On some sums involving the integral part function

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On some sums involving the integral part function

Kui Liu, Jie Wu, Zhishan Yang

To cite this version:

Kui Liu, Jie Wu, Zhishan Yang. On some sums involving the integral part function. 2021. �hal- 03331946�

KUI LIU, JIE WU & ZHISHAN YANG

Abstract. Denote by τk(n), ω(n) and µ2(n) the number of representations of n as product ofk natural numbers, the number of distinct prime factors ofn and the char- acteristic function of the square-free integers, respectively. Let [t] be the integral part of real numbert. Forf =ω,2^ω, µ2, τk, we prove that

X

n6x

fhx n

i

=xX

d>1

f(d)

d(d+ 1) +Oε(x^θf+ε)

forx→ ∞, whereθ_ω= ₁₁₀⁵³,θ₂ω = ₁₉⁹, θ_µ2 = ²₅,θ_τk = _10k−1^5k−1 andε >0 is an arbitrarily small positive number. These improve the corresponding results of Bordell`es.

1. Introduction

Denote by [t] the integral part of the real number t. Recently Bordell`es, Dai, Heyman, Pan and Shparlinski [3] established an asymptotic formula of

S_f(x) := X

n6x

fhx n

i

under some simple assumptions of f. Subsequently, Wu [9] and Zhai [10] improved their results independently. In particular, if f(n)_ε n^ε for any ε > 0 and all n > 1, then [9, Theorem 1.2(i)] or [10, Theorem 1] yields the following asymptotic formula

(1.1) Sf(x) = xX

d>1

f(d)

d(d+ 1) +Oε(x^1/2+ε)

as x → ∞. Ma and Wu [7] observed that if f is factorizable in certain sense, then it is possible to break the ¹₂-barrier for the error term of (1.1). In particular, for the von Mangoldt function Λ(n), they proved, with the help of the Vaughan identity and the technique of one-dimensional exponential sum, that

(1.2) SΛ(x) =xX

d>1

Λ(d)

d(d+ 1) +Oε(x^35/71+ε) (x→ ∞).

Using similar idea to the classical divisor function τ(n), Ma and Sun [6] showed that

(1.3) Sτ(x) = xX

d>1

τ(d)

d(d+ 1) +Oε(x^11/23+ε) (x→ ∞).

Subsequently, with the help of a result of Baker on 2-dimensional exponential sums [1, Theorem 6], Bordell`es [2] sharpened the exponents in (1.2)-(1.3) and also studied some new examples. Denote byτ_k(n),ω(n) andµ₂(n) the number of representations ofn as product

Date: September 2, 2021.

2010Mathematics Subject Classification. 05A17, 11N37, 11P82.

Key words and phrases. partition function, asymptotic formula, polynomial, saddle-point method.

1

of k natural numbers, the number of distinct prime factors of n and the characteristic function of the square-free integers, respectively. His results can be stated as follows:

(1.4) X

n6x

fhx n

i

=xX

d>1

f(d)

d(d+ 1) +O_ε(x^ϑf+ε), with

f Λ ω 2^ω µ₂ τ τ₃ τ_k (k>4)

ϑf 97 203

455 914

97 202

1919 4268

19 40

283 574

4k³−k−2 8k³−2k−2

Very recently, Liu, Wu and Yang [5] succeeded to improve Bordell`es’ exponent ₂₀₃⁹⁷ to ₁₉⁹ for Λ. Their principal tool is a new estimate on 3-dimensional exponential sums.

The aim of this paper is to propose better exponents than all others cases of the above table.

Theorem 1. Under the previous notation, the asymptotic formula (1.4) holds with f ω 2^ω µ₂ τ_k (k>2)

ϑ_{f 110}⁵³ ₁₉^{9 2}_{5 10k−1}^5k−1

The key point of proving Theorem 1 is to establish good bounds for S^f_δ(x, D) := X

D<d62D

f(d)ψ x d+δ

.

where ψ(t) := {t} − ¹₂ and {t} means the fractional part of real number t. Let P be the set of all primes and define

(1.5) 1_P(n) :=

(1 if n ∈P,

0 otherwise, 1(n)≡1, µ(n) :=˜

(µ(d) if n =d², 0 otherwise.

Then we have the following relations:

ω=1_P∗1, τ_k=τk−1∗1, 2^ω =µ₂∗τ, µ₂ = ˜µ∗1.

