On the Waring-Goldbach problem on average

ON THE WARING-GOLDBACH PROBLEM ON AVERAGE ALESSANDRO LANGUASCO

arXiv:1806.05861v1 [math.NT] 15 Jun 2018

Abstract. Let s, ℓ be two integers such that 2 ≤ s ≤ ℓ − 1, ℓ ≥ 3. We prove that a suitable Ís asymptotic formula for the average number of representations of integers n = i=1 pℓi , where pi , i = 1, . . . , s, are prime numbers, holds in short intervals.

1. Introduction Let s ≥ 2, ℓ ≥ 1 be two integers, N be a sufficiently large integer and 1 ≤ H ≤ N an integer. Let Õ rs,ℓ (n) = log p1 · · · log ps, (1) i=1,...,s n=pℓ1 +···+pℓs

be the number of representation of an integer as a sum of exactly s summands each one is a ℓ-th prime power. The problem of obtaining an asymptotic formula for rs,ℓ (n) is usually called the Waring-Goldbach problem. The history about the results on such a problem is a very long one; we refer to the surveys of Vaughan-Wooley [11] and Kumchev-Tolev [3] for an overview. A simpler problem is to study the order of magnitude for an average of rs,ℓ (n) because the averaging procedure let us to gain non-trivial information in cases in which the classical approaches fail, for example when s ≤ 4ℓ(log ℓ + (1/2) log log ℓ + O (1)) for ℓ large or s ≤ H(ℓ) for 4 ≤ s ≤ 10, where H(4) ≤ 14, H(5) ≤ 21, H(6) ≤ 33, H(7) ≤ 47, H(8) ≤ 63, H(9) ≤ 83, H(10) ≤ 107, according to [11], page 20. Recently Cantarini, Gambini and Zaccagnini [1] Í N+H proved that a suitable asymptotic formula in short intervals holds for n=N+1 rs,ℓ (n) when s = ℓ + 1, ℓ ≥ 2 and s = ℓ, ℓ ≥ 2, thus generalizing previous results by Languasco-Zaccagnini [7] (s = 4, ℓ = 3) and [4] (s = 2, ℓ = 2). Here we restrict our attention to the more difficult case in which we have less summands, i.e., when 2 ≤ s ≤ ℓ − 1, ℓ ≥ 3. We also remark that a more general binary problem is treated by Languasco-Zaccagnini in [6] and [8]. Our first result is Theorem 1. Let s, ℓ be two integers such that 2 ≤ s ≤ ℓ − 1, ℓ ≥ 3, N ≥ 2, 1 ≤ H ≤ N be integers. Then, for every ε > 0, there exists C = C(ε) > 0 such that N+H Õ

rs,ℓ (n) =

n=N+1

log N 1/3 Γ(1 + 1/ℓ)s HN s/ℓ−1 + Os,ℓ HN s/ℓ−1 exp −C Γ(s/ℓ) log log N

as N → ∞,

uniformly for N 1−5/(6ℓ)+ε ≤ H ≤ N 1−ε , where Γ is Euler’s function. As an immediate consequence of Theorem 1 we can say that, for N sufficiently large, every interval of size larger than N 1−5/(6ℓ)+ε contains the expected amount of integers which are a sum of exactly s summands, 2 ≤ s ≤ ℓ − 1, each one is a ℓ-th prime power, ℓ ≥ 3. 2010 Mathematics Subject Classification. Primary 11P32; Secondary 11P55, 11P05. Key words and phrases. Waring-Goldbach problem, Hardy-Littlewood method. 1

2

ALESSANDRO LANGUASCO

We remark that the uniformity level for H in Theorem 1 is the expected optimal one given the known density estimates for the non-trivial zeros of the Riemann zeta-function. Assuming the Riemann Hypothesis holds, we can improve the uniformity range of H since in this case Lemma 4 below holds in the whole unit interval for ξ. Theorem 2. Let s, ℓ be two integers such that 2 ≤ s ≤ ℓ − 1, ℓ ≥ 3, N ≥ 2, 1 ≤ H ≤ N be integers and assume the Riemann Hypothesis (RH) holds. Then N+H Õ Γ(1 + 1/ℓ)s HN s/ℓ−1 + Os,ℓ (H 2 N s/ℓ−2 + H 1/2 N s/ℓ−1/(2ℓ)−1/2 (log N)3 ) rs,ℓ (n) = Γ(s/ℓ) n=N+1 as N → ∞, uniformly for ∞(N 1−1/ℓ (log N)6 ) ≤ H ≤ o (N), where f = ∞(g) means g = o ( f ) and Γ is Euler’s function.

