On the Atkinson Formula for the Ζ Function

Home , Erable

arXiv:2104.14253v2 [math.NT] 21 Jun 2021 envletheorem. value mean xoetaecalled are exponent oee ssilupoe,b to o nepii form. Lindelöf hy explicit in [20]. not and or [8] it in be given unproved, are still bounds is however subconvex explicit and [15], aebe nw o ogtm,a es ic adu(e 1,Vl 2 Vol. [13, (see Landau since least at time, for long Currently, a 906]). for known been have h erhfrmaigu onsfor bounds meaningful for search The Introduction 1 h ucinleuto of equation functional the h hwdthat showed who hnacnuy h lsia ojcueon conjecture classical The century. a than ttsthat states 2 1 nteAkno oml o the for formula Atkinson the On < τ < onso order of Bounds nteohrhand, other the On e od n phrases and words Key 2010 22) oevr eetn h tisnfruat h rang the to formula Atkinson the r extend a on we improving Moreover, formula, Titchma Atkinson (2020). of the approach of the version and explicit work an his both Following line. cal uafrtesur enitga au fteReanzt fu zeta Riemann the of value integral mean square the for mula n nepii on o h qaema nerlvleof value integral mean square the for bound explicit an ing on yHlgt n h uhr 21) euemsl class (1918). conve mostly Backlund explicit use by the We given and function (2019). equation authors functional approximate the the and Helfgott by bound ahmtc ujc Classification Subject Mathematics and 1 hnst tisn(98,w nwtefis w em fteas the of terms two first the know we (1938), Atkinson to Thanks

ζ Z 0 | 2 1 T ζ + (

ζ τ ζ ail oa eata uiaAlterman Zuniga Sebastian Dona, Daniele it + ucneiybounds subconvexity | τ 1 2 2 1 t

it | = + 1 + = ) imn eafunction, zeta Riemann : − L 2 | τ it 2 1 it = 2 o + ε eko that know we onsaeese ooti.Casclnnepii versions non-explicit Classical obtain. to easier are bounds ε ζ

( o

| hsipisi atclrthat particular in implies this , ε = t 2 r called are ( | ε dt | t o o any for ) | ε 2 1 = ( − | t ue2,2021 23, June τ T | + 13 84 11M06. : log( ε ζ + Abstract o 0 for ) ( h urn etbudi u oBugi [6], Bourgain to due is bound best current The . s ovxt bounds convexity ε nterne0 range the in ) .Epii ovxbud r ie n[]and [4] in given are bounds convex Explicit ). > ε T 1 (2 + ) L ;b aaadstreln hoe and theorem three-line Hadamard’s by 0; ∞ < τ < L 2 ons aldthe called bounds, om ensur ons xlctbounds, explicit bounds, square mean norm, γ − 1 2 1 . − n onswt vnlower even with bounds and , log(2 < ℜ cn on ySimoniˇc by bound ecent iybud ftezeta the of bounds xity ζ ( s 18) epresent we (1986), rsh π s | nction e ζ n mrvn na on improving and )) ) ( ℜ < T cltos uhas such tools, ical τ ζ ( + s + a pne more spanned has 1 Lindel¨of hypothesis ) function ζ it mttcfor- ymptotic ∈ E ntecriti- the on ) ( 0–1,905– 806–819, , | T 1 4 (1.1) ) = ohssitself pothesis , 4 3 o giv- , ε ( | t | ε for ) , 35 +ε 1 for some function (T ) = O(T 108 ) [10, (15.14)] and (T ) = Ω(T 4 ) [9]. Explicit versions of (1.1) haveE appeared more recently in [7] and [24],E both based on the approximate 3 functional equation for ζ: the error term (T ) in the latter, of order T 4 log(T ), was the record in the explicit case. E 1 3 p Moreover, for 2 <τ < 4 , Matsumoto [16] proved that T 2 ζ(2 2τ) 2 2τ ζ(τ + it) dt = ζ(2τ)T + − T − + (T ) (1.2) | | (2 2τ)(2π)1 2τ Eτ Z0 − − 1 2 3 τ for τ (T ) = Oτ (T 1+4τ log (T )) and τ (T )=Ωτ (T 4 − ); later, (1.2) was extended to 1 E E 2 <τ < 1 by Matsumoto and Meurman [18]. An explicit version of (1.2) has appeared 2 2τ in [7], whose error term of order max T − log(T ), √T absorbs the second main term, and it was the record in the explicit case.{ Any bound of} the form (1.2) can be extended 1 to the range 0 <τ < 2 using the functional equation of ζ(s), and vice versa. In the present paper, we give an explicit version of (1.1) based on a procedure elab- orated by Atkinson [2] and Titchmarsh [27, §7.4], improving on the order of (T ) to √T log2(T ). Moreover, following the same procedure, we give an explicit versionE of (1.2) 3 2 1 1 2 2τ in the range 4 τ < 2 with an error term τ (T ) of order T − log (T ), and then in ≤ E 1 2 1 3 2τ 2 the range 2 < τ 4 with an error term of order T − log (T ). We have already≤ mentioned the O notation and its derivates: for two real-valued functions f,g, the notation f(x)= o(g(x)) means that for any C > 0 there is x0 such that for all x > x0 we have f(x) < Cg(x); an indexed oε indicates that the constant x0 may depend on the variable| ε. Following| the Hardy-Littlewood convention, f(x) = Ω(g(x)) means instead that there is C > 0 such that for any x0 there is some x > x0 with f(x) > Cg(x). | | However, for our purposes we shall use more generally the complex O and O∗ notation. Let f : C C. We write f(s)= O(g( (s), (s))) as s z (z = is allowed) for a real-valued function→ g such that g > 0 inℜ a neighborhoodℑ of→ ( (z), ±∞(z)) to mean that there is an independent constant C such that f(s) Cg( (s)ℜ, (s))ℑ in that neighbor- | |≤ ℜ ℑ hood. We write f(s)= O∗(h( (s), (s))) as s z to indicate that f(s) h( (s), (s)) in a neighborhood of z. ℜ ℑ → | |≤ ℜ ℑ With the notation above at hand, our main result reads as follows. Theorem 1.1. Let T T = 100. Then ≥ 0 T 2 1 2 ζ + it dt = T log(T )+(2γ 1 log(2π))T + O∗(18.169 √T log (T )). 2 − − Z0

Furthermore, if 1 τ < 1 , then 4 ≤ 2 T 2 ζ(2 2τ) 2 2τ 2.215 3 2τ 2 ζ(τ + it) dt = − T − + ζ(2τ)T + O∗ T 2 − log (T ) , 1 2τ 1 2 0 | | (2 2τ)(2π) − τ ! Z − 2 − whereas, if 1 < τ 3 , then 2 ≤ 4 T 2τ 1 2 (2π) − ζ(2 2τ) 2 2τ 4.613 2τ 1 2 ζ(τ + it) dt = ζ(2τ)T + − T − + O∗ T − 2 log (T ) . 1 2 0 | | 2 2τ τ ! Z − − 2 2 For more precise error terms, see Theorem 3.4 and Corollary 3.5. For quantitatively better error terms with higher values of T0 and for results in a wider range of τ, see §4. One might potentially improve on the order of the error term by making later works explicit instead. Atkinson’s later formula [3] oﬀers an estimate for (T ) by way of summations, exact up to error O(log2(T )), based on Vorono¨ı’ssummationE formula for

n X d(n) [28]: it would be feasible to bound such expressions, at the cost of consid- ≤ erable more eﬀort. One could make the estimate for 1 < (s) < 1 of Matsumoto and P 2 Meurman [18] explicit too, and retrieve error bounds for 0 <ℜ (s) < 1 via the functional equation. Other possibilities include following Titchmarsh [26],ℜ Balasubramanian [5], or Ivi´c[10, §15]. For an exposition of some of the aforementioned procedures, we refer the reader to Matsumoto’s survey [17]. Added in proof. Shortly after the appearance the original version of the present paper, Simoniˇcand Starichkova [25] announced that they have given an explicit version 1 5 of (1.1) with an error term of order T 3 log 3 (T ). Their method follows the route through Atkinson’s later paper [3] that we described above: given the constants appearing in 1 their result, the bound we present here for (s)= 2 yields a better error term up to at least T = 1030. ℜ

1.1 Strategy and layout of the presented work As already anticipated, our strategy follows the ideas of Atkinson [2] and Titchmarsh [27, §7.4]. At their core, both results use nothing more than an approximate formula for ζ and several instances of partial summation to estimate a number of weighted sums of the number-of-divisors function d(n). The latter emerge by applying Dirichlet’s convolution to rewrite ζ2, and by appropriately transforming and splitting the integral’s contour via the residue theorem. In particular, we stay closer to Titchmarsh’s ideas in some specific choices of contour 1 for intermediate results (such as Lemma 3.9), which in the case (s)= 2 lead to saving a factor of log(T ) in the error term of one of the main integralsℜ that we estimate (see Proposition 3.1). However, later we diverge from Titchmarsh’s way as many simplifica- ε tions are introduced by applying d(n) = Oε(n ), leading to a final error term of order 1 +ε 2 O(T 2 ). Indeed, since we aim for an error term of order O(√T log (T )), we adopt Atkinson’s approach, by dealing with d(n) by partial summation. We follow essentially the same strategy when working in the range 1 (s) < 1 : in 4 ≤ℜ 2 particular, in the above, we work with the generalized sum-of-divisors functions da : n a R 7→ d n d for a , of which the divisor function d = d0 is a particular case. Our process shows| that Atkinson’s∈ and Titchmarsh’s ideas can be successfully extended outside of the criticalP line while yielding error terms of smaller order than the theoretically predicted two main terms. As a matter of fact, the method applies in principle to the whole critical strip: however, the error terms may be larger than one of the main terms, which is why we decided to concentrate on the regions where this does not happen. The numerical 1 1 estimates improve as well when restricting ourselves to the smaller range 4 (s) < 2 , 1 ≤ℜ when compared to 0 < (s) < 2 . In §2, we collect explicitℜ versions of some classical bounds related to the Riemann ζ

3 function. In §3, we split the integral in Theorem 1.1 into several main pieces; then we estimate each of them in subsequent subsections, in which the relevant weighted sums of da(n) are also bounded. We reserve §4 for commenting about our numerical choices and computations: in it, we also report other versions of the multiplicative constant in the error terms of Theorem 1.1 for diﬀerent choices of T0, as well as showing a result for the 1 whole range 0 < (s) < 2 . For the sake ofℜ rigor, in computing the constants in this article, we have used interval arithmetic implemented by the ARB package [11], which we used via Sage [23]. The necessary code is embedded within the TeX ﬁle of the paper itself via SageTeX.

