An upper bound for error term of Mertens formula

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 = 0.261497212 is the Mertens constant γ = 0.577 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 t3/5. 1.4 1

From 1.3 1.4 by the Abel s identity [1], it is easily given more Et = Oexp a log t 1.5 Et = Oexp a log t 3/5. 1.6 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 3.17 3.20 of [4], 1 log 2 t < Et < 1 log 2 t t > 1. 1.9 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 3.15 3.16 of [4] ϑt = t 1 + θt 1.10 1 log t < θt < 1 log t t 41. 1.11 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 := 1 + 1 2 log p 2 log p log p + log1 + θp, Dp := 1 2 + 2 log p + log1 + θp. 2

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 4 1+ + 2 2.2 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 2. 2.5 p 1 < log log t + b + log t t. 2.6 3

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 1. 3.1. 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 + 2 1 + D 1 t, logt α 0 t := Et + ft f t 1 + 4 logt α 4 t 1 + ft 2 t 8 log 2 t α 1 + 4 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 < 1 + 1 logt α log t. Since t e 14 α 1 1/14 by 1.11, we get logt α = log t + log α 13.925 > 0 2 so + 2 0 t + 1 + D 1 t < 1 logt α log 2 t + 1 2 log t + 4 1 1 + logt α log t + 1 4 log 2 + 1 + 2 t logt α 2 1 1+ logt α log t + 1 log 2 t + 4 1 1+ logt α log t 1 log t + < 0.46 t e 14. 3.1 4

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 > 2 1.6 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. 3.2. 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 0.2577 < 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 0. 3.4 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

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 2. 3.4. 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 = 0. 3.5 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 = 0. 3.6 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. 3.5. 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

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 0 2 1 1/ 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 2 0.9005 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 /2 0.4503 x 2 2 x 1 < β 2 < β 0. Thus we have x 2 e 14, 0 < ε 0 1.3508 x2 0.0015 x 2 e 14. 3.8 3.6. 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 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 = 0. 3.9 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 = 0 3.10 F λ 2 x + x 0 ξ 0 y + F λ 2 x 0 = 0. 3.11 7

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 3.10 3.11 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 > 0. 3.12 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 2 0 1 + ε 0 < δ 0 < 2 ε2 0 1 ε 0. 3.13 3.7. 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 0 1 + ε 0, λ 1 x 1 = 1 4 1 ε 0 + 1 2 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

so δ 1 > ε 0 /2. Therefore we get 1 ε0 2 < δ 1 < ε 0 2. 3.14 3.8. An estimate of ω 0 := λ 1 + λ 2 2 ξ 0 Let λ 0 := λ 1 + λ 2 λ 0. Then from 3.5 3.9 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 0 + 1 λ 2 λ 1 ε 0. ω 0 = ω 1 + ε 0 = 2 δ 0 + 2 ε 0 λ 2 λ 1 ε 0. 3.15 3.9. An estimate of Λ 0 := λ 2 λ 1 Since ξ 0 λ 1 = 1 4 1 ε 0 1 2 δ 1, we have λ 2 λ 1 = 2 λ 0 λ 1 λ 0 λ 0 = = 2 δ 0 + 2 ε 0 + 1 2 1 ε 0 δ 1 λ 2 λ 1 ε 0, hence λ 2 λ 1 = 1 2 + ε 0 1 + ε 0 δ 1 1 + ε 0 + 2 δ 0 1 + ε 0 1 2 < λ 2 λ 1 < 1 2 + 2 ε 0. 3.16 3.10. 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

From this δ 2 < F λ 1 + F λ 2 λ 1 x 1 = F λ 1 = 2 ε 0 1 1 + ε 0 4 1 ε 0 + 1 2 δ 1 < 1 2 ε 0. 1 + 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 0 2 1 ε 0 < δ ε 0 2 < 2 1 + ε 0. 3.17 3.11. 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

