An explicit formula of Atkinson type for the product of the Riemann zeta-function and a Dirichlet polynomial. Author copy

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1 Cent. Eur. J. Math. 9-6 DOI:.478/s Central European Journal of Mathematics An explicit formula of Atkinson type for the product of the Riemann zeta-function and a Dirichlet polynomial Hideaki Ishikawa, Kohji Matsumoto Faculty of Education, Nagasaki University, Nagasaki, Japan Graduate School of Mathematics, Nagoya University, Nagoya, Japan Research Article Received June ; accepted 9 November Abstract: We prove an explicit formula of Atkinson type for the error term in the asymptotic formula for the mean square of the product of the Riemann zeta-function and a Dirichlet polynomial. To deal with the case when coefficients of the Dirichlet polynomial are complex, we apply the idea of the first author in his study on mean values of Dirichlet L-functions. MSC: M6, M4 Keywords: Riemann zeta-function Dirichlet polynomial Atkinson formula Versita Sp. z o.o.. Introduction and statement of results Let s = σ + it be a complex variable, ζs denote the Riemann zeta-function, and As = m M am m s be a Dirichlet polynomial, where M and am C. Let us assume am = Om ε isikawah@nagasaki-u.ac.jp kohjimat@math.nagoya-u.ac.jp

2 H. Ishikawa, K. Matsumoto ε >, where and in what follows ε always denotes a small positive number, not necessarily the same at each occurrence. The mean value T IT, A = ζ + it A + it dt, T, is an interesting object in number theory in connection with power moments and the distribution of zeros of ζs. The asymptotic formula for IT, A was first established by Balasubramanian, Conrey and Heath-Brown []. Denote by k, l the greatest common divisor of k and l, and by = kl/k, l their least common multiple. Let MT, A = k M ak al log k, l T + γ T, πkl where al is the complex conjugate of al and γ is Euler s constant. They proved IT, A = MT, A + ET, A under the condition log M log T the symbol f g means f = Og, where ET, A is the error term. The error estimate proved in [] is ET, A M T ε + T log B T for any B >. The second author [4] proved that if C log / T M µ T ρ T /C log/ T holds where µ, µ <, µ + ρ >, and C is sufficiently large, then ET, A M µ/ T / ρ/+ε + M µ T ρ+ε. The aim of this article is to study ET, A in more detail. For that we will prove an analogue of Atkinson s formula for ET, A. Atkinson s formula was originally obtained [] for the mean square of ζs. It gives an explicit formula for ET defined by T ζ + it dt = T log T + γ log π T + ET. It is known that Atkinson s formula is very useful in the mean square theory of ζs see [7, 8, 3]. Therefore it is natural to search for Atkinson-type formulas for other zeta and L-functions. In [6], Motohashi made a first attempt and applied Atkinson s idea [] to the study of ET, A. He did not obtain an explicit formula of Atkinson type. His idea was to consider a certain exponential integral including ET, A to obtain a sharp estimate. His estimate is O M 4/3 T /3+ε for M T / log 3/4 T actually for the integral from T to T. Steuding made further use of Motohashi s idea and found an interesting result on the distribution of the zeros of ζs in short intervals. The proof of in [4] is also based on a variant of Atkinson s idea cf. Section.7 of Ivić [8]. In the case of ζs, it is clear that T ζ + it dt = T T ζ + it dt. 3 This property was used by Atkinson in the proof of his explicit formula. However, since the am are complex, IT, A is not necessarily equal to T ζ T + it A + it dt. 3

3 An explicit formula of Atkinson type for the product of the Riemann zeta-function and a Dirichlet polynomial This causes some trouble when one tries to prove an Atkinson-type formula for ET, A. The same happens in the case of Dirichlet L-function Ls, χ associated with a complex character χ see []. In [5], the first author introduced a new idea to prove an explicit formula of Atkinson type for T IT, χ = L + it, χ dt, and showed that if χ is a complex primitive character, then IT, χ and the integral of L/ + it, χ from T to do not coincide. The difference of these two quantities is Ω T /4. The fundamental idea in [5], inspired by Hafner and Ivić [3], is to consider the integral over the interval [T, T ] instead of [ T, T ] to avoid the trouble mentioned above. Then, in the course of the proof, there appear some integrals which cannot be treated by Atkinson s original argument. How to deal with those integrals is the most novel point of [5]. In the present paper we apply the idea from [5] to ET, A to obtain an analogue of Atkinson s formula. Throughout the paper we will use the following notation: Let us define κ = k k, l, = l k, l, arcsinh x = log x + x + πu ft, u = T arcsinh T + πut + π u π 4 Σ T, Y = k, n κy ak al / dn I κ n / eπinκ/, ξt, u = T π + u u 4 + ut π,, gt, u = T log T πu T + πu + π 4. πn arcsinh T κ /4 T κ + exp i f T, n πn πn κ κ + π, where dn is the number of positive divisors of n, κ is defined by κκ mod, and Σ T, Y = k, n /κy ak al / dn R κ n The main results in this paper are the following three theorems. / e πinκ/ log T exp i g T, κn. πnκ Theorem.. Let T, Y be positive numbers satisfying C T < Y < C T and T C = max e, C where C, C are fixed constants with < C < C and e = Then we have T T ζ + it A + it dt = MT, A + Σ T, Y + Σ T, ξt, Y 4 MT, A Σ T, Y Σ T, ξt, Y + RT, T, A, where RT, T, A is the error term satisfying RT, T, A M +ε log T + M 5/+ε T /4. From Theorem. it is easy to deduce: 4

