A study of the mean value of the error term in the mean square formula of the Riemann zeta-function in the critical strip 3/4 σ < 1

Transcription

1 Journal de héorie des Nombres de Bordeaux 8 (26), A study of the mean value of the error term in the mean square formula of the Riemann zeta-function in the critical strip 3/4 σ < par Yuk-Kam LAU Résumé. Pour σ dans la bandre critique /2 < σ <, on note E σ ( ) le terme d erreur de la formule asymptotique de ζ(σ + it) 2 dt (pour grand). C est un analogue du terme d erreur classique E( ) (= E /2 ( )). L étude de E( ) a une longue histoire, mais celle de E σ ( ) est assez récente. En particulier, lorsque 3/4 < σ <, on connaît peu d informations sur E σ ( ). Pour en gagner, nous étudions la moyenne E σ(u) du. Dans cet article, nous donnons une expression en série de type Atkinson et explorons quelques une des propriétés de la moyenne comme fonction en. Abstract. Let E σ ( ) be the error term in the mean square formula of the Riemann zeta-function in the critical strip /2 < σ <. It is an analogue of the classical error term E( ). he research of E( ) has a long history but the investigation of E σ ( ) is quite new. In particular there is only a few information known about E σ ( ) for 3/4 < σ <. As an exploration, we study its mean value E σ(u) du. In this paper, we give it an Atkinson-type series expansion and explore many of its properties as a function of.. Introduction Let ζ(s) be the Riemann zeta-function, and let E( ) = ζ(/2 + it) 2 dt (log + 2γ ) 2π denote the error term in the mean-square formula for ζ(s) (on the critical line). he behaviour of E( ) is interesting and many papers are devoted Manuscrit reçu le 2 septembre 24.

2 446 Yuk-Kam Lau to study this function. Analogously, it is defined for /2 < σ <, ( ) E σ ( ) = ζ(σ + it) 2 2σ ζ(2 2σ) dt ζ(2σ) + (2π) 2 2σ 2 2σ. he behaviour of E σ ( ) is very interesting too, and in fact, more delicate analysis is required to explore its properties such as the Atkinson-type series expansion and mean square formula, see ([7]-[2]). Excellent surveys are given in [] and [8]. In the critical strip /2 < σ <, our knowledge of E σ ( ) is not uniform, for example, an asymptotic formula for the mean square is available for /2 < σ 3/4 but not for the other part. In fact, not much is known for the case 3/4 < σ <, except perhaps some upper bound estimates and (.) E σ (t) 2 dt (3/4 < σ < ). (See [7], [2] and [4].) o furnish this part, we look at the mean value E σ(u) du. he mean values of E( ) and E σ ( ) (/2 < σ < 3/4) are respectively studied in [2] and [6], each of which gives an Atkinson-type expansion. Correspondingly, we prove an analogous formula with a good error term in the case 3/4 σ <. Actually, the tight lower bound in (.) is shown in [4] based on this formula. he proof of the asymptotic formula relies on the argument of [2] and uses the tools available in [2] and [9]. But there is a difficulty which we need to get around. In [2], Hafner and Ivić used a result of Jutila [9] on transformation of Dirichlet Polynomials, which depends on the formula b b a n b d(n)f(n) = (log x + 2γ)f(x) dx + d(n) f(x)α(nx) dx, a where α(x) = 4K (4π x) 2πY (4π x) is a combination of the Bessel functions K and Y. It is not available in our case but this can be avoided by using an idea in [9]. In addition, we shall regard the mean value as a function of and study its behaviour; more precisely, we consider (.2) G σ ( ) = n= (E σ (t) + 2πζ(2σ )) dt. (he remark below Corollary explains the inclusion of 2πζ(2σ ).) Unlike the case /2 σ < 3/4, the function G σ ( ) is now more fluctuating. Nevertheless we can still explore many interesting properties, including some power moments, Ω ± -results, gaps between sign-changes and limiting distribution functions, by using the tools in [9], [23], [4], [3], [] and [3]. Particularly, we can determine the exact order of magnitude of the gaps of sign-changes (see heorems 5 and 6). he limiting distribution a

3 Mean value of the error term 447 function is not computed in the case /2 σ < 3/4, perhaps because it is less interesting in the sense that the exact order of magnitude of G σ (t) (/2 σ < 3/4) is known; therefore, the limiting distribution is compactly supported. Here, a limiting distribution P (u) is said to be compactly supported if P (u) = for all u a and P (u) = for all u b, for some constants a < b. (Note that a distribution function is non-decreasing.) However, in our case the distribution never vanishes (i.e. never equal to or ), and we evaluate the rate of decay. 2. Statement of results hroughout the paper, we assume 3/4 σ < to be fixed and use c, c and c to denote some constants which may differ at each occurrence. he implied constants in - or O-symbols and the unspecified positive constants c i (i =, 2,...) may depend on σ. Let σ a (n) = d n da and arsinh x = log(x + x 2 + ). We define Σ (t, X) = ( ) t 5/4 σ 2 ( ) n σ 2σ(n) 2π n 7/4 σ e 2(t, n) sin f(t, n), n X ( ) t /2 σ Σ 2 (t, X) = 2 2π where ( e 2 (t, n) = + πn 2t πn f(t, n) = 2t arsinh B(t, X) = n B(t, X) σ 2σ (n) n σ ) ( ) 2 /4 2t πn πn arsinh, 2t 2t + (2πnt + π2 n 2 ) /2 π 4, g(t, n) = t log t 2πn t + π 4, t 2π + X 2 ( t X 2π + X 4 ( log t ) 2 sin g(t, n), 2πn ) /2 (( t = 2π + X ) /2 X 4 2 heorem. Let σ [3/4, ), and N. We have E σ (t) dt = 2πζ(2σ ) + Σ (, N) Σ 2 (, N) + O(log 2 ). he next result follows with the trivial bounds on Σ (, N) and Σ 2 (, N). Corollary. For all 3, we have E σ (t) dt = 2πζ(2σ ) + O( ) (3/4 < σ < ), ) 2.

