# Divisor Problems Related to Hecke Eigenvalues in Three Dimensions

Authors: Jing Huang; Huafeng Liu
Journal: Journal of Mathematics
 (2021)

---

## Abstract

In this paper, we consider divisor problems related to Hecke eigenvalues in three dimensions. We establish upper bounds and asymptotic formulas for these problems on average.


---

## Body

## 1. Introduction

LetΓ=SL2ℤ be the full modular group. Let Hk∗ denote the set of primitive holomorphic forms with even integral weight k≥2 for Γ. Then Hk∗ consists of common eigenfunctions f of all Hecke operators Tn. The Hecke eigenfunction f has the following Fourier series expansion:(1)fz=∑n=1∞λfnnk−1/2e2πinz,ℑz>0,where λfn denotes the n-th normalized eigenvalue. It is known that λfn, as a function of n, is real-valued and multiplicative. Moreover, for all integers n≥1, Deligne [1] showed that(2)λfn≤dn,where dn is the divisor function. Also, for every prime p,(3)αfp=αfp+βfp,αfpβfp=αfp=βfp=1.Now we introduce some specific automorphicL-functions. Define the Hecke L-function attached to f as(4)Ls,f=∑n=1∞λfnns=∏p1−αfpps−11−αfp−1ps−1,for ℜs>1. Moreover, the Rankin–Selberg L-function attached to f can be defined as(5)Ls,f×f=∏p1−αfp2ps−11−1ps−21−αfp−2ps−1,ℜs>1.Then,Ls,f×f can be rewritten as(6)Ls,f×f=ζ2s∑n=1∞λfn2ns≔∑n=1∞λf×fnns.As usual,ζs denotes the Riemann zeta-function. The symmetric square L-function attached to f can be defined as, for ℜs>1,(7)Ls,sym2f=∏p1−αfp2ps−11−1ps−11−αfp−2ps−1,which can also be expressed in the Dirichlet series(8)Ls,sym2f=∑n=1∞λsym2fnns=∏p1+λsym2fpps+λsym2fp2p2s+λsym2fp3p3s+⋯,ℜs>1.The symmetric squareL-function Ls,sym2f has a functional equation and could be analytic continued to an entire function over the whole complex plane. We refer to works of Hecke in [2], Gelbert and Jacquet [3], Kim [4], and Kim and Shahidi [5,6] for these properties. Therefore, the symmetric square L-function Ls,sym2f could be seen as a general L-function in the sense of Perelli [7].In number theory, considering the properties and average behaviors of the Fourier coefficients is meaningful and interesting. Some classical problems concern mean values of these Fourier coefficients and related problems with the corresponding automorphicL-functions (see [1, 8–29], etc.). Here, we just give a brief history for general divisor problems related to these Fourier coefficients.Letω be an integer greater than one, and(9)λω,fn=∑n=n1n2,…,nωλfn1λfn2,…,λfnω,λω,f×fn=∑n=n1n2,…,nωλf×fn1λf×fn2,…,λf×fnω.In particular, we haveλ1,fn=λfn and λ1,f×fn=λf×fn. In 1927, Hecke [30] showed that(10)∑n≤xλfn≪x1/2.Subsequently, this upper bound was improved by many scholars (for example, see [12, 21, 24]). In this direction, the best result so far was obtained by Wu [24] who showed that(11)∑n≤xλfn≪x1/3logxρ1/2+,where(12)ρ1/2+=102+7212106−2151/2+102−7212106+2151/2−3355,=−0.118….Rankin [20] and Selberg [22] established(13)∑n≤xλf×fn=Cfx+Ofx3/5,where Cf is a positive constant depending on f. Later, Kanemitsu, Sankaranarayanan, and Tanigawa [31] studied general divisor problems and proved(14)∑n≤xλω,fn≪x1−3/2k+2+ɛ,∑n≤xλω,f×fn=Mωx+Ox1−1/2k+ɛ,where ω is an integer and ω≥2; Mωx comes from a residue and takes the form xPω−1logx; Pω−1t denotes a polynomial of t with degree ω−1. Fomenko [32] also proved the same result for the sum ∑n≤xλω,fn. After that, Kanemitsu, Sankaranarayanan, and Tanigawa’s result was improved by Lü [33], and some general results were obtained (see [34–37], etc.).In this paper, we consider divisor problems related to Hecke eigenvalues in three dimensions motivated by the above results and Ivić’s work on three-dimensional divisor problems (see, e.g., [38]). We first introduce some notation. For any fixed integer 1<a<b<c, we define(15)λfa,b,cn≔∑n=n1an2bn3cλfn1λfn2λfn3,λf×fa,b,cn≔∑n=n1an2bn3cλf×fn1λf×fn2λf×fn3.We are interested in studying the average behaviors of sums(16)Sfa,b,c;x≔∑n≤xλfa,b,cn,Sf×fa,b,c;x≔∑n≤xλf×fa,b,cn,which can be seen as divisor problems related to Hecke eigenvalues in three dimensions. We establish the following results.Theorem 1.
Leta,b, and c be fixed integers satisfying 1<a<b<c. Then, for any ϵ>0, one has(17)Sfa,b,c;x≪x1/a−3/27a−b−c+ɛ,ifc≤2a,x1/a−3/25a−b+ɛ,ifb<2a<c,x1/2a+ε,if2a≤b.Theorem 2.
Leta,b, and c be fixed integers satisfying 1<a<b<c. Then, for any ϵ>0, one has(18)Sf×fa,b,c;x=M1x1/a+M2x1/b+M3x1/c+Ox1/a−84/748a−131b+c+ɛ,ifc≤2a,M1x1/a+M2x1/b+Ox1/a−84/486a−131b+ɛ,ifb<2a<c,M1x1/a+Ox5/8a+ɛ,if2a≤b,where(19)M1=Lba,f×fLca,f×fL1,sym2f,M2=Lab,f×fLcb,f×fL1,sym2f,M3=Lac,f×fLbc,f×fL1,sym2f.