4 H. Ishikawa, K. Matsumoto Theorem.. Under the same assumptions as in Theorem., we have ET, A = Σ T, Y + Σ T, ξt, Y + RT, A 5 with RT, A M +ε log 3 T + M +ε log 3/ δ T log log T + M 5/+ε log 3/4+δ T for any δ >. When the coefficients am are real, we do not encounter the problem mentioned above. Therefore in this case the treatment is simpler, and we obtain the following sharper result: Theorem.3. When the coefficients am are real, we have 5 with RT, A M +ε log T + M 5/+ε T /4. Theorems. and.3 are analogues of Atkinson s formula [] for ζs. In fact, the case M =, a = of Theorem.3 gives exactly Atkinson s formula. Our results give an explicit formula for ET, A, by which we can study the behavior of ET, A, especially the oscillation property, quite accurately. In a forthcoming paper we will apply our formula to discuss the difference between IT, A and ζ + it A + it dt analogously to the work of the first author [6], and also power moments of ET, A.. The fundamental decomposition T In this and the next section, we follow Motohashi [6]. Only a brief sketch of the proof is given in [6]. In [7], Steuding explained Motohashi s method, including a treatment of derivatives of the zeta-function, for some special am. Here we will supply more details of Motohashi s method. Let u, v be two complex variables, and we first assume Ru >, Rv >. Since ak ζu Au = = m u k u m= k M q= k M, k q ak q u, we have ζu ζv Au Av = ak q u q= k M, k q r=, l r al r v = + + = B + Bu, v + Bv, u. 6 q=r q<r q>r The term B can be written as B = ak al [k,l] r r u v = ζu + v k M ak al. 7 u+v 5

5 An explicit formula of Atkinson type for the product of the Riemann zeta-function and a Dirichlet polynomial Let us now consider Bu, v. Set bm = ak. k M, k m Then, Bu, v = bm bm + n m u m + n v. m= n= We show that Bu, v can be extended meromorphically to the region Ru >, Rv >, Ru + v >, with the formula Bu, v = Γu Γv ak al l l f= y v e y πif/l In fact, for y >, Ru >, the inner integral on the right-hand side equals h= x u e kx+y πifk/l dx dy. 8 e hky+πifhk/l e hkx x u dx = k u Γu e hky+πifhk/l h u. 9 Hence, assuming further Rv > and Ru + v >, we see that the double integral on the right-hand side of 8 equals Γu k u e πifhk/l h u h= h= y v e hky e y πif/l dy, as the summation and the integration can be interchanged in view of the absolute convergence. The integral on the right-hand side of the above can be expanded as e πifn/l y v e hk+ny dy = Γv e πifn/l hk + n v. n= Therefore the right-hand side of 8 equals ak al k u l h u h= n= hk + n v which, letting hk = m, equals Bu, v. Thus 8 follows. Next, define n= l e πihk+nf/l = ak al k u f= Hz; k, l, f = e kz πifk/l δf kz, h,n l hk+n h u hk + n v, where δf = or depending on whether l divides kf or not. The function Hz; k, l, f is holomorphic at z = and is O min z, for z. Dividing the inner integral on the right-hand side of 8 as δf k x u x + y dx + x u Hx + y; k, l, f dx and noting that x u x + y dx = yu Γu Γ u, < Ru <, 6

6 H. Ishikawa, K. Matsumoto we find that the right-hand side of 8 equals Γ u Γv ak al kl l f= y u+v δf dy + gu, v; A e y πif/l in the region < Ru <, Ru + v >, where gu, v; A = Γu Γv ak al l l f= y v e y πif/l Similarly to 9, we see that the integral in the first term of equals Therefore from 8 and we obtain Γu + v e πimf/l m u v+. m= Γ u Bu, v = Γu + v ak al Γv kl where φs, α = m eπimα m s is the Lerch zeta-function. Since we obtain x u Hx + y; k, l, f dx dy. l δf φ u + v, f + gu, v; A, l l δf φ u + v, f k,l j ζu + v = φ u + v, =, l k, l k, l u+v f= j= f= Γ u Bu, v = Γu + v ζu + v ak al + gu, v; A Γv k, l u+v in the region < Ru <, Ru + v >. 3. The contour integral expression Let C be the contour which comes from + along the positive real axis, rounds the origin counterclockwise, and goes back to + again along the positive real axis. Then it is easy to see that, under the assumptions < Ru <, Ru + v >, gu, v; A can be rewritten as gu, v; A = Γu Γv e πiu e πiv ak al l l f= C y v x u Hx + y; k, l, f dx dy. e y πif/l C However, the right-hand side of is convergent for Ru < and any v C. Hence gives the meromorphic extension of gu, v; A to that region, and therefore is also valid in that region. Combining 6, 7 and, we obtain ζu ζv Au Av = ζu + v k M + Γ v Γu ak al Γ u + u+v Γv Γu + v ζu + v k M Γu + v ζu + v k M ak al k, l u+v ak al + gu, v; A + gv, u; A k, l u+v 7