2 Bounds on functions related to the Riemann Zeta function

Let us recall that the Gamma function Γ is defined for all s C such that (s) > 0 ∞ s 1 t ∈ ℜC as Γ : s 0 t − e− dt. This function can be extended meromorphically to , with simple poles7→ on the set 0, 1, 2, 3,... and vanishing nowhere. Where well-defined, it satisfies theR relationship{ − Γ(s−+1)− = sΓ(} s), so one says that Γ extends the factorial function to the complex numbers. Moreover, this function is closely related to the ζ function, by means of the functional equation, valid for all s C 0, 1 , ∈ \{ } s πs ζ(1 s)= χ(1 s)ζ(s) = 2(2π)− cos Γ(s)ζ(s), (2.1) − − 2 where χ can be extended to a meromorphic function with a simple pole at 1. We will need estimates for the functions involved in the functional equation above. Firstly, concerning the asymptotic behavior of Γ, we have the following. Theorem 2.1 (Explicit Stirling’s formula). Let s = σ + it C ( , 0]. We have ∈ \ −∞ 1 1 1 O∗ s 2 s+ 12s ( 60 s ( s +σ) ) (A1) Γ(s)= √2πs − e− e | | | | ,

π σ 1 1 t σ+ O∗ 1 (t) σ + σ 2 2 12 s 2 ( t=0 60 s ( s +σ) ) (A2) Γ(s) = √2π s − e− | |− | | e { 6 } | | | | | | , | | | | 1 π (A3) (log(Γ(s))) = t log( s ) + sgn(t) σ t ℑ | | − 2 2 − t 1 1 σ 1 + O∗ t=0 (t) σ | | + . − 12 s 2 { 6 } − 2 t 60 s ( s + σ) | | | | | | | |

Moreover, if arg(s) π θ, 0 <θ<π, where arg corresponds to the principal argument of s, then we| have |≤ −

1 Fθ 1 +O∗ s 2 s 12s s 3 (B1) Γ(s)= √2πs − e− e | | ,

3 σ Fθ 1 π 1 σ O∗ 1 (t) | | + σ(1 t=0 (t))+ t=0 3t2 s 3 σ 2 2 t 12 s 2 { 6 } (B2) Γ(s) = √2π s − e− | |e− − { 6 } | | e | | , | | | | 1 π 1 σ (B3) (log(Γ(s))) = t log( s )+ σ sgn(t) t=0 (t) t ℑ | | − 2 2 − { 6 } t −

4 3 t 1 1 σ Fθ + O∗ t=0 (t) σ | | + , − 12 s 2 { 6 } − 2 3 t 3 s 3 | | | | | | 1 where Fθ = 4 θ . 360 sin ( 2 )

Proof. (A1) is given in [22, §2.5 (3”)]; moreover, since s C 0 , s + σ = 0, the estimation is well deﬁned. ∈ \{ } | | 6 Furthermore, by taking real and imaginary parts of the logarithm of Γ, deﬁned through (A1) under the principal complex logarithm log, we derive

log(2π) 1 log(σ2 + t2) σ (log(Γ(s))) = + σ t arg(σ + it) σ + + Ξt , ℜ 2 − 2 2 − − 12(σ2 + t2) σ (2.2) 2 2 t log(σ + t ) 1 t t (log(Γ(s))) = + σ arg(σ + it) t + Ξ′ , (2.3) ℑ 2 − 2 − − 12(σ2 + t2) σ t t 1 where Ξσ , Ξσ′ 60 s ( s +σ) and where arg corresponds to the principal argument func- | | | |≤ | | | | 1 t t tion, which satisfies the identity arg(σ + it) = sgn(t) σ<0 (σ)π + sgn σ arctan |σ| . { } | | 1 Here, sgn corresponds to the sign function and we adopt the conventions sgn 0 = sgn(+ ) = 1 and arctan 1 = arctan(+ )= π . ∞ 0 ∞ 2 π ∞ dt 1 π 1 Now, the estimation arctan (x) = 2 x t2+1 = x=0 (x) 2 + O∗ x , x 0, − { 6 } | | ≥ 1 1 t π σ gives arg(s) = sgn(t) σ<0 (σ)π+ t=0 (t)R sgn σ 2 + O∗ | t| . Thereupon, it is not { } { 6 } | | C 1 1 t π difficult to verify that, for any s ( ,0], sgn( t) σ<0 (σ)π+ t=0 (t)sgn σ 2 = π π ∈ \1 −∞ { } { 6 } sgn(t) , so that t arg(s)= t +O∗ t=0 (t) σ . By using this estimation in (2.2) and 2 2 { 6 } (2.3) (and exponentiating (2.2)),| | we derive (A2)| | and (A3), respectively. On the other hand, set k = 1 in [22, §2.5 (3)] and then observe that µ2 = µ3 can be bounded in C ( , 0] by taking n = 2 in [22, §2.6 (1)] (where ϕ = arg(s)). Now, if \ −∞ 1 1 π θ θ arg(s) π θ ,0 <θ<π, then cos 2 arg(s) = cos 2 arg(s) cos −2 = sin 2 , |whence|≤ the estimation− (B1). | | ≥ t t Fθ Moreover, we can derive (2.2) and (2.3) from (B1), with Ξσ , Ξσ′ s 3 . Finally, | | | | ≤ | | 1 π 1 1 by using the refined estimation arctan (x) = x=0 (x) 2 x + O∗ 3 x 3 , x 0, { 6 } − | | ≥ and proceeding similarly to the obtention of (A2) and (A3), we derive (B2) and (B3), respectively. Secondly, with respect to the complex cosine, we have the following estimation. Proposition 2.2. For s = σ + it C, we may write ∈ π t πs e 2 1 | | O , cos = 1+ ∗ π t 2 2 e | | where cos is the complex cosine function.

5 Proof. For every complex number z, we have the identity sin(z) 2 = cosh2( (z)) cos2( (z)) (for example, combine 4.5.7 and 4.5.54 in [1]). Therefore,| | ℑ − ℜ πs 2 eπ t 1 1 πσ cos = | | 1+ 2+ 4cos2 2 4 eπ t eπ t − 2 | | | | π t π t 2 e | | 2 1 e | | 1 = 1+ O∗ + = 1+ O∗ , (2.4) 4 eπ t e2π t 4 eπ t | | | | | | 2 πσ where we have used that 2 4cos 2 2. The result is concluded by taking square roots in (2.4). − ≤

On the other hand, with respect to ζ itself, Backlund, in equations (53), (54), (56) and (76) of [4], has given an explicit version of a convexity bound for it. It reads as follows. Theorem 2.3 (Explicit convexity bounds of ζ). Let s = σ + it, where t 50. Then ≥ log(t) 0.048 if σ 1, 2 − 1 σ ≥ ζ s t t −2 ( )  t2 4 2π log(t) if 0 σ 1, | |≤  − 1 σ ≤ ≤  t 2 − log(t) if 1 σ 0. 2π − 2 ≤ ≤  t2  As t t2 4 is a decreasing function for t> 2, we immediately deduce 7→ − Corollary 2.4. Let s = σ + it such that t T = 100 and 0 σ 1. Then ≥ 0 ≤ ≤ log(t) if σ 1, 1 σ ≥ ζ(s) − | |≤ (ω t 2 log(t) if 0 σ 1, 2π ≤ ≤ 2 T0 where ω = ω(T0)= T 2 4 1.001. 0 − ≤ Furthermore, we have the following two explicit estimations for ζ when it takes positive values. Proposition 2.5. For any α> 0 and α =1 we have 6 1 α < ζ(α) < . α 1 α 1 − − Proof. See [19, Cor. 1.14]. Lemma 2.6. Let α R+ and X 1. Then ∈ ≥ 1 1 1 (i) = ζ(α) + O∗ , if α> 0 and α =1, nα − (α 1)Xα 1 Xα 6 n X − X≤ − 1 2 (ii) = log(X)+ γ + O∗ , if α =1, nα 3X n X X≤

6 α+1 α X α (iii) n = + O∗(X ), if α 0. α +1 ≥ n X X≤ Proof. By [7, Lemma 2.9], [7, Lemma 2.8] and [21, Lemma 3.1] we derive (i), (ii), (iii), respectively. Finally, we introduce the elementary bounds below, proved by means of Taylor ex- pansions. Lemma 2.7. Let t 1 and 0 <α< 1. Then ≥ t 1 1 1 (a) tα 1+ α(t 1), (b) 1 − t, (c) t + log 1+ 1, ≤ − ≤ log(t) ≤ 2 t ≥ where in (b) we mean for the inequalities to hold for t 1+. →

1 3 R 3 The mean value of the Zeta function in 4, 4 + i

1 In order to derive our main result, we proceed as in [2]. Let τ 0, 2; as ζ(s) = ζ(s), we have that ∈ T 0 ζ (τ + it) 2 dt = ζ (τ + it) 2 dt. 0 | | T | | Z Z− Therefore, we can write

T T 1 τ+iT 1 1 − ζ (τ + it) 2 dt = ζ (τ + it) 2 dt = ζ(1 s)ζ(2τ 1+ s)ds. 0 | | 2 T | | 2i 1 τ iT − − Z Z− Z − − 1 Let 0 <λ< 2 be a parameter. Denote the contour formed by the three lines joining the points 1 τ iT , 2 2τ + λ iTC, 2 2τ + λ + iT , 1 τ + iT . The function s ζ(1 s)−ζ(2τ− 1+ s−) is meromorphic− − and has simple poles− at s = 2 2τ and at7→s = 0,− both with− residue ζ(2τ 1). The only pole inside the region deﬁned− by [1 τ iT, 1 τ + iT ] corresponds− to s =2 2τ. Hence, by residue theorem, C ∪ − − − − T 1 ζ (τ + it) 2 dt = πζ(2τ 1)+ ζ(1 s)ζ(2τ 1+ s)ds. (3.1) 0 | | − − 2i − − Z ZC Now, by (2.1), the integral in the right hand side of (3.1) may be written as

1 1 ζ(1 s)ζ(2τ 1+ s)ds = χ(1 s)ζ(s)ζ(2τ 1+ s)ds. (3.2) 2i − − 2i − − ZC ZC Moreover, as 2 2τ 1, for all s such that (s) > 2 2τ, we have the identity − ≥ ℜ −

1 1 d1 2τ (n) ζ(s)ζ(2τ 1+ s)= = − . (3.3) − ns n2τ 1+s ns n ! n − ! n X X X 7 On the other hand, let X 1 be a parameter. The right hand side of (3.2) can be expressed as I + J, where ≥

1 d1 2τ (n) I = χ(1 s) − ds, (3.4) 2i −  ns  n X ZC X≤   1 d1 2τ (n) J = χ(1 s) ζ(s)ζ(2τ 1+ s) − ds. (3.5) 2i −  − − ns  n X ZC X≤   T In the course of our reasoning, we shall choose X = 2π . The estimation we present for (3.4) is the following. Proposition 3.1. Assume that T T = 100 and X = T . If τ = 1 , then ≥ 0 2π 2

I = T log(T )+(2γ 1 log(2π))T + O∗(19.275 √T log(T )), − − whereas, if 1 τ < 1 , then 4 ≤ 2

ζ(2 2τ) 2 2τ 1.99 3 2τ I = − T − + ζ(2τ)T + O∗ + 35.354 T 2 − log(T ) . (2 2τ)(2π)1 2τ 1 τ − − 2 − The reader should remark that the error term in Proposition 3.1 introduces a saving of a factor of log(T ) with respect to the corresponding estimation presented in [2] when 1 τ = 2 . As for the integral J in (3.5), we may write it as

2 2τ+λ iT 1 τ+iT 1 − − − d1 2τ (n) + χ(1 s) ζ(s)ζ(2τ 1+ s) − ds + K, 2i −  − − ns  1 τ iT 2 2τ+λ+iT ! n X Z − − Z − X≤   J1+J2 (3.6) | {z }

where J1 and J2 are the integrals in the intervals [1 τ iT, 2 2τ + λ iT ] and [2 2τ + λ + iT, 1 τ + iT ] respectively, and where − − − − − − 2 2τ+λ+iT 1 − d1 2τ (n) K χ s − ds d n K , = (1 ) s = 1 2τ ( ) n 2i 2 2τ+λ iT − n ! − Z − − n>XX n>XX 2 2τ+λ+iT 1 − χ(1 s) K ds. n = −s 2i 2 2τ+λ iT n Z − − The expression for K has been derived with the help of identity (3.3) in the ﬁrst equality, and the dominated convergence theorem in the second (as s χ(1 s) is continuous in the 7→ − d1 2τ (n) compact set [2 2τ +λ iT, 2 2τ +λ+iT ] C and − ζ(1+λ)ζ(2 2τ +λ) − − − ⊂ n>X ns ≤ − in that set). P With the deﬁnitions above, it will become clear in §3.2 why we will end up selecting 1 λ of order log(T ) . With that choice, J1, J2 and K are estimated as follows.