## 2. Some Lemmas

To prove our theorems, we need to quote some lemmas, which include the individual and averaged subconvexity bounds for the Riemann zeta-function and the symmetric squareL-function. And from the following Lemma 1, we know that the Rankin–Selberg L-function Ls,f×f could be decomposed into the product of the Riemann zeta-function and the corresponding symmetric square L-function.Lemma 1.
Forℜs>1, one has(20)Ls,f×f=ζsLs,sym2f.Proof.
By comparing the Euler products of two sides of (20) and applying (3), we can get this lemma. This lemma can also be found in [33].Lemma 2.
For anyϵ>0, one has(21)∫1TL12+it,f2dt∼CTlogT,uniformly for T≥1, and the subconvexity bound in the critical strip(22)Lσ+it,f≪1+t21−σ/3+ε,if12<σ≤1,1,ifσ>1,where t≥1.Proof.
These results are proved by Good [11].Lemma 3.
For anyɛ>0, one has(23)∫1Tζ12+it12dt≪T2+ɛ,uniformly for T≥1, and the subconvexity bound in the critical strip(24)ζσ+it≪1+t13/421−σ+ɛ,if12≤σ≤1,1,ifσ>1,where t≥1.Proof.
The twelfth mean value estimate (23) is due to Heath-Brown [39]. The subconvexity bound (24) is due to Bourgain [40].Lemma 4.
For anyϵ>0, one has(25)∫1TLσ+it,sym2f2dt≪T31−σ+ε,uniformly for T≥1, and the subconvexity bound in the critical strip(26)Lσ+it,sym2f≪1+t5/41−σ+ε,if12≤σ≤1,1,ifσ>1,where t≥1.Proof.
The first result follows from Perelli’s mean value theorem withLs,sym2f (see [7]), and the second one is due to Nunes [19].


