The Distribution and Moments of the Error Term in the Dirichlet Divisor Problem

Transcription

1 he Distribution and Moments of the Error erm in the Dirichlet Divisor Problem D.R. Heath-Brown Magdalen College, Oxford OX 4AU Introduction his paper will consider results about the distribution and moments of some of the well known error terms in analytic number theory. o focus attention we begin by considering the error term (x) in the Dirichlet divisor problem, which is defined as (x) = n x d(n) x(log x + 2γ ). We shall investigate the distribution of the function x /4 (x) as x tends to infinity and prove:- heorem he function x /4 (x) has a distribution function f(α) in the sense that, for any interval I we have X mes{x [, X] : x /4 (x) I} f(α)dα as X. he function f(α) and its derivatives satisfy the growth condition d k dα k f(α) A,k ( + α ) A for k =,, 2,... and any constant A. Moreover f(α) extends to an entire function on C. heorem 2 For any exponent k [, 9] the mean value X X k/4 (x) k dx converges to a finite limit as X tends to infinity. Moreover the same is true for the odd moments X X k/4 (x) k dx (.) for k =, 3, 5, 7 or 9. I

2 An examination of the proof shows that the range [, 9] may be extended somewhat. For k = 2 it was shown by Cramér [3] that X X 3/2 (x) 2 dx (6π 2 ) Moreover when k = one has X n= (x)dx = o(x 5/4 ), d(n) 2 n 3/2. (.2) as follows from the work of Voronoi [2]. he cases k = 3 and 4 of (.) have been handled very recently by sang [] who gives the value of the limit explicitly, as the sum of an infinite series. When k = 3 the limit is positive, contrasting with the case k =. he distribution function f(α) of heorem must therefore be skewed towards positive values of α. It should also be noted that the methods of the above papers all give reasonably good estimates for the rate of convergence, whereas our approach will not. he method applies equally well to certain other error terms. We have:- heorem 3 he conclusions of heorems and 2 apply verbatim with (x) replaced by the error term P (x) in the circle problem, namely P (x) = n x r(n) πx; or by the error term E( ) for the mean value of the Riemann Zeta-function, namely E( ) = ζ( 2 + it) 2 dt (log + 2γ ). 2π Moreover heorems and 2 also hold for the error term 3 (x) in the Pilz divisor problem, providing that x /4 (x) is replaced by x /3 3 (x), and k is restricted to the range [, 3) in heorem 2. Of course for 3 (x) we only handle the odd moment, analogous to (.), for k =. Again we may note that X X 3/2 P (x) 2 dx (3π 2 ) n= r(n) 2 n 3/2, (Cramér [3]), X X k/4 P (x) k dx c k, (k = 3, 4), (sang []), 3/2 E(t) 2 dt 2 3 (2π) /2 n= d(n) 2 n 3/2, (Heath-Brown [4]), (.3) 2

3 and k/4 E(t) k dt c k, (k = 3, 4), (sang []), X X 5/3 3 (x) 2 dx = (π 2 ) d 3 (n) 2 n 4/3 + O(X /9+ε ), (.4) n= (ong [], Ivić [6; (3.43)]). It appears that the error terms k (x) for k 4 cannot be handled by our methods, since no result corresponding to (.2) and (.4) is available. Our argument will consider a general function F (t) which is approximated in the mean by an oscillating series as follows. Hypothesis (H): Let a (t), a 2 (t),... be continuous real valued functions of period, and suppose that there are non-zero constants γ, γ 2,... such that lim lim sup N min{, F (t) n N a n (γ n t) }dt =. his condition already suffices to obtain a distribution for F (t), in a weak sense. heorem 4 If F (t) satisfies (H) then the mean value p(f (t))dt converges to a limit as, for any continuous piecewise differentiable function p(α) for which p(α) dα, ˆp(α) dα <. When the constants γ i in Hypothesis (H) are linearly independent over Q and the functions a i (t) are suitably behaved we can say rather more. heorem 5 Let F (t) satisfy (H) and suppose that the constants γ i are linearly independent over Q. Suppose moreover that a n (t)dt = n= (n N), a n (t) 2 dt <, and that there is a constant µ > for which max a n(t) n µ t [,] 3

