Dan Buchanan. Master of Science. in the. Mathematics Program. YOUNGSTOWN STATE UNIVERSITY May, 2018

Transcription

1 Analytic Number Theory and the Prime Number Theorem by Dan Buchanan Submitted in Partial Fulllment of the Requirements for the Degree of Master of Science in the Mathematics Program YOUNGSTOWN STATE UNIVERSITY May, 208

2 Analytic Number Theory and the Prime Number Theorem Dan Buchanan I hereby release this thesis to the public. I understand that this thesis will be made available from the OhioLINK ETD Center and the Maag Library Circulation Desk for public access. I also authorize the University or other individuals to make copies of this thesis as needed for scholarly research. Signature: Dan Buchanan, Student Date Approvals: Dr. Eric Wingler, Thesis Advisor Date Dr. Thomas Smotzer, Committee Member Date Dr. Padraic Taylor, Committee Member Date Dr. Salvatore A. Sanders, Dean of Graduate Studies Date

3 ABSTRACT Analytic Number Theory is the cross between Real and Comple Analysis as well as Number Theory. We will eamine results involving a function whose power series epansion has Fibonacci coecients and another with Catalan number coecients. The divisor function will help in nding an approimation of the number of divisors of a number. Our main focus will be on the Prime Number Theorem and the techniques needed to prove it. We will nish by eamining how this proof helped to shape modern comple analysis, as well as discussing the Riemann Hypothesis. iii

4 ACKNOWLEDGEMENTS I would like to thank Dr. Eric Wingler for his constant help, support, guidance and leadership. I would also like to thank my thesis committee members, Dr. Thomas Smotzer and Dr. Padraic Taylor, as well as the entire YSU Department of Mathematics & Statistics for pushing me to new levels, and always keeping my mind open to new ideas. Also, thank you to my mother Virginia Buchanan and girlfriend Cassie Dodson for their help and support along the way. iv

5 Contents Introduction. Background Information Introductory Problems 4 2. Fibonacci Numbers and The Golden Ratio A Problem Involving the Catalan Numbers The Divisor Function Introduction to The Prime Number Theorem 2 3. History and Motivation Elementary Method Riemann Zeta Function Idea of the Proof Analytic Proof The Contour Integral The Riemann Zeta Function Near s The Final Step Conclusion 56 References 59 v

6 Introduction Analytic Number Theory is the combination of two branches of Mathematics: Analysis and Number Theory. Elementary Number Theory is a course that is often taught at an undergraduate level. Topics include the following: Introducing the student to the notion of modulus, working with prime numbers, several functions and algorithms that reveal properties of the positive integers, and looking at some historic and important results. The study of positive integers (natural numbers N) is (arguably) the oldest branch of Mathematics. When one ponders why Mathematics came about, it becomes clear that its origins were not centered around nding properties of a triangle or nding an equation for the oscillation of a vibrating string. Mathematics came to be because people needed a convenient way to describe and categorize what they needed in their day-to-day lives. Throughout time, there have been several alterations to how we count and understand numbers, but there is now an almost universal understanding of numbers. In trying to gure out how to represent quantities, people developed what was the beginning of Number Theory. In this tet, we will dene the set of natural numbers (or positive integers) to be N {, 2, 3,...}. These are sometimes called the counting numbers as they are the traditional ones we use to count how many elements a given set (or collection of objects) has. Most people have been working with natural numbers since they were kids. One of the rst things we are taught is how to count and how to use and interpret these numbers. We will not focus on constructing the natural numbers, but the reader is encouraged to study how they came about. Number Theory deals with studying properties of natural numbers and functions whose domain is the natural numbers (arithmetical functions). Analytic Number Theory also focuses on these topics, but uses analytic techniques to gure out solutions or approimate certain results. Many mathematicians enjoy Number Theory because of the seemingly simple statement of certain problems. However, most questions in Analytic Number Theory

7 have a similarly simple statement, but the proof (or attempt of a proof) is anything but Elementary. Some of the most popular unsolved problems in mathematics come from Number Theory. Listed below are some of these unsolved problems (note that some of these can be eplained to most students in a pre-calculus class, however, have not yet been solved by even the best mathematicians). Is there an even number greater than 2 that is not the sum of two primes? Are there innitely many primes of the form 2 +? Are there innitely many primes of the form 2 p (Mersenne Primes)? Are there innitely many twin primes (prime numbers only two numbers apart)? Note that many of the unsolved problems in Number Theory deal with prime numbers. This is because there is no known formula that generates all of the prime numbers (and only the prime numbers) in a sequential order. Obviously a function such as f() has a range that contains all of the prime numbers, but there are many other values in the range other than the primes. So this is not a good way to represent the primes. Thus, the diculty with most of these problems is how to get around the need for a formula that produces all of the prime numbers. When we eamine the Prime Number Theorem, we will have to do a lot of work to come up with an epression that is equivalent but doesn't involve the need for a formula that produces all of the prime numbers in order. This is something that has puzzled Mathematicians for centuries, and will continue to do so until a formula is found. 2

