A Short and Simple Proof of the Riemann's Hypothesis

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A Short and Simple Proof of the Riemann's Hypothesis A Short and Simple Proof of the Riemann’s Hypothesis Charaf Ech-Chatbi To cite this version: Charaf Ech-Chatbi. A Short and Simple Proof of the Riemann’s Hypothesis. 2021. hal-03091429v10 HAL Id: hal-03091429 https://hal.archives-ouvertes.fr/hal-03091429v10 Preprint submitted on 5 Mar 2021 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. A Short and Simple Proof of the Riemann’s Hypothesis Charaf ECH-CHATBI ∗ Sunday 21 February 2021 Abstract We present a short and simple proof of the Riemann’s Hypothesis (RH) where only undergraduate mathematics is needed. Keywords: Riemann Hypothesis; Zeta function; Prime Numbers; Millennium Problems. MSC2020 Classification: 11Mxx, 11-XX, 26-XX, 30-xx. 1 The Riemann Hypothesis 1.1 The importance of the Riemann Hypothesis The prime number theorem gives us the average distribution of the primes. The Riemann hypothesis tells us about the deviation from the average. Formulated in Riemann’s 1859 paper[1], it asserts that all the ’non-trivial’ zeros of the zeta function are complex numbers with real part 1/2. 1.2 Riemann Zeta Function For a complex number s where ℜ(s) > 1, the Zeta function is defined as the sum of the following series: +∞ 1 ζ(s)= (1) ns n=1 X In his 1859 paper[1], Riemann went further and extended the zeta function ζ(s), by analytical continuation, to an absolutely convergent function in the half plane ℜ(s) > 0, minus a simple pole at s = 1: s +∞ {x} ζ(s)= − s dx (2) s − 1 xs+1 Z1 ∗One Raffles Quay, North Tower Level 35. 048583 Singapore. Email: [email protected]. The opinions of this article are those of the author and do not reflect in any way the views or business of his employer. 1 Where {x} = x − [x] is the fractional part and [x] is the integer part of x. There is another way [2] to analytically continue ζ(s) to the region ℜ(s) > 0. The idea is to observe that for ℜ(s) > 1: +∞ +∞ 2 1 2 (1 − )ζ(s) = − (3) 2s ns (2n)s n=1 n=1 X X Thus, +∞ 2 (−1)n+1 ζ(s) = (1 − )−1 (4) 2s ns n=1 X 2 = (1 − )−1η(s) (5) 2s The Dirichlet eta function η(s) converges conditionally when ℜ(s) > 0 2kπ and s 6= 1+ i ln(2) . η(s) is used as analytical continuation of the Zeta function on the domain where ℜ(s) > 0. Riemann also obtained the analytic continuation of the zeta function to the whole complex plane. Riemann[1] has shown that Zeta has a functional equation1 s s πs ζ(s) = 2 π −1 sin Γ(1 − s)ζ(1 − s) (7) 2 Where Γ(s) is the Gamma function. Using the above functional equa- tion, Riemann has shown that the non-trivial zeros of ζ are located sym- metrically with respect to the line ℜ(s) = 1/2, inside the critical strip 0 < ℜ(s) < 1. Riemann has conjectured that all the non trivial-zeros are located on the critical line ℜ(s) = 1/2. In 1921, Hardy & Littlewood[2,3, 6] showed that there are infinitely many zeros on the critical line. In 1896, Hadamard and De la Vall´ee Poussin[2,3] independently proved that ζ(s) has no zeros of the form s =1+ it for t ∈ R. Some of the known results[2, 3] of ζ(s) are as follows: • ζ(s) has no zero for ℜ(s) > 1. • ζ(s) has no zero of the form s =1+ iτ. i.e. ζ(1 + iτ) 6= 0, ∀ τ. • ζ(s) has a simple pole at s = 1 with residue 1. • ζ(s) has all the trivial zeros at the negative even integers s = −2k, k ∈ N∗. • The non-trivial zeros are inside the critical strip: i.e. 0 < ℜ(s) < 1. • If ζ(s) = 0, then 1 − s,s ¯ and 1 − s¯ are also zeros of ζ: i.e. ζ(s) = ζ(1 − s)= ζ(¯s)= ζ(1 − s¯) = 0. Therefore, to prove the “Riemann Hypothesis” (RH), it is sufficient to prove that ζ has no zero on the right hand side 1/2 < ℜ(s) < 1 of the critical strip. 