4 and lim n nµ a n (t) 2 dt =. hen F (t) has a distribution function f(α) with the properties described in heorem. Lastly, in order to establish heorem 2 we shall prove:- heorem 6 Let F (t) have a distribution in the sense of heorem 4, and suppose that F (t) K dt for some positive exponent K. hen the limits for real k [, K), and for odd integers k [, K), all exist. lim F (t) k dt, lim F (t) k dt, Results of the above nature have been obtained by Kueh [8], who considered both the general situation and, as particular cases, the distribution of ζ( + it) and (on the Riemann Hypothesis) of n /2 Λ(n) 2x /2. n x However Kueh s method appears not to apply to our examples. (Note that in Kueh s heorem, the ambiguous expression for any p > must be read as for every p >, rather than as for some p >.) It appears to be an open question whether x /2 M(x) = x /2 n x µ(n) has a distribution function. o prove this one would want to assume the Riemann Hypothesis and the simplicity of the zeros, and perhaps also a growth condition on M(x). 4

5 2 Preliminary Results In this section we prove two results about the mean value of certain almostperiodic functions. If f(t) is a continuous function from R to C, we define m (f) = f(t)dt, and if m (f) converges as tends to infinity we write L(f) for the resulting limit. As usual we shall define e(x) = exp(2πix). We first prove:- Lemma Let b i (t), ( i k) be continuous functions of period from R to C. hen m (e(γt)b (γ t)... b k (γ k t)) converges for any real γ, γ,..., γ k. Moreover, if γ is not an integral linear combination of the γ i, then the limit is zero. he proof is by induction on k. When k = we merely observe that m (e(γt)) {, γ =,, γ, ( ). Suppose now that the lemma holds for a particular value of k. For ease of notation we shall write, temporarily, f(t) = e(γt)b (γ t)... b k (γ k t). We begin by observing that the function b k+ has a Fourier series which converges to it in the mean. hus where lim N s N (t) = b k+ (t) s N (t) dt =, N n= N In general, if b(t) is a continuous function of period, then m ( b ) [ ] for. It follows that +[ ] c n e(nt). (2.) b(t) dt = + [ ] m ( b ) 2m ( b ) [ ] m ( b k+ (γ k+ t) s N (γ k+ t) ) (N ) 5

6 uniformly for γ k+. Now, since f(t) is bounded, by B say, we have m (f(t){b k+ (γ k+ t) s N (γ k+ t)}) for some N = N(ε) and any γ k+. However m (f(t)s N (γ k+ t)) = Bm ( b k+ (γ k+ t) s N (γ k+ t) ) < ε, (2.2) N c n m (e({γ + nγ k+ }t)b (γ t)... b k (γ k t)). N Our induction assumption therefore yields m (f(t)s N (γ k+ t)) L ε ( ) for some limit L ε depending on N(ε). Moreover, if γ is not an integral linear combination of γ,..., γ k+, then γ + nγ k+ cannot be an integral combination of γ,..., γ k, whence L ε =. We now have for (ε). It follows from (2.2) that for (ε) and hence that m (f(t)s N (γ k+ t)) L ε < ε m (f(t)b k+ (γ k+ t)) L ε < 2ε (2.3) m (f(t)b k+ (γ k+ t)) m (f(t)b k+ (γ k+ t)) < 4ε for, (ε). he Cauchy convegence criterion is therefore satisfied, and so m (f(t)b k+ (γ k+ t)), converges to a limit L, say, as required. It follows from (2.3) that L L ε 2ε. However our linear independence condition would imply that L ε =, and hence L =, since ε is arbitrary. his completes the proof of Lemma. We now deduce:- Lemma 2 Let b i (t), ( i k) be continuous functions of period from R to C. hen m (b (γ t)... b k (γ k t)) converges for any real constants γ,..., γ k. Moreover, if the numbers γ i, are linearly independent over Q, then the limit is L(b i ). i k 6

7 For the proof we again use induction on k, the results being trivial for k =. We now write f(t) = b (γ t)... b k (γ k t), for short. Suppose the lemma holds for a particular value of k. As before we have m (f(t){b k+ (γ k+ t) s N (γ k+ t)}) < ε, for some N = N(ε) and any γ k+. Moreover by Lemma, and m (f(t)s N (γ k+ t)) L(f(t)s N (γ k+ t)), L(f(t)s N (γ k+ t)) = c L(f), in the notation (2.), providing the linear independence condition holds. Clearly whence m (b k+ ) c L(f) = c L(b... b k ) = b k+ (t)dt = c = L(b k+ ), k+ L(b i ), by our induction assumption. he proof of Lemma 2 may now be completed along the same lines as that of Lemma 3 Proof of heorems 4 and 6 We begin by using Lemma 2 to prove heorem 4. Since p(α) is continuous and piecewise differentiable we have Moreover, since we have A p(f (t)) = lim ˆp(α)e(αF (t))dα. A A α >A ˆp(α) dα <, ˆp(α) dα < ε for A A(ε). For such an A we therefore deduce that A m (p(f )) ˆp(α)m {e(αf (t))}dα < ε. (3.) A 7

