On the mean square of the Epstein zeta-function

We consider the second power moment of the Epstein zeta-function and construct the asymptotic formula in special case, when ϕ₀(u,v) = u² + Av² , A > 0, A ≡ 1, 2(mod 4) and ϕ0(u,v) belongs to the one-class kind G₀ of the quadratic forms of discriminant −4A.

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Опубліковано в: :	Algebra and Discrete Mathematics
Дата:	2005
Автори:	Savastru, O.V., Varbanets, P.D.
Формат:	Стаття
Мова:	English
Опубліковано:	Інститут прикладної математики і механіки НАН України 2005
Онлайн доступ:	https://nasplib.isofts.kiev.ua/handle/123456789/156605
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Назва журналу:	Digital Library of Periodicals of National Academy of Sciences of Ukraine
Цитувати:	On the mean square of the Epstein zeta-function / O.V. Savastru, P.D. Varbanets // Algebra and Discrete Mathematics. — 2005. — Vol. 4, № 1. — С. 105–121. — Бібліогр.: 16 назв. — англ.

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spelling	Savastru, O.V. Varbanets, P.D. 2019-06-18T17:48:07Z 2019-06-18T17:48:07Z 2005 On the mean square of the Epstein zeta-function / O.V. Savastru, P.D. Varbanets // Algebra and Discrete Mathematics. — 2005. — Vol. 4, № 1. — С. 105–121. — Бібліогр.: 16 назв. — англ. 1726-3255 2000 Mathematics Subject Classification: 11N37, 11R42. https://nasplib.isofts.kiev.ua/handle/123456789/156605 We consider the second power moment of the Epstein zeta-function and construct the asymptotic formula in special case, when ϕ₀(u,v) = u² + Av² , A > 0, A ≡ 1, 2(mod 4) and ϕ0(u,v) belongs to the one-class kind G₀ of the quadratic forms of discriminant −4A. en Інститут прикладної математики і механіки НАН України Algebra and Discrete Mathematics On the mean square of the Epstein zeta-function Article published earlier
institution	Digital Library of Periodicals of National Academy of Sciences of Ukraine
collection	DSpace DC
title	On the mean square of the Epstein zeta-function
spellingShingle	On the mean square of the Epstein zeta-function Savastru, O.V. Varbanets, P.D.
title_short	On the mean square of the Epstein zeta-function
title_full	On the mean square of the Epstein zeta-function
title_fullStr	On the mean square of the Epstein zeta-function
title_full_unstemmed	On the mean square of the Epstein zeta-function
title_sort	on the mean square of the epstein zeta-function
author	Savastru, O.V. Varbanets, P.D.
author_facet	Savastru, O.V. Varbanets, P.D.
publishDate	2005
language	English
container_title	Algebra and Discrete Mathematics
publisher	Інститут прикладної математики і механіки НАН України
format	Article
description	We consider the second power moment of the Epstein zeta-function and construct the asymptotic formula in special case, when ϕ₀(u,v) = u² + Av² , A > 0, A ≡ 1, 2(mod 4) and ϕ0(u,v) belongs to the one-class kind G₀ of the quadratic forms of discriminant −4A.
issn	1726-3255
url	https://nasplib.isofts.kiev.ua/handle/123456789/156605
citation_txt	On the mean square of the Epstein zeta-function / O.V. Savastru, P.D. Varbanets // Algebra and Discrete Mathematics. — 2005. — Vol. 4, № 1. — С. 105–121. — Бібліогр.: 16 назв. — англ.
work_keys_str_mv	AT savastruov onthemeansquareoftheepsteinzetafunction AT varbanetspd onthemeansquareoftheepsteinzetafunction