4 448 Yuk-Kam Lau and E 3/4 (t) dt = 2πζ(/2) + O( log ). Remark. It suggests that E σ (t) is a superimposition of the constant 2πζ(2σ ) and an oscillatory function, say, Eσ(t). (Indeed this viewpoint had appeared in [8].) Define G σ (t) = t E σ(t) dt, which is (.2). hen, G σ (t) t /2 (3/4 < σ < ) and G 3/4 (t) t /2 log. Integrating termwisely with partial integrations, one gets (2.) 2 G σ (t) dt = o( +(5/4 σ) ) (3/4 σ < ). In addition, we have the following higher power moments. heorem 2. Let σ [3/4, ) and. We have () (2) 2 2 G σ (t) 2 dt = B(σ) G σ (t) 3 dt = C(σ) 2 2 (t/(2π)) 5/2 2σ dt + O( 3 2σ ), (t/(2π)) 5/4 3σ dt + O( (3 8σ)/3 ), where B(σ) and C(σ) are defined by B(σ) = σ 2σ (n) 2 n 2σ 7/2 = ζ(7/2 2σ)ζ(3/2 + 2σ)ζ(5/2) 2 ζ(5), C(σ) = 3 2 n= s= µ(s) 2 s 2/4 3σ a,b= σ 2σ (sa 2 ) a 7/2 2σ σ 2σ (sb 2 ) b 7/2 2σ σ 2σ (s(a + b) 2 ) (a + b) 7/2 2σ. (3) for any real k [, A ) and any odd integer l [, A ) where A = (σ 3/4), and G σ (t) k dt α k (σ) +k(5/4 σ) G σ (t) l dt β l (σ) +l(5/4 σ). for some constants α k (σ) > and β l (σ) depending on σ. (A denotes when σ = 3/4.)

5 Mean value of the error term 449 Remark. We have no information about the value of β l (σ), which may be positive, negative or even zero. (See [6] for possible peculiar properties of series of this type.) It is expected that G σ (t) is oscillatory and its order of magnitude of G σ (t) is about t 5/4 σ. (2.) shows a big cancellation between the positive and negative parts, but heorem 2 (2) suggests that it skews towards negative. his phenomenon also appears in the case /2 σ < 3/4. Now, we look at its distribution of values from the statistical viewpoint. heorem 3. For 3/4 σ <, the limiting distribution D σ (u) of the function t σ 5/4 G σ (t) exists, and is equal to the distribution of the random series η = n= a n(t n ) where a n (t) = 2 µ(n)2 n 7/4 σ ( ) nr σ 2σ(nr 2 ) sin(2πrt π/4) r= r 7/2 2σ and t n s are independent random variables uniformly distributed on [, ]. Define tail(d σ (u)) = D σ (u) for u and D σ (u) for u <. hen (2.2) exp( c exp( u )) tail(d 3/4 (u)) exp( c 2 exp( u )), exp( c 3 u 4/(4σ 3) ) tail(d σ (u)) exp( c 4 u 4/(4σ 3) ) for 3/4 < σ <. Remark. D σ (u) is non-symmetric and skews towards the negative side because of heorem 2 (2). Again it is true for /2 σ < 3/4. But in the case /2 σ < 3/4, the closure of the set {u R : < D σ (u) < } is compact and it differs from our case. o investigate the oscillatory nature, we consider the extreme values of G σ (t) and the frequency of occurrence of large values. hese are revealed in the following three results. heorem 4. We have G 3/4 ( ) = Ω ( log log ) and G 3/4 ( ) = Ω + ( log log log ). For 3/4 < σ <, G σ ( ) = Ω ( 5/4 σ (log ) σ 3/4 ) and ( G σ ( ) = Ω + 5/4 σ exp ( (log log ) σ 3/4 ) ) c 5 (log log log ) 7/4 σ.

6 45 Yuk-Kam Lau heorem 5. For σ [3/4, ) and for every sufficiently large, there exist t, t 2 [, +c 6 ] such that Gσ (t ) c 7 t 5/4 σ and G σ (t 2 ) c 7 t 5/4 σ 2. In particular, G σ (t) has (at least) one sign change in every interval of the form [, + c 8 ]. heorem 6. Let σ [3/4, ) and δ > be a fixed small number. hen for all sufficiently large (δ), there are two sets S + and S of disjoint intervals in [, 2 ] such that. every interval in S ± is of length c 9 δ, 2. the cardinality of S ± c δ 4( σ), 3. ±G σ (t) (c δ 5/2 2σ )t 5/4 σ for all t I with I S ± respectively. Remark. heorems 5 and 6 determine the order of magnitude of the gaps between sign-changes. 3. Series representation his section is to prove heorem and we need two lemmas, which come from [2, Lemma 3] and [9, Lemma ] with [22] respectively. Lemma 3.. Let α, β, γ, a, b, k, be real numbers such that α, β, γ are positive and bounded, α, < a < /2, a < /(8πk), b, k,, ( t U(t) = 2πk + ) /2 πk, V (t) = 2arsinh 4 2t, L k (t) = (2ki ( π) t /2 V (t) γ U(t) /2 U(t) ) α ( U(t) + ) β 2 2 ( exp itv (t) + 2πikU(t) πik + πi ), 4 and J( ) = 2 b a ( y α ( + y) β log + y ) γ y exp(it log( + /y) + 2πiky) dy dt. hen uniformly for α ɛ, k +, we have J( ) = L k (2 ) L k ( ) + O(a α ) + O( k b γ α β ) + O((/k) (γ+ α β)/2 /4 k 5/4 ). In the case k in place of k, the result holds without L k (2 ) L k ( ) for the corresponding integral.

