# Hybrid Moments of the Riemann Zeta-Function

Authors: Aleksandar Ivić
Journal: Journal of Numbers
 (2015)

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## Abstract

The “hybrid” moments∫T2Tζ1/2+itk∫t-Gt+Gζ1/2+ixldxmdtTε≪G=GT≪T of the Riemann zeta-function ζs on the critical line Res=1/2 are studied. The expected upper bound for the above expression is Oε(T1+εGm). This is shown to be true for certain specific values of k,l,m∈N, and the explicitly determined range of G=G(T;k,l,m). The application to a mean square bound for the Mellin transform function of ζ1/2+ix4 is given.


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## Body

## 1. Introduction

Power moments represent one of the most important parts of the theory of the Riemann zeta-functionζ(s), defined as
(1)ζ(s)=∑n=1∞‍n-s(σ=Res>1)
and otherwise by analytic continuation. Of particular significance are the moments on the “critical line” σ=1/2, and a large literature exists on this subject (see, e.g., the monographs [1–5]). Let us define
(2)Ik(T)=∫0Tζ12+it2kdt,
where k∈R is a fixed, positive number. Naturally one would want to find an asymptotic formula for Ik(T) for a given k, but this is an extremely difficult problem. Conditional bounds, under the Riemann Hypothesis (that all complex zeros of ζ(s) have real parts 1/2, see Riemann [6]), have been obtained by Soundararajan [7] and Radziwill and Soundararajan [8].Except whenk=1 and k=2, no asymptotic formula for Ik(T) is known yet, although there are plausible conjectures for such formulas (see, e.g., the work of Conrey et al. [9–11]). In the absence of asymptotic formulas for Ik(T), one would like then to obtain good upper bounds for Ik(T). A simple bound for |ζ(1/2+it)|k is (see, e.g., [4])
(3)ζ12+itk≪log⁡t∫t-1t+1ζ12+ixkdx+1,
where k∈N is fixed. The use of (3) allows one to replace a power of |ζ(1/2+it)| by its integral over a suitable (short) interval. In employing this procedure one obviously loses something, but on the other hand one gains flexibility from the fact that explicit upper bounds for Ik(T+G)-Ik(T-G) are known only in the case when k=1 (see Lemma 5) and k=2 (see [3]). In this way bounds for Ik+ml(T) are reduced to the so-called “hybrid” moments of the type
(4)∫T2Tζ12+itk∫t-Gt+Gζ12+ixldxmdt00000000000000000000000000k,l,m∈N,
where k, l, and m are assumed to be fixed and 1≪G=G(T)≪T. The expected bound for the expression in (4) (this is consistent with the hitherto unproved Lindelöf hypothesis that ζ(1/2+it)≪ɛ|t|ɛ) is clearly
(5)OɛT1+ɛGm.
Here and later ɛ (>0) denotes arbitrarily small constants, not necessarily the same ones at each occurrence, and a=Oɛ(b) (same as a≪ɛb) means that the implied constant depends only on ɛ. The problem is to find, for given k, l, and m, the range of
(6)G=GT;k,l,m
for which the integral (4) is bounded by (5), and naturally one would like the lower bound for G to be as small as possible. Note that from general results (e.g., see Ramachandra’s monograph [4]) one obtains that the expression in (4) is, for log⁡log⁡T≪G≪T,
(7)≫Gmlog⁡Tl2m/4∫T2Tζ12+itkdt≫TGmlog⁡T(l2m+k2)/4.
This shows that, up to “ɛ,” the bound in (5) is indeed best possible. The (less difficult) case k=0 in (4) was investigated by the author in [12] (l=4) and [13] (l=2). In particular, the former work contains a proof of the bound
(8)∫T2TJ2m(t,G)dt≪ɛT1+ɛ
for T1/2+ɛ⩽G⩽T if m=1,2, for T4/7+ɛ⩽G⩽T if m=3, and for T3/5+ɛ⩽G⩽T if m=4, where
(9)JkT,G:=1πG∫-∞∞ζ12+iT+iu2ke-(u/G)2du000000000000000000000000k>0,1≪G≪T.
The bound (8) in the above range was obtained in [12, 13] by employing Motohashi’s explicit formula (e.g., see [3, 14]) for J2(t,G), which contains quantities from the spectral theory of the non-Euclidean Laplacian.As for the applications of bounds for (4), note that the case (this is k=l=4,  m=1) of the hybrid integral
(10)∫T2Tζ12+it4∫t-Gt+Gζ12+ix4dxdt
appeared in [15] in connection with mean square bounds for the Mellin transform function, defined initially by
(11)Z2(s)=∫1∞ζ12+ix4x-sdx(Rs=σ>1)
and otherwise by analytic continuation. The function Z2(s) was introduced in [16] by Motohashi. The functions Zk(s) (in the general case |ζ(1/2+ix)|4 is replaced by |ζ(1/2+ix)|2k for σ>σ(k) (>1) with suitable σ(k)) are of great importance in the theory of power moments of ζ(1/2+it) (see, e.g., [15]). It was shown by the author in [17] that
(12)∫1T‍Z2σ+it2dt≪ɛT15-12σ/5+ɛ56⩽σ⩽54,
which is the sharpest bound for the range in question.We will obtain results on the integral in (4) when k,l equal 2 or 4, which is logical, since it is in these cases that we have good information on Ik(T). Namely, let, for k∈N fixed,
(13)Ik(T)=∫0Tζ12+it2kdt=TPk2(log⁡T)+Ek(T),
where for some suitable coefficients aj,k one has
(14)Pk2(y)=∑j=0k2aj,kyj,
and Ek(T) is to be considered as the error term in (13). An extensive literature exists on Ek(T), especially on E1(T)≡E(T) (see Atkinson’s classical paper [18]), and the reader is referred to [1] for a comprehensive account. It is known that (γ=-Γ′(1)=0.5772157⋯ is Euler’s constant)
(15)P1y=y+2γ-1-log⁡2π,
and P4(y) is a quartic polynomial in y whose leading coefficient equals 1/(2π2). This was obtained in Ingham’s classical work [19]. For an explicit evaluation of all the coefficients of P4(y) see, for example, Conrey’s work [20]. For the (conjectural) general form of Pk2(y) see [9, 10]. One hopes that
(16)Ek(T)=o(T)(T⟶∞)
will hold for each fixed integer k⩾1, which implies the Lindelöf hypothesis that ζ(1/2+it)≪ɛ|t|ɛ. So far (16) is known to be true only in the cases k=1 and k=2, when Ek(T) is a true error term in the asymptotic formula (13). In particular we have (see [1, 3, 4, 21]) E(T)≪ɛTθ+ɛ for some θ satisfying 1/4⩽θ<1/3, and E(T)=Ω±(T1/4). We also have E2(T)=Ω±(T) and the bounds (see [1, 3, 4, 21])
(17)E2(T)≪T2/3log⁡8T,∫0TE22(t)dt≪T2log⁡22T.
As usual, f(x)=Ω±(g(x)) for a given g(x) (>0 for x>x0>0) means that
(18)limsup⁡x→∞f(x)g(x)>0,liminf⁡x→∞f(x)g(x)<0.


