APPROXIMATIONS FOR SOME FUNCTIONS OF PRIMES

Transcription

1 APPROXIMATIONS FOR SOME FUNCTIONS OF PRIMES Fataneh Esteki B. Sc., University of Isfahan, 2002 A Thesis Submitted to the School of Graduate Studies of the University of Lethbridge in Partial Fulfillment of the Requirements for the Degree MASTER OF SCIENCE Department of Mathematics and Computer Science University of Lethbridge LETHBRIDGE, ALBERTA, CANADA c Fataneh Esteki, 2012

2 ABSTRACT APPROXIMATIONS FOR SOME FUNCTIONS OF PRIMES Fataneh Esteki Department of Mathematics and Computer Science University of Lethbridge M. Sc. Thesis, 2012 Let π) = # {p ; p prime}, θ) = p log p, and ψ) = p k,k 1 log p. This thesis studies different methods in establishing estimates for π), θ), and ψ). This is a summary of the main result of the thesis. i) A detailed eposition of a theorem of Rosser on the estimation of ψ) is given. The theorem is written using parameters instead of the specific constants. So it conveniently will produce new estimates for ψ) whenever new improvements in the values of the parameters occur. As an eample, our theorem, with current known values of parameters gives < ψ) < , for e 20. ii) Different techniques for establishing eplicit upper and lower bounds for θ) are studied. It is proved that θ) < 1, for > iii) The following sharp lower bounds for ψ) θ) are established. ψ) θ) >, for 121 e 145.5, i

3 and ψ) θ) > , for 121. iv) Several tables of upper and lower bounds for θ) based on different methods are generated. v) Different methods for establishing inequalities of the form ψ) θ) < c 1 1/2 + c 2 1/3 are studied and specific numerical eamples are generated. As an eample it is shown that and ψ) θ) < ) 1/ /3, for e 100. vi) Different methods for finding upper and lower bounds for π) of the forms are considered. log a, a > 0, 1 + 1! log log + 2! l 1)! log log l 1 + c log l ), l N, c > 0, Also specific numerical eamples are generated. As a sample it is established that for 10 11, log < π) < log , log log ) log 2 < π) < log log ) log 2. ii

4 ACKNOWLEDGMENTS I am very grateful to all those who have supported and guided me in my graduate studies. In particular I would like to thank my supervisors, Dr. Amir Akbary and Dr. Habiba Kadiri for their ecellent guidance and academic support. I would also like to thank them for the financial aid they have provided throughout my studies. I would like to thank Dr. Dennis Connolly, Dr. Dave Morris, and Dr. Nathan Ng for their kind supports. I wish to thank the other members of the eamination committee Dr. Hadi Kharaghani and Dr. Wei Xu. I would also like to thank all my friends for making me happy in the last two years. Many thanks to my parents and my siblings for their love, encouragement, and guidance along the way. iii

5 Dedicated to My Parents iv

6 Contents 1 Introduction Chebyshev functions Relations between ψ), θ), and π) Eplicit bounds for π), θ), and ψ) Statement of the results Bounds for ψ) Elementary estimates of ψ) Analytic estimates of ψ) Comparing ψ) with its average Relating ψ) and the zeros of the zeta function Eplicit bounds for the sums over the zeros of the zeta function Proof of Theorem Bounds for θ) Introduction Upper bounds for θ) for > Upper bounds for θ) for e b The inequality ψ) θ) > d Lower bounds for θ) for e b The inequality ψ) θ) < c 2 1/2 + c 3 1/ Bounds for π) Introduction Upper bounds over finite ranges v

7 4.3 Upper bounds for π) Lower bounds over finite ranges Lower bounds for π) Tables 93 vi

8 Chapter 1 Introduction 1.1 Chebyshev functions One of the important achievements of the ninetieth century mathematics is the proof of the Prime Number Theorem. This theorem gives the asymptotic density of prime numbers among integers. More precisely, let π) = # {p, p prime}. Then the Prime Number Theorem asserts that The notation f g means log )π) lim f) lim g) = 1. = 1. Using this notation the Prime Number Theorem can be written as as. π) log, The Prime Number Theorem was originally conjectured in 1791 by Gauss and later on, in 1798, independently by Legendre. Gauss looked at the list of primes less than 3, 000, 000 and observed that π) can be approimated very closely by the function 1

9 li), which is defined by li) = 0 dt 1 η ) dt log t = lim + η η log t. By integration by parts on the integral defining li) one can show that the above integral is asymptotically equivalent to /log. Legendre in his tetbook in number theory asserted that π) can be approimated by log Hence Gauss and Legendre s observations announce the Prime Number Theorem. See [1] for more historical information. The first attempt for proving the Prime Number Theorem was done by Chebyshev in Among other results, he established that if the ratio log )π)/ tends to a limit as approaches to infinity then that limit must be one. Moreover he showed see [11, p. 15]) that for all sufficiently large values of, ) π) ) log log. 1.1) For proving the above assertions Chebyshev introduced two new functions θ and ψ. These two functions played an important role in the development of the prime number theory. They can be defined as θ) = p log p, and ψ) = log p. p n,n 1 Chebyshev observed that proving any of θ) or ψ), as, would imply the Prime Number Theorem. However he was not able to establish either of the above asymptotics. 2

10 A revolutionary idea for proving the Prime Number Theorem was introduced by Riemann in In his approach he used Euler s identity n=1 n s = p 1 + p s + p 2s + ) = p 1 p s ) 1, where the products run over all primes p. Riemann studied the above identity where s is a comple variable. He considered ζs) = n s, n=1 for Res) > 1 and verified that ζs) has an analytic continuation to the whole comple plane with the eception of a simple pole with residue 1 at s = 1. Moreover he proved that ζs) satisfies the functional equation s π s/2 Γ ζs) = π 2) 1 s)/2 Γ 1 s 2 ) ζ1 s), where Γs) = 0 s 1 e d see [11, Chapter III]). By using this equation one can show that ζs) vanishes at all the negative even integers. These zeros are called the trivial zeros of the zeta function. Riemann observed that all other zeros of ζs) are situated in the critical strip 0 Res) 1 see [11, p. 58]). We denote these so called non-trivial zeros by ρ = β + iγ where 0 β 1 and γ R. Riemann observed that the distribution of the non-trivial zeros of ζs) had an important role in the proof of the Prime Number Theorem and designed a program for the proof of the Prime Number Theorem based on certain properties of the zeros of the Riemann zeta function. However he failed to deduce the Prime Number Theorem. More precisely, he suggested that to prove the Prime Number Theorem it was sufficient to prove that ζs) does not vanish on the line Res) = 1. Moreover, he went one step further and conjectured that all non-trivial zeros of ζs) lie on the line Res) = 1/2. This is the celebrated Riemann Hypothesis and it has been remained unsolved up to this day. The Prime Number Theorem was finally proven in 1896 by Hadamard and independently by de La Vallée Poussin, following Riemann s suggested program. 3

