The Prime Number Theorem and the Zeros of the Riemann Zeta Function

Transcription

1 The Prime Number Theorem and the Zeros of the Riemann Zeta Function John Nicolas Treuer Advisor: Adam Fieldsteel A thesis in Mathematics submitted to the faculty of Wesleyan University in partial fulfillment of the requirements for the degree of Master of Arts Wesleyan University Middletown, Connecticut Spring, 5

2 Dedication To my parents: my dad who always asks me to explain what I am writing and my mom who doesn t know what a function is. Acknowledgements I would like to thank my parents: my dad who always asks me to explain what I am writing and my mom who doesn t know what a function is. Without your love and support I would not have written this thesis. I would also like to thank my sister. Next, I would like to thank Professor Fieldsteel. You have been a great advisor and I have enjoyed our long meetings. Next, I would like to thank Wesleyan University s mathematics department. I have enjoyed my time here and the department has inspired me to pursue a Ph.D. Next, I would like to thank my fellow math graduate students who have made this year enjoyable. Finally, I would like to thank everyone else who has helped me including Mary Jane and Tommy. i

3 Contents Introduction The Prime Number Theorem. Notation and Basic Facts Chevbyshev s psi function The Euler Product Formula Dirichlet Series Preliminaries to the Prime Number Theorem The Prime Number Theorem Further Results and Comments Functional Equation of the Zeta Function The Gamma Function Functional Equation of the Zeta Function The Zeros of the Zeta Function in the Critical Strip Functions of Finite Growth, Hadamard Factorization Theorem, and Stirling s Formula Order of growth of ξ(s) Proof that ζ(s) has infinitely many zeros in the critical strip The Zeros of the Zeta Function on the Critical Line 5 5. Further Results ii

4 Introduction In this thesis I will prove the Prime Number Theorem, derive what is called functional equation for the Riemann zeta function and provide separate proofs that the Riemann zeta function has infinitely many zeros in what is called the critical strip and the critical line of the zeta function. I will assume that the reader has completed the equivalent of MATH 53 and MATH 54; that is, one semester of complex analysis and one semester measure theory at Wesleyan University. For the proof of the Prime Number Theorem, I have given all of the details with the hope that the reader will be able to understand the proof in its entirety without having to reference any additional sources. In the later chapters, in order for this thesis to be a manageable length for my master s thesis committee, I will assume that the reader is either familiar with or believes the Poisson Summation Formula, the Hadamard Factorization Theorem, and Stirling s formula for the gamma function. This thesis is expository and all of the theorems in this text were proved at least years ago. This thesis is my adaptation of theorems presented in a variety of sources. At times, some of my sources I believe were wrong and I have corrected them.

5 The Prime Number Theorem Let π(n) be the number of primes less than or equal to n. The Prime Number Theorem π(n) log(n) states that lim n n =, where log is the natural logarithm; that is, the number of primes less than or equal to a number n is approximately n log(n). Computation of the values of π(n) and n log(n) for large integers suggests that this approximation is true. For instance, when n =,,, π(n) = 78, 498 and When n =,,, π(n) = 664, 579 and n log(n) n log(n) = 7, 38. = 6, 4. However, a rigorous proof of the prime number theorem is non-trivial and eluded mathematicians for almost the entirety of the 9th century. In 85, Chebyshev made some progress towards the Prime Number Theorem (Jameson, pg. ) by considering what are now called the Chebyshev theta and psi functions: θ(n) = log(p) where p is prime p n and ψ(n) = p a n log(p) where p a is a power of a prime p. Specifically, using θ(n), Chebyshev showed that for sufficiently large n, cn log(n) π(n) Cn log(n) where c = log() + ɛ (Jameson, pg. 8) and C 3.log(4) (Jameson, pg. 37). It turns out that ψ(n) plays a central role in most proofs of the Prime Number Theorem (of the proofs that I have read, ψ(n) has always been essential while θ(n) has not) and using only elementary methods, one can show, as we will do, that if ψ(n) is approximately n, then π(n) is approximately n log(n).

6 After Chebyshev, little progress was made towards solving the Prime Number Theorem until a bold, new idea emerged in 859. In 859, Riemann suggested that researchers use the zeta function, ζ(s) = n= n s, with domain C rather than R to attack the Prime Number Theorem. Riemann started with a well-known identity, which is the result of the Euler Product Formula, ζ(s) = p prime p s and stated that one could use ideas of complex analysis along with this identity to answer questions about prime numbers. Using Riemann s suggestion and the Riemann zeta function, after what was probably 37 years of hard work by various mathematicians, in 896, the Prime Number Theorem was proved independently by Jacques Hadamard and Charles-Jean de la Vallée Poussin. This thesis will begin by presenting a proof of the Prime Number Theorem. I will not present the original proofs of Hadamard and Poussin, but an adaptation of the proof presented by Elias Stein in his book, []. I will also make use of books written by Jameson, [4], and Widder, []. Throughout this thesis, I will indicate generally which source I am using.. Notation and Basic Facts Throughout this thesis, unless otherwise stated, assume that s is a complex number with σ = Re(s) and t = Im(s). The following proposition will be used repeatedly throughout this thesis: 3

7 Proposition. For all r R, r s = r σ. If r >, then r σ = r σ. Proof: r s = e log(rs) = e slog(r) = e (σ+it) log(r) = e σ log(r) e itlog(r) = e σ log(r) = r σ. If r >, then r σ = r σ. In this thesis, the following will be used: Proposition. For r R with r >, r s = r s. Proof: r s = e (σ it) log(r) = e σ log(r) e itlog(r) = e σ log(r) [cos(tlog(r)) isin(tlog(r))] = e σ log(r) [cos(tlog(r)) + isin(tlog(r))] = r s. Almost all of the thesis concerns the zeta function which is defined as follows: Definition.3 For s with Re(s) >, the zeta function is defined by ζ(s) = n= We will show that ζ(s) is well-defined in the region Re(s) >. An immediate consequence of the definition is the following: n s. Proposition.4 For s with Re(s) >, ζ(s) = ζ(s). Proof: 4

8 ζ(s) = n s = n s = n s = ζ(s). n= n= n= The following facts will be used several times throughout this thesis: Proposition.5 Suppose f(t) is non-negative, integrable, and monotonically decreases on [m, n] where m, n Z. Then Proof: n k=m+ f(k) n m f(t)dt n k=m f(k). Fix r Z with m < r n. Note that f(r) r r f(t)dt f(r ) and extend this to a finite sum to get the desired inequality. Corollary.6 For s R with s >, ζ(s) = n= n s + x s dx Proof: Since ζ(s) monotonically decreases on s (, ), we can apply the previous proposition to show that N n= n s + N x s dx for any N. Letting N, since n= n s desired result. and x s dx converge, we get the. Chevbyshev s psi function Definition.7 The relation between two functions f(x) and g(x) is defined by f(x) f(x) g(x) if and only if lim x g(x) =. We immediately note without proof that is an equivalence relation. 5