## 2. Statement of Results

Before we state explicitly our results note that we have the bounds(19)∫T-GT+Gζ12+it2dt≪Glog⁡T0000T1/3≪G=GT≪T,∫T-GT+Gζ12+it4dt≪ɛGTɛ(T2/3≪G=G(T)≪T).
This easily follows from estimates on E(T) and E2(T) mentioned at the end of the last section. It means that we can restrict ourselves to the range G≪T1/3 when l=2 in (4), and to the range G≪T2/3 when l=4. This will be implicitly assumed in the proofs of our results, which are contained in the following.Theorem 1.
One has(20)∫T2Tζ12+it2∫t-Gt+Gζ12+ix2dxdt≪TGlog⁡2T0000000000000000000000000Tɛ≪G=GT≪T,(21)∫T2Tζ12+it2∫t-Gt+Gζ12+ix4dxdt≪ɛT1+ɛG0000000000000000000T3/10≪G=GT≪T,(22)∫T2Tζ12+it2∫t-Gt+Gζ12+ix2dx2dt≪ɛT1+ɛG2000000000000000000000T1/7+ɛ≪G=GT≪T,(23)∫T2Tζ12+it2∫t-Gt+G‍ζ12+ix2dx3dt≪ɛT1+ɛG300000000000000000000000T1/5+ɛ≪G=GT≪T,
and for T7/12log⁡CT≪G=G(T)≪T one has
(24)∫T2Tζ12+it2∫t-Gt+Gζ12+ix4dx2dt≪TG2log⁡9T.Theorem 2.
One has, for1≪G=G(T)≪T and some C>0,
(25)∫T2Tζ12+it4∫t-Gt+G‍ζ12+ix4dxdt00000≪log⁡CTTG+min⁡T5/3,T2G-1,(26)∫T2Tζ12+it4∫t-Gt+Gζ12+ix2dx2dt≪ɛT1+ɛG2000000000000000000000000T7/24⩽G=GT≪T,(27)∫T2Tζ12+it4∫t-Gt+Gζ12+ix4dx2dt≪ɛT1+ɛG2000000000000000000000000T5/9⩽G=GT≪T.To assess the strength of our results note, for example, that (3) and (25) give
(28)∫T2Tζ12+it8dt≪log⁡T∫T2Tζ12+it4∫t-Gt+Gζ12+ix4dxdt+∫T2Tζ12+it4dt≪T3/2log⁡CT(G=T1/2).
The bound in (28), which follows easily by the Cauchy-Schwarz inequality for integrals from estimates of the fourth and twelfth moment of |ζ(1/2+it)| (see [22] and [1, Chapter 8]), is the strongest known bound for the eighth moment of |ζ(1/2+it)|.Our last result concerns an improvement of (12). Let ρ be such a constant for which
(29)∫0Tζ12+it8dt≪ɛTρ+ɛ
holds. At present we have 1⩽ρ⩽3/2. The lower bound follows from general principles (see [1, Chapter 9]). The upper bound is a consequence of (28), and its improvements would be very significant. We will prove, using (21), the following.Theorem 3.
IfZ2(s) is defined by (11) and ρ is defined by (29), then
(30)∫1TZ2σ+it2dt≪ɛT4ρ+4-8σ/3ρ-1+ɛ00000000000000005+ρ8⩽σ⩽1+ρ2.Corollary 4.
One has(31)∫1TZ21316+it2dt≪ɛT1+ɛ1316=0.8125,∫1T‍Z21+it2dt≪ɛT4/7+ɛ47=0.571428⋯.Note that (12) gives
(32)∫1TZ256+it2dt≪ɛT1+ɛ,∫1TZ21+it2dt≪ɛT3/5+ɛ,
while (31) improves both of these bounds, since 13/16<5/6 and 4/7<3/5.