Thus we have η 0 η 0 = η 2 η 1 ε 0 + 2 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 0 1 + ε 0 η 0 ξ 0 + ξ 0 λ 1 < < 2 ε 0 η 2 ξ 0 + ε 0 2. 2 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 0 1 2 ε 0 4 1 ε 0 1 2 δ 1 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 = 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 0 1 + 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 0. 3.18 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 = = 1 1 2 1 + ε 0 4 1 ε 0 1 2 δ 1 + δ 2 < 11

< 1 4 + 2 ε 0 2 ξ 0 η 2 η 1 = ξ 0 η 1 η 2 ξ 0 < < ξ 0 η 1 < 1 4 + 2 ε 0. Therefore, since η 2 + η 1 2 ξ 0 ξ 0 x 1 < 1 < 4 + 2 ε 0 1 2 1 ε 0 < 1 8 + ε 0, we have Q 0 < 1 8 + ε 0. 3.19 3.12. An estimate of H 0 := ξ 0 η 1 ξ 0 x 1 Since ξ 0 η 1 = 1 4 1 ε 0 1 2 δ 1 + δ 2, we easily have ξ 0 x 1 = 1 2 1 ε 0, 1 8 ε 0 < H 0 < 1 8 + ε 0. 3.20 IV. New equality inequality In this section we make a new equality from 3.5 3.6, derive a new inequality from the estimates for the various points discussed in the section III. 4.1. A new equality Now we add 3.5 3.6, 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 0. 4.1 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

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 = 0. 4.2 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 4.2 4.3 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 0. 4.3 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 1 + 13

+A 2 η 2 + η 1 2 ξ 0 ξ 0 x 1 = O1/t 2. 4.5 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 22. 4.6 4.2. 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 1 + +3 A 1 B 0 ξ 0 η 1 ξ 0 x 1, 14

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 0 1 + x 2 x 2 a 2 + b 2 1 + 0.9005 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 0 1 4 < 1 8 + 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 1. 4.7 Now we intend to obtain the estimates for the lower bound of Ω 0 the upper bound of Ω 1 respectively. 4.3. 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 0 0.4789 x 2 2 x 2 x 2 e 14 15

Thus we have Put d µ 2 g ξ 0 = g ξ 0 1 + µ 2 gµ 2 1 0 µ 2 log µ 2 g x 2 0.3711 1 0.2577 x 2 gx 2 x 2 2 x 2 e 14. Ω 0 1 0.4789 8 x 4 + 0.3711 2 x 3 2 x 0.0462 2 x 3 1 x x 1 e 14. 1 Ω 0 := 0.0462 x 3 1 x 1 x 1 e 14. 4.8 4.4. 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 8 1 2 logt α 3 log 2 t α From 1.9 1.11, 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 2 1 + 1 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 1 + 4 logt α 0.6437 x 1 e 14, x 1 D 1 t := d t gt 1 + log t Et + t D 0 t G 1 t 0.1685 t e 14, 16

G 2 t := g t gt = 1 4 t 2 1 8 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 1.3307 x 1 e 14, x 1 = 1 t 3 3 8 1 2 logt α 3 log 2 t α 0.3751 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 3.7650 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 1.000001 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 1 0.1085 x 1 x 1 e 14, a 2 := d t g t 1 = r 0 D 1 t G 2 t 1 0.0439 x 2 1 a 3 := d t g t 1 = r 0 D 1 t G 3 t 1 0.0633 x 3 1 b 2 := d t g t 1 = r 0 D 2 t G 1 t 1 0.8566 x 2 1 b 3 := d t g t 1 = r 0 D 3 t G 1 t 1 2.4236 x 3 1 b 1 := d t g t 1 = r 0 D 2 t G 2 t 1 0.3460 x 3 1 M 0 A 1 + A 2 Q 0 a 2 + 2 b 2 Q 0 0.2223 x 2 1 x 1 e 14, M 1 M 0 1 + 0.9005 0.2225 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 14. 4.5. 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

