x On the Seventh Power Moment of ∆( ) Jinjiang Li Department of Mathematics, 6 China University of Mining and Technology, 1 0 Beijing 100083,P. R. China 2 Email: jinjiang.li.math@gmail.com n a Abstract: Let∆(x)betheerrortermoftheDirichletdivisorproblem. Inthispaper,weestablish J 1 an asymptotic formula of the seventh-power moment of ∆(x) and prove that 2 T 7(5s (d) 3s (d) s (d)) ] ∆7(x)dx= 7;3 − 7;2 − 7;1 T11/4+O(T11/4−δ7+ε) T Z2 2816π7 N with δ =1/336, which improves the previous result. 7 . h Keywords: Dirichlet divisor problem; higher-power moment; asymptotic formula. t a m 1 Introduction and main result [ 1 v Throughout this paper, let d(n) denote the Dirichlet divisor function. In 1838, Dirichlet proved that 5 the error term 1 5 ∆(x):= d(n) xlogx (2γ 1)x, with x>2, − − − 5 n6x X 0 . satisfies ∆(x) x1/2. Here γ is Euler’s constant. Since then the determination of the exact order of 1 ≪ 0 ∆(x) has been called Dirichlet’s divisor problem. Many writers have sharpened Dirichlet’s bound for 6 ∆(x). The latest result is due to Huxley [5], who proved that 1 : v ∆(x) x131/416(logx)26957/8320. i ≪ X r For a survey of the history of this problem, see Krätzel [10]. a In the opposite direction, Hardy [3] proved that Ω x1/4(logx)1/4loglogx , + ∆(x)=  Ω (cid:0)x1/4 . (cid:1) − The best results in this direction to date are(cid:0) (cid:1) ∆(x)=Ω x1/4(logx)1/4(loglogx)(3+log4)/4exp( c logloglogx) + − (cid:16) p (cid:17) 1 and ∆(x)=Ω x1/4exp(c(loglogx)1/4(logloglogx)−3/4 − (cid:16) (cid:17) for some constant c>0, due to Hafner [2] and [1] respectively. It is conjectured that ∆(x) x1/4+ε ≪ is true for every ε > 0. The evidence in support of this conjecture has been given by Tong [14] and Ivic [6], who proved, respectively, that T ζ4(3) ∆2(x)dx= 2 T3/2+O(T log5T) (1.1) 6π2ζ(3) Z2 and T 35 ∆(x)Adx ε T1+A4+ε, for 06A6 (1.2) | | ≪ 4 Z2 and any ε>0. On the other hand , Vorono¨ı[16] proved that T ∆(x)dx=T/4+O(T3/4), (1.3) Z2 whichin conjunction with (1.1) and(1.2) showsthat ∆(x) has a lotof signchangesandcancellations between the positive and negative portions. Tsang [15] first studied the third and fourth-power moments of ∆(x). He proved that T 3c ∆3(x)dx= 1 T7/4+O(T7/4−δ3+ε), (1.4) 28π3 Z2 T 3c ∆4(x)dx= 2 T2+O(T2−δ4+ε), (1.5) 64π4 Z2 where δ =1/14,δ =1/23 and 3 4 c := (αβ(α+β)) 3/2h 9/4 µ(h)d(α2h)d(β2h)d((α+β)2h), 1 − − | | α,β,h N X∈ c2 := (nmkℓ)−3/4d(n)d(m)d(k)d(ℓ), n,m,k,ℓ N X∈ √n+√m=√k+√ℓ and µ(h) is the Mo¨bius function. In [18], Zhai proved that (1.4) holds for δ = 1/4. Ivić and Sargos [8] proved that (1.4) holds 3 for δ = 7/20. Following the approach of Tsang [15], Zhai [18] proved that the equation (1.5) holds 3 for δ = 2/41. This approach used the method of exponential sums. In particular, if the exponent 4 pair conjecture is true, namely, if (ε,1/2+ε) is an exponent pair, then the equation (1.5) holds for δ = 1/14. Moreover, in [8], Ivić and Sargos proved a substantially better result that the equation 4 (1.5) holds for δ =1/12. Later, combining the method of [8] and a deep result of Robert and Sargos 4 [12], Zhai [20] proved that the equation (1.5) holds for δ = 3/28. Recently, Kong [9] proved that 4 δ =1/8. 