Higher moments of primes in short intervals I

a r X i v :m a t h /0409530v 2 [m a t h .N T ] 22 N o v 2004Higher moments of primes in short intervals ITsz Ho Chan February 1,2008AbstractIn this article,we prove an “equivalence”between two higher even moments of primes in short intervals under Riemann Hypothesis.We also provide numerical evidence in support of these asymptotic formulas.1IntroductionRecently,Montgomery and Soundararajan [5]studied the momentsM k (N ;h ):=Nn =1(ψ(n +h )−ψ(n )−h )kwhere k is a positive integer,ψ(x )=n ≤x Λ(n )and Λ(n )is von Mangoldt lambda function.They proved that,under a strong form of Hardy-Littlewood prime-k tuple conjecture,for small ǫ>0,there is a δ>0,M k (N ;h )=µk h k/2 N1(logxGoldston and Montgomery[2]showed that,under Riemann Hypothesis(RH), the stronger form of the Pair Correlation Conjecture as formulated by Mont-gomery[3]is equivalent to an asymptotic formula for(2)in Xǫ≤h≤X1−ǫor an asymptotic formula for(3)in X−1+ǫ≤δ≤X−ǫwhen k=2.The au-thor generalized these to include the second main terms in[1](again only when k=2).So,the main purpose of this paper is to prove the“equivalence”between an asymptotic formula for(2)and an asymptotic formula for(3)in appropriate ranges of h andδfor any positive even integer k.Roughly speaking,we have Theorem 1.1.Let k be a positive even integer.Assuming RH,the following are equivalent:(i) X1(ψ(x+h)−ψ(x)−h)k dx∼µk h k/2+1 X/h E(log xk Eδ k/2 holds uniformly for X−1+ǫ≤δ≤X−ǫ.Here E=2πe C0−1.Our method of proof replaces the brute-force calcula-tions in[1].We will assume RH throughout this paper and k being a positive even integer unless stated otherwise.This work is part of the author’s2002PhD thesis with some improvements. 2Some preparationsFirst of all,ψ(x)=x+O(x1/2log2x)by RH(see[7]).One has the following: X(ψ(x+δx)−ψ(x)−δx)k dx≪X k/2+1log2k X(4) 1for0≤δ≤1,andX(ψ(x+h)−ψ(x)−h)k dx≪X k/2+1log2k X(5) 1for0≤h≤X.Also,estimating trivially,we haveX(ψ(x+δx)−ψ(x)−δx)k dx≪ X1(δx log X)k dx≪δk X k+1log k X(6) 1for0≤δ≤1,andX(ψ(x+h)−ψ(x)−h)k dx≪ X1(h log X)k dx≪h k X log k X(7) 1for0≤h≤X.We also need some lemmas.2Lemma2.1.For any differentiable function f(u)and any0≤η≤1,(1+η)T T f(u)du=ηT f(T)+O η2T2maxT≤t≤(1+η)T|f′(t)| .Proof:By mean-value theorem,we have(1+η)TTf(u)du=ηT f(T+ξT)where0≤ξ≤η=ηT f(T)+ξT f′(T+ξ′T) where0≤ξ′≤ξ, and the lemma follows.Lemma2.2.For any positive integer k,we havex k−y k=(x−y)P(x,y)+(x−y)kwhere P(x,y)is some homogeneous polynomial of degree k−1.Proof:By Factor Theorem,z−1divides z k−1−(z−1)k.So,(z−1)P(z)=z k−1−(z−1)kfor some integer polynomial P(z)of degree k−1.Set z=xE k/2dx+O k(h k/2X1−ǫ1)(8)3holds uniformly for Xǫ≤h≤X1−ǫ,thenX 1(ψ(x+δx)−ψ(x)−δx)k dx=µk2+1X k/2+1δk/2 log1xdx dh+ ∆V/20 V V/2f(x,h)kxdx= 1x2dx=µk h k/2+1 1E k/2dx−1E k/2dx+µk h k/2+1 V U1E k/2du dx+O k(U−ǫ1h k/2) =T1+T2+O k(U−ǫ1h k/2)as long as Vǫ≤h≤U1−ǫwith U≤V≤2U.T2=µk h k/2 V/h U/h y E log u y)=µk h k/2 −h E k/2du+h E k/2du+ V/h U/h(log y/E)k/2x dx=µk2+1h k/2 log V Eh k/2+1 +O k(U−ǫ1h k/2)4as long as V ǫ≤h ≤U 1−ǫwith U ≤V ≤2U .Thus,for I 1and I 2to work,we needV ǫ≤∆V 2 1−ǫ,and V ǫ≤h ≤ h 2≤h ≤∆V.