8 Proposition 3.2. Assume that T T = 100, X = T and λ = 1.501 . If τ = 1 , then ≥ 0 2π log(T ) 2 J 0.47 √T log2(T )+2.825 √T log(T ), | 2|≤ whereas, if 1 τ < 1 , then 4 ≤ 2

0.173 3 2 2τ 1 τ 2 J2 2 +6.76 T − log(T )+0.1 T − log (T ). | |≤ 1 τ ! 2 − On replacing T by T > 0, the same bound may be derived for J . − 1 Proposition 3.3. Assume that T T = 100, X = T and λ = 1.501 . If τ = 1 , then ≥ 0 2π log(T ) 2 K 3.097 √T log2(T )+40.116 √T log(T ), | |≤ whereas, if 1 τ < 1 , then 4 ≤ 2

0.4 3 2τ 2 K + 20.072 T 2 − log (T ). | |≤ 1 τ 2 − The following sections consist of the proof of Propositions 3.1, 3.2 and 3.3: we will analyze I in §3.1, J1 and J2 in §3.2, and K in §3.3. Consequently, by combining them, we derive our main result, which reads as follows. Theorem 3.4. Assume that T T = 100. Then ≥ 0 T 1 2 ζ + it dt = T log(T )+(2γ 1 log(2π))T 2 − − Z0 2 + O∗(4.037 √T log (T )+65.076 √T log(T )),

and if 1 τ < 1 , then 4 ≤ 2 T 2 ζ(2 2τ) 2 2τ ζ (τ + it) dt = − T − + ζ(2τ)T | | (2 2τ)(2π)1 2τ Z0 − − 0.284 3 2 2τ 2 + O∗ 2 + 30.893 T − log (T ) . 1 τ ! ! 2 − By using the functional equation of ζ, we can derive a bound in the other half of the critical strip. Corollary 3.5. Assume that T T = 100. Then if 1 < τ 3 , then ≥ 0 2 ≤ 4 T 2τ 1 2 (2π) − ζ(2 2τ) 2 2τ ζ (τ + it) dt = ζ(2τ)T + − T − | | 2 2τ Z0 − 0.627 1 2τ 2 2 + O∗ 2 + 63.767 T − log (T ) . τ 1 ! ! − 2 9 1 1 Proof. By expressing τ = 1 τ ′, with 4 τ ′ < 2 , observing that ζ(s) = ζ(s) and recalling (2.1), we derive − ≤ T T T 2 2 2 ζ (τ + it) dt = ζ (1 τ ′ it) dt = χ(1 τ ′ it)ζ(τ ′ + it) dt. (3.7) | | | − − | | − − | Z0 Z0 Z0 π Note that τ ′ + i[T0,T ] belongs to the angular sector deﬁned by arg(s) < 2 . We π 1 | | can then use Theorem 2.1(B2) with t > 0 and θ = 2 (Fθ = 90 ) and the estimation 2 4 τ ′ τ ′ log( s ) = log(t)+ 2 + O∗ 4 to obtain that | | 2t 4t

V(τ ,T ) 1 π ′ 0 τ ′ t O∗ 2 Γ(τ ′ + it) = √2πt − 2 e− 2 e t , (3.8) | | 1 1 1 1 1 1 where, by using that s 2 t2 , t3 T t2 and that 4 τ ′ < 2 , | | ≤ ≤ 0 ≤ 2 4 3 1 τ ′ τ ′ τ ′ τ ′ 1 V(τ ′,T )= τ ′ + + + + < V′(T )=0.115. 0 2 − 2 4T 2 12 3 90T 0 0 0 2V (T ) 2V′(T0) ′ 0 O∗ 2 W(T0) T 2 t 2 0 Moreover, observe that e =1+ O∗ 2 , where W(T )= T (e 1). t 0 0 − Thus, from Proposition 2.2 and (3.8), we have 2 2 2 1 2τ ′ 2τ ′ 1 1 W(T0) χ(1 τ ′ it) = (2π) − t − 1+ O∗ 1+ O∗ | − − | eπt t2 Z(T ) = (2π)1 2τ ′ t2τ ′ 1 1+ O 0 , − − ∗ t2 t2 where we have used that t T0 and that the function t T0 eπt is decreasing, ≥ ≥ 7→ 2 T 2 T 2 Z T . W T 0 W (T0) 1 W T 0 W (T0) deﬁning ( 0)=0 459 2 ( 0)+ eπT0 + eπT0 + T 2 ( 0)+ eπT0 + eπT0 . ≥ 0 We conclude from (3.7) that T ζ (τ + it) 2 dt equals 0 | | T0 R T 2 1 2τ ′ 2τ ′ 1 2 Z(T0) ζ (τ + it) dt + (2π) − t − ζ(τ ′ + it) 1+ O∗ dt. (3.9) | | | | t2 Z0 ZT0 T0 2 By a rigorous numerical estimation, we have 0 ζ (τ + it) dt i = 319.388 (see §4 for more details). | | ≤ R For the second term in (3.9), since ζ is holomorphic in C 1 and t ζ (τ + it) 2 \{ } 7→ | | is continuous in [T0,T ], by the fundamental theorem of calculus we have that for any υ, T T t υ 2 υ d 2 t ζ(τ ′ + it) dt = t ζ(τ ′ + it′) dt′ dt | | dt | | ZT0 ZT0 ZT0 T T t υ 2 υ 1 2 = T ζ(τ ′ + it′) dt′ υt − ζ(τ ′ + it′) dt′ dt. (3.10) | | − | | ZT0 ZT0 ZT0 t 2 Furthermore, by Theorem 3.4, for any t T0, we can write ζ(τ ′ + it) dt = Mτ (t) ≥ T0 | | ′ − M (T )+ O (2E (t)), where τ ′ 0 ∗ τ ′ R

ζ(2 2τ ′) 2 2τ ′ Mτ ′ (t)= − t − + ζ(2τ ′)t, (2 2τ )(2π)1 2τ ′ − ′ −

10 3 0.284 2τ ′ 2 E 2 τ ′ (t)= 2 + 30.893 t − log (t). 1 τ ! 2 − ′ Therefore, by assuming that υ < 0, υ = (2 2τ ′), υ = 1 and using integration by parts, we conclude from (3.10) that 6 − − 6 −

T T T υ 2 υ υ υ 1 t ζ(τ ′ + it) dt = t M (t)′dt + O∗ 2T E (T )+ υ t − E (t)dt | | τ ′ τ ′ | | τ ′ ZT0 ZT0 ZT0 ! T υ υ ζ(2 2τ ′) 2 2τ ′+υ 2 2τ ′+υ = t Mτ ′ (t)′dt + O∗ (2T0 Eτ ′ (T )) = − (T − T0 − ) (2 2τ + υ)(2π)1 2τ ′ − ZT0 − ′ − ζ(2τ ′) υ+1 υ+1 υ + (T T )+ O∗ (2T E (T )) , υ +1 − 0 0 τ ′ where we have used that υ = υ, that E is an increasing function and that T υ T υ. | | − τ ′ ≤ 0 In particular, when υ 2τ ′ 1, 2τ ′ 3 we can estimate (3.9) as ∈{ − − }

1 2τ ′ 1 2τ ′ (2π) − ζ(2τ ′) 2τ ′ 2(2π) − Z(T0) ζ(2 2τ ′)T + T + O∗ 1+ Eτ (T ) − 2τ T 1 2τ ′ T 2 ′ ′ 0 − 0 ! 3 2τ ′ 2 Z(T0) T 2 − log (T ) + O∗ i + z + z , (3.11) 2τ ′ 2 2 2τ ′ 3 2τ | | T0 | − | 2 − ′ 2 T0 log (T0)!

3 2τ ′ since t t 2 − log(t) is increasing in [T ,T ], where we have deﬁned 7→ 0

1 2τ ′ (2π) − ζ(2τ ′) 2τ ′ z = ζ(2 2τ ′)T + T . υ − 0 υ 0

1 1 Using Proposition 2.5, τ ′ , and T 2π, we have ∈ 4 2 0 ≥ 1 z z 2τ ′ 2 2τ ′ < ζ(2 2τ ′)T0 < 1+ 2 T0, ≤ − − 8 1 τ ! 2 − ′ ζ(2τ ′) 1 1 z 2τ ′ ζ(2 2τ ′)+ T0 > T0 > 1+ 2 T0, ≥ − 2τ ′ −2τ ′ − 8 1 τ ! 2 − ′ so we have a bound on z2τ , z2 2τ that we can plug into (3.11). | ′ | | − ′ | The result is concluded by replacing τ ′ by 1 τ and calculating the constants in the error term. −

3.1 The integral I We readily derive that

1 χ(1 s) I = d1 2τ (n) − ds = d1 2τ (n) In, − 2i ns − n T ZC n T X≤ 2π X≤ 2π

11 1 τ+iT 1 − χ(1 s) In = −s ds, 2i 1 τ iT n Z − − where in the ﬁrst equality we have used the ﬁniteness of the summation and in the second that the function s χ(1 s) is holomorphic, so that its residues vanish. First, we establish7→ here− some results about the average of arithmetical functions involving the function da. Proposition 3.6. Let X 1 and 0 <σ< 1. Then ≥ ζ(2 2σ) 2 2σ 1 σ 1 d1 2σ(n)= ζ(2σ)X + − X − + O∗ AσX − if σ = , − 2 2σ 6 2 n X − X≤ 1 d0(n)= X log(X)+(2γ 1)X + O∗ A 1 √X if σ = , − 2 2 n X X≤ 1+2σ 1 16 1 1 where Aσ =4+ σ(2 2σ) for σ = 2 , and A 1 = 3 . In particular, Aσ 8 for σ 4 , 2 . − 6 2 ≤ ∈ Proof. By the hyperbola method, we have

a a da(n)= m + m 1 n X n X m n m>√X d X ≤ ≤ | ≤ m X X mX√X X X ≤ = ma 1+ ma (3.12) m √X d X d √X √X

1 2σ 1 1 2σ m − 1= X + O∗ m − m2σ   m √X d X m √X m √X X≤ X≤ m X≤ X≤ 3 σ 1 σ  X 2 − 1 σ X − 1 σ = ζ(2σ)X + O∗(X − )+ O∗ + X 2 − − 2σ 1 2 2σ − − 3 σ X 2 − 5 4σ 1 σ = ζ(2σ)X + O∗ − X − , (3.13) − 2σ 1 2 2σ − − 1 where the last line holds for X 1. For 2 <σ< 1, by Lemma 2.6(i)-(iii) we obtain the same estimate as in (3.13): in fact≥ one more term ζ(2σ 1) emerges, but it is negative 1 σ X − − 1 and bounded in absolute value by 2 2σ by Proposition 2.5 and X 1. For σ = 2 , by Lemma 2.6(ii), we get instead − ≥ 1 1 5 1= X + O∗(√X)= X log(X)+ γX + O∗ √X . (3.14) m 2 3 m √X d X m √X X≤ X≤ m X≤

12 We also use Lemma 2.6 for the inner sum of the second term of (3.12), and derive

2 2σ 1 σ 1 2σ 1 2σ 1 X − X − X − 1 σ m − = + O∗ + X 2 − . (3.15) 2 2σ d − 2 2σ d X ! √X

3 σ 1 2σ ζ(2 2σ) 2 2σ X 2 − 1 σ m − = − X − + + O∗ B X − (3.17) 2 2σ 2σ 1 σ d √X √X 1 and 0 < σ 2 . Recall the deﬁnition of Aσ given in ≥ 1 ≤ Proposition 3.6. Then, if 0 <σ< 2 ,

d1 2σ(n) ζ(2 2σ) 3 2σ 1 σ − =2ζ(2σ)√X + − X 2 − + O∗(Cσ X 2 − ), √n 3 2σ n X 2 X≤ − 2 2σ where Cσ =3+ 1−2σ Aσ, and − d0(n) =2√X log(X) 4(1 γ)√X + O∗ C 1 log(X) , √n − − 2 n X X≤ 1 1 where C 1 = A 1 + (A 1 +3 2γ) . 2 2 2 2 − log(X0)

