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The Quarterly Journal of Mathematics
, Volume Advance Article (2) – Sep 26, 2017

21 pages

/lp/ou_press/gaps-between-zeros-of-the-riemann-zeta-function-c99EmeXnO2

- Publisher
- Oxford University Press
- Copyright
- © The Author 2017. Published by Oxford University Press. All rights reserved. For permissions, please e-mail: journals.permissions@oup.com
- ISSN
- 0033-5606
- eISSN
- 1464-3847
- D.O.I.
- 10.1093/qmath/hax047
- Publisher site
- See Article on Publisher Site

Abstract We prove that there exist infinitely many consecutive zeros of the Riemann zeta-function on the critical line whose gaps are greater than 3.18 times the average spacing. Using a modification of our method, we also show that there are even larger gaps between the multiple zeros of the Riemann zeta-function on the critical line (if such zeros exist). 1. Introduction Let ζ(s) denote the Riemann zeta-function. In this paper, we investigate the vertical distribution of the non-trivial zeros of ζ(s), for the most part restricting our attention to the zeros located on the critical line Re(s)=1/2. In particular, we show that there are gaps between consecutive zeros on the critical line that are much larger than the average spacing. We also show that there are even larger gaps between the multiple zeros of ζ(s) (if such zeros exist). 1.1. Gaps between zeros of the zeta-function Let 0<γ1≤γ2≤⋯≤γn≤⋯ denote the ordinates of the non-trivial zeros of the Riemann zeta-function in the upper half-plane, and let tn denote the imaginary part of the nth zero of ζ(s) on the critical line above the real axis. Here, if a zero of the zeta-function has multiplicity m, then its ordinate is repeated m times in either sequence {tn} or {γn}. The Riemann Hypothesis (RH) states that all the non-trivial zeros of the zeta-function are on the critical line and so we expect that γn=tn for all n≥1. Our main result applies to the sequence {tn} and is unconditional, but if we assume RH, then of course this result applies to the sequence {γn}, as well. It is known that, for T≥10, N(T)≔∑0<γn≤T1=TL2π−T2π+O(L),where, here and throughout this paper, we set L≔logT2π.Therefore, the average size of γn+1−γn2π/logγn is 1 as n→∞, and so defining λ≔limsupn→∞γn+1−γn2π/logγnandΛ≔limsupn→∞tn+1−tn2π/logtn,it follows that Λ≥λ≥1. Our first result proves that there are gaps between successive ordinates of zeros of ζ(s) on the critical line that are much larger than the average spacing. Theorem 1.1. We have Λ>3.18. Consequently, assuming RH, we have λ>3.18. It is widely believed that the zeros of ζ(s) are all simple, though this remains an open problem. Modifying our proof of Theorem 1.1, we prove the existence of even larger gaps between multiple zeros of the zeta-function on the critical line (if such zeros exist). We let tn⋆ denote the imaginary part of the nth multiple zero of ζ(s) on the critical line above the real axis. If there are infinitely many multiple zeros of ζ(s), we set Λ⋆≔limsupn→∞tn+1⋆−tn⋆2π/logtn⋆and otherwise we set Λ⋆=∞. Note that Λ⋆≥Λ and that the value of Λ⋆ is unaffected by whether or not we choose to count the sequence {tn⋆} with multiplicity. Trivially, we have Λ⋆≥2, since there are at most N(T)/2 multiple zeros in the strip 0<Im(s)≤T. More generally, let Ns(T) denote the number of simple zeros of ζ(s) on the critical line with imaginary part in the interval (0,T]. Then if Ns(T)≥(C+o(1))N(T)as T→∞, it