first_indexed	2025-11-25T23:55:28Z
last_indexed	2025-11-25T23:55:28Z
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fulltext	Algebra and Discrete Mathematics RESEARCH ARTICLE Number 1. (2005). pp. 105 – 121 c© Journal “Algebra and Discrete Mathematics” On the mean square of the Epstein zeta-function O. V. Savastru and P. D. Varbanets Communicated by V. V. Kirichenko Dedicated to Yu.A. Drozd on the occasion of his 60th birthday Abstract. We consider the second power moment of the Ep- stein zeta-function and construct the asymptotic formula in special case, when ϕ0(u, v) = u2 + Av2, A > 0, A ≡ 1, 2(mod 4) and ϕ0(u, v) belongs to the one-class kind G0 of the quadratic forms of discriminant −4A. 1. Introduction and statement of result Let ζ(s) be the Riemann zeta-function. In 1926 Ingham [7] proved the relation ∫ T 0 \|ζ(1 2 + it)\|4 dt = T 2π2 log4 T +O(T log3 T ) In series this result was improved. In 1979 Heath-Brown [6] proved that ∫ T 0 \|ζ(1 2 + it)\|4 dt = T 4 ∑ j=0 aj logj T + E2(T ), where E2(T ) = O(T 7/8+ǫ). A.İvič [9] calculated the coefficients aj , j = 1, 2, 3, 4. Heath-Brown’s bound for E2(T ) was improved to E2(T ) = O(T 2/3 logc T ), (c > 0) 2000 Mathematics Subject Classification: 11N37, 11R42. Key words and phrases: Epstein zeta-function, approximate functional equa- tion, asymptotic formula, second power moment. 106 On the mean square of the Epstein zeta-function in [10] İvič and Motohashi. In this paper we shall consider the second power moment of the Ep- stein zeta-function. The function of divisor d(n) and the function rϕ(n) (number of repre- sentations of n by the positive quadratic form ϕ(u, v)) are close. There- fore we can expect that their Dirichlet series have like the mean value. Let ϕ(u, v) denotes positive definite quadratic form ϕ(u, v) = au2 + 2buv + cv2, a, b, c ∈ Z, (a, b, c) = 1, D = ac− b2 > 0. For real numbers α, β, γ, δ and a complex variable s, define the Epstein zeta-function for Res > 1 Zϕ( ∣ ∣ ∣ ∣ α β γ δ ∣ ∣ ∣ ∣ ; s) = ∑ (u,v)∈Z2 (u,v) 6=(−γ,−δ) e(αu+ βv)(ϕ(u+ γ, v + δ))−s. It is known that this function possesses an analytic continuation to the whole complex plane, with the possible exception of a simple pole with residue π√ D at s = 1 which occurs if and only if (α, β) ∈ Z2 (see Epstein [5]). Moreover, one has a functional equation Zϕ( ∣ ∣ ∣ ∣ α β γ δ ∣ ∣ ∣ ∣ ; s) = = e(−αγ − βδ) ( π√ D )−1+2sΓ(1 − s) Γ(s) Zψ( ∣ ∣ ∣ ∣ −γ −δ α β ∣ ∣ ∣ ∣ ; 1 − s). (1) Let rϕ(λ) be the number of the representations λ in the form λ = ϕ(u+ γ, v + δ), and let rϕ(λ;α, β) = ∑ ϕ(u+γ,v+δ)=λ e(αu+ βv). We denote ψ(u, v) = cu2 − 2buv + av2, A = B = √ D π , an = ∑ u,v∈Z ϕ(u+γ,v+δ)=λn e(αu+ βv), bn = e(−αγ − βδ) ∑ u,v∈Z ϕ(u+α,v+δ)=µn e(−γu− δv), 0 < λ1 < λ2 < . . . , 0 < µ1 < µ2 < . . .. By (1) we have AsΓ(s)Φ(s) = B1−sΓ(1 − s)Ψ(1 − s), where Φ(s) = ∞ ∑ n=1 an λsn = Zϕ( ∣ ∣ ∣ ∣ α β γ δ ∣ ∣ ∣ ∣ ; s), Ψ(s) = ∞ ∑ n=1 bn µsn = e(−αγ − βδ)Zϕ( ∣ ∣ ∣ ∣ −γ −δ α β ∣ ∣ ∣ ∣ ; s). We are now prepared to formulae our results. O. V. Savastru, P. D. Varbanets 107 Theorem 1. Let 0 ≤ Res = σ ≤ 1, \|Ims\| = \|t\| ≥ 10, 1 ≤ x, y, xy = ( t √ D π )2 . Then the approximate functional equation Zϕ( ∣ ∣ ∣ ∣ α β γ δ ∣ ∣ ∣ ∣ ; s) = ∑ λn≤x an λsn + χϕ(s) ∑ µn≤y bn µ1−s n +Rϕ(s, x) holds, with χϕ(s) = (√ D π )1−2s Γ(1 − s) Γ(s) ; Rϕ(s, x) ≪ \|t\|1/2x−σ min(1, x \|t\|) log \|t\| log( \|t\| √ D x + x \|t\| √ D )+ +x1−σ(\|t\| √ D)−1(1 + \|t\| √ D x )min(xǫ + log \|t\|, yǫ + log \|t\|). Theorem 2. Let rϕ(n) denotes the number of the representations of n by form ϕ(u, v). Then for any positive ǫ ∫ T 0 \|Zϕ( ∣ ∣ ∣ ∣ 0 0 0 0 ∣ ∣ ∣ ∣ ; 1 2 + it)\|2 dt = 2T ∑ n≤T √ D π r2ϕ(n) n − 2π√ D ∑ n≤T √ D π r2ϕ(n)+ +2 ∑ mn≤T2D π2 rϕ(m)rϕ(n)√ mn (m n )it ( i log m n )−1 +O((T √ D)1/2+ǫ). Theorem 3. Let l, q ∈ N, (l, q) = 1. Then T ∫ 0 Re s=1 2 \| 1 q2s ∑ l1,l2(mod q) ϕ(l1,l2)≡l(mod q) Zϕ( ∣ ∣ ∣ ∣ 0 0 l1 q l2 q ∣ ∣ ∣ ∣ ; s) − ∑ (u,v)∈B ϕ(u, v)−s\|2 dt≪ (T √ D)1+ǫ q1−ǫ , where B denotes the set of points (u, v) for which ϕ(u, v) ≡ l(mod q) and 0 < ϕ(u, v) < 2q. Theorem 4. Let ϕ0(u, v) = u2 + Av2, A > 0, A ≡ 1, 2(mod 4) and let ϕ0(u, v) belongs to the one-class kind G0 of the quadratic forms of discriminant −4A. Then for any ǫ > 0 ∫ T 0 \|Zϕ0( 1 2 + it)\|2 dt = E0T log2 T + E1T log T + E2T +O(T 7/8+ǫ), where E0 > 0, E1 are the computable constants which depends on A. 108 On the mean square of the Epstein zeta-function We shall use the following notation. The Vinogradov symbol X ≪ Y means X = O(Y ). We use ǫ for a positive exponent which may be taken arbitrary close to zero; the constant implied by ≪(or O) may be depend on ǫ. exp(x) = ex, e(x) = e2πix, eq(x) = e(xq ) for x ∈ R; (−Ad ) is symbol Jacoby; Γ(z) is Gamma function. 2. Proof of theorem 1 and theorem 2 Assume first that σ > 1. We shall evaluate the integral I = 1 2πi c+i∞ ∫ c−i∞ sxw−s w(s− w) Zϕ( ∣ ∣ ∣ ∣ α β γ δ ∣ ∣ ∣ ∣ ;w)dw, (1 < c < σ) in two ways. In the above integral we replace Zϕ( ∣ ∣ ∣ ∣ α β γ δ ∣ ∣ ∣ ∣ ;w) by the series ∞ ∑ n=1 an λwn . We then integrate termwise and move the line of integration to Rew = −∞ if λn ≤ x, and to Rew = +∞ if λn > x. By the theorem of residues we obtain ∑ λn≤x 1 2πi c+i∞ ∫ c−i∞ sxw−s w(s− w) an λwn dw = x−s ∑ λn≤x an, ∑ λn>x 1 2πi c+i∞ ∫ c−i∞ sxw−s w(s− w) an λwn dw = ∑ λn>x an λsn . (2) Hence, I = x−s ∑ λn≤x an + ∑ λn>x an λsn = Zϕ( ∣ ∣ ∣ ∣ α β γ δ ∣ ∣ ∣ ∣ ; s) − ∑ λn≤x an λsn + x−s ∑ λn≤x an. (3) In the second evaluation of the integral I we appeal to the analytic con- tinuability and the functional equation of the function Zϕ( ∣ ∣ ∣ ∣ α β γ δ ∣ ∣ ∣ ∣ ; s). We move the line of integration to Rew = −b (0 < b < 1 2), set z = 1−w, and use the functional equation (1): I = 1 2πi 1+b+i∞ ∫ 1+b−i∞ sx1−z−s (1 − z)(s− 1 + z) Zϕ( ∣ ∣ ∣ ∣ α β γ δ ∣ ∣ ∣ ∣ ; 1 − z)dz +R(z) = O. V. Savastru, P. D. Varbanets 109 = e(−αγ − βδ) 1 2πi 1+b+i∞ ∫ 1+b−i∞ sx1−z−s (1 − z)(s− 1 + z) Γ(z) Γ(1 − z) ( π√ D )−(−1+2z) × ×Zψ( ∣ ∣ ∣ ∣ −γ −δ α β ∣ ∣ ∣ ∣ ; z)dz +R(z), where R(z) = resw=0,1 ( sxw−s w(s− w) Zϕ( ∣ ∣ ∣ ∣ α β γ δ ∣ ∣ ∣ ∣ ;w) ) . The series Zψ( ∣ ∣ ∣ ∣ −γ −δ α β ∣ ∣ ∣ ∣ ; z) is absolutely convergent on the line Re z = 1 + b. Integration termwise we obtain I = sx1−s ∞ ∑ n=1 bn 1 2πi 1+b+i∞ ∫ 1+b−i∞ π√ D Γ(z) ( π√ D √ µnx )−2z Γ(1 − z)(1 − z)(s− 1 + z) dz +R(z). (4) We have the Mellin pair J1(x)x −1 and 2z−2 Γ( 1 2 z) Γ(2− 1 2 z) (here J1(x) is Bessel function).Whence for v > 0: J1(v)v −1 = 1 2πi +i∞ ∫ −i∞ 2z−2Γ(1 2z) Γ(2 − 1 2z) v−zdz = = 1 2πi +i∞ ∫ −i∞ 22w−1Γ(w) Γ(1 − w)(1 − w) v−2wdw. Multiplying this by v1−2s and integrating over the interval [2π √ µnx D ,∞) we arrive at the formula ∞ ∫ 2π √ µnx D J1(v)v −2sdv = = 