7 An explicit formula of Atkinson type for the product of the Riemann zeta-function and a Dirichlet polynomial in the region Ru <, Rv <. Note that, changing k and l in the third double sum, we can combine the second and the third term on the right-hand side as Γ u + Γv Γ v Γu Γu + v ζu + v k M ak al k, l u+v. Now assume < Ru <, and take the limit v u. The singularities coming from ζu + v and Γu + v cancel with each other, and the result is for < Ru <. ζu ζ u Au A u = k M ak al + gu, u; A + g u, u; A Γ Γ k, l u + u + log + γ log π Γ Γ kl Next we prove another expression of gu, u; A in the region Ru <. Let R be a large positive integer and C R = C R k, l, f = x = y + πif + π R + e iθ θ < π. l k Then we can easily see that Hx + y; k, l, f if x C R k, l, f. Hence which tends to as R if Ru <. Hence C R Hx + y; k, l, f x u dx C Hx + y; k, l, f x u dx = πi n π R Ru R dθ R Ru, 3 Resn Hx + y; k, l, f x u 4 if Ru <, where Res n denotes the residue of the function at x = y+πil f +k n and n means the summation running over all integers n kf/l. Since Res n Hx + y; k, l, f x u = f y + πi k l + n u, k if we can interchange the summation and integration, from and 4 we obtain where In; k, l, f = C gu, u; A = πi Γu Γ u e πiu e πiu y u f y + πi e y πif/l l + n u dy = e πiu k In case n > kf/l, we can write ak al kl f y + πi l + n f = e πi y + πe πi/ k l + n. k l In; k, l, f, 5 f= n y u f y + πi e y πif/l l + n u dy. k 8

8 H. Ishikawa, K. Matsumoto Hence, setting f y = π l + n e πi/ η, arg η = π k, we obtain In; k, l, f = e πiu e πiu i η u + η u exp πi f + dη n l k η + f l = e πiu e πiu i f η u + η u exp πim l + n η + f dη, k l m= 6 where the second equality is valid if the summation and the integration can be interchanged. In case n < kf/l, since setting f y + πi l + n = e πi y + πe 3πi/ f k l + n k, y = π f l + n k eπi/ η, arg η = π, we find that In; k, l, f has an expression similar to that in 6 with the integral over the interval [, i, instead of [, i. Substituting these results into 5 and using the formula Γu Γ u = π/ sin πu, we obtain where J ± k, l = f= gu, u; A = ak al J + k, l + J k, l, 7 kl l n m= ±i f η u + η u exp πim l + n η + f dη, k l and the summation with respect to n runs over all n > kf/l resp. all n < kf/l for J + resp. J. Using the notations κ and, we have f exp πim l + n η + f = exp πi mf mfκ + n exp πi η. k l l k, lκ Put h = fκ + n. If n > kf/l, then h >. On the other hand, for any positive integer h, we can find integers f and n such that f l and h = fκ + n. In fact, since κ, = we find integers x, y with xκ + y =. Then f = hx + ν and n = hy νκ for any integer ν satisfy h = fκ + n. Choosing ν suitably we have f l, as desired. Therefore J + k, l = m= h= f exp πi mf i η u + η u exp πi l mh k, lκ η dη, where the innermost sum runs over all f satisfying f l and there exists an integer n with h = fκ + n. Recall that κ is an integer satisfying κκ mod. Then h = fκ + n implies f hκ mod. The number of f satisfying this congruence condition and f l is l/ = k, l. Hence f exp πi mf l k,l = j= exp πi m l hκ + j = exp πi mhκ k,l l j= exp πi mj, k, l 9

9 An explicit formula of Atkinson type for the product of the Riemann zeta-function and a Dirichlet polynomial and the inner sum equals k, l if k, l m and otherwise. Therefore, setting m = k, l µ, we obtain J + k, l = k, l Setting further µh = n, we obtain Similarly, µ= h= J + k, l = k, l J k, l = k, l exp πi µhκ i η u + η u exp πi µh κ η dη. i dn e πiκn/ η u + η u e πinη/κ η dη. 8 n= i dn e πiκn/ η u + η u e πinη/κ dη. 9 n= The integrals on the right-hand sides of 8 and 9 are estimated as O n Ru, so the infinite series including those integrals are absolutely convergent for Ru <. Hence we can justify the above interchanges of summation and integration. Lastly, we can rotate the paths of integration from [, ±i to [,. Substituting the resulting expressions into 7, we obtain for Ru <, where gu, u; A = ak al d n e πiκn/ h u, n κ hu, x = n y u + y u expπixy dy. Formulas, 3, are stated by Motohashi in [6, p. 4]. The method used above to deduce these formulas was originally sketched in [6], and it is a variant of the idea presented by Motohashi in [5]. See also [, Section and 5]. Formula is a generalization of the formula stated by Atkinson in [, p. 357]. It is also possible to deduce along the same line as in Atkinson [], using Euler Maclaurin s formula and slightly modified Poisson s summation formula, though we will not give further details here. 4. The analytic continuation In this section we discuss the analytic continuation of gu, u; A. Our method is a direct generalization of Atkinson s original method []. In his argument, Atkinson used the asymptotic formula for n x dn. For our present purpose, we need an asymptotic formula for D x, κ = dn e πiκn/, n x or D x, κ/, whose expression is similar to that of Dx, κ/ but if x is an integer then the last term of the sum is to be halved. These sums have been studied by Jutila []. Define the error term x, κ/ by the formula D x, κ = x log x + γ log x + E, κ + x, κ, where E, κ/ is the constant term which is the value at s = of the Estermann zeta-function. If Dx, κ/ is replaced by D x, κ/ in, we denote the corresponding error term by x, κ/. Clearly x, κ/ = x, κ/ if x is not an integer. Then we have cf. [, Theorem.6 and.5.]