(11)Since ∆=X µ,for X µ+ǫ≤V ≤X ,one can check that (11)are satisfied.Therefore,I 1=µk 2+1∆V ∆V/2hk/2 log V E ∆k/2+1dh +O k (∆k/2+1V k/2+1−ǫ1),I 2=µk 2+1∆V/20hk/2 log VEhk/2+1dh+O k (V (k/2+1)ǫ(log V )k/2+1)+O k (∆k/2+1V k/2+1−ǫ1),I 3≪V ǫV kǫlog k V,Combining these,(10)equalsµk2+1∆Vh k/2 logV Ehk/2+1dh− log1log 2],XX/2M∆0ψ(x +δx )−ψ(x )−δx kdδdx =µk 2+1∆Xhk/2logX kE ∆k/2+1(∆X )k/2+1+O k (X (k +1)ǫlog k X )+O k (∆k/2+1X k/2+1−ǫ1)=µk 2+1 ∆Xh k/2 log XBut,since µ≤1−2ǫ−ǫ1,∆k/2+1X k/2+1−ǫ1=∆k +1X k/2+1+(k/2)µ−ǫ1=∆k +1X (k +1)µ+(k/2+1)(1−µ)−ǫ1≥∆k +1X (k +1)µ+(k/2+1)(2ǫ+ǫ1)≫∆k +1(X µ+ǫ)k +1log k X.(14)Combining (12),(13)and (14),we have∆X1(ψ(x +δx )−ψ(x )−δx )k dx dδ=µk 2+1∆Xh k/2 logX kEhk/2dh +O k (∆k/2+1X k/2+1−ǫ1)=µk2+1(∆X )k/2+1log 1∆)k/2+O k (∆k/2+1X k/2+1−ǫ1).(16)Let g (x,δx )=f (x,∆x )for ∆≤δ≤(1+η)∆.Then one can easily check that f (x,δx )−g (x,δx )=f (1+∆)x,(δ−∆)x .So,(1+η)∆∆X 1f (x,δx )−g (x,δx ) kdxdδ=η∆η∆)k/2(17)by (15),the choice of ηand the range of ∆.Thus,by Lemma 2.3,(16)and (17),(1+η)∆∆X1g (x,δx )k dx dδ≪k|f (x,δx )|k +|f (x,δx )−g (x,δx )|k≪k ηX k/2+1∆k/2+1(log1By Lemma 2.2and Holder’s inequality,(1+η)∆∆X1f (x,δx )k −g (x,δx )k dx dδ=(1+η)∆∆X1P (f,g )(f −g )+(1+η)∆∆X1(f −g )k ≪|P (f,g )|k/(k −1) (k −1)/k|f −g |k1/k+|f −g |k=J (k −1)/k1J 1/k2+J 2where P (x,y )is a homogeneous polynomial of degree k −1.J 2≪k ηk/2+1X k/2+1∆k/2+1(log1η∆)k/2+∆k/2+1X k/2+1−ǫ1by (16)and (18)≪kηX k/2+1∆k/2+1(log1η∆)k/2.(19)Therefore,by (19)and (16),η∆X1(ψ(x +∆x )−ψ(x )−∆x )kdx =g k=f k+OX k/2+1∆k/2+1η3/2(log1kE ∆k/2η+O k (∆k/2+1X k/2+1−ǫ1)+O kX k/2+1∆k/2+1η3/2(log1Dividing through by η∆,we haveX1(ψ(x +∆x )−ψ(x )−∆x )k dx=µk 2+1X k/2+1∆k/2log 1η+O k X k/2+1∆k/2η1/2(log14.4(ii)⇒(i)Theorem 4.1.Assume RH.If,for some small ǫ>ǫ1>0(small in terms of k ), X 1(ψ(x +δx )−ψ(x )−δx )k dx =µk 2+1X k/2+1δk/2log 1Ek/2dx +O k (h k/2X 1−ǫ2)(21)holds uniformly for X2ǫ+ǫ1≤h ≤X1−(k/2+1)ǫ−2ǫ1/2with some ǫ2>0.Proof:Let f (x,h )=ψ(x +h )−ψ(x )−h .Let X 2ǫ≤H ≤X 1−(k/2+1)ǫ−2ǫ1).Say H =X µfor some 2ǫ≤µ≤1−(k/2+1)ǫ−2ǫ1.First,we calculateV V/2H(ψ(x +h )−ψ(x )−h )k dh dx.(22)Substituting δ=h2)−1+ǫVV/2+(VkEδk/2V k/2+2−Uk/2+2+O k (δk/2V k/2+2−ǫ1)8as long as U −1+ǫ≤δ≤V −ǫwith U ≤V ≤2U .In order for this to work for I 1and I 2,we need V V ≤V −ǫ,and V δ −ǫfor H V .