## 3. The Necessary Lemmas

In this section we will state some lemmas that are necessary for the proofs of our theorems. The first is an explicit formula for an integral involving|ζ(1/2+it)|2.Lemma 5.
ForTɛ⩽G=G(T)⩽T1-ɛ, one has
(33)1πG∫-∞∞ζ12+iT+iy2e-(y/G)2dy=O(log⁡T)+2∑n=1∞‍-1nd(n)n-1/2T2πn+141/2-12-1/2×exp⁡-G2arsinhπn2T2sinf(T,n),
where d(n) is the number of divisors of n, arsinhz=log(z+z2+1), and
(34)fT,n=2Tarsinhπn2T+2πnT+π2n2-14π=-14π+22πnT+162π3n3/2T-1/2+a5n5/2T-3/2+a7n7/2T-5/2+⋯
for 1⩽n≪T, where a2m-1 are suitable constants.Proof of Lemma5.
The proof of (33) (see also [12]) is based on Motohashi’s exact formula [14] and in particular [3, Theorem 4.1]. It states that
(35)∫-∞∞ζ12+it2g(t)dt=∫-∞∞ReΓ′Γ12+it+2γ-log⁡2πg(t)dt+2πReg12i+4∑n=1∞dn×∫0∞yy+1-1/2gclog⁡1+1ycos⁡(2πny)dy,
where
(36)gc(x):=∫-∞∞g(t)cos⁡(xt)dt
is the cosine Fourier transform of g(t). One requires the function g(r) in (35) to be real-valued for r∈R and that there exists a large constant A>0 such that g(r) is regular and ≪(|r|+1)-A for |Imr|⩽A. The choice
(37)g(t)=1πGe-(T-t)2/G2,gcx=e-1/4(Gx)2cos⁡(Tx)
is permissible, and then the integral on the left-hand side of (35) becomes (see (9)) J1(T,G). The first integral on the right-hand side of (35) is O(log⁡T), and the second one is evaluated by the saddle-point method (see, e.g., [1, Chapter 2]). A convenient result to use is [1, Theorem 2.2 and Lemma 15.1], due originally to Atkinson [18] for the evaluation of exponential integrals ∫ab‍φ(x)exp⁡(2πiF(x))dx. In the latter, only the exponential factor exp⁡(-1/4G2log(1+1/y)) is missing. In the notation of [1, 18] we have that the saddle point x0 (root of F′(x)=0) satisfies
(38)x0=U-12=T2πn+141/2-12,
and the presence of the above exponential factor makes it possible to truncate the series in (35) at n=TG-2log⁡T with a negligible error. Furthermore, in the remaining range for n we have (in the notation of [18])
(39)Φ0μ0F0-3/2≪nT-3/4,
which makes a total contribution of O(1), as does error term integral in Theorem 2.2 of [1]. The error terms with Φ(a), Φ(b) vanish for a→0+,  b→+∞, and (33) follows. Finally note that by using Taylor’s formula it is seen that the error made by replacing
(40)T2πn+141/2-12-1/2exp⁡-G2arsinhπn2T2
with
(41)T2πn-1/4exp⁡-πG2n2T
in (33) is ≪1 for G⩾T1/5log⁡CT.We remark that the series in (33) can be truncated at T1/3. Namely, the contribution for n⩽T1/3 is, by trivial estimation,
(42)≪T-1/4∑n⩽T1/3dnn-1/4≪log⁡T,
and this is absorbed by the O-term in (33).Lemma 6.
LetN denote the number of solutions in integers m, n, and k of the inequality
(43)m+n-k⩽δM(δ>0)
with M′<n⩽2M′, M<m⩽2M, k∈N, and M′⩽M. Then,
(44)N≪ɛMɛM2M′δ+MM′1/2.Lemma 7.
Letk⩾2 be a fixed integer and let δ>0 be given. Then, the number of integers n1,n2,n3,n4 such that N<n1,n2,n3,n4⩽2N and
(45)n11/k+n21/k-n31/k-n41/k<δN1/k
is, for any given ɛ>0,
(46)≪ɛNɛ(N4δ+N2).Lemma6 was proved by Sargos and the author [23], while Lemma 7 is due to Robert and Sargos [24]. They represent powerful arithmetic tools which are essential in the analysis when the cube or biquadrate of exponential sums involving n appears.Lemma 8.
ForHU≫T1+ɛ and Tɛ≪U⩽1/2T, one has
(47)∫TT+HEx+U-Ex2dx=14π2∑n⩽T/2Ud2(n)n3/2×∫TT+H‍x1/2exp⁡2πiUnx-12dx+OɛT1+ɛ+HU1/2Tɛ.This result was proved by Jutila [25]. The analogous formula also holds with E(T) replaced by
(48)Δx:=∑n⩽xd(n)-x(log⁡x+2γ-1),
the error term in the classical Dirichlet divisor problem. From (47), Jutila deduced (a≍b means that a≪b≪a)
(49)∫T2TEx+U-Ex2dx≍TUlog⁡3TUTɛ≪U⩽12T.
The author sharpened (49) to an asymptotic formula. Namely, it was proved in [26] that, with suitable constants ej (e3>0) and Tɛ≪U⩽1/2T,
(50)∫T2TEx+U-Ex2dx=TU∑j=03ejlog⁡jTU+OɛT1/2+ɛU2+OɛT1+ɛU1/2.


