ISSN2075-9827 e-ISSN2313-0210 http://www.journals.pu.if.ua/index.php/cmp Carpathian Math. Publ. 2016, 8 (1), 16–20 Карпатськi матем. публ. 2016, Т.8, №1, С.16–20 doi:10.15330/cmp.8.1.16-20 BASIUK Y.V., TARASYUK S.I. FOURIER COEFFICIENTS ASSOCIATED WITH THE RIEMANN ZETA-FUNCTION We study the Riemann zeta-function ζ(s) by a Fourierseries method. The summation of log ζ(s) | | with the kernel 1/ s 6 on the critical line Re s = 1 is the main result of our investigation. Also we | | 2 obtain a new restatement of the Riemann Hypothesis. Key words and phrases: Fourier coefficients, the Riemann zeta-function, Riemann Hypothesis. Ivan Franko National University, 1 Universytetska str., 79000, Lviv, Ukraine E-mail: [email protected] (Basiuk Y.V.), [email protected] (Tarasyuk S.I.) INTRODUCTION ∞ 1 It is known that the integral log ζ 2 + it dt, where ζ(s) is the Riemann zeta-function, ∞ | | diverges. M. Balazard, E.Saias, M.−R Yor [1] summed log ζ(s) on the critical line with the kernel | | 1 2 z 1 < 3 1/ s . Using the fact that f (z) = 1 z ζ 1 z , z 1, belongs to the Hardy space H and the | | − − | | result of Bercovici and Foias [2] on the factorization of f (z), they have proved the following theorem. Theorem ([1]). 1 log ζ(s) ρj | 2 | ds = ∑ log , 2π Re s= 1 s | | 1 ρ Z 2 > 1 j | | Re ρj 2 − where ρ is the sequence of non-trivial zeroes of ζ(s). { j} In particular, the Riemann Hypothesis holds if and only if 1 log ζ(s) | 2 | ds = 0. 2π Re s= 1 s | | Z 2 | | A. Kondratyuk, P. Yatsulka [6], using the method of Fourier series, have established the following fact. Theorem ([6]). Let ρ be the sequence of non-trivial zeroes of ζ(s). Then { j} 1 log ζ(s) ρj | | ds = 1 γ + 2 ∑ log 2π Re s= 1 s 4 | | − 1 ρ Z 2 > 1 j | | Re ρj 2 − ( ρ 2 Reρ )( 2Reρ 1) | j| − j j − + ∑ 2 , > 1 ρj(ρj 1) Re ρj 2 | − | УДК 517.53 2010 Mathematics Subject Classification: Primary 11M06, 30D99; Secondary 11M41. c Basiuk Y.V., Tarasyuk S.I., 2016 FOURIER COEFFICIENTS ASSOCIATED WITH THE RIEMANN ZETA-FUNCTION 17 where γ is the Euler constant. The Riemann Hypothesis holds if and only if 1 log ζ(s) | 4 | ds = 1 γ. 2π Re s= 1 s | | − Z 2 | | We make the next step studying the behaviour of the Riemann zeta-function on the critical line. The summation of log ζ(s) with the kernel 1/ s 6 on the critical line Re s = 1 is the main | | | | 2 result of our research. 1 SECTION WITH RESULTS Our result is the following. Theorem 1. Let ρ be the sequence of non-trivial zeroes of ζ(s). Then { j} 2 1 log ζ(s) 7 γ1 γ ρj | 6 | ds = 4γ + − + 6 ∑ log 2π Re s= 1 s | | 2 − 2 1 ρ Z 2 > 1 j | | Re ρj 2 − ( ρ 2 Reρ )(2Reρ 1 ) | j| − j j − + 4 ∑ 2 (1) > 1 ρj(ρj 1) Re ρj 2 | − | Re( ρ 2 ρ )2(2Reρ 1)(2 ρ 2 2Reρ + 1) 1 | j| − j j − | j| − j + ∑ 4 , 2 > 1 ρj(ρj 1) Re ρj 2 | − | where γ is the Euler constant, 1 log2 N γ1 = lim ∑ log m . N ∞ − m N m − 2 ! → ≤ Also we obtain a new restatement of the Riemann Hypothesis. Theorem 2. The Riemann Hypothesis holds if and only if 2 1 log ζ(s) 7 γ1 γ | 6 | ds = 4γ + − . (2) 2π Re s= 1 s | | 2 − 2 Z 2 | | Proof of Theorem 1. Observe that the conformal map z = 1 1/s transforms the domain − s : Re s > 1 onto the unit disc z : z < 1 . Consider the function 2 { | | } n o z 1 f (z) = (s 1) ζ(s) = ζ . − 1 z 1 z − − We have (s 1) ζ(s) = 1 + γ(s 1)+ γ (s 1)2 + + γ (s 1)k+1 + ..., (3) − − 1 − · · · k − where ( 1)k 1 logk+1 N γ = − lim ∑ logk m , k N, k ∞ k! N m N m − k + 1 ! ∈ → ≤ ([5, p.4]). Therefore f (z) is holomorphic in the unit disk. It was showed in [3] that the function f (z) belongs to the Hardy class Hp, 0 < p < 1. Earlier it was established in [1] and [2] that the 18 BASIUK Y.V., TARASYUK S.I. 1 function f (z) belongs to the Hardy class H 3 and σ = 0, where σ is the singular measure from the factorization (see [4]) 2π iϕ 2π iϕ 1 e + z 1 e + z iϕ f (z) = B(z) exp(iC) exp iϕ dσ(ϕ) exp iϕ log f (e ) dϕ , · · −2π Z0 e z 2π Z0 e z | | − − (4) where aj aj z B(z)= ∏ | | − a 1 a z j j − j is the Blaschke product, a is the sequence of zeros of f (z) and C = Im f (0) is a real constant. { j} Consider the Fourier coefficient of log f (reiθ ) : | | 2π 1 ikθ iθ ck(r, f ) = e− log f (re ) dθ, r 1. 