8 However, if we set S N (t) = n N a n (γ n t), then e(αf (t)) e(αs N (t)) 2πA min{, F (t) S N (t) }. uniformly for α A. Moreover, Hypothesis (H) implies that min{, F (t) S N (t) }dt < δ, for any N N(δ) and any = (δ, N). It follows that m {e(αf (t))} m {e(αs N (t))} 2πAδ for N N(δ) and (δ, N), and hence that We now choose A A ˆp(α)m {e(αf (t))}dα ˆp(α)m {e(αs N (t))}dα A A A < 2πAδ ˆp(α) dα. (3.2) A hen (3.) and (3.2) yield A δ = ε{2πa ˆp(α) dα}. A A m (p(f )) ˆp(α)m {e(αs N (t))}dα < 2ε (3.3) A for A A(ε), N N(ε, A) and (ε, A, N). We are now ready to apply Lemma 2, taking whence converges to a limit L(α, N), say. independent over Q, we have b i (t) = e(αa i (t)), m {e(αs N (t))} Moreover, if the numbers γ i are linearly L(α, N) = χ n (α), n N where χ n (z) = e(za n (t))dt. (3.4) 8

9 Since ˆp(α)m {e(αs N (t))} ˆp(α) for any, we may apply Lebesgue s Dominated Convergence heorem and deduce that A A as tends to infinity. hus ˆp(α)m {e(αs N (t))}dα A A ˆp(α)L(α, N)dα, A A ˆp(α)m {e(αs N (t))}dα ˆp(α)L(α, N)dα < ε, (3.5) A A for = (ε, A, N) say, so that for m {p(f (t))} m {p(f (t))} < 6ε, max( {ε, A(ε), N(ε, A(ε))}, {ε, A(ε), N(ε, A(ε))}), by (3.3) and (3.5). We now see that m {p(f (t))} converges as tends to infinity, by the Cauchy convergence criterion. his completes the proof of heorem 4. he proof of heorem 6 is now straightforward. Let A > be given. We choose a continuous non-negative function of compact support, p(α) say, which is twice continuously differentiable except possibly at α =, and which satisfies { = α p(α) k, α A, α k, α > A. hen for every α, whence p(α) α k A k K α K, m {p(f (t))} m ( F (t) k ) A k K m ( F (t) K ). he hypothesis of heorem 6 therefore yields m {p(f (t))} m ( F (t) k ) < ε for sufficiently large A = A(ε), uniformly for >. he conditions on the function p clearly ensure that p(α) dα <. 9

10 Moreover, we have for some θ >, since and A e(αt)p(t)dt = 2πiα A e(αt)t k dt α k e(αt)p (t)dt = e(αa) 2πiα p (A) 2πiα hus the second condition ˆp(α) dα < e(αt)p (t)dt α θ, A e(αt)p (t)dt α. of heorem 4 is also satisfied. hus m {p(f (t))} converges as tends to infinity, and we see that m ( F (t) k ) satisfies the Cauchy convergence criterion. he first part of heorem 6 now follows. o handle the odd moments we proceed in exactly the same way after making the obvious modifications in the choice of the function p(α). 4 Proof of heorem 5 o prove heorem 5 we shall require the following properties of the functions χ n given by (3.4). Lemma 3 Let a n (t) be continuous real valued functions of period, such that and n= a n (t)dt = (n N) (4.) Suppose further that there is a constant µ > for which a n (t) 2 dt <. (4.2) max a n(t) n µ (4.3) t [,] and lim n nµ a n (t) 2 dt =. (4.4)

11 Define χ(z) = χ n (z), (4.5) n= where χ n (z) is given by (3.4). hen the product (4.5) converges absolutely and uniformly for z in any compact set K K = {z = x + iy C : y min(, x /(µ ) )}. he function χ(z) is therefore holomorphic on K. Moreover for any real positive constant A, and χ(z) A e A z (z K) (4.6) d k dx k χ(x) A,k e A x for any real x, any integer k, and any real positive constant A. for Since e(α) = + 2πiα + O( α 2 ) for any α, we see from (4.) that χ n (z) = + O( z 2 a n (t) 2 dt) n z /(µ ). (4.7) hus (4.2) implies the absolute convergence of the product (4.5), uniformly on any compact set. Similarly, since e(α) = + 2πiα 2π 2 α 2 + O( α 3 ), for α, we have χ n (z) = 2π 2 z 2 a n (t) 2 dt + O( z 3 a n (t) 3 dt) = 2π 2 z 2 { + O(n µ z )} for n in the range (4.7), by (4.3). Moreover a n (t) 2 dt for large enough z K, and R(z 2 ) z 2 2 α exp{ R(α) + O( α 2 )} for α. hus, if n N, where N = c z /(µ ) and c is a suitably large constant, we have χ n (z) exp( z 2 a n (t) 2 dt).