## 4. The Proof of Theorem1

We begin with the bound in (20). The left-hand side equals, by the defining relation of E(T) ((13) with k=1),
(51)∫T2Tζ12+it2OGlog⁡T+Et+G-Et-Gdt≪GTlog⁡2T+∫T2T‍ζ12+it2Et+G-Et-Gdt≪GTlog⁡2T+∫T2Tζ12+it4dtllllllllllllllllllllllllllllll×∫T2TEt+G-Et-G2dt1/2≪GTlog⁡2T+Tlog⁡4T·TGlog⁡3T1/2≪GTlog⁡2T
for G⩾Tɛ, as asserted. Here we used the Cauchy-Schwarz inequality for integrals and (49). Note that the upper bound in (20) is the best possible, as it coincides with the lower bound in (7). An interesting, but difficult problem, would be to obtain an asymptotic formula for the integral in (20).To discuss (21), we first exchange the order of integration in the relevant integrals. It follows that the left-hand side of (21) does not exceed
(52)∫T-G2T+Gζ12+ix4∫x-Gx+Gζ12+it2dtdx=∫T-G2T+G‍ζ12+ix4OGlog⁡Tllllllllllllllllllllllllllllllllllllllll+Ex+G-Ex-GOGlog⁡Tdx≪ɛGTlog⁡5T+T547/416+ɛ,
which immediately gives the bound which is somewhat weaker than the one in (21), since 13/10=1.3<547/416=0.314903⋯. Here, we used the sharpest known bound E(T)≪ɛT131/416+ɛ, 131/416=0.314903⋯ of Watt [27], while the related sharpest bound ζ(1/2+it)≪ɛt32/205+ɛ (32/205=0.15609⋯) is due to Huxley [28]. These results have been obtained by a variation of the so-called Bombieri-Iwaniec method for the estimation of exponential sums (see [29, 30]).To obtain the sharper bound asserted by (21) we will use results on the moments of E*(t) (see Section 5), and hence the proof of the bound in question will be completed there.For the proof of (22) we start from (3) which gives, for T/2⩽t⩽5T/2,
(53)ζ12+it2≪log⁡T∫t-Tɛt+Tɛζ12+ix2dx+1,
and we use the trivial inequality
(54)∫t-Gt+Gζ12+ix2kdx=∫-GGζ12+it+iu2kdu⩽e∫-∞∞ζ12+it+iu2ke-(u/G)2du=πeGJkt,G
in the notation of (9), where T/2⩽t⩽5T/2, 1≪G≪T, and k∈N is fixed. This gives
(55)∫T2Tζ12+it2∫t-Gt+Gζ12+ix2dx2dt≪Tɛlog⁡T∫T/25T/2φtJ1t,Tɛ×∫-∞∞ζ12+it+iu2e-(u/G)2du2dt+G2Tlog⁡4T,
following the proof of (20), where φ(t) (⩾0) is a smooth function supported in [T/2,5T/2], such that φ(t)=1 for T⩽t⩽2T and φ(r)(t)≪rT-r for t∈R and any r∈N. For J1(t,Tɛ) we use Lemma 5, writing sinz=(eiz-e-iz)/(2i) and integrate by parts exp⁡(i22πnt). In this way it is seen that
(56)∫T/25T/2φt∑n=1∞-1nd(n)n-1/2⋯exp⁡ift,n×∫-∞∞⋯2dt=∫T/25T/2∫-∞∞‍⋯2φ(t)t∑n=1∞‍i2πn-1nd(n)n-1/2⋯lllllllllllllllllll×exp⁡-i14π+i162π3n3/2t-1/2lllllllllllllllllllllllllllllll+a5in5/2t-3/2+⋯-i14π+i162π3n3/2t-1/2∫-∞∞‍⋯2′dt.
Note that, for G=Tɛ, t≍T, we have
(57)exp⁡-G2arsinhπn2t2′=G2πn/2arsinhπn/2tt3/21+πn/2texp⁡-G2arsinhπn2t2≍G2nt2exp⁡-G2arsinhπn2t2,(58)(φ(t)t)′≪1T,
and that (here G≠Tɛ)
(59)∫-∞∞ζ12+it+iu2e-(u/G)2du2′=2∫-∞∞ζ12+it+iu2e-(u/G)2du×∫-∞∞ddtζ12+it+iu2e-(u/G)2du.
Integrating by parts, we have
(60)∫-∞∞ddtζ12+it+iu2e-(u/G)2du=∫-∞∞dduζ12+it+iu2e-(u/G)2du=2∫-∞∞uG-2ζ12+it+iu2e-(u/G)2du.
Observe that the integrals in (60) can be truncated at |u|=Glog⁡T with a negligible error. Therefore, after an integration by parts, we get an integral with the same type of exponential factor (i.e., f(t,n) in the exponential), but there will be in the integrand a smooth factor of the order ≪G-1T/n. Hence, after a large number of integrations by parts it follows that the contribution of n satisfying n>T1+ɛG-2 will be negligible (i.e., less than T-A for any given A>0 and ɛ=ɛ(A)). This truncation of the series over n is the crucial point in the proof, as the ensuing expression will be quite similar to the expressions for J1(t,G), only in the exponential factor in (57) we will have G=Tɛ. Thus, the proof reduces to the estimation of
(61)TɛG2∫T/25T/2‍φ(t)∑1∑22dt,
where
(62)∑2:=∑n⩽T1+ɛG-2-1nd(n)n-1/2t2πn+141/2-12-1/2×exp⁡-G2arsinhπn2t2sinf(t,n),
and ∑1 is the same expression with G=Tɛ in the exponential factor. The other two terms, which arise after the squaring of the right-hand side of (33), are clearly less difficult to deal with. Note that
(63)∑1∑22=∑m⩽T1+ɛG-2‍-1md(m)⋯sinf(t,m)×∑n⩽T1+ɛG-2‍-1nd(n)⋯sinf(t,n)×∑k⩽T1+ɛG-2‍-1kdk⋯sinft,k,
and write the sines as exponentials. For G⩾T1/7+ɛ, we use Taylor’s formula to remove the terms a7n7/2T-5/2+⋯ from all functions f coming from Lemma 5. Namely, we can truncate the tails of series after sufficiently many terms to obtain a negligible error term. There remain only finitely many terms, but the exponentials are identical, so it suffices to treat the first terms only. Then, we integrate by parts many times, as was done in the previous part of the proof. Thus, we are left with sums containing the exponential
(64)ei(Δt1/2+Et-1/2+Ft-3/2)=eif(t),
say, where we set
(65)Δ:=8πm+n-k,E:=162π3mm+nn-kk,F:=a5m2m+n2n-k2k.
Namely, the terms with
(66)m+n+k,-m-n-k
are clearly negligible by sufficiently many integrations by parts. Thus, only the combination of signs as in (65) is relevant. Here, we suppose that
(67)M′<n⩽2M′,M<m⩽2M,K<k⩽2K,M′⩽M,M,N,K⩾T1/3.
