On the mean square formula of the error term in the Dirichlet divisor problem

Transcription

1 On the ean square forula of the error ter in the Dirichlet divisor proble Y.-K. Lau & K.-M. Tsang Departent of Matheatics, The University of Hong Kong, Pokfula Road, Hong Kong e-ail: 1. Introduction Let d(n) denote the divisor function. In 1849, Dirichlet introduced an eleentary but iportant ethod to show a quite precise asyptotic forula for its suatory function, (1.1) d(n) = t(log t + γ 1) + (t) n t with (t) = O(t 1/ ). Since then, there are plenty of works in the literature devoted to exploring this error ter on various aspects, including for instance, its high oents, distribution function, gaps between sign-changes and oega results (see [8], [0], [5], [6], [18], etc). In this paper we are concerned with the size of reainder ter F (x) in the ean square forula (1.) x (t) dt = 1 6π =1 d() 3/ x3/ + F (x). The forula (1.) with F (x) = O(x 5/4+ɛ ) was first proved by Craér[] in 19. It reveals (t) t 1/4 on average, and suggests the conjecture that (t) t 1/4+ɛ (t ) holds for any positive ɛ. This is a difficult conjecture and the best exponent of t to date is 131/416(= 1/4 + 7/416), due to Huxley [7]. The size of F (x) governs the quality of the ean square of (x) over short intervals x+l x (t) dt cx 1/ L, which follows fro (1.) once F (x) = o(x 1/ L). Such results for short intervals show the local behaviour of (t) (see [15] and [1] for relevant works). Craér s estiate on F (x) is not sharp. In 1955, Tong[19] iproved it to F (x) = O(x log 5 x), which then reained unchallenged for the next three decades, until it was subsequently refined to (1.3) F (x) = O(x log 4 x) by Preissann[17]. Further progress towards its size ay be hard because, as shown in [11], F (x) = Ω (x log x). In other words, the roo for further iproveent is not ore than log x. It is anticipated that the oega result is closer to the true agnitude, in virtue of the alost all result in [1]. Very recently, though unable to iprove the bound for F (x), Nowak[16] pushed forth the record in a closely analogous situation - the circle proble. Let P (t) denote the error ter in the circle proble. Owing to the siilarity between their Voronoi series, the ethodology (based on these series) to handle (t) usually applies to P (t) and vice versa. Let (1.4) Q(x) := x P (t) dt 1 3π n=1 r(n) x3/ n3/ where r(n) denotes the nuber of ways of writing n as a su of two squares. In [17], Preissann showed that Q(x) = O(x log x) along the sae line of arguent in (1.3). On the other side we 1

2 know Q(x) = Ω(x) by [10], so again the roo for iproveent on the bound of Q(x) is at ost log x. Rearkably, in [16] Nowak was able to shave off one-half fro the log-power and got Q(x) = O(x(log x) 3/ log x). Here and in the sequel, log r denotes the r-th iterated logarith, thus log x = log log x. Like Preissann[17], Nowak utilized the Hilbert-type inequality of Montgoery and Vaughan to bound Q(x). But specializng to this case, he ade an insightful use of the frequent vanishing of r(n), and the wide spacing between integers that are sus of two squares. Thus he successfully gained the extra saving. Unfortunately the divisor function d(n) does not possess this special vanishing property, and hence the arguent in [16] does not work for (t). Our goal here is to propose a novel and flexible way of refineent by eans of the correlated su of the arithetical function instead of the vanishing property. This will be carried out in two scenarios by the authors and the anonyous referee respectively. Both give iproveents upon the estiate in (1.3) but indeed the referee s ethod is ore efficient and yields a stronger result. As the ideas behind see to be on different basis, we include both approaches and their resulting consequences (see Theores 1 and below). Theore 1 is due to the authors, whose idea is to group ters with a siilar value of d(n) and then apply Hilbert s inequality to the products of these new subsus. Section 3 is devoted to its proof. The sharper result in Theore is accoplished in Section 4 with the ingenious proof of the anonyous referee. Our coprehension of the subtlety is to pinch the adjacent ters to ake a ore effective use of Hilbert s inequality and the trade-off is soe error ters involving the correlated su of the divisor function. Theore 1. Let x and define F (x) as the reainder ter in (1.). Then, we have F (x) = O(x(log x) 7/ (log x) 5/ ). Theore. The result in Theore 1 can be iproved to F (x) = O(x(log x) 3 (log x)). These ethods work for the error ter P (x), and plainly we deduce the following along the proof of Theore. Theore 3. For x, we have Q(x) = O(x(log x)(log x)) where Q(x) is defined as in (1.4). As is custoary in this circle of research, the novel idea in the divisor proble can also be applied to the error function E(T ) := T 0 ζ( 1 + it) dt T log T (γ 1)T, T > 0. π Atkinson [1] gave an approxiation to E(T ) in ters of two finite sus. The first su is very siilar to the Voronoi series for (x), while the second su is usually not difficult to handle. By using Atkinson s forula, Heath-Brown [3] proved that the reainder ter (1.5) F 1 (T ) := T 0 E(t) dt ct 3/, (c = 3 (π) 1/ d(n) n 3/ ) satisfies F 1 (T ) T 5/4 log T. Later Meuran [13] derived a soothened version of Atkinson s forula (see (5.1) below) in which the error ter is uch sharper, thereby got the better bound O(T log 5 T ) for F 1 (T ). (By a different approach, this sae result was also obtained independently by Motohashi.) Then Preissann (see [9, p.73]) applied the Montgoery-Vaughan inequality and further reduced this bound to O(T log 4 T ). Now we can refine to the following. n=1 Theore 4. Let T > 0 and let F 1 (T ) be defined in (1.5). We have An outline of the proof is given in Section 5. F 1 (T ) = O(T (log T ) 3 (log T )).