12 For other values of n we use the trivial bound It then follows that χ n (y) exp{2π I(z) max t [,] a n(t) } exp{o( z /(µ ) n µ )} exp{o( z /(µ ) )}. χ(z) exp{o( z /(µ ) N) z 2 n N According to (4.4), for any value of B we have a n (t) 2 dt Bn µ for n N and N N(B). Hence (4.8) is at most exp(o( z ) z 2 B N µ ) exp( B z ) a n (t) 2 dt}. (4.8) for constant B, B which tend to infinity with B. hus (4.6) holds. It remains to estimate the derivatives of χ on the real axis. his can be done in the familiar way, using Cauchy s integral formula for the k-th derivative, choosing a circular contour of radius proportional to z /(µ ), and estimating χ(z) by means of (4.6). We now re-examine the proof of heorem 4. It follows from Lemma 3 that L(α, N) χ(α) as N, and since L(α, N), Lebesgue s Dominated Convergence heorem shows that lim lim A N A A ˆp(α)L(α, N)dα = aken in conjunction with (3.3) and (3.5) this produces ˆp(α)χ(α)dα. L{p(F (t))} = = ˆp(α)χ(α)dα p(α) ˆχ(α)dα, the conditions on the function χ needed for Parseval s identity following from Lemma 3. he function L{p(F (t))} is real and non-negative whenever p is. Moreover ˆχ(α) is continuous. hus using a suitable test function p we conclude that ˆχ(α) is necessarily real and non-negative. We now choose a finite interval 2

13 I, and write ψ(x) for its characteristic function. We apply heorem 4 with p chosen so that ψ(x) p(x) for all x R and such that hen lim sup ψ(x) p(x) dx < ε. m (ψ(f (t))) lim m (p(f (t))) Since ε is arbitrary we conclude that Similarly one finds that = < = lim sup m (ψ(f (t))) lim inf m (ψ(f (t))) I p(α) ˆχ(α)dα ψ(α)ˆχ(α)dα + ε max ˆχ(α) α ˆχ(α)dα + ε max ˆχ(α). α he required result, with f(α) = ˆχ(α), now follows. 5 Proof of heorems and 2 I I ˆχ(α)dα. ˆχ(α)dα. In this section we shall show how heorems 4, 5 and 6 may be applied to (x). Our starting point is the formula (x) = x/4 π 2 n X d(n) n 3/4 cos{4π nx π 4 } + O(Xε ), (5.) for any fixed ε >, which holds uniformly for X x 4X. (See itchmarsh [9; (2.4.4)], for example.) We shall define a n (t) = µ2 (n) n 3/4 π 2 F (t) = t /2 (t 2 ), r= d(nr 2 ) r 3/2 cos{(2πrt) π 4 }, 3

14 and In particular we see that F (t) = π 2 γ n = 2 n. n 2 d(n) n 3/4 cos{4πt n π 4 } + O( /2+ε ), uniformly for t 2. hen for any integer N /2 we have F (t) n N a n (γ n t) n 2 n 2 d(n) n cos{4πt n π 3/4 4 } + n 3/4 n N d(n) n 3/4 e(2t n) + N /2 /2+ε r>/ n d(nr 2 ) r 3/2, uniformly for t 2, where indicates that n is restricted to have square-free kernel greater than N. Moreover, if one calculates the mean value termwise, as in Ivić [6; heorem 3.5] for example, one finds that 2 n 2 It therefore follows that 2 F (t) n N d(n) n 3/4 e(2t n) 2 dt = n 2 n>n d(n)2 n 3/2 + O( ε ) d(n) 2 n 3/2 + O( ε ) N ε /2 + ε a n (γ n t) 2 dt ( ε /2 N /2 ) 2 + N ε /2 + ε N ε /2 if ε is small enough. hus Cauchy s inequality implies that lim sup 2 F (t) n N a n (γ n t) dt N ε /4, (5.2) and Hypothesis (H) follows. Moreover heorem is now an immediate consequence of heorem 5, if one takes µ = 5/3, for example. o prove heorem 2 we require an estimate of the form X (x) K dx X +K/4+ε. (5.3) 4

