The number of Goldbach representations of an integer

We prove the following result: Let $N \geq 2$ and assume the Riemann Hypothesis (RH) holds. Then \[ \sum_{n=1}^{N} R(n) =\frac{N^{2}}{2} -2 \sum_{\rho} \frac{N^{\rho + 1}}{\rho (\rho + 1)} + O(N \log^{3}N), \] where $\rho=1/2+i\gamma$ runs over the non-trivial zeros of the Riemann zeta function $\zeta(s)$.


Introduction
Let Λ be the von Mangoldt function and be the counting function for the Goldbach numbers. This paper is devoted to study the behaviour of the average order of magnitude of R(n) for n ∈ [1, N], where N is a large integer. We have the following Theorem 1. Let N ≥ 2 and assume the Riemann Hypothesis (RH) holds. Then where ρ = 1/2 + iγ runs over the non-trivial zeros of the Riemann zeta function ζ(s).
The first result of this kind was proved in 1991 by Fujii who subsequently improved it (see [4]- [5]- [6]) until reaching the error term O (N log N) 4/3 . Then Granville [8]- [9] gave an alternative proof of the same result and, finally, Bhowmik and Schlage-Puchta [2] were able to reach the error term O N log 5 N ; in [2] they also proved that the error term is Ω(N log log N).
Our result improves the upper bound in Bhowmik and Schlage-Puchta [2] by a factor log 2 N. In fact, this seems to be the limit of the method in the current state of the circlemethod technology: see the remark after the proof.
If one admits the presence of some suitable weight in our average, this loss can be avoided. For example, using the Fejér weight we could work with L(N; The key property is that, for 1/N < |α| ≤ 1/2, the function L(N; α) decays as α −2 instead of |α| −1 and so the dissection argument in (26) is now more efficient and does not cause any loss of logs. Such a phenomenon is well-known from the literature about the existence of Goldbach numbers in short intervals, see, e.g., Languasco and Perelli [12].
In fact we will obtain Theorem 1 as a consequence of a weighted result. Letting ψ(x) = m≤x Λ(m), we have Theorem 2. Let 2 ≤ y ≤ N and assume the Riemann Hypothesis (RH) holds. Then The key reason why we are able to derive Theorem 1 from Theorem 2 via partial summation is that the exponential weight in (1) just varies in the range [e −1/N , e −1 ] and so it does not change the order of magnitude of the functions involved.
We will use the original Hardy and Littlewood [10] circle method setting, i.e., the weighted exponential sum where e(x) = exp(2πix), since it lets us avoid the use of Gallagher's Lemma (Lemma 1 of [7]) and hence, in this conditional case, it gives slightly sharper results, see Lemma 1 below. Such a function was also used by Linnik [13,14]. The new ingredient in this paper is Lemma 5 below in which we unconditionally detect the existence of the term −2 ρ N ρ+1 /(ρ(ρ + 1)) by solving an arithmetic problem connected with the original one (see eq. (11) below). In the previously mentioned papers this is obtained applying the explicit formula for ψ(n) twice. The ideas that lead to Theorem 1 and 2 work also for the sum of k ≥ 3 primes, i.e., for the function We can prove the following Theorem 3. Let k ≥ 3 be an integer, N ≥ 2 and assume the Riemann Hypothesis (RH) holds. Then where ρ = 1/2 + iγ runs over the non-trivial zeros of ζ(s).
The proof of Theorem 3 is completely similar to the one of Theorems 1 and 2. We just remark that the main differences are in the use of the explicit formula for where j is a non-negative integer, and of the following version of Lemma 1 of [11]: Another connected problem we can address with this technique is a short-interval version of Theorem 1. We can prove the following where ρ = 1/2 + iγ runs over the non-trivial zeros of ζ(s).
Also in this case we do not give a proof of Theorem 4; we just remark that the main difference is in the use of the exponential sum N +H n=N e(nα) instead of N n=1 e(nα). Acknowledgments. We would like to thank Alberto Perelli for a discussion.