9 Definition.8 Define π : N N by π(n) equals the number of primes less than or equal to n. With these definitions, the Prime Number Theorem states that π(n) n log(n). We begin by discussing Chebyshev s psi function. The proofs presented in this section are based on the proofs given by Stein. Definition.9 Define Chebyshev s psi function by ψ(x) = p a x log(p) where p is prime and p a is a power of p. Theorem. If ψ(x) x, then π(x) x log(x). Proof: π(x) x log(x) is equivalent to π(x) log(x) x, so we want to show that π(x) log(x) π(x) log(x) lim lim. x x x x Note that, ψ(x) = p a x log(p) = p x log p (x) log(p) because log p x gives the number of powers of a prime p that are less than x. This equals Thus, ψ(x) x log(x) log(p) log(p) log(x) log(p) = π(x) log(x). log(p) p x p x π(x) log(x) ψ(x) ψ(x) x. If ψ(x) x, then = lim x x = lim x x. Thus, π(x) log(x) lim. x x For the other inequality, note that for all α with < α <, ψ(x) log(p) p x x α p x log(p). 6

10 This last quantity is greater than or equal to the product of the number of primes between x α and x and the infimum of the possible summands, which is (π(x) π(x α )) log(x α ) = α(π(x) π(x α )) log(x). Thus, απ(x) log(x) ψ(x) + απ(x α ) log(x) Since π(x α ) < x α, we get that απ(x) log(x) ψ(x) + αx α log(x). Dividing both sides of the above by x we get that απ(x) log(x) x ψ(x) x + αxα log(x). x Taking the limit supremum of both sides, απ(x) log(x) ψ(x) lim lim x x x x + lim αx α log(x). x x ψ(x) αx Since ψ(x) x, lim x x = and since α <, lim α log(x) x x =. Thus, απ(x) log(x) lim. x x π(x) log(x) This holds for all α <, so it also holds for α = ; that is lim x x, which is the other desired inequality. We have reduced the Prime Number Theorem to showing that ψ(x) x. We now reduce the problem further. Definition. Define ψ (x) = x ψ(u)du. 7

11 Proposition. If ψ (x) x, then ψ(x) x. Proof: This proof will employ the same strategy that the previous proof employed; ψ(x) ψ(x) we will show that lim x x lim x x. Note that since ψ(x) = p a x log(p) is a monotonically increasing function, if α <, then x ψ(u)du ( α)x αx ψ(x)[x αx] = ψ(x)( ) ( α)x because [x αx] is the length of the interval of integration and ψ(x) is the supremum of the integrand over that interval. Likewise, if β >, then βx ψ(u)du (β )x x ψ(x)[βx x] = ψ(x). (β )x Applying the definition of ψ (x) to the above inequality, we get that ψ(x) (β )x [ψ (βx) ψ (x)]. After multiplying the first term in the brackets by β β by x, we get that ψ(x) x (β )x [ψ (βx)β β ψ (x)]. and both sides of the inequality After distributing the x and taking the limit supremum of both sides, we get that ψ(x) lim x x lim x (β ) [ψ (βx)β β x ψ (x) x ]. Since ψ (x) x, we get that ψ(x) lim x x (β ) [β ] = )(β + ) [(β ] β 8

12 = β +. ψ(x) Since β can take any value greater than, the above holds when β =, so lim x x. ψ(x) We can use a symmetrical argument and (*), to conclude that lim x x. We have now reduced proving the Prime Number Theorem to showing that ψ (x) x. On the surface, this does not seem to be an easier problem at all. Until we see a way of attacking this reduced problem, we should investigate Riemann s suggestion and use the Euler Product Formula to connect the theory of primes with complex analysis..3 The Euler Product Formula We now derive the Euler Product Formula. This entire section is based on Jameson s derivation of the Euler Product formula, which can be found in the beginning of chapter of [4]. I also prove the theorems concerning double sums and infinite products that Jameson proves in appendices B and C of [4]. Lemmas concerning double sums and infinite products Before we get to the Euler Product Formula, it is necessary to derive the following lemmas concerning double sums and infinite products. Lemma.3 Suppose that j= k= a j,k converges. Then S R, so that for all sequences (c n ) n= that are in bijection with the elements of {a j,k : j, k N}, n= c n = S. Proof: First suppose that j, k N that a j,k and suppose that j= k= a j,k, which equals k= a j,k, equals S. Then J, K N, since the summands are all non-negative, J K j= k= a j,k S. Again since a j,k and k= j= a j,k converges, j N, k= a j,k converges to some value call it L j. 9

13 Fix ɛ >. Since k= j= a j,k converges and a j,k, J so that J J, J k= j= a j,k = J j= L j S ɛ. Additionally, since for each j J, k= a j,k converges and J is finite, K so that j J, K k= a j,k L j ɛ J. Thus, J K J a j,k (L j ɛ J ) = J J ɛ L j J = J L j ɛ S ɛ ɛ = S ɛ. j= k= j= j= j= j= Now let (c n ) n= be a sequence which is in bijection with the elements of {a j,k : j, k N}. We want to show that for sufficiently large N, N n= c n S < ɛ. Since N N, for sufficiently large J and K, the summands of N n= c n must appear in J K j= k= a j,k and k, j N, a j,k, N J K c n a j,k S. n= j= k= Thus, it suffices to show that for sufficiently large N, N n= c n S ɛ. Let N be so large that {c,...c N } contains the summands of J j= Kk= a j,k. Then since j, k N, a j,k, N N, N N J K c n c n a j,k S ɛ. n= n= j= k= So we have proved that if j, k N, a j,k and j= k= a j,k = j= k= a j,k = S, then for any rearrangement of the summands in the double sum into a single series n= c n, that n= c n = S. Now, suppose we remove the restriction a j,k. We want to show that if n= c n and n= d n are rearrangements of the terms in {a j,k : j, k N} then both n= c n

14 and n= d n converge to the same finite value. Consider the terms and a + j,k = a j,k if a j,k otherwise a j,k = if a j,k. otherwise a j,k Since j= k= a j,k converges and j, k N, a + j,k, a j,k, by the comparison test, j= k= a + j,k and k= j= a j,k converge. Thus, by what we have proved, if n N, if c n c n c + n =, otherwise c n c n =, c n otherwise and d + n and d n are defined similarly, then we have that c + n = d + n < n= n= and c n = d n <. n= n= Thus, n= c n = n= (c + n c n ) = n= c + n n= c n converges. Likewise, n= d n converges and n= c n = n= d n. As a result, we have the following: Corollary.4 Suppose j= a j, k= b k,..., m= l m denote finitely many absolutely