11 Despite the fact that the Prime Number Theorem determines the asymptotic behavior of π), it does not give eplicit upper and lower bounds for log )π)/. In many applications we need to have such eplicit bounds. As an eample Chebyshev established upper and lower bound for π) given in 1.1). This enabled him to prove Bertrand s postulate which states that for sufficiently large, there is a prime between and 2. To prove this, it suffices to show that for large π2) π) 1. Using 1.1), for sufficiently large, we obtain π2) π) log log. For sufficiently large, the above can be written as π2) π) log + 1/3 log log. Since 3/4)1.842) > 0 Bertrand s postulate follows. This thesis studies different methods in establishing eplicit estimates for π), θ), and ψ). 1.2 Relations between ψ), θ), and π) In order to study the counting function π), we use Chebyshev functions ψ) and θ) since they are easier to deal with in comparison to π). We also introduce another function T ) = n log n = log[]!), where [] denotes the greatest integer function. The following lemma [16, p. 104] gives the basic relations between ψ), θ), T ), and π). Lemma 1.1. Let > 0. We have ψ) = p n,n 1 log p = p 4 [ ] log log p, 1.2) log p

12 and, for every ε > 0, T ) = log[]!) = ψ = n) n 1 k,n 1 θ n ) 1/k ), 1.3) ψ) log θ) ψ), 1.4) θ) log π) θ) 1 ε) log + 1 ε. 1.5) Proof. We start by proving 1.2). Using definitions of ψ) and θ) we see that ψ) = θ 1/n ) = log p = n 1 n 1 p 1/n Since the summand is zero when n > log / log p, we have ψ) = [ ] log log p. log p p p n,n 1 log p. Net we prove 1.3). To do this we require to show that, for every 1, []! = p p αp, where α p = [ To establish 1.6) it suffices to show that ) log[]!) = log p αp = [ ] ) n log p, p k p n where n = []. We have p k=1 k=1 p k ]. 1.6) log[]!) = log d) = log d = log p = log p 1 d n d n d n p k d p k n d n,p k d [ ] ) n log p. = p n k=1 p k From which we deduce 1.6). 5

13 It follows from 1.6) that log[]!) = α p log p = [ ] log p p k p p k=1 = 1 log p = log p. 1.7) p k=1 n p k k,n 1 p n) 1/k On the other hand ψ n) n 1 = ψ n) n = log p = log p. 1.8) n p k n,k 1 k,n 1 p n) 1/k Moreover, following the definition of θ, we find that ) ) 1/k θ = log p. 1.9) n k,n 1 k,n 1 p n) 1/k The proof of 1.3) is immediate by putting together 1.7), 1.8), and 1.9). We verify 1.4) by noting that ψ) θ) = n 2 θ 1/n ) = n 2 It follows that ψ) θ) 0. Moreover we have log p This implies p n 2 p n ) log log p = log p p 2 n log log p p log p. p n log p 1 log log. ψ) log θ). We now prove the left-hand side of 1.5). We have θ) = p log p log p 1 = log )π). 6

14 For the right-hand side of 1.5) we observe that θ) 1 ε p log p log 1 ε ) 1 ε p 1 = 1 ε)log )π) π 1 ε )). This together with the trivial bound π 1 ε ) 1 ε achieve the proof. The relations between ψ), θ), and π) in Lemma 1.1 suggest that by establishing an upper or lower) bound for any of these functions we will be able to deduce an upper or lower) bound for the other functions. This leads us to the following lemma. Lemma 1.2. Suppose that f) and g) are two distinct functions in the set { log )π), θ), ψ) }. If b f) a when is large enough then, for ε > 0, we have for large values of. b ε g) a + ε, Proof. We consider the case f) = π), and g) = θ), the proofs for other cases are similar. Assume that π) a/ log, when 0. From this, by the left-hand side of 1.5) we see that θ) a when 0. We net assume that b/ log π) when 0. We combine this, with the right-hand side of 1.5) to obtain 1 ε ) b log ) θ), ε for ε > 0. Since log / ε decreases to zero, for large we have 1 ε ) b ε ) θ). This completes the proof. 7

15 1.3 Eplicit bounds for π), θ), and ψ) It is not difficult to find numerical upper and lower bounds for π) and the Chebyshev functions. We net establish a numerical lower bound for π). The main argument is due to Nair [15, p. 126]. Lemma 1.3. For 4, log π). Proof. We let d n = lcm 1 m n {m}, where lcm denotes the least common multiple, and Since we have I = 1 0 r=0 I = ) n = n 1 ) n d. n ) n 1) r r, r r=0 n ) n 1) r n+r d = r n ) n 1) r 1 r n + r ) r=0 Since [0, 1] we have 1 ) 1/4. Therefore I 1 4 n. 1.11) Observe that every denominator in 1.10) is not greater than 2n + 1, therefore Id 2n+1 is a positive integer. So together with 1.11) we find that 4 n d 2n ) Let α N such that p α is the eact power of p which divides d 2n+1. Thus p α 2n + 1. This leads to d 2n+1 p 2n+1 We combine 1.12) and 1.13) and take the logarithm to obtain n log 4 log d 2n+1 p 2n+1 log2n + 1) log p 8 p log2n+1) log p. 1.13) log p = log2n + 1)π2n + 1).

16 It follows that Observe that 2n 2) log 2 log 2n 2n log 2 log2n + 1) < 2n 2) log 2 log2n 1) π2n + 1). 1.14) π2n 1) π2n). Therefore 2n 2) log 2 log 2n π2n). 1.15) Putting together 1.14), and 1.15) gives that for every natural number N 2, N 2) log 2 log N πn). 1.16) In order to obtain a lower bound for π) for R, we proceed as follows. We combine 1.16) with the facts that [], and 1 [] to deduce 3) log 2 log [] 2) log 2 log[] π[]) π). Thus 3) log 2 log π). We conclude by checking that, for 4, we have 1 3 ) ) 3) log 2 log 2. log 4 log The left-hand side of the above inequality gives the desired constant Net we present a lemma which gives an upper bound for ψ) see [16, p. 118]). The following argument is due to Mertens. Lemma 1.4. If > 1 then Proof. By 1.3), we may write ψ) < 2. log[]!) = ψ. n) n 1 9

17 From this we find that [ ] log[]!) 2 log!) = ψ/n) 2 ψ/2n). 2 n 1 n 1 Hence [ ] log[]!) 2 log!) = 2 n 1 1) n+1 ψ. n) Since ψ) is an increasing function, from the last identity we have [ ] ψ) ψ/2) < log[]!) 2 log!). 1.17) 2 Let > 14. We define the integer N such that 2N 1) < 2N. We observe that [ ] ) 2N)! log[]!) 2 log!) < log2n)!) 2 logn 1)!) = log. 1.18) 2 N 1)!) 2 By induction we find that Thus 2N)! N 1)!) 2 < e2n 1), for N 8. Putting together 1.17), 1.18), and 1.19) gives ) 2N)! log < 2N 1). 1.19) N 1)!) 2 ψ) ψ/2) <. 1.20) Observe that ψ) = n 0 [ ) )] ψ ψ. 2 n 2 n+1 This combined with 1.20) give ψ) n 0 1 = 2, for > 14. 2n We check that the inequality also holds for 1 < 14. This completes the proof. Having established relations between π), ψ), and θ) in Lemma 1.1, a lower bound for π) in Lemma 1.3 and an upper bound for ψ) in Lemma 1.4, we can deduce the following theorem. 10

18 Theorem 1.5. i) For > 1 we have θ) ψ) < 2. ii) For e 40 we have iii) For e 10 we have π) < log θ) ψ). Proof. We begin with the upper bound for θ). Lemma 1.4 combined with 1.4) yields θ) < 2, for > 1. Using this upper bound for the right-hand side of 1.5) gives π) 2 log 1 ε + log ). 1.21) ε Let ε = 1/10. Since log )/ 1/10 decreases for > e 10, then 1.21) gives π) < log, for e40. Net we put together Lemma 1.3 and the right-hand side of 1.5) to derive 1 ε) log ) θ). ε With ε = 9/10 the last inequality is transformed into From this, by 1.4), we deduce θ), for e ψ), for e 10. The above theorem establishes eplicit upper and lower bounds for the Chebyshev functions and π), however these bounds are rather poor and with more work one can replace them with more precise bounds. Our main goal in this thesis is to develop techniques which enable us to derive sharp eplicit bounds for ψ), θ), and π). 11