We suppose Δ>0, since the case Δ<0 is analogous and the case Δ=0 is easy. By the first derivative test ([1, Lemma 2.1]) it is seen that the contribution is ≪ɛT1+ɛ (G⩾T1/7) if K<AM or K>BM with suitable positive constants A, B (when Δ≫M or Δ≫K). Thus, we may assume that K≍M. Consider now the contribution from the triplets (m,n,k) (∈N3) satisfying Δ⩽Tɛ-1/2. By using the bound (44) of Lemma 6 (with δ=Tɛ-1/2M-1/2), it is seen that the corresponding portion of the integral in (61) is
(68)≪ɛT1+ɛmax⁡K,M,M′⩽T1+ɛG-2,K≍MT-3/4M-3/4(M5/2T-1/2+M)+T1+ɛ≪ɛT1+ɛmax⁡K,M,M′⩽T1+ɛG-2,K≍M(T-5/4M7/4+T-3/4M1/4)+T1+ɛ≪ɛT1+ɛ(T1/2G-7/2+1)≪ɛT1+ɛ
for G⩾T1/7+ɛ, as asserted.Now we proceed analogously as was done in the author’s work [31]. Suppose Δ⩾Tɛ-1/2. We may assume that E>0, since the other case is analogous. Let
(69)Tɛ-1/2⩽Δ⩽Δ0,
where Δ0, which will be determined later, does not depend on m, n, and k. Further suppose that
(70)2-jΔ0<Δ⩽21-jΔ01⩽j⩽J≍log⁡⁡Δ0T1/2-ɛ.
If FT-3/2≪ET-1/2 or FT-3/2≪ΔT1/2 with suitable ≪-constants, then in eif(t) either Δt1/2 or Et-1/2 dominates in size. Hence, we can use the method of [31]. If we have Δ>C1E/T with a sufficiently large C1>0, then f′(t)≫Δ/T in [T,3T]. Also if Δ⩾Tɛ-1/2 and Δ<C2E/T with a sufficiently small C2>0, then f′(t)≫ET-3/2. In both cases we estimate the integral of eif(t) by the first derivative test, and then the sum over m, n, and k by Lemma 6.IfΔ⩾Tɛ-1/2 and Δ≍E/T, then there may exist a saddle point t0=E/Δ (root of f′(t0)=0) in [T,3T] if ΔT≍E. Hence, by the saddle-point method (see [10, Chapter 2]) or by the use the second derivative test, making first the change of variable t=u, we obtain f′′(t0)=1/2Δ5/2E-3/2≍ΔT-3/2. Hence, by the second derivative test (see Lemma 2.2 of [1]) the corresponding portion of the integral in (61) is
(71)≪∑j⩽J‍2j/2Δ0-1/2T3/4T-3/4M-3/4(M5/2Δ02-j+M)≪Tɛ(M7/4Δ01/2+M1/4T1/4+T)≪ɛT1+ɛ
for M7/4Δ01/2≪T or
(72)Δ0≪T2M-7/2.
But since trivially Δ0≪M and M⩾T1/3 it follows that
(73)T2M-7/2⩾T5/6>M≫Δ0,
which is needed.Finally, if|F|T-3/2≫ET-1/2 and |F|T-3/2≫ΔT1/2, then
(74)f′t≫FT-5/2,Δ≪FT-2.
The first derivative test shows that the contribution is, since M⩽T1+ɛG-2,
(75)≪T5/2F-1M-3/4T-3/4M5/2FT-2+M≪T-1/4M7/4+M1/4T7/4F-1≪ɛT1+ɛT1/2G-7/2+M1/4T3/4F-1≪ɛT1+ɛ
for G⩾T1/7+ɛ, provided that
(76)M1/4T3/4F-1≪1.
If, however, |F|≪M1/4T3/4, then ΔT2≪|F|≪M1/4T3/4 and this implies
(77)Δ≪M1/4T-5/4<T-1/2,
and this case has been already dealt with. This completes the proof of (22).The proof of (23) is similar to the proof of (22). The major difference is that, instead of (61), now we will have to bound
(78)TɛG3∫T/25T/2‍φ(t)∑1∑23dt.
We use then Hölder’s inequality to deduce that the integral in (78) does not exceed
(79)∫T/25T/2‍φ(t)∑14dt1/4∫T/25T/2‍φ(t)∑24dt3/4.
Both integrals in (79) are estimated similarly. Here, we have
(80)T1/5+ɛ≪G=G(T)≪T1/3.
Therefore, by using Taylor’s theorem, instead of the exponential
(81)ei(Δt1/2+Et-1/2+Ft-3/2)
we will have the simpler function ei(Δt1/2+Et-1/2)=eiH(t), say, where now Δ, E have a different meaning than in the proof of (22), namely,
(82)Δ=8πm+n-k-l,E=162π3(mm+nn-kk-ll).First note that the sum overn in ∑1 is split into O(log⁡T) subsums where M<n⩽M′⩽2M, with M≪T1+ɛG-2. Instead of Lemma 6 we use (46) of Lemma 7 (with δ=ΔM-1/2), supposing first that Δ>0 and that Δ⩽Tɛ-1/2. Afterwards the integral is estimated trivially. The contribution to the relevant integral in (79) will be
(83)≪ɛT1+ɛmax⁡M≪T1+ɛG-2T-1M-1(M4Tɛ-1/2M-1/2+M2)+T1+ɛ≪ɛT1+ɛ(TG-5+1)≪ɛT1+ɛ
for G⩾T1/5+ɛ, as asserted.Now supposeΔ>Tɛ-1/2. If Δ>0, we may also assume that E>0, for otherwise all derivatives of H(t) have the same sign. Let Δ>C1E/T for some suitable C1>0. Then, H′(t)≫ΔT-1/2, and supposing that
(84)Δ≍2jTɛ-1/2(j=1,2,…),
we obtain by the first derivative test that the contribution is
(85)≪∑j⩾1‍M-1T-12-jT-ɛM7/22jTɛ-1/2+M2≪ɛM5/2Tɛ-3/2+Tɛ≪ɛT1+ɛ.
Similar arguing is used if Δ<C2E/T. There remains the case when Δ≍E/T, in which case (after the substitution u=t) it is seen that
(86)Ht-1/2≍ΔT-3/2.
Let, as in the proof of (22),
(87)2-jΔ0<Δ⩽21-jΔ01⩽j⩽J≍log⁡Δ0T1/2-ɛ.
For M⩽T1/2 the contribution is
(88)≪∑j⩽J‍M-1T-12j/2Δ0-1/2T3/4M7/22-jΔ0+M2≪ɛM5/2T-1/4Δ01/2+MTɛ≪ɛT1+ɛ,
for Δ0⩽Tɛ, when
(89)M5/2T-1/4Δ01/2⩽TΔ01/2≪ɛT1+ɛ/2.
But if Δ0⩾Tɛ, then since Δ≍E/T, successive integrations by part of eiΔt show that the contribution is negligible.Let now(90)T1/2<M⩽T1+ɛG-2⩽T3/5G⩾T1/5+ɛ.
Like in the preceding case, the contribution will be
(91)≪ɛM5/2T-1/4Δ01/2+MTɛ≪ɛT1+ɛ
for
(92)Δ0⩽T5/2+ɛM-5.
If (92) does not hold, then we have Δ0>Tɛ, since M>T1/2. Again we integrate by parts eiΔt sufficiently many times. Each time we get a factor in the integrand which is
(93)≪1TΔ+TΔ·T-3/2E≪T-1/2-ɛ+T-ɛ-1M3/2≪T-1/2-ɛ+T-ɛ-1T9/10<T-1/10,
so that the contribution will be negligible. This proves (23).To prove (24), note that the left-hand side equals (see (13) with k=2)
(94)∫T2Tζ12+it2lll×OGlog⁡4T+E2t+G-E2t-G2dt≪TG2log⁡9T+∫T2T‍ζ12+it4dt∫T/33T‍E24tdt1/2≪TG2log⁡9T+T13/6log⁡CT≪TG2log⁡12T
for G⩾T7/12log⁡CT, as asserted. Here we used the bound, which follows from (17); namely,
(95)∫0T‍E2tAdt≪T2+2/3(A-2)log⁡CT000000A⩾2,C=22+8A-2
with A=4. This completes the proof of Theorem 1.