follows that there are at most (1−C+o(1))N(T)/2 multiple zeros on the critical line up to height T, and thus Λ⋆≥2/(1−C). Since Conrey [12] has shown that C=2/5 is admissible, we know that Λ⋆≥10/3. We prove the following stronger lower bound for Λ⋆. Theorem 1.2. We have Λ⋆>4.05. We remark that Conrey’s result can be slightly improved [8], but this improvement only implies that Λ⋆>3.366. Theorem 1.2 can be improved conditionally. For instance, the result of Bui and Heath-Brown [10] on the proportion of simple zeros of ζ(s) implies that Λ⋆≥27/4 assuming RH. 1.2. Previous results and conjectures The study of the gaps between the zeros of the Riemann zeta-function is an old problem that has received a great deal of attention. We briefly discuss some of the previous results and conjectures to place Theorem 1.1 in context. In 1946, Selberg [30] remarked that he could prove Λ>1. More recently, Bredberg [5] proved the quantitative estimate Λ>2.76. Bredberg’s proof and our approach in the present paper are variations of a method of Hall [21] (see also [19, 20]) who had previously shown that Λ>2.63. We discuss Hall’s method in Section 2, in particular pointing out the novelties in our approach to proving Theorems 1.1 and 1.2. We remark that Hall’s method has also been adapted to study gaps between zeros of zeta and L-functions other than ζ(s), see [2, 5, 9, 32]. A different method of Mueller [26] has been used in a number of papers to prove lower bounds for λ conditional upon RH and its generalizations, see [7, 11, 14, 15, 17, 25, 27, 28]. Our result in Theorem 1.1 that λ>3.18 assuming RH supersedes all of these previous results. Prior to this paper, the strongest known bounds using Mueller’s method was that λ>2.9 assuming RH and that λ>3.072 assuming the Generalized RH for Dirichlet L-functions. These results are established in [6, 17]. It is believed that Λ⋆=Λ=λ=∞. This conjecture is stated by Montgomery [24] in his original paper on the pair correlation of the zeros of ζ(s). Montgomery arrives at this conjecture from the stronger hypothesis that, appropriately normalized, statistics of the non-trivial zeros of the zeta-function should asymptotically behave like the statistics of eigenvalues of large random matrices from the Gaussian Unitary Ensemble (GUE). Indeed, this GUE hypothesis suggests that the gaps γn+1−γn should get as large as 1/logγn. In this direction, Ben Arous and Bourgade [3, Section 1.3] have proposed the more precise conjecture that limsupn→∞(γn+1−γn)logγn32=1. Unconditionally, Littlewood has shown that γn+1−γn=O(1/logloglogγn) as n→∞ while Goldston and Gonek [18], sharpening another result of Littlewood, proved that limsupn→∞(γn+1−γn)loglogγn≤πassuming RH. These results appear to be the best-known upper bounds for gaps between consecutive zeros of the zeta-function. 2. Inequalities, mean value estimates and numerical calculations Modifying an argument of Hall [21], using classical Wirtinger type inequalities, we reduce the problem of detecting large gaps between zeros of the Riemann zeta-function on the critical line to estimating certain mean values of ζ(s) and its derivatives. 