Riemann Hypothesis and Euler Function

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1 Riemann Hypothesis Euler Function Choe Ryong Gil April 5, 207 Department of Mathematics, University of Sciences, Unjong District, Pyongyang, D.P.R.Korea, ; Abstract: The aim of this paper is to show the proof of the Riemann hypothesis RH by the primorial number. We find a new sufficient condition SC for the RH from well-known Robin theorem prove that the SC holds unconditionally. Keywords: Euler function, Primorial number, Riemann hypothesis. I. Introduction Main result of paper Let N be the set of the natural numbers. The function ϕn = n p p n is called the Euler function of n N []. Here ϕ = p n denotes p is the prime divisor of n. The function σn = d n d is the divisor function of n N. Here d n denotes d is the divisor of n. Robin showed in his paper [4] also see [2]. [Robin Theorem] If the Riemann hypothesis RH is false, then there exist constants c > 0 0 < β < /2 such that σn n eγ log log n + c log log n log n β. holds for infinitely many n N, γ = is the Euler constant []. From this we have [Theorem ] If for any n 2 n ϕn eγ log log24 n ρn.2 holds, then the RH is true, For n Nn we define the function ρn = exp log n log log n 2. Then we give [Theorem 2] For any n 2 we have Φ 0 n 24. [Corollary] For any n 5 we have Φ 0 n = expexpe γ n/ϕn..3 n ρn n ϕn log log eγ n2 log log n log n

2 II. Proofs of Theorem Theorem Proof of Theorem It is clear that σn ϕn n 2 for any n 2. If.2 holds, but the RH is false, then by the Robin Theorem, e γ log log n + c log log n log n β σn n n ϕn eγ log log24 n ρn holds for infinitely many n N. On the other h, since log + t t t > 0, we have log log24 n ρn = loglog 24 + log n + log n log log n 2 = log 24 log log n2 = log log n + log + + log n log n log log n + log 24 log n + log log n2 log n eγ c log 24 log n β log log n + eγ c log log n log n /2 β 0 n, 2. but it is a contradiction Reduction to the primorial number Let p = 2, p 2 = 3, p 3 = 5, be the first consecutive primes. Then p m m N is m- th prime number. The number p p m is called the primorial number [3, 7]. Assume that n = q λ qλm m is the prime factorization of n N. Here q,, q m are distinct primes, λ,, λ m are nonnegative integers ωn := m is the number of distinct prime factors of n N[6]. Put I m := p p m, then it is clear that n I m, n m ϕn = m q i p i = I m ϕi m i= i= 2.2 so Φ 0 n Φ 0 I m. This shows that the boundedness of the function Φ 0 n for n N n is reduced to one for the primorial numbers. Now we put C m := Φ 0 I m m Proof of Theorem 2 Let n = q λ,, qλm m be the prime factorization of any n 2. Then Φ 0 n C m. If m 4, we see from the computation by MATLAB see table C 3 = C 2 = C = expexpe γ 2 2 exp log 2 log log 2 2 = , expexpe γ 2 3/2 2 3 exp log2 3 log log2 3 2 = 23.56, expexpe γ 2 3/2 5/ exp log2 3 5 log log = , 2

3 expexpe γ 2 3/2 5/4 7/6 C 4 = exp log log log = If m 5, then we have C m < by the Lemma also see below. Therefore we have Φ 0 n Φ 0 I m = C m max m {C m} Proof of Corollary From the theorem, then for any n 5 we have n ϕn eγ log log n + e γ log 24 log n = e γ log log n + e γ log 24 + log n log log n 2 e γ log log n log log n2 log n = log log n2 log n. log log n2 log n III. Lemma for Boundedness of Sequence {C m } In the proof of the theorem 2, it was used that the sequence {C m } is bounded. So we here would prove the following Lemma. [Lemma ] For any m 5 we have C m <. For the proof of the Lemma, we need some estimates. 3.. Some symbols It is known that by [5], p = log log t + b + Et, 3. p t t is a real number p is the prime number, Et = Oexp a log t a > 0 [5] b = γ + p log /p + /p = is the Mertens constant [8]. Put F m := I m /ϕi m, then m m logf m = log /p i + /p i + /p i = i= i= From this we have = log log p m + γ + Ep m + εp m, εp m = p>p m log /p + /p = O/p m. e γ F m = log p m e 0, expe γ F m = p m e 0, 3.2 3