we have more Thus we have U 0 + A 1 = d τ 0 g τ 0 d µ 2 g ξ 0 + + 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 0 + + 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 0 + + 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 11 8 3 a 2 + b 2 2 + a 1 a 3 + b 3 + 2 a 1 b 1 0.4334 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 2 1 2 < Λ 0 = λ 2 λ 1 < 1 2 + 2 ε 0 < 0.5034, L 21 4 ε 0 1 ε 0 a 2 + b 2 M 0 + 8 1 ε 0 M 0 M 1 + + 3 2 a 2 + b 2 M 0 0.6978 x 1 e 14. V 0 := α 2 α 1 ξ 0 λ 1 < 1 8 + ε 0 < 0.1267 18

we also have U 1 3 U 0 a 3 + b 3 + 3 a 2 + b 2, R 1 a 3 + b 3 + 3 a 2 + b 2 M 0 V 0 Similarly, we have respectively Consequently we obtain 0.1462 x 1 e 14. K 11 3 a 2 + b 2 M 0 + M 1 1.2017 x 1 e 14, L 11 a 3 + b 3 a 1 V 0 0.0342 x 1 e 14, R 21 a 2 + b 2 M 0 λ 2 λ 1 0.1008 x 1 e 14, A 1 a 2 + b 2 M 1 0.2004 x 1 e 14, K 12 4 b 2 M 1 0.7625 x 1 e 14. Ω 1 K 11 + L 21 + R 1 + K 11 + L 11 + R 21 + A 1 + K 12 0.4334 + 0.0342 + 0.6978 + 0.1008 3.5770 + 0.1462 + 0.2004 + 1.2017 x 4 + 1 + 0.7625 x 1 e 14. 4.9 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 4.8 4.9 we get 1 3.5770 0.0462 x 1 0.08 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 4 1 + logp m α 2 pm, 6.1 19

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 29 93109 p m 93118. 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 93109. The algorithm for H m by MATLAB is as follows: Function EMF-Index, clc, b=0.261497212847643; format long, P = [2, 3, 5, 7,, 1202609]; 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 2 1 + 4/V /2/R; m, pm, H m = E1 d B 2/R, end. T able 1 T able 2 m p m H m 1 2 4.92781518770647 2 3 3.79708871931795 3 5 1.82084025624172 4 7 1.24240415973621 5 11 1.05892911097784 6 13 0.82377421885520 7 17 0.75298886049588 8 19 0.60562813217931 9 23 0.56797602737022 10 29 0.59342397038654 m p m H m 93109 1202477 0.00169503567169 93110 1202483 0.00168790073361 93111 1202497 0.00168788503420 93112 1202501 0.00167897043350 93113 1202507 0.00167183566691 93114 1202549 0.00169673633799 93115 1202561 0.00169494084824 93116 1202569 0.00168958602998 93117 1202603 0.00170736660136 93118 1202609 0.00170023228562 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 2. 7.1. 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

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 < 1.0052 p e 14, e 1 < 1.075 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 0.0052 p e14 n=2 r n n! 1 + r + r 2 2 1 r 1 + r + 0.503 r2, e 1 e 1 = expr + log p e 1 1 1 + h + h 2 2 1 h, h = 1 + log p r + 0.503 log p r 2 0.1125 p e 14. e 1 e 1 1 1 + log p Ep + 0.6 1 + log p 2 Ep 2 e 1 > 1, p e 14. 7.3. 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 1 0 1 log p 1. 0 expt < 1 + t + 1 2 t 2 1 t 0 < t < 1, 21

we have hence µ e 1 + 0.505 log p p µ e 1 1 e 1 e 1 1 + 0.55 log p p e 1 > 1, p e 14 e 1 > 1, p e 14. 7.4. 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 4 1 +. 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 0 + + 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 0 1 1 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 3/2 4 + 1. 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 2 4 + 1 logp α 22

log3 p p α 2 1 + log 2 1 2 log p 4 + 1 1 log p + log α p log p 0.01 p log p p e 14. 7.5. 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 + 4 3 log 3 p Sp 1 log p 4 3 log 3 p. If p is a first prime e 14, then p = 1202609 it is 93118-th prime. And we have 0.070 Sp 0.072. 7.6. 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 = 1202609 then we have Now assume p e 14 D m 1 a m 1. Then D 93118 = 0.0103 1 13 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 0 + 1 N 0 log p 0 µ e 1 1 a m 1 + b m 1, 23