13 Proof. Let X Mσ(X) be the main term defined by Proposition 3.6, according to 1 7→ 1 whether σ < 2 or σ = 2 . Since X 1, by Lemma 3.6, we can write n X d1 2σ(n)= ≥ ≤ − M (X)+O A X1 σ so that, by summation by parts, we conclude that, for any υ > 0, σ ∗ σ − P X d1 2σ(n) 1 σ υ 1 σ υ − = (M (X)+ O∗(A X − ))X− (M (t)+ O∗(A t − ))(t− )′dt nυ σ σ − σ σ n X 1 X≤ Z X X dt = M (t) t υdt + M (1) + O A X1 σ υ + υA . (3.19) σ ′ − σ ∗ σ − − σ tσ+υ Z1 Z1 ! 1 1 If 0 <σ< 2 and υ = 2 , by Proposition 2.5, we bound the constant arising from the first two terms of (3.19) as 1 1 1 1 1 < ζ(2σ) ζ(2 2σ) < 1 < 3. −2 − − − 3 2σ − 2 2σ 1 2σ − 2( 3 2σ)(2 2σ) 2 − − − 2 − − Thereupon, we keep the first two terms of highest order in (3.19) and we merge the 1 σ remaining ones to the order X 2 − , obtaining Cσ. 1 On the other hand, if σ = υ = 2 , we derive from (3.19)

d0(n) log(X) =2√X log(X) 4(1 γ)√X +3 2γ + O∗ A 1 1+ (3.20) √n − − − 2 2 n X X≤ log(t) where we have used that dt = 2√t(log(t) 2). Similarly, we can deﬁne C 1 by √t − 2 merging the error term of (3.20) to the order log(X). R Finally, we are going to need the mean estimation below.

1 Lemma 3.8. Let T T0 = 100 and 0 < σ 2 . Recall the deﬁnition of Aσ given in Proposition 3.6. Then,≥ for υ 1 , 1 σ , ≤ ∈ 2 − d1 2σ(n) F1 ,σ,υ log(T ) F2,σ,υ F3,σ,υ − + + υ 2σ 1+υ 2σ 1+υ σ 1 +υ n (T 2πn) ≤ T − T − T 2 T √T − n − − ≤X2π F log(T ) F F + 4,σ,υ + 5,σ,υ + 6,σ,υ , T σ+υ T σ+υ T 1 where for 0 <σ< 2 ζ(2 2σ) ζ(2 2σ) √π F1,σ,1 σ = − , F 1 = −3 , F 1 1 = , 1 σ 1,σ, 2 2σ 1, 2 , 2 − (4π) − (4π) 2 − 2√2 2ζ(2 2σ) 2ζ(2 2σ) F2,σ,1 σ = − , F 1 = 3 − , F 1 1 =0, 1 σ 2,σ, 2 2σ 3 2, 2 , 2 − (4π) − (1 σ) (4π) 2 ( 2σ) − − 2 − 2Aσ F3,σ,1 σ =2Aσ, F 1 = 1 , F 1 1 =2A 1 , 3,σ, 2 σ 3, 2 , 2 2 − (2π) 2 − 3A 1 2 F4,σ,1 σ = (2 σ)(1 σ)Aσ , F4,σ, 1 =0, F4, 1 , 1 = , − − − 2 2 2 4

14 3(1 σ)Aσ F5,σ,1 σ =0, F 1 = 1 − , F 1 1 =0, 5,σ, 2 σ 5, 2 , 2 − (6π) 2 (1 2σ) − − T0 T0 (2γ 1)T0 F6,σ,1 σ = , F6,σ, 1 = , F6, 1 , 1 = − . − T 2π 2 T 2π 2 2 T 2π 0 − 0 − 0 − Proof. By recalling Proposition 3.6, for any 0 < σ 1 and any t 1, we have that ≤ 2 ≥ n t d1 2σ(n) = Mσ(t) + Ξσ(t), where the function Ξσ satisﬁes ≤ − P ζ(2 2σ) 1 1 σ 1 ζ(2σ) 2 −2σ if 0 <σ< 2 , Ξσ(t) Aσt − , Ξσ(1) = 1 Mσ(1) = − − − | |≤ − 2 2γ if σ = 1 . ( − 2 Therefore, by summation by parts, for υ > 0 we have

T √T − d1 2σ(n) 2π Mσ(t)′dt − = +r (T ), (3.21) nυ(T 2πn) tυ(T 2πt) σ,υ T √T 1 n − − Z − ≤X2π with

T √T T √T 1 Ξ (1) Ξσ −2π −2π 1 ′ T σ t dt. rσ,υ( )= − + υ Ξσ( ) υ T 2π T √T √ − 1 t (T 2πt) − −2π T Z − As the only zero of the derivative of t 1 is υT , for υ 1 , 1 σ , we 7→ tυ (T 2πt) (υ+1)2π ∈ 2 − obtain − 1 σ υ υT − − 1 σ υ (υ+1)2π 1 Ξσ(1) Aσ T √T Aσt − − rσ,υ(T ) | − | + − | |≤ T 2π √T 2π − T 2πt ! 1 − − υT T √T T √T 1 σ υ −2π − (υ+1)2π (1 σ)A dt A t − − 2π (1 σ)A dt σ σ σ . + σ+υ− + σ+υ− 1 t (T 2πt) T 2πt υT − υT t (T 2πt) Z (υ+1)2π Z (υ+1)2π − − − a t T Since the function t T 2πt is positive and increasing for t < 2π and any a > 0, in the inequality above7→ we can− dismiss the third and sixth term, as they are negative and, 1 1 moreover, we can bound it as follows. For υ = and σ = , we conclude that r 1 (T ) 2 2 σ, 2 is bounded by 6 | |

1 σ 1 Ξ (1) A 3(1 σ)A T 2 − 1 A | − σ | + σ + − σ 6π − + σ 1 σ 1 1 σ T 2π (2π) 2 T σ 2T σ (2π) 2 T σ − − 2 − − 1 Ξ (1) T 2A 3(1 σ)A | − σ | 0 + σ + − σ , (3.22) 1 σ 1 σ 1 +σ ≤ (T0 2π)T (2π) 2 T σ (6π) 2 (1 2σ)T 2 − − − − where in the obtention of the ﬁrst term above we used that T T . Similarly, for ≥ 0 υ =1 σ, rσ,1 σ(T ) is bounded by − | − | 1 Ξ (1) A (2 σ)(1 σ)A (1 σ)T A | − σ | + σ + − − σ log − + σ T 2π √T T (2 σ)2π √T − −

15 1 Ξσ(1) T0 2Aσ log(T ) | − | + + (2 σ)(1 σ)Aσ , (3.23) ≤ (T0 2π)T √T − − T − (1 σ)T (2 σ)2π where we ignored the negative term coming from log (2 −σ)2π = log(T ) log −1 σ . − − − 1 In both (3.22) and (3.23), we have also 1 Ξσ(1) <1 for 0 <σ< 2 , by Proposition 2.5. 1 | − | If 0 <σ< 2 , the main term in (3.21) equals

T √T T √T ζ(2σ) −2π dt ζ(2 2σ) −2π dt + − . (3.24) 2π tυ T t 2π t2σ 1+υ T t Z1 2π − Z1 − 2π − Since ζ(2σ) < 0, in order to obtain an upper bound, we can dismiss the ﬁrst in- T tegral in (3.24). Subsequently, we can divide the interval of integration into 1, 4π T T √T T T 2σ 1+υ and , − ; then, bounding t in the ﬁrst denominator and t − 4π 2π 2π − ≥ 4π ≥ 2σ 1+υ T h − i 4π in the second, the second term of (3.24) is bounded by

T 2σ 1+υ T √T ζ(2 2σ) 4π 4π dt 4π − −2π dt − 2σ 1+υ + T 2π T 1 t − T T t! Z Z 4π 2π − T 2 2σ υ 2σ 1+υ 2ζ(2 2σ) − − 1 ζ(2 2σ)(4π) − √T = − 4π − + − log T 2 2σ υ 2πT 2σ 1+υ 2 − − − ! 2ζ(2 2σ) ζ(2 2σ) log(T ) < − + − . (3.25) (4π)2 2σ υ(2 2σ υ)T 2σ 1+υ (4π)2 2σ υT 2σ 1+υ − − − − − − − − 1 On the other hand, if σ = υ = 2 , the main term given by (3.21) may be bounded as

T √T T √T 1 −2π (log(t)+2γ)dt log(T ) −2π dt < T T 2π 1 √t t 2π 1 √t t Z 2π − Z 2π − T √T − log(T ) 2√2π 2πt 2π √2 log(T ) arctan(1) √π log(T ) = arctan < = , (3.26) 2π " √T T ! √π√T 2√2√T r 1

where we have used that log(t)+2γ < log(T ) log(2π)+2γ < log(T ). The result is concluded by putting (3.22), (3.23),− (3.25) and (3.26) together. Lemma 3.9. Let U > 0. For any n Z such that n< U , we have ∈ >0 e22π 1 1 + 1 2 iU 24U 30U2 1 1 − χ(1 s) √π e 1+ πU e 1 − ds e + . i ns e2U U 2 iU ≤ √2 √2πn log √U log 2 ! Z−∞− 2πn e 2πn

Proof. By Proposition 2.2, (2.1) and Theorem 2.1(A2) we have the following general estimation:

1 χ(1 s) πs s − = Γ(s)cos (2πn)− 2i ns 2

16 σ 1 √2π 1 O∗ 2 σ + | | + σ σ 2 π t 12 s 2 60 s ( s +σ) = (2πn)− s − 1+ O∗ e− | | e | | | | | | | | . (3.27) 2 | | 1 1 σ 1 σ 2 σ 2 1 In particular, if σ 2 and t = 0, we have that s − t − , |s 2| 2 t and ≤ 6 | | ≤ | | | | ≤ | | 1 s σ 2 s ( s +σ) = |t2|−s t2 . Thus, from (3.27), we conclude that | | | | | | ≤

1 χ(1 s) √2π 1 2 σ + 1 + 1 σ σ 2 π t 24 t 30t2 − (2πn)− t − 1+ e− | | e | | | | . (3.28) 2i ns ≤ 2 | |

Hence, from (3.28), we readily see that

1 1 + 1 1 2 iU 24U 30U2 1 2 1 − χ(1 s) √2π e 1+ πU ds e πn σU σe2 σ dσ. −s (2 )− | | 2i iU n ≤ 2√U Z−∞− Z−∞

The result is concluded by splitting the integral above at σ = 0 and dismissing the negative term that arises in the range σ 0, 1 . ∈ 2 The following result is crucial since, rather than providing an estimation, it exhibits an asymptotic formula. As it turns out, it is the main term of this formula that will give the main term and secondary term of the moment of order 2 of the zeta function in the critical strip. Z T Lemma 3.10. Let T T0 = 100. For any n >0 such that n 2π , we have the following estimation: ≥ ∈ ≤

I =2π + O∗ (2 A +2 B +2 C ) , n | n| | n| | n| where

G1 G2 G1 =0.29, An + , | |≤ √n √T G2 =0.471,

T √T and, if n − , ≤ 2π

H H H H1 =1.001, H2 =0.02, B 1 + 2 + 3 , (3.29) n T T T n | |≤ √n log 2πn T √n log 2πn √ H3 =0.04, 1 τ 1 τ H4 T 2 − (2πn) 2 − C − , H =0.501; (3.30) n 1 τ 1 τ T 4 | |≤ (2π) 2 − n − log 2πn