8 . Background Information Before reading this paper, the reader should be familiar with concepts from Calculus and Algebra and have been eposed to problems in Real Analysis, Comple Analysis, Number Theory, Measure Theory, and Abstract Algebra courses (however, many of the results needed from these disciplines will be eplained when needed). First, one should be familiar with what a function is. Note that we say a function f : X Y has domain X and codomain Y. We will be working with functions involving comple numbers in a later section. Net, one should be comfortable with comple numbers. The set of all comple (or imaginary) numbers, C, is dened as C {a + ib a, b R, and i }. One can also think of the comple numbers as points in the plane, but the -ais is replaced with the real-ais, and the y-ais is replaced with the imaginary-ais. So the comple number a + ib can be thought of as the point (a, b). Note that from this, we can see a one-to-one correspondence from C onto R 2, where R 2 {(, y), y R}. There will be several theorems used from Comple Analysis, Real Analysis and Measure Theory in this tet. The term prime number has been used several times, but what is a prime number? A prime number, p, is an integer greater than such that the only divisors of p are p and (i.e., for any n N other than p and, p/n is not an integer). Some eamples of prime numbers are 2, 3, 5, 7,, 3, 7, and 9. It has been proven (in several dierent ways) that there are innitely many prime numbers. The most universal, and commonly taught, proof was done by Euclid [300 b.c.] in Book IX of Euclid's Elements. Throughout this tet, we will be calling on results and assistance from other books, articles, and websites. The main book that will be used is Introduction to Analytic Number Theory by Tom M. Apostol []. We will also call on the help of certain tets on Analysis for reference on important theorems or denitions, namely [4] and [5]. Some websites and clips of articles will be used to epand on some ideas and how they have historically come about. While it is assumed that the reader can follow along with some results from Calculus, we will make note of results as needed. 3

9 2 Introductory Problems In the following subsections, we will do some problems that are number theoretic in nature but require analysis techniques to ease into the subject. The rst problem involves an interesting result connecting the Fibonacci Numbers and the Golden Ratio, ϕ The net problem introduces the Catalan Numbers and a series associated with these coecients. Lastly, we will introduce a function known as the divisor function which counts the number of divisors of any number. 2. Fibonacci Numbers and The Golden Ratio We start o by dening the Fibonacci Numbers. Denition. Let F, and F 2. Dene F n+ F n + F n. These are called the Fibonacci Numbers. So the rst few numbers are,, 2, 3, 5, 8, 3, 2, 34, 55,... These numbers, interestingly, arise in nature from time to time. They are sometimes referred to as Natures numbering system [2]. We will show a relation involving the Fibonacci Numbers and the Golden Ratio, dened as ϕ Like the Fibonacci numbers, the Golden Ratio sometimes pops up in nature, and life. Given that these two ideas arise in nature and are used to sometimes describe similar objects / events, one might conjecture they have a connection. We will show that they do in fact have a correlation, using an innite series with Fibonacci coecients. Now dene f(z) n F nz n for z C. We wish to nd an eact epression for f(z). First we must show that f(z) is convergent in some neighborhood of 0 with radius R > 0. First, we see that 4

10 F n+ F n + F n F n + F n, since the Fibonacci numbers are an increasing sequence 2F n 2(F n + F n 2 ) 2 2 (F n ), by the same argument as above () 2 3 (F n 2 )... 2 n F 2 n. We can use this to show that this series must have a positive radius of convergence. Now the radius of convergence R is, R lim sup n n F n lim sup, since the Fibonacci numbers are non-negative n n F n lim sup, by the argument from () n n 2n lim sup n 2 n/n (2) 2. So we know that there eists some R /2 such that this series must converge for all z with z < R. Knowing this we can legally rearrange terms if needed, by properties of absolute convergence from [4] and [5]. Now we can nd an eact epression for f(z) by noticing that 5

11 f(z) F n z n n z + z 2 + z + z 2 + F n z n n3 (F n + F n 2 )z n n3 z + z 2 + F n z n + F n 2 z n n3 n n3 z + z 2 + F n+ z n+2 + F n z n+2 z + z 2 + z n F n+ z n+ + z 2 n n F n z n (3) z + z 2 + z(f(z) z) + z 2 f(z) z + zf(z) + z 2 f(z). Solving for f(z) we see that f(z) z z z 2. Since f is analytic at 0, its power series at 0 will converge in a disc whose radius is the distance from 0 to the closest singularity of f. We can see that a problem arises when z ± 5 2 by solving z z 2 0 using the quadratic formula. Since this value of z will dene our radius of convergence, we are only concerned with the smaller of the two numbers 5 2 and Noting that tells us that our radius of convergence is R 5 2 convergence, since ϕ 0.68 > ϕ. Note that this agrees with our estimate for the radius of So we have shown that f(z) n F nz n z z z 2 and has radius of convergence R ϕ.. Note that with this nding, we can nd an approimation for the nth Fibonacci 6

12 number. We see that R ϕ lim sup. n n F n (4) So by taking reciprocals and raising to the n, we get that F n O(ϕ n ). Using other techniques, we will show that F n ( 5 ϕ n ) + ( )n. Here we wish to demonstrate how comple analysis can be used to address this problem. Consider f(z) n F nz n. We have shown above that f(z) z z z 2. We also found that the roots of z z 2 are z ϕ and z ϕ. Note that ϕ + ϕ 5. Also note ϕ n that for z <, Therefore z z n + z + z 2 + z 3 + z n0 z z z 2 z z ( 2 + z ) z (z + ϕ)(z ϕ ) ( ) ϕ ϕ z z + ϕ + ϕ +ϕ z ϕ ( ) ϕ(ϕ+ϕ z ) + z/ϕ ϕ (ϕ +ϕ) z/ϕ (( ) ( ) ) z ϕ 5 + z/ϕ + ϕ 5 z/ϕ ( ϕ ( z n z ( ) n + 5 ϕ) ϕ ) z n ϕ n 5 n0 n ( ( ϕ z z ) 5 ϕ + z2 ϕ 2 z3 ϕ ϕ ( )) + zϕ + z 2 ϕ 2 + z 3 ϕ (( ϕ ) + ϕ z + z ( ϕ 2 ) ( 5 5 ϕ 2 + z2 ϕ 3 + ) ( 5 ϕ 3 + z3 ϕ 4 ) ) 5 ϕ z n( ϕ n + ( ) n (ϕ ) n). 5 n (5) We now make use of the following theorem: 7

