An upper bound for error term of Mertens formula

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1 An upper bound for error term of Mertens formula Choe Ryong Gil June 21, 2017 Abstract: In this paper, it is obtained a new estimate for the error term Et of Mertens formula p t p 1 = log log t + b + Et, t > 1 is a real number, p is the prime number b is the well-known Mertens constant. We, first, provide an upper bound, not a lower bound, of Ep for any prime number p 3, next, give one in the form as Et < log t/ t for any real number t 3. This is an essential improvement of already known results. Such estimate is very effective in the study of the distribution of the prime numbers. Keywords: Mertens formula; Chebyshev s function; Riemann Hypothesis. I. Introduction The formula p 1 = log log t + b + Et 1.1 p t is called the Mertens formula, t > 1 is a real number, p is the prime number, b = γ + p log1 1/p + 1/p = is the Mertens constant γ = is the Euler constant [1, 2]. As usual, we here will call Et the error term of the Mertens formula 1.1. Much papers have been contributed in estimating the orders of the magnitude of Et in the various approximations. It is already well-known that 1 Et = O. 1.2 log t This 1.2 could be found in many books for instance, see [1. 2]. A classic epochal result appeared as the Theorem 23 on p. 65 of Ingham [3] see also on p. 66 of [4] in the form πt = p t 1 = t 2 du log u +Ot exp a log t a > 0, 1.3 a is a positive absolute constant. And the improvement form of the Ingham s result was given in Vinogradov [6] see also on p. 229 of [5] as πt = t 2 du log u + Ot exp a log t3/

2 From by the Abel s identity [1], it is easily given more Et = Oexp a log t 1.5 Et = Oexp a log t 3/ It is obvious that the inverse process is also possible. Of course, these are not the best possible results. It is well-known that the Riemann Hypothesis RH[2] is equivalent to that πt = t 2 du log u + O t log t. 1.7 holds see the Equivalence 5.5 on p. 47 of [2]. Therefore the best one for Et is as the form log t Et = O. 1.8 t In deed, the 1.8 is another one of the forms equivalent to the RH. Unlike above such estimates, Rosser Schoenfeld showed, practically it is very useful, the widely applicable approximations in of [4], 1 log 2 t < Et < 1 log 2 t t > We recall the Chebyshev s function ϑt = p t log p [1]. In the work to obtain the estimate for Et, the function ϑt is used as good tool. By the prime number theorem [1], by of [4] ϑt = t 1 + θt log t < θt < 1 log t t In this paper we show a new estimate for Ep including θp for any prime number p 3. And using it, we give an upper bound, not a lower bound, for Et in the form as Et < log t/ t for any real number t 3. This is an essential improvement of already known results. Such estimate is very effective in the study of the RH the distribution of the prime numbers by Et. II. Main result of paper We give the following theorem. [Theorem 1] For any prime number p 3 we have Bp Ep + Dp θp p 2, 2.1 Bp := log p 2 log p log p + log1 + θp, Dp := log p + log1 + θp. 2

3 We will call 2.1 the condition d below. It is not difficult to see that the condition d is equivalent to that 1+log p Ep dp log2 p α 2 p logp α p holds for any prime number p 3, dp := α := 1 + θp, p log p Ep p θp p log 2. p α For the convenient in the work, we take any prime p 3 introduce following functions. ft = t log t Et t θt, gt = t log 2 t α, dt = ft gt α = 1 + θp is a positive constant such that t [p, p + 1], 1 1/ log p α 1 + 1/ log p. Then both ft gt are continuously differentiable function on the interval p, p + 1. In fact, since the functions log p are constants on p, p + 1, we have E t = p t p 1 b, ϑt = p t 1 t log t, E t is the derivative of Et so on. Hence we obtain θ t = 1 t θt, 2.3 t f t = 1 + log t Et. 2.4 Thus the function dt is also continuously differentiable on p, p + 1. Moreover, dt has the right h derivative at the point t = p. Put d p := lim t p+0 d t. Then we could rewrite Theorem 1 as [Theorem 1 ] For any prime number p 3 we have Also it is clear that 2.5 is equivalent to 2.2. From the Theorem 1 we obtain following important theorem. [Theorem 2] For any real number t 3 we have p t d p gp p p 1 < log log t + b + log t t

4 Rewriting 2.6, then for any real number t 3, Et < log t t 2.7 holds. This 2.7 is a new estimate for Et. Unsatisfactorily, this 2.7 is to give only the upper bound of Et, however it also gives a possibility to get a lower bound for one. Here we accentuate that it is very useful not only 2.7 but also 2.2. III. Some Preparations for Theorem 1 From the section III to the section VI we would hle the Theorem 1. First, we make ready for the proof of the Theorem A Condition d If the Theorem 1 does not hold, then there exists a prime number p 3 such that d p gp p > 2. We fix such prime p. Then from the table 1 the table 2 we see p e 14, because H m 0 for any 3 p m e 14 see 6.1 below. Here 2.2 is equivalent to H m 0 for p m there. Now we define the function Then Gt := d t gt t, t [p, p + 1]. G t = 1 0 t 1 1 t log t D 1 t, logt α 0 t := Et + ft f t logt α 4 t 1 + ft 2 t 8 log 2 t α logt α D 1 t := d t gt = f t dt g t. Hence G t < 0 is equivalent to 2 0 t+ 1 + D 1 t < logt α log t. Since t e 14 α 1 1/14 by 1.11, we get logt α = log t + log α > 0 2 so t D 1 t < 1 logt α log 2 t log t logt α log t log t logt α logt α log t + 1 log 2 t logt α log t 1 log t + < 0.46 t e