## 3. Proof of Theorem1

In this section, we shall complete the proof of Theorem1. Note that(27)Las,fLbs,fLcs,f=∑n=1∞λfa,b,cnns.Then, by (27) and Perron’s formula (see Proposition 5.54 in [41]), we can obtain(28)Sfa,b,c;x=12πi∫η−iTη+iTLas,fLbs,fLcs,fxssds+Ox1/a+εT,where s=σ+it,η=1/a+ɛ and 1≤T≤x is a parameter to be chosen later.We shift the line of the integral of (28) to the line ℜs=1/2a. Then, Cauchy’s residue theorem shows that(29)Sfa,b,c;x=12πi∫1/2a−iT1/2a+iT+∫1/2a+iTη+iT+∫η−iT1/2a−iTLas,fLbs,fLcs,fxssds+Ox1/a+εT≔G1+G2+G3+Ox1/a+εT.The following work is to estimateG1,G2, and G3. The estimates for the integrals over the horizontal segments are similar, so we handle G2 and G3 first. To get this goal, we consider three cases c≤2a, b<2a<c, and 2a≤b.We first consider the casec≤2a. To estimate G2 and G3, we divide the integral interval into the following eight intervals I1,…,I4, some of which may be empty, and use Lemma 2.IntervalI1:(30)I1≔s=σ+iT:12≤aσ≤1,b2a≤bσ≤1,c2a≤cσ≤1=s=σ+iT:12a≤σ≤1c.In this interval, we have(31)T−1×∫I1xσLaσ+iat,fLbσ+ibt,fLcσ+ict,fdσ≪max1/2a≤σ≤1/cxσT2/31−aσ+2/31−bσ+2/31−cσ+εT−1≪max1/2a≤σ≤1/cT1+εxT2/3a+b+cσ≪x1/cT1/3−2a+b/3c+ε+x1/2aT2/3−b+c/3a+ε.IntervalI2:(32)I2≔s=σ+iT:12≤aσ≤1,b2a≤bσ≤1,1<cσ≤cη=s=σ+iT:1c<σ≤1b.In this interval, we have(33)T−1×∫I2xσLaσ+iat,fLbσ+ibt,fLcσ+ict,fdσ≪max1/c<σ≤1/bxσT2/31−aσ+2/31−bσ+εT−1≪max1/c<σ≤1/bT1/3+εxT2/3a+bσ≪x1/bT−1/3−2a/3b+ε+x1/cT1/3−2a+b/3c+ɛ.IntervalI3:(34)I3≔s=σ+iT:12≤aσ≤1,1<bσ≤bη,c2a≤cσ≤1=s=σ+iT:1b<σ≤1c.This interval is an empty set noting that1/b>1/c.IntervalI4:(35)I4≔s=σ+iT:12≤aσ≤1,1<bσ≤bη,1<cσ≤cη=s=σ+iT:1b<σ≤1a.In this interval, we have(36)T−1×∫I4xσLaσ+iat,fLbσ+ibt,fLcσ+ict,fdσ≪max1/b<σ≤1/axσT2/31−aσ+ɛT−1≪x1/aT−1+ε+x1/bT−1/3−2a3b+ɛ.Thus, by (31)–(36), we have(37)G2+G3≪T−1∫1/2aηxσLaσ+iat,fLbσ+ibt,fLcσ+ict,fdσ=T−1∫I1∪⋯∪I4xσLaσ+iat,fLbσ+ibt,fLcσ+ict,fdσ≪x1/2aT2/3−b+c/3a+ε+x1/bT−1/3−2a/3b+ɛ+x1/cT1/3−2a+b/3c+ɛ+x1/a+ɛT−1+ɛ.Next, we handleG1. We have(38)G1≪x1/2a∫1TL12+iat,fLb2a+ibt,fLc2a+ict,ft−1dt+x1/2a≪x1/2alogTmaxT1≤TT1−1∫T1/2T1L12+iat,fLb2a+ibt,fLc2a+ict,fdt+x1/2a.Then, by Lemma2 and applying Cauchy’s inequality, we can