15 Such a bound has been given by Ivić [6; heorem 3.9], who showed that one may take K = 35/4. If one injects the estimate (x) x 7/22+ε of Iwaniec and Mozzochi [7] into Ivić s argument one finds that (5.3) holds even for K = 28/3. he deduction of heorem 2 is now completed by means of the following result, used in conjunction with heorem 6. Lemma 4 Suppose that (5.3) holds for some K > 2. hen for any positive k < K. In fact we shall show that X (x) k dx X +k/4, 2 F (t) k dt, from which Lemma 4 immediately follows. We choose an even integer L > K and we define N = 2 (2L+2). (5.4) hen a n (γ n t) = n N = π 2 π 2 n N 4 n N 4 d(n) n 3/4 cos{4πt n π 4 } +O( n N n 3/4 d(nr2 ) r ) 3/2 r N 2 n /2 d(n) n 3/4 cos{4πt n π 4 } + O(), where Σ restricts n to have square-free kernel at most N. hus 2 n N a n (γ n t) 2L dt 2 n N 4 d(n) n 3/4 cos{4πt n π 4 } 2L dt +. (5.5) We now have recourse to the following result, which we shall prove later. 5

16 Lemma 5 Let n,..., n 2L be positive integers all at most N 4. hen n ±... ± n 2L N 22L+, unless the product n... n 2L is square-full. hus if n... n 2L is not square-full we have 2 n ±... ± n 2L /4, by (5.4), if is large enough. However, so that 2 n N 4 sin πt/4 e(αt)( ) 2 dt =, ( α /4 ), t/4 d(n) n cos{4πt n π 3/4 4 } 2L dt n N 4 d(n) n cos{4πt n π sin πt/4 3/4 4 } 2L ( ) 2 dt t/4 d(n )... d(n 2L ), (5.6) (n... n 2L ) 3/4 where the final sum is for all sets of positive n i N 4 whose product q, say is square-full. Now, for a given value of q there are O(q ε ) possible sets of factors n i. Moreover d(n )... d(n 2L ) q ε. It follows from (5.5) and (5.6) that 2 n N a n (γ n t) 2L dt q 3/4+2ε, where q runs over square-full integers only. he infinite sum above converges, since there ore only O(Q /2 ) square-full integers q Q. We conclude therefore that whence also 2 q= n N a n (t) 2L dt, 2 n N a n (t) k dt, 2 n N a n (t) K dt, (5.7) by Hölder s inequality. 6

17 On the other hand, if 2 < k < K, 2 F (t) n N a n (γ n t) k dt 2 F (t) a n (γ n t) 2 dt n N again by Hölder s inequality. Here by (5.2), while 2 2 F (t) n N a n (γ n t) K dt 2 (K k)/(k 2) F (t) a n (γ n t) K dt n N F (t) n N a n (γ n t) 2 dt N ε /2, 2 (k 2)/(K 2) 2 F (t) K dt + a n (γ n t) K dt +ε, n N by our hypothesis in Lemma 4, together with the second of the bounds (5.7). Hence 2 F (t) n N a n (γ n t) k dt, since N, given by (5.4) is a positive power of. his bound, taken in conjunction with the first of the estimates (5.7) completes the proof of Lemma 4, since values of k 2 can be handled by applying Hölder s inequality to values k > 2. We now prove Lemma 5. If p is a prime not dividing any of the integers m,..., m k then p cannot be a rational linear combination of m,..., m k. his follows from a classical result of Besicovich, to the effect that the square roots of distinct square-free integers are linearly independent over the rationals. Now suppose that the product n... n 2L has a prime factor p which occurs to the first power only, so that p divides n, say, and no other n i. hen any linear relation ± n ±... ± n 2L = (5.8) would yield n2 n2l p =..., m m, 7

18 where m = p n. As noted above such a relation is impossible, so that (5.8) can hold only when the product n... n 2L is square-full. We now set P = σ 2L { i= σ i ni }, where σ runs over all vectors (σ,..., σ 2L ) with σ i = ±. hen P is an integer, and P can only be zero when n... n 2L is square-full. If S is any factor in P then S N 2. Hence, if S is the particular factor occuring in Lemma 5, we have P S (N 2 ) 22L, providing that P. Lemma 5 now follows. 6 Proof of heorem 3 for P (x) and E( ) he method of the previous section requires little modification to handle the functions P (x) and E( ). For P (x) we replace (5.) by the estimate P (x) = x/4 π n X r(n) n 3/4 cos{2π nx + π 4 } + O(Xε ), which holds uniformly for X x 4X. his formula may be proved by the method of itchmarsh [9; 2.4], starting with the function r(n)n s. We can then establish Hypothesis (H) for an appropriate F (t), using the method of 5. o prove the analogue of heorem 2 we also require a bound X P (x) K dx X +K/4+ε of the type given by Ivić [6; heorem 3.2]. If one inserts the estimate P (x) x 7/22+ε of Iwaniec and Mozzochi [7] into Ivić s argument one finds that one may take K = 28/3. his allows for the range [, 9] when one forms the analogue of heorem 2 for P (x). We turn now to the function E( ). Our starting point here is the formula of Atkinson [2], which approximates E( ) by a sum of functions A n ( ) = 2 ( ) n d(n)( n 2π + n2 4 ) /4 { sinh ( πn 2 )/2} cos f(n, ), where f(n, ) = 2 sinh ( πn 2 )/2 + (π 2 n 2 + 2πn ) /2 π 4. 8