5 This shows that the function Gt is decreasing on the interval [p, p + 1]. Thus there exists a point t 1 such that p < t 1 < p + 1 Gp + 1 = Gp + Gp + 1 Gp = = Gp + G t 1 p + 1 p > p. For the convenient discussion, we put x 1 = p, x 2 = p + 1. Then, since Gt Gp + 1, for any t x 1, x 2 we have d t gt t > 2 1 1/ t. 3.2 We will call 3.2 the condition d. For the proof of the Theorem 1, we must obtain a contradiction from the condition d Proof of d t < 0 For any t x 1, x 2 we here have d t < 0. In fact, since d t = 1 t gt 0 t 1 1 log t we easily see that d t < 0 is equivalent to 0 t < 1 + 1/ log t, 0 t < 1 t e 14., 3.3. Function F t F t Put F t := d 2 dt g t gt g 1 d t, t x 1, x 2. Then it is clear g 1 := gx 1 = x2 x 1 F tdt = 0, 3.3 lim gt, d 2 := dx 2 = lim dt. t x 1+0 t x 2 0 Hence there exists a point ξ 0 such that x 1 < ξ 0 < x 2 x2 so x 1 F tdt = F ξ 0 x 2 x 1 = 0 d 2 dξ 0 g ξ 0 = gξ 0 g 1 d ξ We here have F t < 0 for any t x 1, 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 1 d t < 2 d t g t. And there exists a point t 1 such that x 1 < t 1 < t gt gx 1 = g t 1 t x 1 g t 1 5

6 Hence g t 1 g x 1 g 1.01 g t t e 14. x 2 gt g 1 d t 1.01 g t 1 + t gt 1 log t 0t 2 1 1/ t g t t gt 2 d t g t t e 14. Moreover, we note that F t > 0 holds for any t x 1, x An estimate of the point ξ 0 From x2 x 1 F tdt = ξ0 x 1 F tdt + x2 there exist λ 1, λ 2 such that x 1 < λ 1 < ξ 0 < λ 2 < x 2 Then, since F t < 0 t x 1, x 2, we have ξ 0 F tdt = 0, F λ 1 ξ 0 x 1 + F λ 2 x 2 ξ 0 = F λ 1 > F ξ 0 = 0 > F λ 2. Put x 0 := x 1 + x 2 ξ 0. Then from 3.5 we have F λ 1 x 2 x 0 + F λ 2 x 0 x 1 = This 3.6 shows that the line passing the points x 1, F λ 1 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 1 η 2 such that x 1 < η 1 < ξ 0 < η 2 < x 2 dx 2 dξ 0 = d η 2 x 2 ξ 0, gξ 0 gx 1 = g η 1 ξ 0 x 1. Since the function g t is decreasing on x 1, x 2 from the condition d, we have g ξ 0 g η 1, d t > 0 t x 1, x 2. Here if x 0 ξ 0, then x 2 ξ 0 ξ 0 x 1 by 3.4 we get d ξ 0 = d η 2 x2 ξ 0 ξ 0 x 1 g ξ 0 g η 1 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 1 = d 2 d 1 g x 1 = F x 1 F ξ 0 = F β 0 ξ 0 x 1, x 1 < β 0 < ξ 0. Also since ξ 0 x 1 = 1 ε 0 /2 F t = d 2 dt g t gt g 1 d t 2 d t g t, 3.7 6

7 we have d β 0 g β 0 ε 0 = T 1 + T 2, T 1 = d β 0 g β 0 d 2 d 1 g x 1, T 2 = d 2 dβ 0 g β 0 gβ 0 g 1 d β 0 ξ 0 x 1. By the condition d, we get d β 0 g β 0 ε / β 0 g β 0 gβ 0 β 0 T 1 = d β 0 g β 0 d 2 d 1 g x 1 = = d β 0 g β 0 d β 1 g x 1 = ε 0 ε 0 x 2 x 2 = d β 1 g β 0 β 0 β 1 + d β 1 g β 0 β 0 x 1 a 2 + b x 2 1 x 1 < β 1 < x 2, β 0 < β 1 < β 1, x 1 < β 0 < β 0 x 1 e 14, a 2 := d t 1 g t 2, b 2 := d t 1 g t 2 see section 4.4 below. We also have T 2 = d 2 dβ 0 g β 0 gβ 0 g 1 d β 0 ξ 0 x 1 = = d β 2 g β 0 x 2 β 0 g β 2 d β 0 β 0 x 1 ξ 0 x 1 β 0 < β 2 < x 2, a 2 + b 2 / x 2 2 x 1 < β 2 < β 0. Thus we have x 2 e 14, 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 1, F λ 1 λ 2, F λ 2 intersects the line y = 0 at the point λ 0, then we obtain F λ 1 λ 2 λ 0 + F λ 2 λ 0 λ 1 = Since the function F t is decreasing convex on x 1, x 2 under the condition d, it is clear ξ 0 < λ 0. And the equation of the line passing ξ 0, F λ 1 λ 0, 0 is one passing ξ 0, F λ 2 x 0, 0 is F λ 1 x + λ 0 ξ 0 y F λ 1 λ 0 = F λ 2 x + x 0 ξ 0 y + F λ 2 x 0 =