19 1.4 Statement of the results Although the Prime Number Theorem states that ψ) is asymptotic to, it does not give any information about the size of the error term ψ). We are interested in finding the size of this error term. This is done in Theorem 2.4 of Chapter 2 which gives an eposition of Rosser s method [17] for finding an eplicit error term in the Prime Number Theorem. To eplain our theorem we introduce the following notations. Let NT ) be the number of non-trivial zeros ρ = β +iγ of ζs) such that 0 < γ T. Let A be the height that the Riemann Hypothesis has been verified. In other words if ζβ + iγ) = 0 for 0 γ A then β = 1/2. We assume there eists r > 0 such that for γ > A we have We define β < 1 1 r log γ. F T ) = T 2π log T 2π T 2π + 7 8, RT ) = d 1 log T + d 2 log log T + d 3, and assume that NT ) F T ) < RT ), for T 2 and reals d 1, d 2, and d 3. Moreover suppose that F A) NA). Also we define c 1,, c 6 as c 1 = rlog A) 2, c 2 = r log A, c 3 = RA), c 4 = 1 2π + d 1 A log A 2π c 5 = + d 2 Alog A)log A 2π ), 1 loga/2π), 12

20 and c 6 = loga/2π). In Chapter 2 we give a detailed proof of the following theorem. Theorem 2.4. Let m N and a > 0 such that Suppose that and log a < c 1m 2 m + c 5. ρ 1 k, γ m+1 δ = 2 a 1/2 c 3 k + A m+1 a + c mc 6 ) 1/c 2 1 m+c 5) log a m 2 A m a 1/c 2 If 1 + mδa < a we have for a. ε = δ δ) m c 1 m 2 ) ) m + m). 1 ε) log 2π < ψ) < 1 + ε) 1 2 log 1 1/ 2), 1/m+1) This theorem is essentially Theorem 21 of [17], however it is stated using parameters instead of the specific constants. So it conveniently will produce new estimates for ψ) whenever new improvements in the values of the parameters occur. The above theorem plays a fundamental role in effective estimations of the functions of primes such as Chebyshev functions and π). By employing this theorem, one can generate tables to provide bounds for ψ) in various ranges. We say some words on the proof of Theorem 2.4. Riemann s eplicit formula represents the error term in the Prime Number Theorem as a sum over the non-trivial zeros of the Riemann zeta function. More precisely, we have, ψ) = ρ ρ ρ log 2π 1 2 log1 2 ), 13

21 see [5, p. 60]). So φ) = ψ) + log 2π log 1 2) = ρ ρ ρ measures the error term in the Prime Number Theorem. In order to bound φ) we require to estimate ρ ρ /ρ. This is not most convenient since the sum is not absolutely convergent. To resolve this difficulty we introduce K m, h) = h 0 h dy 1 dy 2 0 h 0 φ + y 1 + y y m ) dy m, where m N and h > 0. Net we will bound ψ) in terms of K m, h). We have 1 ε 1 ) log 2π 1 2 log 1 2) ψ) 1 + ε 2 ) log 2π 1 2 log 1 2), where and 0 < δ < 1)/m. have ε 1 = K m, δ) ) m+1 δ + mδ m 2, ε 2 = K m, δ) m+1 δ + mδ m 2, We note ψ) can be related to the non-trivial zeros of ζs). More precisely we 1 ψu)du = 2 2 ρ ρ+1 ρρ + 1) 0) ζ ζ0) + ζ 1) ζ 1) 1 2r 2r2r 1), see [11, p. 73]). This allows us to derive an epression for K m, h) in terms of the zeros of ζs). We have K m, h) = ρ 1 ρρ + 1) ρ + m) r=1 m ) m 1) ) j+m+1 + jh) ρ+m. j Net by employing the properties of the zeros of ζs), we shall bound K m, h) by j=0 K = ρ β 1 / γ m+1. Finally we will prove that ε 1, ε 2 < δ δ) m ) m + m),

22 provided that we have K δ/2) m+1. This will complete the proof of the theorem. In Chapter 3 we derive upper and lower bounds for θ). Upper bounds for θ) can be obtained by means of the lower bound for ψ) θ) together with an estimation table for ψ). More precisely, we proceed by considering intervals 0, e b 1 ], [e b 1, e b 2 ], and [e b 2, ), where b 1 and b 2 are fied positive constants, and do the following. We numerically check that θ) <, for 0, e b 1 ]. When [e b 1, e b 2 ], we estimate θ) by means of a lower bound for ψ) θ). We give here some of the known inequalities for ψ) θ) that can be employed to obtain such lower bounds. ψ) θ) ψ 1/2 ) + θ 1/3 ), ψ) θ) ψ 1/2 ) + ψ 1/3 ) + ψ 1/5 ) ψ 6 ), and ψ) θ) ψ 1/2 ) + ψ 1/3 ) + ψ 1/5 ) θ 1/6 ) ψ 1/30 ). Finally when [e b 2, ), we bound θ) by using the trivial inequality θ) ψ). We then consider the maimum of the upper bounds for θ) derived over the intervals 0, e b 1 ], [e b 1, e b 2 ], [e b 2, ) as the upper bound of θ) for > 0. To eplain our results more specifically, let A b) < ψ) < A + b), for e b. In other words A + b) and A b) are upper and lower bounds for ψ)/ on the interval [e b, ). 15

23 θ). The following theorem shows that how one can improve a given upper bound for Theorem Let b 1 and b 2 be positive constants such that 0 < b and b 1 < b 2. Let c 1 = { ma A + b 1 ) A b 1 /2) A b 1 /3) A b 1 /5) + c } 0 [e b 1,e b 2 ] 1/2 2/3 4/5 + A+ b 1 /30), 5/6 29/30 where c 0 is an upper bound for θ)/ when > 0. Then where c 0 = ma{c 1, A + b 2 )}. θ) < c 0, for > 0, The above will allow us to establish an upper bound for θ) valid for all > 0 which surpasses [7, Proposition 5.1]. Eample θ) < ), for > 0. We point out that in the process of establishing upper bounds for θ), we need lower bounds for ψ) θ). It is known that ψ) θ) >, e 36.8, see [18, p. 73]). In Chapter 3, we will etend the above range significantly to obtain Theorem ψ) θ) >, 121 e We net employ upper bounds for ψ) θ) to generate several lower bounds tables for θ) over different ranges. Finally, in Chapter 3, we present some techniques which allow us to derive inequalities in the form ψ) θ) < c 2 1/2 + c 3 1/3, for c 2, c 3 > 0. We will prove the following theorem. Theorem Suppose that for e b there is a positive constant ε such that ε > A + b/2) 1 > 0 and e b 4A + b/5) 5 A + b/2) 1 ε) 16 ) 10 3.