7 Lemma 3.2. Let 2σ (t) = n t σ 2σ(n) Mean value of the error term 45 ( ζ(2 2σ) ζ(2σ)t + 2 2σ t2 2σ ) ζ(2σ ) 2 where the sum n t counts half of the last term only when t is an integer. Define 2σ (ξ) = ξ 2σ(t) dt ζ(2σ 2)/2. Assuming 3/4 σ <, we have for ξ, 2σ (ξ) = C ξ 5/4 σ n= σ 2σ (n)n σ 7/4 cos(4π nξ + π/4) + C 2 ξ 3/4 σ σ 2σ (n)n σ 9/4 cos(4π nξ π/4) + O(ξ /4 σ ) n= where the two infinite series on the right-hand side are uniformly convergent on any finite closed subinterval in (, ), and the values of the constants are C = /(2 2π 2 ), C 2 = (5 4σ)(7 4σ)/(64 2π 3 ). In addition, we have for 3/4 σ <, 2σ (v) v σ, 2σ (ξ) ξ r log ξ, x x 2σ (v) 2 dv x log x, 2σ (v) 2 dv x 7/2 2σ where < r = (4σ 2 7σ + 2)/(4σ ) /2. Proof of heorem. From [7, (3.4)] and [2, (3.)], we have t t ζ(σ + iu) 2 du = 2ζ(2σ)t + 2ζ(2σ )Γ(2σ ) sin(πσ) σ t2 2σ 2i σ+it σ it g(u, 2σ u) du + O(min(, t 2σ )). (Note that the value of c 3 in [2, (3.)] is zero.) Hence, we have E σ (t) = i Define σ+it (3.) h(u, ξ) = 2 σ it g(u, 2σ u) du + O(min(, t 2σ )). y u ( + y) u 2σ cos(2πξy) dy.

8 452 Yuk-Kam Lau Assume A X and X is not an integer where < A < is a constant. hen, following [9, p ], we define (3.2) G (t) = n X σ 2σ (n) G 2 (t) = 2σ (X) G 3 (t) = σ+it σ it X G 4(t) = 2σ (X) σ+it σ it σ+it σ it h(u, n) du, h(u, X) du, (ζ(2σ) + ζ(2 2σ)ξ 2σ )h(u, ξ) dξ du, σ+it G 4 (t) = 2σ (ξ) X σ it σ+it h (u, X) du, ξ σ it 2 h (u, ξ) du dξ. ξ2 hen, we have (3.3) 2 2 E σ (t) dt = i i 2 2 G (t) dt + i G 4(t) dt i 2 2 G 2 (t) dt i G 3 (t) dt G 4 (t) dt + O() ) Evaluation of 2 G (t) dt. By Lemma 3. with γ =, α = β = σ, we have from (3.), 2 σ+it σ it Noting that 2 h(u, n) du dt = 4i = 2i Im (y( + y)) σ (log( + /y)) sin(t log(( + y)/y)) cos(2πny) dy dt 2 (y( + y)) σ (log( + /y)) { exp(i(t log(( + y)/y) + 2πny)) } + exp(i(t log(( + y)/y) 2πny)) dy dt = 2i Im (L n (2 ) L n ( )) + O( 3/4 σ n σ 9/4 ) L n (t) = (i 2) (t/(2π)) 5/4 σ ( ) n n σ 7/4 e 2 (t, n) exp(i(f(t, n) + π/2)),

9 Mean value of the error term 453 we get with (3.2) that 2 G (t) dt = ( ) t 5/4 σ 2i ( ) n σ 2σ(n) 2π n 7/4 σ e 2(t, n) sin f(t, n) n X (3.4) + O( 3/4 σ ). 2) Evaluation of 2 G 2(t) dt. he treatment is similar to G. From (3.2) and Lemma 3., 2 σ+it σ it h(u, X) du dt = 2i Im(L X (2 ) L X ( )) + O( 3/4 σ X σ 9/4 ). Since L X (t) t 5/4 σ X σ 7/4 /2 for t = or 2, we have (3.5) 2 3) Evaluation of 2 G 3 (t) = 2iπ (ζ(2σ) + ζ(2 2σ)X 2σ ) G 2 (t) dt 2σ (X) /2 /2 σ. G 3(t) dt. Using [7, (4.6)], we have y σ ( + y) σ (log( + /y)) sin(2πxy) sin(t log( + /y)) dy + ( 2σ)π ζ(2 2σ)X 2σ y ( + y) 2σ sin(2πxy) σ+it ( ) + y u (u + 2σ) du dy. y σ it Direct computation shows that for y >, σ+it σ it ( ) + y 2σ (u + 2σ) ( + /y) u du = 2πi y ( σ+it it ) + + ( + /y) u (u + 2σ) du. +it σ it hen, we have 2 (3.6) G 3 (t) dt = 2i( 2σ)ζ(2 2σ) X 2σ I 2iπ (ζ(2σ) + ζ(2 2σ)X 2σ )I 2 + π ( 2σ)ζ(2 2σ)X 2σ I 3 2

10 454 Yuk-Kam Lau where I = I 2 = I 3 = 2 y 2σ sin(2πxy) dy 2 y σ ( + y) σ (log( + /y)) sin(2πxy) sin(t log( + /y)) dy dt, y ( + y) 2σ sin(2πxy) ) ( + /y) u (u + 2σ) du dy dt. ( σ+it it + +it σ it hen, I = 2 2σ 2 π 2σ X 2σ /(Γ(2σ) sin(πσ)) which is the main contribution. Interchanging the integrals, we have I 2 = y σ ( + y) σ( log( + /y) ) 2 sin(2πxy) cos(t log( + /y)) t=2 t= We split the integral into two parts c + c for some large constant c >. Expressing the product sin( ) cos( ) as a combination of exp(i(t log(+/y)±2πxy)), since (d/dy)(t log(+/y)±2πxy) = ±2πX t/(y( + y)) X for y c (recall t = or 2 ), the integral c is X by the first derivative test. Applying the mean value theorem for integrals, we have c c y σ ( + y) sin(2πxy) cos(t log( + /y)) dy. c Integration by parts yields that the last integral c (3.7) t ( y σ sin(2πxy) sin(t log( + /y)) c c c c c equals ) O(y σ sin(2πxy) + y σ X) dy. Hence I 2. For I 3, the extra integration over t is in fact not necessary to yield our bound. hus, we write I 3 = 2 (I 3 +I 32 ) dt, separated according to the integrals over u. I 3 and I 32 are treated in the same way, so we work dy.