obtain(39)G1≪x1/2alogTmaxT1≤TT1−1T12/31−b/2aT12/31−c/2a∫T1/2T1L12+iat,f2dt1/2×∫T1/2T11dt1/2+x1/2a≪x1/2alogTmaxT1≤TT14/3−b+c/3a+ɛ≪x1/2aT4/3−b+c/3a+ɛ.According to (29), (37), and (39), we have(40)Sfa,b,c;x≪x1/2aT4/3−b+c/3a+ɛ+x1/bT−1/3−2a/3b+x1/a+ɛT−1+ɛ.By takingT=x3/27a−b−c in (40), we can obtain(41)Sfa,b,c;x≪x1/a−3/27a−b−c+ɛ,which proves the first result of Theorem 1.For the caseb<2a<c, to estimate G2 and G3, we still divide the integral interval into four corresponding short intervals I1″,…,I4″, which are different from ones for the case c≤2a. In fact, the corresponding short intervals I1″,I3″ become empty sets in the current situation. However, we still can estimate G2+G3 similar to the corresponding parts of the case c≤2a and get(42)G2+G3≪x1/2aT−b/3a+ɛ+x1/bT−1/3−2a/3b+ɛ+x1/a+ɛT−1+ɛ.The estimate ofG1 becomes the following at the current case by noting c/2a>1.(43)G1≪x1/2a∫1TL12+iat,fLb2a+ibt,fLc2a+ict,ft−1dt+x1/2a≪x1/2alogTlimT1≤TT1−1∫T1/2T1L12+iat,fLb2a+ibt,fLc2a+ict,fdt+x1/2a≪x1/2alogTlimT1≤TT1−1+2/31−b/2a∫T1/2T1L12+iat,f2dt1/2∫T1/2T11dt1/2+x1/2a≪x1/2alogTlimT1≤TT1−1+2/31−b/2a+1/2+1/2+ɛ+x1/2a≪x1/2aT2/3−b/3a+ɛ.Recalling (29), we have(44)Sfa,b,c;x≪x1/2aT2/3−b/3a+ɛ+x1/bT−1/3−2a/3b+ɛ+x1/a+ɛT−1+ɛ.By takingT=x3/25a−b in (44), we can get(45)Sfa,b,c;x≪x1/a−3/25a−b+ɛ,which proves the second result of Theorem 1.For the case2a≤b, to estimate G2 and G3, we also divide the integral interval into four corresponding short intervals I1′,…,I4′, which are different from ones for the case c≤2a. In fact, the corresponding short intervals I1′,I2′,I3′ become empty sets in the current situation. However, we still can estimate G2+G3 similarly to the corresponding parts of the case c≤2a and get(46)G2+G3≪x1/2aT−2/3+ɛ+x1/a+ɛT−1+ɛ.The estimate ofG1 becomes the following by noting c/2a>b/2a>1.(47)G1≪x1/2a∫1TL12+iat,fLb2a+ibt,fLc2a+ict,ft−1dt+x1/2a≪x1/2alogTmaxT1≤TT1−1∫T1/2T1L12+iat,fLb2a+ibt,fLc2a+ict,fdt+x1/2a≪x1/2alogTmaxT1≤TT1−1∫T1/2T1L12+iat,f2dt1/2∫T12T11dt1/2+x1/2a≪x1/2alogTmaxT1≤TT1−1+1/2+1/2+ɛ+x1/2a≪x1/2aTɛ.Recalling (29), we have(48)Sfa,b,c;x≪x1/2aTɛ+x1/a+ɛT−1+ɛ.By takingT=x1/2a in (48), we can get(49)Sfa,b,c;x≪x1/2a+ɛ,which proves the third result of Theorem 1.