## 5. The Proof of Theorem2

We have first, similarly as in (94),
(96)∫T2Tζ12+it4∫t-Gt+Gζ12+ix4dxdt=∫T2Tζ12+it4OGlog⁡4Tllllllllllllllllllllllllllllllllllll+E2t+G-E2t-GOGlog⁡4Tdt≪TGlog⁡8T+T5/3log⁡CT,
where we used the first bound in (17). This implies that the left-hand side of (96) is ≪(TG+T5/3)log⁡CT in the whole range 1≪G=G(T)≪T. The bound in question was actually proved by Ivić et al. [15] in connection with mean square estimates for Z2(s). Hence, the main task is to prove the other bound in (25), for which we need the second bound in (17). Note that the left-hand side of (25) is majorized by a multiple (L=log⁡T) of
(97)∫T2Tζ12+it4∫-∞∞ζ12+it+iu4e-(u/G)2dudt≪TGL8+∫T2Tζ12+it4×∫-∞∞ddtE2t+ue-(u/G)2dudt≪TGL8+2∫T2Tζ12+it4×∫-∞∞E2t+uuG-2e-(u/G)2dudt≪TGL8+LG∫T-GL2T+GL|E2(u)|∫u-GLu+GLζ12+it4dtdu≪LC∫u-GL2u+GL2E2tdtduTG+T3/2+G-2∫T-GL2T+GLE2ullllllllllllllllllllllllllllllllllllllllllllllllll×∫u-GL2u+GL2E2tdtdu.
Here we used the fact that
(98)ddtE2(t+u)=dduE2t+u,
and integrated by parts. We used a similar procedure later, namely (Q4=P4+P4′ (see (13)-(14)) is a suitable polynomial of degree four, C is henceforth a generic positive constant, u≍T):
(99)∫u-GLu+GLζ12+it4dt≪∫-∞∞ζ12+iu+iv4e-(v/GL)2dv=∫-∞∞‍Q4log⁡u+v+E2′u+ve-(v/GL)2dv=O(GLC)+2∫-∞∞vG-2E2(u+v)e-(v/GL)2dv≪LCG+G-1∫u-GL2u+GL2‍E2tdt.
We also used the bound, which follows by the Cauchy-Schwarz inequality for integrals from the mean square bound in (17); namely,
(100)∫0TE2tdt≪T3/2log⁡CT.
To complete the proof it remains to note that
(101)∫T-GL2T+GLE2u∫u-GL2u+GL2E2tdtdu2≪∫T-GL2T+GLE22udu∫T-GL2T+GL∫u-GL2u+GL2E2tdt2du≪T2LC∫T-GL2T+GLGL2∫u-GL2u+GL2E22tdtdu≪T2GLC∫T-GTɛ2T+TɛE22(t)∫t-GL2t+GL2dudt≪T4LCG2.
When we take the square root in (101) and insert the resulting bound in (97) we are left with the bound
(102)OLCTG+T3/2+T2G-1
for the left-hand side of (25). But as T3/2⩽TG for G⩾T1/2 and T3/2⩽T2G-1 for G⩽T1/2, this means that in the bound above the term T3/2 may be omitted, and (25) follows. We point out yet another estimate, namely (131), for the integral in (25). This was derived for the proof of Theorem 3 and does not contain the (expected) term TGlog⁡CT, but terms which are reasonably small when G is “about” T131/416.The proof of (26) is based on the use of (98) and the fourth moment of the function E*(t), defined by
(103)E*(t):=E(t)-2πΔ*t2π,
where (see (48))
(104)Δ*x:=-Δ(x)+2Δ(2x)-12Δ4xllllllllll=12∑n⩽4x-1nd(n)-xlog⁡x+2γ-1,
and Δ(x) is the error term in the Dirichlet divisor problem. The function E*(t) was investigated by several authors, including Jutila [32, 33], who introduced it and the author [12, 17, 31, 34–37]. Among other things, the author ([34–37, Part II] and [31]) proved that
(105)∫0TE*t5dt≪ɛT2+ɛ
and that ([34–37, Part IV, Corollary 2] and [31])
(106)∫0T‍E*t4dt≪ɛT7/4+ɛ.
The advantage of working with E*(t) instead of E(t) is that the former is, in the mean power sense, smaller than the latter (for this see [1, Chapter 15] and [34–37]).For our proof, we need from [13] the elementary formula (cf. (44))
(107)J1t,G=2πG3∫-Glog⁡TGlog⁡TxE*t+xe-(x/G)2dx+Olog⁡2T,
which is valid for Tɛ⩽G=G(T)⩽T1/3, T/2⩽t⩽5T/2. First observe that the left-hand side of (26) is majorized (φ(t) is as in the proof of (22)) by
(108)G2∫T/25T/2ζ12+it4φ(t)J12(t,G)dt=G2∫T/25T/2Q4log⁡t+E2′tφ(t)J12(t,G)dt=OTG2log⁡8T-G2∫T/25T/2‍E2tφtJ12t,G′dt.
Namely, by the Cauchy-Schwarz inequality for integrals and the classical bound
(109)∫0Tζ12+it4dt≪Tlog⁡4T,
we have
(110)G2∫T/25T/2Q4(log⁡t)φ(t)J12(t,G)dt≪Glog⁡4T∫T/25T/2∫-∞∞ζ12+it+iu4e-(u/G)2dudt≪Glog⁡4T∫T/33Tζ12+ix4∫x-GLx+GLe-(x-t)2/G2dtdx≪TG2log⁡8T.
We also use φ′(t)≪1/T  (the contribution of this derivative is easily handled) and (107). Hence, with suitable g(t) we obtain