Thus when f = ω, τ_k,2^ω, µ₂, we can use these relations to decompose S^f_δ(x, D) into bilinear forms

S^ω_δ(x, D) := X

D<d`62D

1_P(d)ψ x d`+δ

, (1.6)

S^τ_δ^k(x, D) := X

D<d`62D

τ_k−1(d)ψ x d`+δ

, (1.7)

S²_δ^ω(x, D) := X

D<d²`62D

µ(d)τ(`)ψ x d<sup>2</sup>`+δ

, (1.8)

S^µ_δ²(x, D) := X

D<d²`62D

µ(d)ψ x d²`+δ

. (1.9)

In the third section, we shall use the Fourier analyse and the technique of multiple expo- nential sums to establish our bounds for these bilinear forms.

2. Preliminary lemmas

In this section, we shall cite three lemmas, which will be needed in the next section.

The first one is [5, Proposition 3.1].

Lemma 2.1. Let α >0, β >0, γ >0 and δ ∈R be some constants. For X >0, H >1, M >1 and N >1, define

(2.1) S_δ=S_δ(H, M, N) := X

h∼H

X

m∼M

X

n∼N

a_h,mb_ne

XM^βN^γ H^α

h^α m^βn^γ+δ

,

wheree(t) := e^2πit, the ah,m andbnare complex numbers such that|ah,m|61and|bn|61, and m∼M means that M < m62M. For any ε >0 we have

S_δ (X^κH^2+κM^2+κN^1+κ+λ)^1/(2+2κ)+HM N^1/2 + (HM)^1/2N +X^−1/2HM N X^ε uniformly forM >1,N >1, H 6M^β−1N^γ and06δ61/ε, where(κ, λ)is an exponent pair and the implied constant depends on (α, β, γ, ε) only.

The second one is the Vaughan identity [8, formula (3)].

Lemma 2.2. There are six real arithmetic functionsα_k(n)verifyingα_k(n)τ(n) log(2n) for (n>1, 16k 66) such that for D>1 and any arithmetical function g, we have

(2.2) X

D<d62D

Λ(d)g(d) = S₁+S₂+S₃+S₄, where τ(n) is the classic divisor function and

S₁ := X

m6D^1/3

α₁(m) X

D<`mn62D

g(mn),

S2 := X

m6D^1/3

α2(m) X

D<`mn62D

g(mn) logn,

S₃ := X X

D^1/3<m,n6D^2/3 D<`mn62D

α₃(m)α₄(n)g(mn),

S₄ := X X

D^1/3<m,n6D^2/3 D<`mn62D

α₅(m)α₆(n)g(mn).

The same result also holds for the M¨obius function µ.

The third one is due to Vaaler (see [4, Theorem A.6]).

Lemma 2.3. Let ψ(t) := {t} − ¹₂, where {t} means the fractional part of real number t.

For x>1 and H >1, we have

ψ(x) =− X

16|h|6H

Φ h H+ 1

e(hx)

2πih +R_H(x),

where Φ(t) := πt(1− |t|) cot(πt) +|t| and the second term R_H(x) satisfies

(2.3) |R_H(x)|6 1

2H+ 2 X

06|h|6H

1− |h|

H+ 1

e(hx).

3. bilinear forms

The aim of this section is to establish some non-trivial bounds for the bilinear forms given by (1.6), (1.9), (1.7) and (1.8). For the convenience, we consider

(3.1) S^Λ∗1_δ (x, D) := X

D<d`62D

Λ(d)ψ x

d`+δ

,

instead ofS^ω_δ(x, D). The following proposition gives the required estimate forS^Λ∗1_δ (x, D).

Proposition 3.1. For any ε >0, we have

(3.2) S^Λ∗1_δ (x, D)_ε (x¹⁹D⁸⁷)^1/133+ (x¹⁷¹D⁶²⁴)^1/1026 x^ε

uniformly for x > 3, D 6 x^57/103 and 0 6 δ 6 ε⁻¹, where the implied constant depends on ε only.

Proof. Applying the Vaughan identity (2.2) withg(d) = ψ _d`+δ^x

, it follows that (3.3) S^Λ∗1_δ (x, D) = S^Λ∗1_δ,1 +S^Λ∗1_δ,2 +S^Λ∗1_δ,3 +S^Λ∗1_δ,4 ,

where

S^Λ∗1_δ,1 :=X

`6D

X

m6(D/`)^1/3

α₁(m) X

D<`mn62D

ψ x

`mn+δ

,

S^Λ∗1_δ,2 :=X

`6D

X

m6(D/`)^1/3

α2(m) X

D<`mn62D

ψ x

`mn+δ

logn, S^Λ∗1_δ,3 :=X

`6D

X X

(D/`)<sup>1/3</sup><m,n6(D/`)^2/3 D<`mn62D

α₃(m)α₄(n)ψ x

`mn+δ

,

S^Λ∗1_δ,4 :=X

`6D

X X

(D/`)<sup>1/3</sup><m,n6(D/`)^2/3 D<`mn62D

α₅(m)α₆(n)ψ x

`mn+δ

.