T √T T otherwise, if −

H1′ √T H2′ H3′ H4′ H1′ =1.501, H2′ =0.5, Bn + + + , (3.31) | |≤ √n √n √T √n T √n H3′ =0.02, H4′ = H3, 1 τ 1 H5′ T 2 − C τ , H′ = H . (3.32) n 1 τ 1 τ 5 4 | |≤ 2 − π 2 n (2 ) − −

17 T Remark 3.11. Observe that (3.29) and (3.30) are not as sharp as n approaches 2π . Indeed, if 2πn>T √T then, by Lemma 2.7(b), 1 2πn √T , and thus it log T T 2πn − ( 2πn ) ≥ − ≫ is better to consider the bounds (3.31) and (3.32), respectively. Instead, if 2πn T √T , we have that 1 T √T , and thus it is better to consider the bound≤ − (3.29) log T T 2πn ( 2πn ) ≤ − ≪ 1 τ 1 τ T 2 − (2πn) 2 − 1 2 τ over the one given in (3.31); moreover, as − T τ T − , it is also better in log( 2πn ) ≪ this case to consider the bound (3.30) over (3.32). Proof. Let U >T . By using (2.1), the residue theorem and Theorem 2.1(A2) we conclude, as in [27, §7.4], that for any U > 0 we have 2π = In +An An +Bn Bn +Cn Cn, where − − −

1 2 iU 1 − πs s An = Γ(s)cos (2πn)− ds, i iU 2 −∞− Z 1 2 iT 1 − πs s Bn = Γ(s)cos (2πn)− ds, i 1 iU 2 Z 2 − 1 τ iT 1 − − πs s Cn = Γ(s)cos (2πn)− ds. i 1 iT 2 Z 2 − 2 Furthermore, by selecting U = 2e T and using that T T0, An can be estimated with the help of Lemma 3.9, giving ≥

√ 1 + 1 2π 48e2T 4 2 1 e 1 0 120e T0 An e 1+ 2 4 + | |≤ 2 e2πe T0 √2πn log 2e T √2e√T log T 2πn πn ! √ 1 + 1 2π 48e2T 120e4 T 2 1 e 1 0 0 e 1+ 2 + , (3.33) ≤ 2 e2πe T0 √ π e4 √n √ e √T 2 log (2 ) 2 log (2) T where we have used that n 2π . ≤ 1 With respect to Bn, observe that 2 i[U,T ] is a subset of the angular sector deﬁned π − π by arg(s) < 2 ; hence, we can use Theorem 2.1(B2)-(B3) with σ 0, t< 0 and θ = 2 | | 1 ≥ (so that Fθ = 90 ), along with the deﬁnition of the complex cosine and write

πi 1 πs s e− 2 log( Γ(s) )+i (log(Γ(s))) π is π is log(2πn)s Γ(s)cos (2πn)− = e | | ℑ e 2 + e− 2 e− , i 2 2 x2 which, upon using that log(1 + x)= x + O∗ for x> 0, so that log( s ) = log( t )+ 2 | | | | σ2 σ4 1 1 σ2 1 1 2t2 + O∗ 4t4 , that s 2 = t2 + O∗ t4 and that s 3 t 3 , may be rewritten as | | | | ≤ | | 1 σ πi (2π) 2 − e− 4 σ 1 if(σ,t) πσi πt r (σ,t)+ir (σ,t) t − 2 e (1 + e− e )e 1 2 , (3.34) 2nσ | |

18 σ4 1 σ3 1 σ2 1 r (σ, t) e (σ, t)= + σ + + . (3.35) | 2 |≤ 2 4 − 2 3 90 12 t 3 | | Observe that, if t T T , then, for any σ 0, we have the following bounds: | |≥ ≥ 0 ≥ e (σ, T )T 2 e (σ, T )T 2 e (σ, t) min e (σ, T ), 1 0 0 , e (σ, t) min e (σ, T ), 2 0 0 . 1 ≤ 1 0 t2 2 ≤ 2 0 t2 Therefore, by the complex series representation of the exponential function and using that r (σ, t)+ ir (σ, t) r (σ, t) + r (σ, t) , we may write | 1 2 | ≤ | 1 | | 2 | O∗(E(σ, T0)) er1(σ,t)+ir2(σ,t) =1+ , (3.36) t2 where

E(σ, T )= T 2(ee1(σ,T0)+e2(σ,T0) 1). (3.37) 0 0 − 1 On the other hand, with the help of (3.34) and (3.36) with σ = 2 , and using that πσi πt πT e− e e− 0 if t T T , we derive | |≤ ≤− ≤− 0 T πi − e 4 1 πi 1 1 if( ,t) πt r1( ,t)+ir2( ,t) Bn = e 2 (1 + e− 2 e )e 2 2 dt | | 2e2T 2√n Z− 2 2 1 2e T 2e T if( 2 , t) 1 1 if( 1 , t) 1 e − 1+ e 2 − dt +E ,T0 dt , (3.38) πT0 2 ≤ 2√n e T 2 T t ! Z Z where we have used the change of variables t t in both integrals above and where, 1 t 1 ↔ − for t> 0, f 2 , t = t log 2πn + t 24t . The second integral in (3.38) can be readily − 1 −1 − 1 bounded by 1 2e2 T ; concerning the ﬁrst one, we may write f 2 , t = g(t)+ h(t), − t 1 − with g(t) = t log + t, h(t) = . Moreover, g′(t) = 0 for t (T, ), since − 2πn − 24t 6 ∈ ∞ g (t) = log t > log T and, by hypothesis, T 1. We may use then the ′ 2πn 2πn 2πn |following| identity: ≥ W i(g(t)+h(t)) W W i(g(t)+h(t)) i(g(t)+h(t)) if( 1 , t) e e h′(t) e g′′(t) e 2 − dt = dt, (3.39) g (t) − g (t) − g (t)2 ZV ′ V ZV ′ ′

V valid for any V, W such that W > V and n< 2π , and derive

W W if( 1 , t) 1 1 1 1 e 2 − dt + + + dt W V 2 t 2 t V ≤ log log V 24t log t log Z 2πn 2πn Z 2πn 2πn 2 1 1 1 1 + . (3.40) ≤ log V 24 V − W log V 2πn 2πn 2 T √T T Thus, by using (3.40) with V = T , W = 2e T and n −2π < 2π , we derive from (3.38) that B is bounded by ≤ | n| 2 1 2e 1 1 2 1 1 E ,T0 − 1+ + + 2 . (3.41) 2√n eπT0 2e2 1 48e2T log T 2e2T − 2πn ! 19 T √T As pointed out in Remark 3.11, we adopt the bound (3.41) only for n −2π since, T √T T ≤ otherwise, it becomes too big in magnitude. When −2π

1 (2e2 1)√T 1 1 1 1 1 2+ − − √T + + √T 3√T +1+ 1 , ≤ 24 e2 √T T 2 ≤ 24 − 2e2 √T 2 ( + 1) !! using (3.40) with V = T + √T , W =2e2T . Therefore B is also bounded by | n| 2 1 2e 1 1 3√T +1 1 E ,T0 − 1+ + + 2 . (3.42) 2√n eπT0 2e2 1 48e2√T 2e2T − ! 1 With respect to Cn, observe that [ 2 , 1 τ] iT is also a subset of the angular sector π − − deﬁned by arg(s) < 2 . By recalling (3.34), (3.36) and the fact that the function | | 1 3 σ E(σ, t) is increasing for σ > 2 , so that E(1 τ,T0) E 4 ,T0 , we obtain that for any7→s [ 1 , 1 τ] iT , − ≤ ∈ 2 − − 3 σ 1 πs s 1 E 4 ,T0 π T Γ(s)cos (2πn)− 1+ 1+ , (3.43) i 2 ≤ eπT0 T 2 2T 2πn 0 ! r

so that, upon integrating the bound given by (3.43) on the variable σ, we obtain that C is at most either | n| 1 3 T 2 τ 1 E ,T0 1 − 1 T √T 1+ 1+ 4 2πn − , if n − , or eπT0 T 2 2√n log T ≤ 2π 0 ! 2πn (3.44) 3 1 τ 1 1 E ,T0 1 T 2 − T √T T τ 1+ 1+ 4 , if −

T T σ The bound (3.45) is correct since, whenever n 2π , the function σ 2πn is increas- ≤ 7→1 ing. Furthermore, observe that (3.44) and (3.45) vanish when τ = 2 . Subsequently, we 3 1 1 E( 4 ,T0) deﬁne the constant H4 as 1+ πT0 1+ 2 . 2 e T0 Finally, in order to derive the constants of the statement, we combine and evaluate T √T either estimations (3.33), (3.41) and (3.44), if n −2π , or estimations (3.33), (3.42) T √T T ≤ and (3.45), when −

20 Proof of Proposition 3.1. We have that I = n T d1 2τ (n) In. As per Lemma 3.10 ≤ 2π − T √T and Remark 3.11, we split that sum into two parts,P according to whether n −2π or T √T T ≤ −2π < n 2π . For the ﬁrst interval we use (3.29) and (3.30) with the simpliﬁcation T ≤T 2πn log 2πn −T from Lemma 2.7(b), and in the second interval we use (3.31) and (3.32). Thus,≥ I is equal to

d1 2τ (n) 2π d1 2τ (n)+2O∗ η1(T ) − + η2(T ) d1 2τ (n) −  √n − n T n T n T X≤ 2π X≤ 2π X≤ 2π  d1 2τ (n) d1 2τ (n) + η3(T ) − + 1 1 (τ)η4(T, τ) − τ< 2 1 τ √n(T 2πn) { } n − (T 2πn) T √T T √T n − − n − − ≤X2π ≤X2π

d1 2τ (n) d1 2τ (n) +η5(T ) − + 1 1 (τ)η6(T, τ) − , (3.46) τ< 2 1 τ  √n { } n − T √T T T √T T −

H3 H3′ η1(T )= G1 + , η3(T )= H1T + H2, η5(T )= H′ √T + H′ + , T 1 2 √T 3 τ 1 τ G H T 2 1 H T 2 η (T )= 2 , η (T, τ)= 4 − , η (T, τ)= τ 4 − . 2 4 1 τ 6 1 τ √T π 2 2 − π 2 (2 ) − (2 ) − 1 Assume first that 0 <τ < 2 . For the main term in (3.46), we use Proposition 3.6, obtaining ζ(2τ) ζ(2 2τ) A d n T T 2 2τ O τ T 1 τ . 1 2τ ( )= + − 2 2τ − + ∗ 1 τ − − 2π (2 2τ)(2π) − (2π) − n T − X≤ 2π The first error term in (3.46) may be bounded by Proposition 3.7. For the second error term we use Proposition 3.6 again. For the third and fourth error terms, we apply 1 directly Lemma 3.8 with υ = 2 and υ = 1 τ respectively. Finally, the fifth and sixth summations in (3.46) can be bounded through− Proposition 3.6 using

υ d1 2τ (n) 2π − < d1 2τ (n) nυ T √T − T √T T T √T T −

21 1 Assume now that τ = 2 . In this case, the fourth and sixth error term in (3.46) disappear. For the main term in (3.46), we use Proposition 3.6, obtaining

A 1 T log(T ) 2γ 1 log(2π) 2 d(n)= + − − T + O∗ √T . 2π 2π √2π n T X≤ 2π The ﬁrst error term in (3.46) may be bounded by Proposition 3.7. We can forgo the negative term therein since, as T T0, this term is smaller in absolute value than the ≥ T positive one; moreover, we can also bound log 2π by log(T ). In order to estimate the second error term in (3.46), we use Proposition 3.6 and the bound log T log(T ) again, giving that it is at most 2π ≤ 1 2γ 1 A 1 T log(T )+ − T + 2 √T. 2π 2π √2π

1 For the third error term, we apply directly Lemma 3.8 with σ = υ = 2 . Finally, the ﬁfth summation in (3.46) can be bounded through Proposition 3.6 by

d(n) √2π < d(n) √n 1 T √T T 1 √T T √T T −

q 1 √T0 log(2π)+2γ 1 +2A 1 log(T ) √2π √T 1 2 + 0− − − , ≤ √2π 1 1 1 1 − √T0 − √T0 q q where we used that √T log 1 1 √T √T0 by Lemma 2.7(b) and that √T √T 1 √T0 1 − − ≤ − ≤ − T T0. ≥By combining all terms, and merging all error terms to the order √T log(T ), we conclude the result.