1 4 ( 2π √ µnx D )2−2s 1 2πi +i∞ ∫ −i∞ Γ(w) ( 4π2µnx D )−w Γ(1 − w)(1 − w)(s− 1 + w) dw. (5) The path of integration we can move to Rew = 1 + b. Now from (4)-(5) we infer I = sx1−s ∞ ∑ n=1 bn π√ D ( 4π2µnx D )s−1 ∞ ∫ 2π √ µnx D J1(v)v −2sdv +R(z). (6) 110 On the mean square of the Epstein zeta-function Hence, by (2),(6) we obtain Zϕ( ∣ ∣ ∣ ∣ α β γ δ ∣ ∣ ∣ ∣ ; s) − ∑ λn≤x an λsn + x−s ∑ λn≤x an = = 4s ( 4π2 D )s−1 ∞ ∑ n=1 π√ D bn µ1−s n ∞ ∫ 2π √ µnx D J1(v)v −2sdv +R(z). (7) Further, resw=0 ( sxw−s w(s− w) Zϕ( ∣ ∣ ∣ ∣ α β γ δ ∣ ∣ ∣ ∣ ;w) ) = −x−se−2πi(αγ+βδ), resw=1 ( sxw−s w(s− w) Zϕ( ∣ ∣ ∣ ∣ α β γ δ ∣ ∣ ∣ ∣ ;w) ) = ǫ(α, β) sx1−s s− 1 π√ D , where ǫ(α, β) = { 0 if (α, β) /∈ Z2, 1 if (α, β) ∈ Z2. Thus from (7) we obtain Zϕ( ∣ ∣ ∣ ∣ α β γ δ ∣ ∣ ∣ ∣ ; s) = ∑ λn≤x an λsn + χϕ(s) ∑ µn≤x bn µ1−s n − −x−s   ∑ λn≤x an − ǫ(α, β) π√ D x  + χϕ(s) ∑ µn≤y bn µ1−s n un+ + ∑ µn>y sD π2 ( π2 D )s ∞ ∫ 2π √ µnx D J1(v)v −2sdv + ǫ(α, β) x1−s s− 1 π√ D , (8) where un = χϕ(1 − s) sD π2 ( π2 D )s ∞ ∫ 2π √ µnx D J1(v)v −2sdv − 1. From (8) we have Zϕ( ∣ ∣ ∣ ∣ α β γ δ ∣ ∣ ∣ ∣ ; s) = ∑ λn≤x an λsn + χϕ(s) ∑ µn≤y bn µ1−s n +Rϕ(s, x). In order to calculate the integral In(s) = ∞ ∫ 2π √ µnx D J1(v)v −2sdv O. V. Savastru, P. D. Varbanets 111 we can apply lemma 1 [11] or lemma III.1.2 [12]. Then after the calcula- tion of In(s) (by Jutila’s method [11]) we have Rϕ(s, x) ≪ \|t\|1/2x−σ min(1, x \|t\|) log \|t\| log( \|t\| √ D x + x \|t\| √ D )+ +x1−σ(\|t\| √ D)−1(1 + \|t\| √ D x )min(xǫ + log \|t\|, yǫ + log \|t\|). Forthemore, from (8) we have for x = y = t √ D π = τ , 0 ≤ σ ≤ 1, χϕ(1 − s)Rϕ(s, τ) = − √ 2τ− 1 2 ∆ϕ(τ) +O(t− 1 4D 1 8 ), (9) where ∆ϕ(x) = ∑ u,v∈Z ϕ(u+γ,v+δ)≤x e(αu+ βv) − ǫ(α, β) π√ D x. Remark 1. The estimate of ∆ϕ(x) can be obtained by Perron’s formula for Zϕ( ∣ ∣ ∣ ∣ α β γ δ ∣ ∣ ∣ ∣ ; s). The same reasoning as in the circle problem we easy obtain ∆ϕ(x) = −(Dx) 1 4 π ∑ λn≤N an λ 3 4 n cos(2π √ nx D + π 4 ) +O(xǫ + ( x D ) 1 2 +ǫ N− 1 2 ). Trivially we have ∆ϕ(x) ≪ x 1 3 +ǫD 1 2 . Thus from (9) we obtain the estimate for Rϕ(s, x) in case x = y = t √ D π Rϕ(s, x) ≪ τ− 1 6 +ǫ. However, the error term in the asymptotic formula in the approximate functional equation, which we obtain, is large for the construction of an asymptotic formula for T ∫ 0 Re s=1 2 \|Zϕ( ∣ ∣ ∣ ∣ α β γ δ ∣ ∣ ∣ ∣ ; s)\|2 dt. Thus we build a formula for \|Zϕ( ∣ ∣ ∣ ∣ α β γ δ ∣ ∣ ∣ ∣ ; s)\|2 in which the error term is sufficiently small. We shall use by the idea of D.R. Heath-Brown [6]. Let α = β = γ = δ = 0. We define f(w) =: { ( π√ D )−2w Γ(w + it)Γ(w − it)Zϕ(w + it)Zψ(w − it) } . 112 On the mean square of the Epstein zeta-function Since Zϕ( ∣ ∣ ∣ ∣ 0 0 0 0 ∣ ∣ ∣ ∣ ; s) =: Zϕ(s) = ∑ u,v∈Z (u,v) 6=(0,0) 1 ϕ(u, v)s = Zψ(s) = ∑ u,v∈Z (u,v) 6=(0,0) 1 ψ(v, u)s we have f(1 − w) = f(w), f(1 2 − w) = f(1 2 + w). Moreover f(w) is meromorphic on the complex plane, the only pole being at w = ±it and w = 1 ± it. We consider the integral J = 1 2πi 1+i∞ ∫ 1−i∞ f( 1 2 + z)ez 2/T dz z . If we move the path of integration to Re z = −1 and set w = −z , then we obtain J = −J + resz=0 ( f( 1 2 + z)ez 2/T 1 z ) + resz=± 1 2 ±it ( f( 1 2 + z)ez 2/T 1 z ) We can show that for 1 2T ≤ t ≤ 5T resz=± 1 2 ±it ( f( 1 2 + z)ez 2/T 1 z ) ≪ T 2e− t2 T −πt. Hence, f( 1 2 ) = 2J +O(T 2e− t2 T −πt). (10) Now we have