15 convergent series, and they converge to A, B,...,L and converge absolutely to A, B,..., L. If n= c n is any series formed by arrranging the summands of k= j= m= a j b k l m into a single series, then n= c n converges absolutely and converges (not absolutely) to AB L. Proof: We show the statement for two series j= a j and k= b k. The statement for any finite number of series follows by induction. For all j, k N, define a j,k = a j b k. a j,k = a j b k = a j b k = a j B = B a j = A B. j= k= j= k= j= k= j= j= Thus, by the previous lemma, all rearrangements of j= k= a j,k as a single series converge to the same value. Since j= k= a j b k = j= (a j ( k= b k )) = AB, after writing j= k= a j b k as a single series we see that n= c n converges to AB. To see that n= c n converges absolutely, we change n= a n and n= b n to n= a n and n= b n and reapply this theorem to get that n= c n converges. Lemma.5 Suppose n N, a n C, a n, and n, ɛ n and ɛ n a n. If n= a n converges, then n= ( + ɛ n a n ) converges to a non-zero value. Proof: Define {p n } n= by p n = n k= ( + a k ) and suppose that n= a n = S. Observe that n with n, p n = ( + a n )p n, which implies that p n p n = ( + a n )p n p n = a n p n. Also note that since the exponential function is greater than the function f(x) = x +

16 for non-negative x, we have that n N, + a k + a k e a k. Thus, n N, n n p n = ( + a k ) e ak n = e k= ak e S, k= k= where the last inequality follows because n k= a k k= a k = S. Thus, p n p n = a n p n a n e S. Furthermore, since p n p n = a n p n a n e S = e S a n e S a n = e S S, n= n= n= n= n= n= p n p n and n= a n p n converge because their partial sums form a monotonically increasing and bounded series. Notice that N N, N n= (p n p n ) = p N p. Since n= p n p n converges, n= (p n p n ) also converges, say to L. As N, p N p (p n p n ) = L, n= which implies that as N, N p N = ( + a k ) L + p = L + ( + a ). k= Thus, k= ( + a k ) converges. We want to show that k= ( + a k ) also converges. Define {b k } k= by b k = a k +a k. This sequence is well-defined because k N, a k. Since k= a k converges, as k, a k, so + a k. Thus, for sufficiently large k, + a k, which 3

17 implies that for sufficiently large k, b k = a k + a k a k = a k. Thus, k= b k k= a k. Since the latter series converges, by the comparison test, k= b k converges. Thus, k= b k converges, so we can rerun the first part of the proof to show that k= (+( b k )) converges. Since b k = +a k +a k a k +a k = +a k, we have shown that k= ( + a k ) converges. Since k= ( + a k ) and k= ( + a k ) = k= +a k both converge, neither must converge to. So, we have shown that if k= a k conveges and k N, a k, then k= ( + a k ) converges to a nonzero value. Since k N, ɛ k, k= ɛ k a k converges by the comparison test. Since ɛ k a k, by what we have just proved, k= (+ɛ k a k ) converges to a nonzero value as desired. Euler Product Formula We will derive the Euler Product Formula in this section. Definition.6 Let P be the set of all primes, P[N] be the set of all primes less than or equal to N, E[N] be the set of numbers expressible as product of powers of primes in P[N], and E*[N] = N - E[N]. Definition.7 A function a is arithmetic if its domain is N. 4

18 Definition.8 An arithmetic function a is completely multiplicative if m, n N, a(mn) = a(m)a(n). As a result, a() = because for the function to be well-defined, we must have that a(n) = a(n) = a(n)a() = a(n). We say that a function is completely multiplicative as opposed to just multiplicative because a multiplicative function is only required to satisfy the condition a(mn) = a(m)a(n) when gcd(m, n) =. Proposition.9 Let F be a finite set of primes and E F be the set of all products of powers of primes in F and. If a(n) is completely multiplicative and p F, a(p) <, then n E F a(n) = p F a(p) Proof: If a(p) <, then a(p) = n= a(p) n. Let T F = p F = p F n= a(p) n. a(p) When geometric series converge, they do so absolutely; thus, by Corollary.4, p F n= a(p) n can be rewritten as a single series with the summands appearing in any order and the expression will still converge to p F n= a(p) n. So it suffices to show that n E F a(n) is a representation of p F n= a(p) n as a single series. Since all numbers in E F can be uniquely written as a product of primes in F, if we suppose that n E F and n = p α pα k k, then a(n) = a(pα pα k k ). Since the function is completely multiplicative, a(p α pα k k ) = a(pα ) a(pα k k ), which is a summand of n= p F a(p) n after the product has been expanded out into a single sum. Thus, every summand of n E F a(n) is a summand of p F n= a(p) n. 5

19 Similarly, every summand in p F n= a(p) n is of the form a(p ) k a(p m ) km = a(p k pkm m ). Since p k pkm m E F, a(p k pkm m ) is a summand of n E F a(n). Thus, n E F a(n) is a representation of p F n= a(p) n as a single series. We now adapt the above proposition to prove the Euler Product Formula. The Euler Product Formula is essentially the infinite case of the above proposition. As we will see, the above proposition fails to prove the Euler Product Formula because the set F is no longer a finite set of primes, so n= p F a(p) n contains summands such as a() a(3) = a( 3 ). The infinite number 3 is not represented in E F, so the above proof fails in the infinite case. Theorem. Euler Product Formula Suppose a(n) is completely multiplicative and that n= a(n) converges. Then n= a(n) is non-zero and n= a(n) = p P a(p). if n is prime Proof: For all n N, let ɛ(n) =. otherwise Since n= a(n) converges, as n, a(n). So for any prime p as k, a(p k ). Since a(p k ) = a(p) k, it must be that a(p) <. Thus, n, a(n) <. Moreover, since a(n) is completely multiplicative a() =. Thus, n N, a(n). Moreover, n N, ɛ(n)a(n) because a(n) only equals at n = while ɛ() =. Hence, by Lemma.5, n= ( + ɛ(n)a(n)) = p P ( a(p)) converges to a nonzero value. Thus, p P a(p) also converges to a nonzero value. Thus, if n= a(n) = p P a(p), then n= a(n) is non-zero. 6

20 As in the previous proposition, we can argue that for all finite N, a(p) = a(n). p P [N] E[N] We now have to show that this relation holds as N. We do this by showing that as N, E[N] a(n) n= a(n). It will then follow that as N, p P [N] a(p) n= a(n) as desired. Note that since n= a(n) converges absolutely, a(n) a(n) = a(n). n= n E[N] n E [N] Thus, a(n) a(n) = a(n) a(n) a(n) n= n N n E[N] n E [N] n E [N] and this last quantity goes to as N because n= a(n) converges absolutely. This implies that as N, n E[N] a(n) n= a(n) as desired. We now wish to look at the direct applications of the Euler Product Formula to the zeta function. We will look at the region for s C with Re(s) >. Let a(n) = n. It s is clear that a(n) is completely multiplicative. To apply the Euler Product Formula, it remains to show that ζ(s) is absolutely convergent in this region. Proposition. For all s C with Re(s) >, ζ(s) is absolutely convergent. Proof: For all s C, n s = e slog(n) = e (σ+it) log(n) = e σ log(n). eitlog(n) 7