4 By a unified approach, Zhai [19] proved that the asymptotic formula T ∆k(x)dx=CkT1+k/4+O T1+k/4−δk+ε (1.6) Z1 (cid:16) (cid:17) 2 holds for 3 6 k 6 9, where C and 0 < δ < 1 are explicit constants. He gives δ = 1/64,δ = k k 5 6 35/4742,δ =17/6312,δ =8/9433,δ =13/75216.Theasymptoticformula(1.6)improvedtheresult 7 8 9 of Heath-Brown [4]. When k = 5, the asymptotic formula (1.6) holds for δ = 1/64, which improved 5 an earlier exponent δ =5/816provedin [18] by the approachof Tsang [15]. In [21], Zhang and Zhai 5 improved the previous result of the case k = 5 and proved δ = 3/80. Meanwhile, Wang [17] studied 5 the case k =6 and proved δ =3/248, which improved the result of Zhai [19], i.e. δ =35/4742. 6 6 The aimofthis paperistoimprovethe valueofδ =17/6312,whichisachievedbyZhai[19]. The 7 main result is the following Theorem 1.1 We have T 7(5s (d) 3s (d) s (d)) ∆7(x)dx= 7;3 − 7;2 − 7;1 T11/4+O(T11/4−δ7+ε) 2816π7 Z2 with δ =1/336, where 7 d(n)d(m)d(k)d(ℓ)d(r)d(s)d(q) s (d)= , 7;3 (nmkℓrsq)3/4 n,m,k,Xℓ,r,s,q∈N∗ √n+√m+√k+√ℓ=√r+√s+√q d(n)d(m)d(k)d(ℓ)d(r)d(s)d(q) s (d)= , 7;2 (nmkℓrsq)3/4 n,m,k,Xℓ,r,s,q∈N∗ √n+√m+√k+√ℓ+√r=√s+√q d(n)d(m)d(k)d(ℓ)d(r)d(s)d(q) s (d)= . 7;1 (nmkℓrsq)3/4 √n+√mn+,m√,kk,+Xℓ,√r,ℓs+,q√∈rN+∗√s=√q Notations. Throughout this paper, x denotes the distance from x to the nearest integer, i.e., k k x = min x n. [x] denotes the integer part of x; n N means N < n 6 2N; n N means k k n Z| − | ∼ ≍ C N 6 n∈6 C N with positive constants C ,C satisfying C < C . ε always denotes an arbitrary 1 2 1 2 1 2 smallpositiveconstantwhichmaynotbethesameatdifferentoccurances. Weshallusetheestimates d(n) nε. Suppose f :N R is any function satisfying f(n) nε, k >2 is a fixed integer. Define ≪ → ≪ f(n )f(n ) f(n ) s (f):= 1 2 ··· k , 16ℓ1isalargeparameter, k;ℓ and we define f(n )f(n ) f(n ) s (f;y):= 1 2 ··· k , 16ℓ3,(i ,i , ,i ) 0,1 k 1 such that 1 2 k 1 − ··· − ∈{ } √n1+( 1)i1√n2+( 1)i2√n3+ +( 1)ik−1√nk =0. − − ··· − 6 3 Then we have √n1+( 1)i1√n2+( 1)i2√n3+ +( 1)ik−1√nk max(n1,n2, ,nk)−(2k−2−2−1). | − − ··· − |≫ ··· Proof. See Lemma 2.2 of [19]. Lemma 2.2 If g(x) and h(x) are continuous real-valued functions of x and g(x) is monotonic, then b v g(x)h(x)dx max g(x) max h(x)dx . Za ≪(cid:18)a6x6b| |(cid:19)(cid:18)a6u10,α,β R,2K 1/2 6 α K1/2 and 0<δ <1/2. Then we have − ∈ | |≪ # k K : β+α√t <δ Kδ+K1/2+ε. { ∼ k k }≪ Proof. See Lemma 4 of [20]. Lemma 2.5 Let a,δ be real numbers, 0 < δ < a/4, and let k be a positive integer. There exists a