19 Specifically we have E(t) = n X A n (t) + Σ 2 (t) + O(log 2 t) uniformly for X t 4X, where 4X X Σ 2 (t) 2 dt X(log X) 4, by Heath-Brown [4; Lemma 3]. We choose an integer N X /6, and define S = {n = mr 2 : µ 2 (m) =, m N, mr 2 N 4 }. hen the non-diagonal terms of 4X X { n X, n S A n (t)} 2 dt contribute a total O( +ε ), as in the proof of Heath-Brown [4; Lemma ], whereas the diagonal terms are X 3/2 d(n) 2 n 3/2 n S X 3/2 d(n) 2 n 3/2 n>n X 3/2 N ε /2, since A n (t) d(n)x /4 n 3/4. It follows that 4X X E(t) n S A n (t) dt X(log X) 2 + 4X X n X, n S X(log X) 2 + X 5/4 N ε /4 X 5/4 N ε /4, by Cauchy s inequality. We therefore conclude that 4X A n (t) dt + Σ 2 (t) dt X 4X X 5/4 E(t) A n (t) dt N ε /4. n S However for n S and X t 4X we have X A n (t) = B n (t) + O(n 3/4+ε X /4 ), 9

20 where hus B n (t) = ( 2t π )/4 ( ) n d(n) n 3/4 cos{ 8πnt π 4 }. 4X X 5/4 E(t) B n (t) dt N ε /4 + X /2 (N 4 ) 7/4+ε N /8. n S X aking X = 2, and we conclude that F (t) = t /2 E(t 2 ) C n (t) = t /2 B n (t 2 ) 2 F (t) C n (t) dt N /8 (6.) n S uniformly in. However, if with then a n (t) = µ 2 (n)( 2 π )/4 sup C n (t) t n S n N r= ( ) nr d(nr2 ) (nr 2 ) cos{2πrt π 3/4 4 }, γ n = 2n π, a n (γ n t) m N m N r>n 2 / m m /2+ε N d(mr 2 ) (mr 2 ) 3/4 N /2+ε. (6.2) It follows from (6.) that 2 F (t) a n (γ n t) dt N /8, n S from which Hypothesis (H) may be deduced in the usual way. We can now proceed to prove versions of heorems and 2 for E( ), just as before. We shall require the bound E(t) K dt +K/4+ε 2

21 with K = 28/3. However this follows from the argument given by Ivić [6; heorem 5.7], on inserting the estimate E( ) 7/22+ε of Heath-Brown and Huxley [5] into the argument. 7 Proof of heorem 3 for 3 (x) For the function 3 (x) it would be nice to use the estimate 3 (x) = x/3 π 3 n 3 /x d 3 (n) n 2/3 cos{6π 3 nx} + O(X +ε / ), of Atkinson [], as quoted by itchmarsh [9; (2.4.6)]. Unfortunately itchmarsh omits the condition x /2+ε x 2/3 ε required by Atkinson for the proof of his formula. he upper bound on prevents us getting an error term o(x /3 ) in the formula, and there appears to be no simple way of extending the range for. We must therefore return to first principles. We consider a non-negative function ω(x), supported in [ 2, 7 2 ], with derivatives of all orders and which satisfies ω(x) = on [, 8]. Our aim will be to show that lim sup F (t) n N a n (γ n t) 2 ω((t/ ) 3 )dt as N tends to infinity, where F (t) = t 3 (t 3 ) and a n (t) = ε(n) π 3 r= d 3 (nr 3 ) cos{6πrt}, (nr 3 ) 2/3 γ n = 3 n, with ε(n) = if n is cube free, and ε(n) = otherwise. On integration by parts the asymptotic formula (.4) yields where F (t) 2 ω((t/ ) 3 )dt = 6π 2 { d 3 (n) 2 n 4/3 }J + O( 2/3+ε ), (7.) We proceed to estimate J = n= ω((t/ ) 3 )dt. n N a n (γ n t) 2 ω((t/ ) 3 )dt, 2