21 Since t and log(n) are real numbers, this equals e σ log(n) = n σ = n σ. Thus, for all s C with Re(s) >, n= n s = n= n σ, which absolutely converges by the p-test from single variable calculus. We can now apply the Euler Product Formula to ζ(s). Corollary. For all s C with Re(s) >, ζ(s) = p P ( p s ) and for all s C with Re(s) >, ζ(s). Proof: By the Euler Product Formula, ζ(s) = p P = ( p p s ) s p P and n= a(n) = n= n s is nonzero. This last corollary will prove to be essential later on in the proof of the Prime Number Theorem and the proof that ζ(s) has infinitely many zeros in what is called the critical strip (a proof is provided in chapter 4.) We now know that ζ(s) is connected to the primes by this formula. It makes sense to learn more about it. We will do this by studying Dirichlet series, which is a large set of series that includes ζ(s)..4 Dirichlet Series A Dirichlet Series of an arithmetic function a(n) is a series n= a(n)exp( λ n s) where λ, i N, λ i λ i+, and lim n λ n =. If we let λ n = log(n), then we get the more commonly seen "ordinary" Dirichlet series: a(n) n= n. If n N, a(n) =, then s 8

22 n= a(n) n = ζ(s). s The zeta function is the most important Dirichlet series we will see in this thesis, but not the only one. Using Widder s book, [], we determine when Dirichlet series are convergent and differentiable, so that we do not have to check for these properties whenever we encounter a new Dirichlet series. For the reader with prior exposure to only ordinary" Dirichlet series, it may seem odd to define a Dirichlet series as done by Widder, but Widder s definition has the advantage that we can apply the following theorems to the theta function, ϑ(u) = n= e πnu, which will appear in subsequent chapters. Theorem.3 Let f(s) = n= a(n)exp( λ n s). Suppose for some s C that f(s ) converges absolutely. Then s C with σ > σ, f(s) converges absolutely. Proof: Note that since λ n R, exp( λ n s) = exp( λ n σ). Thus, if σ < σ, then a(n)exp( λ n s) a(n) exp( λ n σ) a(n) exp( λ n σ ) n= n= n= = a(n)exp( λ n s ). n= Since n= a(n)exp( λ n s ) converges absolutely, n= a(n)exp( λ n s) also converges. Hence f(s) converges absolutely as desired. Corollary.4 There is a value σ a R called the abscissa of absolute convergence where s with σ > σ a f(s) converges absolutely, and s with σ < σ a, f(s) does not converge absolutely. 9

23 Proof: Consider K = {σ R : t so that f(σ + it) converges absolutely} and let σ a = inf(k). For a fixed s with σ > σ a, σ K with σ < σ. Since σ K, t so that f(σ + it ) converges absolutely. Thus, by the theorem, f(s) converges absolutely. For a fixed s with σ < σ a, σ K because σ < inf(k) = σ a. Thus, f(s ) cannot converge absolutely. The notation σ a is standard and it will be used in this thesis. We show a similar result for ordinary convergence. First we show a lemma, which is rather computational. Lemma.5 Let S N (s) = N n= a(n)exp( λ n s). Suppose that there exists M R and s C so that N N, S N (s ) M. Then s with σ > σ, N > N n= a(n)exp( λ n s) M s s σ σ [exp[ λ (σ σ )] exp[ λ N (σ σ )]] +Mexp[ λ N (σ σ )]. Proof: N N a(n)exp( λ n s) = S (s )exp[ λ (s s )]+ (S n (s ) S n (s ))exp[ λ n (s s )] n= n= N N = S (s )exp[ λ (s s )] + S n (s )exp[ λ n (s s )] S n (s )exp[ λ n (s s )] n= n= = S (s )exp[ λ (s s )] + = N n= N n= N S n (s )exp[ λ n (s s )] S n (s )exp[ λ n+ (s s )] [S n (s )(exp[ λ n (s s )] exp[ λ n+ (s s )])] + S N (s )exp[ λ N (s s )]. n=

24 Note that, λn+ exp[ λ n (s s )] exp[ λ n+ (s s )] = (s s ) exp[ t(s s )]dt. λ n Thus, N n= N λn+ a(n)exp( λ n s) = (s s ) S n (s ) exp[ t(s s )]dt λ n n= +S N (s )exp[ λ N (s s )]. So, N n= N λn+ a(n)exp( λ n s) s s S n (s ) exp[ t(s s )] dt λ n n= s s M s s M λn + S N (s ) exp[ λ N (s s )] λ exp[ t(s s )] dt + M exp[ λ N (s s )] λn λ exp[ t(σ σ )]dt + Mexp[ λ N (σ σ )] s s M (σ σ ) [exp[ λ N(σ σ )] exp[ λ (σ σ )]] + Mexp[ λ N (σ σ )] = s s M (σ σ ) [exp[ λ (σ σ )] exp[ λ N (σ σ )]] + Mexp[ λ N (σ σ )]. Theorem.6 Let f(s) = n= a(n)exp( λ n s). Suppose for some s C that f(s ) converges. Then s C with σ > σ, f(s) converges. Proof: Since n= a(n)exp( λ n s ) converges, the series and its partial sums are bounded by some M R. If we consider consider an arbitrary m N, then m a(n)exp( λ n s ) = a(n)exp( λ n s ) a(n)exp( λ n s ) n=m n= n= n= m a(n)exp( λ n s ) + a(n)exp( λ n s ) n=

25 M + M = M. If we consider arbitrary N, N N with < N < N <, then N a(n)exp( λ n s ) = a(n)exp( λ n s ) a(n)exp( λ n s ) n=n n=n n=n a(n)exp( λ n s ) + a(n)exp( λ n s ) n=n n=n M + M = 4M. So for an arbitrary m N, the partial sums of the Dirichlet series n=m a(n)exp( λ n s ) are bounded by 4M; thus, by the previous lemma for all s C, with σ > σ, for all M > m, M n=m a(n)exp( λ n s) 4M s s σ σ [exp[ λ m (σ σ )] exp[ λ M (σ σ )]] +4Mexp[ λ M (σ σ )]. Since f(x) = e x monotonically decreases as x increases, the above is less than or equal to 4M s s σ σ [exp[ λ m (σ σ )] exp[ λ m (σ σ )]] + 4Mexp[ λ m (σ σ )]. Since s is fixed, the above can be made arbitrarily small by making m sufficiently large. Thus, we have shown that n= a(n)exp( λ n s) is Cauchy, so it converges. Corollary.7 There is a value σ c R called the abscissa of convergence where s with σ > σ c, f(s) converges, and s with σ < σ c, f(s) does not converge. Proof: The proof is identical to the proof of Corollary.4. As with σ a, the notation σ c is standard.