24 We let Then h) = A + b/2) 1 ε ) 1/6 + A + b/3) + A + b/5) 2/15. ψ) θ) < 1 + ε) 1/2 + he b ) 1/3, for e b. In Chapter 4 we demonstrate some techniques which give sharp estimates for π) over different ranges. Note that by the Prime Number Theorem with the remainder we have π) = li) + Oe c log ), for some constant c. It follows from [13, p. 55] that if 2 < then π) < li). Therefore we will be able to establish upper bounds for π) for by establishing upper bounds for li). For eample we prove the following inequality. ) Corollary 4.5. π) < , for log log Net we assume that we have positive constants β and η k and a natural number k such that We let 0 β, and introduce J, η k ) = π 0 ) θ 0) log 0 + θ) < η k, for β. log k log + η k log k log 2 y + η ) k dy, log k+2 y We shall verify J, η k ) < π) < J, η k ), for k 1 and 0. This enables us to estimate π) by employing J, ±η k ). Here we give a sample of our results on estimation of π). In Chapter 4, we prove that for 0, there eist positive numbers d 1, d 2, and d 3 such that π) < π) < log 1 + d ) 1, log log log + d 2 log 2 ), 17

25 π) < log log + 2 log 2 + d ) 3 log 3. Admissible values for d 1, d 2, and d 3 are given in Table 1.1. Table 1.1: 0 d 1 d 2 d Also we will prove that for 10 11, π) > π) > log ), log log log ) log 2, and for 10 10, π) > log log + 2 log ) log 3. We also establish another form of bounds for π). As a sample of our results we will show that log < π) < log , for

26 Chapter 2 Bounds for ψ) 2.1 Elementary estimates of ψ) Around 1850 Chebyshev introduced the function ψ) and obtained numerical bounds for it. More precisely he proved that for every ɛ > 0 and sufficiently large, where ν ɛ) ψ) ɛ)ν, ) 2 1/2 3 1/3 5 1/5 ν = log = ) 30 1/30 Note that 6/5)ν = Over the years many authors attempted to improve the Chebyshev estimate and also make the above estimation effective. For eample Erdős [8] proved that ψ) < log 4 = , for > 0. This bound has been improved later by Grimson and Hanson [9] who obtained ψ) < log 3 = , for > 0. Another improvement was given by Deshouillers [6]. He verified that ψ) < , for > 0, 19

27 and < ψ), for 59. All the above estimates are obtained by elementary means and without employing the analytic properties of the Riemann zeta function. Later in this chapter we study the analytic estimates of ψ), however before doing this we give an eplicit version of Chebyshev s theorem as given by Landau in [14, Vol. 1, p.88]. Recall that by Lemma 1.3) we have T ) = n Using this we verify the following. log n = Lemma 2.1. Let > 0 and ) ) ) α) = T ) T T T + T n=1 ψ. 2.2) n) 30). Then ν 5log + 1) α) ν + 5log + 1). 2.3) Proof. By partial summation formula [11, Theorem A, p.10] we have T ) = n log n = [] log 1 [t] t dt, where [] is the greatest integer less than or equal to. Writing [] = {}, we have T ) = {}) log 1 t {t} dt. t By splitting the integral at = 2 we find that T ) = log + U), 2.4) where U) = 2 {t} dt {} log log t 20

28 By using the fact that 0 {} < 1 we observe that We combine 2.4) and 2.5) to obtain U) log ) T ) log + log + 1. From the definition of α) and the above inequality we have α) log + 2 log log log 5 30 log log log log log log + 1 = 5log + 1). 30 Recognizing ) 2 1/2 α) log 3 1/3 5 1/5, 30 1/30 in the left-hand side of the last inequality implies the proof. In the net lemma we establish upper and lower bound for ψ) in terms of α). Lemma 2.2. For 2 we have ) α) < ψ) < ψ + α). 2.6) 6 Proof. Using 2.2) we find that α) = ) ) ) ) ψ ψ ψ ψ + ψ n) 2n 3n 5n 30n n=1 n=1 n=1 n=1 ) ) = ψ) ψ + ψ ψ + ψ ψ + ψ ) 11) 12) 13) ψ + ψ ψ + ψ ψ + ψ 15) 17) 18) 19) 20) 23) ψ + ψ ψ +. 24) 29) 30) n=1 We claim that α) = n=1 A n ψ, 2.7) n) 21

29 where 1 if gcdn, 30) = 1, A n = 1 if gcdn, 30) has at least two distinct prime divisors, 0 otherwise. In order to verify this, we split integers in types: 1) gcdn, 30) = 1 which gives A n = = 1. 2) Only one of 2, 3 and 5 divides n. This gives A n = = 0. 3) Only two of 2, 3 and 5 divides n. Hence we have A n = = 1. 4) 30 n which gives A n = = 1. We now let c 0 = 1 < c 1 < c 2 < be the sequence of all integers n such that A n 0. We observe that A cn = 1) n. Therefore 2.7) can be written as follows. ) α) = 1) n ψ. It follows that ) ψ) ψ 6 from which the proof follows. Theorem 2.3. For 2 we have n=1 ) ) ) = ψ ψ < α) < ψ = ψ), c 0 c 1 c 0 ν 5log + 1) < ψ) < 6 ν + 3 log + 5) log + 1), 2.8) 5 where ν being given by 2.1). Proof. We begin with the right-hand side of 2.8). deduce Hence for all n N ) ψ 6 n c n ) ψ) ψ < ν + 5log + 1). 6 ) ψ ν 6 n log ) n n ν + 5 log + 1). 6n 22 We combine 2.3) and 2.6) to

30 Summing on n = 0, 1,, [log / log 6], the sum on the left-hand side gives a telescoping series and we obtain proof. ν ψ) = [ log log 6] n=0 1 6 n + 5 [ log log 6] n=0 ) )) ψ ψ 6 n 6 n+1 [ log log 6 < 6 5 ν + 5 log 6 log + 5 ] ) + 1 log + 1) ) log + 1). We obtain the announced upper bound for ψ) by noting that 5/ log 6 < 3. Now 2.3) combined with 2.6) gives the left-hand side of 2.8). This completes the to Note that a careful reading of the above proof allows us to tighten the upper bound 1.2ν log log Analytic estimates of ψ) As we saw in the previous section, the elementary method of Chebyshev will establish the eplicit inequality ν ɛ) ψ) 6 5 ν + ɛ), for ɛ > 0 and large values of. From the Prime Number Theorem we know that 1 ɛ) < ψ) < 1 + ɛ), for large values of. To establish the latter type inequality one needs to find eplicit estimates for ψ), the so called the error term of the Prime Number Theorem. Rosser [17] was the first who considered this problem. Rosser used numerical verification of the Riemann hypothesis together with an eplicit zero-free region in study of the error term and developed an analytical method which allowed him to derive eplicit estimates for functions involving primes. Rosser s estimates were more precise 23

31 than the ones which have been established by elementry methods. His results were later improved by Rosser and Schoenfeld [18], [19] and Schoenfeld [20]. Among many other results, Rosser and Schoenfeld [18] proved that ψ)/ takes its maimum at = 113. More precisely ψ) < , for > 0. Eplicit upper and lower bounds for ψ) can be obtained by means of Theorem 21 of [17]. In this chapter we will give a detailed proof of this theorem. We start by setting up our notations. We define φ) = ψ) + log 2π log 1 1/ 2). φ) can be considered as the error term in the Prime Number Theorem. In Theorem 2.4, we give an upper and lower bounds for this error term. Let m be a positive integer and and h be positive reals. For > 1 and +mh > 1, we define the multiple integral K m, h) = h 0 h dy 1 dy 2 0 h 0 φ + y 1 + y y m ) dy m, and f m,n,a, h, z) = K m, h) + 1 h n 2 nha zh a ) Since for h > 0, y i [0, h], and so we have +y 1 + +y m >. Thus φ+y y m ) eists provided that > 1. Also for h < 0, y i [h, 0] we have + y y m > + mh. Hence φ + mh) eists provided that + mh > 1. So the above definitions are well defined. Net we review some facts about the zeros of the Riemann zeta function. known that ζs) has zeros at all negative even numbers see [11, p. 49]). This zeros are called the trivial zeros of ζs). It is It is known that all the other zeros of ζs) are in the strip 0 < Res) < 1 see [11, p. 58]). We denote the non-trivial zeros of ζs) by ρ = β + iγ, where 0 < β < 1 and γ R. Let NT ) be the number of non-trivial zeros of ζs) such that 0 < γ T. 24