Using Lemma 2.3 and splitting the intervals of summation into the dyadic intervals, we can write

(3.4)

S^Λ∗1_δ,j =− 1 2πi

X

H⁰

X

L

X

M

X

N

S^Λ∗1_δ,j,[(H⁰, L, M, N) +S^Λ∗1_δ,j,[(H⁰, L, M, N)

+X

L

X

M

X

N

S^Λ∗1_δ,j,†(L, M, N), where 16H⁰ 6H 6D, a_h := ^H_h⁰Φ _H^h₊₁

1 and S^Λ∗1_δ,1,[ =S^Λ∗1_δ,1,[(H⁰, L, M, N) := 1

H⁰ X

h∼H⁰

X

`∼L

X

m∼M

X

n∼N D<`mn62D

a_hα₁(m)e hx

`mn+δ

,

S^Λ∗1_δ,1,† =S^Λ∗1_δ,1,†(L, M, N) :=X

`∼L

X

m∼M

X

n∼N D<`mn62D

α₁(m)R_H x

`mn+δ

,

and S^Λ∗1_δ,j,[(H⁰, L, M, N), S^Λ∗1_δ,j,†(L, M, N) can be defined similarly for j = 2,3,4. We have

(3.5) LM N D, M 6(D/L)^1/3 for j = 1,2

and

(3.6) LM N D, (D/L)^1/3 6M 6(D/L)^1/2 6N 6(D/L)^2/3 for j = 3,4.

In the last case, we have considered the symmetry of m and n.

A. Bounds of S^Λ∗1_δ,j for j = 1,2.

Since we can remove the factor lognby a simple partial integration, we only treatS^Λ∗1_δ,1 . IfL6D^1/5, applying the exponent pair (¹₆,⁴₆) to the sum over n, we have

S^Λ∗1_δ,1,[ x^ε 1 H⁰

X

h∼H⁰

X

`∼L

X

m∼M

xH⁰ LM N²

1/6

N^4/6+ xH⁰ LM N²

−1

x^ε (xHL⁵M⁵N²)^1/6+x⁻¹(LM N)² . In view ofL6D^1/5 and (3.5), we derive that

(3.7) S^Λ∗1_δ,1,[ (x⁵D¹⁷H⁵)^1/30+x⁻¹D² x^ε for H⁰ 6H 6D and (L, M, N) verifying (3.5).

Next we suppose that D^1/5 6 L 6 D^9/19. Firstly we remove the extra multiplicative condition D < `mn62D at the cost of a factor (logx)² and then apply Lemma 2.1 with α=β =γ = 1, (X, H, M, N) = (xH⁰/LM N, H⁰, LM, N) and (κ, λ) = (¹₂,¹₂) to get

(3.8)

S^Λ∗1_δ,1,[ (xL⁴M⁴N³)^1/6+LM N^1/2 + (LM)^1/2N +x^−1/2(LM N)^3/2 x^ε (x³D¹⁰L²)^1/18+ (D²L)^1/3 +DL^−1/2+ (x⁻¹D³)^1/2

x^ε (x¹⁷¹D⁶²⁴)^1/1026+D^9/10+ (x⁻¹D³)^1/2

x^ε

for H⁰ 6 H 6 D^20/57 (6 (D/L)^2/3 6 N) and (L, M, N) verifying (3.5) and D^1/5 6 L 6 D^9/19.

When D^9/196L6D, we apply the exponent pair (¹₆,⁴₆) to the sum over ` to get S^Λ∗1_δ,1,[ x^ε 1

H⁰ X

h∼H⁰

X

m∼M

X

n∼N

xH⁰ L²M N

1/6

L^4/6+ xH⁰ L²M N

−1

x^ε (xHL²M⁵N⁵)^1/6 +x⁻¹H⁰⁻¹(LM N)² . From this we deduce that

(3.9) S^Λ∗1_δ,1,[ (x¹⁹D⁶⁸H¹⁹)^1/114+x⁻¹D² x^ε for H⁰ 6H 6D and (L, M, N) verifying (3.5) and D^9/196L6D.