Theorem 2. Let ϕ(u, v) = au2 + 2buv + cv2, (a, b, c) = 1 and rϕ(n) denote the number of the representations of n by form ϕ(u, v). Then T ∫ 0 \|Zϕ( 1 2 + it)\|2 dt = 2T ∑ n≤T √ D π r2ϕ(n) n − 2π√ D ∑ n≤T √ D π r2ϕ(n)+ +2 ∑ mn≤T2D π2 rϕ(m)rϕ(n) (mn)1/2 (m n )iT ( i log m n )−1 +O((T √ D)1/2+ǫ). (11) Proof. We have ϕ(u, v) = ψ(−v,−u). Hence, rϕ(n) = rψ(n), Zϕ(s) = Zψ(s). Now from (10) we obtain uniformly for T ≤ t ≤ 2T \|Zϕ( 1 2 + it)\|2 = √ π \|Γ(1 2 + it)\|2 f( 1 2 ) = 2 1 2πi 1+i∞ ∫ 1−i∞ √ π \|Γ(1 2 + it)\|2 π− 1 2 −z× O. V. Savastru, P. D. Varbanets 113 ×Γ( 1 2 +z+it)Γ( 1 2 +z−it)Zϕ( 1 2 +z+it)Zϕ( 1 2 +z−it)e z 2 T dz z +O(T−2) = = 2 ∞ ∑ m,n=1 rϕ(m)rϕ(n) (mn)1/2 (m n )iT I(mn, t) +O(T−2), (12) where I(n, t) =: 1 2πi 1+i∞ ∫ 1−i∞ ( πn√ D )−z G(z, t)e z2 T dz z , G(z, t) =: Γ(1 2 + z + it)Γ(1 2 + z − it) \|Γ(1 2 + it)\|2 . Therefore, by Stirling’s series for log Γ(z), I(n, t) = 1 2πi 1+i∞ ∫ 1−i∞ ( t √ D πn )z e z2 T dz z +O ( T− 1 6 e −T 8 log2 ( t √ D πn )) . (13) Further, we have for ∣ ∣ ∣log t √ D πn ∣ ∣ ∣≫ T− 1 2 log T I(n, t) =    1 +O(e −T 8 log2 ( t √ D πn ) ), if n < t √ D π O(e −T 8 log2 ( t √ D πn ) ), if n > t √ D π . (14) For ∣ ∣ ∣ log t √ D πn ∣ ∣ ∣ ≪ T− 1 2 log T I(n, t) << log T. (15) (In detail, see ([6], lemma 1)). Now, by (12)-(15) we infer for any T1, T2 with T ≤ T1 < T2 ≤ 2T T2 ∫ T1 \|Zϕ( 1 2 + it)\|2 dt = 2 ∑ n2≤cT 2D r2ϕ(n) n T2 ∫ T1 H(n2, t) dt+2 ∑ mn≤cT2D, m6=n rϕ(m) rϕ(n) (mn)1/2 × × T2 ∫ T1 H(mn, t) (m n )iT dt+O((T √ D)1/2+ǫ), (16) where H(n, t) = { 1, if n < t √ D π , 0, if n > t √ D π . (17) 114 On the mean square of the Epstein zeta-function Therefore, from (17) T2 ∫ T1 H(m2, t)dt =    2(T2 − T1), if m < T1 π , 2(T2 − πm), if T1 π ≤ m ≤ T2 π , 0, if m > T2 π . and for m 6= n T2 ∫ T1 H(mn, t) (m n )iT dt = ( m n it ) ( i log m n )−1 H(mn, t) ∣ ∣ ∣ ∣ T2 T1 + +O((T √ D)1/2+ǫ). Now we can obtain the following correlation by taking T1 = T0, T2 = 2T0, T0 = T 2n and summing for 2 ≤ 2n ≤ T : T ∫ 0 \|Zϕ( 1 2 + it)\|2 dt = 2T ∑ n≤T √ D π r2ϕ(n) n − 2π√ D ∑ n≤T √ D π r2ϕ(n)+ +2 ∑ mn≤T2D π2 , m6=n rϕ(m) rϕ(n) (mn)1/2 (m n )iT ( i log m n )−1 +Oǫ((T √ D)1/2+ǫ). Remark 2. Since rϕ(n) << d(n), we can obtain instead the third sum such estimate T √ D log3(TD). To this end it suffices to use lemma 4 [3]. Bellow we will obtain more precise result. 3. Proof of theorem 3 In order to prove theorem 3 we shall need several auxiliary assertions. Lemma 1. Let the Dirichlet series Φ(s) = ∞ ∑ n=1 an λsn , Ψ(s) = ∞ ∑ n=1 bn µsn , s = σ + it, O. V. Savastru, P. D. Varbanets 115 be absolutely convergent for Re s > 1, and assumed that Φ(s), Ψ(s) can be continued analytically over whole s- plane ( except at the finite number singular points ), moreover the functional equation AsΓ(ms+ v)Φ(s) = B1−sΓ(m(1 − s) + v)Ψ(1 − s), (A,B are constants) holds. Then, for every τ ∈ C, arg τ = ( π 2 − 1 t ) sign t, and for any fixed strip a ≤ σ ≤ b uniformly for \|t\| ≥ t0, A, B, τ , the approximate functional equation Φ(s) = ∑ anλ −s n F (s, λnτ m A ) + ∑ z 6=s res { ( A τm )z−s Γ(mz + v)Φ(z) z − s } + B1−sΓ(m(1 − s) + v) AsΓ(ms+ v) ∑ µn≤y log y bnµ s−1 n F (1− s, µnτ −m B ) +O(x−M + y−M ) holds, where M > 0 is any fixed constant, F (w,X) = 1 Γ(mw + v) 1 2πi ∫ (∆) Γ(m(w + z) + v) Xs z dz, ∆ is such that in region Re s ≥ ∆ there are no singularities of the