32 Suppose that A > 0 is such that for 0 < γ A we have β = ) Also assume that r > 0 is such that for γ > A we have β < 1 1 r log γ. 2.11) and We define and assume that F T ) = T 2π log T 2π T 2π + 7 8, RT ) = d 1 log T + d 2 log log T + d 3, NT ) F T ) < RT ), 2.12) for T 2 and real numbers d 1, d 2, and d 3. Moreover suppose that F A) NA). We also define c 1,, c 6 as follows. c 4 = 1 2π + d 1 A log A 2π c 1 = rlog A) 2, 2.13) c 2 = r log A, 2.14) c 3 = RA), 2.15) + d ) Alog A)log A ), 2π and c 5 = 1 loga/2π), 2.17) c 6 = loga/2π). 2.18) We will prove the following theorem. Theorem 2.4. Let m N and a > 0 such that log a < c 1m 2 m + c 5. 25

33 Suppose that and ρ 1 k, γ m+1 δ = 2 a 1/2 c 3 k + A m+1 a + c mc 6 ) 1/c 2 1 m+c 5) log a m 2 A m a 1/c 2 If 1 + mδa < a we have for a. ε = δ δ) m c 1 m 2 ) ) m + m). 1/m+1) 1 ε) log 2π < ψ) < 1 + ε) 1 2 log 1 1/ 2), 2.19) Before presenting the proof of this theorem, we give an overview of the three main steps of the argument. Step 1: We first find a lower and upper bound for the error term of the Prime Number Theorem ψ), in terms of K m, h). This is done in Proposition 2.11 by the following inequalities. Km, δ) ) m+1 δ + mδ ) log 2π 12 1 m 2 log 1 ) ψ), 2 and Km, δ) ψ) m+1 δ m where 0 < δ < 1)/m. + mδ ) log 2π log 1 ), 2 Step 2: In the net step we will establish the following eplicit bound for K m, h), K m, ±δ) < m δ) m ) m K,, where K = ρ β 1 γ m+1. Combining this with Proposition 2.11, in Proposition 2.15 we will derive lower and upper bounds for ψ) in terms of the sum K. 26

34 Step 3: Net we bound K by splitting the sum between the zeros on the left of the 1/2-line and those on the left of the zero-free region: where K 1 2 ρ 1 γ m+1 + A<γ fγ) = 1 r log γ γ m+1. fγ), This is done in Lemma 2.16 by means of 2.11) and properties of the zeros of the zeta function. To find an effective upper bound for K it suffices to eplicitly bound A<γ fγ). This has been done in Lemma 2.17 by applying partial summation formula on A<γ fγ). This lemma gives an upper bound for A<γ fγ). It is shown that fγ) < A<γ c 3 A m+1 1/c 2 + c 4 A fy) log y 2π dy. Finally Lemma 2.18 furnishes an eplicit upper bound for fy) logy/2π)dy. Combining all these will establish the desired upper bound for A K. In the net three sections we will describe in details each step Comparing ψ) with its average In this section we compare ψ) to its average K m, h). Proposition 2.11 is the main result of this section and is obtained by using Lemma 2.7 and 2.9. These two lemmas give upper and lower bounds for φ) depending on m-average K m, h). For proving them we need Lemma 2.5 and 2.6. Lemma 2.5 establishes a recurrence relation for f m,n,a. Lemma 2.5. We have h 0 f m,n,a, h, y 1 + y y n ) dy n = f m,n 1,a+1, h, y 1 + y y n 1 ). 27

35 Proof. We use 2.9) to compute the integral h 0 f m,n,a, h, y 1 + y y n ) dy n h Km, h) = + 1 ) 2 nha y 1 + y y n )h a 1 dy n 0 h n = K m, h) + n 1 )h a+1 y h n y n 1 ) h a 2 = f m,n 1,a+1, h, y 1 + y y n 1 ). Lemma 2.6. We have K m, h) = h 0 h dy 1 dy 2 0 h Proof. We prove this lemma by induction on n. 0 f m,n,a, h, y 1 + y y n ) dy n. 2.20) For n = 1 we use 2.9) and compute the right-hand side of 2.20) to get h 0 We assume f m,1,a, h, y)dy = K m, h)h h K m, h) = h 0 h dy 1 dy 2 0 h ha a 1 h2.h h 2 = K m, h). f m,n,a, h, y 1 + y y n ) dy n, and together with Lemma 2.5 we obtain h h h ) f m,n+1,a, h, y y n+1 ) dy n+1 dy 1 dy n 0 h = h 0 f m,n,a+1, h, y y n ) dy 1 dy n = K m, h). We now define f m, h, z) = f m,m,1, h, z) = K m, h) h m 28 + mh 2 z.

36 Lemma 2.7. If h > 0 then there eists z [0, mh] such that φ + z) f m, h, z). 2.21) Proof. Assume to the contrary that, for all z [0, mh], we have Then h 0 h 0 φ + z) > f m, h, z). φ + y 1 + y y m ) dy 1 dy 2 dy m > h 0 h 0 f m, h, y y m ) dy 1 dy 2 dy m. We recognize K m, h) in the left-hand side of the above inequality. However by Lemma 2.6 the right-hand side of it equals K m, h). This is a contradiction. Remark 2.8. Formula 2.21) states that there eists z 1 [0, mh] such that where ξ 1 = 1 h m h 0 Observe that 1 h m h 0 h 0 h 0 φ + z 1 ) + z 1 ξ 1, φ + y 1 + y y m ) + y 1 + y y m )) dy 1 dy 2 dy m. y 1 + y y m ) dy 1 dy 2 dy m = m h h m = mh 2. Lemma 2.9. If h < 0 then there eists z [mh, 0] such that 0 h 0 y 1 dy 1 dy m f m, h, z) φ + z). 2.22) Proof. Assume to the contrary that for all z [mh, 0] we have φ + z) < f m, h, z). 29

37 Then 0 < h 0 h 0 h 0 φ + y 1 + y y m ) dy 1 dy 2 dy m h f m, h, y y m ) dy 1 dy 2 dy m. By definition the left-hand side of the above inequality is 1) m K m, h). However by Lemma 2.6 the right-hand side of that equals 1) m K m, h). contradiction. Remark Formula 2.22) states that there eists z 2 [0, mh] such that where ξ 2 = 1 h m h 0 h 0 φ z 2 ) z 2 ξ 2, Thus we get a φ + y 1 + y y m ) y 1 + y y m )) dy 1 dy 2 dy m. We are now ready to give a bound for ψ) in terms of the K m s. Proposition If 0 < δ < 1)/m, then ε 1 = K m, δ) ) m+1 δ + mδ m 2, and ε 2 = K m, δ) m+1 δ + mδ m 2, 2.23) 1 ε 1 ) log 2π 1 2 log ) ψ) 1 + ε 2 ) log 2π 1 2 log ). 2.24) Proof. To prove the right-hand side of 2.24) we use the definitions of φ + z) and f m, h, z). Lemma 2.7 yields that there eists z [0, mh] such that ψ + z) + z) + log 2π + 1 ) 1 2 log 1 K m, h) + mh + z) 2 h m 2 For z 0, we have the inequalities log 1 + z) 2 ) 30 z. 2.25) log 1 ), 2.26) 2