26 If we consider ζ(s) = n= n s we note that ζ() diverges because it is the harmonic series and s R with s >, ζ(s) converges by the p-test from single-variable calculus. Thus, for the Riemann zeta function, σ c =. Since for s R, n s = n s, σ a =. We now show that Dirichlet Series are holomorphic in their regions of convergence. Theorem.8 Suppose that n= a(n)exp( λ n s) defines a Dirichlet series with a finite abscissa of convergence σ c. Then in its region of convergence, the Dirichlet series is holomorphic with its kth derivative given by n= a(n)exp( λ n s)( λ n ) k. Proof: Fix s in the plane of convergence and consider a ball of radius r centered at s called B(s, r) that also satisfies that B(s, r) is in the region of convergence. We want to show that s B(s, r) that n= a(n)exp( λ n s) converges uniformly inside B(s, r). It then follows from the uniformity that inside B(s, r), d k ds k a(n)exp( λ n s) = n= n= d k ds k a(n)exp( λ ns) = a(n)exp( λ n s)( λ n ) k. n= Since the point s was selected arbitrarily in the region of convergence, the theorem will be proved. We want to show that ɛ >, N N so that m N, s B(s, r), n=m a(n)exp( λ n s) < ɛ. Fix ɛ >. Fix s C so that σ c < σ < σ and t = t. Let d be the distance from s to B(s, r). Let x = s s. Since n= a(n)exp( λ n s ) converges, the series and its partial sums are bounded by some M R. Thus, by the proof of Theorem.6 for an arbitrary m N, the partial sums of the Dirichlet series n=m a(n)exp( λ n s ) are bounded by 4M. So, by Lemma 3

27 Figure : We want to show that the n= a(n)exp( λ n s) converges uniformly inside the disk..5, for all s C, with σ > σ, after noting that as N that exp[ λ N (σ σ )], n=m a(n)exp[( λ n s)] 4M s s σ σ exp[ λ m (σ σ )] Note that σ σ > d and x < r; thus, by the triangle-inequality, s s < σ σ +x < σ σ + r. So, 4M s s exp[ λ m (σ σ )] M(σ σ + r) exp( λ m d). σ σ d This quantity is independent of the choice of s. So let m be so large that it is less than or equal to ɛ. An immediate application to the previous theorem is that the Riemann zeta function is holomorphic in its region of convergence. We state it below for reference. Corollary.9 The Riemann zeta function, ζ(s), is holomorphic in its region of convergence. 4

28 We have studied Dirichlet series as a subject on their own and have derived conditions for convergence and differentiability. This should be reminiscent of the study of power series that most mathematicians are exposed to at some point during graduate school. Dirichlet series can be thought of as the less famous cousin of power series, and one can continue to compare and contrast them. For those interested, on page 48 of Widder s book, [], Widder provides a nice table that summarizes the differences between them..5 Preliminaries to the Prime Number Theorem From here on out, the motivating ideas behind the proof to the Prime Number Theorem become much harder to understand. This may be expected because it took almost 4 years between the time Riemann correctly outlined an approach to the Prime Number Theorem and when the first proof of it was published. To prove the Prime Number Theorem, we will first use the Euler Product Formula and the fact that a Dirichlet series is holomorphic in its region of convergence to derive a formula for ζ (s) ζ(s) and log(ζ(s)). Then we will meromorphically extend ζ(s) to the line Re(s) = and use log(ζ(s)) to prove that ζ(s) has no zeros on the line Re(s) =. Next, we will derive upper bounds for ζ(s) and ζ (s) ζ(s) and reduce the Prime Number Theorem to deriving asymptotic bounds for a new integral. Finally, we will use everything just mentioned to derive those asymptotic bounds. Expression for ζ (s) ζ(s) : The material in this section can be found in Jameson s book, [4]. We first derive an expression for log(ζ(s)) and then take its derivative to get an expression for ζ (s) ζ(s). 5

29 Lemma.3 For z <, let h(z) = m= z m m. Then h(z) is a logarithm of z ; that is, e h(z) = z. Additionally, for z <, h(z) z. Proof: Since h(z) is a power series that converges for z with z <, it can be differentiated term-by-term. So h (z) = z m = z m = z. m= m= Note that d dz ( z)eh(z) = e h(z) + ( z)h (z)e h(z) = e h(z) + z z eh(z) = So ( z)e h(z) is a constant say c. So, for all z with z <, ( z)e h(z) = c. If z =, then c = e h() = e m m= m = e =. Thus, ( z)e h(z) =, which is equivalent to e h(z) = z. Thus, h(z) is by definition a logarithm of z, which concludes the first part of the lemma. For the second part of the lemma, first note that if z <, then z z < z because z < z [ z ] > > z > z z. Thus, when z <, h(z) zm m z m m= m= which concludes the proof. = z m= z m = z z < z, Recall from Definition.6 that P is the set of primes and P [N] is the set of primes 6

30 less than or equal to N. Theorem.3 Let h(z) be defined as in the lemma. For Re(s) >, p P h( p s ) is a logarithm for ζ(s) and p P h( p s ) = p P m= mp ms = n= c(n) n s where m c(n) = n = pm for some p P and m N. otherwise As a result, we write that log(ζ(s)) = n= c(n) n s. Proof: Recall from the lemma the definition of the function h: z C, with z <, h(z) = m= z m m. For s C with Re(s) >, for all p P, p s = p σ <. So by the previous lemma, p P, h( p s ) is a logarithm of p s = [ p s ]. So e h( p s ) = [ p s ]. We want to show that p P[ p s ] = p P e h( p s ) = e p P h( p s ), which makes p P h( p s ) a logarithm of p P [ p s ], which equals ζ(s) by the Euler Product Formula. The first part of the theorem will then be established. Using laws of exponents, since for all primes, e h( p s ) = [ p ], it is also true that for s all N, p P [N] [ p s ] = e p P [N] h( p s ). To extend this to the infinite case, we want to show that as N, p P [N] h( p s ) p P h( p s ). 7

31 Note that, h( p s ) = m= mp ms m= = h( p σ ) mp ms = m= mp mσ Since p, p, p < σ. Thus, by the lemma the above is less than or equal to ( p ). σ Thus, h( p s ) p P p P p σ, which converges by the p-test from single variable calculus because σ >. Thus, p P h( p s ) converges, which proves that p P [N] h( p s ) p P h( p s ). Note that it has also been shown that p P h( p s ) converges absolutely. For the second part of the theorem, since p P h( p s ) = p P m= mp ms converges absolutely, by Corollary.4, it may be rewritten as a single series that converges to the same value as m= p P mp. ms We rewrite the series as follows: p P m= mp ms = n= c(n) n s where m c(n) = n = pm for some p P and m N. otherwise We define the Von-Mangoldt function as follows: 8

32 log(p) n = p m for some p P and m N Definition.3 Define Λ (n) =. otherwise Theorem.33 d ds (log(ζ(s))) = n= Λ(n) n s. Proof: Since for Re(s) > ζ(s) is well-defined and nonzero log(ζ(s)) converges for s with Re(s) >. Since log(ζ(s)) is a Dirichlet series, by Theorem.8, it is holomorphic in this region and where as before, d ds (log(ζ(s))) = d c(n) ds n s n= = n= c(n) log(n) n s m c(n) = n = pm for some p P and m N. otherwise Moreover, c(p m ) log(p m ) = (m) log(p) = log(p). m Thus, n= c(n) log(n) n s = n= Λ (n) n s. We now show that as one would expect that the derivative of the log(ζ(s)) is ζ (s) ζ(s) Corollary.34 For s with Re(s) >, ζ (s) ζ(s) = n= Λ(n) n s. Proof: If g(s) is differentiable with e g(s) = f(s) for some function f, then g (s)e g(s) = f (s) g (s) = f (s) f(s). 9