11 out I 3 only. Using integration by parts over u, Mean value of the error term 455 I 3 = y ( + y) 2σ (log( + /y)) sin(2πxy) exp(it log( + /y)) { ( + /y) α α=σ σ } α + 2σ + it + ( + /y) α dα (α + 2σ + it) 2 dy. hen we consider α= y (+y) 2σ (log(+/y)) sin(2πxy) exp(it log(+/y))(+/y) α dy. Again, we split the integral into c + c. hen c X. If α 2, then c trivially; otherwise, we have (see (3.7)) c c y α ( + y) sin(2πxy) exp(it log( + /y)) dy. c herefore, I 3 and so I 3. Putting these estimates into (3.6), we get (3.8) 2 G 3 (t) dt = i2 2σ 2σ ( 2σ)ζ(2 2σ) π + O() Γ(2σ) sin(πσ) = 2πiζ(2σ ) + O(). 4) Evaluation of 2 G 4(t) dt. From [9, Section 4], we obtain 2 G 4(t) dt = 4i 2σ (X)((2σ )I + I 2 σi 3 I 4 ) where by Lemma 3., (recall L X (t) /2 X /2 for t = or 2 ) I = X 2σ 2 2 = X 2 2 I 2 = X 2σ t X X /2 cos(2πy) sin(t log( + X/y)) y σ (X + y) σ log( + X/y) dy dt cos(2πxy) sin(t log( + /y)) y σ ( + y) σ dy dt X 3/2 log( + /y) sup 2 2 cos(2πy) cos(t log( + X/y)) y σ (X + y) σ+ dy dt log( + X/y) 2 and similarly I 3, I 4 X 3/2. With Lemma 3.2, (3.9) 2 cos(2πxy) cos(t log( + /y)) y σ ( + y) σ+ log( + /y) G 4(t) dt r /2 log log. dy dt

12 456 Yuk-Kam Lau 5) Evaluation of 2 (3.) G 2 4 (t) dt. [9, (3.6) and Section 5] gives G 4 (t) dt = 4iI + 4iI 2 + 4iI 3. I, I 2 and I 3 are defined as follows: write (3.) w(ξ, y) = 2σ (ξ)ξ 2 y σ ( + y) σ 2 (log( + /y)) cos(2πξy), then I = I 2 = I 3 = 2 X 2 t 2 w(ξ, y) sin(t log( + /y)) dy dt dξ t X 2 X w(ξ, y)h (y) cos(t log( + /y)) dy dt dξ w(ξ, y)h (y) sin(t log( + /y)) dy dt dξ where H (y) and H (y) are linear combinations of y µ (log( + /y)) ν with µ+ν 2 and µ+ν respectively. (Remark: It is stated in [9] µ+ν 2 only for both H (y) and H (y).) When ξ X t and µ + ν 2, we have (3.2) (3.3) 2 exp(it log( + /y)) cos(2πξy) y σ µ ( + y) σ+2 dy dt, (log( + /y)) ν+ exp(it log( + /y)) cos(2πξy) y σ µ ( + y) σ+2 (log( + /y)) ν+ dy /2. he estimate (3.3) can be seen from [9, p.368]. o see (3.2), we split the inner integral into c + c. First derivative test gives c ξ. For c, we integrate over t first and plainly c 2. Using (3.2) and Lemma 3.2, we have I 3 X 2σ (ξ)ξ 2 dξ /4 σ. Applying integration by parts to the t-integral, we find that I 2 3/4 σ with (3.2) and (3.3). (Here we have used µ + ν for H (y).) Since 2 t 2 sin(t log( + /y)) dt = t 2 (log( + /y)) cos(t log( + /y)) + 2t(log( + /y)) 2 sin(t log( + /y)) (log( + /y)) 2 sin(t log( + /y)) dt, the last two terms contribute 3/4 σ and /4 σ in I respectively by using (3.3) and (3.2). Substituting into (3.), we get with [7, Lemma 3] (or

13 [5, Lemma 5.]) and (3.) 2 (3.4) G 4 (t) dt = 4it 2 X + O( 3/4 σ ) = iπ /2 t 5/2 + O( 3/4 σ ) X Mean value of the error term 457 w(ξ, y)(log( + /y)) cos(t log( + /y)) dy dξ 2σ (ξ) cos(tv + 2πξU πξ + π/4) ξ 3 V 2 U /2 (U /2) σ (U + /2) σ+2 dξ 2 t=2 where U and V are defined as in Lemma 3. with k replaced by ξ. Applying the argument in [9, Section 6] to (3.4), we get 2 (3.5) G 4 (t) dt ( ) t /2 σ ( σ 2σ (n) = 2i 2π n B(t, n σ log X) + O(log ). t ) 2 sin g(t, n) 2πn t= t=2 t= (Remark: he σ in [9, Lemma 4] should be omitted, as mentioned in [8].) Inserting (3.4), (3.5), (3.8), (3.9), (3.5) into (3.3), we obtain (3.6) 2 E σ (t) dt = 2πζ(2σ ) + Σ (t, X) 2 Σ 2(t, X) 2 + O(log ). 6) ransformation of Dirichlet Polynomial. Let X, X 2 (both are not integers) and denote B = B(, X ) and B 2 = B(, X 2 ). Assume X < X 2. Write F (x) = x σ ( log then we have (3.7) B(, X 2 )<n B(, X ) ) 2 exp(i( log 2πx 2πx + 2πx + π 4 )), σ 2σ (n)n σ (log(/(2πn))) 2 sin g(, n) = Im B 2 <n B σ 2σ (n)f (n).