38 and ψ) ψ + z). An application of this fact and 2.26) in 2.25) and replacing h by δ in 2.25) yield ψ) + z) + log 2π log 1 1 ) K m, δ) + mδ z. 2 δ) m 2 By simplifying the above inequality we get ψ) 1 + K m, δ) m+1 δ + mδ ) log 2π 12 1 m 2 log 1 ). 2 This proves the inequality in the right-hand side of 2.24). For proving the inequality in the left-hand side of 2.24), note that by Lemma 2.9 there eists z [mh, 0] such that ψ + z) + z) + log 2π log 1 Note that for z 0 we have 1 2 log 1 ) 1 K m, h) + mh z. 2.27) + z) 2 h m 2 ) z) 2 2 log 1 ), 2 and ψ) ψ + z). We replace h by δ in 2.27) to obtain ψ) + z) + log 2π log 1 1 ) K m, δ) + mδ z. 2 δ) m 2 Simplifying the above inequality yields Km, δ) 1 ) m+1 δ + mδ )) log 2π 12 1 m 2 log 1 ) ψ), 2 which is the desired result Relating ψ) and the zeros of the zeta function In this section we find an eplicit bound for K m, h) in terms of the zeros of the zeta function. This is done by using Proposition 2.13 which provides an eplicit formula. Then Proposition 2.14 gives a bound for K m, ±δ) in terms of K = ρ β 1 / γ m+1. The last result of this section is Proposition 2.15 which gives bounds for ψ) in terms of δ. 31

39 Proposition We have h 0 φ + z)dz = ρ 1 ρ+1 + h) ρ+1). 2.28) ρρ + 1) Proof. Using the definition of φ) we have φ + z) = ψ + z) + z) + log 2π log 1 ) 1. + z) 2 We split φ + z) into two parts and calculate the integral from 0 to h of them separately. By change of the variable we have h 0 ψ + z)dz = +h ψu)du = +h 1 ψu)du We now employ the classical eplicit formula [11, page 73], 1 ψu)du = 2 2 ρ and apply it on the right-hand side of 2.29) to get h 0 ψ + z)dz = h2 2 + h ρ Net we consider +h 1 ψu)du. 2.29) ρ+1 ρρ + 1) 0) ζ ζ0) + ζ 1) ζ 1) 1 2r 2r2r 1), + h) ρ+1 ρ+1 ρρ + 1) r=1 h ζ 0) ζ0) r=1 + z log2π) 12 )) log 1 1 dz. + z) 2 By using the Taylor epansion ) 1 log 1 = + z) 2 it follows that r=1 1 r + z) 2r, + h) 1 2r 1 2r. 2r2r 1) 2.30) ) h + z log2π) dz 0 2 r + z) 2r r=1 = h + h2 2 h log 2π 1 + h) 1 2r 1 2r. 2.31) 2 r2r 1) r=1 32

40 From [14, page 317] we have ζ 0) ζ0) = log 2π. Having this we now subtract 2.30) from 2.31) to obtain 2.28). Proposition We have K m, h) = ρ 1 ρρ + 1) ρ + m) m ) m 1) ) j+m+1 + jh) ρ+m. 2.32) j j=0 Proof. The proof follows by induction on m. By using the definition of K m, h) and Proposition 2.12 we find that K 1, h) = h 0 φ + z)dz = ρ 1 ρ+1 + h) ρ+1) ρρ + 1) = 1 ) 1 1 1) ) j+2 + jh) ρ+1. ρρ + 1) j ρ j=0 We now assume K m, h) = ρ We need to prove that K m+1, h) = ρ 1 m ) m 1) ) j+m+1 + jh) ρ+m. 2.33) ρρ + 1) ρ + m) j j=0 1 ρρ + 1) ρ + m + 1) m+1 By using the definition of K m+1, h) we have K m+1, h) = h 0 0 h ρ 0 h 0 ) m + 1 1) ) j+m+2 + jh) ρ+m+1. j j=0 φ + y y m+1 ) dy 1 dy m ) dy m+1 j=0 = h 0 K m + y m+1, h)dy m+1. Net we apply 2.33) on the above equality to obtain h 1 m ) ) m K m+1, h) = 1) j+m+1 + y m+1 + jh) ρ+m dy m+1. ρ ρ + m) j 33

41 Interchanging the sum and the integral is legitimate since 1 m ρ ρ + m) ρ j=0 = ρ ) ) m 1) j+m+1 + y m+1 + jh) ρ+m dy m+1 j + y m+1 ) ρ+m ρ ρ + m) m j=0 ) m 1 ± jδ j + y m+1 ρ + h) m+1 γ m+1 m j=0 ) ρ+m ) ) m )1 + jδ) m+1, j noting that ρ 1/ γ m+1 is convergent for m 1. Thus we obtain K m+1, h) = 1 m ) m h 1) j+m+1 + y m+1 + jh) ρ+m dy m+1. ρ ρ + m) j ρ 0 j=0 It is now easy to compute the integral: h 0 + y m+1 + jh) ρ+m dy m+1 = + jh + h)ρ+m+1 + jh) ρ+m+1. ρ + m + 1 Therefore the sum over j equals m ) m + j + 1)h) 1) j+m+1 ρ+m+1 + jh) ρ+m+1. j ρ + m + 1 j=0 The above can be rewritten as: 1 m ) m 1) j+m+1 + j + 1)h) ρ+m+1 ρ + m + 1 j j=0 1 m m 1) j+m+1 ρ + m + 1 j Changing the variable j to j 1 in the first sum of the above formula gives m+1 1 ) m 1) j+m + jh) ρ+m+1 ρ + m + 1 j 1 j=1 1 ρ + m + 1 j=0 m m 1) j+m+1 j j=0 ) + jh) ρ+m+1. ) + jh) ρ+m+1. 34

42 The last epression can be transformed into since 1 ρ + m + 1 m j=1 ) m + 1 1) j+m + jh) ρ+m+1 j 1 + 1) 2m+1 + m + 1)h) ρ+m+1 1) m+1 ρ+m+1), 2.34) ρ + m + 1 ) m + j 1 ) m = j m + 1 j ), for 1 j m. To complete the proof we observe that 2.34) can be rewritten as m+1 1 ) m + 1 1) ) j+m+2 + jh) ρ+m+1. ρ + m + 1) j j=0 This leads to K m+1, h) = ρ m+1 1 ) m + 1 1) ) j+m+2 + jh) ρ+m+1. ρρ + 1) ρ + m + 1) j j=0 Proposition If δ 0 then K m, ±δ) < m δ) m ) m K, 2.35) where K = ρ β 1 γ m+1. Proof. Replacing h by ±δ in Proposition 2.13 yields K m, ±δ) = ρ+m m ) m 1) )1 j+m+1 ± jδ) ρ+m. ρρ + 1) ρ + m) j ρ From ρ+m = β+m and ρ = β 2 + γ 2 γ, we find that ρ+m ρρ + 1) ρ + m) < β+m m+1 m+1 < m ) γ γ 35 j=0