## 4. Proof of Theorem2

In this section, we shall prove Theorem2, the process of which is more complicated than Theorem 1. Note that(50)Las,f×fLbs,f×fLcs,f×f=∑n=1∞λf×fa,b,cnns.Then, by (50) and Perron’s formula (see Proposition 5.54 in [41]), we have(51)Sf×fa,b,c;x=12πi∫η−iTη+iTLas,f×fLbs,f×fLcs,f×fxssds+Ox1/a+ɛT,where s=σ+it,η=1/a+ɛ and 1≤T≤x is a parameter which will be decided later.In view of (20), the points s=1/a,s=1/b and s=1/c are the only three possible simple poles of the integrand in the rectangle IT=s=σ+it:1/2a≤σ≤η,t≤T. The corresponding possible residues at s=1/a,s=1/b, and s=1/c are equal to(52)M1≔Lba,f×fLca,f×fL1,sym2f,M2≔Lab,f×fLcb,f×fL1,sym2f,M3≔Lac,f×fLbc,f×fL1,sym2f,respectively.We move the integral line of the integral in (28) to the parallel segment with ℜs=1/2a. We first consider the case c≤2a. In this situation, the points s=1/a, s=1/b, and s=1/c are all poles of the integrand in the rectangle IT. Therefore, by Cauchy’s residue theorem, we can obtain(53)Sf×fa,b,c;x=Ress=1/a+Ress=1/b+Ress=1/cLas,f×fLbs,f×fLcs,f×fxss+12πi∫1/2a−iT1/2a+iT+∫1/2a+iTη+iT+∫η−iT1/2a−iTLas,f×fLbs,f×fLcs,f×fxssds+Ox1/a+εT≔M1x1/a+M2x1/b+M3x1/c+J1+J2+J3+Ox1/a+εT.Now, the remaining work is to handle these three termsJ1,J2, and J3. In addition, the estimates of these integrals on the horizontal parts are analogous, and so we deal with J2 and J3 firstly. To estimate J2 and J3, similarly to the method estimating G2 and G3, we still divide the integral interval into the following four short intervals I1∗,…,I4∗ and apply Lemmas 3 and 4.IntervalI1∗:(54)I1∗≔s=σ+iT:12≤aσ≤1,b2a≤bσ≤1,c2a≤cσ≤1=s=σ+iT:12a≤σ≤1c.In this interval, we have(55)T−1×∫I1∗xσζaσ+iatLaσ+iat,sym2fζbσ+ibtLbσ+ibt,sym2f⋅ζcσ+ictLcσ+ict,sym2fdσ≪max1/2a≤σ≤1/cxσT13/42+5/41−aσT13/42+5/41−bσT13/42+5/41−cσT−1+ɛ≪max1/2a≤σ≤1/cT309/84+ɛxT131/84a+b+cσ≪x1/cT89/42−131a+b/84c+ɛ+x1/2aT487/168−131b+c/168a+ɛ.IntervalI2∗:(56)I2∗≔s=σ+iT:12≤aσ≤1,b2a≤bσ≤1,1<cσ≤cη=s=σ+iT:1c<σ≤1b.In this interval, we have(57)T−1×∫I2∗xσζaσ+iatLaσ+iat,sym2fζbσ+ibtLbσ+ibt,sym2f⋅ζcσ+ictLcσ+ict,sym2fdσ≪max1/c<σ≤1/bxσT13/42+5/41−aσT13/42+5/41−bσT−1+ε≪max1/c<σ≤1/bT89/42+ɛxT131/84a+bσ≪x1/bT47/84−131a/84b+ɛ+x1/cT89/42−131a+b/84c+ɛ.IntervalI3∗:(58)I3∗≔s=σ+iT:12≤aσ≤1,1<bσ≤bη,c2a≤cσ≤1=s=σ+iT:1b<σ≤1c.This interval is an empty set noting that1/b>1/c.IntervalI4∗:(59)I4∗≔s=σ+iT:12≤aσ≤1,1<bσ≤bη,1<cσ≤cη=s=σ+iT:1b<σ≤1a.In this interval, we have(60)T−1×∫I4∗xσζaσ+iatLaσ+iat,sym2fζbσ+ibtLbσ+ibt,sym2f⋅ζcσ+ictLcσ+ict,sym2fdσ≪max1/b<σ≤1/axσT13/42+5/41−aσT−1+ɛ≪x1/aT−1+ε+x1/bT47/84−131a/84b+ɛ.From (55)–(60), we can obtain(61)J2+J3≪T−1∫1/2aηxσζaσ+iatLaσ+iat,sym2fζbσ+ibtLbσ+ibt,sym2f⋅ζcσ+ictLcσ+ict,sym2fdσ=T−1∫I1∗∪⋯∪I4∗xσζaσ+iatLaσ+iat,sym2fζbσ+ibtLbσ+ibt,sym2f⋅ζcσ+ictLcσ+ict,sym2fdσ≪x1/2aT487/168−131b+c/168a+ɛ+x1/bT47/84−131a/84b+ɛ+x1/cT89/42−131a+b/84c+ɛ+x1/a+ɛT−1+ɛ.Now, we turn to estimateJ1, and we have(62)J1≪x1/2a∫1TL12+iat,f×fLb2a+ibt,f×fLc2a+ict,f×ft−1dt+x1/2a≪x1/2a+x1/2alogTmaxT1≤TT1−1∫T1/2T1ζ12+iatL12+iat,sym2f⋅ζb2a+ibtLb2a+ibt,sym2fζc2a+ictLc2a+ict,sym2fdt.Then, using Lemmas3 and 4 and Hölder’s inequality, we obtain(63)J1≪x1/2a+x1/2alogTmaxT1≤TT1−1T113/421−b/2aT15/41−b/2aT113/421−c/2aT15/41−c/2a×∫T1/2T1ζ12+iatL12+iat,sym2fdt≪x1/2a+x1/2alogTmaxT1≤TT1−1+εT113/421−b/2aT15/41−b/2aT113/421−c/2aT15/41−c/2a×∫T1/2T1ζ12+iat12dt1/12∫T1/2T1L12+iat,sym2f2dt1/2∫T1/2T11dt5/12≪x1/2a+ɛT145/42−131b+c/168a+ɛ.By