14 458 Yuk-Kam Lau Stieltjes integration gives B 2 <n B σ 2σ (n)f (n) (3.8) = B B 2 B B 2 F (t)(ζ(2σ) + ζ(2 2σ)t 2σ ) dt + 2σ (t)f (t) 2σ (t)f (t) dt = I + I 2 I 3, say. Now, since (d/dt)(g(, t) + 2πt) = 2π /t < c when B 2 < t < B, we have I = B B 2 (ζ(2σ) + ζ(2 2σ)t 2σ ) σ. t σ (log(/(2πt))) 2 exp(i(g(, t) + 2πt)) dt By Lemma 3.2, I 2. Direct computation gives F (t) = i(2π t )tσ ( log ) 2 exp(i( log 2πt 2πt +2πt + π 4 ))+O(tσ 2 ) where B 2 t B. As B B 2 2σ (t) t σ 2 dt σ log, we have by (3.8) that σ 2σ (n)f (n) B 2 <n B (3.9) he integral B 2 B B B 2 = i exp(i( log 2π + π B 4 )) 2σ (t)(2π t )tσ ( log ) 2 exp(i(2πt log t)) dt + O(). 2πt in (3.9) is, after by parts, 2σ (t)(2π ( t )tσ log 2πt B B 2 2σ (t) d dt {(2π t )tσ ( log 2πt B 2 ) 2 B 2 exp(i(2πt log t)) B ) 2 exp(i(2πt log t))} dt

15 Mean value of the error term 459 he first term is σ /2 log by Lemma 3.2. Besides, computing directly shows that for B 2 t B, ( d dt { } = i(2πt )2 t σ 3 log ) 2 exp(i(2πt log t)) + O(t σ 2 ). 2πt reating the O-term with Lemma 3.2, (3.9) becomes B 2 <n B σ 2σ (n)f (n) (3.2) = exp(i( log 2π + π 4 )) B B 2 2σ (t)(2πt ) 2 t σ 3 ( log ) 2 exp(i(2πt log t)) dt + O( σ /2 log ). 2πt Inserting the Voronoi-type series of 2σ (t) (see Lemma 3.2) into (3.2), we get B 2 <n B σ 2σ (n)f (n) (3.2) = exp(i( log 2π + π 4 )) { σ 2σ (n) C n 7/4 σ J (n) + C 2 n= + O( σ /2 log ) n= } σ 2σ (n) n 9/4 σ J 2(n) where J (n) = J 2 (n) = B B 2 (2πt ) 2 t 7/4 cos(4π nt + π 4 ) dt B B 2 (2πt ) 2 t 9/4 cos(4π nt π 4 ) dt. ( log ) 2 exp(i(2πt log t)) 2πt ( log ) 2 exp(i(2πt log t)) 2πt Applying the first derivative test or bounding trivially, we have J 2 (n) /4 for n c, J 2 (n) 3/4 for c < n < c and /4 n /2 for

16 46 Yuk-Kam Lau n c. hus, the second sum in (3.2) is (3.22) /4 + 3/4 n c + /4 σ /2. n c After a change of variable t = x 2, J (n) = c <n<c σ 2σ (n)n σ /4 B B2 (2πx 2 ) 2 x 5/2 ( log σ 2σ (n)n σ 9/4 ) 2 2πx 2 { exp ( i(2πx 2 2 log x + 4π nx + π 4 )) + exp ( i(2πx 2 2 log x 4π nx π 4 ))} dx. hen we use [5, heorem 2.2], with f(x) = x 2 π log x, Φ(x) = x 3/2, F (x) =, µ(x) = x/2 and k = ±2 n. hus, ( ) 3/4 J (n) = δ n 2π 2 e 2 (, n) exp(i(f(, n) log 2π 2π + πn + 3π 4 )) + O ( δ n /4) + O ( 3/4 exp( c n c ) ) + O ( 3/4 min(, X ± n ) ) + O ( 3/4 min(, X 2 ± n ) ) where δ n = if B 2 < x < B and k >, or δ n = otherwise. (x = /(2π) + n/4 n/2 is the saddle point.) Note that B2 < x < B is equivalent to X < n < X 2. hus, for the first term in (3.2), we have C exp ( i( log 2π + π 4 )) n= σ 2σ (n) n 7/4 σ J (n) = ( ) 3/4 σ 2σ(n) 2 2π n 7/4 σ e 2(, n) exp(i(f(, n) πn + π)) X <n<x 2 + O( σ /2 log )

17 Mean value of the error term 46 ogether with (3.22), (3.2) and (3.7), we obtain σ 2σ (n)n σ (log /(2πn)) 2 sin g(, n) B(, X 2 )<n B(, X ) (3.23) = 2 ( 2π ) 3/4 + O( σ /2 log ). X <n<x 2 ( ) n σ 2σ(n) n 7/4 σ e 2(, n) sin f(, n) We can complete our proof now. aking X = [ ] /2 in (3.6), we have Σ i (t, X) Σ i (t, ) log for i =, 2 and t =, 2 ; hence 2 E σ (t) dt = 2πζ(2σ ) + Σ (t, t) 2 Σ 2(t, t) 2 ( (Σ (2, 2 ) Σ(2, ) ) + ( Σ 2 (2, ) Σ 2 (2, 2 ) )) + O(log ). Choosing X and X 2 in (3.23) to be half-integers closest to and 2 respectively, then (Σ (2, 2 ) Σ(2, )) + (Σ 2 (2, ) Σ 2 (2, 2 )) log. Hence, E σ (t) dt = 2πζ(2σ ) + Σ (, ) Σ 2 (, ) + O(log 2 ). he extra log in the O-term comes from the number of dyadic intervals. Suppose N. We apply (3.23) again with X = [N] + /2 and X 2 = [ ] + /2 to yield our theorem. 4. he second and third power moments he proof of the second moment is quite standard, see [9], [2] or [5] for example. Part () of heorem 2 follows from that for N, 2 2 Σ 2 (t, N) 2 dt, Σ (t, N) 2 dt = B(σ) Moreover, one can show 2 2 Σ (t, N)Σ 2 (t, N) dt log, ( ) t 5/2 2σ dt + O( 3 2σ ). 2π Lemma 4.. Define Σ M,M (t) = Σ,M (t) Σ,M (t) for M M. hen, we have 2 Σ M,M (t) 2 dt 7/2 2σ M 2σ 5/2. he next result is of its own interest and will be used in the proof of the third moment.