4 Similarly, we easily have e 0 = expep m + εp m, e 0 = explog p m e 0. e γ F m = log p m e, expe γ F m = p m e, 3.3 e = expep m + εp m, e = explog p m e. We recall the Chebyshev s function ϑt = p t log p[]. By the prime number theorem [], ϑp m = p m + θp m, 3.4 Then we see Now put Then we have θp m = Oexp a 2 log p m a 2 > 0 [5]. log I m = p m α 0, log I m = p m α, α 0 = + θp m, α = + θp m. N i := log I m i log log I m i 2 i = 0,. N 0 = p m α 0 log 2 p m α 0, N = p m α log 2 p m α An estimate of e e We put p = p m, p 0 = p m below. For the theoretical calculation we assume p e 4. The discussion for p e 4 is supported by MATLAB. Since e γ F m = log p e < log p + log p p by 3.30 of [5], we respectively have e <.0052 p e 4, e <.075 p e 4, e e <.08 p e An estimate of e e Since if e then e, we have e e. On the other h, it is known that / log 2 t Et / log 2 t t > 3.7 by of [5]. Hence, since εp < 0, if e >, then 0 < r := Ep + εp < log 2 p p e4 e = + r + n=2 r n n! + r + r 2 2 r + r r2, 4

5 Therefore we have e e h 2 = expr + log p e + h + 2 h, h = + log p r log p r p e 4. e e + log p Ep + εp log p 2 Ep + εp 2 e >, p e An estimate of V 0 := p 0 e 0 α 0 p e α It is clear that p 0 α 0 p α = log p 0, since Et = p t p log log t b, we have From this Thus we have Since we have Ep 0 Ep = log p0 log, p 0 log p εp 0 εp = log p0. p 0 e 0 log p = +, e log p 0 p 0 e 0 e = p log p e exp p 0 p 0 V 0 = p e p0 e 0 p e log p 0 = log p 0 µ e, 3.9 µ = p exp log p 0 log p e p 0. µ e + 2 log p e log p e p p e log p p, e >, p e 4 µ e e e log p p. e >, p e An estimate of G 0 := log p 0 RI m N 0 N /N 0 Here RI m := log log I m 2 2 log 4 +. I m log log I m It is known that p 2 k+ p p k for p k 7 by 246p of [6] hence log p 0 log I m < 2 p e4. 5

6 Since log + t t t 2 /2 for any t 0 < t < /2, we have N 0 N = log I m I m log log I m log I m log log I m 2 log log I m 2 log p 0 2 log log log I m 2 + I m +2 log I m log log I m log + log p 0 log I m log p 0 2 log log log I m 2 + I m + log p 0 2 log log I m log p 0 log Im 2 log I m G 0 N 0 log p 0 2 log Im log Im log log I m 2 + log 2 p 0 + log I m log log I 3/2 m log 2 p 0 log log I m 2 log I m 3/ log log I m And it is known that p 2 k+ 2 p2 k for p k 7 by 247p. of [6] / log t < θt < / log t t 4 3. by of [5]. So log p 0 log p + log 2. log p Since p e 4, we have α /4 the function log 3 t/t is decreasing on the interval e 3, +. Therefore we get log 2 p 0 G 0 log I m log log I m log3 p p α 2 + log 2 2 log p 4 + log p + log α p log p 0.0 p log p 3.6. An estimate of Sp := p p /p log p Put st := p = log log t + b + Et. p t Then by the Abel s identity [], we have Sp = + p log t dst = p e dt p log t t log t + det 6

7 log p Ep + log p + p t log 4 t dt log p + log 3 p 3 log 3 t + p = = log p log 3 p 3.3 Sp log p 4 3 log 3 p. If p is a first prime e 4, then p = it is 938-th prime. And we have 3.7. A Condition d Put Sp ft = t log t Et t θt, dt = ft gt gt = t log 2 t α, 3 p t < p + 2, 3.5 t is a real number α = + θp is dependent on p a positive constant such that / log p α + / log p. Then both ft gt are continuously differentiable function on the interval p, p + 2. In fact, since the functions log p are constants on p, p + 2, we have E t = p t p b, ϑt = p t t log t, E t is the derivative of Et so on. Hence we obtain θ t = t θt, 3.6 t f t = + log t Et. 3.7 Thus the function dt is also continuously differentiable on p, p + 2, since gt > 0. Moreover, dt has the right h derivative at the point t = p, that is, + dp := Put d p = + dp. We here assume that lim t p+0 dt dp. t p d p gp p 2 p we will call 3.8 the condition d below. If the condition d holds, then for any prime p 3 we have +log p Ep dp log2 p α 2 p logp α p 7

8 3.8. Proof of Lemma Now we are ready for the proof of the lemma. Let D m := p m e 0 α 0 pm α 0 log 2 p m α 0 m Then C m < is equivalent to D m <. And we here have D m < for any p m e 4, D m a m := 3 Sp m 3.2 for any p m e 4. In fact, D m < is also equivalent to R m < 0 it is easy to see that R m := loge γ F m log loglog I m + log I m log log I m 2 < for p m e 4 by MATLAB see the table the table 2. Next, we will prove D m a m for any p m e 4 by the mathematical induction with respect to m. If p = then we have D 938 = Sp 0.09 < Now assume p e 4 D m a m. Then D m = N 0 p e α + V 0 = D m N N 0 + V 0 N 0 a m N N 0 + N 0 log p 0 µ e a m +b m, 3.24 b m = N 0 log p 0 µ e a m N 0 N By the assumption D m a m, we get p α log 2 e p α α + a m p by taking logarithm of both sides log 2 p α = α + a m p α log e = log p e θp + a m log2 p α p α. From this Thus e + θp + a m log p log2 p α, p α Ep + εp log p θp + a m log2 p α p α. log p Ep θp a m log2 p α p α log p εp 8