13 Theorem 2. Let f(z) be a comple-valued function. Suppose that f(z) n0 a nz n and f(z) n0 b nz n. Then a n b n for all n 0,, 2, 3,... equal. Since we have two series representations of f(z) z z z 2 their coecients must be 5 5 Note when n, then F and ϕ+ϕ ( When n 2, then F 2 and ϕ2 ϕ 2 5 When n 3, then F 3 2 and ϕ 3 + ϕ 3 5 ( + ) )( ) ϕ ϕ+ ϕ ϕ 5 ( 5 ) (6) ( This will work for all n, 2, 3, 4,..., and each case can be solved by epanding ( 5 ) n and 2 using the Binomial Theorem. Therefore F n ϕn + ( ) n ( ϕ ) n 5. + ) n 5 2 8

14 2.2 A Problem Involving the Catalan Numbers Consider a set with a binary operation dened on it. Assume that is not associative, meaning that the values of (a b) c and a (b c) may be dierent. Let c n be the number of possible values of a a 2... a n. Then c n is the number of ways of placing parentheses around these elements in order to specify the order in which the operations are to be performed. Clearly, c and c 2 are both, since no parentheses are required. Now consider c 3. So we see that there are two ways to uniquely put parentheses around a list of 3 numbers. Specically, they are (.) (a b) c, and (2.) a (b c). Note that (a b c) (a b) c. So c 3 2. Now consider c 4. Here, we see that there are 5 ways of putting parentheses around 4 numbers. Notably, (.) ((a b) c) d, (2.) (a (b c)) d, (3.) a (b c) d, (4.) a ((b c) d), and (5.) a (b (c d)). So c 4 5. Computing c 5, c 6,... by hand can become time consuming. So we will simply list them as follows: c c 2 c 3 2 c 4 5 c 5 4 c 6 42 c 7 32 We can see that these numbers grow very rapidly, and are seemingly arbitrary. However, if we look more carefully at each number and the process to obtain the net, we will see a pattern develop. Notice that c c 2 c c 9

15 c 3 2 c c 2 + c 2 c c 4 5 c c 3 + c 2 c 2 + c 3 c c 5 4 c c 4 + c 2 c 3 + c 3 c 2 + c 4 c c 6 42 c c 5 + c 2 c 4 + c 3 c 3 + c 4 c 2 + c 5 c c 7 32 c c 6 + c 2 c 5 + c 3 c 4 + c 4 c 3 + c 5 c 2 + c 6 c... c n n i c ic n i. This formula arises in the way we compute c n for large n. Consider {a, a 2,..., a n }. Then we can break this set up as {a } and {a 2, a 3,..., a n }, or {a, a 2 } and {a 3, a 4,..., a n }, etc. Thus in general, we can break up {a, a 2,..., a n } as {a,..., a k } and {a k+,..., a n }. Note that {a, a 2,..., a k } corresponds to c k which we have already determined. So we can use the previous results to help us determine c n. So we have found a recursive formula for the Catalan numbers, c n. Now we want to nd an eact epression for a function of the form f() c n n, n where the coecients are the Catalan numbers. To do this, we will rst do some manipulations with the series n c n n that may or may not be legal. We will assume that f() is analytic at 0, and use this assumption to nd an epression for f(). 0

16 ( )( ) (f()) 2 c n n c n n n n (c + c c c )(c + c c c ) c c 2 + (c c 2 + c 2 c ) 3 + (c c 3 + c 2 c 2 + c 3 c ) c c c (7) (c + c 2 c 2 + c c ) c f(). Hence, (f()) 2 f() + 0. Solving this quadratic equation for f() we get f() ± 4 2. Now to gure out whether the numerator should be + 4 or 4, we will use the fact that f(0) 0 (since f(0) c (0) + c 2 (0) ). So it must be that f() 4. 2 The above function is analytic in the disc { : < /4} and this disc can be no larger because of the singularity at /4. Hence, it will have a power series n a n n with radius of convergence /4. Because of our initial assumption, we will need to verify that a n c n for all n. In order to nd a Taylor Series for the function f() 4 2, we will rst nd a power series epression for. Using the formula g() g (n) (0) n0 n! n, for some analytic function g(), we get the following: 2 2! n ! (2n 3) 2 n n. n! ! (8) Now, 4 n (2n 3) 2 n n. (9) n!

17 Thus, 4 ( n n (2n 3) 2 n n n! (2n 3) 2 n n. n! (0) Dividing by 2, we get that 4 2 n n n (2n 3) 2 n n n! (2n 2)! 2 n (n )! n! 2n n (2n 2)! (n )! n! n. () Now, we need to show that the coecients (2n 2)! (n )! n! are the Catalan numbers. Proof. Let a n (2n 2)! (n )!n!. We have the following: 4 2 a n n ; n ( ) 2 a n n ( n n n2 k a k a n k ) n ; Hence, n 4 2 a n n 4 2 n2 + ( n a k a n k ) n ; k n2 ( n a k a n k ) n, k Equating the coecients, we get a and a n n k a ka n k for each n >. Note that a c. Since a n n i a ia n i and c n n k c kc n k (they are dened the same way) and a c, it follows that a n c n. 2