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 1 1 + 1 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 4 1 + + 2. logp α p If e 1 > 1, then, since 0 < a m 1 1 1 1/14 α 1 + 1/14, we also have 1 + log p 2 Ep 2 log4 p α 1 p α 2 + 2 logp α + 2 2 α log 2 p α 0.4287 log4 p α p α p e 14 log p 0 µ e 1 1 a m 1 N 0 N 1 log p 0 1 + log p Ep a m 1 N 0 N 1 + 0.55 log2 p 0 p + 0.6 log p 0 1 + log p 2 Ep 2 24

a m 1 G 0 N 0 + 0.55 log2 p 0 + 0.2572 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 0 + 0.55 +0.2572 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 + 1 + log 2/ log p α 3/2 + 2 1 + log 2 log α α log p + log α 2 1 + log α/ log p2 + p log p 1 p log p 0.01 p log p + 0.01 p log p + 10.421 p log p + 2.203 p log p 13 p log p p e14. b m 1 0.55 log2 p 0 p N 1 0.01 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 4 1 + + 2. logp α p + 2 logp α + 2 < 0.935 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

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 93109 m 93118. 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=0.261497212847643; format long P = [2, 3, 5, 7,, 1202609]; 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 1 2 0.11488663599975 2 3 0.15649580772857 3 5 0.42381139039502 4 7 0.48652145124644 5 11 0.59198165652987 6 13 0.57080225951896 7 17 0.58721773344997 8 19 0.56143838923050 9 23 0.55912374510489 10 29 0.56745957678030 m p m Z m 11 31 0.54628472181999 12 37 0.54636660696827 13 41 0.53633985985008 14 43 0.51944301813516 15 47 0.50956392869224 16 53 0.50518392944789 17 59 0.50037701659240 18 61 0.48761926443285 19 67 0.48260095592014 20 71 0.47441536685574 T able 5 T able 6 m p m D m 1 2 25.62071141247196 2 3 8.97923715714347 3 5 1.91868003953127 4 7 1.04417674546040 5 11 0.65533994162650 6 13 0.50557929260089 7 17 0.40546150815241 8 19 0.35633549506425 9 23 0.31785034607111 10 29 0.27811621292050 m p m D m 93109 1202477 0.01038794622881 93110 1202483 0.01038795465981 93111 1202497 0.01038796309106 93112 1202501 0.01038797210397 93113 1202507 0.01038798158228 93114 1202549 0.01038798943309 93115 1202561 0.01038799740043 93116 1202569 0.01038800571686 93117 1202603 0.01038801287098 93118 1202609 0.01038802049040 26

References [1] Tom M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, New York, Heidelberg Berlin, 1976. [2] P. Borwein, S. Choi, B. Rooney, A. Weirathmueller, The Riemann Hypothesis, Springer, Berlin Heidelberg, 2007. [3] A. E. Ingham, The distribution of prime numbers, Cambridge Tract No.30, Cambridge University Press, 1932. [4] J. B. Rosser, L. Schoenfeld, Approximate formulas for some functions of prime numbers, IIlinois J. Math. 6 1962, 64-94. [5] J. Sor, D. S. Mitrinovic, B. Crstici, Hbook of Number theory 1, Springer, 2006. [6] I. M. Vinogradov, Novaya ocenka funkcii ζ1 + it, Izv. Akad. Nauk SSSR, Ser. Mat., vol. 221958, pp. 161-164. 2010 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 Email; ryonggilchoe@star-co.net.kp 27