function ϕ(y) which is k times continuously differentiable and such that ϕ(y)=1, for y 6a δ, | | −   0<ϕ(y)<1, for a−δ <|y|a+δ, and its Fourier transform  | | + ∞ Φ(x)= e( xy)ϕ(y)dy − Z−∞ satisfies the inequality k 1 1 k Φ(x) 6min 2a, , . | | π|x| π|x| (cid:18)2π|x|δ(cid:19) ! Proof. See [11] or [13]. Lemma 2.6 Let d(n) denote the divisor function. Then we have s (d) s (d;y) y 1/2+ε, 16ℓ3;(i ,i , ,i ) 0,1 k 1,(i ,i , ,i )=(0,0, ,0); 1 2 k 1 − 1 2 k 1 ··· − ∈{ } ··· − 6 ··· 10. Hence the quantity of | | A (N,M,K,L,R,S,Q;∆) does not exceed the number of solutions of 1 2ℓ1/2(√n+√m+√k √r √s)+(√n+√m+√k √r √s)2+ℓ q 2L 1/2, and D the number of solutions of (2.3) with α <2L 1/2. By Lemma 2.4, we get − 2 − | | | | D (∆Q1/2L+L1/2+ε)NMKRS 1 ≪ ∆Q1/2NMKLRS+NMKRSL1/2+ε. ≪ Now we estimate D . From α < 2L 1/2, we can get K R. If √n+√m+√k = √r+√s, then 2 − | | ≍ from (2.2), we get ℓ=q. This contradicts to the fact that √n+√m+√k+√ℓ √r √s√q >0. | − − | Therefore,wehave√n+√m+√k =√r+√s.ByLemma2.1,wehave √n+√m+√k √r+√s 6 | − |≫ S 15/2 for any such (n,m,k,r,s). By a splitting argumentand Lemma 2.7, there exists a δ satisfying − S 15/2 δ L 1/2, which holds − − ≪ ≪ D log2S 1 2 ≪ · δ<√n+√m+X√k √r+√s62δ | − | log2S (δS1/2NMKR+NMKR) ≪ · L 1/2S1/2+εNMKR+NMKRSε − ≪ NMKRSε, ≪ which can be absorbed into the estimate of D . This completes the proof of Lemma 2.8. 1 Lemma 2.9 Suppose 1 6 N 6 M 6 K 6 L,1 6 R 6 S 6 Q,L Q and 0 < ∆ Q1/2. Let ≍ ≪ A (N,M,K,L,R,S,Q;∆) denote the number of solutions of the inequality 2, ± 0< √n+√m+√k+√ℓ+√r √s √q <∆ | ± − | with n N,m M,k K,ℓ L,r R,s S,q Q. Then ∼ ∼ ∼ ∼ ∼ ∼ ∼ A (N,M,K,L,R,S,Q;∆) ∆Q1/2NMKLRS+NMKRSL1/2+ε. 2, ± ≪ In particular, if ∆Q1/2 1, then ≫ A (N,M,K,L,R,S,Q;∆) ∆Q1/2NMKLRS. 2, ± ≪ Proof. The proof of Lemma 2.9 is similar to that of Lemma 2.8, so we omit the details. Lemma 2.10 Suppose N >2 (j =1,2,3,4,5,6,7) are real numbers, ∆>0. let j A (N .N ,N ,N ,N ,N ,N ;∆) denote the number of solutions of the inequality , 1 2 3 4 5 6 7 ±± 0< √n +√n +√n +√n √n √n √n <∆ 1 2 3 4 5 6 7 | ± ± − | with n N ,(j =1,2,3,4,5,6,7),n N . Then we have j j j ∗ ∼ ∈ 7 A (N .N ,N ,N ,N ,N ,N ;∆) ∆1/7N13/14+N5/7 Nε. ±,± 1 2 3 4 5 6 7 ≪ j j j jY=1(cid:16) (cid:17) 6 Proof. Taking a = 6∆/5,δ = ∆/5 in Lemma 2.5, there exists a function ϕ (y), which is ℓ = 1 [7log(N N N N N N N )] times continuously differentiable such that 1 2 3 4 5 6 7 ϕ (y)=1, if y 6∆, 1 | |   0<ϕ1(y)<1, for ∆<|y|<7∆/5, ϕ (y)=0, for y >7∆/5. 