As an immediate consequence of Theorem 2 we can say that, for N sufficiently large, every interval of size larger than N 1−1/ℓ+ε contains the expected amount of integers which are a sum of exactly s summands, 2 ≤ s ≤ ℓ − 1, each one is a ℓ-th prime power, ℓ ≥ 3. We remark that in this case the H-level is essentially optimal given the spacing of the sequence. In both the proofs of Theorems 1-2 we will use the circle method with the original HardyLittlewood generating functions to exploit the easier main term treatment they allow (comparing with the one which would follow using Lemmas 2.3 and 2.9 of Vaughan [10]). Key tools in the proofs of Theorems 1-2 are Lemma 5 and 4, which, respectively, give suitable estimates on the truncated mean-square average for the exponential sum over prime powers Seℓ (α) and for the error term in its first order approximation which, by Lemma 2, is directly connected with the non-trivial zeros of the Riemann zeta-function. 2. Settings Let s ≥ 2, ℓ ≥ 1 be two integers, N be a sufficiently large integer, 1 ≤ H ≤ N an integer, e(α) = e2πiα , α ∈ [−1/2, 1/2], L = log N, z = 1/N − 2πiα, ∞ ∞ Õ Õ ℓ ℓ eℓ (α) := Seℓ (α) := Λ(n)e−n /N e(nℓ α) and V log p e−p /N e(pℓ α). n=1

p=2

We remark that

|z| −1 ≪ min N, |α| −1 and Seℓ (α) ≪ℓ N 1/ℓ, (2) where the second inequality is a direct consequence of the Prime Number Theorem. We further set H Õ U(α, H) := e(mα) m=1

and, moreover, we also have the well known numerically explicit inequality |U(α, H)| ≤ min H; |α| −1 .

We list now the needed preliminary results.

eℓ (α)| ≪ℓ N 1/(2ℓ) . Lemma 1 (Lemma 3 of [4]). Let ℓ ≥ 1 be an integer. Then | Seℓ (α) − V

Lemma 2 (Lemma 2 of [7]). Let ℓ ≥ 1 be an integer, N ≥ 2 and α ∈ [−1/2, 1/2]. Then Γ(1/ℓ) 1 Õ −ρ/ℓ z Γ(ρ/ℓ) + Oℓ (1), − Seℓ (α) = ℓ ρ ℓz 1/ℓ

(3)

WARING-GOLDBACH PROBLEM IN SHORT INTERVALS

3

where ρ = β + iγ runs over the non-trivial zeros of ζ (s) and Γ is Euler’s function. Lemma 3. Let N be a positive integer and µ > 0. Then, uniformly for n ≥ 1 and X > 0, we have ∫ X 1 n µ−1 + Oµ , z −µ e(−nα) dα = e−n/N Γ(µ) nX µ −X where Γ is Euler’s function. Proof. We remark that the proof is identical to the one of Lemma 4 of [5] but in that case we just stated the lemma in the particular case X = 1/2. Now we need its full strength and hence, for completeness, we rewrite its proof. We start with the identity ∫ 1 D s−1 e−aD eiDu du = , 2π R (a + iu)s Γ(s) which is valid for σ = ℜ(s) > 0 and a ∈ C with ℜ(a) > 0 and D > 0. Letting u = −2πα and taking s = µ, D = n and a = N −1 we find ∫ ∫ n µ−1 e−n/N e(−nα) −µ dα = z e(−nα) dα = . −1 − 2πiα) µ Γ(µ) R R (N For 0 < X < Y let ∫ Y eiDu du. I(X, Y ) = µ X (a + iu) An integration by parts yields ∫ h 1 eiDu µ Y eiDu i Y I(X, Y ) = du. + iD (a + iu) µ X D X (a + iu) µ+1 Since a > 0, the first summand is ≪ µ D−1 X −µ , uniformly. The second summand is ∫ µ Y du ≪ ≪ µ D−1 X −µ . D X u µ+1 The result follows.

Lemma 4 (Lemma 4 of [7]). Let ε be an arbitrarily small positive constant, ℓ ≥ 1 be an integer, N be a suﬃciently large integer and L = log N. Then there exists a positive constant c1 = c1 (ε), which does not depend on ℓ, such that ∫ ξ L 1/3 Γ(1/ℓ) 2 e 2/ℓ−1 exp − c1 dα ≪ℓ N Sℓ (α) − log L ℓz 1/ℓ −ξ uniformly for 0 ≤ ξ < N −1+5/(6ℓ)−ε . Assuming RH we get ∫ ξ Γ(1/ℓ) 2 e Sℓ (α) − dα ≪ℓ N 1/ℓ ξ L 2 1/ℓ ℓz −ξ

uniformly for 0 ≤ ξ ≤ 1/2.