43 Therefore by using the fact that 1 + jδ) < 1 + δ) j for δ > 0 one can deduce m ) m m ) m 1 1) j+m+1 1 ± jδ) ρ+m < ) + δ) j m+1 j j j=0 j=0 m ) m 1 = ) + δ) m+1 j = 1 + δ) m ) m. j j=0 Thus m m 1) j+m+1 j j=0 ) 1 + jδ) ρ+m 1 + δ) m ) m. 2.37) The proof follows by putting together 2.32), 2.36), and 2.37). Proposition If δ 2K 1/m+1) and ɛ = δ ) 1 + δ) m+1 m m), 2 2 then ε 1 < ɛ, ε 2 < ɛ, 2.38) where ε 1 and ε 2 are defined in 2.23). Proof. It follows from Proposition 2.14 that ε 1, ε δ)m+1 + 1) m K δ m + mδ ) Since δ 2K 1/m+1) we have K < δm+1. Applying this upper bound for K, 2.39) is 2 m+1 transformed into ε 1, ε 2 < 1 + δ)m+1 + 1) m δ m+1 δ m 2 m+1 from which the proof follows. + mδ 2 = δ δ) m ) m + m), 36

44 2.2.3 Eplicit bounds for the sums over the zeros of the zeta function In this section we will derive an upper bound for K by finding an upper bound for A<γ fγ) where fγ) = 1/r log γ /γ m+1. Lemma We have K 1 2 ρ 1 γ m+1 + A<γ fγ). 2.40) Proof. For β 1/2 we have β 1 1/2. Hence β 1 2 β 1 γ m+1 β 1 2 1/2 γ m+1 1/2 ρ ) γ m+1 If β > 1/2 it follows from 2.10), that γ > A. We now use 2.11) to deduce β 1 < 1 r log γ. From this, by the fact that both β + iγ and β iγ are zeros of ζs), we find that β> 1 2 β 1 γ m+1 = 2 γ m+1 2 β 1 β> 1,γ>0 2 1 r log γ γ β> 1 2,γ>A Since β + iγ and 1 β + iγ = β + iγ are zeros of ζs), we observe that 2 1 r log γ β> 1 2,γ>A γ m+1 = 1 r log γ β> 1 2,γ>A = 1 r log γ β> 1 2,γ>A γ m+1 + Hence 2.42) can be transformed into β> 1 2 γ m r log γ γ m+1 β < 1 2,γ>A β 1 γ m+1 γ>a The proof is now complete by adding 2.41) to 2.43) r log γ γ m+1 β> 1 2,γ>A γ>a m ) 1 r log γ γ m+1. 1 r log γ. 2.43) γm+1

45 Net we find an upper bound for A<γ fγ). In order to do this we use the partial summation formula for A<γ fγ). We also make use of the upper bound for Ny) given in 2.12). Lemma If log c 1 m + 1) we have fγ) < A<γ c 3 A m+1 1/c 2 + c 4 A fy) log y 2π dy, where c 1, c 2, c 3, and c 4 are defined in 2.13), 2.14), 2.15), and 2.16). Proof. By employing the partial summation formula, we see that We have fγ) = A<γ f y) = ep A ) log r log y y m+2 Ny)f y)dy NA)fA). 2.44) ) log m + 1). 2.45) rlog y) 2 Since y > A and log c 1 m + 1) = rlog A) 2 m + 1), we find that log m + 1) log rlog y) 2 m + 1) < 0. rlog A) 2 We combine this with 2.45) to deduce that f y) > 0. Ny) F y) + Ry) in 2.44). We obtain fγ) < A<γ We use integration by parts on the above integral to derive A A F y) + Ry)) f y)dy = F A) + R A)) fa) From this and 2.46) we find that fγ) < A<γ A This enables us to use F y) + Ry)) f y)dy NA)fA). 2.46) A F y) + R y)) fy)dy. F y) + R y)) fy)dy + RA)fA), 2.47) 38

46 noting that F A) NA). Observe that RA)fA) = and F y) = 1 logy/2π). This implies 2π Since A < y we have Therefore A R y) = d 1 y + d 2 y log y RA) A m+1 1 r log A F y)fy)dy = 1 2π < 1 y < d1 log y 2π A log A 2π A log y 2π. A log A 2π where d 1 d 2 α = + A log A Alog A)log A ). 2π 2π By using the latter result it follows that A R y)fy)dy < α + A = c 3 A m+1 1 c 2, 2.48) fy) log y dy. 2.49) 2π ) d 2 log y 2π Alog A)log A ) = α log y 2π, 2π log y fy)dy. 2.50) 2π Putting together 2.47), 2.48), 2.49), and 2.50) completes the proof. The net lemma provides an upper bound for fy) logy/2π)dy. A Lemma If log < c 1 m 2 /m + c 5 ), then A fy) log y 2π dy < 1 + c 6 m ), 1 m+c 5) log c 1 m m A m 1/c 2 2 where c 1, c 2, c 5, c 6 are defined in 2.13), 2.14), 2.17) and 2.18). Proof. Let I = A fy) log y 2π dy = A ) 1 y log log ym+1 2π )ep dy. r log y 39

47 Choosing u = ) log y 2π u = 1 y m+1 ) log and v = ep yields r log y m log y ) ) log log, v = ep y m m 2 2π r log y rylog y). 2 Using integrating by parts we find that ) 1 + m log A ) 2π log ) log 1 + m log y I = ep + ep 2π) log dy A m m 2 r log A A r log y rlog y) 2 m 2 y ) m+1 ) ) 1 + m log A ) log 2π log ep log y ) 1 + m log r log y 2π log y 2π = ep + dy A m m 2 r log A A y m+1 m 2 rlog y) ) 2 ) 1 + m log A ) 1 2π log + m log log y 2π = ep + fy) log y A m m 2 r log A m 2 rlog y) 2 2π dy. Since A < y we have 1 log y 2π + m log y) 2 which gives ) 1 + m log A ) 2π log I < ep + A m m 2 r log A 1 log A 2π 1 + m log A), 2 log A 2π ) + m log m 2 rlog A) 2 Following the definition of I, from 2.51) we find that ) 1 + m log log I 1 A 2π < m 2 rlog A) 2 A A ) 1 + m log A 2π ep A m m 2 fy) log y dy. 2.51) 2π ) log. r log A It follows that for that I < log < rlog A)2 m 2 m + 1 log A 2π ) 1 + m log A 2π. A m m 2 1 = c 1m 2 m + c 5, ep ) log r log A «1 log 2π A +m log m 2 rlog A) 2. 40

48 2.2.4 Proof of Theorem 2.4 In this section we give the proof of Theorem 2.4. By use of Proposition 2.11, Proposition 2.15, Lemma 2.16, Lemma 2.17 and Lemma 2.18 we deduce Theorem 2.4 which gives eplicit bounds for ψ). Proof. We begin with proving the right-hand side of 2.19). Observe that by Proposition 2.11 we have 1 ε 1 ) log 2π 1 2 log ) ψ) 1 + ε 2 ) log 2π 1 2 log ). Noting that log 2π < 0 it remains to verify that It follows from 2.38) that ε 2 < ε = δ 2 provided that K δ/2) m+1. Recall that by Lemma 2.40 we have K 1 2 ε 2 < ε. 2.52) 1 + δ) m ρ 1 2 γ m+1 + A<γ ) m + m), fγ). We combine this with Lemma 2.17 and Lemma 2.18 to deduce K 1 2 ρ 1 γ m+1 + c 3 A m+1 1 c 2 + From this, by ρ 1/ γm+1 k, we find that Since a we may infer K 1 2 k + c 3 A m+1 1 c 2 + K a 1 2 k + c 3 A m+1 a 1 c 2 + c mc 6 ) ) 1 m+c 5 log ) c 1 m m 2 A m 1 2 c mc 6 ) ) 1 m+c 5 log ) c 1 m m 2 A m 1 2 c 2. c mc 6 ) ) 1 m+c 5 log a) c 1 m m 2 A m a c 2. c 2.