putting (53), (61), and (63) together, we have(64)Sf×fa,b,c;x=M1x1/a+M2x1/b+M3x1/c+Oxɛx1/2aT145/42−131b+c/168a+x1/bT47/84−131a/84b+x1/cT89/42−131a+b/84c+x1/aT−1+ɛ.By takingT=x84/748a−131b+c in (64), we can obtain(65)Sf×fa,b,c;x=M1x1/a+M2x1/b+M3x1/c+Ox1/a−84/748a−131b+c+ɛ,which proves the first result in Theorem 2.For the caseb<2a<c, we use a similar argument to the first corresponding case. In this case, the points s=1/a and s=1/b are the two simple poles of the integrand in the rectangle IT. Then, by Cauchy’s residue theorem we have(66)Sf×fa,b,c;x=Ress=1/a+Ress=1/bLas,f×fLbs,f×fLbs,f×fxss+12πi∫1/2a−iT1/2a+iT+∫1/2a+iTη+iT+∫η−iT1/2a−iTLas,f×fLbs,f×fLcs,f×fxssds+Ox1/a+εT=M1x1/a+M2x1/b+J1〞+J2〞+J3〞+Ox1/a+εT.To estimateJ2″+J3″, we still divide the integral interval into four short intervals I1∗∗,…,I4∗∗, which are different from ones for the case c≤2a. In fact, the corresponding short intervals I1∗∗,I3∗∗ become empty sets in this situation. However, we still can estimate J2″+J3″ by following a similar argument to the corresponding parts of the case c≤2a and get(67)J2″+J3″≪x1/2aT75/56−131b/168a+ɛ+x1/bT47/84−131a/84b+ɛ+x1/a+ɛT−1+ɛ.The estimate ofJ1′′ becomes the following at the current case by noting c/2a>1>b/2a.(68)J1′≪x1/2a∫1Tζ12+iatL12+iat,sym2fζb2a+ibtLb2a+ibt,sym2f⋅ζc2a+ictLc2a+ict,sym2ft−1dt+x1/2a≪x1/2alogTmaxT1≤TT1−1+εT1131/841−b/2a∫T1/2T1ζ12+iatL12+iat,sym2fdt+x1/2a≪x1/2alogTmaxT1≤TT1−1+εT1131/841−b/2a∫T1/2T1ζ12+iat12dt1/12×∫T1/2T1L12+iat,sym2f2dt1/2∫T1/2T11dt5/12+x1/2a≪x1/2aT159/84−131b/168a+ɛ.Thus, recalling (67), we have(69)Sf×fa,b,c;x=M1x1/a+M2x1/b+Ox1/2aT159/84−131b/168a+ɛ+x1/bT47/84−131a/84b+ɛ+x1/a+ɛT−1+ɛ.By takingT=x84/486a−131b in (69), we have(70)Sf×fa,b,c;x=M1x1/a+M2x1/b+Ox1/a−84/486a−131b+ɛ,which proves the second result of Theorem 2.For the case2a≤b, we use a similar argument to the first corresponding case. In this situation, the point s=1/a is the only simple pole of the integrand in the rectangle IT. Then, Cauchy’s residue theorem shows(71)Sf×fa,b,c;x=Ress=1/aLas,f×fLbs,f×fLcs,f×fxss+Ox1/a+εT+12πi∫1/2a−iT1/2a+iT+∫1/2a+iTη+iT+∫η−iT1/2a−iTLas,f×fLbs,f×fLcs,f×fxssds=M1x1/a+J1′+J2′+J3′+Ox1/a+εT.To estimateJ2′+J3′, we still divide the integral interval into four short intervals I1∗∗∗,…,I4∗∗∗, which are different from ones for the case c≤2a. In fact, the corresponding short intervals I1∗∗∗,I2∗∗∗,I3∗∗∗ become empty sets in this situation. However, we still can estimate J2′+J3′ by following a similar argument to the corresponding parts of the case c≤2a and get(72)J2′+J3′≪x1/2aT−37/168+ɛ+x1/a+ɛT−1+ɛ.The estimate ofJ1′ becomes the following by noting c/2a>b/2a>1.(73)J1′≪x1/2a∫1Tζ12+iatL12+iat,sym2fζb2a+ibtLb2a+ibt,sym2f⋅ζc2a+ictLc2a+ict,sym2ft−1dt+x1/2a≪x1/2alogTmaxT1≤TT1−1+ε∫T1/2T1ζ12+iatL12+iat,sym2fdt+x1/2a≪x1/2alogTmaxT1≤TT1−1+ε∫T1/2T1ζ12+iat12dt1/12×∫T1/2T1L12+iat,sym2f2dt1/2∫T1/2T11dt5/12+x1/2a≪x1/2aT1/3+ɛ.Thus, recalling (71), we have(74)Sf×fa,b,c;x=M1x1/a+Ox1/2aT1/3+ε+x1/a+εT−1.By takingT=x3/8a in (74), we have(75)Sf×fa,b,c;x=M1x1/a+Ox5/8a+ɛ,which proves the third result of Theorem 2.



---

*Source: 9928233-2021-04-20.xml*