The following two lemmas hold for a real index k instead of an integral one ℓ. The new ingredient we are using here is based on a Tolev’s lemma [9]. Lemma 5. Let k > 1, n ∈ N and τ > 0. We have ∫ ∫ τ 3 2 1/k 2/k−1 | Sek (α)| dα ≪k τN + N L and −τ

τ

−τ

ek (α)| 2 dα ≪k τN 1/k + N 2/k−1 L 3 . |V

4


Proof. We just prove the first part since the second one follows immediately by remarking that the primes are supported on a thinner set than the prime powers. Let P = (2N L/k)1/k . A direct Í k estimate gives Sek (α) = n≤P Λ(n)e−n /N e(n k α) + Ok (L 1/k ). Recalling that the Prime Number Í Theorem implies Sk (α; t) := n≤t Λ(n)e(n k α) ≪ t, a partial integration argument gives ∫ P Õ k k −n k /N k t k−1 e−t /N Sk (α; t) dt + Ok (L 1/k ). Λ(n)e e(n α) = − N 1 n≤P Using the inequality (|a| + |b|)2 ≪ |a| 2 + |b| 2 , Cauchy-Schwarz inequality and interchanging the integrals, we get that ∫ τ ∫ P ∫ τ 2 k 1 2 e t k−1 e−t /N Sk (α; t) dt dα + L 2/k | Sk (α)| dα ≪k −τ N 1 −τ ∫ τ ∫ P ∫ P 1 k−1 −t k /N k−1 −t k /N ≪k 2 |Sk (α; t)| 2dα dt + L 2/k . t e t e dt N −τ 1 1 Í Lemma 7 of∫Tolev [9] in the form given in Lemma 5 of [2] on Sk (α; t) = n≤t Λ(n)e(n k α) τ implies that −τ |Sk (α; t)| 2 dα ≪k τt + t 2−k (log t)3 . Using such an estimate and remarking ∫P k that 1 t k−1 e−t /N dt ≪k N, we obtain that ∫ τ ∫ P 1 k 2 e | Sk (α)| dα ≪k τt + t 2−k t k−1 e−t /N (log t)3 dt + L 2/k N 1 −τ ≪k τN 1/k + N 2/k−1 L 3 by a direct computation. This proves the first part of the lemma. The last lemma is a consequence of Lemma 5.

Lemma 6. Let N ∈ N, k > 1, u ≥ 1 and N −u ≤ ω ≤ N 1/k−1 /L. Let further I(ω) := [−1/2, −ω] ∪ [ω, 1/2]. We have ∫ ∫ 2/k−1 N 2/k−1 3 2 dα e ek (α)| 2 dα ≪k N | Sk (α)| ≪k L and |V L3 . |α| ω |α| ω I(ω) I(ω) Let further assume the Riemann Hypothesis, ℓ ≥ 1 be an integer and N −u ≤ η ≤ 1/2. Then ∫ Γ(1/ℓ) 2 dα e ≪ℓ N 1/ℓ L 3 . Sℓ (α) − ℓz 1/ℓ |α| I(η)

Proof. By partial integration and Lemma 5 we get that ∫ 1/2 ∫ ξ ∫ 1/2 ∫ 1/2 ∫ dξ 1 ω e 2 dα 2 2 e e | Sek (α)| 2 dα 2 | Sk (α)| | Sk (α)| dα + | Sk (α)| dα + ≪ α ω −ω ξ −ξ ω ω −1/2 ∫ 1/2 L3 ξ N 1/k + N 2/k−1 ≪k dξ ωN 1/k + N 2/k−1 + N 1/k L 3 + L 3 ω ξ2 ω N 2/k−1 3 N 2/k−1 3 ≪k N 1/k L 3 | log(2ω)| + L ≪k L ω ω

since N −u ≤ ω ≤ N 1/k−1 /L. A similar computation proves the result in [−1/2, −ω] too. The ek (α) can be obtained analogously. The third estimate requires Lemma 4 instead estimate on V of Lemma 5 but follows analogously.


5

3. Proof of Theorem 1 Let s, ℓ be two integers such that 2 ≤ s ≤ ℓ − 1 and ℓ ≥ 3. Let further H > 2B, where B = Nε.

(4)

Letting I(v) := [−1/2, −v] ∪ [v, 1/2], where 0 < v < 1/2, and recalling (1), we have ∫ 1/2 N+H Õ −n/N eℓ (α)s U(−α, H)e(−Nα) dα V e rs,ℓ (n) = −1/2

n=N+1

=

∫

B/H

e Sℓ (α) U(−α, H)e(−Nα) dα + s

−B/H

+

∫

1/2

−1/2

∫

I(B/H)