18 This shows that f() n c n n is analytic in a neighborhood of 0 since f() 4 2. Since R 4 lim sup n n c n, then c O(4 n ). However, this is a rather crude estimate, and we can nd a better one using the fact that c n (2n 2)! (n )! n!. For this we will use Stirling's approimation to estimate n!. This approimation says n! n n e n 2πn. So, c n (2n 2)! (n )! n! (2n 2) 2n 2 e 2 2n 2π(2n 2) (n ) n e n 2π(n ) n n e n 2πn (2n 2)2n 2 e 2π(2n 2) (n ) n 2π(n ) n n 2πn 22n 2 (n ) n e 2π(2n 2) 2π(n ) n n 2πn (2) 2 2 2n 2 e(n ) n n n 2nπ 22n 2 e(n ) n n n nπ 4n n nπ. This gives us a better estimate for the growth of the Catalan numbers. 3

19 2.3 The Divisor Function We say that a number d divides a number n, written d n, if for some integer k, n dk. Also, when we refer to a number we will mainly be referring to positive integers (or natural numbers). For this problem, we will make use of the divisor function, which is dened as d(n) d n. This function counts the number of divisors of n. For eample, the divisors of 6 are, 2, 3, 6. So d(6) 4. Note that if p is a prime number, then d(p) 2. Clearly the values of this function seem to uctuate. Our goal is to nd an estimate for the number of divisors of n. To do this, we will utilize some graphs involving d(n). First we will plot the graph of d(n). d(n) n We see that the graph of d(n) is oscillatory with ma k n d(k) increasing. Trying to nd a line of best-t will be dicult since the graph of d(n) doesn't appear to be linear. In order to better understand the average behavior of d(n), we will eamine n n d(). Note that this sum resembles a Riemann Sum. Also note that n n d(), d and counting the divisors of is equivalent to counting the number of pairs a, b such that ab. So we are adding one for every pair of numbers (a, b) such that ab. Thus when we eamine n d(), we are adding one for every lattice point (coordinate pair of integers) on the graph of ab or b a. Therefore n d() is summing all lattice points (a, b) such that ab n. Consider the following eample for n 6. 4

20 b 6 a. For n 6 we will be nding all lattice points such that ab 6. Note that the curve is b a Without loss of generality, we are looking at a graph of y n, and counting the number of lattice points on or below this graph. The region each graph represents all numbers and y that multiply to result in n. Since we noted that n n d() resembles a Riemann sum, to nd the average number of divisors of n we will integrate f() n, but we need to consider what endpoints to integrate from. Note that this integral will approimate n i d(i). If we integrate n y n d, we will be considering the following graphical representation (for n 6): n 6): From this picture it is clear that n i d(i) < n + n n d. Now, if we consider n+ n d, we will have the following graphical representation (for 5

21 y We see that this integral gives an underestimate on the value of n n i d(i) n+ n d. So we can combine these results to nd an estimate for n i d(i). So, n+ n d Evaluating these integrals, we get that, n d(i) i n n d + n. i d(i). Hence, n ln(n + ) n d(i) n ln(n) + n. i Dividing by n yields, ln(n + ) n n d(i) ln(n) +. i Note that ln(n + ) ln(n) + ln( n+ n ) ln(n) + ln( + n ). As n grows larger, ln( + n ) will tend to 0. So to represent this, we say that ln(n + ) ln(n) + O( n ), where we dene this big Oh notation as follows: Denition 3. For two functions f(z) and g(z) we say that f(z) O(g(z)) as z z 0 if f(z)/g(z) is bounded in a neighborhood of a point z 0. Dividing our previous result by ln(n), we have, ln(n + ) ln(n) n n i d(i) ln(n) + ln(n). ln(n) 6

22 Therefore, lim n n n i d(i), ln(n) or n n i d(i) ln(n). To more thoroughly show this, we turn to Introduction to Analytic Number Theory by Tom Apostol. He shows that d(n) ln() + (2γ ) + O( ) []. n ( Here, note that γ is Euler's constant, γ lim n ( n k k ) ln(n)) In order to prove this, we will need to use the following results that have been previously proven in []:.) n /d d + O(). 2.) n n ln() + γ + O( ). Proof. (Note that this proof comes from []). Since d(n) d n, we can rewrite n d(n) as, n d(n) n. For any divisor d of n, we can say that n dr, for some positive integer r. So we can rewrite the sum so that it sums over the pairs of positive integers r, and d; that is, d(n). ( ) n n n drn dr d n d n By the argument given above, we are considering all lattice points (, y) N 2 such that y n. We will rst consider for each d, the lattice points on the horizontal line segment 7

23 r /d, and then sum over all d. So we can rewrite the above sum as, /d d(n). n d r So we can use the two previously proven results from [] to see that, d(n) n d ( ) d + O() d d + O() (3) ( ln() + C + O(/) ) + O() ln() + O(). Recall that (*) counts the number of lattice points in the region under the hyperbola. We can use the line y to slice this region in half. So the total number of lattice points in the region is twice the number below this line plus the number on the line segment. y Here the dots represent lattice points. So referring to this picture, we see that n d(i) 2 i d n [ ] n d d + [ n]. 8

24 Using the relation [ n] n + O() and the equations given prior to the proof, we see that d(i) 2 i n d n 2n d n ( ) n d d + O() + O( n) d 2 d n d + O( n) 2n ( log( n) + γ + O ( n )) 2 (n 2 + O( n) ) + O( n) n log(n) + 2nγ + 2nO( n ) n 2O( n) + O( n) (4) n log(n) + (2γ )n + O( n). Note that over the years, the term O( n) has been improved. In 903 Voronoi proved that the error is O(n /3 ln(n)); in 922 van der Corput improved this to O(n 33/00 ). The best estimate to date is O(n (2/37)+ɛ ) for every ɛ > 0, obtained by Kolesnik in 969. [] Figuring out the greatest lower bound for O(n θ ) for all θ is an unsolved problem, called Dirichlet's divisor problem. In 95, Hardy and Landau showed that infemum of θ /4 []. We will now eamine a table of n, d(n), n i d(i) n and ln(n). We will round n i d(i) n and ln(n) to 3 decimal places, for convenience. 9