8 We put 0 := F λ 1 x 0 ξ 0 + F λ 2 λ 0 ξ 0 0, 1 := F λ 1 F λ 2 δ 0. Then y-coordinate of the cross point of above two lines is 1 / 0. Here if δ 0 > 0 then 1 / 0 < 0 1 < 0, if δ 0 < 0 then 1 / 0 > 0 1 > 0. Hence we always have 0 > 0 From this And since by 3.6 we have so F λ 1 x 0 ξ 0 + F λ 2 λ 0 ξ 0 > F λ 2 δ 0 < F λ 1 + F λ 2 x 0 ξ 0. F λ 1 + F λ 2 F λ 2 x 2 x 0 = 1 ε 0, 2 x 0 x 1 = 1 + ε 0, 2 = x 0 x 1 x 2 x 0 1 = 2 ε 0 1 ε 0 δ 0 < 2 ε 2 0/1 ε 0. Similarly, for the lines passing the points ξ 0, F λ 1, x 0, 0 the points ξ 0, F λ 2, λ 0, 0, we have F λ 1 λ 0 ξ 0 + F λ 2 x 0 ξ 0 > 0 from this Therefore we have 1 δ 0 > 1 2 ε 2 0/1 + ε 0. 2 ε ε 0 < δ 0 < 2 ε2 0 1 ε An estimate of δ 1 := λ 1 λ 1 Here λ 1 := x 1 + ξ 0 λ 1. By the same method as in δ 0, for the lines passing x 1, F λ 1, λ 1, 0 x 1, F λ 2, λ 1, 0 we have F λ 1 λ 1 x 1 + F λ 2 λ 1 x 1 > 0. From this, since F λ 1 + F λ 2 F λ 1 = 2 ε ε 0, λ 1 x 1 = ε δ 1, we have δ 1 < ε 0 /2. Also for the lines passing the points x 1, F λ 1, λ 1, 0 x 1, F λ 2, λ 1, 0 we have F λ 1 λ 1 x 1 + F λ 2 λ 1 x 1 > 0. 8

9 so δ 1 > ε 0 /2. Therefore we get 1 ε0 2 < δ 1 < ε An estimate of ω 0 := λ 1 + λ 2 2 ξ 0 Let λ 0 := λ 1 + λ 2 λ 0. Then from we get Hence Put ω 1 := λ 1 + λ 2 x 1 + x 2. Then, since F λ 1 F λ 2 = x 2 ξ 0 ξ 0 x 1 = λ 0 λ 1 λ 2 λ 0 = λ 2 λ 0 λ 0 λ 1 λ 0 = λ 1 ξ 0 x 1 + λ 2 x 2 ξ 0, λ 0 = λ 1 x 2 ξ 0 + λ 2 ξ 0 x 1. λ 0 λ 0 = λ 2 λ 1 ε 0 λ 1 + λ 2 = λ 0 + λ 0 = 2 λ 0 λ 0 λ 0, we have Thus x 1 + x 2 = x 0 + ξ 0 = 2 x 0 x 0 ξ 0, ω 1 = 2 δ λ 2 λ 1 ε 0. ω 0 = ω 1 + ε 0 = 2 δ ε 0 λ 2 λ 1 ε An estimate of Λ 0 := λ 2 λ 1 Since ξ 0 λ 1 = ε δ 1, we have λ 2 λ 1 = 2 λ 0 λ 1 λ 0 λ 0 = = 2 δ ε ε 0 δ 1 λ 2 λ 1 ε 0, hence λ 2 λ 1 = ε ε 0 δ ε δ ε < λ 2 λ 1 < ε An estimate of δ 2 := λ 1 η 1 By the same method as above, for the lines passing the points x 1, F λ 1, λ 1, 0 x 1, F λ 2, η 1, 0 we have F λ 1 η 1 x 1 + F λ 2 λ 1 x 1 > 0. 9