Seℓ (α)s U(−α, H)e(−Nα) dα

eℓ (α)s − Seℓ (α)s U(−α, H)e(−Nα) dα = I1 + I2 + I3, V

(5)

say. We need to split the unit interval in this way because in the unconditional case Lemma 4 works only on a subset of [−1/2, 1/2]; in the remaining part of the interval we need a different estimate based on Lemma 6. By (3), the ideal splitting level should be with B = 1 but that would not be good enough for controlling the estimate in eq. (6) below since the expected order of magnitude of the main term is essentially cs,ℓ HN s/ℓ−1 , where cs,ℓ = Γ(1 + 1/ℓ)s /Γ(s/ℓ). Hence we have to choose a larger B and, in fact, the choice made in eq. (4) is sufficiently good for our purposes. We also remark that I3 collects the difference between the contributions e we’ll see that from the exponential sums over prime powers e S and the one over primes V; such contributions are negligible. Finally, we will see that I1 contains the “main term” of the problem together with several error terms connected with the distribution of zeros of the Riemann zeta-function; the main term will be evaluated using Lemma 3, while the long part needed to estimate the error terms makes use of Lemmas 4 and 5. At the end of the proof we will see how to get rid of the e−n/N weight we have on the left-hand side of (5). Now we evaluate these terms. 3.1. Estimation of I2 . Since s ≥ 2, using (2)-(3) and Lemma 6 with ω = B/H, we have that ∫ dα H s−2 I2 ≪ (max | Seℓ (α)| ) | Seℓ (α)| 2 ≪s,ℓ N s/ℓ−1 L 3, (6) |α| B I(B/H) provided that H ≫ N 1−1/ℓ BL.

3.2. Estimation of I3 . Clearly eℓ (α)s − Seℓ (α)s | ≪s,ℓ |V eℓ (α) − e eℓ (α)| + | Seℓ (α)|)s−1 |V Sℓ (α)|(|V eℓ (α) − e eℓ (α)| s−1 ; | Seℓ (α)| s−1 ). ≪s,ℓ |V Sℓ (α)| max(|V

Hence by Lemma 1 we have ∫ 1/2 1/(2ℓ) eℓ (α)| s−1 + | Seℓ (α)| s−1 |U(−α, H)| dα = N 1/(2ℓ) (K1 + K2 ), |V I3 ≪s,ℓ N

(7)

−1/2

say. If s ≥ 3, by (2)-(3), Lemmas 5-6 with τ = 1/H, we get ∫ 1/2 s−3 e | Seℓ (α)| 2 |U(−α, H)| dα K2 ≪ (max | Sℓ (α)| ) ∫ (s−3)/ℓ ≪ℓ N H

−1/2 1/H

−1/H

| Seℓ (α)| 2 dα +

∫

I(1/H)

| Seℓ (α)| 2

dα ≪ℓ HN s/ℓ−1/ℓ−1 L 3, |α|

(8)

6


provided that H ≫ N 1−1/ℓ L. If s = 2, by (3), the Cauchy-Schwarz estimate, Lemmas 5-6 with τ = 1/H, we get 1/2 1/2 ∫ 1/2 ∫ 1/2 2 e |U(−α, H)| dα | Sℓ (α)| |U(−α, H)| dα K2 ≪ −1/2

∫ ≪ℓ H

≪ℓ H

1/H

| Seℓ (α)| dα + 2

−1/H 1/2 1/ℓ−1/2 2

N

−1/2

∫

I(1/H)

L ,

| Seℓ (α)|

2 dα

|α|

1/2 ∫ H

1/H

−1/H

provided that H ≫ N 1−1/ℓ L. Analogous computations give ( HN s/ℓ−1/ℓ−1 L 3 if s ≥ 3 K1 ≪ℓ H 1/2 N 1/ℓ−1/2 L 2 if s = 2, provided that H ≫ N 1−1/ℓ L. By (7)-(10), we can finally write ( HN s/ℓ−1/(2ℓ)−1 L 3 if s ≥ 3 I3 ≪s,ℓ H 1/2 N 3/(2ℓ)−1/2 L 2 if s = 2,

dα +

∫

I(1/H)

dα 1/2 |α|

(9)

(10)

(11)

provided that H ≫ N 1−1/ℓ L. 3.3. Evaluation of I1 . Since Γ(1 + 1/ℓ) = Γ(1/ℓ)/ℓ, we let ds,ℓ := Γ(1 + 1/ℓ)s and cs,ℓ := Γ(1 + 1/ℓ)s /Γ(s/ℓ), say, and we have that ∫ B/H ∫ B/H ds,ℓ ds,ℓ s e I1 = Sℓ (α) − s/ℓ U(−α, H)e(−Nα) dα U(−α, H)e(−Nα) dα + s/ℓ z −B/H z −B/H = J1 + J2, (12) say. By Lemma 3 and a direct calculation we obtain N+H cs,ℓ Õ s/ℓ−1 H H s/ℓ = n + Os,ℓ H 2 N s/ℓ−2 + J1 = cs,ℓ n e + Os,ℓ N B e n=N+1 N B n=N+1 cs,ℓ H H s/ℓ = . (13) HN s/ℓ−1 + Os,ℓ H 2 N s/ℓ−2 + e N B N+H Õ