9 the both sides multiply by p p log 2 p α, then dp = p log p Ep p θp p log 2 p α a m p log p εp α p log 2 p α If the condition d holds, then from 3.9 we get + log p Ep a m log2 p α 2 p α 4 + logp α + log p εp + 2 p, because εp < 0 log p log p p e 4, α /4. logp α Thus we see +log p Ep+εp a m log2 p α 2 p α logp α p If e >, then, since 0 < a m /4 α + /4, we also have + log p 2 Ep + εp 2 log4 p α p α logp α α log 2 p α log4 p α p α p e log p 0 µ e a m N 0 N log p 0 + log p Ep + εp a m N 0 N log2 p 0 p a m G 0 N log2 p 0 p log p 0 + log p 2 Ep + εp log p 0 log4 p α p α + 2 log p 0 p. Finally, by the function log 4 t/ t is decreasing on the interval e 8, + we have b m a m G log4 p p log p + log 2 log α 2 p α log p + log α p log p + + log 2/ log p α 3/2 9 + log α/ log p2 + p log p

10 + 2 + log 2 log α 2 α log p + log α p log p 0.0 p log p p log p p log p p log p 3 p log p p e Next, if e then we have b m 0.55 log2 p 0 p N 0.0 p log p p e Algorithm Tables for Sequence {C m } {R m } The table shows the values of C m = Φ 0 I m R m to ωn = m for n N. There are only values of C m R m for m 0 here. But it is not difficult to verify them for 3 p m e 4. Note, if more informations, then it should be taken R m < 0, not C m <, for 263 p m e 4, by reason of the limited values of matlab 6.5. The table 2 shows the values R m for 9309 m 938. Of course, all the values in the table the table 2 are approximate. The algorithm for R m to ωn = m by matlab is as follows: Function ETF-Index, clc, gamma= ; format long P = [2, 3, 5, 7,, ]; M=lengthP; for m = : M; p = P : m; q =./p; F = gamma + logprod./q; N = sumlogp.; N2 = N /2 ; N3 = logn 2 ; N4 = N2 N3; N5 = N + N4; m, pm, C m = expexpexpf / expn/ expn4, R m = F loglogn5, end T able m pm C m R m e e e e e T able 2 m pm R m

11 IV. Lemma 2 for Proof of Condition d In the middle of the proof of the Lemma, we used the condition d. So we here would prove the following Lemma 2, which shows that the condition d holds. [Lemma 2] For any prime number p 3 we have We make ready for the proof of the Lemma 2. d p gp p A Condition d If the Lemma 2 does not hold, then there exists a prime number p 3 such that d p gp p > We fix such prime p. Then from the table 3 the table 4 we see p e 4, because H m < 0 for any 3 p m e 4 see 4.36 below. Now we define the function Gt := d t gt t, t p, p + 2. Then G t = 0 t t log t logt α D t, t = Et + ft 4 t 8 log 2 t α 4 + ft logt α 2 t + f t + 4 logt α D t = d t gt = f t dt g t. Hence G t < 0 is equivalent to 2 0 t+ + D t < + logt α log t. Since t e 4 α /4 by 3., we get so by 3.7, t logt α 2 logt α logt α = log t + log α > 0 2 D t < logt α log 2 t + 2 log t + log t + 4 log t logt α 4 + logt α log t log t + log 2 t log t + < 0.46 t e

12 This shows that the function Gt is decreasing on the interval p, p + 2. Thus there exists a point t such that p < t < p + Gp + = Gp + Gp + Gp = = Gp + G t p + p > 2 2 p, Gp := Gp + 0 = lim Gt = t p+0 d p gp p. For the convenient discussion, we put x = p, x 2 = p +. Then, since Gt Gp +, for any t x, x 2 we have d t gt t > 2 / t. 4.5 We will call 4.5 the condition d. For the proof of the lemma 2, we must obtain a certain contradiction from the condition d Proof of d t < 0 For any t x, x 2 we here have d t < 0. In fact, since d t = t gt 0 t log t we easily see that d t < 0 is equivalent to 0 t < + / log t, 0 t < t e 4., 4.3. Function F t F t Put F t := d 2 dt g t gt g d t, t x, x 2. Then it is clear x2 x F t dt = 0, 4.6 g = gx, d 2 = dx 2. Hence there exists a point ξ 0 such that x < ξ 0 < x 2 x2 x F t dt = F ξ 0 x 2 x = 0 so d 2 dξ 0 g ξ 0 = gξ 0 g d ξ We here have F t < 0 for any t x, x 2 under the condition d. In fact, since d t > 0 from the condition d, for F t < 0 it is sufficient to show gt g d t < 2 d t g t. And there exists a point t such that x < t < t gt gx = g t t x g t 2