10 From this δ 2 < F λ 1 + F λ 2 λ 1 x 1 = F λ 1 = 2 ε ε ε δ 1 < 1 2 ε ε 0 Similarly, for the lines passing the points x 1, F λ 1, η 1, 0 x 1, F λ 2, λ 1, 0 we have F λ 1 λ 1 x 1 + F λ 2 η 1 x 1 > 0. From this Consequently, we have δ 2 > F λ 1 + F λ 2 F λ 2 λ 1 x 1 > 1 2 ε 0 1 ε 0. 1 ε ε 0 < δ ε 0 2 < ε An estimate of Q 0 := η 2 + η 1 2 ξ 0 ξ 0 x 1 First, we will find the lower bound of Q 0. If the line passing η 1, F η 1, η 2, F η 2 intersects y = 0 at η 0, then from this From 4.10 we have F η 1 η 2 η 0 + F η 2 η 0 η 1 = 0 F λ 1 η 2 η 0 + F λ 2 η 0 η 1 + W 0 = 0, W 0 = F η 1 F λ 1 η 2 η 0 + F η 2 F λ 2 η 0 η 1. η 0 = η 1 + η 2 η 1 x 2 ξ 0 + W 1, W 1 = x 2 ξ 0 F λ 1 W 0. Since F t is convex on x 1, x 2 η 1 < ξ 0 < η 2, we have η 0 > ξ 0 From this Here if W 1 < 0 then η 1 + η 2 η 1 x 2 ξ 0 + W 1 > ξ 0. Q 0 > η 2 ξ 0 ε 0 W 1. Q 0 > ε 0 if W 1 > 0 then we put η 0 := η 1 + η 2 η 0. Then by same way as above F λ 1 η 0 η 1 + F λ 2 η 2 η 0 + W 0 = 0 η 0 = η 1 + η 2 η 1 ξ 0 x 1 W 1. 10

11 Thus we have η 0 η 0 = η 2 η 1 ε W 1 so η 0 η 0 > 0. Similarly, for the lines passing the points λ 1, F λ 1, η 0, 0 λ 1, F λ 2, η 0, 0 we have On the other h, since η 2 > η 0 F λ 1 η 0 λ 1 + F λ 2 η 0 λ 1 > 0. ξ 0 λ 1 < 1 4, we have η 0 η 0 < F λ 1 + F λ 2 F λ 1 η 0 λ 1 < Hence < 2 ε ε 0 η 0 ξ 0 + ξ 0 λ 1 < < 2 ε 0 η 2 ξ 0 + ε W 1 = η 0 η 0 η 2 η 1 ε 0 < < 2 ε 0 η 2 ξ 0 + ε 0 2 η 2 ξ 0 ε 0 ξ 0 η 1 ε 0 ε 0 η 2 ξ 0 + ε 0 2 ξ 0 λ 1 ε 0 δ 2 ε 0 = = ε 0 η 2 ξ 0 + ε ε ε δ 1 δ 2 ε 0 < < ε 0 η 2 ξ 0 + ε ε2 0, since x 2 > η 2 x 2 ξ 0 = 1 + ε 0 /2, Q 0 > η 2 ξ 0 ε 0 W 1 > > 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 ε 0 ε 0 8 ε2 0 > ε 0. Therefore, generally, we have Q 0 = η 2 + η 1 2 ξ 0 ξ 0 x 1 > ε Next, we will find the upper bound of Q 0. It is easy to see that η 2 + η 1 2 ξ 0 = η 2 ξ 0 ξ 0 η 1 < < x 2 ξ 0 ξ 0 η 1 = = ε ε δ 1 + δ 2 < 11

12 < ε 0 2 ξ 0 η 2 η 1 = ξ 0 η 1 η 2 ξ 0 < < ξ 0 η 1 < ε 0. Therefore, since η 2 + η 1 2 ξ 0 ξ 0 x 1 < 1 < ε ε 0 < ε 0, we have Q 0 < ε An estimate of H 0 := ξ 0 η 1 ξ 0 x 1 Since ξ 0 η 1 = ε δ 1 + δ 2, we easily have ξ 0 x 1 = ε 0, 1 8 ε 0 < H 0 < ε IV. New equality inequality In this section we make a new equality from , derive a new inequality from the estimates for the various points discussed in the section III 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 F λ 1 + F λ 2 = F λ 1 F λ 2 ε 0 F λ 1 + F λ 2 d η 2 g ξ 0 = = F λ 1 F λ 2 d η 2 g ξ 0 ε F λ 1 + F λ 2 = F λ 2 F ξ 0 F ξ 0 F λ 1 = = F α 2 λ 2 ξ 0 F α 1 ξ 0 λ 1 = = F α 2 F α 1 ξ 0 λ 1 + F α 2 λ 2 ξ 0 ξ 0 λ 1 = = F τ 0 α 2 α 1 ξ 0 λ 1 + F α 2 λ 1 + λ 2 2 ξ 0 12

