A Necessary Condition for a Nontrivial Zero of the Riemann Zeta Function via the Polylogarithmic Function Lazhar Fekih-Ahmed To cite this version: Lazhar Fekih-Ahmed. A Necessary Condition for a Nontrivial Zero of the Riemann Zeta Function via the Polylogarithmic Function. 2012. hal-00693281v1 HAL Id: hal-00693281 https://hal.archives-ouvertes.fr/hal-00693281v1 Preprint submitted on 2 May 2012 (v1), last revised 18 Mar 2013 (v2) HAL is a multi-disciplinary open access L’archive ouverte pluridisciplinaire HAL, est archive for the deposit and dissemination of sci- destinée au dépôt et à la diffusion de documents entific research documents, whether they are pub- scientifiques de niveau recherche, publiés ou non, lished or not. The documents may come from émanant des établissements d’enseignement et de teaching and research institutions in France or recherche français ou étrangers, des laboratoires abroad, or from public or private research centers. publics ou privés. May 2, 2012 12:14 WSPC/INSTRUCTION FILE Fekihahmed˙polylogarithm˙v0.1 International Journal of Number Theory c World Scientific Publishing Company A Necessary Condition for a Nontrivial Zero of the Riemann Zeta Function via the Polylogarithmic Function LAZHAR FEKIH-AHMED∗ Ecole´ Nationale d’Ing´enieurs de Tunis Le Belv´ed`ere 1002 , Tunis, Tunisia. [email protected] Received (May 2, 2012) Accepted (May 2,2012) Communicated by xxx We provide a new series expansion of the polylogarithm of complex argument Lis(x) = ∞ xn Pn=1 ns . From the new series, we define a new entire function Z(s,x) which is related to Lis(x) but processes several advantages over the initial polylogarithmic series. For example, the limit of Z(s,x) when x → 1 is convergent to (s − 1)ζ(s) for all complex numbers s while le limit of Lis(x) converges only when Re(s) > 1. As an application of the expansion of Z(s,x), we derive of a necessary condition for a non-trivial zero of the Riemann zeta function. Keywords: Number Theory; Polylogarithm function; Riemann Zeta function; Riemann nontrivial zeros Mathematics Subject Classification 2010: 33B30,11M26, 11M06 1. Appel and Jonqui`ere Integrals The polylogarithm Lis(x) is defined by the power series ∞ xn Li (x)= . (1.1) s ns nX=1 The definition is valid for all complex values s and all complex values of x such that x < 1. The series is convergent for x = 1 only when Re(s) > 1. | | Using the identity ∞ 1 1 −nt s−1 s = e t dt, (1.2) n Γ(s) Z0 ∗Ecole´ Nationale d’Ing´enieurs de Tunis, BP 37, Le Belv´ed`ere 1002 , Tunis, Tunisia. 1 May 2, 2012 12:14 WSPC/INSTRUCTION FILE Fekihahmed˙polylogarithm˙v0.1 2 Lazhar Fekih-Ahmed equation (1.1) can be rewritten as x ∞ ts−1 Lis(x)= dt. (1.3) Γ(s) Z et x 0 − The integral in (1.3) is called Appell’s integral or Jonqui`ere’s integral. It defines Li (x) not only in the unit circle but also in the whole slit plane C [1, ) provided s \ ∞ that Re(s) > 0 To obtain a formula valid for practically every complex number s, we use Han- kel’s device which consists in replacing the real integral by a contour integral. The contour is denoted by and is called Hankel contour. It consists of the three parts C C = C− C C : a path which extends from ( , ǫ), around the origin counter ∪ ǫ ∪ + −∞ − clockwise on a circle of center the origin and of radius ǫ and back to ( ǫ, ), − −∞ where ǫ is an arbitrarily small positive number. The integral (1.3) becomes Γ(1 s) xts−1 Lis(x)= − dt. (1.4) 2πi Z e−t x C − Equation (1.4) now defines Lis(x) for any x in the cut plane and any s not a positive integer. 