Combining (3.7), (3.8) and (3.9), we obtain

(3.10) S^Λ∗1_δ,1,[ (x¹⁷¹D⁶²⁴)^1/1026+D^9/10+ (x⁻¹D³)^1/2 + (x¹⁹D⁶⁸H¹⁹)^1/114+ (x⁵D¹⁷H⁵)^1/30

x^ε

for H⁰ 6 H 6 D^20/57 and (L, M, N) verifying (3.5), where we have used the fact that x⁻¹D² 6(x⁻¹D³)^1/2. The same bound holds for S^Λ,†_δ,2. Therefore for j = 1,2, we have (3.11) S^Λ∗1_δ,j (x¹⁷¹D⁶²⁴)^1/1026+D^9/10+ (x⁻¹D³)^1/2

+ (x¹⁹D⁶⁸H¹⁹)^1/114+ (x⁵D¹⁷H⁵)^1/30 x^ε for H 6D^20/57 and (L, M, N) verifying (3.5).

B. Bounds of S^Λ∗1_δ,j for j = 3,4.

In this case, we should estimate S^Λ∗1_δ,3,[ =S^Λ∗1_δ,3,[(H⁰, L, M, N) := 1

H⁰ X

h∼H⁰

X

`∼L

X

m∼M

X

n∼N D<`mn62D

a_hα₃(m)α₄(n)e hx

`mn+δ

,

S^Λ∗1_δ,3,†=S^Λ∗1_δ,3,†(L, M, N) := X

`∼L

X

m∼M

X

n∼N D<`mn62D

α₃(m)α₄(n)R_H x

`mn+δ

,

for (L, M, N) verifying (3.6).

Firstly we bound S^Λ∗1_δ,3,[(H⁰, L, M, N). We remove the extra multiplicative condition D/` < mn 6 2D/` at the cost of a factor logx. By Lemma 2.1 with α = β = γ = 1, (X, H, M, N) = (xH⁰/LM N, H⁰, M, LN) and (κ, λ) = (¹₂,¹₂), we can derive

(3.12) S^Λ∗1_{δ,3,[ ε} (xL³M⁴N³)^1/6+M(LN)^1/2+M^1/2LN + (x⁻¹DH⁰)^1/2 x^ε _ε (x²D⁷L⁻¹)^1/12+ (D⁵L)^1/6+ (x⁻¹DH⁰)^1/2

x^ε,

for H⁰ 6H 6D^1/2 (6 (LD)^1/2 6LN) and (L, M, N) verifying (3.6), where we have re- moved the term (D³L⁻¹)^1/4 (6(D⁵L)^1/6). On the other hand, we can apply the exponent pair (¹₆,⁴₆) to the sum over ` (similar to (3.7)) to get

(3.13) S^Λ∗1_δ,3,[ (xD⁵HL⁻³)^1/6+x⁻¹D² x^ε for H⁰ 6H 6D and (L, M, N) verifying (3.6).

From (3.12) and (3.13), we deduce that

(3.14) S^Λ∗1_{δ,3,[ ε} (x²D⁷L⁻¹)^1/12+ (xD²⁰H)^1/24+ (x⁻¹DH)^1/2+x⁻¹D² x^ε,

for H 6D^1/2 and (L, M, N) verifying (3.6), where we have used the following estimate min{(D⁵L)^1/6,(xD⁵HL⁻³)^1/6}6((D⁵L)^1/6)¹⁸²⁴((xD⁵HL⁻³)^1/6)²⁴⁶ = (xD²⁰H)^1/24. Secondly we boundS^Λ∗1_δ,3,†. Using (2.3) of Lemma 2.3, we have

S^Λ∗1_δ,3,†x^εX

`∼L

X

m∼M

X

n∼N

R_H x

`mn+δ

x^ε H

X

`∼L

X

m∼M

X

n∼N

X

06|h|6H

1− |h|

H+ 1

e hx

`mn+δ

x^ε DH⁻¹+ max

16H⁰6H

eS^Λ∗1_δ,3,†(H⁰, L, M, N)

,

where

Se^Λ∗1_δ,3,†(H⁰, L, M, N) := 1 H

X

`∼L

X

m∼M

X

n∼N

X

h∼H⁰

1− |h|

H+ 1

e hx

`mn+δ

.