13 F λ 1 F λ 2 = F λ 1 F ξ 0 F λ 2 F ξ 0 = = F α 1 ξ 0 λ 1 F α 2 λ 2 ξ 0 = = F α 2 F α 1 ξ 0 λ 1 F α 2 λ 2 ξ 0 + ξ 0 λ 1 = = F τ 0 α 2 α 1 ξ 0 λ 1 F α 2 λ 2 λ 1, λ 1 < α 1 < ξ 0 < α 2 < λ 2 α 1 < τ 0 < α 2. From 3.4, there exist the points η 1 η 2 such that x 1 < η 1 < ξ 0 < η 2 < x 2 d 2 dξ 0 g ξ 0 gξ 0 g 1 d ξ 0 = = d η 2 g ξ 0 x 2 ξ 0 g η 1 d ξ 0 ξ 0 x 1 = = d η 2 g ξ 0 g η 1 d ξ 0 ξ 0 x 1 + +d η 2 g ξ 0 x 0 ξ 0 = Here also there exist µ 1, µ 2 such that η 1 < µ 1 < ξ 0 < µ 2 < η 2 We put Then since from 3.7 we have from we get d η 2 g ξ 0 g η 1 d ξ 0 = = d η 2 d ξ 0 g ξ 0 + g ξ 0 g η 1 d ξ 0 = = d µ 2 g ξ 0 η 2 ξ 0 + g µ 1 d ξ 0 ξ 0 η 1 = = d µ 2 g ξ 0 + g µ 1 d ξ 0 ξ 0 η 1 + +d µ 2 g ξ 0 η 2 + η 1 2 ξ A 0 := d η 2 g ξ 0, A 1 := d µ 2 g ξ 0 + g µ 1 d ξ 0, A 2 := d µ 2 g ξ 0, B 0 := d α 2 g α 2, B 1 := d 2 dα 2 g α 2 gα 2 g 1 d α 2, U 0 := d τ 0 g τ 0 + d τ 0 g τ 0, U 1 := d 2 dτ 0 g τ 0 gτ 0 g 1 d τ 0. F t = d 2 dt g t gt g 1 d t 3 d t g t + d t g t, 4.4 F α 2 = B 1 2 B 0, F τ 0 = U 1 3 U 0, M 0 := A 0 ε 0 = A 1 ξ 0 η 1 ξ 0 x

14 +A 2 η 2 + η 1 2 ξ 0 ξ 0 x 1 = O1/t Thus the left side of 4.1 is L 0 := F λ 1 + F λ 2 d η 2 g ξ 0 = L 1 + L 2, L 1 = F τ 0 d η 2 g ξ 0 α 2 α 1 ξ 0 λ 1 = L 11 L 12, L 2 = F α 2 d η 2 g ξ 0 λ 1 + λ 2 2 ξ 0 = L 21 L 22. L 11 = U 1 A 0 α 2 α 1 ξ 0 λ 1 = O1/t 4, L 12 = 3 U 0 A 0 α 2 α 1 ξ 0 λ 1 = O1/t 3. L 21 = 2 A 0 B 1 δ 0 4 A 0 B 0 δ 0 + +B 1 2 λ 2 λ 1 M 0 = O1/t 4, L 22 = 2 2 λ 2 λ 1 B 0 M 0 = O1/t 3. Similarly, the right side of 4.1 is R 0 := F λ 1 F λ 2 M 0 = R 1 R 2, R 1 = F τ 0 M 0 α 2 α 1 ξ 0 λ 1 = = U 1 3 U 0 M 0 α 2 α 1 ξ 0 λ 1 = O1/t 4. And R 2 := F α 2 M 0 λ 2 λ 1 = R 21 R 22, R 21 = B 1 M 0 λ 2 λ 1 = O1/t 4, R 22 = 2 B 0 M 0 λ 2 λ 1 = O1/t 3. Thus 4.1 is equivalent to L 0 = R 0, that is, we have a new equality L 11 + L 21 L 12 + L 22 = R 1 R 21 R A new inequality Put K 1 := L 12 + L 22 + R 22, K 2 := L 11 + L 21 R 1 + R 21, then, from the equality 4.6, we have K 1 = K 2. And by M 0 of 4.5 we have K 1 = A 0 + K 11 + K 12, A 0 = A 1 B 0 ξ 0 η 1 ξ 0 x 1, K 11 = 3 U 0 A 0 α 2 α 1 ξ 0 λ A 1 B 0 ξ 0 η 1 ξ 0 x 1, 14

15 K 12 = 4 A 2 B 0 η 2 + η 1 2 ξ 0 ξ 0 x 1. By the condition d we have d α 2 g α 2 d η 2 g ξ x 2 x 2 a 2 + b x1 x 1 e 14 M 1 := B 0 ε 0 = d α 2 g α 2 d η 2 g ξ 0 M 0 = O1/t 2. And also we see A 2 > 0, M 1 > 0. Thus from 3.18 we get K 12 K 12 := 4 A 2 M 1 = O1/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 1, x 2, so we have A 0 > 0, A 1 > 0, B 0 > 0, U 0 < 0. On the other h, hence by 3.20, We also have from 3.20 From this we have a new inequality α 2 α 1 ξ 0 λ 1 < λ 2 λ 1 ξ 0 λ 1 < < Λ < ε 0 K 11 K 11 + K 11, K 11 = 3 8 U 0 A 0 + A 1 B 0 = O1/t 4 see below, K 11 = 3 ε 0 U 0 A 0 A 1 B 0 = O1/t 4. A 0 Ω 0 A 1, Ω 0 = 1 8 A 1 B 0, A 1 = A 1 M 1 = O1/t 4. Ω 0 Ω 1 := L 11 +L 21 +R 21 R 1 K 12 +K 11 +K 11+A Now we intend to obtain the estimates for the lower bound of Ω 0 the upper bound of Ω 1 respectively Lower bound of Ω 0 := A 1 B 0 /8 Using the condition d, we get B 0 = d α 2 g 1 α 2 x 2 x 2 g µ 1 d ξ x 2 2 x 2 x 2 e 14 15