10 H. Ishikawa, K. Matsumoto Lemma 4.. For any x >, x, κ = x / where Y, K are standard notation for Bessel functions and, moreover, n= 4π nx dn n / e πiκn/ Y + 4π nx π eπiκn/ K, 3 x, κ = O /3 x /3+ε 4 if x and x. Formula 3 is the Voronoï-type formula for x, κ/. Let = κ,, whose value we will specify later, and write dn e πiκn/ h u, n = κ n> By rotating the path of integration of to [, i we can easily see that for Ru <. Hence, if Ru <, as x. Therefore by integration by parts we have x h u, dd x, κ. 5 κ hu, x = O x Ru 6 x h u, D x, κ κ dn e πiκn/ h u, n = h u, D, κ κ κ n> hu, x/κ D x, κ dx. 7 x Substituting into the integral on the right-hand side of 7, and applying integration by parts once more, we obtain for Ru <, where dn e πiκn/ h u, n = g u g u + g 3 u g 4 u κ n= g u = dn e πiκn/ h u, n, 8 κ n g u = h u,, κ, 9 κ g 3 u = g 4 u = h u, hu, x/κ x x log x + γ log dx, κ x, κ dx.

11 An explicit formula of Atkinson type for the product of the Riemann zeta-function and a Dirichlet polynomial Similarly, we have hence from we have n= dn e πiκn/ h u, n κ = g u g u + g 3 u g 4 u, for Ru <. gu, u; A = k M ak al g u g u + g 3 u g 4 u + g u g u + g 3 u g 4 u 3 In view of 6, g u and g u are holomorphic in the region Ru <. Also, by [, p. 359] we have hu, x = O x Ru. x This estimate and 4 imply that g 4 u is holomorphic for Ru < /3. As for g 3 u, similarly to the argument in [, pp ], we obtain g 3 u + g 3 u = κ π log + γ log y u + y u sin π y dy κ + κ y u + y u sin π y dy. πu κ The two integrals on the right-hand side are convergent uniformly in Ru ε. expression of gu, u; A valid for Ru < /3. By the same argument we have which is valid for Ru > /3. g u, u; A = ak al g u g u + g 3 u g 4 u + g u g u + g 3 u g 4 u 5. The decomposition of the mean square 3 Hence 3 with 3 gives an Now we start to consider the mean square integral of ζ/ + it A/ + it. The rest of the proof of Theorem. is an analogue of the argument in [5]. Let C, C, and C be the same as in the statement of Theorem.. Assume C T < Y < C T, T C, and let = κ, = κy. Let u = / + it in 3, and integrate both sides on the interval [T, T ]. By Stirling s formula we obtain T Γ T Γ + it + Γ Γ it dt = t=t log Γ i + it log Γ it = t log t t t=t + O. t=t t=t Using this, 3, 3 and 49 below, we obtain T T ζ + it A + it dt = MT, A MT, A + I I + I 3 I 4 + O M ε, 33 3

12 H. Ishikawa, K. Matsumoto where I ν = k M T R ak al g ν + it + g ν T it dt 34 for ν 4. Let C l y = y / + y l log + y y. Substituting 8 and 9 with into 34 for ν =,, and changing the order of integration, we get where and Using 3 we have where and I = S t; n, k, l n T t=t, I = S t; k, l ak al S t; n, k, l = I dn e πiκn/ e πiny/κ C / y eit log+y/y dy ak al + I dn e πiκn/ e πiny/κ e it log+y/y dy C / y = S t; n, k, l + S t; n, k, l, ak al S t; k, l = I, κ e πiy/κ C / y eit log+y/y dy ak al + I, κ e πiy/κ e it log+y/y dy. C / y T t=t, 35 I 3 = R ak al κ log + γ log I3 k, l + κ iπ iπ I 3k, l, 38 I 3 k, l = sin πy/κ e it log +y/y T y C / y sin πy/κ /+it I 3 k, l = y /+it u t=t dy u + y du dy. As for I 4, similarly to [, p. 36], we first carry out the integration with respect to t, then put xy = η, change the order of differentiation and integration and differentiate with respect to x, and then again put η/x = y. Changing the variable temporarily to η is necessary to ensure the interchange of differentiation and integration. The result is that where I 4 = k M y S 4 t; k, l T t=t, 39 S 4 t; k, l = S 4 S 4 + S 4 S 43 3

13 An explicit formula of Atkinson type for the product of the Riemann zeta-function and a Dirichlet polynomial with ak al S 4 = R ak al S 4 = I x, κ/ x x, κ/ x e πixy/κ C 3/ y t eit log +y/y dy dx e πixy/κ C 3/ y eit log +y/y, + log + y/y dy dx, and S 4 resp. S 43 is similar to S 4 resp. S 4 with x, κ/ and e πixy/κ replaced by their complex conjugates. We decompose S 4 T t=t = J k, l + J k, l J 3 k, l, where and Therefore ak al J k, l = R ak al J k, l = R ak al J 3 k, l = R x, κ/ x x, κ/ x x, κ/ x e πixy/κ e πixy/κ C 3/ y T eit log +y/y dy dx, C 3/ y T eit log +y/y dy dx, e πixy/κ C 3/ y T eit log +y/y dy dx. 4 I 4 = k, l + J k, l J 3 k, l + J S 4 + S 4 S 43 T. 4 t=t In order to evaluate exponential integrals appearing in the above formulas, here we quote the following lemma, due to the first author [5]. Lemma 5. [5, Lemma 5 and Remark 5]. i Let α, β, γ, a, b, k, T be real numbers such that α, β, γ are positive and bounded, α, < a < /, a < T /8πk, k >, T, and b max T, k, T. πk Then, b a exp ± i T log + y/y + πky y α + y β log γ dy = T / exp ± i T V + πku πk + π/4 + y/y kπ / V γ U / U / α U + / β a α b γ α β + O + O + RT, k + O e CT T k + O e C kt T γ α β+ + T + O k e CkT b γ α β + T γ α β 4 uniformly for α > ε, where C is an absolute constant, U = RT, k 4 + T πk, πk T, V = arcsinh T γ α β/ /4 k γ α β/ 5/4 if < k T, T / α k α if k T. 4