2.1. Wirtinger type inequalities Theorem 2.1. Let f and f′be complex-valued continuous functions on the interval [a,b]. If f(a)=f(b)=0, then ∫ab∣f(t)∣2dt≤(b−aπ)2∫ab∣f′(t)∣2dt. If f(a)=f(b)and ∫abf(t)dt=0, then ∫ab∣f(t)∣2dt≤(b−a2π)2∫ab∣f′(t)∣2dt. Proofs of (i) and (ii) can be found in [22, Theorems 257 and 258], where it is shown that these inequalities hold for functions from [a,b]↦R. The theorems can be extended to complex-valued functions in a straightforward manner by applying the inequalities for real-valued functions to the real and imaginary parts of f separately and then adding. The inequality (i) is sometimes referred to as Wirtinger’s inequality in the literature. It is not clear how old these inequalities are or who first proved them. For instance, a proof of (i) was given by Scheeffer [29] in 1885 and a proof of (ii) was given by Almansi [1] in 1905. 2.2. Reduction of Theorem 1.1 to mean value estimates Suppose, for the sake of contradiction, that Λ≤κ. (2.1)Let M(s) be a Dirichlet polynomial (chosen to ‘amplify’ the zeta-function on the critical line) and let F(t,v,κ,M)≔eivtLζ(12+it)ζ(12+it+iκπL)M(12+it),where v∈R is a bounded real number to be chosen later. The factor eivtL makes F(t,v,κ,M) mimic a real-valued function when T≤t≤2T for a certain choice of v (depending on M). In order to simplify a later calculation, we have chosen to use the linear function tL2 in the exponent in place of factor θ(t)=Im(logΓ(14+it2))−(logπ)t2which appears in the definition of the Hardy Z-function, Z(t)=eiθ(t)ζ(12+it), used in [5, 21]. Denote the zeros of F in the interval [T,2T] by t˜1≤t˜2≤⋯≤t˜N. In view of our assumption (2.1), we have t˜n+1−t˜n≤(1+o(1))κπLfor 1≤n≤N−1 as T→∞ and so inequality (i) in Theorem 2.1 implies that ∫t˜nt˜n+1∣F(t,v,κ,M)∣2dt≤(t˜n+1−t˜nπ)2∫t˜nt˜n+1∣F′(t,v,κ,M)∣2dt≤(1+o(1))κ2L2∫t˜nt˜n+1∣F′(t,v,κ,M)∣2dt.Summing over n, we derive that ∫t˜1t˜N∣F(t,v,κ,M)∣2dt≤(1+o(1))κ2L2∫t˜1t˜N∣F′(t,v,κ,M)∣2dt.Now, by (2.1), we see that t˜1−T and 2T−t˜N are ≪1. Moreover, our choice of M(s) will ensure that these integrals are ≫T and ∣F(k)(t,v,κ,M)∣2≪k,ɛ(∣t∣+1)1−ɛ for ɛ>0, so it follows that ∫T2T∣F(t,v,κ,M)∣2dt≤(1+o(1))κ2L2∫T2T∣F′(t,v,κ,M)∣2dt.Therefore, if h1(v,κ,M)≔limsupT→∞L2κ2∫T2T∣F(t,v,κ,M)∣2dt∫T2T∣F′(t,v,κ,M)∣2dt>1,then we have contradicted (2.1) and we may conclude that Λ>κ. 2.3. Reduction of Theorem 1.2 to mean value estimates We first note that if a and b are multiple zeros of F, then ∫abF′(t,v,κ,M)dt=0andF′(a,v,κ,M)=0=F′(b,v,κ,M).Therefore, inequality (ii) in Theorem 2.1 implies that ∫ab∣F′(t,v,κ,M)∣2dt≤(b−a2π)2∫ab∣F″(t,v,κ,M)∣2dt.Now suppose that Λ⋆≤κ. Summing over the multiple zeros of F in [T,2T] and arguing as in Section 2.2, it follows that we derive a contradiction if h2(v,κ,M)≔limsupT→∞4L2κ2∫T2T∣F′(t,v,κ,M)∣2dt∫T2T∣F″(t,v,κ,M)∣2dt>1,in which case we can conclude that Λ⋆>κ. Comparing with h1(v,κ,M), we see that h2(v,κ,M) has an extra factor of 4 in the numerator, but the ratio of integrals ends up being smaller. Nevertheless, we are able to derive a stronger lower bound for Λ⋆ than for Λ. 2.4. Remarks We now point out some of the novelties of our approach. Hall [21] essentially chooses M(s)=1 and v=2 while Bredberg [5] chooses M(s)=∑h≤y1/hs with y=Tϑ and ϑ<1/11. We improve upon their results and these choices in a number of ways: We choose a more general amplifier of the form M(s)≔M(s,P)=∑h≤ydr(h)P[h]hs, (2.2)where y=Tϑ, 0<ϑ<1/4, r∈N, dr(h) are the coefficients the Dirichlet series of ζ(s)r, and P[h]≔P(logy/hlogy)for 1≤h≤y, where P(x)=∑j≥0bjxj is a certain polynomial. By convention, we set P[h]=0 for h≥y. Note that with this definition, we have P[h]=∑j≥0bjj!