16 Thus we have Put d µ 2 g ξ 0 = g ξ µ 2 gµ µ 2 log µ 2 g x x 2 gx 2 x 2 2 x 2 e 14. Ω x x 3 2 x x 3 1 x x 1 e Ω 0 := x 3 1 x 1 x 1 e Some estimates For any t x 1, x 2 it is easy to see that And it is also clear that g t = log2 t α 2 t 4 1 +, logt α g t = log2 t α 4 t t 8 1 log 2 t α, g t = log2 t α 3 t 2 t logt α 3 log 2 t α From , we respectively have 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 = 1 + log t Et, f t = 1 t Et 1 1, log t f t = 1 t log 2 t Et. D 0 t := log t Et θt 2 t e 14, log t G 1 t := g t gt = 1 2 t logt α x 1 e 14, x 1 D 1 t := d t gt 1 + log t Et + t D 0 t G 1 t t e 14, 16

17 G 2 t := g t gt = 1 4 t log 2 t α 0.26 x 2 1 x 1 e 14, D 2 t := d t gt f t + t D 0 t G 2 t + 2 D 1 t G 1 t G 3 t := g t gt x 1 e 14, x 1 = 1 t logt α 3 log 2 t α x 3 1 x 1 e 14 D 3 t := d t gt f t + t D 0 t G 3 t+ +3 D 2 t G 1 t + D 1 t G 2 t x 2 1 x 1 e 14. From this for any t, t 1 x 1, x 2 we have r 0 := gt 1 gt 1 + g t 0 gt t < t 0 < t 1, By 4.5, we get a 1 := d t g t 1 = r 0 D 1 t G 1 t x 1 x 1 e 14, a 2 := d t g t 1 = r 0 D 1 t G 2 t x 2 1 a 3 := d t g t 1 = r 0 D 1 t G 3 t x 3 1 b 2 := d t g t 1 = r 0 D 2 t G 1 t x 2 1 b 3 := d t g t 1 = r 0 D 3 t G 1 t x 3 1 b 1 := d t g t 1 = r 0 D 2 t G 2 t x 3 1 M 0 A 1 + A 2 Q 0 a b 2 Q x 2 1 x 1 e 14, M 1 M x1 x 2 1 x 1 e 14. x 1 e 14, x 1 e 14, x 1 e 14, x 1 e 14, x 1 e Upper bound of Ω 1 The upper bound of Ω 1 is obtained as follows. First we will show the upper bound of K 11 = 3 8 U 0 A 0 + A 1 B 0. Since U 0 A 0 + A 1 B 0 = U 0 + A 1 A 0 A 0 B 0 A 1, 17

18 we have more Thus we have U 0 + A 1 = d τ 0 g τ 0 d µ 2 g ξ d τ 0 g τ 0 g µ 1 d ξ 0 = = 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 1 d ξ 0 b 1 + b 3 + b 1 + 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 < a 2 + b 2 A 0 = d η 2 g ξ 0 < a 1, A 1 = g µ 1 d ξ 0 + d µ 2 g ξ 0 < a 2 + b 2. K a 2 + b a 1 a 3 + b a 1 b Next, we would show the upper bound of x 1 e 14. L 21 = 2 A 0 B 1 δ 0 4 A 0 B 0 δ 0 + Since we have And since +B 1 2 λ 2 λ 1 M 0 = O1/t 4. A 0 ε 0 = M 0, B 0 ε 0 = M 1 B 1 = d 2 dα 2 g α 2 gα 2 g 1 d α 2 < a 2 + b < Λ 0 = λ 2 λ 1 < ε 0 < , L 21 4 ε 0 1 ε 0 a 2 + b 2 M ε 0 M 0 M a 2 + b 2 M x 1 e 14. V 0 := α 2 α 1 ξ 0 λ 1 < ε 0 <

19 we also have U 1 3 U 0 a 3 + b a 2 + b 2, R 1 a 3 + b a 2 + b 2 M 0 V 0 Similarly, we have respectively Consequently we obtain x 1 e 14. K 11 3 a 2 + b 2 M 0 + M x 1 e 14, L 11 a 3 + b 3 a 1 V x 1 e 14, R 21 a 2 + b 2 M 0 λ 2 λ x 1 e 14, A 1 a 2 + b 2 M x 1 e 14, K 12 4 b 2 M x 1 e 14. Ω 1 K 11 + L 21 + R 1 + K 11 + L 11 + R 21 + A 1 + K x x 1 e V. Proof of Theorem 1 We take arbitrarily the prime number p 3. If 3 p e 14, then we could confirm that the condition d holds from the table 1 the table 2. Hence if the Theorem 1 does not hold, then there exists a prime number p e 14 such that the condition d holds. For such prime p, we have Ω 0 Ω 1, finally, from we get x x 1 e 14, but it is a contradiction. This shows that the condition d is not valid. Consequently, the Theorem 1 holds for any prime number p 3. VI. Algorithm Tables for Sequence {H m } Here dp m log2 p m α 2 p m H m := 1 + log p m Ep m logp m α 2 pm,