14 H. Ishikawa, K. Matsumoto The term O e CT is to be omitted if < α or < α < with k T. When we replace k by k on the left-hand side of 4, then the explicit term and RT, k for k T on the right-hand side are to be omitted. ii If we replace the assumption k > by k B, where B is an absolute constant satisfying < B, then the last three error terms on the right-hand side of 4 are absorbed into the other error terms. When k B and k on the left-hand side of 4 is replaced by k, then the explicit term and the last three error terms on the right-hand side of 4 are to be omitted. This is a modified version of the saddle point lemma of Atkinson []. The important point of this modification is that it is available for small k. 6. Evaluation of I, I and I 3 In this section we apply Lemma 5. to I, I and I 3. From 35, we have I = k M n S T ; n, k, l k M <n S T ; n, k, l k M Similarly to Section 8 of [5], we apply part i of Lemma 5. to I 3 to obtain S T ; n, k, l = I I I n I 3 = Σ T, Y + O T /4 ak al κ 5/4. 44 The same type of formula, replacing Σ T, Y with Σ T, Y, holds for I. Next, again similarly to [5, Section 8], applying 4 note that T C and part ii of Lemma 5. to 37, we obtain I ak al κ /3+ε +ε T /6+ε. 45 The quantity I 3 can be treated similarly to the argument in [, pp ]. In fact, we have I 3 k, l T /. 46 As for I 3, we decompose the integral with respect to y at y =, and denote the contributions of the intervals [, ] and [, by I 3 and I 3, respectively. Then I 3 T /, I 3 T log T. 47 Note that in [], the term corresponding to I 3 is shown to be equal to π i + OT /. In [] the integration with respect to u is from / it to / + it, and the term π i appears from the residue at u = when the path of integration is deformed. In the present situation, however, the integration with respect to u is from / + it to / + it, hence the residue does not appear. Therefore the first estimate of 47 follows. From 38, 46 and 47, we obtain I 3 k M ak al κ log κt T /. 48 5

15 An explicit formula of Atkinson type for the product of the Riemann zeta-function and a Dirichlet polynomial Using the equality = kl/k, l and putting k, l = h, k = k h, l = l h, we have h M h k,l M/h k l log 3 M. 49 Applying this estimate and, we find that the error term on the right-hand side of 44 is O M 5/+ε T /4. Similarly, from 45 and 48 we have I M 4/3+ε T /6+ε, I 3 M +ε T / log T. Collecting all the results obtained in this section, we have 7. Evaluation of I 4 I I + I 3 = Σ T, Y Σ T, Y I + O M 5/+ε T /4 + M 4/3+ε T /6+ε. 5 Now we proceed to the evaluation of I 4. Applying part Lemma 5.ii to the inner integral of S 4, we have ak al S 4 / x, κ/ κ ak al dx = κ / x x By the Cauchy Schwarz inequality, each integral on the right-hand side is j j x 3 dx The second factor of the above is estimated using U which is implied by [, Theorem.]. Hence U / j j j j= j x, κ / dx. x, κ dx U 3/ + U +ε, U, ak al S 4 κ / ak al and the same estimate holds also for S 4, S 43. Next consider J 3 k, l. Here we use the formula j= x 3/ j / j 3/4 + j /+ε κ /4 3/4 T /4 + κ ε +ε T /+ε, x, κ dx. 5 x, κ = / x /4 π n= 33/ 3 π x /4 dn 4π nx n 3/4 e πiκn/ cos π 4 dn 4π nx n 5/4 e πiκn/ sin π 4 n= + O 5/ x 3/4, 5 which can be easily obtained from 3 by using asymptotic expansion formulas for Bessel functions. 6

16 H. Ishikawa, K. Matsumoto We proceed similarly to the argument in [5, pp ]. First applying Lemma 5.ii to the inner integral on the right-hand side of 4, and estimating the error term by the same argument as 5. Then applying 5 to the explicit term we obtain where and with ak al J 3 k, l = J 3 k, l J 3 k, l + J 33 k, l + O κ /4 3/4 T /4 + κ ε +ε T /+ε, 53 J 3 k, l = T π R ak al κ /4 3/4 /κ n= dn κn n 3/4 e πiκn/ cos 4πx π 4 ψ 3/ T, x exp i ft, x πx + π dx, J 3 k, l = 3T 3π R ak al κ /4 5/4 dn κn /κ n 5/4 e πiκn/ sin 4πx π 4 n= ψ 5/ T, x exp i ft, x πx + π dx, Since ψ T, x/κ κ/x, we have ak al J 33 k, l = O T 5/ x 7/4 ψ T, x/κ dx π T ψ α T, x = x arcsinh α x T πx T πx + /4. 4 ak al J 33 k, l κ 3/4 7/4 T /4. 56 As for J 3 k, l, we consider a finite truncation of the integral and apply [5, Lemma 6]. Here we note that on the left-hand side of 4 in [5], the factor g 3/ T, x cos exp i dx is to be on the numerator. Similarly to 43 in [5], we obtain where J 3 k, l = W k, l + W k, l + W 3 k, l + W 4 k, l, 57 W k, l = T π R ak al κ /4 3/4 dn κn /4 4π exp igt, κn/ n 3/4 e πiκn/ T log T /πκn n Z here we put Z = /κ ξt, /κ = /κ ξt, Y, W k, l W 3 k, l ak al ak al κ /4 3/4 T / dn κn n 3/4 n Z κ /4 3/4 T /4 dn min n3/4 n= /4 T /π κn/ /,, κn/ ξt, Y, 58 and W 4 k, l ak al κ /4 3/4 n= / dn e CT. n3/4 κn 7