Clearly we can bound Se^Λ∗1_δ,3,†(H⁰, L, M, N) in the same way as S^Λ∗1_δ,3,[(H⁰, L, M, N) and get (3.15) S^Λ∗1_{δ,3,† ε} DH⁻¹+ (x²D⁷L⁻¹)^1/12+ (xD²⁰H)^1/24+ (x⁻¹DH)^1/2+x⁻¹D²

x^ε for H 6D^1/2 and (L, M, N) verifying (3.6). Thus (3.14) and (3.15) imply that (3.16) S^Λ∗1_{δ,j ε} (x²D⁷)^1/12+x⁻¹D²+DH⁻¹+ (xD²⁰H)^1/24+ (x⁻¹DH)^1/2

x^ε, for j = 3,4, H 6D^1/2 and (L, M, N) verifying (3.6).

C.End of proof of (3.2).

Inserting (3.11) and (3.16) into (3.3) , we find that

S^Λ∗1_δ (x, D) (x¹⁷¹D⁶²⁴)^1/1026+ (x²D⁷)^1/12+D^9/10+ (x⁻¹D³)^1/2

+DH⁻¹ + (x¹⁹D⁶⁸H¹⁹)^1/114+ (x⁵D¹⁷H⁵)^1/30+ (xD²⁰H)^1/24 x^ε for H 6D^20/57, where we have removed the term (x⁻¹DH)^1/2 6(x⁻¹D³)^1/2. Optimising H over [1, D^20/57], it follows that

S^Λ∗1_δ (x, D) (x¹⁷¹D⁶²⁴)^1/1026+ (x²D⁷)^1/12+D^9/10+ (x⁻¹D³)^1/2 + (x¹⁹D⁸⁷)^1/133+ (x⁵D²²)^1/35+ (xD²¹)^1/25+D^9/19 + (x¹⁹D⁶⁸)^1/114+ (x⁵D¹⁷)^1/30+ (xD²⁰)^1/24

x^ε

which implies the required inequality, since the 5th term dominates the 7th one provided D6x^171/308 and the first term dominates the others ones provided D6x^57/103.

The second proposition gives the required estimate for S^τ_δ^k(x, D) defined as in (1.7).

Proposition 3.2. Let k >2 be an integer. For any ε >0, we have (3.17) S^τ_δ^k(x, D)_k,ε (x^kD^4k−1)^1/6kx^ε

uniformly for x> 3, 1 6D 6 xmin{4k/(5k+1),k/(2k−2)} and 06 δ 6 ε⁻¹, where the implied constant depends on k and ε only.

Proof. Noticing thatτ_k =τk−1∗1, using Lemma 2.3 and splitting the interval of summation into the dyadic intervals, we can write

(3.18)

S^τ_δ^k(x, D) = − 1 2πi

X

H⁰

X

M

X

N

S^τ_δ,[^k(H⁰, M, N) +S^τ_δ,[^k(H⁰, M, N)

+X

M

X

N

S^τ_δ,†^k(M, N),

where H 6D,M N D (i.e. DM N D),N >D^1/k,a_h := ^H_h⁰Φ _H+1^h

1 and S^τ_δ,[^k(H⁰, M, N) := 1

H⁰ X

h∼H⁰

a_h X

m∼M

X

n∼N D<mn62D

τk−1(m)e hx mn+δ

,

S^τ_δ,†^k(M, N) := X

m∼M

X

n∼N D<mn62D

τk−1(m)R_H x mn+δ

.

Firstly we boundS^τ_δ,[^k(H⁰, M, N). Applying the exponent pair (¹₂,¹₂) to the sum overn, it follows that

(3.19)

S^τ_δ,[^k(H⁰, M, N) x^ε H⁰

X

h∼H⁰

X

m∼M

n xh mN²

1/2

N^1/2+ xh mN²

⁻¹o

(xH⁰M N⁻¹)^1/2 +x⁻¹D² x^ε.

On the other hand, we remove the extra multiplicative condition D < mn 6 2D at the cost of a factor logx, and then apply Lemma 2.1 with α = β = γ = 1, (X, H, M, N) = (xH⁰/M N, H⁰, M, N) and (κ, λ) = (¹₂,¹₂) to get

S^τ_δ,[^k(H⁰, M, N) (xM⁴N³)^1/6+M N^1/2+H^0−1/2M^1/2N + (x⁻¹H⁰⁻¹M³N³)^1/2 X^ε

provided H⁰ 6H 6D^1/k (6N). Using M N D and N >D^1/k, we can derive that (3.20) S^τ_δ,[^k(H⁰, M, N) (x^kD^4k−1)^1/6k+D^1−1/(2k)+H^0−1/2M^1/2N+ (x⁻¹D³)^1/2

x^ε (x^kD^4k−1)^1/6k+H^0−1/2M^1/2N

x^ε

for H⁰ 6 H 6 D^1/k. In the last inequality, we can remove D^1−1/(2k) and (x⁻¹D³)^1/2 because they can be absorbed by (x^kD^4k−1)^1/6k for D 6 xmin{4k/(5k+1),k/(2k−2)}. From (3.19) and (3.20), we deduce that