20 α := 1 + θp m, dp m := p m log p m Ep m θp m pm log 2. p m α The table 1 2 show the values of H m for 2 p m p m Note that 1.11 holds for any p m 3 p m 41, the condition d holds if only if H m 0 for any m 2. It is easy to verify 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 = 1 : M; p = P 1 : m; E = sum1./p b loglogpm; E1 = 1+logpm E; V 1 = sumlogp.; Q = V 1/pm 1; R = pm 1/2 ; V = logv 1; g = R V 2 ; f = pm logpm E Q; d = f/g; B = V /V /2/R; m, pm, H m = E1 d B 2/R, end. T able 1 T able 2 m p m H m m p m H m VII. Some Preparations for Theorem 2 From the section VII to the section IX we would hle the Theorem 2. As in the section III, we make ready for the proof of the Theorem Some symbols Let p 1 = 2, p 2 = 3, p 3 = 5, be the first consecutive primes. Then p m m N is m-th prime number. We arbitrary choose the prime number p m 3 fix it. We put p 0 = p m, p = p m 1 below. For the theoretical calculation we assume p e 14. The discussion for p e 14 is supported by MATLAB. Put e 0 := expep 0, e 0 := explog p 0 e 0 1, e 1 := expep, e 1 := explog p e 1 1, α 0 := 1 + θp 0, N 0 := p 0 α 0 log 2 p 0 α 0, α := 1 + θp, N 1 := p α log 2 p α. 20

21 7.2. An estimate of e 1, e 1 e 1 e 1 From 1.9 we respectively have e 1 < exp1/ log 2 p < p e 14, e 1 < p e 14, e 1 e 1 < 1.08 p e 14. Since if e 1 1 then e 1 1, we have e 1 e 1 1. If e 1 > 1, then Therefore we have e 1 = 1 + r + 0 < r := Ep < 1 log 2 p p e14 n=2 r n n! 1 + r + r r 1 + r r2, e 1 e 1 = expr + log p e h + h h, h = 1 + log p r log p r p e 14. e 1 e log p Ep log p 2 Ep 2 e 1 > 1, p e An estimate of V 0 := p 0 e 0 α 0 p e 1 α By 1.10 it is clear that p 0 α 0 p α = ϑp 0 ϑp = log p 0, since we have From this Thus we have Since e 0 e 1 Et = p t p 1 log log t b, Ep 0 Ep = 1 log p0 log. p 0 log p e 0 log p = expp 1 0 e 1 log p, 0 = p exp log p e 1 expp 1 0. p 1 0 V 0 = p e p0 e 0 1 p e 1 log p 0 = log p 0 µ e 1 1, 1 µ = p exp log p e 1 expp log p 1. 0 expt < 1 + t t 2 1 t 0 < t < 1, 21

22 we have hence µ e log p p µ e 1 1 e 1 e log p p e 1 > 1, p e 14 e 1 > 1, p e An estimate of G 0 := log p 0 Rp α N 0 N 1 /N 0 Here Rp α := log2 p α 2 p α logp α It is known that p 2 k+1 p 1 p k for p k 7 on p. 246 of [6] hence log p 0 logp α < 1 2 p e14. Since log1 + t t t 2 /2 for any t 0 < t < 1/2, we have N 0 N 1 = p 0 α 0 p α log 2 p 0 α p α log 2 p 0 α 0 log 2 p α log p 0 2 p 0 α 0 log 2 p α+ +2 p α logp α log 1 + log p 0 p α + log p 0 log p 0 2 p log 2 p α+ 0 α 0 2 logp α p α 1 log p 0 2 p α G 0 N 0 log p log 2 p α+ 2 p α p0 α 0 + log2 p 0 logp α p α 3/2 log 2 p 0 log2 p α 1 p α 3/ logp α And it is known that p 2 k+1 2 p2 k for p k 7 on p. 247 of [6] so log p 0 log p 1 + log 2. log p Since p e 14, we have α 1 1/14 the function log 3 t/t is decreasing on the interval e 3, +. Therefore we get G 0 log2 p 0 1 p α logp α 22

23 log3 p p α log log p log p + log α p log p 0.01 p log p p e An estimate of Sp := p p 1/p log p Put st := p 1 = log log t + b + Et. p t Then by the Abel s identity [1], we have + Sp 1 + = p log t dst = 1 dt p log t t log t + det 1 log p Ep + log p + dt p t log 4 t 1 log p + 1 log 3 p 1 3 log 3 t + p = = 1 log p log 3 p Sp 1 log p 4 3 log 3 p. If p is a first prime e 14, then p = it is th prime. And we have Sp Lemma For any m 1 we put D m := p m e 0 α 0 pm α 0 log 2 p m α 0. Then we are ready for the proof of the following lemma. [Lemma] For any m 5 we have D m < 1. Proof. First, for any 11 p m e 14 we see D m < 1 by MATLAB see the table 5 the table 6 below. Next, we will prove that D m a m := 1 13 Sp m holds for any p m e 14 by the mathematical induction with respect to m. If p = then we have Now assume p e 14 D m 1 a m 1. Then D = Sp 0.1 < 1. D m = 1 N 0 p e 1 α + V 0 = D m 1 N1 N 0 + V 0 N 0 a m 1 N1 N N 0 log p 0 µ e 1 1 a m 1 + b m 1, 23