Setting of the circle method
For brevity, throughout the paper we write The first lemma is a L 2 -estimate for the difference S(α) − 1/z.
Lemma 1 (Languasco and Perelli [12]). Assume RH. Let N be a sufficiently large integer and z be as in (3).
This follows immediately from the proof of Theorem 1 of [12] since the quantity we would like to estimate here is Lemma 1 is the main reason why we use S(α) instead of its truncated form S(α) = N n=1 Λ(n)e(nα) as in Bhowmik and Schlage-Puchta [2]. In fact Lemma 1 lets us avoid the use of Gallagher's Lemma [7] which leads to a loss of a factor log 2 N in the final estimate (compare Lemma 1 with Lemma 4 of [2]). For a similar phenomenon in a slightly different situation see also Languasco [11].
The next four lemmas do not depend on RH. By the residue theorem one can obtain Lemma 2 (Eq. (29) of [12]). Let N ≥ 2, 1 ≤ n ≤ N and z be as in (3). We have uniformly for every n ≤ N.
Lemma 3. Let N be a sufficiently large integer and z be as in (3). We have Proof. By the Parseval theorem and the Prime Number Theorem we have Recalling that the equation at the beginning of page 318 of [12] implies Proof. We recall the that the function w/(e w − 1) has a power-series expansion with radius of convergence 2π (see for example Apostol [1], page 264). In particular, uniformly for |w| ≤ 4 < 2π we have w/(e w − 1) = 1 + O(|w|). Since z satisfies (3) Lemma 5. Let N be a large integer, 2 ≤ y ≤ N and z be as in (3). We have We remark that Lemma 5 is unconditional and hence it implies, using also Lemma 6, that the ability of detecting the term depending on the zeros of the Riemann ζ-function in Theorem 1 does not depend on RH. Proof. Writing R(α) = S(α) − 1/z, by Lemma 4 we have since, by the Parseval theorem and Lemma 3, the error term above is Again by Lemma 4, we have and hence (7) implies The Cauchy-Schwarz inequality and the Parseval theorem imply that By (8)-(9), we have Now, by (2) and (4), we can write so that since the condition m 1 + m 2 = n implies that both variables are < n. Now ψ(n) = ψ(n − 1) + Λ(n), so that Remarking that since the integral on (0, 2] gives a contribution O(1). For t ≥ 2 we will use the explicit formula (see eq. (9)-(10) of §17 of Davenport [3]) where The term − ζ ′ ζ (0) − 1 2 log 1 − 1 t 2 gives a contribution O(M) to the integral over [2, M] in (12). We need now a L 1 estimate of the error term defined in (14). Let The first term in (14) gives a total contribution to E(M, Z) which is The second term in (14) gives a total contribution to E(M, Z) which is Combining (15)-(17), for Z = M log 2 M we have that Inserting now (13) and (18) into (12) we obtain The lemma follows from (19), by remarking that

Proof of Theorem 1
We will get Theorem 1 as a consequence of Theorem 2. By partial summation we have Theorem 1 now follows inserting Lemma 6 and the identity N n=1 n = N 2 /2 + O(N) in (21).

Proof of Theorem 2
Let 2 ≤ y ≤ N. We first recall the definition of the singular series of the Goldbach problem: S(k) = 0 for k odd and for k even. Hence, using the well known estimate R(n) ≪ nS(n) ≪ n log log n, we remark that y n=1 R(n) − (2ψ(n) − n) e −n/N ≪ y n=1 n log log n ≪ y 2 log log y.
So it is clear that (1) say.
Evaluation of I 1 (y). By Lemma 2 we obtain Estimation of I 2 (y). By (6) of Lemma 5 we obtain Estimation of I 3 (y). Using (5) and Lemma 1 we have that End of the proof. Inserting Combining (22) and (27) we get that Theorem 2 is proved.
Remark. Let Then f satisfies both Lemma 1 and Lemma 3 in the sense that for y = N 1/2 and sufficiently large N. This means that the crucial bound for I 3 (y) in (26) is essentially optimal in the present state of knowledge, and that it can not be improved without deeper information on S(α) − z −1 , such as the stronger analogue of Lemma 1 that follows from a suitable form of Montgomery's Pair-Correlation Conjecture.