18 462 Yuk-Kam Lau Proposition 4.. Let A < (σ 3/4). hen, we have 2 G σ (t) A dt +A(5/4 σ). Proof. he case A 2 is proved by Hölder s inequality and part () of heorem 2. Consider the situation 2 < A < (σ 3/4). hen, for t 2 and N, we have Σ 2 (t, N) /2 and hence 2 Σ 2(t, N) A dt A/2. We take N = 2 R and write M = 2 r. hen Σ (t, N) r R Σ M,2M(t). By Hölder s inequality, we have Σ (t, N) A αr A Σ M,2M (t) A α A/(A ) r r R r R A. aking α r = M ( A(σ 3/4))/(2A) with the trivial bound Σ M,2M (t) 5/4 σ M σ 3/4, we have (4.) 2 by Lemma 4.. Σ (t, N) A dt A (5/4 σ)(a 2) r R A +A(5/4 σ) 2 αr A M (σ 3/4)(A 2) Σ M,2M (t) 2 dt Proof of heorem 2 (2). We have, with M = [δ /3 ] for some small constant δ >, (4.2) 2 G σ (t) 3 dt = 2 2 Σ,M (t) 3 dt + O( Σ M, (G σ (t) 2 + Σ 2,M(t)) dt). Proposition 4. and (4.) yields that the O-term is O( (3 8σ)/3 ). he integral on the right-sided of (4.2) is treated by the argument in [23]. hen the result follows. 5. Limiting distribution functions We first quote some results from [, heorem 4.] and [3, heorem 6]. Let F be a real-valued function defined on [, ), and let a (t), a 2 (t),... be real-valued, continuous and of period such that a n(t) dt = and n= a n(t) 2 dt <. Suppose that there are positive constants γ,

19 Mean value of the error term 463 γ 2,... which are linearly independent over Q, such that lim lim sup N min(, F (t) n N a n (γ n t) ) dt =. Fact I. For every continuous bounded function g on R, we have lim g(f (t)) dt = g(x)ν(dx), where ν(dx) is the distribution of the random series η = n= a n(t n ) and t n are independent random variables uniformly distributed on [, ]. Equivalently, the distribution function of F, P (u) = µ{t [, ] : F (t) u}, converges weakly to a function P (u), called the limiting distribution, as. Fact II. If F (t) A dt, then for any real k [, A) and integral l [, A), the following limits exist: lim F (t) k dt and lim F (t) l dt. Now, let us take F (t) = t 2σ 5/2 G σ (2πt 2 ), γ n = 2 n and (5.) a n (t) = 2 µ(n)2 n 7/4 σ ( ) nr σ 2σ(nr 2 ) r 7/2 2σ sin(2πrt π/4). r= Following the computation in [3, p.42] with Lemma 4., we get 2 (F (t) n N a n (2 nt)) 2 dt N 2σ 5/2 (N ). hen heorem 2 (c) and the first part of heorem 3 are immediate consequence of Facts I and II with Proposition 4.. We proceed to prove the lower bounds in (2.2) with the idea in [, Section V]. Lemma 5.. Let n be squarefree. Define A n = {t [, ] : a n (t) > B σ 2σ (n)n σ 7/4 } where B = 4A( r= r4σ 7 ) and A = 2 r= σ 2σ(r 2 )r 2σ 7/2. hen, we have µ(a n ) /(AB) where µ is the Lebesgue measure. he proof makes use of the fact that a+ n (t) dt = a n (t) dt where a ± n (t) = max(, ±a n (t)), and a + n (t) 2 dt + a n (t) 2 σ 2σ (nr 2 ) 2 dt = n 7/2 2σ r 7 4σ. he readers are referred to [] for details. r=

20 464 Yuk-Kam Lau Proof of lower bounds in heorem 3. By Markov s inequality, we have Pr( a m (t m ) 2 K) a m (t) 2 dt 3 4K 4 m=n+ m= where Pr(#) denotes the probability of the event # and K = a m (t) 2 dt < +. m= Consider the set { E n = (t, t 2,...) : t m A m for m n and m=n+ where A m = [, ] if m is not squarefree. hen, n Pr(E n ) = Pr(A m )Pr( a m (t m ) 2 K) m= m=n+ a m (t m ) 2 } K 3 4(AB) n due to Pr(A m ) = µ(a m ) and Lemma 5.. When (t, t 2,...) E n, we have a m (t m ) σ 2σ (m) B m 7/4 σ 2 K m= m n m squarefree { log n if σ = 3/4, n σ 3/4 if 3/4 < σ <. Our result for D σ (u) follows after we replace n by [e u ] if σ = 3/4 and by [u 4/(4σ 3) ] if 3/4 < σ <. he case of D σ ( u) can be proved in the same way. o derive the upper estimates, we need a result on the Laplace transform of limiting distribution functions [3, Lemma 3.]. Lemma 5.2. Let X be a real random variable with the probability distribution D(x). Suppose D(x) > for any x >. For the two cases: (i) ψ(x) = x log x and φ(x) = log x, or (ii) ψ(x) = x 4/(7 4σ) and φ(x) = x (4σ 3)/4, there exist two positive numbers L and L such that (a) if lim sup λ ψ(λ) log E(exp(λX)) L, then lim sup x log( D(φ(x))) L, x (b) if lim sup λ ψ(λ) log E(exp( λx)) L, then lim sup x log D( φ(x)) L. x