17 An explicit formula of Atkinson type for the product of the Riemann zeta-function and a Dirichlet polynomial In the case of J 3 k, l, it is not necessary to consider the finite truncation of the integral, because the infinite series on the right-hand side of 55 is absolutely convergent. Hence, we can apply [5, Lemma 6] to J 3 k, l to obtain J 3 k, l = W 5 k, l + W 6 k, l + W 7 k, l + W 8 k, l, 59 where W 5 k, l = 3T 64π R ak al κ /4 5/4 dn κn 3/4 4π exp i gt, κn/ n 5/4 e πiκn/, T log T /πκn T /π κn/ n Z and W 6 k, l, W 7 k, l, W 8 k, l are similar to W k, l, W 3 k, l, W 4 k, l, respectively, just replacing the factor κ /4 3/4 by κ /4 5/4, the factor n 3/4 by n 5/4, in the case of W 6 k, l T /π κn/ / by T /π κn/ 3/, and in the case of W 7 k, l T /4 by T /4. It is immediate that Also, replacing the factor min by, we get Next we note that, when n Z, the inequalities hold. Since T ξt, /κ < T, we have By using these facts, we can show ak al W 4 k, l κ /4 5/4 e CT, 6 ak al W 8 k, l κ 3/4 7/4 e CT. 6 akó, al W 7 k, l κ /4 5/4 T /4. 6 T π κn T, log T πκn / dn T n log + T. / κ κ n Z W k, l, W 5 k, l W 6 k, l ak al ak al T / log + T κ T 3/ log + T κ, Consider W 3 k, l. Split the sum on the right-hand side of 58 as = S + S + S 3 + S 4 + S 5. n Z/ Z/<n Z Z Z Z<n Z+ Z Z+ Z<n Z + n>z 8

18 H. Ishikawa, K. Matsumoto If Z, we can estimate these sums similarly to the argument on the last page in []. We have S, S 5 /4 T /4 log + T κ κ and, S, S 4 κ κ /4 S 3 T /4 log + T κ /4 T /4 log + T κ., Hence W 3 k, l ak al log + T κ + κ / T /4 log + T κ 65 if Z. For Z <, we decompose the sum on the right-hand side of 58 as. n Z + n>z The first sum is obviously O, and the second sum is easily estimated as O /κ /. Hence, in this case, we have where the second inequality follows from / ak al W 3 k, l κ /4 3/4 T /4 ak al + κ /, 66 κ Combining 65 and 66, we obtain the conclusion that κ κ T ξt, Y = Z <. κ ak al W 3 k, l log + T + κ / T /4 log + T κ κ Now, from 53, 56, 57, 59, 6, 6, 6, 63, 64 and 67, we obtain J 3 k, l = W k, l ak al + O κ / T /4 log + T + log + T + κ / κ κ + κ ε +ε T /+ε + κ /4 3/4 + κ /4 5/4 + κ 3/4 7/4 T /4. + κ /. 67 The quantity J k, l can be treated similarly. Note that the term κ /4 5/4 T /4 on the right-hand side can be omitted, because κ /4 5/4 κ /4 3/4 + κ 3/4 7/4. Since W k, l = Σ T, ξt, Y, substituting the obtained results into 4 we get I 4 = Σ T, ξt, Y + Σ T, ξt, Y + J k, l + O ak al 68, 69 where means the same as in the curly parentheses on the right-hand side of 68. Using and 49, we find that the above error term is O M +ε log T + M 7/4+ε T /4, and hence I 4 = Σ T, ξt, Y + Σ T, ξt, Y + J k, l + O M +ε log T + M 7/4+ε T /4. 7 9

19 An explicit formula of Atkinson type for the product of the Riemann zeta-function and a Dirichlet polynomial 8. The cancellation Substituting 5 and 7 into 33, we obtain T T ζ + it A + it dt = MT, A MT, A + Σ T, Y Σ T, Y + Σ T, ξt, Y Σ T, ξt, Y I J k, l + O M 5/+ε T /4 + M 4/3+ε T /6+ε + M +ε log T. 7 Hence the remaining task is to consider I and J k, l. In this section we will show that there is a big cancellation between these two quantities and consequently I = k M J k, l + O M 5/+ε T /4 + M 4/3+ε T /6+ε. 7 This type of result was proved, in the case of Dirichlet L-functions, by the first author in [5, Section ]. Our proof of 7 follows an argument similar to that in [5]. Remark 8.. In the case of Dirichlet L-functions, this cancellation argument is the most novel part of [5]. When Hafner and Ivić [3] studied the asymptotic behavior of the integral of ET, the same type of cancellation also happened, and they proved this fact by applying Jutila s transformation method [9]. However, their argument does not seem to be applicable in the general L-function case and also to our present case, so the first author developed an alternative method of proving the cancellation process in [5], which we also use here. Let and rewrite 36 with t = T as ak al S 3 t; n, k, l = I dn e πiκn/ e πiny/κ C / y e it log +y/y dy S T ; n, k, l = S T ; n, k, l S 3 T ; n, k, l + S 3 T ; n, k, l + S T ; n, k, l. 73 Substitute this expression into the definition of I. Denote by R the contribution of the last two terms on the right-hand side of 73. As for the first two terms, we have e πiny/κ C / y e it log +y/y e it log +y/y dy = e πiny/κ /+it + y / it u + y /+it du dy = here the second equality is justified because hu, n/κ is uniformly convergent in Ru <, so I = k M /+it I / it ak al <n We express the inner sum as a Stieltjes integral, as in 5, and apply to obtain y / it h u, n du κ dn e πiκn/ h u, n du + R. 74 κ <n dn e πiκn/ h u, n = x h u, log x + γ log dx κ κ + h u, x x, κ κ hu, x/κ x x, κ dx. 75