(logy)j12πi∫(1)(yh)sdssj+1 (2.3)for h∈N (and y≠h if j = 0) where here, and throughout the article, the notation ∫(c) means ∫c−i∞c+i∞. This general amplifier has been used previously in theory of the Riemann zeta-function, for instance in studying gaps between zeros [11, 27] and in establishing lower bounds for moments on the critical line [13, 31]. In addition to choosing a more general amplifier, we also take the advantage of a longer admissible Dirichlet polynomial in the twisted fourth moment of the zeta-function as a consequence of the recent work of Bettin et al. [4]. That paper evaluates the integral from T to 2T of mean fourth power of the zeta-function on the critical line times the mean square of a Dirichlet polynomial of length Tϑ for ϑ<1/4. Bredberg used a result of Hughes and Young [23] that is valid for ϑ<1/11. Another novel aspect of our work is that we express our mean value estimates in a more concise and much simpler form. Bredberg’s asymptotic formulae took five pages to display while ours are derived from one multiple integral formula (Theorem 2.2). Among other things, this helps facilitate numerical calculations. Finally, our proof of Theorem 1.2 seems to be the first approach that uses inequality (ii) in Theorem 2.1 to study the zeros of ζ(s). 2.5. A smoothing argument To use the result on the twisted fourth moment of the Riemann zeta-function from [4] directly (see Theorem 5.1), we introduce a smooth function w(t) with support in the interval [1,2] and satisfying w(j)(t)≪j,ɛTɛ for any j≥0 and ɛ>0. For r∈N, we also define the constant Ar via the well-known asymptotic formula ∑n≤xdr(n)2n∼Ar(logx)r2(r2)!,as x→∞, so Ar=∏pprime((1−1p)r2∑ℓ=0∞dr(pℓ)2pℓ).The smoothed mean values of ∣F(t,v,κ,M)∣2, ∣F′(t,v,κ,M)∣2 and ∣F″(t,v,κ,M)∣2 are given by the following theorem. Theorem 2.2. Suppose that ϑ<1/4. Then, for T large, we have ∫−∞∞∣F(j)(t,v,κ,M)∣2w(tT)dt=cj(v,κ)Ar+2(logy)r2+4rL4+2j2(r2−1)!((r−1)!)4w^(0)T+O(TL(r+2)2+2j−1)for j=0,1,2, where cj(v,κ)=∫[0,1]9x+x1+x2≤1x+x3+x4≤1eiϑκπ(x2−x4−(x3−x4)t3+(x1−x2)t4)−iκπ((1−ϑ(x1+x3))t1−(1−ϑ(x2+x4))t2)(t3−t4)(1−ϑ(x1+x3))(1−ϑ(x2+x4))(ϑ(x1−x2)+(1−ϑ(x1+x3))t1−(1−ϑ(x2+x4))t2)(ϑ(x3−x4)+(1−ϑ(x1+x3))t1−(1−ϑ(x2+x4))t2)xr2−1(x1x2x3x4)r−1(v−ϑ(x+x1+x2+x3+x4)−(1−ϑ(x1+x3))t1−(1−ϑ(x2+x4))t2)2jP(1−x−x1−x2)P(1−x−x3−x4)dx1dx2dx3dx4dxdt1dt2dt3dt4. 2.6. Numerical calculations It is a standard exercise to deduce from Theorem 2.2 the unsmoothed mean values ∫T2T∣F(j)(t,v,κ,M)∣2dt=cj(v,κ)Ar+2(logy)r2+4rL4+2j2(r2−1)!((r−1)!)4T+O(TL(r+2)2+2j−1).Hence, we deduce from the analysis in Sections 2.2 and 2.3 that h1(v,κ,M)=c0(v,κ)κ2c1(v,κ)andh2(v,κ,M)=4c1(v,κ)κ2c2(v,κ).A numerical calculation with the values ϑ=14,r=1,v=1.26andP(x)=1−5.8x+6.4x2yields h1(1.26,3.18,M)>1.0002,while the values ϑ=14,r=1,v=1.25andP(x)=1−5.2x+5.5x2numerically give h2(1.25,4.05,M)>1.0048.This implies that Λ>3.18 and Λ⋆>4.05 and, therefore, Theorems 1.1 and 1.2 follow from Theorem 2.2. 3. A shifted mean value result Rather than working directly with the mean squares of F(t,v,κ,M), F′(t,v,κ,M) and F″(t,v,κ,M), we instead consider the shifted mean value Iα̲,β̲(M)=∫−∞∞ζ(12+α1+it)ζ(12+α2+it)ζ(12+β1−it)ζ(12+β2−it)M(12+α3+it)M(12+β3−it)w(tT)dt, (3.1)where the shifts αj,βj≪L−1 and the Dirichlet polynomial M(s) is defined in (2.2). Our main goal in the rest of the paper is to prove the following lemma. Lemma 3.1. Suppose ϑ<1/4. Then we have Iα̲,β̲(M)=c(α̲,β̲)Ar+2(logy)r2+4rL42(r2−1)!