24 b m 1 = 1 N 0 log p 0 µ e 1 1 a m 1 N 0 N 1. By the assumption D m 1 a m 1, we get p α log 2 e p α 1 α + a m 1 p by taking logarithm of both sides log 2 p α = α 1 + a m 1 p α log e 1 = log p e 1 1 θp + a m 1 log2 p α p α. From this Thus e θp + a m 1 log p log2 p α, p α Ep 1 log p θp + a m 1 log2 p α p α. log p Ep θp a m 1 log2 p α p α 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 1. α From the Theorem 1 we get 1 + log p Ep a m 1 log2 p α 2 p α logp α p If e 1 > 1, then, since 0 < a m /14 α 1 + 1/14, we also have 1 + log p 2 Ep 2 log4 p α 1 p α logp α α log 2 p α log4 p α p α p e 14 log p 0 µ e 1 1 a m 1 N 0 N 1 log p log p Ep a m 1 N 0 N log2 p 0 p log p log p 2 Ep 2 24

25 a m 1 G 0 N log2 p log p 0 log4 p α + 2 log p 0. p p α p Finally, by the function log 4 t/ t is decreasing on the interval e 8, + we have b m 1 a m 1 G log4 p p Next, if e 1 1 then we have log p 1 + log 2 log α 1 2 p α log p + log α p log p log 2/ log p α 3/ log 2 log α α log p + log α log α/ log p2 + p log p 1 p log p 0.01 p log p p log p p log p p log p 13 p log p p e14. b m log2 p 0 p N p log p p e14. VIII. Proof of Theorem 2 First, we would show that 2.6 holds for any prime number p 3. Since expt > 1 + t 1 < t < 1, for any prime number p 3 we have p log p Ep p θp = = p 1 + log p Ep p 1 + θp p e 1 p α. From this, by the Lemma, we have dp m D m for any m 6. Hence dp < 1 for any p 11 by 2.1 of the Theorem 1 Ep Here if p 71 then 1 log 2 p α log p 1 + log p 2 1 log 2 p α 1 + log p 2 p logp α p + 2 logp α + 2 < p 71, if 3 p 71 then we get 2.6 see the Table 3 the Table 4 below. Thus Ep < log p p. p 3. holds for any prime number p 3. Next, for any real number t 3, there exists a prime number p 3 such that p t < p 0. Put Zt := Et log t t t [p, p 0, 25

26 then Z t < 0 t p, p 0 so Zt Zp < 0. This shows that 2.6 holds for any real number t 3. IX. Algorithm Tables for Sequence {Z m } {D m } The table 3 the table 4 show the values of Z m := Ep m log p m pm for 1 m 20 to m. Note that 2.6 holds for m 1 if only if Z m < 0. And the table 5 shows the values of D m. There are only values of D m for 1 m 10 here. But it is not difficult to verify them for 31 p m e 14. The table 6 shows the values D m for m Of course, all the values in the tables are approximate. The algorithm for Z m D m to m by matlab is as follows: Function EMF-Index, clc, b= ; format long P = [2, 3, 5, 7,, ]; M=lengthP; for m = 1 : M; p = P 1 : m; Epm = sum1./p b loglogpm; Apm = logpm/ pm; ϑpm = sumlogp.; α 0 = ϑpm/pm; R = pm 1/2 ; V = logϑpm; g = R V 2 ; e 0 = explogpm expepm 1, m, p m, Z m = Epm Apm, D m = pm e 0 α 0 /g, end T able 3 T able 4 m p m Z m m p m Z m T able 5 T able 6 m p m D m m p m D m

27 References [1] Tom M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, New York, Heidelberg Berlin, [2] P. Borwein, S. Choi, B. Rooney, A. Weirathmueller, The Riemann Hypothesis, Springer, Berlin Heidelberg, [3] A. E. Ingham, The distribution of prime numbers, Cambridge Tract No.30, Cambridge University Press, [4] J. B. Rosser, L. Schoenfeld, Approximate formulas for some functions of prime numbers, IIlinois J. Math , [5] J. Sor, D. S. Mitrinovic, B. Crstici, Hbook of Number theory 1, Springer, [6] I. M. Vinogradov, Novaya ocenka funkcii ζ1 + it, Izv. Akad. Nauk SSSR, Ser. Mat., vol , pp Mathematics Subject Classification; 11M26, 11N05. Address Dr. Choe Ryong Gil Department of Mathematics University of Sciences Unjong District, Gwahak 1-dong Pyongyang, D.P.R.Korea ; ryonggilchoe@star-co.net.kp 27