20 H. Ishikawa, K. Matsumoto Substitute this expression into 74 and denote the contribution of the first resp. second term on the right-hand side of 75 to 74 by R resp. R 3. We have I = k M /+it I / it ak al hu, x/κ x x, κ dx du + R + R + R Change the order of integration on the right-hand side and use /+it / it hu, x/κ x du = x e πixy/κ log +y/y y / + y 3/ eit it log + y/y T dy log + y/y t= T which is similar to the formula given in [, p. 36]; cf. 39. Let us decompose e it log +y/y it T it log +y/y = it e log + y/y t= T it e it log +y/y + e it log +y/y T log + y/y and denote the contribution of the second resp. third term on the right-hand side of 77 to 76 by R 4 resp. R 5. Since the contribution of the first term on the right-hand side of 77 to 76 is J k, l, we obtain I = k M J k, l + We estimate R j, j 5. By using Lemma 5.ii, and 49, we have R k M <n t= T, 77 5 R j. 78 j= 5/4 ak al κ dn T /4 M 5/+ε T /4. 79 n Carrying out the integration with respect to u in the definition of R 3, we find that the resulting expression can be treated similarly to the case of I. Hence, as in Section 6, we have R 3 M 4/3+ε T /6+ε. 8 We can treat R 4 similarly to S 4, and treat R 5 similarly to S 4, S 43. Hence the same estimate in 5 holds for R 4 and R 5, which implies R 4, R 5 M +ε T /4. 8 Finally consider R. Changing the order of integration and applying Lemma 5.ii to the inner integral, we have R = R + R, where R = k M ak al I G x; G x; κ, e ifx;κ, dx 8

21 An explicit formula of Atkinson type for the product of the Riemann zeta-function and a Dirichlet polynomial with and It is easy to see that Fx; κ, = ft, x/κ πx κ + 3π 4, G x; = log x + γ log, G x; κ, = R = k M To estimate R, we quote the following arcsinh πx 4T κ T x πκ + x 4κ /4, ak al I G x; RT, x/κ dx. R ak al κ / log + log M +ε T / log T. 83 T Lemma 8. Heath-Brown [4]. Let Fx, G j x, j J, be continuous functions defined on an interval [a, b], which are monotone. Assume F x is also monotone, F x M and G j x M j, j J, on [a, b]. Then In the present case, we see that b a G x G J x e ifx dx J+3 J M j. j= 4πT π F x; κ, = xκ + π κ κ, which is positive and decreasing. Therefore Fx; κ, is also monotone and we can take M = F ; κ,. Hence by Lemma 8., we find that the integral on the right-hand side of 8 is log + log G κt / log κt. F ; κ, This implies that R M +ε T / log T, and this with 83 gives R M +ε T / log T. 84 From 78, 79, 8, 8 and 84 we obtain the assertion of 7. Substituting 7 into 7, and noting that M 4/3+ε T /6+ε = M /+ε/ M 5/6+ε/ T /6+ε M +ε + M 5/3+ε T /3+ε M +ε log T + M 5/+ε T /4, we complete the proof of Theorem..

22 H. Ishikawa, K. Matsumoto 9. Proof of Theorem. Let L be a positive integer satisfying L T C. 85 Then replacing T resp. Y by j T resp. j Y in 4 makes the formula valid for j L. Adding up, we obtain T ζ + it A + it dt = MT, A + Σ T, Y + Σ T, ξt, Y M L T, A Σ L T, L Y Σ L T, ξ L T, L Y L L L + O M +ε log T j T + M 5/+ε j T /4 + ζ + it A + it dt. j= The condition 85 is equivalent to j= 86 [ ] log T log C L, 87 log where [x] is the integer part of x. Since T C e, we have log log T. Therefore the choice satisfies 87 for any positive α. This choice implies Using 49 and 88, we have Next, since arcsinh x x for small x, we have [ ] log T log C α log log T L = log T log α T L T log α T. 88 M L T, A M ε T L log T L Mε log α T log log T. 89 Σ L T, L Y / T M ε κ dn L n k, n κy / L M ε log α/ T 9 κ log κy / L M +ε log α/ T log log T k, and Also, Σ L T, L Y M ε k, κ / n /κy / L dn n / M+ε log α/ T log log T. 9 M +ε M 5/+ε L log j T M +ε L log T M +ε log 3 T, 9 j= L j T /4 M 5/+ε T /4 L/4 M 5/+ε log α/4 T. 93 j= 3