21 Mean value of the error term 465 Proof of upper bounds in heorem 3. We take N = λ if σ = 3/4, and N = λ 4/(7 4σ) if 3/4 < σ <. When n N, we use exp(±λa n (t)) dt exp(λa σ 2σ(n)µ(n) 2 n 7/4 σ ). Recall that A = 2 r= σ 2σ(r 2 )r 7/2 2σ. Now consider n > N. If λaσ 2σ (n) < n 7/4 σ, then by the inequality e x + x + x 2 for x, and a n(t) dt =, we have exp(±λa n (t)) dt exp((λa) 2 σ 2σ(n) 2 µ(n) 2 n 7/2 2σ ). Otherwise, λaσ 2σ (n) n 7/4 σ, it is obvious that herefore, log E(exp(±λX)) is exp(±λa n (t)) dt exp(λa σ 2σ(n)µ(n) 2 n 7/4 σ ) exp((λa) 2 σ 2σ(n) 2 µ(n) 2 n 7/2 2σ ). λa σ 2σ (n) n 7/4 σ + (λa)2 n N n>n { λ log λ if σ = 3/4, λ 4/(7 4σ) if 3/4 < σ <. he proof is complete with Lemma Ω ± -results σ 2σ (n) 2 n 7/2 2σ his section is devoted to prove heorem 4. We apply the methods in [2] or [7], but beforehand, we transform G σ (t) into a simple finite series by convolution with the kernel ( ) sin 2πBu 2 K(u) = 2B. 2πBu Similarly to [5], we have, for B L /4 /6, L (6.) t 2σ 5/2 G σ (2π(t + u) 2 )K(u) du = S B (t) + O(B 4σ 5 ) where L (6.2) S B (t) = 2 n B 2 ( ) n ( n B )σ 2σ(n) n 7/4 σ sin(4π nt π 4 ). o prove the Ω -result, we use Dirichlet s heorem to align the angles. More specifically, for any small δ >, we can find l [ /,

22 466 Yuk-Kam Lau ( + δ B2 ) / ] such that l n < δ. aking B δ log, we have l [ /, /5 ] and S B (l) = n ( ) n ( B )σ 2σ(n) n 7/4 σ n B 2 + O (δ n ) ( B )σ 2σ(n) (6.3) n 7/4 σ. n B 2 A simple calculation shows that 2 2σ n B 2 ( ) n ( n B )σ 2σ(n) n 7/4 σ = ( n B 2 n B )σ 2σ(n) n 7/4 σ. We thus infer S B (t) = Ω (log log t) if σ = 3/4, and Ω ((log t) σ 3/4 ) if 3/4 < σ <. We proceed to prove the Ω + -result with the method in [2]. ake x = δ log log log log log and B = / (L = B 4 ) for a small number δ >. We consider the convolution of S B (t) with a kernel involving the function x (u) = ( + cos(4π qu)) = ( + e4πi qu + e 4πi ) qu 2 q Q x q Q x where Q x is the set of positive squarefree integers whose prime factors are odd and smaller than x. he convolution will pick out terms with the desired frequencies, ( ) sin ɛπu 2 ɛ S B (t + u) x (u) du ɛπu = 2 n B 2 n Qx ( ) n ( n B )σ 2σ(n) n 7/4 σ sin(4π nt π 4 ). o maximize the right-hand side, we apply Dirichlet s theorem again to find a number l [ /, ( + δ Qx ) / ] so that the right-side is { ( σ 2σ (p)) log x if σ = 3/4, + exp(cx σ 3/4 /(log x)) if σ > 3/4. p Q x p 7/4 σ his follows from the the estimation of p x pσ 7/4 for σ = 3/4 and σ > 3/4 respectively. he cardinality of Q x is O(exp(cx/ log x)) for some positive constant c. Our choice of x ensures that l is of a size of a small

23 power of. Consequently, we obtain Mean value of the error term 467 sup S B (u) / u /4 { log log log if σ = 3/4, exp(c(log log ) σ 3/4 (log log log ) σ 7/4 ) if σ > 3/4. 7. Occurrence of large values Proof of heorem 5. Define K τ (u) = ( u )( + τ sin(4παu)) where τ = or + and α is a large constant. Following the argument in [4], we derive that L (t + u) 2σ 5/2 G σ (2π(t + u + v) 2 )K(v) dvk τ (u) du L = τ 2 ( B ) cos(4πt π/4) + O(α 2 ) + O(B 4σ 5 ). where δ,n = if n = and otherwise. Our assertion follows by choosing B and α (L = B 4 ) sufficiently large, and 4t /8 with t [, +]. (Note that τ can be + or at our disposal.) o prove heorem 6, we need the next lemma which is the key. Lemma 7.. For 5/2 H /2, 2 max (G σ(t + h) G σ (t)) 2 dt H 5 4σ h H where the implied constant depends on σ. Proof. Following the arguments in [8], we have (7.) 2 (G σ (t + h) G σ (t)) 2 dt h 5 4σ min((σ 3/4), log(/h 2 )) where log 2 h. Let b = /24 and H = 2 λ b. hen, as in [4], we can show max G σ(t + h) G σ (t) max G σ(t + jb) G σ (t) + O( 2( σ)/3+ɛ b h H j 2 λ for any fixed t. Let us take j = j (t) 2 λ such that G σ (t + j b) G σ (t) = max j 2 λ G σ(t + jb) G σ (t). hen we can express j = 2 λ µ S t 2 µ for some set S t of non-negative integers. Hence, G σ (t + j b) G σ (t) = µ S t G σ (t + (ν + )2 λ µ b) G σ (t + ν2 λ µ b)

