Riemann Hypothesis Euler Function Choe Ryong Gil April 5, 207 Department of Mathematics, University of Sciences, Unjong District, Pyongyang, D.P.R.Korea, Email; ryonggilchoe@star-co.net.kp 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 = 0.577 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 + 2.483..4 log n
II. Proofs of Theorem Theorem 2 2.. 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. 2.2. 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. 2.3. 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 = 9.6680, expexpe γ 2 3/2 2 3 exp log2 3 log log2 3 2 = 23.56, expexpe γ 2 3/2 5/4 2 3 5 exp log2 3 5 log log2 3 5 2 = 7.7386, 2
expexpe γ 2 3/2 5/4 7/6 C 4 = 2 3 5 7 exp log2 3 5 7 log log2 3 5 7 2 = 0.837. 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} 24. 2.3 2.4. 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 + 2.483 + 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 = 0.2649722 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
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. 3.2. 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 2 3.5 by 3.30 of [5], we respectively have e <.0052 p e 4, e <.075 p e 4, e e <.08 p e 4. 3.6 3.3. 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 3.7 3.20 of [5]. Hence, since εp < 0, if e >, then 0 < r := Ep + εp < log 2 p 0.0052 p e4 e = + r + n=2 r n n! + r + r 2 2 r + r + 0.503 r2, 4
Therefore we have e e h 2 = expr + log p e + h + 2 h, h = + log p r + 0.503 log p r 2 0.25 p e 4. e e + log p Ep + εp+ +0.6 + log p 2 Ep + εp 2 e >, p e 4. 3.8 3.4. 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 + 0.503 log p p, e >, p e 4 µ e e e +0.55 log p p. e >, p e 4. 3.0 3.5. 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
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 2 + + 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 3/2 4 +. 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 3.5 3.6 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 2 4 + 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 4. 3.2 + dt p log t t log t + det 6
log p Ep + log p + p t log 4 t dt log p + log 3 p 3 log 3 t + p = = log p + 4 3 log 3 p 3.3 Sp log p 4 3 log 3 p. If p is a first prime e 4, then p = 202609 it is 938-th prime. And we have 3.7. A Condition d Put 0.070 Sp 0.072. 3.4 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 3 3.8 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 4 + + 2. 3.9 logp α p 7
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 5. 3.20 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 < 0. 3.22 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 = 202609 then we have D 938 = 0.00 0.06 3 Sp 0.09 <. 3.23 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. 3.25 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
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. 3.26 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 2 4 + + log p p e 4, α /4. logp α Thus we see +log p Ep+εp a m log2 p α 2 p α 4 + + 2. 3.27 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 2 + 2 logp α + 2 2 α log 2 p α 0.4287 log4 p α p α p e 4 3.28 log p 0 µ e a m N 0 N log p 0 + log p Ep + εp a m N 0 N + 0.55 log2 p 0 p a m G 0 N 0 + 0.55 log2 p 0 p + 0.6 log p 0 + log p 2 Ep + εp 2 + 0.2572 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 0 + 0.55 +0.2572 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
+ 2 + log 2 log α 2 α log p + log α p log p 0.0 p log p + 0.0 p log p + 0.42 p log p + 2.203 p log p 3 p log p p e4. 3.29 Next, if e then we have b m 0.55 log2 p 0 p N 0.0 p log p p e4. 3.30 3.9. 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=0.577256649053286060; format long P = [2, 3, 5, 7,, 202609]; 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 2 9.668063388849 2 3 23.56879826350 0.73259862957209 3 5 7.73864609733096 0.4633620860732 4 7 0.837792006862 0.00636499588 5 0.0428273904 0.09308687002330 6 3.029966548700e 004 0.2730939385590 7 7 3.8342599453073e 007 0.50773685433 8 9.39756045763582e 009 0.59609230879 9 23 2.82898264763264e 02 0.6627880559 0 29 2.085428922468e 05 0.745284347098 T able 2 m pm R m 9309 202477 0.0547993387 930 202483 0.054786567870 93 202497 0.0547820949 932 20250 0.054775835370 933 202507 0.054770468282 934 202549 0.05476503339 935 20256 0.054759738330 936 202569 0.054754372957 937 202603 0.0547490094 938 202609 0.05474364485 0
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 2. 4.. A Condition d If the Lemma 2 does not hold, then there exists a prime number p 3 such that d p gp p > 2. 4. 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 + + 2 logt α D t, 4.2 0 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, + + + 0 t + + 4 logt α 2 logt α logt α = log t + log α 3.925 > 0 2 D t < logt α log 2 t + 2 log t + log t + 4 log 2 + + 2 t logt α 4 + logt α log t log t + log 2 t + 2 4.3 log t + < 0.46 t e 4. 4.4
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. 4.2. 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 0.2577 < 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 0. 4.7 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
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 4. 4.8 Moreover, we note F t > 0 holds for any t x, x 2. 4.4. 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 = 0. 4.0 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 = 0. 4. 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. 4.5. 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
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 2 0.9005 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 /2 0.4503 x 2 2 x < β 2 < β 0. Thus we have x 2 e 4, 0 < ε 0.3508 x2 0.005 x 2 e 4. 4.3 4.6. 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 = 0. 4.4 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 = 0 4.5 F λ 2 x + x 0 ξ 0 y + F λ 2 x 0 = 0. 4.6 4