3 3. Preparations To prove Theore 1, we need the following variant of Hilbert s inequality. Lea.1. Let A and B be two sets of (distinct) real nubers such that either A = B or A B =. Define δ A B (λ) := in λ ν. ν A B ν λ Given any two sequences {a λ } λ A and {b µ } µ B of coplex nubers, we have, for all T 1, ( ) 1/ T a λ a λ b µ e((λ µ)u) du 1/ b µ. 0 δ A B (λ) δ A B (µ) µ B λ A, µ B λ µ λ A Proof. For A = B, this is erely a reforulation of Lea B in [16], whose proof is based on Hilbert s inequality in [14, Theore ]. When A and B are two disjoint sets, the sae arguent reduces to showing (.1) λ A, µ B λ µ ( a λ b µ λ µ λ A ) 1/ a λ δ A B (λ) µ B b µ δ A B (µ) To this end, we consider the two sequences {â ν } ν A B and { b ν } ν A B defined as { { aλ if ν = λ A, bµ if ν = µ B, â ν = and bν = 0 otherwise, 0 otherwise. 1/. Then (.1) is plainly a consequence of Hilbert s inequality applied to {â ν } and { b ν }. Lea.. Let y and assue 1 h y 5/6. Then d()d( + h) σ 1 ( h )y log y where σ 1 (h) = d h d 1. y Proof. The case of positive h follows iediately fro [4], or see [11, p.83]. Replacing + h by then yields the other case. 3. Proof of Theore 1 Let us fix any sufficiently large nuber x, and define F (x) as in (1.). Then, by equation (.) in [11], we have (3.1) where F (x) = S(x) + S 1 (x) + O(x) S(x) = (4π ) 1 d()d(n) (n) 3/4 n x 7 x u cos(4π( n) u) du and S 1 (x) = (4π ) 1 d()d(n) x u sin(4π( + n) u) du. (n) 3/4,n x 7 (This is shown by using the good approxiation to (x) developed by Meuran[13] with soe standard arguents.)

4 4 We show that the double su S 1 (x) is negligible in coparison with our result. Indeed, by integrating by parts, the integral is x( + n) 1. Hence we have S 1 (x) x d() d(n) 3/4 n x 5/4 log3 x. n x 7 Thus, (3.) F (x) = S(x) + O(x log 3 x). Now take J := [30 log x] and for 0 j J, define C j := {n x 7 : j 1 < d(n) j } and C J+1 := {n x 7 : d(n) > J }. Then we can decopose S(x) into (3.3) where (3.4) S ij (x) = C i, n C j n J+1 S(x) = (4π ) 1 d()d(n) (n) 3/4 x i,j=0 S ij (x) u cos(4π( n) u) du. Applying a change of variable and the ean value theore for definite integrals, we see that for soe x 1 x x, x d()d(n) S ij (x) x x 1 (n) e(( n)v) dv 3/4 C i, n C j n x 1/ d() (3.5) d(n) 3/ δ Ci ij () n 3/ δ ij (n) n C j by applying Lea.1. Here we abbreviate δ ij () for δ Ci C j () = in l l C i C j l. Our ain task is to bound the last product of two sus. The distance function δ ij () has a quite trivial lower bound (3.6) δ ij () 1/ (which will also be explained below). Following fro (3.6) and the well-known fact y d() y log 3 y, one deduces that (3.7) C i d() 3/ δ ij () d() (log x)4. x 7 Inserting this into (3.5) and then (3.3), (3.) yields the result of Preissann[17] odulo a factor of (log x) (caused by the dyadic division in (3.3)). Our novel idea refines the estiate for the su in (3.7) when i j. Let 0 i j J + 1 and consider first the case i J. We split C i d() 3/ δ ij ()

