The analysis of the Riemann hypothesis Bahattin Gunes To cite this version: Bahattin Gunes. The analysis of the Riemann hypothesis. 2019. hal-02077752v2 HAL Id: hal-02077752 https://hal.archives-ouvertes.fr/hal-02077752v2 Preprint submitted on 14 Apr 2019 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. The analysis of the Riemann hypothesis Bahattin Gunes Abstract. Here, we put forth two different proofs for the Riemann hypothe- sis. The first one was presented by using one of the integral representatives of the Riemann zeta function. The second one was claimed by using the Dirichlet s lambda function and the function q(s) = 2 (s). Keywords: Riemann hypothesis, proof, Riemann zeta function 1. Introduction Here, we will present two different proofs which prepared very rigorously. First, let us introduce some basic properties of the Riemann zeta function and some well- known arts as below to make the arguments also accessible to a nonspecialist. We hope that you will read this quite-fluent file nicely. The Riemann zeta function is defined as below in the half-plane with the com- plex variable s = + it. It is an absolutely convergent series by 1 < (s). The (s) is the real part of the complex variable. < < 1 1 1 1 1 1 (s) = = + + + for 1 < (s) (1.1) ns 1s 2s 3s 4s n=1 ··· < X As shown by Riemann, (s) extends to the whole complex plane C by analytic continuation as satisfies functional equation ([6], p2 ) with a simple pole at s = 1: s 1 s s 1/2 (s) = (1 s) for s = 1 (1.2) 2 2 6 Meanwhile, if we call another functional equation ([9], p146 ) here; s s 1 s (s) = 2 sin( )(1 s)(1 s) for s = 1 (1.3) 2 6 we can see that the Riemann zeta function (s) has zeros at the negative even integers ( 2, 4, 6, ... ) on it . These are called as the trivial zeros. The Riemann hypothesis which proposed in 1859 is concerned with the non-trivial zeros [1]. Let us denote these non-trivial zeros as s = + it . The Riemann hypothesis: The real part of any non-trivial zeros of the Riemann zeta function is 1/2. It means, (s ) = 0 = 1/2. ) 2000 Mathematics Subject Classification. Primary 11M26. 1 2 BAHATTINGUNES 1.1. Some other well-known information/features related the RH. The Riemann hypothesis (RH) is one of the most important unsolved • problems in mathematics which will help to illuminate the distribution of the prime numbers. In 1914 Hardy [5] proved that (1/2+it) has infinitely many zeros. Beside, • you can review our other work about the zeros of the zeta function [4]. If (s) is non-trivially equal to zero, the real part must satisfy the • following condition which had been already proven ([9], p5 and [11], p30 ): 0 < (s ) < 1 or 0 < < 1 for s = + it < Remark 1. The gamma function (s) is defined and analytic for any complex numbers ´ s´ except the non-positive integers. In the region 0 < (s) < 1, we know that (s) and (1 s) have no zeros, no poles and they are finite< ([8], p5). Let us present some other well-known features of the gamma function: i-) It is expressed with the positive real part via the improper integral: 1 s 1 x (s) = x e dx for 0 < (s) (1.4) < Z0 ii-) It satisfies the functional equations as: (s)(1 s) = for all ´s´ except the integers (1.5) sin(s) Remark 2. If one of the Riemann zeta functions which in (s), (1 s), (s) and (1 s) is non-trivially equal to zero; then all of them are equal to zero: (s) = 0 (1 s) = 0 (s) = 0 (1 s) = 0 (1.6) , , , Proof. In fact, it is well-known. However, it will be useful for us to show it again. If we set the functional equation in 1.3 equal to zero: s s 1 s (s) = 0 2 sin( )(1 s)(1 s) = 0 ) 2 The following result is obtained by 0 < R(s) < 1: (1 s)(1 s) = 0 And we know (1 s) = 0 from Remark 1. Consequently, (1 s) = 0. On the other hand, the Schwarz6 reflection