25 n d(n) n i d(i) n ln(n) Looking at this table reveals that the two columns furthest to the right seem to be asymptotic, which is what we veried above. 20

26 3 Introduction to The Prime Number Theorem Before diving into the proof and analysis of The Prime Number Theorem (PNT), we will begin by giving a brief overview of the problem, its history, strategies we will use, and some facts that will be needed during the proof. 3. History and Motivation Prime Numbers have been an area of study and wonder for hundreds of years. How frequently do they occur? Are there innitely many of a certain form? Do they have an average distribution? All of these seemingly simple questions are very dicult to answer, and many remain unsolved to this day. One of the biggest problems is the lack of a formula to represent them or an algorithm to check if a given number is prime. So many of the results that help us understand prime numbers are estimates, or approimations. The Prime Number Theorem is arguably the most important result as far as development and application of Analytic Number Theory. The theorem states that the number of primes less than a positive integer (denoted π()) is approimately ln(). There have been several proofs of this theorem. We will be eamining two of them: An Elementary Proof and an analytic proof involving Comple Analysis techniques. The Prime Number Theorem was rst hypothesized by Gauss (792) and Legendre (798) when they wrote down a table of primes less than or equal to a number. They made an estimate indicating that π() seems to be trending similar to / ln(). The table below shows some of these values: 2

27 π() ln() Their result was rst veried by Hadamard and de la Vallee Poussin in 896, using techniques from comple analysis and properties of the Riemann Zeta Function. Many thought that an elementary proof didn't eist and that the only way to prove the theorem was by analysis. However, in 949, an elementary proof was presented by Erdös and Selberg. Their proof didn't involve any techniques from analysis and can be read and understood by a calculus student; however, their proof is very detailed and requires the denition of many other functions and the verication of many other results. We will give a detailed proof of the analytic method and will skim through the elementary method, as some results from this are needed for the analytic proof, while referring the reader to [] and [3] for detailed proofs of the necessary results. 22

28 3.2 Elementary Method For the Elementary proof, our goal is to show that π() π() ln() lim lim. ln() However, in order to do this, we will need to dene several functions and state several results. We will focus on the main ideas and results needed for the analytic method, while giving a sketch of the elementary method. Dene the Mangoldt function as ln(p), if n p m where p is prime and m ; Λ(n) 0, otherwise. We will later show that if Λ(n) as, n then the prime number theorem will hold. Let ψ() Λ(n). n Our goal is to show that lim ψ() and lim π() ln() that since Λ(n) 0 if n p m, then we can rewrite ψ() as are equivalent. Note ψ() Λ(n) Λ(p m ) n m p m m p /m ln(p). Note that since we are only considering n, we are taking the sum over nitely many terms, so the sum on m is nite. So we can say that m p /m ln(p) 23 k m p /m ln(p),

29 where k is the highest power of a prime p that is needed. Note that ψ() 0 if < 2 since we have dened Λ() 0. Note that /m for m 2, 3, 4,..., k, if 2. Thus this is true for /m < 2. In this case ln( /m ) < ln(2), so m So we can say that, ψ() log 2 () m ln() < ln(2), thus m > ln() ln(2) log 2(). p /m ln(p). Denition 4. For > 0, dene Chebyshev's ϑ (theta) function by ϑ() p ln(p), where p runs through all primes less than or equal to. With this new notation, we can say that ψ() From [], we have the following theorem: log 2 () m ϑ( /m ). Theorem 5. For > 0, ( ψ() lim ϑ() ) 0 This result will help with a future theorem. We now wish to nd a relation between ϑ() and π(). It is worth noting that both of these functions have jumps at the primes (hence they behave like step functions) but π() jumps by at each prime, and ϑ() jumps by ln(p) at each prime. In order to nd a relation between ϑ() and π() we use Abel's Identity. Denition 6. Abel's Identity: For any arithmetical function a(n) let A() n a(n), where A() 0 if <. 0 < < y. Then, <n y Assume f has a continuous derivative on the interval [, y], y a(n)f(n) A(y)f(y) A()f() A(t)f (t)dt. 24

30 In the following theorem, we will relate π() and ϑ() using Abel's Identity. Theorem 7. Let 2. Then ϑ() π() ln() 2 π(t) dt t and π() ϑ() ln() + ϑ(t) 2 t ln 2 t dt. The proof of this comes from Abel's Identity, letting a(n) (the arithmetic function) be X p (n), the characteristic function of the primes, dened by, if n is prime; X p (n) 0, otherwise. With this we can rewrite π() as π() X p (n), p2 <n and ϑ() as ϑ() ln(p) X p (n) ln(n). p2 <n So by Abel's Identity, letting f() ln() and y, we get the result for ϑ(). Then, letting f() ln() and y 3/2, we get the result for π(). Now that we have a relation between ψ() and ϑ(), and between ϑ() and π(), we can state the following theorem. 25

31 Theorem 8. The following are equivalent: π() ln(). lim ϑ() 2. lim 3. lim (5) (6) ψ(). (7) Note that this theorem does not prove the Prime Number Theorem, but this shows that in order to prove the Prime Number Theorem, it suces to show that either (2) or (3) holds. Proof. Note that by Theorem 5, we have already seen that (2) holds if and only if (3) holds. () (2) Assume that π() ln() lim. By the previous theorem, we have that ϑ() π() ln() 2 π(t) dt, t and In order to show that we need to show that π() ln() ϑ() + ln() ϑ() lim, 2 ϑ(t)dt t ln 2 t. ( ϑ() π() ln() lim lim 2 ) π(t) dt. t Since we are assuming () holds, it suces to show only that lim 2 π(t) dt 0. t 26