(3.21) S^τ_δ,[^k(H⁰, M, N)(x^kD^4k−1)^1/6kx^ε

for H⁰ 6 H 6 D^1/k and M N = D 6 xmin{4k/(5k+1),k/(2k−2)}, where we have used the following estimates

min{(xH⁰M N⁻¹)¹², H⁰⁻¹²M¹²N}6((xH⁰M N⁻¹)¹²)¹³(H⁰⁻¹²M¹²N)²³

= (xH⁰⁻¹M³N³)¹⁶ 6(xD³)^1/6

and max{(xD³)^1/6, x⁻¹D²}6(x^kD^4k−1)^1/6k for D6x^7k/(8k+1). On the other hand, (2.3) of Lemma 2.3 allows us to derive that

S^τ_δ,†^k(M, N)

6 X

m∼M

X

n∼N

R_H x mn+δ

6 1

2H+ 2 X

06|h|6H

1− |h|

H+ 1 X

m∼M

X

n∼N

e xh

d²mn+δ

.

When h6= 0, we can bound the triple sums as before. Thus (3.22) S^τ_δ,†^k(M, N) DH⁻¹+ (x^kD^4k−1)^1/6k

x^ε for H 6D^1/k and M N D6xmin{4k/(5k+1),k/(2k−2)}.

Inserting (3.21) and (3.22) into (3.18) and takingH =D^1/k, we can obtain the required

inequality.

The third proposition gives the required estimate forS²_δ^ω(x, D) defined as in (1.8).

Proposition 3.3. For any ε >0, we have

S²_δ^ω(x, D)_ε (x²D⁷)^1/12x^ε

uniformly for x > 3 and 1 6 D 6 x^8/11 and 0 6 δ 6 ε⁻¹, where the implied constant depends on ε only.

Proof. Firstly we write

S²_δ^ω(x, D) = X

d6√ 2D

µ(d) X

D/d²<`62D/d²

τ(`)ψ x/d²

`+δ/d²

= X

d6√ 2D

µ(d)S^τ_δ/d2(x/d², D/d²).

Secondly it is easy to see that D6 x^8/11 implies D/d² 6 (x/d²)^8/11. Thus we can apply

Proposition 3.2 with k= 2 to get the required result.

The last proposition gives the required estimate for S^µ_δ²(x, D) defined as in (1.9).

Proposition 3.4. For any ε >0, we have

S^µ_δ²(x, D)_ε (xD³)^1/7+ (xD²)^1/6 x^ε

uniformly for x>3, 16D6x^8/11 and 06δ6ε⁻¹, where the implied constant depends on ε only.

Proof. Using Lemma 2.3 and splitting the interval of summation [1,2D] into the dyadic intervals (L,2L], we can write

(3.23) S^µ_δ² =− 1 2πi

X

L

S^µ_δ,[²(L) +S^µ_δ,[²(L)

+X

L

S^µ_δ,†²(L), where H 6D and

S^µ_δ,[²(L) := X

h6H

1

hΦ h H+ 1

X

`∼L

X

(D/`)<sup>1/2</sup><d6(2D/`)^1/2

µ(d)e hx d²`+δ

,

S^µ_δ,†²(L) :=X

`∼L

X

(D/`)<sup>1/2</sup><d6(2D/`)^1/2

µ(d)R_H x d²`+δ

.

Inverting the order of summations and applying the exponent pair (¹₆,⁴₆) to the sum over

`, it follows that

(3.24)

S^µ_δ,[²(L) X

h∼H

1 h

X

(D/2L)^1/2<d6(2D/L)^1/2

xh d²L²

1/6

L^4/6+ xh d²L²

−1

(xD²H)^1/6+x⁻¹D² x^ε.

On the other hand, (2.3) of Lemma 2.3 allows us to derive that S^µ_δ,†²(L)

6 X

(D/2L)^1/2<d6(2D/L)^1/2

X

`∼L

R_H x d²`+δ

6 1

2H+ 2 X

06|h|6H

1− |h|

H+ 1

X

(D/2L)^1/2<d6(2D/L)^1/2

X

`∼L

e xh d²`+δ

.