1 Let  | | + ∞ Φ (x)= e( xy)ϕ (y)dy, 1 1 − Z−∞ then it satisfies ℓ 12∆ 1 1 5ℓ Φ (x) 6min , , , (2.4) 1 | | 5 π|x| π|x|(cid:18)2π|x|∆(cid:19) ! and + ∞ ϕ (y)= e(xy)Φ (x)dx. (2.5) 1 1 Z−∞ Set R = ϕ (√n +√n +√n +√n √n √n √n ). , 1 1 2 3 4 5 6 7 ±± ± ± − j=1n,2jX,∼3,N4,j5,6,7 By the definition of ϕ (y), we get 1 A (N .N ,N ,N ,N ,N ,N ;∆)6R . (2.6) , 1 2 3 4 5 6 7 , ±± ±± We estimate R first. By (2.5), we have , −− R = ϕ (√n +√n +√n +√n √n √n √n ) , 1 1 2 3 4 5 6 7 −− − − − j=1n,2jX,∼3,N4,j5,6,7 + ∞ = e(x(√n +√n +√n +√n √n √n √n ))Φ (x)dx. 1 2 3 4 5 6 7 1 − − − j=1n,2jX,∼3,N4,j5,6,7Z−∞ Let S(x;N):= e(x√n), we have n N ∼ P + ∞ R = S(x;N )S(x;N )S(x;N )S(x;N )S(x;N )S(x;N )S(x;N )Φ (x)dx, , 1 2 3 4 5 6 7 1 −− Z−∞ if we notice that S(x;N)= e( x√n). Applying Ho¨lders inequality, we get − n N ∼ P 7 + 1/7 R 6 ∞ S(x;N )7 Φ (x)dx . (2.7) , j 1 −− | | | | jY=1(cid:18)Z−∞ (cid:19) Let + T(N):= ∞ S(x;N)7 Φ (x)dx. 1 | | | | Z0 7 It is sufficient to estimate T(N), where N =N for some j 1,2,3,4,5,6,7 .Let j ∈{ } 100ℓ2 K := , ℓ=[7log(N N N N N N N )]; K :=N1/2. 1 2 3 4 5 6 7 0 ∆ Using the trivial estimate S(x;N) N and the estimate ≪ ℓ 1 5ℓ Φ (x) 6 , 1 | | π x 2π x∆ | |(cid:18) | | (cid:19) we have ℓ ∞ S(x;N)7 Φ (x)dx N7 ∞ Φ (x)dx N7 5ℓ ∞ 1 dx | | | 1 | ≪ | 1 | ≪ 2π∆ xℓ+1 ZK ZK (cid:18) (cid:19) ZK N75ℓ ℓ ℓ N75ℓ 1 ≪ ℓ 2πK∆ ≪ ℓ ℓℓ (cid:18) (cid:19) N7(N N )7log5 1 7 ··· 1. (2.8) ≪ (N1···N7)7log7(N1···N7)7loglog(N1···N7) ≪ For the mean square of S(x;N), we have K0 K0 S(x;N)2dx = e(x(√n √m))dx | | − Z0 Z0 n Nm N X∼ X∼ K0 = e(x(√n √m))dx − n Nm NZ0 X∼ X∼ K0 = + e(x(√n √m))dx. − (cid:26)n Nm N n Nm N(cid:27)Z0 X∼n=mX∼ X∼n=mX∼ 6 If we use the trivial estimate S(x;N) N, then ≪ K0 e(x(√n √m))dx N3/2. − ≪ n Nm NZ0 X∼n=mX∼ For the case n=m, we have 6 K0 e(x(√n √m))dx − n Nm NZ0 X∼n=mX∼ 6 1 1 N1/2 ≪ √n √m ≪ n m n Nm N | − | n Nm N | − | X∼n=mX∼ X∼n=mX∼ 6 6 1 N1/2 N3/2logN. ≪ r ≪ n N16r N X∼ X≪ Therefore, we have K0 S(x;N)2dx N3/2logN. (2.9) | | ≪ Z0 8 If we notice Φ (x) ∆ from (2.4) and the trivial estimate S(x;N) N, then 1 | |≪ ≪ K0 S(x;N)7 Φ (x)dx 1 | | | | Z0 K0 ∆N5 S(x;N)2dx ≪ | | Z0 ∆N13/2logN. (2.10) ≪ If K 6K , then from (2.8) and (2.10) we get 0 T(N) ∆N13/2logN. (2.11) ≪ Now suppose K 10,N =T,y =T1/4, we have 2T 7(5s (d) 3s (d) s (d)) R17dx= 7;3 −28176;2π7 − 7;1 T11/4+O(T11/4−1/336+ε). (3.2) ZT Proof. Let d(n)d(m)d(k)d(ℓ)d(r)d(s)d(q), if n,m,k,ℓ,r,s,q6y, (nmkℓrsq)3/4 g :=g(n,m,k,ℓ,r,s,q)=  0, otherwise. According to the elementary formula 1 cosa cosa cosa = cos a +( 1)i1a + +( 1)ih−1a , 1 2··· h 2h 1 1 − 2 ··· − h − (i1,i2,···,ihX−1)∈{0,1}h−1 (cid:0) (cid:1) we can write R7 =S (x)+S (x)+S (x)+S (x)+S (x)+S (x)+S (x), 1 1 2 3 4 5 6 7 where 35 π S (x) := cos gx7/4, 1 64 4 n,m,k,ℓ,r,s,q6y X √n+√m+√k+√ℓ=√r+√s+√q 35 S (x) := gx7/4 2 64 n,m,k,ℓ,r,s,q6y X √n+√m+√k+√ℓ=√r+√s+√q 6 π cos 4π √n+√m+√k+√ℓ √r √s √q √x , × − − − − 4 21 (cid:16)3π(cid:16) (cid:17) (cid:17) S (x) := cos gx7/4, 3 64 4 n,m,k,ℓ,r,s,q6y X √n+√m+√k+√ℓ+√r=√s+√q 21 S (x) := gx7/4 4 64 n,m,k,ℓ,r,s,q6y X √n+√m+√k+√ℓ+√r=√s+√q 6 3π cos 4π √n+√m+√k+√ℓ+√r √s √q √x , × − − − 4 (cid:18) (cid:16) (cid:17) (cid:19) 10

On the Seventh Power Moment of $\Delta(x)$

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