22 where N /6, by termwise integration. Since Y { cos{6πy 3 n} cos{6πy 3 Y/2 + O(), m = n, m}dy = O( 3 m 3 n ), m n, an integration by parts yields (cos{6πt 3 n}) 2 ω((t/ ) 3 )dt = J + O(), (7.2) 2 and cos{6πt 3 n} cos{6πt 3 m}ω((t/ ) 3 )dt 3 m 3 n (m n). (7.3) Moreover, if m n, we have { 3 m 3 n (mn) /3 m n, m n m, min(m /3, n /3 ), otherwise. (7.4) A straightforward estimate then shows that where n N a n (γ n t) 2 ω((t/ ) 3 )dt = n N S n + O(N /3+ε ), (7.5) S n = J ε(n) 6π 2 d 3 (nr 3 ) 2 (nr 3 ) 4/3. r= We now consider the cross terms and we therefore examine F (t)a n (γ n t)ω((t/ ) 3 )dt, 3 (x) cos{6π 3 nx}ω(x/x) dx x = I(n, X), say, where X = 3. Let γ( ) denote a path from 2 i to 2 + i, passing to the left of s =. hen Perron s formula shows that ζ 3 (s) xs 2πi s ds γ( ) tends to D 3 (x) as tends to infinity, for all non-integer values of x. Moreover the discrepancy is bounded uniformly in for all x [X/2, 7X/2]. hus Lebesgue s Bounded Convergence heorem shows that I(n, X) = lim 2πi γ( ) ζ 3 (s)k(s) ds s, (7.6) 22

23 where K(s) = x s cos{6π 3 nx}ω(x/x)dx. Clearly K(s) is an entire function. If we integrate by parts k times we find that K(σ + it) = ( ) k ( + it)(2 + it)... (k + it) k+it dk x dx k {xσ cos{6π 3 nx}ω(x/x)}dx k,σ,ω ( + t ) k X σ (nx) k/3. (7.7) he path of integration in (7.6) may therefore be moved to R(s) =, giving where I(n, X) = ζ() 3 K() + 2πi = ζ() 3 K() + 2πi +i i +i i ζ 3 (s)k(s) ds s χ(s) = 2 s π s sin( πs )Γ( s). 2 ζ 3 ( s)χ 3 (s)k(s) ds s he bound (7.7) yields K(). We now expand ζ 3 ( s) as a Dirichlet series d3 (m)m s and integrate termwise. When m > X /2 the contribution is d 3 (m)m 2 { (nx) /3 d 3 (m)m 2 n 3/2 X /2, ( + t) 7/2 X dt + t 3/2 n 5/3 X 2/3 dt} (nx) /3 on using (7.7) with k = and k = 5. he total contribution to I(n, X) arising from terms m > X /2 is therefore O(n 3/2 X ε ) = O(X /4+ε ). For the remaining terms we move the line of integration to R(s) = 6 + ε. hen 2πi +i i = m s χ 3 (s)k(s) ds s cos{6π 3 nx}ω(x/x){ 2πi In the inner integral we note that /6+ε+i /6+ε i χ(s) 3 s = 2π 33s /2 χ(3s){ + O( s )} = 2π 33s /2 χ(3s) + O( s ε ) (mx) s χ 3 (s) ds }dx (7.8) s 23

24 on the line of integration, by Stirling s approximation. Moreover 2πi σ+i σ i x s χ(s)ds = 2x cos(2πx) for 2 < σ < and any real positive x. It follows that 2πi /6+ε+i /6+ε i = 2πi = (mx)/3 π 3 (mx) s χ 3 (s) ds s /2+3ε+i /2+3ε i 6π 3 (3(mx)/3 ) s χ(s)ds + O((mx) /6+ε ) cos{6π(mx) /3 } + O((mx) /6+ε ). he error term above contributes O(m ε 5/6 X ε+/6 ) to (7.8), and hence, on summing over m X /2, a total O(X /4+2ε ) to I(n, X). o handle the main term we note that cos{6π 3 nx} cos{6π 3 mx}x 2/3 ω(x/x)dx = 3 cos{6πt 3 n} cos{6πt 3 m}ω((t/ ) 3 )dt, on recalling that X = 3. In view of (7.2), (7.3) and (7.4) the contribution to (7.8) is 3 2π n 2/3 J + O(n 2/3 ), for m = n, m n, if m n and m n m, and m 2/3 min(m /3, n /3 ) otherwise. When we sum over m X /2 the contribution to I(n, X) arising from the error terms is therefore O(X ε ), and we conclude, finally, that 3 I(n, X) = 2π d 3(n)n 2/3 J + O(X /4+ε ). A change of variable now shows that F (t) cos{6πt 3 n}ω((t/ ) 3 )dt = 2π 3 d 3(n)n 2/3 J + O( 3/4+ε ), 24