24 468 Yuk-Kam Lau where ν = ν t,µ < 2 µ is an integer. By Cauchy-Schwarz s inequality and inserting the remaining ν s, we get (G σ (t + j b) G σ (t)) 2 ( ) 2 ( σ)µ 2 ( σ)µ (G σ (t + (ν + )2 λ µ b) G σ (t + ν2 λ µ b)) 2 µ S t µ S t 2 ( σ)µ (G σ (t + (ν + )2 λ µ b) G σ (t + ν2 λ µ b)) 2 µ S t ν<2 µ as µ S t 2 ( σ)µ. Integrating over [, 2 ] and using (7.), we see that 2 µ S t max (G σ(t + h) G σ (t)) 2 dt h H 2 +ν2 λ µ 2 ( σ)µ b ν<2 µ +ν2 λ µ b H 5 4σ µ S t H 5 4σ. ν<2 µ 2 (4 3σ)µ his complete the proof of Lemma 7.. (G σ (t + 2 λ µ b) G σ (t)) 2 dt + 7/2+ɛ Proof of heorem 6. Define G ± σ (t) = max(, ±G σ (t)). By heorem 2 (c), we have 2 G σ(t) 3 dt +3(5/4 σ). Hence, Cauchy-Schwarz inequality gives ( G σ (t) dt) 2 G σ (t) dt G σ (t) 3 dt. we have 2 G σ(t) dt +(5/4 σ). ogether with (2.), 2 G± σ (t) dt 2 c 2 t5/4 σ dt. Consider K ± (t) = G ± σ (t) (c 2 ɛ)t 5/4 σ where ɛ = δ 5/2 2σ, we have 2 K ± (t) dt ɛ 2 t 5/4 σ dt and K ± (t + h) K ± (t) = G ± σ (t + h) G ± σ (t) + O( /4 σ h). Since G ± σ (t + h) G ± σ (t) G σ (t + h) G σ (t), it follows that together with Lemma 7., 2 max h H K± (t + h) K ± (t) dt H 5/2 2σ + 5/4 σ H. Define ω ± (t) = K ± (t) max h H K ± (t + h) K ± (t). aking H = c ɛ 2/(5 4σ) (= c δ ) for some sufficiently small constant c >, we

25 Mean value of the error term 469 have ω ± (t) dt ɛ t 5/4 σ dt max h H K± (t + h) K ± (t) dt ɛ +(5/4 σ). Let I ± = {t [, 2 ] : ω ± (t) > }. hen 2 ω ± (t) dt ω ± (t) dt K ± (t) dt I ± I ± ( ) /2 ( 2 /2 dt K ± (t) dt) 2. I ± We infer I ± ɛ 2 as 2 K± (t) 2 dt 2 G σ(t) 2 dt + 7/2 2σ. When t I ±, we have K ± (t) max h H K ± (t + h) K ± (t). Hence, K ± (u) for all u [t, t + H], i.e. G ± σ (t) (c 2 ɛ)t 5/4 σ. he number of such intervals is not less than I ± /H c 3 δ 4( σ). Acknowledgement he author would thank the referee for the valuable comments. Also, he wishes to express his sincere gratitude to Professor Kohji Matsumoto for his warm encouragement. Hearty thanks are due to Charlies u for unfailing supports. References [] P.M. Bleher, Z. Cheng, F.J. Dyson, J.L. Lebowitz, Distribution of the Error erm for the Number of Lattice Points Inside a Shifted Circle. Comm. Math. Phys. 54 (993), [2] J.L. Hafner, A. Ivić, On the Mean-Square of the Riemann Zeta-Function on the Critical Line. J. Number heory 32 (989), 5 9. [3] D.R. Heath-Brown, he Distribution and Moments of the Error erm in the Dirichlet Divisor Problem. Acta Arith. 6 (992), [4] D.R. Heath-Brown, K. sang, Sign Changes of E( ), (x) and P (x). J. Number heory 49 (994), [5] A. Ivić, he Riemann Zeta-Function. Wiley, New York, 985. [6] A. Ivić, Mean values of the Riemann zeta function. Lectures on Math. 82, ata Instit. Fund. Res., Springer, 99. [7] A. Ivić, K. Matsumoto, On the error term in the mean square formula for the Riemann zeta-function in the critical strip. Monatsh. Math. 2 (996), [8] M. Jutila, On the divisor problem for short intervals. Ann. Univ. urkuensis Ser. A I 86 (984), [9] M. Jutila, ransformation Formulae for Dirichlet Polynomials. J. Number heory 8 (984), [] I. Kiuchi, On an exponential sum involving the arithmetical function σ a(n). Math. J. Okayama Univ. 29 (987), [] I. Kiuchi, K. Matsumoto, he resemblance of the behaviour of the remainder terms E σ(t), 2σ (x) and R(σ + it). Un Sieve Methods, Exponential Sums, and their Applications in Number heory, G.R.H. Greaves et al.(eds.), London Math. Soc. LN 237, Cambridge Univ. Press (997),

26 47 Yuk-Kam Lau [2] Y.-K. Lau, On the mean square formula for the Riemann zeta-function on the critical line. Monatsh. Math. 7 (994), 3 6. [3] Y.-K. Lau, On the limiting distribution of a generalized divisor problem for the case /2 < a <. Acta Arith. 98 (2), [4] Y.-K. Lau, On the error term of the mean square formula for the Riemann zeta-function in the critical strip 3/4 < σ <. Acta Arith. 2 (22), [5] Y.-K. Lau, K.-M. sang, Ω ± -results of the Error erm in the Mean Square Formula of the Riemann Zeta-function in the Critical Strip. Acta Arith. 98 (2), [6] Y.-K. Lau, K.-M. sang, Moments of the probability density functions of error terms in divisor problems. Proc. Amer. Math. Soc. 33 (25), [7] K. Matsumoto, he mean square of the Riemann zeta-function in the critical strip. Japan J. Math. 5 (989), 3. [8] K. Matsumoto, Recent Developments in the Mean Square heory of the Riemann Zeta and Other Zeta-Functions. In Number heory, rends Math., Birhkäuser, Basel, (2), [9] K. Matsumoto,. Meurman, he mean square of the Riemann zeta-function in the critical strip III. Acta Arith. 64 (993), [2] K. Matsumoto,. Meurman, he mean square of the Riemann zeta-function in the critical strip II. Acta Arith. 68 (994), [2]. Meurman, On the mean square of the Riemann zeta-function. Quart. J. Math. Oxford (2) 38 (987), [22]. Meurman, he mean square of the error term in a generalization of Dirichlet s divisor problem. Acta Arith. 74 (996), [23] K.-M. sang, Higher power moments of (x), E(t) and P (x). Proc. London Math. Soc. (3) 65 (992), Yuk-Kam Lau Department of Mathematics he University of Hong Kong Pokfulam Road, Hong Kong yklau@maths.hku.hk