23 An explicit formula of Atkinson type for the product of the Riemann zeta-function and a Dirichlet polynomial Finally, using the well-known mean square estimate for ζs [8, Theorem 7.3], we obtain L T ζ + it A + it dt M +ε T log T L L M+ε log α T log log T. 94 Substituting into 86, we obtain RT, A M +ε log 3 T + M +ε log α T log log T + M +ε log α/ T log log T + M 5/+ε log α/4 T. This is the general form of the estimation of RT, A we have proved. Theorem. is the special case α = 3 4δ.. Proof of Theorem.3 In this section, we discuss the case when am R for all m briefly. In this case the analogue of 3 holds for ζsas. Therefore, as an analogue of 33, we can show IT, A = T T ζ + it A + it dt = MT, A + I I + I3 I4 + O, where I ν, ν 4, are almost the same as I ν, just replacing the integral from T to T by that from T to T. The analogues of 35, 38, 39 hold for I ν, replacing T resp. T by T resp. T. First consider I. We see that I = I 3 I 3, where I 3 is the same as in 43, and I 3 is almost the same as I 3, just replacing T by T. From 44 and the corresponding result for I 3, we obtain I = Σ T, Y + O T /4 ak al κ 5/4. Next, it is easy to see that exactly the same estimate 45 for I holds for I. As for I 3, define I 3 k, l as an analogue of I 3 k, l replacing T resp. T by T resp. T. Similarly, we define I3 k, l, and further define I 3 k, l and I 3 k, l splitting the integral at y =. Then, similarly to 46 and 47, the estimates I 3k, l T /, I 3 k, l T log T hold. Since the inner integral of I 3 k, l is from / it to / + it, this time a residue appears in the course of the argument, so I 3k, l = π i + O T /. Hence, corresponding to 48, we obtain I 3 = π k M ak al κ + O ak al κ log κt T /. At last, we consider S4 T ; k, l S 4 T ; k, l. I 4 = k M 4

24 H. Ishikawa, K. Matsumoto As in Section 5, we decompose S 4 ±T ; k, l = S 4 ±T S 4 ±T + S 4 ±T S 43 ±T. The term S 4 T is exactly the same as J 3 k, l defined by 4, hence 68 can be used for S 4 T. Therefore, the term Σ T, ξt, Y appears from this part. As for S4 T, S 4 T, S 43 T, we use the estimate 5. We obtain S 4 T ; k, l = Σ T, ξt, Y + E T, where E T is the error term satisfying the same estimate as the error term on the right-hand side of 69. Similarly, the term Σ T, ξt, Y again appears from the sum of S4 T, and so S 4 T ; k, l = Σ T, ξt, Y + E T, where E T satisfies the same estimate as that of E T. We thus obtain I 4 = Σ T, ξt, Y + E T + E T. Collecting the above results, we obtain the conclusion of Theorem.3. References [] Atkinson F.V., The mean-value of the Riemann zeta function, Acta Math., 949, 8, [] Balasubramanian R., Conrey J.B., Heath-Brown D.R., Asymptotic mean square of the product of the Riemann zetafunction and a Dirichlet polynomial, J. Reine Angew. Math., 985, 357, 6 8 [3] Hafner J.L., Ivić A., On the mean-square of the Riemann zeta-function on the critical line, J. Number Theory, 989, 3, 5 9 [4] Heath-Brown D.R., The mean value theorem for the Riemann zeta-function, Mathematika, 978, 5, [5] Ishikawa H., A difference between the values of L/ + it, χ j and L/ + it, χ k. I, Comment. Math. Univ. St. Pauli, 6, 55, 4 66 [6] Ishikawa H., A difference between the values of L/ + it, χ j and L/ + it, χ k. II, Comment. Math. Univ. St. Pauli, 7, 56, 9 [7] Ivić A., The Riemann Zeta-Function, Wiley-Intersci. Publ., John Wiley & Sons, New York, 985 [8] Ivić A., Lectures on Mean Values of the Riemann Zeta Function, Tata Inst. Fund. Res. Lectures on Math. and Phys., 8, Springer, Berlin, 99 [9] Jutila M., Transformation formulae for Dirichlet polynomials, J. Number Theory, 984, 8, [] Jutila M., Lectures on a Method in the Theory of Exponential Sums, Tata Inst. Fund. Res. Lectures on Math. and Phys., 8, Springer, Berlin, 987 [] Katsurada M., Matsumoto K., Asymptotic expansions of the mean values of Dirichlet L-functions, Math. Z., 99, 8, 3 39 [] Katsurada M., Matsumoto K., A weighted integral approach to the mean square of Dirichlet L-functions, In: Number Theory and its Applications, Kyoto, November -4, 997, Dev. Math.,, Kluwer, Dordrecht, 999, 99 9 [3] Matsumoto K., Recent developments in the mean square theory of the Riemann zeta and other zeta-functions, In: Number Theory, Trends Math., Birkhäuser, Basel, 4 86 [4] Matsumoto K., On the mean square of the product of ζs and a Dirichlet polynomial, Comment. Math. Univ. St. Pauli, 4, 53, 5

25 An explicit formula of Atkinson type for the product of the Riemann zeta-function and a Dirichlet polynomial [5] Motohashi Y., A note on the mean value of the zeta and L-functions. I, Proc. Japan Acad. Ser. A Math. Sci., 985, 67, 4 [6] Motohashi Y., A note on the mean value of the zeta and L-functions. V, Proc. Japan Acad. Ser. A Math. Sci., 986, 6, [7] Steuding J., On simple zeros of the Riemann zeta-function in short intervals on the critical line, Acta Math. Hungar.,, 964, [8] Titchmarsh E.C., The Theory of the Riemann Zeta-Function, Clarendon Press, Oxford, 95 6