H H s/ℓ

s/ℓ−1 −n/N

3.4. Evaluation of J2 . From now on, we denote mℓ (z) := Γ(1 + 1/ℓ)z −1/ℓ , so that ds,ℓ /z s/ℓ = mℓ (z)s and Eeℓ (α) := Seℓ (α) − mℓ (z). Using f 2 − g 2 = 2g( f − g) + ( f − g)2 we obtain ( eℓ (α) Ís−1 Seℓ (α)s−1− j mℓ (z) j E if s is odd j=0 s s (14) Seℓ (α) − mℓ (z) = Í s/2−1 s−2−2 j 2 j 2 Sℓ (α) mℓ (z) if s is even. (2mℓ (z)Eeℓ (α) + Eeℓ (α) ) j=0 e We define

E :=

∫

L 1/3 , A(d) := exp d log L B/H

−B/H

S :=

∫

| Eeℓ (α)| 2 dα ≪ℓ N 2/ℓ−1 A(−c1 ),

B/H

−B/H

| Seℓ (α)| 2 dα ≪ℓ N 2/ℓ−1 L 3,

(15) (16) (17)


7

where d is a real constant and in which the estimates follow respectively using Lemma 4 with ξ = B/H and H ≥ BN 1−5/(6ℓ)+ε for (16) and Lemma 5 with τ = B/H and H ≥ BN 1−1/ℓ for (17). Moreover using (2), for every integral u ≥ 2 we get |mℓ (z)| u ≪u,ℓ N (u−2)/ℓ |mℓ (z)| 2 ≪u,ℓ N (u−2)/ℓ | Seℓ (α)| 2 + | Eeℓ (α)| 2 . (18) Hence we have, using also (16)-(17), that ∫ B/H |mℓ (z)| u dα ≪u,ℓ N (u−2)/ℓ (S + E ) ≪u,ℓ N u/ℓ−1 L 3 . M (u) :=

(19)

−B/H

Assume s is odd. Inserting (14) into (12) and using (3) we get that s−1 ∫ B/H s−1 Õ Õ s−1− j j e e J2 ≪ H | Eℓ (α)|| Sℓ (α) ||mℓ (z) | dα = H K j, j=0

−B/H

(20)

j=0

say. If 0 ≤ j ≤ s − 2, using the Cauchy-Schwarz inequality, (2) and (16)-(17) we have ∫ B/H j s−2− j e | Eeℓ (α)|| Seℓ (α)| dα K j ≪ (max |mℓ (z) |)(max | Sℓ (α) |) −B/H

≪s,ℓ N

(s−3)/ℓ

E

1/2

S

1/2

≪s,ℓ N

s/ℓ−1

A(−c1 /4),

(21)

provided that H ≥ BN 1−5/(6ℓ)+ε . If j = s − 1, using the Cauchy-Schwarz inequality, (16) and (19) we have ∫ B/H | Eeℓ (α)||mℓ (z)| s−1 dα ≪ E 1/2 M (2s − 2)1/2 ≪s,ℓ N s/ℓ−1 A(−c1 /4), Ks−1 ≪ (22) −B/H

provided that H ≥ BN 1−5/(6ℓ)+ε . Summarizing, using (20)-(22), we obtain, for s odd, that J2 ≪s,ℓ HN s/ℓ−1 A(−c1 /4),

(23)

provided that H ≥ BN 1−5/(6ℓ)+ε . Assume s is even. Inserting (14) into (12) and using (3) we get that J2 ≪ H

s/2−1 Õ ∫ B/H j=0

+H

−B/H

eℓ (α)|| Seℓ (α)s−2−2 j ||mℓ (z)2 j+1 | dα |E

s/2−1 Õ ∫ B/H j=0

−B/H

| Eeℓ (α)| 2 | Seℓ (α)s−2−2 j ||mℓ (z)2 j | dα = H

s/2−1 Õ

(U j + Vj ),

(24)

j=0

say. Now we estimate the Vj ’s. Recalling (2) and (16), we have

Vj ≪ (max |mℓ (z)2 j |)(max | Seℓ (α)s−2−2 j |) E ≪s,ℓ N s/ℓ−1 A(−c1 ),

(25)

provided that H ≥ BN 1−5/(6ℓ)+ε . Now we estimate the U j ’s. If s ≥ 4 and 0 ≤ j ≤ s/2 − 2, recalling (2), using the Cauchy-Schwarz inequality and (16)-(17), we obtain as in (21) that ∫ B/H 2 j+1 s−3−2 j eℓ (α)|| Seℓ (α)| dα e |E U j ≪ (max |mℓ (z) |)(max | Sℓ (α) |) −B/H