3.2 The integrals Ji c Choice of parameter. In Propositions 3.2 and 3.3, our choice will be λ = log(T ) , where c=1.501 optimizes the arising constants. A particular result that we need in this section is the following.

1 Proposition 3.12. Let X 1 and 0 < σ 2 . Recall the deﬁnition of Aσ given in 1 ≥ ≤ Lemma 3.6. Then, if σ < 2 ,

d1 2σ(n) − = ζ(2 2σ) log(X)+ O∗(D ), n2 2σ − σ n X − X≤

22 1 1 where Dσ =2Aσ + 2 2σ + (1 2σ)2 , whereas − −

d0(n) 1 2 = log (X)+2γ log(X)+ O∗ D 1 . n 2 2 n X X≤ where D 1 =2A 1 +2γ 1. 2 2 − 1 Proof. Consider (3.19) with υ = 2 2σ. Assume ﬁrst that 0 <σ< 2 . Similar to the proof of Proposition 3.7, after integrating− we keep the ﬁrst term of (3.19) and merge 2σ 1 the remaining ones to a constant order: note that the second term of order X − is of smaller order than a constant, unlike in Proposition 3.7. By Proposition 2.5 we have

3+6σ 4σ2 ζ(2 2σ) ζ(2σ) ζ(2σ) 1 − − < ζ(2σ)+ − + < 1+ , (1 2σ)2(2 2σ) 2 2σ 1 2σ − (1 2σ)X1 2σ 1 2σ − − − − − − − 1 1 and both sides are bounded in absolute value by 2 2σ + (1 2σ)2 ; moreover, as X 1, − − ≥ the remainder coming from (3.19) is bounded by 2Aσ. Therefore, Dσ may be deﬁned as in the statement. 1 On the other hand, if σ = , we readily deﬁne D 1 upon observing from (3.19) and 2 2 Proposition 3.6 that

d0(n) 1 2 = log (X)+2γ log(X)+2γ 1+ O∗ 2A 1 . n 2 − 2 n X X≤

Recall the deﬁnition of J1,J2 given in (3.6). We are now ready to bound them.

Proof of Proposition 3.2. From (3.6), we may write that J = L L , where 2 1 − 2 1 τ+iT 1 − L1 = χ(1 s)ζ(s)ζ(2τ 1+ s)ds, 2i 2 2τ+λ+iT − − Z − 1 τ+iT 1 − d1 2τ (n) L χ s − ds. 2 = (1 ) s 2i 2 2τ+λ+iT − n Z − n T X≤ 2π First, note that [1 τ + iT, 2 2τ + λ + iT ) belongs to the angular sector deﬁned by arg(s) < π . We can− then use− Theorem 2.1(B2) with σ [1 τ, 2 2τ + λ), t = T > 0 | | 2 ∈ − − π 1 σ2 σ4 and θ = (so that F = ), along with the estimation log( s ) = log(T )+ 2 +O∗ 4 2 θ 90 | | 2T 4T to obtain that

R(σ,T ) 1 π 0 σ T O∗ 2 Γ(s) = √2πT − 2 e− 2 e T , (3.47) | | 1 1 1 1 where, by using that s 2 T 2 and that T 3 T T 2 , | | ≤ ≤ 0 1 σ2 σ4 σ σ3 1 R(σ, T )= σ + + + + . 0 − 2 2 4T 2 12 3 90T 0 0

23 Moreover, as 1 1 τ σ < 2 2τ + λ 3 + c , we have the uniform bound 2 ≤ − ≤ − ≤ 2 log(T0) 3 c 3 R(σ, T0) R + ,T0 R′ , c,T0 = 4.393, and speciﬁcally R′(1, c,T0) ≤ 2 log(T0) ≤ 2 ≤ 1.614. Therefore, by (2.1), (3.47) and Proposition 2.2, we conclude that for any s = σ + it [1 τ + iT, 2 2τ + λ + iT ), ∈ − −

R (3/2,c,T ) 1 1 ′ 0 s πs σ 1 σ O∗ 2 χ(1 s) = 2(2π)− cos Γ(s) = (2π) 2 − 1+ O∗ T − 2 e T , | − | 2 eπT

so that, since T T , we derive the uniform bound on [1 τ + iT, 2 2τ + λ + iT ), ≥ 0 − − σ 1 T − 2 χ(1 s) κ(c,T ) , (3.48) | − |≤ 0 2π

R′(3/2,c,T0) R′(3/2,c,T0) 2 2 πT0 where κ(c,T )=1.001 e T0 + e T0 − . 0 ≥ Secondly, we may derive an upper bound for L1 by using the convexity bounds of ζ and the deﬁnition of ω given in Corollary 2.4. Together with (3.48), we conclude that, for all s [1 τ + iT, 2 2τ + λ + iT ), ∈ − − 1 2τ 1 2 2− 2 1 2 T − χ(1 s)ζ(s)ζ(2τ 1+ s) κ(c,T0) log (T ) [1 τ,1)(σ) ω | − − |≤ − 2π σ (2τ 1) 1 − 2 − σ 2 1 T 1 T − + [1,2 2τ)(σ)ω + [2 2τ,2 2τ+λ)(σ) (3.49) − 2π − − 2π   so that, by integrating (3.49), we have

1 2 τ κ(c,T0) 2 log(2π) T − L1 ω τ log(T )+2ω 1+ 1 | |≤ 2 log T0 2π − 2π ! ! 1 τ 1 τ log(2π) T 2 − T − +(ec 1) 1+ log(T ) (3.50) − log T0 2π 2π 2π ! !

λ where we have used that T

Let us give the following tail estimation of an arithmetical function involving da and the parameter λ. Proposition 3.13. Let X 1, 0 < σ 1 and λ> 0. Then, if σ< 1 , ≥ ≤ 2 2 1 1 d1 2σ(n) ζ(2 2σ + λ) 2 2σ + (1 2σ)(1 2σ+λ) − O − − − 2 2σ+λ = − λ + ∗ λ n − λX X n>XX 1 ζ(2 2σ + λ)+ 1 2σ+λ +1 − − , + 1+λ X ! whereas 2 d(n) log(X) ζ(1 + λ)+ γ ζ(1 + λ)+ 3λ 1 = + O∗ 1+ + . n1+λ λXλ λ X Xλ n>XX Proof. Let 0 < σ 1 . Observe that ≤ 2 2 2σ+λ d1 2σ(n) 1 d − − ζ λ ζ σ λ . 2 2σ+λ = 1+λ = (1 + ) (2 2 + ) (3.52) n − d n − n n d n X X X| On the other hand, by Lemma 2.6(i), we have

1 2σ+λ 2 2σ+λ d1 2σ(n) 1 d − d − − = ζ(2 2σ + λ) + O∗ n2 2σ+λ d1+λ − − (1 2σ + λ)X1 2σ+λ X2 2σ+λ n X − d X − − X≤ X≤ − 1 1 = ζ(2 2σ + λ) ζ(1 + λ) + O∗ − − λXλ X1+λ

1 1 1 1 2σ + O∗ d − . − (1 2σ + λ)X1 2σ+λ d2σ X2 2σ+λ  − d X − d X − X≤ X≤ 1   1 Then, for 0 <σ< 2 we use Lemma 2.6(i)-(iii) on the remaining sums, while for σ = 2 we use Lemma 2.6(ii) and d X 1 X. Subsequently, we subtract the resulting expression ≤ ≤ P 25 1 from (3.52). When 0 <σ< 2 , the negative summand coming from the ﬁrst term of Lemma 2.6(i) is smaller in absolute value than the positive summand coming from the second term, by Proposition 2.5: thus, when combining everything into the error term, we can forget about the former. The result follows. Lemma 3.14. Let T T = 100 and 0 < σ 1 . Then ≥ 0 ≤ 2

d1 2σ(n) 2 P3,σ P4,σ log(T ) P5,σ − P log (T )+ P log(T )+ + + , 1 2σ 1,σ 2,σ 1 σ 1 σ 1 σ n − (2πn T ) ≤ T 2 − T − T − T +√T

1 with Aσ as in Proposition 3.6, whereas for σ = 2

A 1 √2 1 2γ log(π) 2 1 P1, 1 = , P2, 1 = − , P3, 1 = 1+ , 2 4π 2 4π 2 √π s √T0

A 1 2 P4, 1 = , P5, 1 =4A 1 √π. 2 2√2π 2 2 Proof. We follow the same reasoning as in Lemma 3.8. By Proposition 3.6, we write T +√T 1 σ √ d (n) = M (t) M + Ξ (t) with Ξ (t) 2A t and T + T

T T ζ(2σ) π dt ζ(2 2σ) π dt T + − T 2π T +√T t1 2σ t 2π T +√T t Z 2π − − 2π Z 2π − 2π T T π ′ Ξσ π 1 1 + Ξσ(t) dt. (3.53) T 1 2σ T +√T 1 2σ T − T − 2π t − t ! π Z 2π − 2π Since ζ(2σ) < 0, the ﬁrst term in (3.53) can be ignored. The integral in the second term 1 is equal to 2 log(T ). As for the terms involving Ξσ, they can be bounded by

T π T 2A 1 2A tσ 1 π 2A (1 σ) σ + σ + σ − dt σ 1 σ T T +√T 1 σ T π T − 2π t √ 2π t t " 2π T + T Z 2π − 2π − 2π − T 2A A (T + √T )σ 2σ 1 π 2A (1 σ) < σ σ π 1 σ σ dt σ 1 σ + (2 ) − 1 σ + 1 σ −T π T − π √T − T − ! 2π T − T +√T t 2π Z 2π − 2π 2A 2A A (1 σ) log(T ) < σ + σ + σ − , (3.54) σ 1 σ σ 1 σ σ 1 σ π T − (2π) T 2 − (2π) T −

σ σ 1 σ σ σ 1 where we used Lemma 2.7(a) to show that (T + √T ) T (1 + σT − 2 )

26 1 We can proceed similarly for σ = 2 . By Proposition 3.6, the sum in the statement is bounded as

T T π π 2A 1 √tdt 4A 1 √π 1 (log(t)+2γ)dt 1 2 2 T + 2 + . (3.55) 2π T +√T t 2π T +√T t T √T Z 2π − 2π Z 2π − 2π The ﬁrst integral, coming from the main term, can in turn be easily bounded since by deﬁnition log(t)+2γ log(T )+2γ log(π) for t T , which implies ≤ − ≤ π T T 1 π (log(t)+2γ)dt log(T )+2γ log(π) T π T − log t 2π T +√T t ≤ 2π − 2π T +√T Z 2π 2π 2π − 1 2γ log(π) = log2(T )+ − log(T ). (3.56) 4π 4π For the second integrand in (3.55) we use instead that, for x>a> 0, √xdx √x 1 √x + √a = log . (x a)2 a x − 2√a √x √a Z − − − Then, the second term in (3.55) is equal to

T 2πt π A 1 √ +1 2 t π T T log π − − 2T q 2πt  t 2π 1 − r T T +√T − 2π  q  1 √2+1 1 log 1+ 1+ 7 log A 1 2√2 √T0 log(T ) log √2 1 2 2 1+ + + + 3 − q ≤√2π  s √T0 − √T √T 2√T √T − √T    A 1 √2 1 A 1 log(T ) < 2 1+ + 2 , (3.57) √π s √T0 2√2π √T where in the second line we used the mean value theorem to obtain that T + √T √T √T 3 1 1 − 7≥ > 7 for T T0 in order to bound log 1+ √ 1 by 2plog(T ) + log 3 , 2√T +√T ≥ − T − and where in the last line we dropped all the termsq of order 1 since they amount to a √T negative contribution to the bound, given the choice of T0. The result is concluded by putting (3.56) and (3.57) back into (3.55). Lemma 3.15. Let T >T = 100 and λ = c . For any n Z such that n > T , 0 log(T ) ∈ >0 2π we have the following estimation:

with Q Y , n 2 2τ+λ | |≤ n −

27 3 2τ 1 2τ R T 2 R T T + √T X 1 − + 2 − if n> , (3.58) n 2 2τ+λ 2πn n2 2τ+λ π | |≤ n − log T − 2 R T 2 2τ R T 1 2τ T T + √T X 1′ − 2′ −

Q =0.766, R1 =2.147, R2 =1.949, R1′ =3.081, R2′ = R2, whereas for 1 τ < 1 4 ≤ 2

Q =0.766, R1 =2.649, R2 =5.322, R1′ =3.583, R2′ = R2.