((r−1)!)4w^(0)T+O(TL(r+2)2−1),where c(α̲,β̲)is given by ∫[0,1]9x+x1+x2≤1x+x3+x4≤1y−(α3+β3)x−α3(x1+x2)−β3(x3+x4)−β1x1−β2x2−α1x3−α2x4−(α2−α1)(x3−x4)t3−(β2−β1)(x1−x2)t4(Ty−x1−x3)−(α1+β1)t1−(α2−α1)t1t3−(β2−β1)t1t4(Ty−x2−x4)−(α2+β2)t2+(α2−α1)t2t3+(β2−β1)t2t4(1−ϑ(x1+x3))(1−ϑ(x2+x4))(ϑ(x1−x2)+(1−ϑ(x1+x3))t1−(1−ϑ(x2+x4))t2)(ϑ(x3−x4)+(1−ϑ(x1+x3))t1−(1−ϑ(x2+x4))t2)xr2−1(x1x2x3x4)r−1P(1−x−x1−x2)P(1−x−x3−x4)dx1dx2dx3dx4dxdt1dt2dt3dt4 (3.2)uniformly for αj,βj≪L−1. We prove this lemma in Section 5. We conclude this section by proving that Theorem 2.2 follows from Lemma 3.1. When j=0, we have ∣F(t,v,κ,M)∣=∣ζ(12+it)ζ(12+it+iκπL)M(12+it)∣and hence, c0(v,κ)=c(α̲,β̲)∣α1=α3=β1=β3=0α2=iκπ/L,β2=−iκπ/L.In the case j=1, we have F′(t,v,κ,M)ieitvL=vLζ(12+it)ζ(12+it+iκπL)M(12+it)+(ddα1+ddα2+ddα3)ζ(12+α1+it)ζ(12+α2+it+iκπL)M(12+α3+it)∣α̲=0=LQ(1L(ddα1+ddα2+ddα3))ζ(12+α1+it)ζ(12+α2+it+iκπL)M(12+α3+it)∣α̲=0,where Q(x)=v+x.Hence, ∫−∞∞∣F′(t,v,κ,M)∣2w(tT)dt=L2Q(1L(ddα1+ddα2+ddα3))Q(1L(ddβ1+ddβ2+ddβ3))Iα̲,β̲(M)∣α1=α3=β1=β3=0α2=iκπ/L,β2=−iκπ/L. (3.3)Similarly, we have ∫−∞∞∣F″(t,v,κ,M)∣2w(tT)dt=L4Q(1L(ddα1+ddα2+ddα3))2Q(1L(ddβ1+ddβ2+ddβ3))2Iα̲,β̲(M)∣α1=α3=β1=β3=0α2=iκπ/L,β2=−iκπ/L. (3.4) We obtain the constants c1(v,κ) and c2(v,κ) by applying the above differential operators to c(α̲,β̲). Since I(α̲,β̲) and c(α̲,β̲) are holomorphic with respect to each variable αj and βj in a small disc centred at 0, the derivatives appearing in (3.3) and (3.4) can be expressed using Cauchy’s integral formula as integrals of radii ≍L−1 around the points α1=α3=β1=β3=0,α2=iκπ/L,β2=−iκπ/L.Since the asymptotic formula in Lemma 3.1 holds uniformly on these contours, each derivative adds a factor that is O(L) to the error term that holds for I(α̲,β̲). Therefore, we can use Lemma 3.1 and (3.3) to prove Theorem 2.2 in the case j=1 with an error of O(TL(r+2)2+1), and similarly use Lemma 3.1 and (3.4) to prove Theorem 2.2 in the case j=2 with an error of O(TL(r+2)2+3). To see that applying the above differential operators to c(α̲,β̲) does indeed give the constants c1(v,κ) and c2(v,κ), note that Q(1L(ddα1+ddα2+ddα3))jX1α1X2α2X3α3=Q(logX1+logX2+logX3L)jX1α1X2α2X3α3.Using this expression and (3.2), we have Q(1L(ddα1+ddα2+ddα3))jQ(1L(ddβ1+ddβ2+ddβ3))jc(α̲,β̲)=∫[0,1]9x+x1+x2≤1x+x3+x4≤1y−(α3+β3)x−α3(x1+x2)−β3(x3+x4)−β1x1−β2x2−α1x3−α2x4−(α2−α1)(x3−x4)t3−(β2−β1)(x1−x2)t4(Ty−x1−x3)−(α1+β1)t1−(α2−α1)t1t3−(β2−β1t1)t4(Ty−x2−x4)−(α2+β2)t2+(α2−α1)t2t3+(β2−β1)t2t4(1−ϑ(x1+x3))(1−ϑ(x2+x4))(ϑ(x1−x2)+(1−ϑ(x1+x3))t1−(1−ϑ(x2+x4))t2)(ϑ(x3−x4)+(1−ϑ(x1+x3))t1−(1−ϑ(x2+x4))t2)xr2−1(x1x2x3x4)r−1Q(−ϑ(x+x1+x2+x3+x4)−(1−ϑ(x1+x3))t1−(1−ϑ(x2+x4))t2)2jP(1−x−x1−x2)P(1−x−x3−x4)dx1dx2dx3dx4dxdt1dt2dt3dt4.Theorem 2.2 now follows by setting α1=α3=β1=β3=0,α2=iκπ/L,β2=−iκπ/L and simplifying. 4. Two additional lemmas Lemma 4.1. Let j≥0, n≥1and r≥1be integers. Let y>0, and let Kj(α,β)=12πi∫(L−1)(yn)uζr(1+α+u)ζr(1+β+u)duuj+1 (4.1)with y≠nif j=0. Then we have Kj(α,β)=(logy/n)j+2r((r−1)!)2j!∫∫x1+x2≤10≤x1,x2≤1(yn)−αx1−βx2(x1x2)r−1(1−x1−x2)jdx1dx2+O((logy)j+2r−1)uniformly for α,β≪(logy)−1. Proof By a standard application of the residue theorem, we can replace the contour in the integral on the right-hand side of (4.1) by a small circle with radius ≍(logy)−1 around the origin plus an error term of size O(1). This integral is trivially bounded by O((logy)j+2r). Since ζ(1+s)=1s+O(1)for s near zero, taking the first terms in the Laurent series of ζ(1+α+u) and ζ(1+β+u) gives Kj(α,β)=12πi∮qu1(α+u)r(β+u)rduuj+1+O((logy)j+2r−1),where q=y/n. We apply the identity 1(α+u)r=1(r−1)!∫1/q1aα+u−1(log1a)r−1da+q−α−u∑k=0r−1(logq)kk!