principle in complex analysis [2] says: (s) = (s) (1.7) Thus, we can write accordingly: (s) = 0 (s) = 0 , THERIEMANNHYPOTHESIS 3 2. The First Proof of The RH Step 1: Let us call the integral representative of the zeta function ([7], p6 ) which valid in the critical strip 0 < (s) < 1. < sin (s/2) 1 1 s 1 1 (s) = x + dx (2.1) (s 1)1 s sin2 ix x2 Z0 Let us define two new functions as below: sin (s/2) 1 1 f(s) = and g(x) = + (2.2) (s 1)1 s sin2 ix x2 Now, let us consider (s) and (1 s) as partial with a new variable ´u´ (as u R which is suffi ciently large) by denoting them as below: 2 u u 1 s s (s) = f(s) x g(x)dx ; (1 s) = f(1 s) x g(x)dx (2.3) u u Z0 Z0 If we handle them with the non-trivial zeros (s ) by Remark 2 as: u u 1 so so lim f(so) x g(x)dx = 0 ; lim f(1 so) x g(x)dx = 0 (2.4) u u !1 Z0 !1 Z0 s s 1 s Step 2: Let us call equation 1.3 as (s) = 2 sin( )(1 s)(1 s). 2 We know from Remark 1 that (1 s) has no zeros and poles and it is finite in the critical strip (0 < (s) < 1). < Remark 3. Thus, the following function will be entire/ analytic/ continuous and finite and nonzero in the critical strip. Naturally, we can see that the non- trivial zeros the Riemann zeta function have no any privilege or exception on it. We can also observe the similar approaches in Schipani’s work ([10], p100). s s 1 s h(s) = 2 sin( )(1 s) (2.5) 2 Meanwhile, we know z = z for any ´z´ complex number. Thus, let us write equation 1.3 again, but thisj j timej j as below: s s 1 s (s) = (s) = 2 sin( )(1 s)(1 s) (2.6) j j j j 2 Now, by noticing Remark 3, let us write the following one. Here s refers the values which are different than the non-trivial zeros of the Riemann zeta function. Depending on Remark 2; if s=s , we see (s) = 0 and (1 s) = 0. 6 6 6 (s ) s s s 1 = 2 sin( )(1 s) (2.7) (1 s ) 2 4 BAHATTINGUNES With the equation in 2.1 accordingly (for f(s) and g(x) see equation 2.2): 1 1 s f(s) x g(x)dx 0 Z = h(s) (2.8) 1 j j f(1 s ) xs g(x)dx 0 Z Right after, let’swrite it in the following new form with the equations in 2.3: u 1 s lim f(s) x g(x)dx u !1 0 Z u = h(s) (2.9) s j j lim f(1 s) x g(x)dx u !1 0 Z Because the limit result will be stay as the same, we can write it as below: u 1 s f(s) x g(x)dx 0 lim Z u = h(s) (2.10) u j j !1 f(1 s ) xs g(x)dx 0 Z Now, let us handle equation 2.10 with the non-trivial zeros of the Riemann zeta function under the limit approach as u . However this time, the left side will appear in the indeterminate form as 0/!0 1(see the results in 2.4). On the other hand, with the non-trivial zeros, the existence of a limit-result will keep on being in the same way (see Remark 3): u 1 so f(so) x g(x)dx 0 s s 1 s lim Z u = 2 sin( )(1 s ) (2.11) u 2 !1 f(1 s ) xso g(x)dx o 0 Z At this stage, let’sask ourselves the following critical question: Could we find the non-trivial zeros of the Riemann zeta function which provide the above- equation? Or at least, any feature about them? Step 3: At moment, I think that we can apply L’Hopital rule to the left side of the above-equation with respect to the parameter ´u´; because: i-) The limit of it will appear as ´0/0´ with any non-trivial zeros. ii-) The important thing here we know that there is a limit result as one of the finite positive real numbers as h(so) . It’s values only depend on the non-trivial zeros of the Riemann zetaj function.j We mean: s s 1 s h(so) = 2 sin( )(1 s ) j j 2 THERIEMANNHYPOTHESIS 5 iii-) Both the numerator and the denominator are differentiable. For example, we can get (you can also see it with Wolfram Development Platform [13]): d u 1 1 f(s) xsg(x)dx = f(s)usg(u) by g(x) = + (2.12) du sin2 ix x2 Z0 iv-) Also HERE, we have to show that the denominator shouldn’t equal u so zero as u(1 s) = f(1 so) x g(x)dx = 0 with any u > M.