32 π() ln() π() Since we are assuming that lim, then lim lim ln(). Since π() ( ) O, ln() we have 2 π(t) ( t dt 2 O ln(t) dt ). By L'Hospital's Rule, lim 2 ln(t) dt 0. Thus, lim 2 π(t) t dt 0. (2) (). Now assume that lim ϑ(). By the previous theorem, we must show that ( π() ln() ϑ() lim lim + ln() 2 By assumption, it suces to show that lim ln() ln() since ϑ(t) O(t). Using L'Hospitals rule, we get ln() lim ϑ(t)dt t ln 2 t ( ϑ(t)dt ln() ) t ln 2 t O dt 2 ln 2, t dt ln 2 t lim ln 2 ln() ln 2 ) ϑ(t)dt t ln 2. t 0. Now lim ln() 0. Therefore π() ln() lim. We can now outline the elementary proof. Note that there is much work still needed, and many parts will be omitted, however we want to focus on the analytic proof. The reader 27

33 is encouraged to see [3] and [] for more specics on the elementary proof. Outline of Elementary Proof. This proof will use Selberg's asymptotic formula, and part of Shapiro's Tauberian Theorem. Theorem 9. (Selberg) Let F be a real or comple-valued function dened on (0, ), and let G() ln() n F (/n). Then, F () ln() + ( ) F Λ(n) µ(d)g n d n ( d ), where µ(n) is the Möbius function dened by µ(), and for p i prime ( ) k, if n p a µ(n)...pa k k, and a... a k ; 0, otherwise. Theorem 0. (Sharpiro's) Let {a n } be a non-negative sequence such that a n ln() + O(). n n Then, for all, n a(n) n ln() + O(). Letting F () ψ(), then by Selberg's formula, ψ() ln() + ( ) Λ(n)ψ 2 ln() + O(). n n This relation is a result of a very involved proof using Selberg's formula, Euler's constant γ, and many other theorems from [] and [3]. Then we dene a new function σ() e ψ(e ). In order to show that the prime number theorem holds, one can relate σ() and Selberg's formula, and then show that 28

34 lim sup σ() 0. This, again, is a result of a very involved proof, and the reader is again referred to [] and [3] for further clarication. End of Sketch. As we can see, the proof of this seemingly simple result is anything but simple. The elementary proof uses no results from comple analysis, and some very basic ideas from real analysis and measure theory (moving limits inside of integrals, limsup, etc.), but there are many functions that need to be dened, many rules of calculus that are needed, and many conclusions to be drawn during the proof. The analytic proof, while it involves more dicult and advanced techniques, will prove to be shorter and easier" to follow. We will give a full, and detailed proof of the Prime Number Theorem using the analytic technique in the net section. 3.3 Riemann Zeta Function Denition. We dene the Riemann Zeta Function ζ(s) by, ζ(s) n n s if Re(s) >, and ( ) ζ(s) lim n s s n for 0 < Re(s) <. The number s C is often written as s σ + it. 29

35 The zeta function is connected with the prime numbers in that the following holds: p prime ( p s ), note that this is a geometric series p prime p s ( + p s + p 2s + ) +... p3s p prime ( + 2 s + 2 2s + )( s 3 s + 3 2s + 3 ( + 5 s + 5 2s + ) s + 2 s + 3 s + 2 2s + 5 s + 2 s 3 s + 7 s s + 3 s + 4 s + 5 s + 6 s + 7 s +... n n s ) s (8) ζ(s). In the above argument, we use the fact that every number has a unique prime factorization (Fundamental Theorem of Arithmetic). From these equations for ζ(s) we can get an epression for ζ (s). Since ζ(s) is dened for Re(s) >, we can do a term-by-term dierentiation of ζ(s) within this domain of convergence. So we get ζ (s) d ds n ( n d ds n ) n s ) ( n s ln(n) n s. (9) You can do this because the series n n s is uniformly convergent on {s Re(s) d} for all d >. We later show that this converges for all σ >. For the Prime Number Theorem, we will need to work with ζ (s) ζ(s). This means we will 30

36 want to eamine, ζ (s) ζ(s) n ln(n)n s n n s 0 + ln(2)2 s + ln(3)3 s + ln(4)4 s s + 3 s + 4 s (20) For the purposes of the Prime Number Theorem, we will want to nd the singularities of this. However, this becomes dicult in the form we have it in right now, as that would mean nding the zeros of the Riemann Zeta Function, which is the premise of the Riemann Hypothesis (which has not been proven). This means we need a form that is easier to work with. Consider ( ln(ζ(s)) ln p prime ( p s ) ) ( ) ( ) ln p s p prime ln ( p ) s. p prime (2) From this, we see that ( ) d ln ζ(s) ds ζ (s) ζ(s) d ds ( p prime p prime p prime p prime ln ( p )) s d ( ds ln p ) s ln(p) p s p s ln(p) p s p s. (22) Note that this holds for Re(s) >. When working with the proof of the Prime Number Theorem, we will use many estimates for ζ(s) and ζ (s), as well as properties of ζ(s) near the line σ. Recall from our denition of the Riemann zeta function, it is not dened for σ. We will show that 3