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 > 0. 4.7 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 2 0 + ε 0 < δ 0 < 2 ε2 0 ε 0. 4.8 4.7. 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 0 + 2 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 0 2. 4.9 5
4.8. An estimate of ω 0 := λ + λ 2 2 ξ 0 Let λ 0 := λ + λ 2 λ 0. Then from 4.0 4.4 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 0 + 2 ε 0 λ 2 λ ε 0. 4.20 4.9. 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 0 + 2 ε 0 + 2 ε 0 δ λ 2 λ ε 0, hence λ 2 λ = 2 + ε 0 + ε 0 δ + ε 0 + 2 δ 0 + ε 0 2 < λ 2 λ < 2 + 2 ε 0. 4.2 4.0. 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
= 2 ε 0 + ε 0 4 ε 0 + 2 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 0. 4.22 4.. 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 0 + 2 W 7
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 0 2. 2 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 0 4 + 2 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 0. 4.23 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 < < 4 + 2 ε 0 8
2 ξ 0 η 2 η = ξ 0 η η 2 ξ 0 < < ξ 0 η < 4 + 2 ε 0. Therefore, since η 2 + η 2 ξ 0 ξ 0 x < < 4 + 2 ε 0 2 ε 0 < 8 + ε 0, we have Q 0 < 8 + ε 0. 4.24 4.2. 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 4.0 4., 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 0. 4.25 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 0. 4.26 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
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 = 0 4.27 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 0. 4.28 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 2. 4.30 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
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 22. 4.3 4.4. 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
+ 0.9005 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. 4.32 Now we intend to obtain the estimates for the lower bound of Ω 0 the upper bound of Ω respectively. 4.5. 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 0 0.4789 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 2 0.37 0.2577 x 2 gx 2 x 2 2 x 2 e 4. 22
Thus we have Put Ω 0 0.4789 8 x 4 + 0.37 2 x 3 2 x 0.0462 2 x 3 x x e 4. Ω 0 := 0.0462 x 3 x x e 4. 4.33 4.6. 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 3.7 3., 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 0.6437 x e 4, x D t := d t gt + log t Et + t D 0 t G t G 2 t := g t gt 0.685 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
G 3 t := g t gt = t 3 3 8 2 logt α 3 log 2 t α 0.375 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 3.7650 x 2 x e 4. From this for any t, t x, x 2 we have r 0 := gt gt + g t 0 gt.00000 t < t 0 < t, By 4.24, we get a := d t g t = r 0 D t G t 0.085 x x e 4, a 2 := d t g t = r 0 D t G 2 t 0.0439 x 2 a 3 := d t g t = r 0 D t G 3 t 0.0633 x 3 b 2 := d t g t = r 0 D 2 t G t 0.8566 x 2 b 3 := d t g t = r 0 D 3 t G t 2.4236 x 3 b := d t g t = r 0 D 2 t G 2 t 0.3460 x 3 M 0 A + A 2 Q 0 a 2 + 2 b 2 Q 0 0.2223 x 2 x e 4 M M 0 + 0.9005 0.2225 x x 2 x e 4. x e 4, x e 4, x e 4, x e 4, x e 4. 4.7. 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 0 + + d τ 0 g τ 0 g µ d ξ 0 = 24
= d τ 0 g τ 0 g ξ 0 + d τ 0 d µ 2 g ξ 0 + + 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 0 + + 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 2 8 2 + a a 3 + b 3 + 2 a b Next, we would show the estimate of 0.4334 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 < 2 + 2 ε 0 < 0.5034, L 2 4 ε 0 ε 0 a 2 + b 2 M 0 + 8 ε 0 M 0 M + + 3 2 a 2 + b 2 M 0 0.6978 x 4 x e 4. V 0 := α 2 α ξ 0 λ < 8 + ε 0 < 0.267 25
we also have U 3 U 0 a 3 + b 3 + 3 a 2 + b 2, R a 3 + b 3 + 3 a 2 + b 2 M 0 V 0 Similarly, we have respectively 0.462 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 0 0.0342 x 4 x e 4, R 2 a 2 + b 2 M 0 λ 2 λ 0.008 x 4 x e 4, A a 2 + b 2 M 0.2004 x 4 x e 4, K 2 4 b 2 M 0.7625 x 4 x e 4. Ω K + L 2 + R + K + L + R 2 + A + K 2 0.4334 x 4 + 0.0342 x 4 + 0.6978 x 4 + 0.008 x 4 3.5770 x 4 + 0.462 x 4 + 0.2004 x 4 +.207 x 4 + + 0.7625 x 4 x e 4. 4.34 4.8. 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 3.5770 0.0462 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 3. 4.9. 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 4.36 26
α = + θp m. The table 3 4 show the values of H m for 2 p m 29 9309 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 9309. The algorithm for H m by MATLAB is as follows: Function EMF-Index, clc, b=0.2649722847643; format long, P = [2, 3, 5, 7,, 202609]; 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 2 4.927858770647 2 3 3.797088793795 3 5.8208402562472 4 7.242404597362 5.058929097784 6 3 0.8237742885520 7 7 0.75298886049588 8 9 0.60562832793 9 23 0.56797602737022 0 29 0.59342397038654 m pm H m 9309 202477 0.006950356769 930 202483 0.006879007336 93 202497 0.0068788503420 932 20250 0.0067897043350 933 202507 0.00678356669 934 202549 0.0069673633799 935 20256 0.0069494084824 936 202569 0.0068958602998 937 202603 0.007073666036 938 202609 0.0070023228562 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 09 2002, 534-543. [3] J. L. Nicolas, Small values of the Euler function the Riemann hypothesis, arxiv: 202.0729v2 [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. 63 984, 87-23. [5] J. B. Rosser, L. Schoenfeld, Approximate formulas for some functions of prime numbers, IIlinois J. Math. 6 962, 64-94. [6] J. Sor, D. S. Mitrinovic, B. Crstici, Hbook of Number theory, Springer, 2006. [7] P. Sole, M. Planat, Robin inequality for 7-free integers, arxiv: 02.067v [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 9 2007, 357-372. 200 Mathematics Subject Classification; M26, N05. 27