(111)J12t,G′=2J1t,G1-2x2G2e-(x/G)2-2πG3∫-Glog⁡TGlog⁡TE*t+xllllllllll000lllll×1-2x2G2e-(x/G)2dx+g′t,
where g(t)≪log⁡2T. The term g′(t) is integrated back by parts, and its contribution is easily seen to be ≪ɛT1+ɛG2 in the required range. The first term on the right-hand side of (111) is inserted in (108). It follows that the main contribution to the left-hand side of (108) is (L=log⁡T) bounded by T1+ɛG2 plus
(112)G2∫T/25T/2E2tφ(t)G-5∫-GLGLE*t+xdx2dt≪G-3∫T/25T/2E2t2dt1/2×∫T/25T/2∫-GLGL‍E*t+xdx4dt1/2≪TLCG-3∫T/25T/2G3∫t-GLt+GLE*u4dudt1/2≪TLCG-3∫T/2-GL5T/2+GLG3E*u4∫u-GLu+GLdtdu1/2≪ɛT15/8+ɛG-1,
where we used Hölder’s inequality for integrals, (98) and (106). Since
(113)T15/8G-1⩽TG2(forG⩾T7/24),
the bound in (26) follows.Finally it remains to prove the bound (27) of Theorem 2. We proceed similarly as in the proof of (26), and we majorize first ∫T2T⋯ by ∫T/25T/2φ(t)⋯. Then, we majorize
(114)∫t-Gt+Gζ12+ix4dx2
by G2J22(t,G) and write
(115)ζ12+it4=Q4log⁡t+E2′t,ζ12+it+iu4=Q4log⁡t+u+E2′t+u.
After this we integrate by parts E2′, using (98) and obtaining E2(t),E2(t+u), but gaining essentially a factor of 1/G each time in the process. The major contribution to the left-hand side of (27) will be ≪ɛT1+ɛG2 plus
(116)G-3∫T/25T/2‍φtE2t∫-GLGLE2t+x4e-x/G2dx2dt≪G-3∫T/25T/2‍φtE2t2dt1/2×∫T/25T/2∫t-GLt+GLE2udu4dt1/2.
Now we use (95) with A=4 and Hölder’s inequality, to obtain that
(117)∫T/25T/2‍∫t-GLt+GLE2udu4dt≪GL3∫T/25T/2∫t-GLt+GLE2u4du=GL3∫T/2-GL5T/2+GLE24(u)∫u-GLu+GLdtdu≪T10/3G4logCT.
Hence, if we insert (117) in (116) and use (98), we obtain that the expression in (116) is
(118)≪T8/3G-1logCT≪ɛT1+ɛG2
for G⩾T5/9, which yields then (27) and completes the proof of Theorem 2.It remains yet to complete the proof of (21). From (52), we have
(119)∫T-G2T+Gζ12+ix4∫x-Gx+Gζ12+it2dtdx=∫T-G2T+Gζ12+ix4lllllllllllllll×OGlog⁡T+Ex+G-Ex-Gdx≪ɛGTɛ+∫T-G2T+Gζ12+ix4llllllllllllllllllllllllll×E*x+G-E*x-Gdx.
Here, we used the defining property of E*(t) together with the elementary bound
(120)Δ*x+G-Δ*x-G≪ɛGTɛx≍T,1≪G≪T,
which is easily obtained, since
(121)Δ*(x)=12∑n⩽4x-1nd(n),d(n)≪ɛnɛ.
At this point we use Hölder’s inequality for integrals, (105) and the bound (see [1, Chapter 8])
(122)∫0Tζ12+it5dt≪ɛT9/8+ɛ,
to deduce that the last integral in (119) is
(123)⩽∫T/33T‍ζ12+ix5dx4/5∫T/33TE*x5dx1/5≪ɛT9/8·4/5+2·1/5+ɛ=T13/10+ɛ,
which yields (21). For small values of G, namely, for 1≪G≪T1/10, the bound in (21) may be improved by using a more general result than (49); namely,
(124)∫TT+HEx+U-Ex2dx≍HUlog⁡3TU,
deduced by Jutila [25] from (47), for HU≫T1+ɛ and Tɛ≪U⩽1/2T. If (124) is applied in conjunction with (52), the Cauchy-Schwarz inequality and the bound (see (28)) ∫0T‍|ζ(1/2+it)|8dt≪ɛT3/2+ɛ, we obtain (21) with Tɛ(GT+T5/4G1/2), and
(125)T5/4G1/2⩽T13/101≪G⩽T1/5.
This is unconditional, but conjectures on the order of E(T) and E*(T) would lead to further improvements, for example, the conjectural bound E(T)≪ɛT1/4+ɛ would replace the exponent 13/10 in (21) by 5/4.We end this section by pointing out that one can improve (8) for the range given in Section 1. Namely, since the integral Jk(t,G) in (9) can be truncated at u=±Glog⁡T with an error which is ≪T-A for any given A>0, it follows that (8) is equivalent to
(126)∫T2T∫t-Gt+Gζ12+ix4dxmdt≪ɛT1+ɛGm.
We proceed as in the proof of (24), using (95), to infer that the integral in (126) is (as before C>0 is a generic constant)
(127)≪TGmlog⁡T4m+∫T/33TE2tmdt≪TGmlog⁡T4m+T2+2/3(m-2)log⁡CT≪TGmlog⁡TC
for
(128)G⩾T2m-1/3mm⩾2,
where incidentally m does not have to be an integer. In particular, it follows then from (128) that (8) holds for G⩾T1/2 (m=2),  G⩾T5/9 (m=3), and G⩾T7/12 (m=4). Since 5/9<4/7 and 7/12<3/5, this means that we have improved the range of G for which (8) holds when m=3,4.