≪s,ℓ N (s−2)/ℓ E 1/2 S 1/2 ≪s,ℓ N s/ℓ−1 A(−c1 /4),

(26)

8


provided that H ≥ BN 1−5/(6ℓ)+ε . If j = s/2 − 1, s ≥ 2, recalling (2), using the Cauchy-Schwarz inequality, (16) and (19), we have as in (22) that ∫ B/H (27) | Eeℓ (α)||mℓ (z)| s−1 dα ≪ E 1/2 M (2s − 2)1/2 ≪s,ℓ N s/ℓ−1 A(−c1 /4), Us/2−1 ≪ −B/H

provided that H ≥ BN 1−5/(6ℓ)+ε . Summarizing, using (24)-(27), we obtain, for s even, that J2 ≪s,ℓ HN s/ℓ−1 A(−c1 /4),

(28)

provided that H ≥ BN 1−5/(6ℓ)+ε . By (12)-(13), (23) and (28) we have that, for every ε > 0, there exists c1 = c1 (ε) > 0 such that cs,ℓ I1 = HN s/ℓ−1 + Os,ℓ (HN s/ℓ−1 A(−c1 /4)), (29) e provided that H ≥ BN 1−5/(6ℓ)+ε . 3.5. Final words. By (5)-(6), (11) and (29) we have that, for every ε > 0, there exists c1 = c1 (ε) > 0 such that N+H Õ

cs,ℓ HN s/ℓ−1 e H H H s/ℓ , + Os,ℓ HN s/ℓ−1 A(−c1 /4) + N s/ℓ−1 L 4 + H 2 N s/ℓ−2 + B N B

e−n/N rs,ℓ (n) =

n=N+1

provided that BN 1−5/(6ℓ)+ε ≤ H ≤ N. The error terms are dominated by the first one assuming H ≤ N 1−ε and remarking B = N ε by (4). Hence we can write that, for every ε > 0, there exists C = C(ε) > 0 such that N+H Õ


n=N+1

cs,ℓ HN s/ℓ−1 + Os,ℓ (HN s/ℓ−1 A(−C)), e

(30)

provided that N 1−5/(6ℓ)+2ε ≤ H ≤ N 1−ε . From e−n/N = e−1 + O (H/N) for n ∈ [N + 1, N + H], 1 ≤ H ≤ N, we get that, for every ε > 0, there exists C = C(ε) > 0 such that N+H Õ

rs,ℓ (n) = cs,ℓ HN

n=N+1

s/ℓ−1

+ Os,ℓ (HN

s/ℓ−1

A(−C)) + Os,ℓ

H N+H Õ N

n=N+1

rs,ℓ (n) ,

provided that N 1−5/(6ℓ)+2ε ≤ H ≤ N 1−ε . Using en/N ≤ e2 and (30), the last error term is ≪s,ℓ H 2 N s/ℓ−2 . Hence we get that, for every ε > 0, there exists C = C(ε) > 0 such that N+H Õ

rs,ℓ (n) = cs,ℓ HN s/ℓ−1 + Os,ℓ (HN s/ℓ−1 A(−C)),

n=N+1

provided that N 1−5/(6ℓ)+2ε ≤ H ≤ N 1−ε . Theorem 1 follows by rescaling ε and recalling (15) and cs,ℓ = Γ(1 + 1/ℓ)ℓ /Γ(s/ℓ).


9

4. Proof of Theorem 2 From now on we assume the Riemann Hypothesis holds. The proof runs analogously to the unconditional case we described in section 3 but we can slightly simplify the approach since in this case Lemma 4 holds in the whole unit interval for ξ. Let s, ℓ be two integers such that 2 ≤ s ≤ ℓ −1 and ℓ ≥ 3. Recalling (1) and ds,ℓ := Γ(1+1/ℓ)s , we have ∫ 1/2 N+H Õ −n/N eℓ (α)s U(−α, H)e(−Nα) dα e rs,ℓ (n) = V n=N+1

−1/2

∫ 1/2 ds,ℓ U(−α, H) s e S (α) − e(−Nα) dα + = ds,ℓ U(−α, H)e(−Nα) dα ℓ z s/ℓ z s/ℓ −1/2 −1/2 ∫ 1/2 eℓ (α)s − Seℓ (α)s U(−α, H)e(−Nα) dα V + ∫