(α+u)r−k, (4.2)which is valid for all α,u∈C and q>0, to the above integral, writing it as the sum of (r+1) terms. The last r terms can be easily seen to vanish. Hence, Kj(α,β)=1(r−1)!∫1/q1aα−1(log1a)r−112πi∮(qa)u1(β+u)rduuj+1da+O((logy)j+2r−1).We use (4.2) again, but with the lower boundary of integration at 1/qa. Again we write the innermost integral as the sum of (r+1) terms where the last r terms vanish. In this way, we derive that Kj(α,β)=1((r−1)!)2∫1/q1∫1/qa1aα−1bβ−1(log1a)r−1(log1b)r−112πi∮(qab)uduuj+1dbda+O((logy)j+2r−1)=1((r−1)!)2j!∫1/q1∫1/qa1aα−1bβ−1(log1a)r−1(log1b)r−1(logqab)jdbda+O((logy)j+2r−1).Making the variable changes a↦q−x1 and b↦q−x2, we obtain the lemma.□ Lemma 4.2. Suppose f is a smooth function. Then we have ∑n≤ydr(n)n1+αf(logy/nlogy)=(logy)r(r−1)!∫01y−αxxr−1f(1−x)dx+O((logy)r−1). Proof This is a standard exercise in partial summation using the formula ∑n≤ydr(n)=y(logy)r−1(r−1)!+O((logy)r−2),as y→∞. See [8, Corollary 4.5].□ 5. Proof of Lemma 3.1 5.1. Reduction to a contour integral We first state the twisted fourth moment of the Riemann zeta-function from [4]. Theorem 5.1 (Bettin et al.). Let G(s)be an even entire function of rapid decay in any fixed strip ∣Re(s)∣≤Csatisfying G(0)=1, and let V(x)=12πi∫(1)G(s)(2π)−2sx−sdss. (5.1)Then, for T large, we have ∑h,k≤yahak¯hk∫−∞∞ζ(12+α1+it)ζ(12+α2+it)ζ(12+β1−it)ζ(12+β2−it)(hk)−itw(tT)dt=∑h,k≤yahak¯hk∫−∞∞w(tT){Zα1,α2,β1,β2,h,k(t)+(t2π)−(α1+β1)Z−β1,α2,−α1,β2,h,k(t)+(t2π)−(α1+β2)Z−β2,α2,β1,−α1,h,k(t)+(t2π)−(α2+β1)Zα1,−β1,−α2,β2,h,k(t)+(t2π)−(α2+β2)Zα1,−β2,β1,−α2,h,k(t)+(t2π)−(α1+α2+β1+β2)Z−β1,−β2,−α1,−α2,h,k(t)}dt+Oɛ(T1/2+2ϑ+ɛ+T3/4+ϑ+ɛ)uniformly for α1,α2,β1,β2≪L−1, where ɛ>0and Zα,β,γ,δ,h,k(t)=∑hm1m2=kn1n21m11/2+αm21/2+βn11/2+γn21/2+δV(m1m2n1n2t2). Remark 5.1. To simplify later calculations, it is convenient to prescribe certain conditions on the function G(s). To be precise, we assume that G(s) vanishes at s=−(αi+βj)/2 for 1≤i,j≤2. Let ϑ<1/4. Recalling the definition of Iα̲,β̲(M) in (3.1), we write Iα̲,β̲(M)=I1+I2+I3+I4+I5+I6+Oɛ(T1−ɛ)corresponding to the decomposition in Theorem 5.1. We first estimate I1 using Lemmas 4.1 and 4.2, and then indicate what changes need to be made in our argument to estimate the integrals I2,…,I6. Observe that I1=∑h,k≤ydr(h)dr(k)P[h]P[k]h1/2+α3k1/2+β3∑hm1m2=kn1n21m11/2+α1m21/2+α2n11/2+β1n21/2+β2∫−∞∞w(tT)V(m1m2n1n2t2)dt.In view of (2.3) and (5.1), we have I1=∑i,jbibji!j!(logy)i+j(12πi)3∫−∞∞∫(1)∫(1)∫(1)w(tT)G(s)(t2π)2syu+v∑hm1m2=kn1n2dr(h)dr(k)h1/2+α3+uk1/2+β3+vm11/2+α1+sm21/2+α2+sn11/2+β1+sn21/2+β2+sduui+1dvvj+1dssdt.Since the sum in the integrand is multiplicative, we can express it as a Euler product and then factor out the poles in terms of ζ(s). In particular, this sum equals ∑hm1m2=kn1n2dr(h)dr(k)h1/2+α3+uk1/2+β3+vm11/2+α1+sm21/2+α2+sn11/2+β1+sn21/2+β2+s=A(α̲,β̲,u,v,s)ζ(1+α1+β1+2s)ζ(1+α1+β2+2s)ζ(1+α2+β1+2s)ζ(1+α2+β2+2s)ζr2(1+α3+β3+u+v)ζr(1+α3+β1+u+s)ζr(1+α3+β2+u+s)ζr(1+β3+α1+v+s)ζr(1+β3+α2+v+s), (5.2)where A(α̲,β̲,u,v,s) is an arithmetical factor (Euler product) converging absolutely in a product of half-planes containing the origin. We first move the u and v contours in (5.1) to Re(u)=Re(v)=δ, and then move the s contour to Re(s)=−δ/2, where δ>0 is some fixed constant such that the arithmetical factor A(α̲,β̲,u,v,s) converges absolutely. In doing so, we cross a pole at s=0 and no other singularities of the integrand. Note that the poles at s=−(αi+βj)/2, 1≤i,j≤2, of the zeta-functions are cancelled by the zeros of G(s) and so the integrand is analytic at these points. On the new line of integration, we bound the integral by absolute values, giving a contribution of ≪ɛT1+ɛy2δT−δ≪ɛT1−ɛ.Hence, I1=w^(0)Tζ(1+α1+β1)ζ(1+α1+β2)ζ(1+α2+β1)ζ(1+α2+β2)∑i,jbibji!j!