(23)Since H =X µ,one can check that (23)are satisfied for X µ+ǫ≤V ≤X .Thus,I 1=µk 2+2 2H/V H/VH 2 k/2+2 δk/2 log 1k2k/2+2H/Vδk/2log12+2),then (22)=νk Hk/2+22H/VH/V1Eδ k/2dδ+νk Vk/2+2H/Vδk/2log12k/2+22H/Vδk/2log 1log 2],XX/2MH(ψ(x +h )−ψ(x )−h )k dh dx=νk Hk/2+22M +1H/XH/X1Eδk/2dδ+νk Xk/2+2H/Xδk/2log12M +1k/2+22M +1H/Xδk/2log1δ2log1Eδk/2dδ+O k (H k/2+1X 1−ǫ1).(24)because,as X 2ǫ≤H ,X 1+(k +1)ǫlog k X ≪H k/2+1X 1−ǫ1.Also,the termsinvolving 2M +1are absorbed into the error term as µ≤1−ǫ−2ǫing (5),X/2M1H 0ψ(x +h )−ψ(x )−h kdh dx ≪H (X µ+ǫ)k/2+1log 2k X.(25)9But,since H =X µ≤X 1−(k/2+1)ǫ−2ǫ1,H (X µ+ǫ)k/2+1log 2k X ≤X 1−(k/2+1)ǫ−2ǫ1H k/2+1X (k/2+1)ǫlog 2k X≪Hk/2+1X1−ǫ1.(26)Combining (24),(25)and (26),we haveHX 1ψ(x +h )−ψ(x )−h kdx dh =νk Hk/2+21/EH/X1Eδk/2dδ+νk Xk/2+2H/Xδk/2log1δ2log1δ2log 1Eδk/2dδ+O k (H k/2+1X 1−ǫ1)=−νk Hk/2+2(1+η)H/X H/X 1Eδk/2dδ+µk ηHk/2+21/EH/X1Eδk/2dδ+νk Xk/2+2(1+η)H/XH/Xδk/2log1H)k/2+O k (H k/2+1X 1−ǫ1)=µk ηHk/2+21/EH/X1Eδ k/2dδ+O kη2H k/2+1X (logXby Lemma2.1.Therefore(1+η)HHX 1ψ(x +h )−ψ(x )−h kdx dh=µk ηHk/2+2X/H EloguH)k/2+O k (H k/2+1X 1−ǫ1).(28)Let g (x,h )=f (x,H )for H ≤h ≤(1+η)H .Again,one can check that f (x,h )−g (x,h )=f (x +H,h −H ).So,(1+η)H HX 1f (x,h )−g (x,h ) kdxdh =ηHX +H1+Hf (x,h )k dxdh≪k ηk/2+1XHk/2+1logX ηH)k/2+H k/2+1X 1−ǫ1.(30)By Lemma 2.2and Holder’s inequality,(1+η)HHX1f (x,h )k −g (x,h )k dxdδ=(1+η)HHX1P (f,g )(f −g )+(1+η)HHX1(f −g )k≪ |P (f,g )|k/(k −1) (k −1)/k |f −g |k 1/k + |f −g |k =K (k −1)/k1K 1/k2+K 2(31)where P (x,y )is a homogeneous polynomial of degree k −1.From (29),K 2≪k ηk/2+1XHk/2+1logX ηH)k/2(33)11by(28)and(30).Consequently,by(31),(32),(33)and Lemma2.3,f k−g k≪kη3/2XH k/2+1 log X)k/2ηH=µkηH k/2+2 X/H E log u)k/2 .ηHDivide through byηH and recallη=X−2ǫ1/3,we get the theorem withǫ2=ǫ1References[1]T.H.Chan,More Precise Correlation of Zeros and Primes in Short Intervals,J.London Math.Soc.(2)68,2003,no.3,579-598.[2]D.A.Goldston and H.L.Montgomery,On pair correlations of zeros andprimes in short intervals,Analytic Number Theory and Diophantine Prob-lems(Stillwater,OK,July1984),Prog.Math.70,Birkauser,Boston,1987, pp.183-203.[3]H.L.Montgomery,The pair correlation of zeros of the zeta function,AnalyticNumber Theory(St.Louis Univ.,1972),Proc.Sympos.Pure Math.24, Amer.Math.Soc.,Providence,1973,pp.181-193.[4]H.L.Montgomery and K.Soundararajan,Beyond pair correlation,PaulErd¨o s and his mathematics,I(Budapest,1999),Bolyai Soc.Math.Stud.11,J´a nos Bolyai Math.Soc.,Budapest,2002,pp.507-514.[5]H.L.Montgomery and K.Soundararajan,Primes in Short Intervals,preprint.[6]B.Saffari and R.C.Vaughan,On the fractional parts of x/n and relatedsequences II,Ann.Inst.Fourier(Grenoble)(2)27,1977,1-30.[7]H.von Koch,Sur la distribution des nombres premiers,Acta Math.24,1901,159-182.Tsz Ho ChanAmerican Institute of Mathematics360Portage AvenuePalo Alto,CA94306USAthchan@13。