inte- grating. Moreover, we have uniformly for all parameters: F (w,X) = l+ +O ( exp ( −\|X\| 1 m \|t\| ) ( \|X\| \|t\|m )Rew+ 1 m Re v )  1 + ∣ ∣ ∣ ∣ ∣ m √ \|t\| − \|X 1 m \| √ \|t\| ∣ ∣ ∣ ∣ ∣ −1   , where l = { 1, if λn ≤ x, µn ≤ y, 0, else, x = mm\|τ \|−1A\|t\|m, y = mm\|τ \|B\|t\|m. This lemma is a special case of Lavrik’s theorem ([13]). Corollary 1. Let Φ(s) = ∞ ∑ n=1 ann −s, Ψ(s) = ∞ ∑ n=1 bnn −s, where an = { rϕ(n), if n ≡ l(mod q), 0, else, bn = 1 q ∑ (u,v)∈Z2, ψ(u,v)=n ∑ l1,l2 (mod q), ϕ(l1,l2)≡l(mod q) eq(l1u+ l2v). (18) 116 On the mean square of the Epstein zeta-function Then for s = 1 2 + it, \|t\| ≥ t0, m=1, v=0, A = B = √ D π q, x = A\|tτ−1\|, y = B\|tτ \|, arg τ = arg s, \|τ \| = 1, we have Φ(s) = ∑ n≤\|s\|q2 √ D π , n≡l(mod q) rϕ(n) n 1 2 +it + ( π2 D )it Γ(1 2 − it) Γ(1 2 + it ∑ n≤ \|s\| √ D π bn n 1 2 −it+ +O(q−1 log(Mq\|t\|)) +O(( √ D\|t\|)−M ), (19) (O- constants can depends on only M, t0 ). The proof of this statement carry out in lemma 5 [15]. Lemma 2. Let l, q ∈ N, 1 ≤ l ≤ q. Then for (l, q) = 1 ∑ l1,l2 (mod q), ϕ(l1,l2)≡l(mod q) eq(l1u+ l2v) ≪ q 1 2 (u, v, q) 1 2d(q), ( here d(q) is the number of divisors of n ). This statement is the well-known Weil’s estimate [16] of a trigono- metric sum along a curve over a finite field. Lemma 3. Let B denotes the set of points (u, v) for which ϕ(u, v) ≡ l(mod q) and 0 < ϕ(u, v) < 2q. Then for 0 < ǫ < 1/2, T > 1, in a rectangle −ǫ ≤ Re s ≤ 1 + ǫ, 1 ≤ \|Ims\| ≤ T , \| 1 q2s ∑ l1,l2(mod q) ϕ(l1,l2)≡l(mod q) Zϕ( ∣ ∣ ∣ ∣ 0 0 l1 q l2 q ∣ ∣ ∣ ∣ ; s) − ∑ (u,v)∈B ϕ(u, v)−s\| = = O ( ( \|t\| √ D ) 2(1+ǫ)(1+ǫ−σ) 1+2ǫ ǫ−2q 1 2− 3 2σ− ǫ 2 1+2ǫ ) , ( The O- constant does not depend on t, σ, ǫ, T). This statement is a corollary of lemma 2 and Phragmen-Lindelöf’s theorem. Now we come to the proof of the theorem 3. If we put T0 = max (t0, q ǫ) with t0 from corollary 1 of lemma 1, then T ∫ 0 Re s=1 2 \| 1 q2s ∑ l1,l2(mod q) ϕ(l1,l2)≡l(mod q) Zϕ( ∣ ∣ ∣ ∣ 0 0 l1 q l2 q ∣ ∣ ∣ ∣ ; s) − ∑ (u,v)∈B ϕ(u, v)−s\|2 dt = O. V. Savastru, P. D. Varbanets 117 = T0 ∫ 0 + T ∫ T0 = I1 + I2, say. By lemma 3 it is easily to see that I1 ≪ q−1+2ǫǫ−2. (20) In order calculate I2 we applay the corollary 1 from lemma 1, and then obtain I2 ≪ T ∫ T0 \| ∑ 2q≤n≤U rϕ(n)n− 1 2 −it\|2dt+ T ∫ T0 \| ∑ n≤V bnn − 1 2 +it\|2dt+ + √ DTq−1 log2(MTq) + ( √ DT0) −M+1, (21) ( here U = V = 1 π \|s\| √ D.) The integrals on the right-hand side of (21) can be estimated by the general scheme of the estimation of the mean values of the Dirichlet series ( see, for example, [14], Chapt. 6 and 7). Hence we get I2 ≪ (T +N0) ∑ 2q<n≤U0 a2 n n + (T + V0) ∑ n≤V0 b2n n , where N0 = ∑ 2q<n≤cqT √ D an 6=0 1 ≪ T √ D; U0 ≪ T √ D,V0 ≪ cT √ D. Since rϕ(n) ≪ d(n) we get ( using the notations (18)): I2 ≪ T √ D q ((TDq)2ǫ + log2(TMq) + ( √ DT0) −M+1). (22) The assertion of the theorem follows from (20) and (22) if we put M = −1 + 1 ǫ . 4. Proof of Theorem 4 Consider a quadratic form ϕ0(u, v) = u2+Av2, A ∈ N. Well-known ( see, for example, [4]) that there is finite number of the negative discriminants of the quadratic form for which a kind consists out of one class. Let A is such number. 