33 Let g(s) = log(ζ(s)). So f(s) = ζ(s). So, d ds (log(ζ(s))) = ζ (s) ζ(s) By the theorem, it follows that n= Λ(n) n s = ζ (s) ζ(s). Meromorphic Continuation of ζ(s) to Re(s) > : We will now derive a meromorphic continuation of ζ(s) to the half-plane to the right of Re(s) =. Over the course of the thesis, we will show an additional proof of this fact. It will be necessary to prove all of the variations because each will be useful for a purpose that the others cannot fulfill. Additionally, we will also show that the zeta function can be meromorphically continued to the entire complex plane. That proof will be long and is the content of chapter 3. Non-existence of zeros on the line Re(s) = : Proposition.35 Suppose f is differentiable over a compact line segment γ with f bounded by A over γ. Then a, b γ, f(b) f(a) A b a. Proof: Fix a, b γ and let γ be the compact subsegment with endpoints a and b. For all ɛ >, for all x γ, there exists a h x so that for all h with h h x and x + h γ, f(x + h) f(x) A h + ɛ h. Let O be the open cover of γ generated by open sets centered at each x γ with diameter equal to h x. Since γ is compact, let O f be a finite subcover and suppose the centers of the elements of O f are x, x,...x N. 3

34 Let α : [, ] γ be a parametrization of γ and for α(t) = x and α(t ) = y, let x < γ y if t < t. Subdivide γ into subsegments with endpoints a = x, x,... x N, x N+ = b so that for all k {,..., N}, x k < γ x k and both x k and x k+ lie in the element of O f centered at x k. This last condition implies that except for k + = N +, x k+ lies in the intersection of the elements centered at x k and x k+. Thus, f(a) f(b) = f(x ) f(x ) + f(x ) f(x ) + f(x N+ ) A x x + ɛ x x + + A x N x N+ + ɛ x N x N+ = A b a + ɛ b a. Letting ɛ, we get the desired result. We also recall the following proposition. This proposition was proved in MATH53, but it is a theorem that will appear multiple times in this thesis and it has not appeared in any of the other textbooks that I have read, so I will reprove it. Proposition.36 Let Ω be an open set in C and let F(z,s) be defined for (z,s) Ω [, ]. Also, suppose that (i) F(z,s) is holomorphic in z for each s (ii) F is continuous on Ω [, ] If we consider the Riemann integral, then the function f defined on Ω by f(z) = F (z, s)ds is holomorphic. Proof: Recall one of the distinguishing characterstics of complex analysis is that if a sequence of holomorphic functions converge uniformly to a function f in every compact subset of Ω, then f is holomorphic in Ω. I say that this is a distinguishing characteristic 3

35 because it is not necessarily true for differentiable real-valued functions. Consider {f n } n= where f n(z) = n nk= F (z, k n ). For all n N, f n is a finite sum of functions that are each holomorphic in Ω; thus, each f n is holomorphic in Ω. Define f by f(z) = lim n n nk= F (z, k n ). This equals the Riemann integral F (z, s)ds. We know that f is well defined because of condition (ii). To show that f is holomorphic on Ω, it suffices to show that {f n } n= f uniformly on any compact disk D Ω. Fix a compact disk D Ω and ɛ >. Since F is continuous over D [, ], which is compact, δ so that s, s, with s s < δ, z D, F (z, s ) F (z, s ) < ɛ. Let n be so large that n > δ which implies that n < δ. Then z D, f n (z) f(z) = n F (z, k n n ) F (z, s)ds. k= Since k n k n = k n k n + n = n, the above equals n n k= k n n k n = F (z, k n )ds F (z, s)ds = n k= k n k n n k= F (z, k n ) F (z, s)ds n k= k n k n k n F (z, k n )ds n Since s [ k n, k n ] and n < δ, k n s < k n k n = n < δ; thus, n k= k n k n F (z, k n ) F (z, s)ds n k= k n k n k n k= k n k n F (z, k ) F (z, s)ds. n ɛ ds = n k= ɛ n = ɛ. F (z, s)ds 3

36 Thus, {f n } f uniformly, which establishes the theorem. The above proposition relies on [,] being compact and the choice of the endpoints and is arbitrary. Thus, the above theorem can be adapted so that it applies to any compact arc in the complex plane because compact arcs in the complex plane are homeomorphic to [,]. We now present an analytic continuation of ζ(s) to Re(s) >. This analytic continuation is employed by Stein in his proof of the Prime Number Theorem. The following proofs are adapted from Stein s book, []. Proposition.37 Let δ n (s) = n+ n Moreover, for s with Re(s) >, n s x s dx. For all n N, δ n is an entire function. δ n (s) s n σ+ and n<n n s N x s dx = n<n δ n (s). Proof: Let F (s, x) = n s x s for (s, x) C [n, n + ]. For all n N, note that for each x, the integrand n s x s is holomorphic for all s C and that n s x s is continuous on C [n, n + ]. Thus by the previous proposition, n N, δ n is entire. Now we prove the upper bound for δ n (s). Fix s with Re(s) >. Let f(x) = x s. So, f (x) s x s s x σ. For x [n, n + ], since x n and σ >, the above is less than or equal to s n σ = s n σ+. 33

37 So for a fixed s, s n σ+ Proposition.35, f(n) f(x) = n s x s is a constant that bounds f (x); thus for x [n, n + ] by s n σ+ n x. Since n x n +, x n so δ n (s) = n+ n n s x s dx n+ s n σ+ n x n s dx = s nσ+ s n σ+. Thus, s with Re(s) >, (n + n) = s nσ+ n σ+. To prove the formula, note that n<n as desired. δ n (s) = = n<n n<n n+ n n s x s dx = n s [n + n] N n<n x s dx = n+ n n<n n s dx n s N n<n x s dx n+ n x s dx Corollary.38 Let H(s) = n= δ n (s). For all s with Re(s) >, H(s) is holomorphic and ζ(s) s = H(s). This provides a meromorphic continuation of ζ(s) to Re(s) > with the only singularity ocurring at s =. Proof: To show that H(s) = n= δ n (s) is holomorphic in Re(s) >, since the partial sums of H(s) are finite sums of holomorphic functions, it suffices to show the partial sums of H(s) converge uniformly to H(s) inside any compact disk D lying entirely in Re(s) >. Fix a disk D centered at s, with radius r, that lies in Re(s) >. For all s D, δ n (s) δ n (s) = δ n (s) δ n (s) s n n= σ+. n N n>n n N n N 34

38 Since s D, s s + r and σ r + σ +. Thus, the above is less than or equal to n N s + r n σ r+. Since σ r + >, this sum can be made arbitrarily small by making N suitably large enough. Thus, we have established uniform convergence. Now, note that N x s = s x s+ ] N = s [N s+ s+ ] = s [N s+ ] = s [ ]. N s When s satisfies Re(s) >, since N s = N σ, which goes to as N, as N, s [ ] N s s = s. Also, recall that as N, n<n n s ζ(s). Thus, as N, for s with Re(s) >, since each term in the formula of the proposition converges to the corresponding term in ζ(s) s = H(s), the formula stated holds. Since H(s) defines a holomorphic function in Re(s) > and s is meromorphic there with a pole at s =, ζ(s) = s + H(s) provides a meromorphic continuation of ζ(s) to Re(s) > with the pole at s =. Now that we have shown that ζ(s) can be continued past the line Re(s) =, we can now investigate whether ζ(s) takes on that line. It turns out that it cannot and the proof of this fact seems bizarre because it hinges on a seemingly unrelated 35