1/2

−1/2

= J1 + J2 + J3,

(31)

say. This time we collect in J3 the difference between the contributions from the exponential e in fact J3 is equal to I3 of section 3 and sums over prime powers Se and the one over primes V; we’ll see that it is a negligible term in this case too. J1 is directly linked with the “main term” of the problem while J2 is connected with the distribution of zeros of the Riemann zeta-function. At the end of the proof we will see how to get rid of the e−n/N weight we have on the left-hand side of (31). Now we evaluate these terms. 4.1. Evaluation of J1 . Recalling cs,ℓ := Γ(1 + 1/ℓ)s /Γ(s/ℓ), by Lemma 3 and a direct calculation we have N+H N+H H c H Õ Õ s,ℓ s/ℓ−1 −n/N = + H 2 N s/ℓ−2 J1 = cs,ℓ n s/ℓ−1 + Os,ℓ n e + Os,ℓ N e n=N+1 N n=N+1 H cs,ℓ = HN s/ℓ−1 + Os,ℓ + H 2 N s/ℓ−2 . (32) e N 4.2. Estimate of J2 . Recalling mℓ (z) := Γ(1+1/ℓ)z −1/ℓ , so that ds,ℓ /z s/ℓ = mℓ (z)s and Eeℓ (α) := Seℓ (α) − mℓ (z), we define ∫ 1/2 | Eeℓ (α)| 2 |U(−α, H)| dα ≪ℓ N 1/ℓ L 3, (33) E := −1/2

S :=

∫

1/2

−1/2

| Seℓ (α)| 2 |U(−α, H)| dα ≪ℓ HN 2/ℓ−1 L 3,

(34)

in which the estimates follow respectively using (3), Lemmas 4 and 6 with ξ = η = 1/H for (33) and Lemmas 5 and 6 with τ = ω = 1/H and H ≥ N 1−1/ℓ L for (34). Moreover, using (18), for every integral u ≥ 2, we have, using also (33)-(34), that ∫ 1/2 |mℓ (z)| u |U(−α, H)| dα ≪u,ℓ N (u−2)/ℓ (S + E) ≪u,ℓ HN u/ℓ−1 L 3, M(u) := (35) −1/2

since H ≥ N 1−1/ℓ L.

10


Using (14) for s odd we can write that s−1 ∫ 1/2 s−1 Õ Õ s−1− j j e e J2 ≪ | Eℓ (α)|| Sℓ (α) ||mℓ (z) ||U(−α, H)| dα = B j, −1/2

j=0

say. If 0 ≤ j ≤ s − 2, using the Cauchy-Schwarz inequality, (2) and (33)-(34) we have ∫ 1/2 j s−2− j | Eeℓ (α)|| Seℓ (α)||U(−α, H)| dα B j ≪ (max |mℓ (z) |)(max | Seℓ (α) |) ≪s,ℓ N

(s−2)/ℓ

(36)

j=0

1/2 1/2

E S

≪s,ℓ H

1/2

N

−1/2 s/ℓ−1/(2ℓ)−1/2 3

L ,

(37)

provided that H ≥ N 1−1/ℓ L. If j = s − 1 using the Cauchy-Schwarz inequality, (33) and (35) we have ∫ 1/2 | Eeℓ (α)||mℓ (z)| s−1 |U(−α, H)| dα ≪ E1/2 M1 (2s − 2)1/2 Bs−1 ≪ −1/2 1/2

≪s,ℓ H

N s/ℓ−1/(2ℓ)−1/2 L 3 .

(38)

Hence, using (36)-(38), we obtain, for s odd, that

J2 ≪s,ℓ H 1/2 N s/ℓ−1/(2ℓ)−1/2 L 3, provided that N 1−1/ℓ L ≤ H ≤ N. Assume s is even. Using (14) we get that s/2−1 Õ ∫ 1/2 eℓ (α)|| Seℓ (α)s−2−2 j ||mℓ (z)2 j+1 ||U(−α, H)| dα |E J2 ≪ j=0

+

(39)

−1/2

s/2−1 Õ ∫ 1/2 j=0

−1/2

eℓ (α)| 2 | Seℓ (α)s−2−2 j ||mℓ (z)2 j ||U(−α, H)| dα = |E

s/2−1 Õ

(C j + D j ),

(40)

j=0

say. Now we estimate the D j ’s. Using (2) and (33) we have

D j ≪ (max |mℓ (z)2 j |)(max | Seℓ (α)s−2−2 j |) E ≪s,ℓ N (s−1)/ℓ L 3 .

(41)

Now we estimate the C j ’s. If s ≥ 4 and 0 ≤ j ≤ s/2 − 2, using the Cauchy-Schwarz inequality, (2) and (33)-(34) we obtain as in (37) that ∫ 1/2 2 j+1 s−3−2 j e | Eeℓ (α)|| Seℓ (α)||U(−α, H)| dα C j ≪ (max |mℓ (z) |)(max | Sℓ (α) |) ≪s,ℓ N (s−2)/ℓ E1/2S1/2 ≪s,ℓ H 1/2 N

−1/2 s/ℓ−1/(2ℓ)−1/2 3

L,

(42)

provided that H ≥ N 1−1/ℓ L. If j = s/2 − 1, s ≥ 2, using the Cauchy-Schwarz inequality, (33) and (35) we have as in (38) that ∫ 1/2 | Eeℓ (α)||mℓ (z)| s−1 |U(−α, H)| dα ≪ E1/2 M(2s − 2)1/2 Cℓ/2−1 ≪ −1/2 1/2

≪s,ℓ H

N s/ℓ−1/(2ℓ)−1/2 L 3 .

