≤N \atop n≡a(mod q)}} λ_{R}(n)λ_{R}(n + jq) = ∑_{r, r′≤R} \frac{μ^{2}(r)μ^{2}(r′)}{ϕ(r)ϕ(r′)} ∑_{{d|r \atop e|r′}} dμ(d)eμ(e) ∑_{n≤N ; {n≡a(mod q) \atop d|n, e|n+jq}} 1.

(4.13)

In the innermost sum, the conditions on n, d, and e imply (q, de) = 1, (d, e) | j, and thus n belongs to a unique residue class modulo [q, d, e]. Hence we have

∑_{n≤N ; {n≡a(mod q) \atop d|n, e|n+jq}} 1 = \frac{N}{[q, d, e]} + O(1).  (4.14)

The contribution of the O(1)-term in (4.14) to (4.13) is

≪ ∑_{r, r′≤R} \frac{μ^{2}(r)μ^{2}(r′)}{ϕ(r)ϕ(r′)} ∑_{{d|r \atop e|r′}} de = (∑_{r≤R} \frac{μ^{2}(r)σ(r)}{ϕ(r)})^{2} ≪ R^{2}  (4.15)

by (3.16), and this is where the O(R^{2})-term in (4.11) and (4.12) comes from. Hence

∑_{{n≤N \atop n≡a(mod q)}} λ_{R}(n)λ_{R}(n+jq) = \frac{N}{q} ∑_{r, r′≤R} \frac{μ^{2}(r)μ^{2}(r′)}{ϕ(r)ϕ(r′)} ∑_{d| \frac{r}{(r,q)} ; {e| \frac{r′}{(r′,q)} \atop (d,e)|j}} μ(d)μ(e)(d, e) +O(R^{2}).

(4.16)

Let (d, e) = δ, d = d′δ, e = e′δ, so that (d′, e′) = 1. The inner sums over d and e become

∑_{{δ|j \atop δ|( \frac{r}{(r,q)} , \frac{r′}{(r′,q)} )}} δ ∑_{d′| \frac{r}{δ(r,q)}} μ(d′) ∑_{{e′| \frac{r′}{δ(r′,q)} \atop (e′,d′)=1}} μ(e′).  (4.17)

Here the innermost sum is

∑_{{e′| \frac{r′}{δ(r′,q)} \atop (e′,d′)=1}} μ(e′) = ∏_{{p| \frac{r′}{δ(r′,q)} \atop p∤d′}} (1 + μ(p)) = \{ \begin{matrix}1 if \frac{r′}{δ(r′,q)} | d′ ; \\ 0 otherwise. \end{matrix}  (4.18)

Next the sum over d′ becomes

∑_{{d′| \frac{r}{δ(r,q)} \atop \frac{r′}{δ(r′,q)} |d′}} μ(d′) = \{ \begin{matrix}μ(\frac{r}{δ(r, q)}) & if \frac{r′}{(r′,q)} = \frac{r}{(r,q)} ; \\ 0 & otherwise, \end{matrix}  (4.19)

so the main term of (4.16) is

\frac{N}{q} ∑_{r, r′≤R_{\frac{r}{(r,q)} = \frac{r′}{(r′,q)}}} \frac{μ^{2}(r)μ^{2}(r′)}{ϕ(r)ϕ(r′)}μ(\frac{r}{(r, q)}) ∑_{δ|( \frac{r}{(r,q)} ,j)} δμ(δ).  (4.20)

Since

∑_{δ|( \frac{r}{(r,q)} ,j)} δμ(δ) = μ((\frac{r}{(r, q)}, j))ϕ((\frac{r}{(r, q)}, j)) ,  (4.21)

the main term is

\frac{N}{q} ∑_{r, r′≤R_{\frac{r}{(r,q)} = \frac{r′}{(r′,q)}}} \frac{μ^{2}(r)μ^{2}(r′)}{ϕ(r)ϕ(r′)}μ(\frac{r}{(r, q)})μ((\frac{r}{(r, q)}, j))ϕ((\frac{r}{(r, q)}, j)).  (4.22)

Writing (r, q) = ℓ, (r′, q) = m, r = ℓs, r′ = ms where (s, q) = 1, (4.22) takes the form

\frac{N}{q} ∑_{ℓ|q} \frac{μ^{2}(ℓ)}{ϕ(ℓ)} ∑_{m|q} \frac{μ^{2}(m)}{ϕ(m)} ∑_{{s≤min( \frac{R}{ℓ} , \frac{R}{m}) \atop (s,q)=1}} \frac{μ(s)}{ϕ^{2}(s)}μ((s, j))ϕ((s, j)).  (4.23)