T Remark 3.16. Observe that if (3.58) is not as sharp as n approaches 2π from the 3 right, for if T < 2πn T + √T , then by Lemma 2.7(b) √T T 2 T , so log 2πn 2πn T ≤ ( T ) ≥ − ≫ that it is better to consider the bound (3.59). Instead, if 2πn>T + √T , we have that √T 2πn√T T and thus, in this case, it is better to consider the bound (3.58) log 2πn 2πn T ( T ) ≤ − ≪ over the one given in (3.59).

Proof. By (2.1) and using that χ(1 s)= χ(1 s), we may write Kn = Xn+Yn Xn Yn, where − − − − T 1 π(2 2τ + λ + it) (2 2τ) λ it X = Γ(2 2τ + λ + it)cos − (2πn)− − − − dt, (3.60) n i − 2 Z1 1 1 π(2 2τ + λ + it) (2 2τ) λ it Y = Γ(2 2τ + λ + it)cos − (2πn)− − − − dt. (3.61) n i − 2 Z0 By Theorem 2.1(B2)-(B3), we can obtain an expression analogous to (3.34): we change variables via t t, which does not change the absolute value of the integrands above, and, since t> 0↔− for all s 2 2τ + λ + i(0,T ], we have ∈ − πi 3 2τ+λ if (t) 1 χ(1 s) e− 4 t 2 − e λ,τ 1 O∗(F(c,T0, τ)) − = 1+ O∗ 1+ , s 3 2τ+λ πt 2 2i n π 2 n2 2τ+λ e t 2(2 ) − − 2πn 1 1 1 where fλ,τ (t)= f(2 2τ + λ, t)= t log t + t + 2 (2 2τ + λ)(1 2τ + λ)+ 6 t and where F(c,T , τ−) is a numerical− upper bound of E(2 −2τ + λ, T ),− deﬁned as 0 − 0 c c 2 e1 2 2τ+ ,T0 +e2 2 2τ+ ,T0 E(2 2τ + λ, T ) T e − log(T0) − log(T0) 1 , − 0 ≤ 0 −

where e1,e2 are defined in (3.35) and E is defined in (3.37), and where we have used that c 1 λ . Since σ E(σ, T0) is increasing, we can define F(c,T0, τ) F(c,T0, ) = log(T0) 4 ≤ 1 7→ ≤ 4.451 and F(c,T0, 2 )=1.63. Therefore, from (3.60) we conclude that

T πi 3 2τ+λ if (t) e− 4 t 2 − e λ,τ 1 O∗(F(c,T0, τ)) Xn = 1+ O∗ 1+ , dt 3 2τ+λ 2 2τ+λ πt 2 | | 1 2(2π) 2 − n − e t Z

28 3 π T T 2τ+λ ifλ,τ (t) e 3 t 2 e +1 2τ+λ ifλ,τ (t) − t 2 − e dt +F(c,T0, τ) dt , π 3 2τ 2 2τ+λ 2 ≤ 2e (2π) 2 − n − 1 1 t ! Z Z (3.62)

1 1 3 3 where we have used that for t [1,T ], eπt eπ and that 2 2τ + λ> 2 2τ. As 2τ 1 0 and λ< 1 , the∈ second integral≤ in (3.62) is− readily bounded− by − ≤ 2 T T 1 2τ 1 2τ+λ 1 2τ 3 +λ T − t− 2 − dt T − t− 2 dt . (3.63) ≤ ≤ 1 λ Z1 Z1 2 − 2πn As for the ﬁrst integral, we may write fλ,τ (t)= g(t)+hλ,τ (t), where g(t)= t log t +t 1 1 1 and hλ,τ (t) = 2 (2 2τ + λ)(1 2τ + λ)+ 6 t , similarly to the obtention of identity − − 2πn (3.39). Moreover, we have that g′(t) = 0 for t (0,T ], since g′(t) = log 6 ∈ | | t ≥ log 2πn and, by hypothesis, 2πn > 1; then, for any V, W such that V < W and T T 2πn > 1, we may use W W W ifλ,τ (t) W l(t)e l(t) ′ l(t)hλ,τ′ (t) l(t)eifλ,τ (t)dt = eifλ,τ (t) + dt, ig (t) − ig (t) g (t) ZV ′ V ZV ′ ′ !

3 2τ+λ where l(t)= t 2 − , and derive

W 3 2τ+λ 1 W 1 2τ+λ 3 2W 2 − (2 2τ + λ)(1 2τ + λ)+ t− 2 − 2 2τ+λ ifλ,τ (t) 6 t − e dt 2πn + − − 2πn dt V ≤ log 2 V log Z W Z t 1 3 2τ+λ (2 2τ + λ)(1 2τ + λ)+ 1 1 W 2 − 2+ − − 6 , (3.64) ≤ 2 V − W log 2πn W where we have used that 3 2τ > 0 and that t l(t) is increasing. Hence, by selecting 2 − 7→ g′(t) V = 1, W = T inside (3.64) and by using that T λ = ec, λ c , we observe from ≤ log(T0) (3.62) and (3.63) that

π c 3 2τ e +1 e 25 T 2 X (2 2τ + λ)(1 2τ + λ)+ − n 3 2τ 2πn | |≤ eπ π 2 n2 2τ+λ 2 − − 6 log 2 (2 ) − − T

F (c,T0, τ) 1 2τ + 1 c T − . (3.65) 2 − log(T0) ! As pointed out in Remark 3.16, we can do better than (3.65) when n T , T +√T . ∈ 2π 2π 3 2τ+λ 3 2τ+λ T 2 − T 2 − 2 2τ+λ i In this range, by Lemma 2.7(c), we have that T − , so log 2πn log T T √T ≤ T √T ≤ − − T T √T T 3 2τ+λ if (t) − 3 2τ+λ if (t) 3 2τ+λ if (t) t 2 − e λ dt t 2 − e λ dt + t 2 − e λ dt 1 ≤ 1 T √T Z Z Z −

1 25 2 2τ+λ 2 2τ+λ (2 2τ + λ)(1 2 τ + λ)+ T − + T − , (3.66) ≤ 2 − − 6

29 where, in the ﬁrst integral above, since 2πn > 1, we used (3.64) with V = 1, W = T √T − T √T and, in the second one, we have bounded trivially. Thus, by using that T λ = ec, λ − c , we plug (3.63) and (3.66) into (3.62) and obtain ≤ log(T0) π c e +1 e 37 2 2τ X (2 2τ + λ)(1 2τ + λ)+ T − n 3 2τ | |≤ eπ π 2 n2 2τ+λ 2 − − 6 2 (2 ) − −

F (c,T0, τ) 1 2τ + 1 c T − . (3.67) ! 2 − log(T0) 1 1 Finally, for t [0, 1], λ< 2 and τ 4 , we can bound Γ(2 2τ + λ + it) by Γ(2) = 1, π(2 2τ+∈λ+it) π ≥ πx π | − | and cos − by e 2 , since e −2 e 2 for x [ 1, 1], so that by (3.61) 2 ≤ ∈ − π e 2 1 Y . n 2 2τ+λ 2 2τ+λ (3.68) | |≤ (2π) − n − The results is concluded by combining (3.65), (3.67) and (3.68) together and using that K 2 X +2 Y . | n|≤ | n| | n|

Proof of Proposition 3.3. We have that K = n> T d1 2τ (n)Kn. As per Lemma 2π − 3.15 and Remark 3.16, we split that sum into three parts, according to whether T < P 2π T +√T T +√T T T n 2π , 2π π . For the ﬁrst interval we use (3.59), in the second ≤ ≤ 2πn 2πn T interval we use (3.58) with the simpliﬁcation log − from Lemma 2.7(b), T ≥ 2πn while in the third interval we use (3.58) as well with log 2πn log(2). Thus, T ≥ d1 2τ (n) d1 2τ(n) K ξ T, τ − ξ T, τ − 1( ) 2 2τ+λ + 2( ) 2 2τ+λ ≤ n − n − T T T +√T n> 2π π 2π X ≤ π X with 1+λ 1 2τ 2R1(2π) 3 2τ ξ (T, τ)=2Q +2R T , ξ (T, τ)= T 2 , 1 2 − 3 ec − 2 2τ 2R1 3 2τ ξ (T, τ)=2R′ T − , ξ (T, τ)= T 2 − . 2 1 4 log(2)

1 1 We bound the ﬁrst summation in (3.69) via Proposition 3.13. For 4 τ < 2 , choosing c ≤ λ = log(T ) , we have

1 + 1 1 d1 2τ (n) ζ(2 2τ + λ) 2 2τ (1 2τ)(1 2τ+λ) ζ(2 2τ + λ)+ 1 2τ+λ +1 − − − − − − 2 2τ+λ − λ + λ + 1+λ n − ≤ T T T n> T λ 2π 2π 2π X2π (2π)λ 3+ λ (2π)λ (2π)λ 2 log2(T )+ log(T )+ ≤ c2ec (1 2τ)cec ec −

30 (2π)1+λ 5 + λ log(T ) (2π)1+λ + 2 + . ecT ecT c λ c 1 where we used T = e and Proposition 2.5. For τ = 2 , we get instead

d(n) log T ζ(1 + λ)+ γ ζ(1 + λ)+ 2 1 2π + 1+ + 3λ n1+λ ≤ T λ λ T T λ n> T λ 2π 2π ! 2π X2π (2π)λ(c + 1) (2π)λ(γ +1 log(2π)) (2π)λ log2(T )+ − log(T )+ ≤ c2ec cec ec 5(2π)1+λ log(T ) (2π)1+λ + + , (3.70) 3cecT ecT where we can then forget the term of order log(T ) since γ + 1 log(2π) < 0. The last summation in (3.69) is bounded analogously, replacing every− 2π with π; this time, 1 γ +1 log(π) > 0 means that we cannot forget the term of order log(T ) when τ = 2 . The− second sum in (3.69) can be bounded through Proposition 3.6 as follows: for 1 τ < 1 , we have 4 ≤ 2 2 2τ+λ d1 2τ (n) (2π) − − d n 2 2τ+λ c 2 2τ 1 2τ ( ) n − ≤ e T − − T

d(n) (2π)1+λ d(n) n1+λ ≤ ecT T

4 Numerical considerations

100 2 Bounding an integral in an area. In Corollary 3.5, we bound 0 ζ(τ + it) dt for 1 3 | | all 2 < τ 4 . This is computed directly via Sage [23]. First, we retrieve bounds for 2≤ 1 R3 ζ(τ + it) in small square areas covering the rectangle R = 2 , 4 + [0, 100]i. Such bounds| deﬁne| a piecewise constant real-valued function p on R, whose integral on any