13 Hence g t g x g.0 g t t e 4. x 2 gt g d t.0 g t + t gt log t 0t 2 / t g t t gt 2 d t g t t e Moreover, we note F t > 0 holds for any t x, x An estimate of the point ξ 0 From x2 x F t dt = ξ0 x F t dt + x2 there exist λ, λ 2 such that x < λ < ξ 0 < λ 2 < x 2 Then, since F t < 0 t x, x 2, we have ξ 0 F t dt = 0, 4.9 F λ ξ 0 x + F λ 2 x 2 ξ 0 = F λ > F ξ 0 = 0 > F λ 2. Put x 0 := x + x 2 ξ 0. Then from 4.0 we have F λ x 2 x 0 + F λ 2 x 0 x = This 4. shows that the line passing the points x, F λ x 2, F λ 2 passes the point x 0, 0. On the other h, by the mean value theorem, there exist the points η η 2 such that x < η < ξ 0 < η 2 < x 2 dx 2 dξ 0 = d η 2 x 2 ξ 0, gξ 0 gx = g η ξ 0 x. Since the function g t is decreasing on x, x 2 from the condition d, we have g ξ 0 g η, d t > 0 t x, x 2. Here if x 0 ξ 0, then x 2 ξ 0 ξ 0 x by 4.7 we get d ξ 0 = d η 2 x2 ξ 0 ξ 0 x g ξ 0 g η d η 2, but it is a contradiction to d t < 0. Thus we have x 0 ξ 0 > 0 under the condition d An estimate of ε 0 := x 0 ξ 0 Since F ξ 0 = 0, we have F x = d 2 d g x = F x F ξ 0 = F β 0 ξ 0 x, x < β 0 < ξ 0. Also since ξ 0 x = ε 0 /2 F t = d 2 dt g t gt g d t 2 d t g t, 4.2 3

14 we have d β 0 g β 0 ε 0 = T + T 2, T = d β 0 g β 0 d 2 d g x, T 2 = d 2 dβ 0 g β 0 gβ 0 g d β 0 ξ 0 x. By the condition d, we get d β 0 g β 0 ε 0 2 / β 0 g β 0 gβ 0 β 0 T = d β 0 g β 0 d 2 d g x = = d β 0 g β 0 d β g x = ε 0 ε 0 x 2 x 2 = d β g β 0 β 0 β + d β g β 0 β 0 x a 2 + b x 2 x < β < x 2, β 0 < β < β, x < β 0 < β 0 x e 4, a 2 := d t g t 2, b 2 := d t g t 2 see below. We also have T 2 = d 2 dβ 0 g β 0 gβ 0 g d β 0 ξ 0 x = = d β 2 g β 0 x 2 β 0 g β 2 d β 0 β 0 x ξ 0 x β 0 < β 2 < x 2, a 2 + b 2 / x 2 2 x < β 2 < β 0. Thus we have x 2 e 4, 0 < ε x x 2 e An estimate of δ 0 := λ 0 x 0 Here a point λ 0 is determined as follows. If the line passing the points λ, F λ λ 2, F λ 2 intersects the line y = 0 at the point λ 0, then we obtain F λ λ 2 λ 0 + F λ 2 λ 0 λ = Since the function F t is decreasing convex on x, x 2 under the condition d, it is clear ξ 0 < λ 0. And the equation of the line passing ξ 0, F λ λ 0, 0 is one passing ξ 0, F λ 2 x 0, 0 is F λ x + λ 0 ξ 0 y F λ λ 0 = F λ 2 x + x 0 ξ 0 y + F λ 2 x 0 =