39 trigonometric inequality. Lemma.39 If θ R, then 3 + 4cosθ + cosθ. Proof: 3 + 4cosθ + cosθ = + 4cosθ + cos θ = ( + cosθ). Proposition.4 If σ >, then ζ(σ) 3 ζ(σ + it) 4 ζ(σ + it). Proof: Note that Re(n s ) = Re(e (σ+it) log(n) ) = e σ log(n) cos(tlog(n)) = n σ cos(tlog(n)) and Re(log(z)) = Re[ln z + iarg(t)] = ln z = log z. Thus, log ζ 3 (σ)ζ 4 (σ + it)ζ(σ + it) ( ) = 3log ζ(σ) + 4log ζ(σ + it) + log ζ(σ + it) = 3Re[log(ζ(σ))] + 4Re[log(ζ(σ + it))] + Re[log(ζ(σ + it))].( ) Recall that by Theorem.3, log(ζ(s)) = n= c(n) n s where m c(n) = if n = pm for some p P and m N. otherwise So (**) equals 3Re( c(n)n σ ) + 4Re( c(n)n (σ+it) ) + Re( c(n)n (σ+it) ) n= n= n= = 3 Re(c(n)n σ ) + 4 Re(c(n)n (σ+it) ) + Re(c(n)n (σ+it) ). n= n= n= 36

40 Since c(n) is always real, the above equals 3 c(n)re(n σ ) + 4 c(n)re(n (σ+it) ) + c(n)re(n (σ+it) ) n= n= n= = 3 c(n)n σ cos((log(n))) + 4 c(n)n σ cos(t(log(n)))+ n= n= c(n)n σ cos(t(log(n))). n= Let θ n = tlog(n). Then the above equals (3 + 4cos(θ n ) + cos(θ n )) c(n)n σ.( ) Since c(n), n σ and (3 + 4cos(θ n ) + cos(θ n )) are always positive, (***) is always positive. n= Since (***) is always positive, (*) is always positive; thus, log ζ(σ) 3 ζ(σ + it) 4 ζ(σ + it), which implies that ζ(σ) 3 ζ(σ + it) 4 ζ(σ + it) as desired. Theorem.4 ζ(s) has no zeros on the line Re(s) =. Proof: Recall that ζ(s) has a meromorphic continuation to the line Re(s) = where ζ(s) is holomorphic everywhere in this region except for a simple pole at s =. Suppose that ζ( + it ) = for some t. In a sufficiently small neighborhood, call it U, around +it, ζ(s) = (s (+it ))g(s) 37

41 where g(s) is a holomorphic function in this neighborhood. So if s U as s approaches + it from the right, ζ(s) = (s ( + it ))g(s) = (σ )g(σ + it) + i(t t )g(σ + it) = (σ )g(σ + it). Since g is holomorphic in U, g(σ + it) is bounded by a constant C. Thus, the above is less than or equal to C σ. Similarly, in a sufficiently small neighborhood around s =, call it V, since ζ(s) has a simple pole at s =, ζ(s) = h(s) s where h(s) is a holomorphic function in V. Since h(s) is holomorphic there, it is bounded there by a constant, call it C. Since Re(s) s, for s V, ζ(s) C s C σ. Thus, as σ from the right in V, ζ(σ) 3 C 3 σ 3. Similarly, ζ(s) is bounded by M in a sufficiently small neighborhood of (σ + it ) because ζ(s) is holomorphic in that neighborhood. Putting these statements together yields that as σ, from the right ζ 3 (σ)ζ 4 (σ + it)ζ(σ + it) C 4 C 3 (σ )4 M (σ ) 3, which contradicts the previous proposition that says that σ >, ζ(σ) 3 ζ(σ + it) 4 ζ(σ + it). Upper bounds for ζ(s), ζ (s), and ζ(s). The two main theorems of this section are adapted from Elias Stein s book, []. How- 38

42 ever, they were written very tersely, were computationally-heavy, and as written I believe they are wrong. The following proof is adapted from Proposition.7 part (i) on page 73 of Stein s book, []. However, Stein appears to be wrong and I have added an extra condition to the theorem s hypothesis to correct his proof. Theorem.4 Suppose σ satisfies σ and ɛ. Then there exists a constant c ɛ R so that s with σ that satisfies σ σ, s t and t, ζ(s) c ɛ t σ +ɛ. Proof: Recall from Corollary.38 and Proposition.37 that s with Re(s) > that H(s) = n N δ n(s) = ζ(s) (s ), δ n(s) δ n (s) n+ n s, δ n σ+ n (s) = n+ n n s n+ x s dx n s + x s dx n n s x s dx. Note that, Since x n, x s = x σ n σ = n s Thus, we have n+ n n s + x s dx n+ n n s dx = n+ n σ dx = n n σ. So, for all ɛ, δ n (s) = ( δ n (s) ) ɛ ( δ n (s) ) ɛ ( s n σ+ )ɛ ( n σ ) ɛ = s ɛ ɛ n ɛ σ+ɛ +σ ɛ σ s ɛ n ɛ +σ Let ɛ = σ + ɛ. Using Corollary.38, ζ(s) = H(s) + s s + n= 39 δ n (s) s + n= s σ +ɛ n σ +ɛ+σ.

43 Since σ σ, n +ɛ n +ɛ+(σ σ ), so the above is less than or equal to Since Im(s) s, s + s σ +ɛ.( ) n+ɛ n= s Im(s ) = Im(s) = t, and since t >, s.( ) Since s is bounded by a constant, n= n +ɛ is a constant dependent on ɛ, say c ɛ, so using (*) and (**), ζ(s) + c ɛ s σ +ɛ c ɛ s σ +ɛ where c ɛ absorbed the constants and c ɛ. Since s t, this is less than or equal to c ɛ[ t ] σ +ɛ. Since σ, the above is less than or equal to c ɛ() +ɛ t σ +ɛ c ɛ t σ +ɛ where c ɛ = c ɛ() ɛ. Note that in the previous theorem we can set σ =, so that the theorem reads that ɛ >, c ɛ so that s with σ, t and s t, ζ(s) c ɛ t ɛ. By doing this, we have sacrificed a slightly stronger result for clarity. We also note that Stein originally tried to prove the result without the hypothesis that s t. However, I have no idea how to justify that c ɛ s σ +ɛ c ɛ t σ +ɛ 4