5 5 into two parts according as δ ij () 1/ log x or not. Clearly, the subsu subject to δ ij () 1/ log x is (log x) d() log x. x 7 It reains to consider the subsu constrained by δ ij () < 1/ log x. For each, there is an l C i C j, not equal to, such that l = δ ij (). The constraint on δ ij () forces < log x if l 4, and in this case, δ ij () l/ 1. For l < 4, we have δ ij () l /, which leads to (3.6) and l log x under the constraint. Thus we further divide this subsu into two pieces, running over < log x or otherwise. The piece over < log x is, by (3.6), d() (log x) 4. log x Before estiating the other piece, we note that the value of l depends on. To get rid of the dependence, we take all possible l with l log x into account. Letting h = l (and thus δ ij () h / ), we bound this piece, up to a constant ultiple, by h 1 d() (3.8) 1 h log x log x (, +h) C(i;j) where C(i; j) := {(a, b) : a C i, b C i C j }. Then, under the assuption i j we observe that d( + h) > i 1 1 d() for C i and + h C i C j. The inner su in (3.8) can thus be replaced by the correlated su, or ore precisely, it is less than d()d( + h) σ 1 ( h ) log 3 x, x 7 by Lea.. As 1 h log x σ 1( h )/ h log x, we infer that the double su in (3.8) is log 3 x log x, and consequently, for 0 i j J + 1, i J, (3.9) d() 3/ δ ij () log3 x log x. C i For the case i = J + 1, a better upper bound follows, due to the sparsity of large values of the divisor function. Indeed, using (3.6) and y d()3 y log 7 y, we have C J+1 d() 3/ δ Jj () d() C J+1 J d()3 C J+1 J x 7 d() 3 J log 8 x 1 by our choice of J, so (3.9) is also valid in this case. In view of the condition 0 i j J + 1, we are able to apply (3.9) to one of the two factors in the product in (3.5) (but not both unless i = j). The other factor is handled by (3.7). It follows that S ij (x) x(log x) 7/ (log x) 1/. By (3.) and (3.3), our proof of Theore 1 is coplete.

6 6 4. Proof of Theore We give essentially verbati the proof provided by the referee, which begins with the following robust lea. Lea 4.1. Suppose a n are coplex nubers for N < n N. Let K N be a positive integer. Then a a n S 1 + S + S 3 n where and Proof. N< n N S 1 = N 3/ K 1 S = N 1/ h N/K h N/K h 1 h N<n N h N<n N h S 3 = KN 1/ N<n N a n a n+h a n. a n a n+h, To prove the lea we divide (N, N] into K disjoint intervals ( k 1 I = K N, k ] K N, K < k K. We write g(n) for the lower end-point of the interval I for which n I, so that g(n) < n g(n) + N/K. We clai that a a n a a n (4.1) = + O(S 1 ) + O(S ). n g() g(n) N< n N N<,n N g() g(n) We first consider ters with n + N/K, so that g() g(n). Then 0 < n g(n) N/K, whence 0 < n g(n) N/K, and siilarly for. We also have n ( n)/ N. Moreover if n N/K then g() g(n) N/K n ( n)/. On the other hand, if N/K n < N/K we have g() g(n) N/K > ( n)/, since g() g(n), as rearked above. In either case we conclude that g() g(n) (g() g(n))/ N ( n)/ N. It follows that 1 1 = + O ( N 3/ ) n g() g(n) K( n) when n + N/K. Thus these O-ters contribute O(S 1 ) to (4.1). Secondly we consider ters in which n < < n+n/k. Here we have n ( n)/ N. Moreover if g() > g(n) we have g() g(n) N/K, whence g() g(n) N/K ( n)/ N. It follows that the ters under consideration contribute O(S ) to (4.1), which establishes the clai.

7 7 We now group ters n fro a particular interval I, all of which have the sae lower end-point g(n). Thus, letting I 1,, I J be the sequence of intervals I we set A(j) = n I j a n and we write g j for the lower end-point of I j. Thus N<,n N g() g(n) a a n = A(j)A(k) g() g(n) gj. g k j k Moreover for j > k we have g j g k (g j g k )/ N N/K. We can now apply Hilbert s inequality (which is (.1) with A = B) to give a bound KN 1/ j J A(j). When we expand A(j) we get ters a n a n+h with n, n + h I j. Thus h N/K. When h 0 we have KN 1/ N 1/ h 1, so that the overall contribution is O(S ). Finally, ters with h = 0 produce a contribution O(S 3 ), which copletes the proof of Lea 4.1. We proceed now to prove Theore. By (3.) and a treatent as in the line before (3.5), there exist a n C for n x 7 such that a n d(n)n 3/4 and F (x) x(log x) 3 + x a a n n. n Let N j = j/ and decopose the su on the left into O((log x) ) subsus S i,j = N i < N i N j <n N j a a n n, where it is understood that n if i = j. When i j + we have n N i, and a trivial bound yields S i,j N 1/4 i j (log x). Thus the total contribution fro all such sus S i,j is O((log x) 3 ). Sus with j i + ay be handled siilarly. We also have a a n S i,i+1 + S i+1,i = S i,i S i+1,i+1 +. n N i< n N i The su on the right ay be bounded by Lea 4.1, as can S i,i and S i+1,i+1. We take K = [N i /(1 + log N i )] and find (via Lea.) that a n a n+h N 1/ (log x) σ 1 (h), N<n N h whence S 1 (log x), S (log x) (log x) and S 3 (log x). It follows that S i,i (log x) (log x) and S i,i+1 + S i+1,i (log x) (log x). Thus on suing over i log x we finally conclude that F (x) x(log x) 3 (log x).