31 path contained in R is an upper bound for the integral of ζ(τ + it) 2 on the same path. Then, we only need to check the finitely many possibilities| that arise| from the definition of p. 1 3 1 If we are interested in widening the range of τ (see below) from 2 , 4 to 2 , 1 , we have to deal separately with the pole of ζ in 1. Considering the Laurent expansion of ζ and the bounds on its coefficients given by Lavrik [14, Lemma 4], we canbound ζ(s) 2 1 τ | | 5 1 − 2 5 by (1 τ)2 for s = τ + it such that 1 s 10 . Thus, 0 ζ(τ + it) dt 1 τ for − | − | ≤ | | ≤ − 1 τ 1 . Then, in the rest of the rectangle we bound the function ζ(τ + it) 2(1 τ) 20 R − ≤ 2 100 2 | 159.694| − numerically as we did in i for ζ(τ + it) , so that 0 ζ(τ + it) dt i′ = 1 τ . | | | | ≤ − Widening the range of τ. As mentioned in §1,R the strategy in proving Theorem 3.4 may be extended to the whole critical strip. We chose not to do so because the error 3 2 2 2τ 1 term of order T − log (T ) in the range 0 <τ < 4 becomes larger than the second main term of order T , and restricting the range of τ allows us to approximate constants more tightly, yielding a better quantitative result. For the interest of the reader, however, we report here a version of the main result valid for the whole strip. If T T = 100 and 0 <τ< 1 , then ≥ 0 2 T 2 ζ(2 2τ) 2 2τ ζ (τ + it) dt = − T − + ζ(2τ)T | | (2 2τ)(2π)1 2τ Z0 − − 0.641 2.682 3 2 2τ 2 + O∗ 2 + + 44.96 T − log (T ) . 1 τ τ ! ! 2 − A process like the one in the proof of Corollary 3.5 holds in this range too. Thus, if T T = 100 and 1 <τ< 1, then ≥ 0 2 T 2τ 1 2 (2π) − ζ(2 2τ) 2 2τ ζ (τ + it) dt = ζ(2τ)T + − T − | | 2 2τ Z0 − 1.4 3.554 2τ 1 2 O . T 2 T . + ∗ 2 + 2 + 90 397 − log ( ) τ 1 (1 τ) ! ! − 2 −

provided that, in its corresponding proof, we use the value i′ above instead of i.

Increasing T0. The choice of T0 = 100 was made for the sake of convenience. Indeed, we needed to choose T 50, because we relied upon Theorem 2.3 to bound ζ(s) on horizontal lines in §3.2,≥ and we have asked for various largeness conditions to| simplify| many computations. For instance, during the proof of Proposition 3.1 we required the negative term from Proposition 3.7 to be smaller in absolute value than the positive one T for X = 2π ; furthermore, for the purpose of properly rounding constants, we asked for some functions such as t log(t) to be decreasing in the interval [T , ) so that we are 7→ √t 0 ∞ able to absorb the terms that are asymptotically of smaller order, via inequalities like log(T ) log(T0) for T T0. √T ≤ √T0 ≥ One can repeat the same calculations with a higher T0 and expect to improve on the error terms in Theorems 1.1 and 3.4. Say that the following are the error terms in the various cases.

32 τ Theorem 1.1 Theorem 3.4 1 e √ 2 m √ 2 m √ 2 1 T log (T ) 11 T log (T )+ 12 T log(T ) e 3 2 m 3 2 1 1 2 2 2τ 21 m 2 2τ 4 , 2 (1/2 τ)2 T − log (T ) (1/2 τ)2 + 22 T − log (T ) − − Then, see the table below for diﬀerent values of T0.

T0 e1(T0) m11(T0) m12(T0) e2(T0) m21(T0) m22(T0) 103 8.452 3.423 34.742 0.845 0.139 11.3 104 5.866 3.154 24.972 0.463 0.089 5.989 106 4.178 2.907 17.557 0.219 0.05 2.692 1010 3.272 2.724 12.624 0.097 0.026 1.138 1015 2.936 2.636 10.349 0.057 0.016 0.655 1020 2.794 2.594 9.251 0.042 0.012 0.48 1030 2.67 2.552 8.177 0.029 0.008 0.347 1040 2.614 2.531 7.65 0.024 0.006 0.296

For intervals of integration with extremum lower than 100, one can estimate it directly using rigorous numerical integration implemented in the ARB package [12]. Computing the integral up to T = 1000, for example, takes a couple of seconds using the function “CBF.integral”. Choice of λ. Another significant choice that we have made concerns the parameter λ appearing in §3.2 and §3.3. First of all, the order of λ as a function of T has been 1 chosen to give the optimal order of error in the main theorem for the case τ = 2 . We 1 could not have chosen λ = o log(T ) , or else, in §3.3, (3.70) would have been too large. 1 Nor could we have chosen λ= o(log( T )), or else, in §3.2, (3.51) would have been too large. It is noteworthy that an error of order √T log2(T ) emerges also as consequence of (3.50), regardless of the choice of λ, as this comes from the use of the convexity bounds described in Corollary 2.4. Upon fixing λ as function of T , it remains to choose the optimal value of c according c to the expression λ = log(T ) . For simplicity, since the optimal c may vary with τ, we 1 chose to optimize only with respect to the case τ = 2 . Hence, we have selected c = 1.501 after numerical experimentation through a computer search. The only constraint we 1 m are facing is that 0 <λ< 2 , so our goal is to minimize the coefficient 11(T0) of the √ 2 1 error term of order T log (T ) inside the more precise Theorem 3.4 (when τ = 2 ), given the fact that we are bounding λ by its maximum value c whenever necessary. For log(T0) this matter, we have considered the range c (0, 2.302) and checked for an optimal 1 N ∈ c 1000 . Moreover, it is clear that changing T0 and τ may change the best c to select. Here∈ follows a table featuring an analogous optimization of c and its effect on the error 1 terms of Theorems 1.1 and 3.4 in the case τ = 2 .

T0 c e1(T0) m11(T0) m12(T0) 103 1.622 8.603 3.401 35.935 104 1.688 5.949 3.107 26.171 106 1.758 4.178 2.828 18.646 1010 1.819 3.205 2.614 13.599

33 The fact that the values e1(T0) are worse than in the previous table is due to our choice of optimizing only m11(T0) in Theorem 3.4, thus not taking into consideration the contribution of the smaller terms.

Acknowledgements

The authors would like to thank Harald Helfgott for his valuable suggestions. Daniele Dona was supported by the European Research Council under Programme H2020-EU.1.1., ERC Grant ID: 648329 (codename GRANT) during his permanence at Georg-August-Universität Göttingen. He was also supported by the Emily Erskine En- dowment Fund and a postdoctoral fellowship from the Einstein Institute of Mathematics during his permanence at the Hebrew University of Jerusalem. Sebastian Zuniga Alterman was supported by the postdoctoral grant of the DAMSI University Centre of Excellence during his permanence at the Nicolaus Copernicus Uni- versity at Toruń.

References

[1] M. Abramowitz, I. A. Stegun, Handbook of Mathematical Functions, Tenth Print- ing, National Bureau of Standards Applied Mathematics Series, Washington D.C., 1972. [2] F. V. Atkinson, The mean value of the zeta-function on the critical line,Q.J. Math. 10(1) (1939), 122–128. [3] F. V. Atkinson, The mean value of the Riemann zeta function, Acta Math. 81 (1949), 353–376. [4] R. J. Backlund, Uber¨ die Nullstellen der Riemannschen Zetafunktion, Acta Math. 41 (1918), 345–375 (in German). [5] R. Balasubramanian, An improvement on a theorem of Titchmarsh on the mean square of ζ( 1 + it) , Proc. Lond. Math. Soc. (3) 36 (1978), 540–576. | 2 | [6] J. Bourgain, Decoupling, exponential sums and the Riemann zeta function, J. Amer. Math. Soc. 30(1) (2017), 205–224. [7] D. Dona, H. Helfgott, S. Zuniga Alterman, Explicit L2 bounds for the Riemann ζ function. arXiv:1906.01097v6 (2019). Submitted. [8] K. Ford, Vinogradov’s integral and bounds for the Riemannn zeta function, Proc. Lond. Math. Soc. (3) 85 (2002), 565–633. [9] A. Good, Ein Ω-Resultat f¨ur das quadratische Mittel der Riemannschen Zetafunk- tion auf der kritische Linie, Invent. Math. 41 (1977), 233–251 (in German). [10] A. Ivi´c, The Riemann zeta-function, John Wiley & Sons, New York (1985).

34 [11] F. Johansson, Arb: a C library for ball arithmetic, ACM Communications in Com- puter Algebra 47(4) (2013), 166–169. [12] F. Johansson, Numerical integration in arbitrary-precision ball arithmetic, in: In- ternational Congress on Mathematical Software, Springer, 2018, 255–263. [13] E. Landau, Handbuch der Lehre von der Verteilung der Primzahlen, Teubner, Leipzig, 1909 (in German). [14] A. F. Lavrik, On the main term of the divisor’s problem and the power series of the Riemann’s zeta function in a neighborhood of its pole, Proc. Steklov Inst. Math. 142 (1976), 165–173 (in Russian). [15] R. S. Lehman, On the distribution of zeros of the Riemann zeta-function, Proc. Lond. Math. Soc. (3) 20 (1970), 303–320. [16] K. Matsumoto, The mean square of the Riemann zeta-function in the critical strip, Jpn. J. Math. 15 (1989), 1–13. [17] K. Matsumoto, Recent Developments in the Mean Square Theory of the Riemann Zeta and Other Zeta-Functions, in: Number Theory, Birkhauser, 2000, 241–286. [18] K. Matsumoto, T. Meurman, The mean square of the Riemann zeta-function in the critical strip III, Acta Arith. 64(4) (1993), 357–382. [19] H. L. Montgomery, R. C. Vaughan, Multiplicative number theory: I. Classical theory, Cambridge University Press, Cambridge, 2007.

1 [20] D. J. Platt, T. S. Trudgian, An improved explicit bound on ζ 2 + it , J. Number Theory 147 (2015), 842–851.

[21] O. Ramar´e, Some elementary explicit bounds for two molliﬁcations of the Moebius function, Funct. Approx. Comment. Math. Volume 49, Number 2 (2013), 229-240. [22] R. Remmert, Classical Topics in Complex Function Theory, Springer-Verlag, New York, 1998. [23] Sage Developers, SageMath, the Sage Mathematics Software System (Version 8.9), http://www.sagemath.org (2019). [24] A. Simoniˇc, Explicit zero density estimate for the Riemann zeta-function near the critical line, J. Math. Anal. Appl. 491(1) (2020). [25] A. Simoniˇc, V. V. Starichkova, Atkinson’s formula for the mean square of ζ(s) with an explicit error term, arXiv:2105.06821 (2021). [26] E. C. Titchmarsh, On van der Corput’s method and the zeta-function of Riemann (V), Q. J. Math. Ser. 5(1) (1934), 195–210. [27] E. C. Titchmarsh, The Theory of the Riemann Zeta-function. 2nd Edition, Oxford University Press, New York, 1986.

35 [28] G. Vorono¨ı, Sur une fonction transcendante et ses applications àla sommation de quelques séries, Ann. Sci. Ec.´ Norm. Supér. (3) 21 (1904), 207–267, 459–533 (in French).

D. Dona, Einstein Institute of Mathematics, The Hebrew University of Jerusalem, Edmond J. Safra Campus Givat Ram, Jerusalem 9190401, Israel. [email protected]

S. Zuniga Alterman, Wydzia lMatematyki i Informatyki, Nicolaus Copernicus University, 12-18 Chopina, 87-100 Torun,´ Poland. [email protected]