The j = 0 case: We rewrite (4.23) as

\frac{N}{q} ∑_{ℓ|q} \frac{μ^{2}(ℓ)}{ϕ(ℓ)} ∑_{m|q} \frac{μ^{2}(m)}{ϕ(m)} ∑_{{s≤min( \frac{R}{ℓ} , \frac{R}{m}) \atop (s,q)=1}} \frac{μ^{2}(s)}{ϕ(s)}.  (4.24)

It is convenient to regard (4.24) as

\frac{N}{q} ∑_{ℓ|q} \frac{μ^{2}(ℓ)}{ϕ(ℓ)} ∑_{m|q} \frac{μ^{2}(m)}{ϕ(m)} ∑_{{s≤ \frac{R}{ℓ} \atop (s,q)=1}} \frac{μ^{2}(s)}{ϕ(s)} − \frac{N}{q} ∑_{ℓ|q} \frac{μ^{2}(ℓ)}{ϕ(ℓ)} ∑_{{m|q \atop m>ℓ}} \frac{μ^{2}(m)}{ϕ(m)} ∑_{{\frac{R}{m} <s≤ \frac{R}{ℓ} \atop (s,q)=1}} \frac{μ^{2}(s)}{ϕ(s)}.

(4.25)

For the first term of (4.25), we observe that

∑_{m|q} \frac{μ^{2}(m)}{ϕ(m)} = \frac{q}{ϕ(q)},  (4.26)

and by Lemma 3 and Lemma 5 we have

\frac{N}{q} ∑_{ℓ|q} \frac{μ^{2}(ℓ)}{ϕ(ℓ)} ∑_{m|q} \frac{μ^{2}(m)}{ϕ(m)} ∑_{{s≤ \frac{R}{ℓ} \atop (s,q)=1}} \frac{μ^{2}(s)}{ϕ(s)}

= \frac{N}{ϕ(q)}(log R + c + v(q)) − \frac{N}{q} ∑_{ℓ|q} \frac{μ^{2}(ℓ)}{ϕ(ℓ)} log ℓ + O(\frac{Nw(q)}{ϕ(q)\sqrt{R}} ∑_{ℓ|q} \frac{μ^{2}(ℓ)\sqrt{ℓ}}{ϕ(ℓ)})

= \frac{N}{ϕ(q)}(log R + c) + O(\frac{N}{ϕ(q)\sqrt{R}} ∏_{p|q}(1 + \frac{1}{\sqrt{p}})(1 + \frac{\sqrt{p}}{p − 1})).  (4.27)

The last product has logarithm

∑_{p|q} log(1 + \frac{1}{\sqrt{p}}) + log(1 + \frac{\sqrt{p}}{p − 1}) ≤ 2 ∑_{p|q} \frac{1}{\sqrt{p}} + O(1) ≪ \frac{\sqrt{log q}}{log log 3q} ,  (4.28)

by (3.6). Hence the first term of (4.25) is

\frac{N}{ϕ(q)}(log R + c) + O(\frac{N}{ϕ(q)R^{\frac{1}{2} −ϵ}})  (4.29)

for any arbitrarily small and fixed ϵ > 0. Using

∑_{{\frac{R}{m} <s≤ \frac{R}{ℓ} \atop (s,q)=1}} \frac{μ^{2}(s)}{ϕ(s)} = \frac{ϕ(q)}{q} log \frac{m}{ℓ} + O(w(q)\sqrt{\frac{m}{R}}),  (4.30)

which is implied by Lemma 3, the second term of (4.25) is expressed as

\frac{Nϕ(q)}{q^{2}} ∑_{ℓ|q} \frac{μ^{2}(ℓ)}{ϕ(ℓ)} ∑_{{m|q \atop m>ℓ}} \frac{μ^{2}(m)}{ϕ(m)} log m − \frac{Nϕ(q)}{q^{2}} ∑_{ℓ|q} \frac{μ^{2}(ℓ)}{ϕ(ℓ)} log ℓ ∑_{{m|q \atop m>ℓ}} \frac{μ^{2}(m)}{ϕ(m)}

+ O(\frac{Nw(q)}{q\sqrt{R}} ∑_{ℓ|q} \frac{μ^{2}(ℓ)}{ϕ(ℓ)} ∑_{{m|q \atop m>ℓ}} \frac{μ^{2}(m)}{ϕ(m)}\sqrt{m}).  (4.31)

Each term of (4.31) is majorized by deleting the restriction m > ℓ. Then, by (4.26)

and (3.8), the first two terms are each ≪ \frac{Nv(q)}{ϕ(q)}, and the error term is the same as

that of (4.27). Hence we have shown (4.11).