## 6. The Proof of Theorem3

The method of obtaining mean square bounds forZ2(s) (see (11)) was developed in [15, 17], so that we will be fairly brief. From [15, pages 337–339], we have
(129)∫T2TZ2σ+it2dt≪ɛT2+ɛX1-2σ+log⁡T×sup⁡T1-ɛ⩽K⩽X∫T/25T/2∫K2Kζ12+ix4x-2σlllllllllllllllllllllllllllllllllllll×∫x-KTɛ-1x+KTɛ-1ζ12+iy4dydxdt≪ɛT2+ɛX1-2σ+Tlog⁡Tsup⁡T1-ɛ⩽K⩽XK-2σ×∫K2Kζ12+ix4∫x-Gx+Gζ12+iy4dydx
with G=KTɛ-1, σ>1/2, and X a parameter to be suitably chosen. For the last integral above one could use (25) of Theorem 2. However, this would not lead to the result of Theorem 3, as we need a bound when G is “about” K1/3, or even slightly smaller. Thus we will proceed differently, and use the elementary inequality
(130)ab⩽12a1/2b3/2+a3/2b1/2a,b⩾0
to obtain
(131)∫T2Tζ12+it4∫t-Gt+Gζ12+ix4dxdt≪∫T2T∫t-Gt+Gζ12+it2ζ12+ix6llllllllllllllllllllllll+ζ12+it6ζ12+ix2dxdt≪∫T-G2T+Gζ12+ix6∫x-Gx+Gζ12+it2dtdx+∫T2Tζ12+it6∫t-Gt+Gζ12+ix2dxdt≪∫T/33Tζ12+it6OGlog⁡Tlllllllllllllllllllllllllllllllllllll+Et+G-Et-GOGlog⁡Tdt≪ɛTɛ+1+ρ/2(G+T131/416).
Here we used the defining relation of E(T) together with the sharpest known bound E(T)≪ɛT131/416+ɛ (see [27]). We also used the bound for the sixth moment which follows from (29) and the Cauchy-Schwarz inequality; namely,
(132)∫0T‍ζ12+it6dt⩽∫0Tζ12+it4dt∫0Tζ12+it8dt1/2≪ɛT1+ρ/2+ɛ.
We use (131) (with G=KTɛ-1) in (129) to obtain
(133)Tsup⁡T1-ɛ⩽K⩽XK-2σ∫K2Kζ12+ix4lllllllllllllllllllllllllll×∫x-Gx+Gζ12+iy4dydx≪ɛT1+ɛsup⁡T1-ɛ⩽K⩽XK1+ρ/2+131/416-2σllllllllllllllllllllllllllllll+T-1K3+ρ/2-2σ≪ɛT1+ɛ+TɛX3+ρ/2-2σ
if
(134)1+ρ4+131832⩽σ⩽3+ρ4.
If (134) holds, then we obtain from (129)(135)∫T2TZ2σ+it2dt≪ɛTɛT+X3+ρ/2-2σ+T2X1-2σ≪ɛTɛT+T6+2ρ-8σ/ρ+1≪ɛT1+ɛ
for σ⩾(5+ρ)/8 if we choose X=T4/(1+ρ). However, for ρ⩽181/104=1.74038⋯, we have
(136)1+ρ4+131832⩽5+ρ8.
With ρ=3/2, we have (5+ρ)/8=13/16, and the first assertion of (31) follows. Note that from (see (29))
(137)∫X2Xζ12+ix8x1-2σdx≪1σ>121+ρ,
one deduces easily that
(138)∫T2TZ2σ+it2dt≪1σ>121+ρ.
Thus, the bound in (30) follows from (135), (138), and the convexity of mean values for regular functions (cf. [1, Lemma 8.3]). Setting σ=1 and ρ=3/2 in (30), we obtain the second bound in (31). We remark that, by [28, equation (3.21)], we have
(139)∫T2Tζ12+it8dt≪ɛT2σ-1∫0T1+ɛZ2σ+it2dt0000000000000000000000000000012<σ⩽1.
The bound (139) links the eighth moment of |ζ(1/2+it)| to the mean square of Z2(s). Thus, the second bound in (31) gives the value ρ=11/7, which is close to the best known bound ρ=3/2.



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