118 On the mean square of the Epstein zeta-function Lemma 4. Let a kind of the quadratic form ϕ0(u, v) = u2 +Av2, A > 0, A ≡ 1, 2(mod 4), consists out of one class and let rϕ0(n) = ∑ u,v∈Z, ϕ0(u,v)=n 1. Then 1 2rϕ0(n) is a multiplicative function if A > 1, and 1 4rϕ0(n) is a multiplicative function if A=1. Proof. Let for some n ∈ N we have n = u2 0 + Av2 0, and let ϕj(u, v) be a primitive quadratic form of discriminant −4A also represent of n,ϕj(u1, v1) = n. We shall show that ϕj is equivalent to ϕ0 (ϕj ∼ ϕ0). Indeed , we take into account the connection between the classes of divi- sors of field Q( √ −A) and the classes of quadratic forms of a discriminant −4A (in a case A ≡ 1, 2(mod 4)). Let a quadratic form ϕj(u, v) represent of n( i.e. n = ϕj(u1, v1)), then in a appropriate class of divisors has a divisor ℜj for which N(ℜj) = n ( norma of ℜj). The quadratic form ϕ0 belongs to main kind G0. Hence the divisor ℜ0 belongs to main kind G0 of divisors, and then by theorem 6 (Ch. III,§ 8) the divisor ℜj also belongs to G0. But the kind G0 consists only one class. Therefore ℜ0 and ℜ1 belongs the same class and hence ϕ0 ∼ ϕj . Further, if A = 1 we have 1 4rϕ0(n) = ∑ d\|n, d is odd (−1) d−1 2 , and hence 1 4rϕ0(n) is a multiplicative function. Let A > 1. Then the field Q( √ −A) contains only two the roots of 1. We assume that the form ϕ0 represent each of numbers n1 and n2, (n1, n2) = 1. Let ℜ1, . . . ,ℜh1 and ℑ1 . . .ℑh2 are all different divisors each of which has a norma n1 or n2 respectively. Then the divisors ℜi, ℑj belongs to the kind G0. But the product n1n2 also can be repre- sented by ϕ0. Hence ℜiℑj ∈ G0, i = 1, . . . , h1, j = 1, . . . , h2 (here h1 = 1 2 rϕ0(n1), h2 = 1 2 rϕ0(n2)). Since ℜiℑj are all different divisors we have 1 2 rϕ0(n1n2) ≥ 1 2 rϕ0(n1) 1 2 rϕ0(n2). On the other hand, any integer divisor C, N(C) = n1n2, can be represented in the form of a product of coprime divisors ℜi, ℑj . Hence 1 2 rϕ0(n1n2) ≤ 1 2 rϕ0(n1) 1 2 rϕ0(n2). Therefore 1 2 rϕ0(n1n2) = 1 2 rϕ0(n1) 1 2 rϕ0(n2). O. V. Savastru, P. D. Varbanets 119 Remark 3. Let ϕ0(u, v) = u2 + Av2 belongs to the one-class kind G0, and let p be prime number. For any k ∈ N rϕ0(p k) =        2(k + 1), if ( −A p ) = 1; 1 + (−1)k, if ( −A p ) 6= 1; 2 , if p \|A. Lemma 5. Let ϕ0(u, v) = u2 + Av2 belongs to the one-class kind G0. Then ∑ n≤x r2ϕ0 (n) = c0x log x+ c1x+O(x1/2+ǫ) with constants, which can depend from A. Proof. For Re s > 1 we have 1 4 ∞ ∑ n=1 r2ϕ0 (n) ns = ∏ p, χ(p)=1 (1 + 4 ps +O ( 1 \|p2s\| ) ) ∏ p \|D (1 + 1 ps +O ( 1 \|p2s\| ) )× ×g0(s) = ∏ p, χ(p)=1 (1 + 1 ps )4 ∏ p \|D (1 + 1 ps )g1(s) = ζ2(s) ∏ p \|D (1 + 1 ps )−1g2(s), where g0(s), g1(s), g2(s) are the regular functions for Re s > 1 2 . Now by the Perron’s formula we easily get our assertion. Lemma 6. Let l, q ∈ N, (l, q) = 1. Then in the conditions of Lemma we have for any ǫ > 0 ∑ n≡l(mod q), n≤x rϕ0(n) = πx√ D 1 q2 Jq(l, A) +O ( x 1 2 +ǫ q 1 4 ) , where Jq(l, A) = ∑ l1,l2(mod q), l1+Al2≡l(mod q) 1. Proof. For Re s > 1 ∞ ∑ n=1, n≡l(mod q) rϕ0(n) ns = ∑ l1,l2(mod q) l21+Al22≡l(mod q) 1 q2s Zϕ( ∣ ∣ ∣ ∣ 0 0 l1 q l2 q ∣ ∣ ∣ ∣ ; s). Hence, for c > 1, T > 1 ∑ n≡l(mod q), n≤x rϕ0(n) = 120 On the mean square of the Epstein zeta-function = 1 2πi c+iT ∫ c−iT ( ∑ l1,l2(mod q) l21+Al22≡l(mod q) 1 q2s Zϕ0( ∣ ∣ ∣ ∣ 0 0 l1 q l2 q ∣ ∣ ∣ ∣ ; s) − ∑ (u,v)∈B ϕ0(u, v) −s) xs s ds+ +O ( xc Tq(c− 1) ) +O(xǫ). After shifting the contour of integration to the line Re s = −ǫ, applying the functional equation