(logy)i+jJi,j+Oɛ(T1−ɛ), (5.3)where Ji,j=(12πi)2∫(1)∫(1)yu+vA(α̲,β̲,u,v,0)ζr2(1+α3+β3+u+v)ζr(1+α3+β1+u)ζr(1+α3+β2+u)ζr(1+β3+α1+v)ζr(1+β3+α2+v)duui+1dvvj+1. Expressing ζr2(1+α3+β3+u+v) as an absolutely convergent Dirichlet series and then changing the order of summation and integration, we obtain Ji,j=∑n≤ydr2(n)n1+α3+β3(12πi)2∫(1)∫(1)(yn)u+vA(α̲,β̲,u,v,0)ζr(1+α3+β1+u)ζr(1+α3+β2+u)ζr(1+β3+α1+v)ζr(1+β3+α2+v)duui+1dvvj+1. (5.4)Note that here we are able to restrict the sum over n to n≤y by moving the u-integral and the v-integral far to the right. We now move the contours of integration to Re(u)=Re(v)≍L−1. Bounding the integrals trivially shows that Ji,j≪Li+j+r2+4r. Hence, from the Taylor series A(α̲,β̲,u,v,0)=A(0̲,0̲,0,0,0)+O(L−1)+O(∣u∣+∣v∣), we can replace A(α̲,β̲,u,v,0) by A(0̲,0̲,0,0,0) in Ji,j with an error of size O(Li+j+r2+4r−1). We next show that A(0̲,0̲,0,0,0)=Ar+2. Letting αj=βj=0 for j=1,2,3 and u=v=s in (5.2), we have A(0̲,0̲,s,s,s)=ζ(1+2s)−(r+2)2∑hm1m2=kn1n2dr(h)dr(k)(hkm1m2n1n2)1/2+s=ζ(1+2s)−(r+2)2∑h=kdr+2(h)dr+2(k)(hk)1/2+s=ζ(1+2s)−(r+2)2∏p∑n≥0dr+2(pn)2pn(1+2s).Hence, A(0̲,0̲,0,0,0)=Ar+2. The u and v variables in (5.4) are now separated so that Ji,j=Ar+2∑n≤ydr2(n)n1+α3+β3Ki(α3+β1,α3+β2)Kj(β3+α1,β3+α2)+O(Li+j+r2+4r−1),where the function Kj(α,β) is defined in Lemma 4.1. This lemma implies that Ji,j=Ar+2(logy)i+j+4r((r−1)!)4i!j!∫∫∫∫x1+x2,x3+x4≤10≤x1,x2,x3,x4≤1(x1x2x3x4)r−1(1−x1−x2)i(1−x3−x4)j∑n≤ydr2(n)n1+α3+β3(yn)−(α3+β1)x1−(α3+β2)x2−(β3+α1)x3−(β3+α2)x4(logy/nlogy)i+j+4rdx1dx2dx3dx4+O(Li+j+r2+4r−1).Using Lemma 4.2, we deduce that Ji,j=Ar+2(logy)i+j+r2+4r(r2−1)!((r−1)!)4i!j!∫01∫∫∫∫x1+x2,x3+x4≤10≤x1,x2,x3,x4≤1y−(α3+β3)x−((α3+β1)x1+(α3+β2)x2+(β3+α1)x3+(β3+α2)x4)(1−x)xr2−1(1−x)i+j+4r(x1x2x3x4)r−1(1−x1−x2)i(1−x3−x4)jdx1dx2dx3dx4dx+O(Li+j+r2+4r−1)=Ar+2(logy)i+j+r2+4r(r2−1)!((r−1)!)4i!j!∫∫∫∫∫x+x1+x2,x+x3+x4≤10≤x,x1,x2,x3,x4≤1y−(α3+β3)x−(α3+β1)x1−(α3+β2)x2−(β3+α1)x3−(β3+α2)x4xr2−1(x1x2x3x4)r−1(1−x−x1−x2)i(1−x−x3−x4)jdx1dx2dx3dx4dx+O(Li+j+r2+4r−1).Inserting this expression back into (5.3), we conclude that I1=Ar+2(logy)r2+4rw^(0)(r2−1)!((r−1)!)4ζ(1+α1+β1)ζ(1+α1+β2)ζ(1+α2+β1)ζ(1+α2+β2)∫∫∫∫∫x+x1+x2,x+x3+x4≤10≤x,x1,x2,x3,x4≤1y−(α3+β3)x−(α3+β1)x1−(α3+β2)x2−(β3+α1)x3−(β3+α2)x4xr2−1(x1x2x3x4)r−1P(1−x−x1−x2)P(1−x−x3−x4)dx1dx2dx3dx4dx+O(Lr2+4r−1). 5.2. Deduction of Lemma 3.1 Note that I2 is essentially obtained by multiplying I1 with T−(α1+β1) and changing the shifts α1⟷−β1, I3 is obtained by multiplying I1 with T−(α1+β2) and changing the shifts α1⟷−β2, I4 is obtained by multiplying I1 with T−(α2+β1) and changing the shifts α2⟷−β1, I5 is obtained by multiplying I1 with T−(α2+β2) and changing the shifts α2⟷−β2, and I6 is obtained by multiplying I1 with T−(α1+α2+β1+β2) and changing the shifts α1⟷−β1 and α2⟷−β2. Hence, Iα̲,β̲(M)=Ar+2(logy)r2+4rw^(0)(r2−1)!