1This is slightly different from the functional equation presented in Riemann’s paper[1]. This is a variation that is found everywhere in the litterature[2,3,4]. Another variant using the cos: s s πs ζ(1 − s) = 21− π− cos Γ(s)ζ(s) (6) 2 2 1.3 Proof of the Riemann Hypothesis Let’s take a complex number s such that s = σ + iτ. Unless we explicitly mention otherwise, let’s suppose that 0 <σ< 1, τ > 0 and ζ(s) = 0. We have from the Riemann’s integral above: s +∞ {x} ζ(s)= − s dx (8) s − 1 xs+1 Z1 We have s 6= 1, s 6= 0 and ζ(s) = 0, therefore: 1 +∞ {x} = dx (9) s − 1 xs+1 Z1 Therefore: 1 +∞ {x} = dx (10) σ + iτ − 1 xσ+iτ+1 Z1 And σ − 1 − iτ +∞ cos(τ ln (x)) − i sin(τ ln (x)) {x} = dx (11) (σ − 1)2 + τ 2 xσ+1 Z1 The integral is absolutely convergent. We take the real part and the imaginary part of both sides of the above equation and define the functions F and G as following: σ − 1 F (σ, τ) = (12) (σ − 1)2 + τ 2 +∞ cos(τ ln (x)) {x} = dx (13) xσ+1 Z1 And τ G(σ, τ) = (14) (σ − 1)2 + τ 2 +∞ sin(τ ln (x)) {x} = dx (15) xσ+1 Z1 We also have 1 − s¯ = 1 − σ + iτ = σ1 + iτ1 a zero for ζ with a real part σ1 such that 0 <σ1 = 1 − σ < 1 and an imaginary part τ1 such that τ1 = τ. Therefore σ1 − 1 F (1 − σ, τ) = 2 2 (16) (σ1 − 1) + τ1 −σ = (17) σ2 + τ 2 +∞ cos(τ ln (x)) {x} = dx (18) x2−σ Z1 3 And τ1 G(1 − σ, τ) = 2 2 (19) (σ1 − 1) + τ1 τ = (20) σ2 + τ 2 +∞ sin(τ ln (x)) {x} = dx (21) x2−σ Z1 Before we move forward, we need to define the following function I(a,σ,τ) for a> 0: a sin(τ ln(x)) I(a,σ,τ) = dx (22) xσ Z1 ln(a) σ x = sin(τx)e(1− ) dx (23) Z0 cos(τ ln(a)) (σ − 1) sin(τ ln(a)) = K(σ, τ) 1 − − (24) aσ−1 τ aσ−1 Where τ K(σ, τ) = (25) (σ − 1)2 + τ 2 Let’s define the function J(a,σ,τ) for a> 0: a cos(τ ln(x)) J(a,σ,τ) = dx (26) xσ Z1 ln(a) σ x = cos(τx)e(1− ) dx (27) Z0 (σ − 1) (σ − 1) cos(τ ln(a)) sin(τ ln(a)) = K(σ, τ) − + (28) τ τ aσ−1 aσ−1 Where K(σ, τ) is defined above in the equation (25). Now, let’s write G(σ, τ) as the limit of a sequence as follows: N {x} sin(τ ln(x)) G(σ, τ) = lim dx (29) N 1+σ →+∞ 1 x Z τ N x σ x ln( ) ǫ(e τ )e− τ = lim dx sin(x) (30) N→+∞ τ Z0 τ ln(N) = lim dxg(x) sin(x) (31) N→+∞ Z0 = lim U(N,σ) (32) N→+∞ Where τ ln(N) U(N,σ) = dxg(x) sin(x) (33) 0 Z x ǫ(e τ ) σ x g(x) = e− τ (34) τ ǫ(x) = {x} (35) 4 And the same for F (σ, τ): N {x} cos(τ ln(x)) F (σ, τ) = lim dx (36) N 1+σ →+∞ 1 x Z τ N x σ x ln( ) ǫ(e τ )e− τ = lim dx cos(x) (37) N→+∞ τ Z0 τ ln(N) = lim dxg(x) cos(x) (38) N→+∞ Z0 = lim V (N,σ) (39) N→+∞ Where τ ln(N) V (N,σ) = dxg(x) cos(x) (40) Z0 Let’s study the function g over R. We have the function g piecewise continuous and and its derivatives are also peicewise continuous and: x ǫ(e τ ) σ x g(x) = e− τ (41) τ 1−σ x e τ x σ g′(x) = ǫ′(e τ ) − g(x) (42) τ 2 τ The interval (0,τ ln(N)) is the union of the intervals (τ ln(n),τ ln(n + 1)) where 1 ≤ n < N. We have ǫ′(x) = 1 for each x ∈ (τ log(n),τ log(n + 1)) for each 1 ≤ n < N. Therefore, thanks to the integration by parts we can write: τ ln(N) U(N,σ) = dxg(x) sin(x) (43) Z0 N−1 = g(0+) cos(0) + lim g(τ ln(n)+ ǫ) cos(τ ln(n)+ ǫ) − g(τ ln(n) − ǫ) cos(τ ln(n) − ǫ) (44) ǫ→0 n=1 X i τ ln(N) −g(τ ln(N)−) cos(τ ln(N)) + dxg′(x) cos(x) (45) Z0 N−1 = lim g(τ ln(n)+ ǫ) − g(τ ln(n) − ǫ) cos(τ ln(n)) (46) ǫ→0 n =1 i X τ N cos(τ ln(N)) ln( ) − + dxg′(x) cos(x) (47) τN σ Z0 N−1 τ N 1 cos(τ ln(n)) cos(τ ln(N)) ln( ) = − − + dxg′(x) cos(x) (48) τ nσ τN σ n=1 0 X Z N τ N τ N 1−σ x 1 cos(τ ln(n)) σ ln( ) ln( ) e τ x = − − dxg(x) cos(x)+ dx ǫ′(e τ ) cos(x) (49) τ nσ τ τ 2 n=1 0 0 ! X Z Z N τ N 1−σ x 1 cos(τ ln(n)) σ ln( ) e τ = − − V (N,σ)+ dx cos(x) (50) τ nσ τ τ 2 n=1 0 X Z 5 Therefore N σ 1 cos(τ ln(n)) 1 U(N,σ)+ V (N,σ) = − + J(N,σ,τ) (51) τ τ nσ τ n=1 X Therefore N cos(τ ln(n)) − J(N,σ,τ) = −τU(N,σ) − σV (N,σ) (52) nσ n=1 X N cos(τ ln(n)) Therefore, the sequence n=1 nσ −J(N,σ,τ) converges N≥1 and its limit is as follows: P N cos(τ ln(n)) τ 2 + σ(σ − 1) lim σ − J(N,σ,τ) = − (53) N→+∞ n (1 − σ)2 + τ 2 n=1 X Therefore we can write the following: N cos(τ ln(n)) τ 2 + σ(σ − 1) = J(N,σ,τ) − + αN (54) nσ (1 − σ)2 + τ 2 n=1 X Where (αN ) is a bounded sequence with limit zero.
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