44 without that extra hypothesis. It should also be noted that the value is arbitrary and it can be replaced by any positive number greater than. At the conclusion of the Prime Number Theorem, the reader can check that such an adjustment will not affect any of the other theorems in this chapter. In Stein s book, [], there is a second part to the above theorem; however, with the extra hypothesis that I added to the second part, the proof no longer holds. Using Jameson s book [4], I present a series of proofs that will ultimately accomplish to prove what Stein had in mind. Proposition.43 Euler s Summation Formula Let f be differentiable on [m, n]. Then Proof: n r=m+ n n f(r) f(t)dt = (t t )f (t)dt. m m When r t < r, t = r ; thus, since t = r only at a single point on [r, r], r r (t t )f (t)dt = r r (t (r ))f (t)dt = r r (t r + )f (t)dt. If we integrate r r (t r + )f (t)dt by parts with u = t r + and dv = f (t)dt, then du = dt, v = f(t) and r r r r (t r + )f (t)dt = (t r + )f(t)] r r f(t)dt = f(r) f(t)dt. r r Thus, r r (t t )f (t)dt = r r r (t r + )f (t)dt = f(r) f(t)dt. r The above can be extended to a finite sum, so n r=m+ r ( r (t t )f (t)dt) = n r=m+ r (f(r) r f(t)dt) 4

45 n (t t )f (t)dt = m n r=m+ n f(r) f(t)dt. m Corollary.44 If f is differentiable on [, ] and both r= f(r) and f(t)dt converge, then r= f(r) f(t)dt = f() + (t t )f (t)dt Proof: By Euler s Summation Formula, n r=m+ n n f(r) f(t)dt = (t t )f (t)dt. m m Let m =. Since the sum and the integral converge, we can let n, so that we get r= f(r) f(t)dt = (t t )f (t)dt. If we add f() to both sides we get r= f(r) f(t)dt = f() + (t t )f (t)dt. Using this corollary, we can provide another analytic continuation of ζ(s) to Re(s) >. This analytic continuation will both help patch the hole in Stein s proof and will be useful in Chapter 4. Corollary.45 For Re(s) >, ζ(s) = s + s x x dx. x s+ Proof: First consider the region Re(s) >. Let f(x) = x s. For Re(s) >, s is well-defined, so x s dx = s x s ]. x s = x σ, so as x, since Re(s) >, x s. Thus, the above equals s s = s. 4

46 Additionally, for Re(s) >, we know that x= x s converges because it is equal to ζ(s). So by the previous corollary, for Re(s) >, ζ(s) = s + s x x x s+ dx.( ) For Re(s) >, we note that except at s =, s converges because for all T >, is holomorphic and x x dx x s+ T x x x s+ dx T x s+ dx = T x σ+ dx. When σ >, this converges to as T goes to infinity. We can use a similar argument as was done in Proposition.37 to show that x x dx x s+ is holomorphic for Re(s) >, which will establish that (*) is another meromorphic continuation of ζ(s) to Re(s) >. Proposition.46 For all N N, s C with s and Re(s) >, ζ(s) = N n= n s + N s s + r N(s) where Moreover, r N (s) = s N r N (s) s σn σ. x x x s+ dx. Proof: By Euler s Summation formula with f(x) = x s and m =, we have f() + N n= n s = f() + N x s dx s N x x x s+ dx 43

47 N n= N n= n s = + N s s N s s n s = + s N s N s s x x x s+ dx x x x s+ dx.( ) Recall from the previous corollary that ζ(s) = s + s x x dx. Subtracting (*) x s+ from this expression we get, ζ(s) N n= n = N s s s N x x x s+ dx Note that ζ(s) = N n= n + N s s s N x x x s+ dx. r N (s) = s N x x x s+ dx s N x σ+ dx = s σx σ ] N = s σn σ. Lemma.47 For N, N n= log(n) n (log(n)) + 8. Proof: Let f(x) = log(x) x. Then f (x) = x( x ) log(x) x = log(x) x. Note that if log(x) x =, then x = e and f (x) < when x > e. Thus, f(x) monotonically decreases over (e, ), so N n=4 log(n) n N 3 log(x) x dx 44

48 If we do u-substitution from single-variable calculus with u = log(x), then du = x dx and the integral above equals (log(n)) (log(3)). It follows by computation that 3 n= log(n) n (log(3)) <. < 8. Thus, N n= log(n) n 3 n= log(n) n (log(3)) + (log(n)) (log(n)) + 8. Proposition.48 For all σ and t, ζ (σ + it) (log(t) + 3). Proof: Fix s that satisfies the proposition s hypotheses. From the previous proposition, N N, ζ(s) = N n= n s respect to s, + N s s s N x x dx. Taking the derivative with x s+ ζ (s) = N n= log(n) n s N s log(n) s +s N N s (s ) (x x ) log(x) x s+ dx.( ) N x x x s+ dx Let N = t. We will now give approximations to each of the terms in the expression listed above and then combine them to give the desired upper bound: ) Since σ, N n= log(n) n s N n= log(n) n σ N n= log(n) n. From the lemma, this is less than or equal to (log(n)) + 8. Since N t, this quantity is less than or equal to (log(t)) + 8. ) Since N > and σ, N s = N σ, so N s log(n) s 45 log(n) s. Since

49 Im(s) < s, this is less than or equal to log(n) t lemma, log(x) x takes a local maximum at x = e and log(e) e N s log(n) < log(t) < s t e <. log(t) t. As shown in the proof of the = e <. Thus, 3) As shown in ), N s and s t ; thus, 4) N N. N s (s ) t = 4. x x dx x s+ N dx. Since σ, this is less than or equal to x σ+ N dx = x 5) N (x x ) log(x) dx log(x) x s+ N dx. Let u = log(x) and dv = x x (σ+) dx. Then σ+ du = x dx and v = σ x σ and Thus, N log(x) log(x) dx = xσ+ σx σ ] N + σ N x σ x = log(n) σn σ + σ N σ. dx = log(n) σn σ + σ N dx x+σ s N (x x ) log(x) x s+ dx s [ log(n) σn σ + σ ] [ σ + it ][log(n) N σ σn σ + σ N σ ]. Since σ, σ, so the above is less than or equal to [ σ + it ][ log(n) σn σ + + ] [ σ + t ][log(n) σn σ σn σ ] = [ + t σ ][log(n) + N σ ]. Since σ and N >, N N σ, so the above is less than or equal to [ + t σ ][log(n) + ]. N 46

50 Since σ, t σ t, so the above is less than or equal to ( + t)[ log(n) + ]. N Note that since N = t and t, t + N + N, so + t (log(n) + ) (log(t) + ) = log(t) +. N Putting the 5 estimates together yields that (*) is bounded by ζ (s) (log(t)) log(t) (log(t)) + log(t) = [(log(t) + 4log(t) + 8] < [log(t) + 6log(t) + 9] = (log(t) + 3). We can weaken the proposition s hypothesis by showing that the upper bound for ζ (σ + it) holds for t and not just for t. Corollary.49 For σ and t, ζ (σ + it) (log(t) + 3). Proof: We first show that ζ (s) = ζ (s). Recall from the proposition that ζ (s) = N n= log(n) n s N s log(n) N s s (s ) N x x x s+ dx+s N (x x ) log(x) x s+ dx Since n s = n s and the integrals are Riemann integrable (so they are limits of sums), it follows that ζ (s) = ζ (s). 47