When h6= 0, as before we apply the exponent pair (¹₆,⁴₆) to the sum over ` and obtain (3.25) S^µ_δ,†²(L) DH⁻¹+ (xD²H)^1/6+x⁻¹D²

x^ε. Inserting (3.24) and (3.25) into (3.23), it follows that

S^µ_δ² DH⁻¹+ (xD²H)^1/6+x⁻¹D² x^ε for H 6D. OptimisingH on [1, D], we find that

S^µ_δ² (xD³)^1/7+ (xD²)^1/6+x⁻¹D² x^ε

This implies the required result, sincex⁻¹D² 6(xD³)^1/7 for D6x^8/11.

4. Proof of Theorem 1

Letf =ω orτ_k or 2^ω orµ₂ and letN_f ∈[1, x^1/2) be a parameter which can be chosen later. First we write

(4.1) S_f(x) =X

n6x

fhx n

i

=S_f^†(x) +S_f^](x) with

S_f^†(x) := X

n6Nf

fhx n

i

, S_f^](x) := X

N_f<n6x

fhx n

i .

In our case, we havef(n)_ε n^ε for any ε >0 and all n>1. Thus

(4.2) S_f^†(x)ε Nfx^ε.

In order to boundS_f^](x), we put d= [x/n]. Noticing that

x/n−1< d6x/n ⇔ x/(d+ 1)< n 6x/d, we can derive that

(4.3)

S_f^](x) = X

d6x/Nf

f(d) X

x/(d+1)<n6x/d

1

= X

d6x/Nf

f(d)x

d −ψx d

− x

d+ 1 +ψ x d+ 1

=xX

d>1

f(d)

d(d+ 1) +R^f1(x, Nf)−R^f0(x, Nf) +O(Nf), where we have used the following bounds

x X

d>x/Nf

f(d)

d(d+ 1) _ε N_fx^ε, X

d6Nf

f(d) ψ x

d+ 1

−ψx d

_εN_fx^ε

and

R^f_δ(x, N_f) = X

Nf<d6x/Nf

f(d)ψ x d+δ

.

Combining (4.1), (4.2) and (4.3), it follows that S_f(x) = xX

d>1

f(d)

d(d+ 1) +O_ε |R^f₁(x, N_f)|+|R^f₀(x, N_f)|+N_fx^ε .

Thus in order to prove Theorem 1, it suffices to show that (4.4) R^f_δ(x, N_f)N_fx^ε (x>1) for

f ω τ_k 2^ω µ₂

Nf x^53/110 x(5k−1)/(10k−1) x^9/19 x^2/5

4.1. Proof of (4.4) with f =ω.

We haveN_ω =x^53/110. In order to apply (3.2) of Proposition 3.1, we must switch from R^ωδ(x, N_ω) = X

Nω<p`6x/Nω

ψ x p`+δ

= X

Nω<p`6x/Nω

Λ(p)

logpψ x p`+δ

to

Re^ωδ(x, Nω, t) :=X

d6t

Λ(d) X

Nω/d<`6(x/Nω)/d

ψ x

d`+δ

.

For this, we need to estimate the contribution of prime powers:

R1 := X

Nω<p^ν`6x/Nω

`6(x/Nω)^53/114, ν>2

Λ(p^ν)

logp^νψ x p^ν`+δ

= X

Nω<p^ν`6x/Nω

`6(x/Nω)^53/114, ν>2

1

νψ x p^ν`+δ

,

R2 := X

Nω<p^ν`6x/Nω

`>(x/Nω)^53/114, ν>2

Λ(p^ν)

logp^νψ x p^ν`+δ

= X

Nω<p^ν`6x/Nω

`>(x/Nω)^53/114, ν>2

1

νψ x p^ν`+δ

.

Firstly we have trivially R1 X

`6(x/Nω)^53/114

(x/N_ω`)^1/2 (x/N_ω)^167/228x^167/440.

On the other hand, applying [10, Lemma 2.4] with (κ, λ) = (¹₂,¹₂), we can derive that R2 = X

p^ν6(x/Nω)^61/114 ν>2

1 ν

X

(x/Nω)^53/114<`6x/Nωp^ν

ψ x p^ν`+δ

X

p^ν6(x/Nω)^61/114 ν>2

max

(x/Nω)^53/114<N16x/Nωp^ν N1<N262N1

N₁

X

N1<`6N2

1

`ψ x p<sup>ν</sup>`+δ

X

p^ν6(x/Nω)^61/114 ν>2

max

(x/Nω)^53/114<N16x/Nωp^ν N1<N262N1

N₁²

(x/p^ν) + (x/p^ν)^1/3

X

p^ν6(x/Nω)^61/114 ν>2

x

N_ω²p^ν + (x/p^ν)^1/3

x/N_ω²+x^1/3(x/Nω)^(61/114)/6 x^167/440

Thus we can write

(4.5)

R^ωδ(x, Nω) = X

Nω<d`6x/Nω

Λ(d) logdψ

x d`+δ

+O x^167/440

=

Z x/Nω

2−

1

logtdRe^ωδ(x, Nω, t) +O x^167/440 max

26t6x/Nω

eR^ωδ(x, N_ω, t)

+x^167/440.