((r−1)!)4∫∫∫∫∫x+x1+x2,x+x3+x4≤10≤x,x1,x2,x3,x4≤1y−(α3+β3)x−α3(x1+x2)−β3(x3+x4)U(x̲)xr2−1(x1x2x3x4)r−1P(1−x−x1−x2)P(1−x−x3−x4)dx1dx2dx3dx4dx+O(Lr2+4r−1),where U(x̲)=y−β1x1−β2x2−α1x3−α2x4(α1+β1)(α1+β2)(α2+β1)(α2+β2)−T−(α1+β1)yα1x1−β2x2+β1x3−α2x4(α1+β1)(−β1+β2)(α2−α1)(α2+β2)−T−(α1+β2)y−β1x1+α1x2+β2x3−α2x4(−β2+β1)(α1+β2)(α2+β1)(α2−α1)−T−(α2+β1)yα2x1−β2x2−α1x3+β1x4(α1−α2)(α1+β2)(α2+β1)(−β1+β2)−T−(α2+β2)y−β1x1+α2x2−α1x3+β2x4(α1+β1)(α1−α2)(−β2+β1)(α2+β2)+T−(α1+α2+β1+β2)yα1x1+α2x2+β1x3+β2x4(α1+β1)(β1+α2)(β2+α1)(α2+β2).We write y−β1x1−β2x2−α1x3−α2x4(α1+β1)(α1+β2)(α2+β1)(α2+β2)=y−β1x1−β2x2−α1x3−α2x4(α1+β1)(−β1+β2)(α2−α1)(α2+β2)−y−β1x1−β2x2−α1x3−α2x4(−β1+β2)(α1+β2)(α2+β1)(α2−α1)and T−(α1+α2+β1+β2)yα1x1+α2x2+β1x3+β2x4(α1+β1)(α1+β2)(α2+β1)(α2+β2)=T−(α1+α2+β1+β2)yα1x1+α2x2+β1x3+β2x4(α1+β1)(−β1+β2)(α2−α1)(α2+β2)−T−(α1+α2+β1+β2)yα1x1+α2x2+β1x3+β2x4(−β1+β2)(α1+β2)(α2+β1)(α2−α1). (5.5)Notice that we can interchange the roles of x1 with x2, or of x3 with x4, in any term of U(x̲) without affecting the value of Iα̲,β̲(M). Applying both changes to the last term on the right-hand side of (5.5), we can replace U(x̲) with the expression y−β1x1−β2x2−α1x3−α2x4(−β1+β2)(α2−α1)(1−(Ty−x1−x3)−(α1+β1)α1+β1)(1−(Ty−x2−x4)−(α2+β2)α2+β2)−y−β1x1−β2x2−α1x3−α2x4(−β1+β2)(α2−α1)(1−(Ty−x2−x3)−(α1+β2)α1+β2)(1−(Ty−x1−x4)−(α2+β1)α2+β1).Using the integral formula 1−z−(α+β)α+β=(logz)∫01z−(α+β)tdt, (5.6)we find that Iα̲,β̲(M)=Ar+2(logy)r2+4rL2w^(0)(r2−1)!((r−1)!)4(−β1+β2)(α2−α1)∫∫∫∫∫∫∫x+x1+x2,x+x3+x4≤10≤x,x1,x2,x3,x4,t1,t2≤1y−β1x1−β2x2−α1x3−α2x4y−(α3+β3)x−α3(x1+x2)−β3(x3+x4)(Ty−x1−x3)−(α1+β1)t1(Ty−x2−x4)−(α2+β2)t2(1−ϑ(x1+x3))(1−ϑ(x2+x4))xr2−1(x1x2x3x4)r−1P(1−x−x1−x2)P(1−x−x3−x4)dx1dx2dx3dx4dxdt1dt2−Ar+2(logy)r2+4rL2(r2−1)!((r−1)!)4(−β1+β2)(α2−α1)∫∫∫∫∫∫∫x+x1+x2,x+x3+x4≤10≤x,x1,x2,x3,x4,t1,t2≤1y−β1x1−β2x2−α1x3−α2x4y−(α3+β3)x−α3(x1+x2)−β3(x3+x4)(Ty−x2−x3)−(α1+β2)t1(Ty−x1−x4)−(α2+β1)t2(1−ϑ(x2+x3))(1−ϑ(x1+x4))xr2−1(x1x2x3x4)r−1P(1−x−x1−x2)P(1−x−x3−x4)dx1dx2dx3dx4dxdt1dt2+O(Lr2+4r−1).Denote the two integrands by V1(x,x1,x2,x3,x4,t1,t2) and V2(x,x1,x2,x3,x4,t1,t2), respectively. We note that Iα̲,β̲(M) is unchanged if we swap any of these pairs of variables x1⟷x2, x3⟷x4, and t1⟷t2 in the integrands. Hence, we can write Iα̲,β̲(M)=Ar+2(logy)r2+4rL2w^(0)2(r2−1)!((r−1)!)4(−β1+β2)(α2−α1)∫∫∫∫∫∫∫x+x1+x2,x+x3+x4≤10≤x,x1,x2,x3,x4,t1,t2≤1(V1(x,x1,x2,x3,x4,t1,t2)+V1(x,x2,x1,x4,x3,t2,t1)−V2(x,x2,x1,x3,x4,t1,t2)−V2(x,x1,x2,x4,x3,t2,t1))dx1dx2dx3dx4dxdt1dt2+O(Lr2+4r−1).Thus Iα̲,β̲(M)=Ar+2(logy)r2+4rL2w^(0)2(r2−1)!((r−1)!)4∫∫∫∫∫∫∫x+x1+x2,x+x3+x4≤10≤x,x1,x2,x3,x4,t1,t2≤1y−(α3+β3)x−α3(x1+x2)−β3(x3+x4)W(x̲,t1,t2)(1−ϑ(x1+x3))(1−ϑ(x2+x4))xr2−1(x1x2x3x4)r−1P(1−x−x1−x2)P(1−x−x3−x4)dx1dx2dx3dx4dxdt1dt2+O(Lr2+4r−1), (5.7)where W(x̲,t1,t2)=1(−β1+β2)(α2−α1)(y−β1x1−β2x2−α1x3−α2x4(Ty−x1−x3)−(α1+β1)t1(Ty−x2−x4)−(α2+β2)t2+y−β1x2−β2x1−α1x4−α2x3(Ty−x2−x4)−(α1+β1)t2(Ty−x1−x3)−(α2+β2)t1−y−β1x2−β2x1−α1x3−α2x4(Ty−x1−x3)−(α1+β2)t1(Ty−x2−x4)−(α2+β1)t2−y−β1x1−β2x2−α1x4−α2x3(Ty−x2−x4)−(α1+β2)t2(Ty−x1−x3)−(α2+β1)t1).Using (5.6) again we see that W(x̲,t1,t2)=y−β1x1−β2x2−α1x3−α2x4(Ty−x1−x3)−(α1+β1)t1(Ty−x2−x4)−(α2+β2)t2(1−(yx3−x4(Ty−x1−x3)t1(Ty−x2−x4)−t2)−(α2−α1)α2−α1)(1−(yx1−x2(Ty−x1−x3)t1(Ty−x2−x4)−t2)−(β2−β1)β2−β1)=L2(ϑ(x1−x2)+t1(1−ϑ(x1+x3))−t2(1−ϑ(x2+x4)))(ϑ(x3−x4)+t1(1−ϑ(x1+x3))−t2(1−ϑ(x2+x4)))y−β1x1−β2x2−α1x3−α2x4(Ty−x1−x3)−(α1+β1)t1(Ty−x2−x4)−(α2+β2)t2∫01∫01(yx3−x4(Ty−x1−x3)t1(Ty−x2−x4)−t2)−(α2−α1)t3(yx1−x2(Ty−x1−x3)t1(Ty−x2−x4)−t2)−(β2−β1)t4dt3dt4.Using this expression in (5.7), we obtain the lemma. 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For permissions, please e-mail: journals.permissions@oup.com This article is published and distributed under the terms of the Oxford University Press, Standard Journals Publication Model (https://academic.oup.com/journals/pages/about_us/legal/notices)

The Quarterly Journal of Mathematics – Oxford University Press

**Published: ** Sep 26, 2017

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