for Zϕ0( ∣ ∣ ∣ ∣ 0 0 l1 q l2 q ∣ ∣ ∣ ∣ ; s) and lemma 3 we obtain ∑ n≡l(mod q), n≤x rϕ0(n) = πx√ D 1 q2 ∑ l1,l2(mod q), l1+Al2≡l(mod q) 1 + ∑ (u,v)∈Z2\(0,0) 1 ϕ0(u, v)1+ǫ × × ∑ l1,l2(mod q), l1+Al2≡l(mod q) e −2πi( l1v+l2u q ) · 1 2πi −ǫ+iT ∫ −ǫ−iT Γ(1 − s) Γ(s) ( π√ D )−1+2sxs s ds+ +O ( xc Tq(c− 1) ) +O(xǫ) +O(T ǫq 1 2 +ǫ). (23) Now trivially estimating the integral and applying lemma 2 we get the assertion of lemma if set T = x 1 2 q 3 4 . Remark 4. A non-trivial estimate the integral in (23) give an estimate of the error term as ≪ x 1 3 +ǫ. Corollary 2. Uniformly for 1 ≤ h ≤ x 5 6 −ǫ there exist constant c0(h) such that ∑ n≤x rϕ0(n) rϕ0(n+ h) = c0(h)x+O(x 5 6 +ǫ), where ǫ is an arbitrarily small, positive constant. Besides, c0(h) ≪ d(h). This statement can be proved similarly the proof of the analogies assertion in [1], [8]. The proof of theorem 4 follows by Heath-Brown’s method [2] from theorem 2 with using lemma 5 and corollary from lemma 6. References [1] G. Belozorov, The Asymptotic formulas for number of solutions of some diofantic equations, Dis., Odessa, 1991 (in Russian). [2] Z. Borevich, J. Shafarevich, Number Theory, M., 1964 (in Russian). O. V. Savastru, P. D. Varbanets 121 [3] J.B.Conrey, The fourth moment of derivatives of the Riemann zeta-function, Quart. J. Math., Oxford (2), 39 (1988), 21-36. [4] L.E. Dikson, Introduction to the theory of numbers, Oxf., 1929. [5] P.Epstein, Zur Theorie allgemeiner Zetafunktionen, Math. Annalen, 56 (1903), 615-644. [6] D.R. Heath-Brown, The fourth power moment of the Riemann zeta-function, Proc. London Math. Soc. (2), 38 (1979), 385-422. [7] A.E. Ingham, Mean-value theorems in the theory of the Riemann zeta-function, Proc. London Math. Soc. (2), 27(1926), 273-300. [8] D. Ismoilov, Additive divisor problems, Tadzic State University, Dushanbe, 1988 (in Russian). [9] A. Ivič, On the fourth moment of the Riemann zeta-function, Public. De L’Inst. Math., Nouvelle Serie, 57 (71), 1995, 101-110. [10] A. Ivič, Y. Motohashi, On the fourth power moment of the Riemann zeta-function, J. Number Theory, 51(1995), 16-45. [11] M.Jutila, On the approximate functional equation for ζ2(s) and other Dirichlet series, Quart. J. Math. Oxford (2), 37 (1986), 193-209. [12] A.A. Karacuba, Bases of Analytic Number Theory, M., 1975 (in Russian). [13] A.F. Lavrik, Approximate functional equation for the Dirichlet L-functions, Trudy Moscov. Math. Obsch., 18 (1968), 91-104 (in Russian). [14] H.L. Montgomery, Topics in Multiplicative Number Theory, Springer-Verlag, 1971. [15] P.D. Varbanets, P. Zarzycki, Divisors of the Gaussian Integers in an Arithmetic Progression, J. Number Theory, 33, No. 2 (1989), 152-169. [16] A. Weil, On some exponential sums, Proc. Nat. Acad. Sci. U.S.A., 24 (1948), 204-207. Contact information O.V. Savastru Department of computer algebra and dis- crete mathematics, Odessa national uni- versity, ul.Dvoryanskaya 2, Odessa 65026, Ukraine E-Mail: savastru@bk.ru P.D. Varbanets Department of computer algebra and dis- crete mathematics, Odessa national uni- versity, ul.Dvoryanskaya 2, Odessa 65026, Ukraine E-Mail: varb@te.net.ua Received by the editors: 08.11.2004 and in final form 21.03.2005.

On the mean square of the Epstein zeta-function

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