15 We put 0 := F λ x 0 ξ 0 + F λ 2 λ 0 ξ 0 0, := F λ F λ 2 δ 0. Then y-coordinate of the cross point of above two lines is / 0. Here if δ 0 > 0 then / 0 < 0 < 0, if δ 0 < 0 then / 0 > 0. > 0. Hence we always have 0 > 0 From this And since we have so F λ x 0 ξ 0 + F λ 2 λ 0 ξ 0 > F λ 2 δ 0 < F λ + F λ 2 x 0 ξ 0. F λ + F λ 2 F λ 2 x 2 x 0 = ε 0, 2 x 0 x = + ε 0, 2 = x 0 x x 2 x 0 = 2 ε 0 ε 0 δ 0 < 2 ε 2 0/ ε 0. Similarly, for the lines passing the points ξ 0, F λ, x 0, 0 the points ξ 0, F λ 2, λ 0, 0, we have F λ λ 0 ξ 0 + F λ 2 x 0 ξ 0 > 0 from this Therefore we have δ 0 > 2 ε 2 0/ + ε 0. 2 ε ε 0 < δ 0 < 2 ε2 0 ε An estimate of δ := λ λ Here λ := x + ξ 0 λ. By the same method as in δ 0, for the lines passing x, F λ, λ, 0 x, F λ 2, λ, 0 we have F λ λ x + F λ 2 λ x > 0. From this, since F λ + F λ 2 F λ = 2 ε 0 + ε 0, λ x = 4 ε δ, we have δ < ε 0 /2. Also for the lines passing the points x, F λ, λ, 0 x, F λ 2, λ, 0 we have F λ λ x + F λ 2 λ x > 0. so δ > ε 0 /2. Therefore we get ε0 2 < δ < ε

16 4.8. An estimate of ω 0 := λ + λ 2 2 ξ 0 Let λ 0 := λ + λ 2 λ 0. Then from we get Hence Put ω := λ + λ 2 x + x 2. Then, since F λ F λ 2 = x 2 ξ 0 ξ 0 x = λ 0 λ λ 2 λ 0 = λ 2 λ 0 λ 0 λ λ 0 = λ ξ 0 x + λ 2 x 2 ξ 0, λ 0 = λ x 2 ξ 0 + λ 2 ξ 0 x. λ 0 λ 0 = λ 2 λ ε 0 λ + λ 2 = λ 0 + λ 0 = 2 λ 0 λ 0 λ 0, we have Thus x + x 2 = x 0 + ξ 0 = 2 x 0 x 0 ξ 0, ω = 2 δ 0 + λ 2 λ ε 0. ω 0 = λ + λ 2 2 ξ 0 = ω + ε 0 = = 2 δ ε 0 λ 2 λ ε An estimate of Λ 0 := λ 2 λ Since ξ 0 λ = 4 ε 0 2 δ, we have λ 2 λ = 2 λ 0 λ λ 0 λ 0 = = 2 δ ε ε 0 δ λ 2 λ ε 0, hence λ 2 λ = 2 + ε 0 + ε 0 δ + ε δ 0 + ε 0 2 < λ 2 λ < ε An estimate of δ 2 := λ η By the same method as above, for the lines passing the points x, F λ, λ, 0 x, F λ 2, η, 0 we have F λ η x + F λ 2 λ x > 0. From this δ 2 < F λ + F λ 2 F λ λ x = 6

17 = 2 ε 0 + ε 0 4 ε δ < 2 ε 0. + ε 0 Similarly, for the lines passing the points x, F λ, η, 0 x, F λ 2, λ, 0 we have F λ λ x + F λ 2 η x > 0. From this Consequently, we have δ 2 > F λ + F λ 2 F λ 2 λ x > 2 ε 0 ε 0. ε 0 2 ε 0 < δ ε 0 2 < 2 + ε An estimate of Q 0 := η 2 + η 2 ξ 0 ξ 0 x First, we will find the lower bound of Q 0. If the line passing η, F η, η 2, F η 2 intersects y = 0 at η 0, then from this From 4.0 we have F η η 2 η 0 + F η 2 η 0 η = 0 F λ η 2 η 0 + F λ 2 η 0 η + W 0 = 0, W 0 = F η F λ η 2 η 0 + F η 2 F λ 2 η 0 η. η 0 = η + η 2 η x 2 ξ 0 + W, W = x 2 ξ 0 F λ W 0. Since F t is convex on x, x 2 η < ξ 0 < η 2, we have η 0 > ξ 0 From this Here if W < 0 then η + η 2 η x 2 ξ 0 + W > ξ 0. Q 0 > η 2 ξ 0 ε 0 W. Q 0 > ε 0 if W > 0 then we put η 0 := η + η 2 η 0. Then by same way as above F λ η 0 η + F λ 2 η 2 η 0 + W 0 = 0 η 0 = η + η 2 η ξ 0 x W. Thus we have η 0 η 0 = η 2 η ε W 7