WritingD_j :=x/(2^jN_ω), we have N_ω 6D_j 6x/N_ω 6 x^57/110 for 06j 6 ^log(x/N_{log 2}^ω2). Thus we can apply (3.2) of Proposition 3.1 to get

|eR^ω_δ(x, N_ω, t)|6 X

06j6log(x/Nω²)/log 2

|S^Λ∗1_δ (x, D_j)|

X

06j6log(x/Nω²)/log 2

(x¹⁹D⁸⁷_j )^1/133+ (x¹⁷¹D⁶²⁴_j )^1/1026 x^ε (x¹⁰⁶N_ω⁻⁸⁷)^1/133+ (x⁷⁹⁵N_ω⁻⁶²⁴)^1/1026

x^ε

for all 26t6x/N_ω. Inserting this into (4.5), we get the required result.

4.2. Proof of (4.4) with f =τ_k.

We have N_τk = x(5k−1)/(10k−1). Let D_j := x/(2^jN_τk), then we easily see that N_τk 6 D_j 6x/N_τk 6xmin{4k/(5k+1),k/(2k−2)} for 06j 6 ^log(x/N

τk2 )

log 2 . Thus Proposition 3.2 gives us

|R^τ_δ^k(x, N_τk)|6 X

06j6log(x/N_τk² )/log 2

|S^τ_δ^k(x, D_j)|

X

06j6log(x/N_τk² )/log 2

(x^kD^4k−1_j )^1/6kx^ε (x^5k−1N_τ^−(4k−1)

k )^1/6kx^ε, which implies the required result.

4.3. Proof of (4.4) with f = 2^ω.

We have N₂^ω = x^9/19. Let D_j := x/(2^jN₂^ω), then N₂^ω 6 D_j 6 x/N₂^ω 6 x^10/19 for 06j 6 ^log(x/N_{log 2}^22ω). Noticing that 2^ω = ˜µ∗τ, we can apply Proposition 3.3 to get

|R²δ^ω(x, N2^ω)|6 X

06j6log(x/N₂²ω)/log 2

|S²_δ^ω(x, Dj)|

X

06j6log(x/N₂²ω)/log 2

(x²D⁷_j)^1/12x^ε (x⁹N₂⁻⁷ω )^1/12x^ε,

which implies the required result.

4.4. Proof of (4.4) with f =µ₂.

We have N_µ2 = x^2/5. Let D_j := x/(2^jN_µ2), then N_µ2 6 D_j 6 x/N_µ2 6 x^3/5 for 06j 6 ^log(x/N

µ22)

log 2 . Noticing that µ₂ = ˜µ∗1, we can apply Proposition 3.4 to get

|R^µ_δ²(x, N_µ2)|6 X

06j6log(x/Nµ²2)/log 2

|S^µ_δ²(x, D_j)|

X

06j6log(x/Nµ²2)/log 2

(xD_j³)^1/7+ (xD²_j)^1/6 x^ε (x⁴N_µ⁻³₂ )^1/7 + (x³N_µ⁻²₂ )^1/6

x^ε, which implies the required result.

Acknowledgement. This work is supported in part by the National Natural Science Foundation of China (Grant Nos. 12071238, 11771252, 11971370 and 12071375) and by the NSF of Chongqing (Grant No. cstc2019jcyj-msxm1651).

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Kui Liu, School of Mathematics and Statistics, Qingdao University, 308 Ningxia Road, Qingdao, Shandong 266071, China

E-mail address: liukui@qdu.edu.cn

Jie Wu, CNRS UMR 8050, Laboratoire d’Analyse et de Math´ematiques Appliqu´ees, Universit´e Paris-Est Cr´eteil, 94010 Cr´eteil cedex, France

E-mail address: jie.wu@u-pec.fr

Zhishan Yang, School of Mathematics and Statistics, Qingdao University, 308 Ningxia Road, Qingdao, Shandong 266071, China

E-mail address: zsyang@qdu.edu.cn