(43)

Summarizing, using (40)-(43), we obtain, for s even, that

J2 ≪s,ℓ N (s−1)/ℓ L 3 + H 1/2 N s/ℓ−1/(2ℓ)−1/2 L 3,

(44)


11

provided that N 1−1/ℓ L ≤ H ≤ N. In conclusion, by (39) and (44), we can write

J2 ≪s,ℓ N (s−1)/ℓ L 3 + H 1/2 N s/ℓ−1/(2ℓ)−1/2 L 3,

(45)

for every s, 2 ≤ s ≤ ℓ − 1, provided that N 1−1/ℓ L ≤ H ≤ N. 4.3. Estimate of J3 . It is clear that J3 = I3 of section 3.2. Hence by (11) we obtain ( HN s/ℓ−1/(2ℓ)−1 L 3 if s ≥ 3 J3 ≪s,ℓ H 1/2 N 3/(2ℓ)−1/2 L 2 if s = 2,

(46)

provided that H ≫ N 1−1/ℓ L. 4.4. Final words. By (31)-(32) and (45)-(46), we have N+H Õ


n=N+1

cs,ℓ HN s/ℓ−1 + Os,ℓ (H 2 N s/ℓ−2 + H 1/2 N s/ℓ−1/(2ℓ)−1/2 L 3 ) e

(47)

which is an asymptotic formula ∞(N 1−1/ℓ L 6 ) ≤ H ≤ o (N). From e−n/N = e−1 + O (H/N) for n ∈ [N + 1, N + H], 1 ≤ H ≤ N, we get N+H Õ

rs,ℓ (n) = cs,ℓ HN s/ℓ−1 + Os,ℓ (H 2 N s/ℓ−2 + H 1/2 N s/ℓ−1/(2ℓ)−1/2 L 3 ) + Os,ℓ

n=N+1

H N+H Õ N

n=N+1

rs,ℓ (n) .

Using en/N ≤ e2 and (47), the last error term is ≪s,ℓ H 2 N s/ℓ−2 + H 3/2 N s/ℓ−1/(2ℓ)−3/2 L 3 . Hence we get N+H Õ rs,ℓ (n) = cs,ℓ HN s/ℓ−1 + Os,ℓ (H 2 N s/ℓ−2 + H 1/2 N s/ℓ−1/(2ℓ)−1/2 L 3 ) n=N+1

uniformly for ∞(N 1−1/ℓ L 6 ) ≤ H ≤ o (N). Theorem 2 follows by recalling cs,ℓ = Γ(1 + 1/ℓ)ℓ /Γ(s/ℓ). References [1] M. Cantarini, A. Gambini, and A. Zaccagnini, On the average number of representations of an integer as a sum of like powers, https://arxiv.org/pdf/1805.09008.pdf, (2018). [2] A. Gambini, A. Languasco, and A. Zaccagnini, A diophantine approximation problem with two primes and one k-power of a prime, J. Number Theory 188 (2018), 210–228. [3] A. Kumchev and D. Tolev, An invitation to additive prime number theory, Serdica Mathematical Journal 31 (2005), 1–74. [4] A. Languasco and A. Zaccagnini, Short intervals asymptotic formulae for binary problems with primes and powers, II: density 1, Monatsh. Math. 181 (2016), 419–435. [5] A. Languasco and A. Zaccagnini, Sum of one prime and two squares of primes in short intervals, J. Number Theory 159 (2016), 1945–1960. [6] A. Languasco and A. Zaccagnini, Short intervals asymptotic formulae for binary problems with prime powers, to appear in Journal de Théorie des Nombres de Bordeaux (2018), http://arxiv.org/abs/1806.05373. [7] A. Languasco and A. Zaccagnini, Sum of four prime cubes in short intervals, https://arxiv.org/abs/1705.04457, submitted, (2017). [8] A. Languasco and A. Zaccagnini, Short intervals asymptotic formulae for binary problems with prime powers, II, in preparation, (2018).

12


[9] D. Tolev, On a Diophantine inequality involving prime numbers, Acta Arith. 61 (1992), 289–306. [10] R. C. Vaughan, The Hardy-Littlewood method, second ed., Cambridge U. P., 1997. [11] R. C. Vaughan and T. D. Wooley, Waring’s problem: a survey, Number Theory for the Millennium, vol. III, A. K. Peters, 2002, pp. 301–340.

Alessandro Languasco, Dipartimento di Matematica “Tullio Levi-Civita”, Via Trieste 63, 35121 Padova, Italy. e-mail: [email protected]