18 so η 0 η 0 > 0. Similarly, for the lines passing the points λ, F λ, η 0, 0 λ, F λ 2, η 0, 0 we have On the other h, since η 2 > η 0 F λ η 0 λ + F λ 2 η 0 λ > 0. ξ 0 λ < 4, we have η 0 η 0 < F λ + F λ 2 F λ η 0 λ < Hence < 2 ε 0 + ε 0 η 0 ξ 0 + ξ 0 λ < < 2 ε 0 η 2 ξ 0 + ε W = η 0 η 0 η 2 η ε 0 < < 2 ε 0 η 2 ξ 0 + ε 0 2 η 2 ξ 0 ε 0 ξ 0 η ε 0 ε 0 η 2 ξ 0 + ε 0 2 ξ 0 λ ε 0 δ 2 ε 0 = = ε 0 η 2 ξ 0 + ε 0 2 ε 0 4 ε 0 2 δ δ 2 ε 0 < < ε 0 η 2 ξ 0 + ε ε2 0, since x 2 > η 2 x 2 ξ 0 = + ε 0 /2, Q 0 > η 2 ξ 0 ε 0 W > > η 2 ξ 0 ε 0 ε 0 2 η 2 ξ 0 ε 0 8 ε2 0 > > 3 2 ε 0 x 2 ξ 0 ε 0 8 ε2 0 = = 3 4 ε 0 + ε 0 ε 0 8 ε2 0 > ε 0. Therefore, generally, we have Q 0 = η 2 + η 2 ξ 0 ξ 0 x > ε Next, we will find the upper bound of Q 0. It is easy to see that η 2 + η 2 ξ 0 = η 2 ξ 0 ξ 0 η < < x 2 ξ 0 ξ 0 η = = 2 + ε 0 4 ε 0 2 δ + δ 2 < < ε 0 8

19 2 ξ 0 η 2 η = ξ 0 η η 2 ξ 0 < < ξ 0 η < ε 0. Therefore, since η 2 + η 2 ξ 0 ξ 0 x < < ε 0 2 ε 0 < 8 + ε 0, we have Q 0 < 8 + ε An estimate of H 0 := ξ 0 η ξ 0 x Since ξ 0 η = 4 ε 0 2 δ + δ 2, we easily have 4.3. A new equality Now we add , then we have both sides multiply by d η 2 g ξ 0, then On the other h, since F ξ 0 = 0, we get ξ 0 x = 2 ε 0, 8 ε 0 < H 0 < 8 + ε F λ + F λ 2 = F λ F λ 2 ε 0 F λ + F λ 2 d η 2 g ξ 0 = = F λ F λ 2 d η 2 g ξ 0 ε F λ + F λ 2 = F λ 2 F ξ 0 F ξ 0 F λ = = F α 2 λ 2 ξ 0 F α ξ 0 λ = = F α 2 F α ξ 0 λ + F α 2 λ 2 ξ 0 ξ 0 λ = = F τ 0 α 2 α ξ 0 λ + F α 2 λ + λ 2 2 ξ 0 F λ F λ 2 = F λ F ξ 0 F λ 2 F ξ 0 = = F α ξ 0 λ F α 2 λ 2 ξ 0 = = F α 2 F α ξ 0 λ F α 2 λ 2 ξ 0 + ξ 0 λ = = F τ 0 α 2 α ξ 0 λ F α 2 λ 2 λ, 9

20 λ < α < ξ 0 < α 2 < λ 2 α < τ 0 < α 2. As mentioned in the section 3.4, there exist the points η η 2 such that x < η < ξ 0 < η 2 < x 2 d 2 dξ 0 g ξ 0 gξ 0 g d ξ 0 = = d η 2 g ξ 0 x 2 ξ 0 g η d ξ 0 ξ 0 x = = d η 2 g ξ 0 g η d ξ 0 ξ 0 x + +d η 2 g ξ 0 x 0 ξ 0 = from 4.7. Here also there exist µ, µ 2 such that η < µ < ξ 0 < µ 2 < η 2 We put Then since by 4.2 we have Thus the left side of 4.26 is d η 2 g ξ 0 g η d ξ 0 = = d η 2 d ξ 0 g ξ 0 + g ξ 0 g η d ξ 0 = = d µ 2 g ξ 0 η 2 ξ 0 + g µ d ξ 0 ξ 0 η = = d µ 2 g ξ 0 + g µ d ξ 0 ξ 0 η + +d µ 2 g ξ 0 η 2 + η 2 ξ A 0 := d η 2 g ξ 0, A := d µ 2 g ξ 0 + g µ d ξ 0, A 2 := d µ 2 g ξ 0, B 0 := d α 2 g α 2, B := d 2 dα 2 g α 2 gα 2 g d α 2, U 0 := d τ 0 g τ 0 + d τ 0 g τ 0, U := d 2 dτ 0 g τ 0 gτ 0 g d τ 0. F t = d 2 dt g t gt g d t 3 d t g t + d t g t, 4.29 F α 2 = B 2 B 0, F τ 0 = U 3 U 0, M 0 := A 0 ε 0 = A ξ 0 η ξ 0 x + +A 2 η 2 + η 2 ξ 0 ξ 0 x = O/t L 0 := F λ + F λ 2 d η 2 g ξ 0 = L + L 2, L = F τ 0 d η 2 g ξ 0 α 2 α ξ 0 λ, L 2 = F α 2 d η 2 g ξ 0 λ + λ 2 2 ξ 0. 20