2π Z0 | | ≤ Note that c k(r, f ) = ck(r, f ). − It follows from (3) that f (0) = 1, and (4) yields 1 c0(1, f ) = log B(0) = ∑ log − | | a j | j| and 1 2π eiϕ + reiθ log f (reiθ ) = log B(reiθ) + Re log f (eiϕ) dϕ. (5) | | | | 2π 0 eiϕ reiθ | | Z − In some neighborhood of the origin, the function F(z)= log f (z), log f (0) = 0, is holomor- 2 phic. Let F(z)= A1z + A2z + . be its Maclaurin expansion. According to (3) γ γ2 A = γ; A = 1 − . 1 2 2 On the other hand, F + F γr(eiϕ + e iϕ) (γ γ2)r2(e2iϕ + e 2iϕ) log f (reiϕ) = Re log f (reiϕ)= = − + 1 − − + ..., | | 2 2 4 where r is sufficiently small. The relation (5) implies, for small r, 2 γ1 γ 2 2 − r = c 2(r, B)+ r c 2(1, f ). 4 − − In [7], the expression for the Fourier coefficient of the Blaschke product was obtained 2 ∞ r 1 4 c 2(r, B)= ∑ aj 1 − 4 a 2 | | − j=1 j for r < a . Thus, | 1| 2 ∞ γ1 γ 1 1 4 c 2(1, f ) = − ∑ aj 1 . (6) − 4 − 4 a 2 | | − j=1 j FOURIER COEFFICIENTS ASSOCIATED WITH THE RIEMANN ZETA-FUNCTION 19 Note that 2π 1 1 2iθ 1 c 2(1, f ) = + e log ζ dθ. (7) − 4 2π 1 eiθ Z0 − Return to the variable s. Taking (6) and (7) into account, we obtain 1 1 1 2 log ζ(s) + 1 | | ds π s 2 4 2 Z − s | | Re s= 1 | | 2 (8) 2 ρ 2(2Reρ 1)(2 ρ 2 2Reρ + 1) γ1 γ 1 j j − | j| − j = − + ∑ 2 4 . 4 4 > 1 (ρj 1) ρj Re ρj 2 − | | Taking the real parts of both sides (8), we get 1 1 log ζ(s) 1 log ζ(s) 1 2 log ζ(s) + | 2 | ds | 4 | ds + Re s | 6 | ds 4 2π Re s= 1 s | |− 2π Re s= 1 s | | 2π Re s= 1 s | | Z 2 | | Z 2 | | Z 2 | | 2 Re( ρ 2 ρ )2(2Reρ 1)(2 ρ 2 2Reρ + 1) γ1 γ 1 | j| − j j − | j | − j = − + ∑ 4 . 4 4 > 1 ρj(ρj 1) Re ρj 2 | − | Note that ∞ 1 log ζ(s) 1 log ζ 2 + it Re s 2 ds = 2 t2 dt | 6 | 3 s | | 0 4 − 1 2 Z 1 | | Z + t Re s= 2 4 ∞ 1 ∞ 1 log ζ 2 + it log ζ 2 + it = 2 dt + dt 2 3 − Z0 1 2 Z0 1 2 4 + t 4 + t log ζ(s) 1 log ζ(s) = | 4 | ds + | 6 | ds . − Re s= 1 s | | 2 Re s= 1 s | | Z 2 | | Z 2 | | Using the results from [1] and [6], we obtain (1). The proof is completed. Proof of Theorem 2. If the Riemann Hypothesis is true, then the series at the right hand side of (1) are absent, and we have (2) 2 1 log ζ(s) 7 γ1 γ | 6 | ds = 4γ + − . 2π Re s= 1 s | | 2 − 2 Z 2 | | Now assume that relation (2) holds. If the Riemann Hypothesis is not true, then in (1) ρ ( ρ 2 Reρ )(2Reρ 1) j | j| − j j − > 6 ∑ log + 4 ∑ 2 0. > 1 1 ρj > 1 ρj(ρj 1) Re ρj 2 − Re ρj 2 | − | Examine carefully the series Re( ρ 2 ρ )2(2Reρ 1)(2 ρ 2 2Reρ + 1) | j| − j j − | j| − j ∑ 4 . > 1 ρj(ρj 1) Re ρj 2 | − | 20 BASIUK Y.V., TARASYUK S.I. We are interested in when all terms of this series are positive. The following conditions appear Re( ρ 2 ρ )2 > 0. | j| − j If 0 < Re ρ < 1 and Im ρ > 1 + 1 , then Re( ρ 2 ρ )2 > 0. j | j | 2 √2 | j| − j It is known (see [8]) that the first 10 22 + 1 non-trivial zeros of the Riemann zeta-function lie on the critical line. In particular, Im ρ1 = 14, 1347 . These facts imply Re( ρ 2 ρ )2 > 0 for all non-trivial zeros ρ that lie inside the critical | j| − j j strip 0 < Re s < 1. Hence, if the Riemann Hypothesis is not true, then 2 1 log ζ(s) > 7 γ1 γ | 6 | ds 4γ + − .