The j ≠ 0 case: Since

∑_{{s=1 \atop (s,q)=1}}^{∞} \frac{μ(s)}{ϕ^{2}(s)}μ((s, j))ϕ((s, j)) = ∏_{{p|j \atop p∤q}}(1 + \frac{1}{p − 1}) ∏_{p∤jq}(1 − \frac{1}{(p − 1)^{2}} ) = 𝔖(jq)\frac{ϕ(q)}{q},

(4.32)

by (3.10) and (3.11), (4.23) is

\frac{N}{ϕ(q)}𝔖(jq) + O(\frac{N}{q} ∑_{ℓ|q} \frac{μ^{2}(ℓ)}{ϕ(ℓ)} ∑_{m|q} \frac{μ^{2}(m)}{ϕ(m)} ∑_{s>min( \frac{R}{ℓ} , \frac{R}{m})} \frac{μ^{2}(s)μ^{2}((s, j))ϕ((s, j))}{ϕ^{2}(s)}).

(4.33)

The sum over s was encountered before in (3.13) and majorized as in (3.9), so the O-term in (4.33) is

≪ \frac{N}{qR} \frac{j d(j)}{ϕ(j)} ∑_{ℓ|q} \frac{μ^{2}(ℓ)}{ϕ(ℓ)} ∑_{m|q} \frac{μ^{2}(m)}{ϕ(m)} max(ℓ, m)

≪ \frac{Ng(q)}{ϕ(q)R} \frac{j d(j)}{ϕ(j)}.  (4.34)

This completes the proof of Proposition 4.

We now return to the proof of the theorem. The relevant contributions to (2.7) are

∑_{{x<n≤2x+h \atop n≡a(mod q)}} λ_{R}^{2}(n)f(n, x, h) = \frac{xh}{ϕ(q)}(log R + c + O(v(q)) + O(R^{− \frac{1}{2} +ϵ})) + O(hR^{2}),

(4.35)

and

∑_{{0<k≤h \atop q|k}} ∑_{{x<n≤2x+h−k \atop n≡a(mod q)}} λ_{R}(n)λ_{R}(n + k)f(n, x, h − k)  (4.36)

= \frac{x}{ϕ(q)} ∑_{0<j≤ \frac{h}{q}} 𝔖(jq) + O(\frac{xh}{Rϕ(q)}g(q) ∑_{j≤ \frac{h}{q}} \frac{jd(j)}{ϕ(j)})

= \frac{1}{2} \frac{xh^{2}}{ϕ^{2}(q)} − \frac{1}{2} \frac{xh}{ϕ(q)} log \frac{h}{q} + O(\frac{xh}{ϕ(q)}(log log 3q)^{3})+ O(\frac{xh^{2}g(q)}{Rqϕ(q)} log \frac{2h}{q})+O(\frac{h^{2}R^{2}}{q}).

Now we put together equations (2.2), (2.5), (2.6), (2.7), (4.3), (4.5), (4.35), (4.36), subject to (1.3), (4.2) and qR^{2} ≤ x, to obtain

I(x, h, q, a) ≥\frac{xh}{ϕ(q)} log \frac{Rq}{h} + O(\frac{xh}{ϕ(q)}(log log 3q)^{3}) + O(\frac{xh^{2}}{Rϕ(q)}(\frac{d(q)}{ϕ(q)} + \frac{g(q)}{q}))

+ O(\frac{x^{\frac{1}{2}} h^{\frac{3}{2}} R log^{2} x}{q^{\frac{1}{2}}}) + O(\frac{x^{\frac{1}{2}} h^{2}R^{\frac{1}{2}} log^{2} x}{ϕ(q)}) + O(\frac{h^{2}R^{2}}{q}). (4.37)

Recall that d(q) and g(q) are both ≪ q^{ϵ}. Here we pick

R = (\frac{x}{hq log^{4} x})^{\frac{1}{2}} .  (4.38)

This choice of R makes all the error terms o(\frac{xh}{q}) provided that h ≤ (xq)^{\frac{1}{3} −ϵ} .

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DAG:Department of Mathematics and Computer Science, San Jose State University, San Jose, CA 95192, USA

E-mail address: goldston@mathcs.sjsu.edu

CYY:Department of Mathematics, Bilkent University, Ankara 06533, Turkey

E-mail address: yalcin@fen.bilkent.edu.tr