21 And L = L L 2, L = U A 0 α 2 α ξ 0 λ = O/t 4, L 2 = 3 U 0 A 0 α 2 α ξ 0 λ = O/t 3. Also L 2 = L 2 L 22, L 2 = 2 A 0 B δ 0 4 A 0 B 0 δ 0 + +B 2 λ 2 λ M 0 = O/t 4, L 22 = 2 2 λ 2 λ B 0 M 0 = O/t 3 Similarly, the right side of 4.26 is R 0 := F λ F λ 2 M 0 = R R 2, R = F τ 0 M 0 α 2 α ξ 0 λ = = U 3 U 0 M 0 α 2 α ξ 0 λ = O/t 4 And R 2 := F α 2 M 0 λ 2 λ = R 2 R 22, R 2 = B M 0 λ 2 λ = O/t 4, R 22 = 2 B 0 M 0 λ 2 λ = O/t 3. Thus 4.26 is equivalent to L 0 = R 0, that is, we have a new equality L + L 2 L 2 + L 22 = R R 2 R A new inequality Put K := L 2 + L 22 + R 22, K 2 := L + L 2 R + R 2, then, from the equation 4.26, we have K = K 2. And by M 0 we have K = A 0 + K + K 2, A 0 = A B 0 ξ 0 η ξ 0 x, K = 3 U 0 A 0 α 2 α ξ 0 λ + +3 A B 0 ξ 0 η ξ 0 x, K 2 = 4 A 2 B 0 η 2 + η 2 ξ 0 ξ 0 x. By the condition d we have d α 2 g α 2 d η 2 g ξ 0 + x 2 x 2 a 2 + b 2 2

22 x x e 4 M := B 0 ε 0 = d α 2 g α 2 d η 2 g ξ 0 M 0 = O/t 2. And also we see A 2 > 0, M > 0. Thus we get K 2 K 2 := 4 A 2 M = O/t 4. It is clear that d t > 0 by the condition d g t > 0, g t < 0, d t < 0 for any t x, x 2, so we have A 0 > 0, A > 0, B 0 > 0, U 0 < 0. On the other h, hence by 4.25, We also have A 0 Ω 0 A, From this we have a new inequality α 2 α ξ 0 λ < λ 2 λ ξ 0 λ < < Λ 0 4 < 8 + ε 0 K K + K, K = 3 8 U 0 A 0 + A B 0 = O/t 4 see below, K = 3 ε 0 U 0 A 0 A B 0 = O/t 4, Ω 0 = 8 A B 0, A = A M = O/t 4. Ω 0 Ω := L +L 2 +R 2 R K 2 +K +K +A Now we intend to obtain the estimates for the lower bound of Ω 0 the upper bound of Ω respectively Lower bound of Ω 0 := A B 0 /8 Using the condition d, we get B 0 = d α 2 g α 2 x 2 x 2 g µ d ξ x 2 2 x 2 x 2 e 4 d µ 2 g ξ 0 = g ξ 0 + µ 2 gµ 2 0 µ 2 log µ 2 g x x 2 gx 2 x 2 2 x 2 e 4. 22

23 Thus we have Put Ω x x 3 2 x x 3 x x e 4. Ω 0 := x 3 x x e Some estimates For any t x, x 2 it is easy to see that g t = log2 t α 2 t 4 +, logt α And it is also clear that g t = log2 t α 4 t t By , we respectively have 8 log 2 t α, g t = log2 t α 3 t 2 t 8 2 logt α 3 log 2 t α d t gt = f t dt g t, d t gt = f t dt g t 2 d t g t, d t gt = f t dt g t 3 d t g t + d t g t. f t = + log t Et, f t = t Et, log t f t = t 2 + log 2 t Et. D 0 t := log t Et θt 2 t e 4, log t G t := g t gt = 2 t + 4 logt α x e 4, x D t := d t gt + log t Et + t D 0 t G t G 2 t := g t gt t e 4, = 4 t 2 8 log 2 t α 0.26 x 2 x e 4, D 2 t := d t gt f t + t D 0 t G 2 t + 2 D t G t.3307 x x e 4, 23

24 G 3 t := g t gt = t logt α 3 log 2 t α x 3 x e 4 D 3 t := d t gt f t + t D 0 t G 3 t+ +3 D 2 t G t + D t G 2 t x 2 x e 4. From this for any t, t x, x 2 we have r 0 := gt gt + g t 0 gt t < t 0 < t, By 4.24, we get a := d t g t = r 0 D t G t x x e 4, a 2 := d t g t = r 0 D t G 2 t x 2 a 3 := d t g t = r 0 D t G 3 t x 3 b 2 := d t g t = r 0 D 2 t G t x 2 b 3 := d t g t = r 0 D 3 t G t x 3 b := d t g t = r 0 D 2 t G 2 t x 3 M 0 A + A 2 Q 0 a b 2 Q x 2 x e 4 M M x x 2 x e 4. x e 4, x e 4, x e 4, x e 4, x e Upper bound of Ω The upper bound of Ω is obtained as follows. First we will show the estimate of K = 3 8 U 0 A 0 + A B 0. Since we have U 0 A 0 + A B 0 = U 0 + A A 0 A 0 B 0 A, U 0 + A = d τ 0 g τ 0 d µ 2 g ξ d τ 0 g τ 0 g µ d ξ 0 = 24