37 ζ (s) ζ(s) s is analytic on the line σ. Essentially, the pole at s is being subtracted o, thus making this epression analytic. The following argument makes use of the Dirichlet Eta function, given by η(s) ( ) n n n s 2 s + 3 s 4 s +..., for a comple number s with Re(s) > 0. Note that η() is dened, as η() gives the alternating harmonic series, which converges to ln(2). Our goal is to show that ζ(s) has a singularity at s. We start by noting that ζ(s) η(s) ( + 2 s + 3 s + 4 ) s +... ( 2 2 s + 4 s + ) 6 s +... ( 2 s + 2 s + 3 ) s +... ζ(s). 2s ( 2 s + 3 s 4 ) s +... (23) Thus, ( η(s) ζ(s) ) 2 s. Therefore, ζ(s) η(s) 2 s. Now we can let s approach and see that ζ(s) has a pole at s. Note that by the coincidence principle, we can legally do the above manipulations. 32

38 We now want to show that ζ(s) has a residue of at the pole s. s lim(s )ζ(s) η() lim s s 2 s 2 s (s ) η() lim s 2 s 2 s s ln(2) lim s 0 2 s s ln(2) lim s 0 2 s ln(2) by denition of a derivative ln(2) (24). Therefore ζ(s) has a residue of at s. To show that ζ (s) ζ(s) s is analytic at s, we use the following Lemma from []: Lemma 2. If f(s) has a pole of order k at s a then the quotient f (s)/f(s) has a rst order pole at s a with residue k. Proof. Note that we can rewrite f(s) as f(s) g(s) (s a) k, where g is analytic at a, and g(a) 0, since f has a pole of order k at s a. Using the quotient rule, we get f (s) g (s)(s a) k k(s a) k g(s) (s a) 2k g (s)(s a) k (s a) 2k k(s a)k g(s) (s a) 2k g (s) (s a) k kg(s) (s a) k+ g(s) [ g (s) (s a) k g(s) k s a ]. (25) Therefore, f (s) f(s) g(s) [ g (s) (s a) k g(s) k ] (s a) k s a g(s) g (s) g(s) k s a. 33

39 Since g(s) 0 and is analytic, it follows that g (s) g(s) f (s) f(s) + k s a too. is analytic at s a and is equal to Thus, since ζ (s) ζ(s) and s their dierence is analytic at s. both have a rst order pole at s with residue, then 3.4 Idea of the Proof For our proof of the Prime Number Theorem, we will rst outline the proof without much elaboration on the ideas used or the results found. The subsequent sections will then give more details. The purpose of this is so that the proof is as straightforward as it can be, while still eplaining all of the results. When we come across a line that needs further eplanation, the reader is referred to a later subsection, where the result will be eplained. Much of this proof comes from [], but some details that are not featured in that tet, will be featured in this. The goal is to have the reader follow as much of the proof as possible. When one does a search on proof of the Prime Number Theorem, there are many results, but almost none of these results give a straightforward proof. They prove many results and ideas, and then the PNT follows from these. Our goal is to rst prove it, and then elaborate when necessary. 34

40 4 Analytic Proof of The Prime Number Theorem Theorem 3. (The Prime Number Theorem) Let π() denote the number of primes less than or equal to. Then π() or equal to. ln() ; that is, there are approimately ln() primes less than Proof. The goal is to show that π() ln() lim. In our discussion on the elementary method, we showed that if the limit does tend to, then the following limits also equal : ψ(n) () lim n n ϑ(n) (2) lim n n. While much of the elementary method focuses on (2), we will work with () for the analytic method. Our goal is to show that ψ(), as. Recall that we dened ψ() as ψ() Λ(n). n Note that ψ() is a step function, growing by ln(p) whenever one reaches a power of a prime. We dene a new function ψ () by ψ () We will show that ψ () 2 2 as. The rst result we get is that, ψ(t)dt. ψ () 2 c+ i s ( ) ζ (s) ds, 2πi c i s(s + ) ζ(s) 35

41 where c >. Note that this is a contour integral along the line Re(s) c. This result is veried and elaborated on in section 4.. Working with this contour integral, we can also show that ψ () 2 ( ) 2 c+ i s ( ζ (s) 2 2πi c i s(s + ) ζ(s) ) ds, s for c >, and (this result is also veried in section 4.). Recall that our goal is to show that as, then ψ () 2 2. So we want to show the above integral is 0. Our main problem with the above integral is when Re(s), since our current denition of ζ(s) is not dened for s. Thus we need some results about ζ(s), ζ (s), ζ(s), and ζ (s) near σ. Section 4.2 will be devoted to nding these relations, but our main result is that ζ( + it) 0, for any t R. This is also proven in section 4.2. Recall from section 3.3 that ζ(s) has a pole at s, but the dierence ζ (s) ζ(s) s is analytic at s. ( ) For convenience, let h(s) s(s+) ζ (s) ζ(s) s. We will devote section 4.3 to showing that ψ () 2 ( ) 2 h( + it)e it ln() dt. 2 2π Note that we get the change in integrals by letting σ in our original integral. It will then be shown that we only need to be concerned with the integral along the line σ. Specics are covered in section 4.3, but our general method is to do a contour integral around a rectangle R {( it ), (c it ), (c + it ), ( + it )} and let T. We show that the integral around R must be 0, by Cauchy's Theorem. We then show that on the upper and lower lines of the rectangle, the integral is 0 and we come to this conclusion by taking bounds on h(s). We then show that the two integrals along the vertical line segments are the same and they cancel each other. In doing so, we show that these integrals (which end up being our h(s) from above) are bounded. We can use the Riemann-Lebesgue Lemma to show that the entire integral must be 0, 36

42 since we have shown that the integrals are bounded. With our ndings on ζ(s) from sections 3.3 and 4.2, we can apply the Riemann-Lebesgue Lemma to the integral, even if c. Therefore, ψ () 2 ( 2. 2 ) Thus as, ψ () 2 2. Recall that ψ () ψ(t)dt. Also recall ψ() 0. By a lemma from [] we can dierentiate both sides of the above to get d ( ψ () ) ψ() d ( ) 2. d d 2 Therefore, as, ψ(). From our result in section 3.2, we get that since ψ() as, then π() ln() as. Hence π() ln(). 37