8 8 5. Proof of Theore 4 To prove Theore 4, we begin with the odified version of Atkinson s forula as given by Meuran in [13]. Using the sae notations in Meuran s paper, (5.1) where 1 (T ) = E(T ) = ( T π (T ) = ) 1/4 1 (T ) + (T ) + π + O(T 1/4 log T ) n (a+u) η(n)( 1) n d(n)n 3/4 e(t, n) cos f(t, n), ξ(t, n) d(n)n 1/ (log T πn ) 1 cos g(t, n). Here T 1/4 U T 1/4, T 1/ a T 1/. The sus 1 (T ) and (T ) are essentially the sae as those in the original forula of Atkinson, except for the soothing functions η(n) and ξ(t, n), which approxiate the characteristic functions on the intervals [0, (a + U) ] and [0, Z(0)] respectively. With the functions η(n) and ξ(t, n), the O-ter in (5.1) is uch sharper. As was shown in [13, p ] (5.) T where for j, k {1, }, Moreover, T E (t) dt = I 11 + I 1 + I + O(T log T + T 1/ I 1/ ) I jk = T T (t) (t) dt. j I 11 = c((t ) 3/ T 3/ ) + M(T ), k I T log 4 T with M(T ) T log 5 T. This bound on M(T ) yields Meuran s result. In fact, apart fro the harless factor ( 1) n η(n), M(T ) is essentially the sae as F (x) in (3.1). Hence the idea in Section 4 proves (5.3) M(T ) T (log T ) 3 (log T ). To coplete the proof of Theore 4, we establish the bounds (5.4) and (5.5) I T log 3 T I 1 T (log T ) 3 (log T ) 1/, which are ore than enough. The proof of (5.4) follows the standard arguent of opening up the square of (t) and then integrating ter by ter. By the copound angle forula, there is a subsu involving cos(g(t, n)+ g(t, )) which, by the ean value theore, is ( d(n) d() ax (ξ(t, n)ξ(t, )) log T n T <t T πn log T ) 1 π n, Z(0) (5.6) ( d(n)n 1/ ) T log T.

9 Note that Z(0) (1 δ) T π for soe δ > 0, depending on a. The other subsu, involving cos(g(t, n) g(t, )), is T d(n) ( log T ) + ( d(n) d() log T n πn n πn log T π log n ) 1, < again by applying the ean value theore. Then partial suation together with the estiate y d(n) y log 3 y shows the first su here is log 3 T. The second su can be further splitted into two pieces. The one in which < n/, n Z(0) is n < d(n) n n d() T log T. In view of the inequality (log(n/)) 1 /(n ), the other piece is ( d(n) d() log T ) n πn n say. The su b is n < = a + b, d(n) d() n d(n) d() n + n n 5/6 h Z(0) 5/6 h 1 Z(0) n n 5/6 <<n d() d( + h) T log 3 T, d(n) d() n by using Lea.. To bound the su a, we first estiate the inner su over by partial suation together with the forula (1.1), and it is log n. Hence a d(n) log n T log 3 T. These together with (5.6) proves (5.4) The proof of (5.5) follows the sae arguent of [13, p.341], where 1 is splitted into two parts, according as n T/A and T/A < n (a + U) respectively. The contribution fro the first part is T log T, as shown in [13]. The contribution of the second part, by using Cauchy-Schwarz s inequality together with (5.4) and the bound for M(T ) in (5.3), is {(T log 3 T + T log 3 T log T )(T log 3 T )} 1/ T log 3 T log 1/ T. This copletes the proofs of (5.4) and (5.5), and hence Theore 3. Acknowledgeent The authors wish to acknowledge the generous referee for providing the result in Theore and its proof. The work described in this paper was fully supported by a grant fro the Research Grants Council of the Hong Kong Special Adinistrative Region, China (HKU 704/04P). 9

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