25 = d τ 0 g τ 0 g ξ 0 + d τ 0 d µ 2 g ξ d τ 0 d ξ 0 g τ 0 + g τ 0 g µ d ξ 0 b + b 3 + b + a 3 A 0 B 0 = d η 2 g ξ 0 d α 2 g α 2 = = d η 2 d α 2 g ξ g ξ 0 g α 2 d α 2 < more Thus we have < a 2 + b 2 A 0 = d η 2 g ξ 0 < a, A = g µ d ξ 0 + d µ 2 g ξ 0 < < a 2 + b 2. K 3 a 2 + b a a 3 + b a b Next, we would show the estimate of x 4 x e 4. L 2 = 2 A 0 B δ 0 4 A 0 B 0 δ 0 + Since we have And since +B 2 λ 2 λ M 0 = O/t 4. A 0 ε 0 = M 0, B 0 ε 0 = M B = d 2 dα 2 g α 2 gα 2 g d α 2 < < a 2 + b 2 2 < Λ 0 = λ 2 λ < ε 0 < , L 2 4 ε 0 ε 0 a 2 + b 2 M ε 0 M 0 M a 2 + b 2 M x 4 x e 4. V 0 := α 2 α ξ 0 λ < 8 + ε 0 <

26 we also have U 3 U 0 a 3 + b a 2 + b 2, R a 3 + b a 2 + b 2 M 0 V 0 Similarly, we have respectively x 4 x e 4. K 3 a 2 + b 2 M 0 + M.207 x 4 x e 4, Consequently we obtain L a 3 + b 3 a V x 4 x e 4, R 2 a 2 + b 2 M 0 λ 2 λ x 4 x e 4, A a 2 + b 2 M x 4 x e 4, K 2 4 b 2 M x 4 x e 4. Ω K + L 2 + R + K + L + R 2 + A + K x x x x x x x x x 4 x e Proof of Lemma 2 If 3 p m e 4, then we could confirm that the condition d holds from the table 3 the table 4. Hence if the Lemma 2 does not hold, then there exists a prime number p m e 4 such that the condition d holds. For such prime p m, we have Ω 0 Ω, finally, we get x 0.08 x e 4, 4.35 but it is a contradiction. This shows that the condition d is not valid. Consequently, the Lemma 2 holds for any prime number p m Algorithm Tables for Sequence {H m } Here H m := + log p m Ep m dp m log2 p m α 2 p m 4 + logp m α 2 pm

27 α = + θp m. The table 3 4 show the values of H m for 2 p m p m 938. Note that the condition d holds if only if H m < 0 for any m 2. It is easy to see that H m < 0 for any 29 p m The algorithm for H m by MATLAB is as follows: Function EMF-Index, clc, b= ; format long, P = [2, 3, 5, 7,, ]; M=lengthP; for m = : M; p = P : m; E = sum./p b loglogpm; E = +logpm E; V = sumlogp.; Q = V /pm ; R = pm /2 ; V = logv ; g = R V 2 ; f = pm logpm E Q; d = f/g; B = V 2 + 4/V /2/R; m, pm, H m = E d B 2/R, end. T able 3 T able 4 m pm H m m pm H m References [] Tom M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, New York, Heidelberg Berlin, 976. [2] J. C. Lagarias, An elementary problem equivalent to the Riemann hypothesis,amer. Math. Monthly , [3] J. L. Nicolas, Small values of the Euler function the Riemann hypothesis, arxiv: v2 [math.nt] 5 Apr 202. [4] G. Robin, Gres valeurs de la fonction Somme des diviseurs et hypothese de Rimann, Journal of Math. Pures et appl , [5] J. B. Rosser, L. Schoenfeld, Approximate formulas for some functions of prime numbers, IIlinois J. Math , [6] J. Sor, D. S. Mitrinovic, B. Crstici, Hbook of Number theory, Springer, [7] P. Sole, M. Planat, Robin inequality for 7-free integers, arxiv: v [math.nt] 3 Dec 200. [8] YoungJu Choie, Nicolas Lichiardopol, Pieter Moree, Patrick Sole, On Robin s criterion for the Riemann hypothesis, Journal de Theorie des Nombers de Bordeaux , Mathematics Subject Classification; M26, N05. 27