49 Thus K 2 a 1 c 3 2 k + 2 A m+1 a 1 c 2 + 1/m+1 m+1 c mc 6 ) ) 1 m+c 5 log a) c 1 m m 2 A m a 1 2 c 2 = ) m+1 δ, 2 which achieves the proof. For the left-hand side of 2.19), having established Proposition 2.11, we require to prove that ε 1 < ε, 2.53) since log 1 1/ 2 ) > 0. The proof of 2.53) is identical to the proof of 2.52). We end this chapter by providing two eamples based on specific numerical values of parameters in Theorem 2.4. The following lemma which provides eplicit upper bounds for ρ 1/ γ m+1 where m 3 will be used to determine values for k in our eamples. Lemma We have ρ 1 γ 2 < , ρ 1 γ 3 < , and ρ 1 γ 4 < Proof. See [17, Lemma 17, p. 225]. Eample We let A = , r = see [17, pp ]) and m = 2. We see that c 1 = , c 2 = , c 3 = , c 4 = , c 5 = , and c 6 = Observe that c 1 m 2 /m + c Thus we choose a = ep20) so that a < ep ). Following the definition of δ and ε we obtain δ = and ε = Hence by using Theorem 2.4 we deduce < ψ) < , for e

50 Eample Let A = Then Wedeniwski has shown that NA) = and the Riemann Hypothesis is true up to height A. We let r = from [12], and m = 2. We observe that c 1 = , c 2 = , c 3 = , c 4 = , c 5 = , and c 6 = We find that c 1 m 2 /m + c Hence we need to choose a < ep ). We let a = ep20). Following the definition of δ and ε we obtain δ = and ε = Hence by using Theorem 2.4 we deduce < ψ) < , for e 20. By comparing the results of the above eamples we conclude that improving the values of A and r gives sharper results. We can use Theorem 2.4 to generate tables of upper and lower bounds for ψ). In this thesis Tables 5.1 and 5.2 give estimations for ψ). The values given in these tables will be used throughout the thesis

51 Chapter 3 Bounds for θ) 3.1 Introduction In this section we assume that we have a table of upper and lower bounds for ψ). Therefore the positive values A + b) and A b) are given such that A b) < ψ) < A + b), for e b. 3.1) The relation between θ) and ψ) given by ψ) = θ 1/k ). 3.2) k=1 plays a key role in our estimates. The above can be derived directly from the definition of ψ) and θ). From 3.2) we deduce the trivial bound ψ) θ), for > 0. by Using the Prime Number Theorem one can show the asymptotic behavior of θ) θ), 3.3) as. However 3.3) does not give any information about numerical estimates for θ)/. We here mention a result, due to Rosser and Schoenfeld [18, 5.4), page 77 ], on the estimation of θ)/ by elementry methods. 44

52 Proposition 3.1. θ) < 1.11 for > 0. Proof. By direct computation one can show that θ) < for 0 < ) For larger values of we proceed as follows. From Chapter 2, Theorem 2.3 we may infer θ) ψ) < 1.2ν + 3 log + 5) log + 1) for 2, where ν is given in 2.1). Hence θ) < 1.2ν + s) ), where s) = 3 log + 5) log + 1). We note that 1.2ν + s)/ is a decreasing function for > Moreover, 1.2ν + s)/ < 1.2ν + s10 8 )/10 8 < 1.11 for > From this, and 3.4) we deduce that θ) < 1.11 for > 0. It is possible to improve the above bound by elementary methods. For eample Hanson [10] has showed that θ) < log 3) < , for > 0. This bound was improved again by Grimson and Hanson [9], who obtained θ) < , for > 0. A breakthrough came in 1989 where Costa Pereira [4] by employing an elementary method proved that θ) < 532 = for >

53 In another front, as described in the previous chapter, Rosser [17] developed an analytic method which allowed him to obtain sharp numerical estimates for ψ)/ and therefore for functions involving primes such as θ)/. Numerical upper and lower bounds for θ)/ were improved later by Rosser and Schoenfeld [18], [19], and Schoenfeld [20] by some refinements in their analytical method. Table 3.1 summarizes the history of numerical upper bounds for θ)/ which are deduced by analytical methods. Table 3.1: Authors c 0 Rosser & Schoenfeld [18] Rosser & Schoenfeld [19] Schoenfeld [20] Dusart[7] θ) < c 0, for > 0. The goal of this chapter is to develop techniques which give sharper estimates for θ)/ compare to those that we have in the literature. We start by giving upper bound for θ)/. By a recent result of Dusart [7, Proposition 5.1, p. 4] we have θ) <, for 0 < e ) We combine 3.5) with the trivial bound θ) ψ) to obtain θ) < A ), for > 0. This leads to the following proposition. Proposition 3.2. θ) < ), for > 0. Thus we can assume that for c 0 = we have θ) < c 0 for > 0. 46

54 In this chapter we present several methods to improve this upper bound for θ). Net we consider the problem of finding a numerical lower bound for θ)/. To do this we consider the identity ψ) θ) = θ 1/k ). k 2 From the above equation we can derive several forms of upper bounds for ψ) θ), and consequently we will be able to deduce tables for lower bounds of θ) in different ranges. Table 3.2 provides the history of numerical lower bounds for θ)/ given by analytical methods. Table 3.2: Authors B b) b Rosser & Schoenfeld[18] Rosser & Schoenfeld[18] Rosser & Schoenfeld [19] Schoenfeld [20] Dusart [7] θ) > B b), for e b. 3.2 Upper bounds for θ) for > 0 The following inequalities play important roles in establishing bounds for θ). Dusart [7, Proposition 5.1] θ) <, for 0 < e ) Rosser-Schoenfeld [18, Theorem 24] 1/2 < ψ) θ), for e ) 47

55 Rosser-Schoenfeld [18, Theorem 19] 2 1/2 < θ), for 0 < and ) The main goal of this section is studying methods for providing sharp upper bounds for θ)/. We point out that a lower bound for ψ) θ) will result in an upper bound for θ) provided that we have estimates for ψ) in different ranges. Using this fact we present the following theorem. Theorem 3.3. Let b 1 and b 2 be positive constants such that b and b 1 < b 2. Let and Then c 1 = A + b 1 ) A b 1 2 ) e b 2/2, c 0 = ma{c 1, A + b 2 )}. θ) < c 0, for > 0. Proof. Let 0 < < e b 1. By 3.6), we have θ) <. We let [e b 1, e b 2 ]. From 3.2) we deduce that ψ 1/2 ) = θ 1/2k ). 3.9) k 1 Therefore ψ) θ) = θ 1/k ) = θ 1/2k ) + θ 1/2k+1 ). 3.10) k 1 k 1 k=2 The previous identity implies that ψ 1/2 ) ψ) θ). 3.11) From 3.11) and 3.1) we obtain, for [e b 1, e b 2 ], ) θ) ψ) ψ 1/2 ) < A + b 1 ) A b 1 2 ). 3.12) e b