25 uniformly for n N /6. It follows that F (t) ε(n) π 3 r d 3(nr 3 ) (nr 3 ) 2/3 cos{6πtr 3 n}ω((t/ ) 3 )dt = ε(n) 6π 2 J r d 3(nr 3 ) 2 (nr 3 ) 4/3 + O(n 2/3+ε 3/4+ε ) = S n + O(n /3+ε /6+ε J) + O(n 2/3+ε 3/4+ε ) = S n + O(n /3+ε 5/6+ε ), where indicates the condition nr 3 /6. Moreover whence a n (t) ε(n) π 3 F (t)a n (γ n t)ω((t/ ) 3 )dt r d 3(nr 3 ) (nr 3 ) 2/3 cos{6πrt} n /3+ε /8+ε, = S n + O(n /3+ε 5/6+ε ) + O(n /3+ε /6+ε F (t) ω((t/ ) 3 )dt). Cauchy s inequality applied to (7.) shows that F (t) ω((t/ ) 3 )dt), whence the second error term above is O(n /3+ε 7/8+ε ). It therefore follows that F (t) n N a n (γ n t)ω((t/ ) 3 )dt = n N S n + O( 35/36+ε ) for N /24. Combining this result with (7.) and (7.5) we see that F (t) n N a n (γ n t) 2 ω((t/ ) 3 )dt = n>n S n + O( 35/56+ε ) N ε /3 + 35/36+ε N /4, for N /24. Hypothesis (H) now follows. Moreover, on applying heorem 5 with µ = 3/2 we obtain a result corresponding to heorem for 3 (x). An analogue of Lemma 4 can be obtained by mimicing the argument of 5. However the bound X 3 (x) K dx X +K/3+ε (7.9) 25

26 corresponding to (5.3), which is needed for the proof of the relevent version of heorem 2, is not immediately available in the literature. heorem 3. of Ivić [6] comes close to what is required, and, as indicated in itchmarsh [9; p.327] the argument can be modified so as to establish (7.9) with K = 3. We complete our treatment of 3 (x) by supplying the necessary details. Lemma 6 For any ε > we have X 3 (x) 3 dx X 2+ε. Note that, in contrast to the situation with (x), the above result does not seem to be susceptible to small improvements through the use of upper bounds for 3 (x). We adapt Ivićs argument [6; pp ] by observing that if 3 (t) V with /2 t, then d 3 (n)n 2/3 e(3 3 nt) /3 (log ) V (7.) M<n 2M for some M = 2 m 2+ε V 3, as is shown by [6; (3.5)]. Moreover (7.) yields M /3 (log M) 2 /3 (log ) V, whence in fact +ε V 3 M 2+ε V 3. (7.) We now divide the points t,..., t R considered by Ivić into sets for which (7.) holds. If each such set has R(M) elements then there will be some value of M for which R R(M) log. he advantage of this procedure is that we may choose the parameter in Ivić s argument to depend on M. Indeed we shall select = c( M) 2/3, with a suitable constant c >. With this choice the exponential sums that occur can all be bounded via the first derivative estimate, and one obtains R(M) ( + / ){( M) 2/3 + 4/3 M /3 V }V 2 ε he bounds (7.) then yield {( M) 2/ /3 M /3 V + 5/3 M /3 V }V 2 ε R(M) { V V 2 }V 2 2ε 2+3ε V 4, whence R 2+3ε V 4. he proof of the lemma is now readily completed as in Ivić s work. 26

27 References [] F.V. Atkinson, A divisor problem, Quart. J. Math. Oxford Ser., 2 (94), [2] F.V. Atkinson, he mean value of the Riemann Zeta-function, Acta Math., 8 (949), [3] H. Cramér, Über zwei Sätze von Herrn G.H. Hardy Math. Zeit., 5 (922), 2-2. [4] D.R. Heath-Brown, he mean value theorem for the Riemann Zetafuction, Mathematika, 25 (978), [5] D.R. Heath-Brown, and M.N. Huxley, Exponential sums with a difference, Proc. London Math. Soc. (3), 6 (99), [6] A. Ivić, he Riemann Zeta-function (Wiley, New York, 985). [7] H.Iwaniec, and C.J. Mozzochi, On the divisor and circle problems, J. Number heory, 29 (988), [8] K.-L. Kueh, he moments of infinite series, J. reine angew. Math., 385 (988), -9. [9] E.C. itchmarsh, he theory of the Riemann Zeta-function, 2nd Edn. (Oxford, 986). [] K.-C. ong, On divisor problems, III Acta Math. Sinica, 6 (956), [] K.-M. sang, Higher power moments of (x), E(t) and P (x), to appear. [2] G.F. Voronoi, Sur une fonction transcendente et ses applications à la sommation de quelques séries, Ann. École Normale (3), 2 (94), &