2. A New Expansion of Lis(x) The integral in (1.4) can be rewritten as t Γ(1 s) xte s−2 Lis(x)= − t dt. (2.1) 2πi Z 1 xet C − By observing that d xtet t 1 = + xet, (2.2) dt1 xet (1 xet)2 1 xet − − − we may integrate by parts (2.1) to obtain + t −∞ t 1 Γ(1 s) xte s−1 d xte s−1 Lis(x)= − t t t t dt. s 1 2πi 1 xe − − Z dt1 xe − − −∞ C − 1 Γ(1 s) t 1 = − + xetts−1 dt, (2.3) −s 1 2πi Z (1 xet)2 1 xet − C − − where − and + are the endpoints of the contour . −∞ −∞ C To have readable equations, we set X =1 xet, (2.4) − t = log(1 X) log x, (2.5) − − where log(1 X) and log x are both real when X < 1 and x> 0 respectively. − May 2, 2012 12:14 WSPC/INSTRUCTION FILE Fekihahmed˙polylogarithm˙v0.1 3 The function between the parenthesis inside the integral (2.3) becomes t 1 log(1 X) 1 log x + = − + (1 xet)2 1 xet − X2 X − X2 − − ∞ Xn−1 log x = , (2.6) − n +1 − X2 nX=1 provided of course that the infinite series on the right hand side of (2.6) is convergent. The radius of convergence of the series is 1, so we require that X = | | 1 xet < 1. | − | On the contour , t can be expressed as t = reiθ, where θ a small number and C − r a positive number, then et = e−ru for some complex u such that u = 1. Hence, | | one needs to find the requirements on x so that 1 xue−r < 1 for all r > 0 and | − | all u on the unit circle. When 1 x < 1, the condition 1 xue−r < 1 is trivially | − | | − | satisfied for all r> 0. The set 1 x < 1 is a disk of center 1 and radius 1. If we want x to avoid the | − | cut [1, ), then it is judicious to restrict x to the set D defined by ∞ D = x C : x < 1 and 1 x < 1 . (2.7) { ∈ | | | − | } In this paper, we will further restrict x to the real interval (0, 1) which is a subset of D since the consideration of complex values of x does not offer any advantages to our analysis. Finally, if we go back to the original variables, (2.3) simplifies to ∞ t n−1 Γ(1 s) (1 xe ) t s−1 (s 1)Lis(x)= − − xe t dt − − 2πi Z n +1 C nX=1 Γ(1 s) et x log x − ts−1 dt. (2.8) − 2πi Z (1 xet)2 C − Now, can we interchange the sum and the integral in (2.8)? The answer is affir- mative if we can show that the series ∞ Γ(1 s) (1 xet)n−1 − − xetts−1 dt (2.9) − 2πi Z n +1 nX=1 C converges absolutely and uniformly along the contour . For this purpose, we C set Γ(1 s) t n−1 t s−1 σn(s, x)= − (1 xe ) xe t dt. (2.10) − 2πi ZC − The series then (2.9) can be written as the function Z(s, x) defined by May 2, 2012 12:14 WSPC/INSTRUCTION FILE Fekihahmed˙polylogarithm˙v0.1 4 Lazhar Fekih-Ahmed Definition 2.1. For x (0, 1) and for s / 1, 2, , we define the series ∈ ∈{ ···} ∞ σ (s, x) Z(s, x) , n . (2.11) n +1 nX=1 Uniform and absolute convergence of (2.9) or (2.11) is a direct consequence of the asymptotic estimate of σn(s, x) in Proposition 3.7 to be proved later in Section 3. However, we will prove the result using simpler but characteristic estimates when Re(s) > 0. Indeed, let’s choose a cut long the negative real axis so that ts−1 is properly −iπ iπ defined. We have t = re along C− and t = re along C+, as r varies from ǫ to s−1 σ−1 −πy s−1 σ−1 πy . We also have t = r e along C− and t = r e along C , where ∞ | | | | + s = σ + iy. Along the contour C, y is obviously bounded. Moreover, the integral along Cǫ is bounded and tends to 0 as ǫ goes to 0 when Re(s) > 0. We omit the details which can be found in [9] for example. Therefore, to prove absolute and uniform convergence along C, it suffices to prove absolute and uniform convergence for the series ∞ ∞ (1 xe−t)n−1 − e−ttσ−1 dt. (2.12) Z n +1 nX=1 0 A straightforward calculation of the derivative shows that the function (1 −t n−1 −t/2 −t 1 − xe ) e is maximized when e = x(2n−1) and that the maximum value is equal to 1 1 K = (1 )n−1 . (2.13) − 2n 1 x(2n 1) − − p Hence, for n 2, ≥ ∞ ∞ (1 xe−t)n−1xe−ttσ−1 dt = (1 xe−t)n−1e−t/2(xe−t/2tσ−1) dt Z0 − Z0 − ∞ K xe−t/2tσ−1 dt ≤ Z0 1 n−1 xΓ(σ) = 1 − x(2n 1) x(2n 1)(1/2)σ − − √x p K′ . (2.14) ≤ √2n 1 − The last inequality implies that each term of the dominating series is bounded by K′√x/(n + 1)√2n 1. Therefore, by the the comparison test the series (2.12) − is absolutely and uniformly convergent, and we can rewrite it as May 2, 2012 12:14 WSPC/INSTRUCTION FILE Fekihahmed˙polylogarithm˙v0.1 5 ∞ t n−1 Γ(1 s) (1 xe ) t s−1 (s 1)Lis(x)= − − xe t dt − − 2πi Z n +1 nX=1 C Γ(1 s) et x log x − ts−1 dt.