43 4. The Contour Integral The following Lemma comes by use of Abel's Identity. Lemma 4. For any arithmetical function a(n), let A() n a(n), where A() 0 if <. Then Proof. By Abel's identity, ( n)a(n) n a(n)f(n) A()f() n A(t)dt. A(t)f (t)dt, if f has a continuous derivative on [, ] (smooth). Let f(t) t. Note that on [, ] f(t) t is smooth. Using Abel's Identity, a(n)f(n) n a(n)n. n Also, Therefore, A()f() a(n). n a(n)f(n) na(n) n n A()f() a(n) n A(t)f (t)dt A(t)dt. (26) Thus, A(t)dt ( n)a(n). (27) n We can use this result to nd a relation for ψ (). 38

44 Theorem 5. ψ () ( n)λ(n). n This is because ψ() n Λ(n), and ψ () ψ(t)dt. So we can use the previous lemma with A(t) ψ(t), and a(n) Λ(n), since the Mangoldt function is an arithmetical function. We would like to epress ψ () in terms of a contour integral, as contour integrals can often be simplied and give a result that can be worked with. The following lemma will be useful in nding the contour integral. Lemma 6. If c > 0, and u > 0, then for every integer k we have c+ i u z 2πi c i z(z + ) (z + k) dz k! ( u)k if 0 < u 0 if u >, where the integral is absolutely convergent. For this proof, we will make use of the Gamma function, dened as Γ(s) Note that for any n N, Γ(n) n!. 0 s e d. Proof. Note that Γ(z + ) zγ(z). So, u z Γ(z) Γ(z + k + )... u z Γ(z) (z + k)γ(z + k) u z Γ(z) (z + k)(z + k )Γ(z + k ) u z Γ(z) (z + k)(z + k ) zγ(z) u z (z + k)(z + k ) z. (28) We can now apply Cauchy's Residue Theorem to u z Γ(z) 2πi C(R) Γ(z+k+) dz, where C(R) is a 39

45 contour we will dene momentarily. Recall that Cauchy's Residue Theorem states that if C is a simple, closed, positively oriented contour in the comple plane, and a function f is analytic ecept for some sequence of points {z, z 2,..., z n } inside C, then C n f(z)dz 2πi Res f (z k ). For this problem, we will have two dierent contours C(R) to consider, depending on the value of u. This is a circle, where we are cutting o at the line of Re(z) c, and integrating the left side counterclockwise (positively). This will be used for 0 < u. k 40

46 This is a circle also cut o at real part Re(z) c, but we are integrating the right side in a counterclockwise direction. This will be used for u >. The reason we choose these contours is so that the circular parts will be zero. We need the circle to miss the poles of the Gamma function, which occur at negative integers. So we need to make sure the circle's radius is strictly greater than 2k + c, as to miss these points. We will show that as the radius R, the integrals along the circular arcs go to zero. Let z + iy and z R. Since u > 0 we have, u z z(z + ) (z + k) u u iy z z + z + k u z z + z + k u c R z + z + k, (29) where the inequality holds because u is increasing on 0 < u and decreasing for u >. 4

47 Let n k. Then we have z + n z n R n R k R/2, since R > 2k + c. Now we have, u z z(z + ) (z + k) u c R z + z + k u c R(R/2) (R/2) u c R(R/2) k. (30) From this we get the following bound for the integral around the circular arc, C: C u z z(z + ) (z + k) dz C 2π 0 2πu c (R/2) k, u z z(z + ) (z + k) Rdz u c R(R/2) k Rdz (3) by directly evaluating the last integral. Letting R, this tends to 0. So we will eamine how these contour integrals behave for dierent values of u. Case : Consider u >. Then inside C(R), the integrand is analytic; therefore C(R) f(z) 0. So letting R, we have that the integral is 0. Case 2: Consider 0 < u. We will use Cauchy's Residue Theorem (stated above). 42

48 Note that this integral has poles at n 0,, 2,..., k. Thus, u z 2πi C(R) z(z + ) (z + k) dz k n0 k Res z n lim z n n0 k n0 k n0 u z z(z + ) (z + k) z u (z + n) z(z + ) (z + k) u n ( n)( n + ) ( ) 2 (k n) u n ( ) n n!(k n)! k u n ( ) n n!(k n)! n0 k ( ) k u n ( ) n n k! n0 k ( ) k ( u) n k! n n0 (32) k! ( u)k. We can now get a contour integral for ψ () 2. Theorem 7. If c > and, then ψ () 2 c+ i s ( ) ζ (s) ds. 2πi c i s(s + ) ζ(s) Proof. By a previously proven theorem, ψ () ( ) n Λ(n). n 43

49 Now by the previous lemma, letting k and u n/ with n we get n ( n/)! 2πi c+ i c i (/n) s (33) s(s + ) ds. Thus, Λ(n) Λ(n) n ( Λ(n) n ) c+ i 2πi c i Λ(n)(/n) s ds. s(s + ) Recall from a previous result that ψ ()/ n Λ(n)( (n/)). So, ψ () n c+ i 2πi c i Λ(n)(/n) s ds s(s + ) n c+ i 2πi c i Λ(n)(/n) s ds. s(s + ) We want to be able to interchange the summation and integration symbols. Note that the partial sums satisfy N k c+ i c i Λ(k)(/k) c s(s + ) ds N k N k Ω c+ i c i c+ i Λ(k) k c N k Λ(k) k c, Λ(k)(/k) c ds s s + c i c s s + ds (34) where Ω is a constant. Since the partial sums are bounded absolutely, the series must converge absolutely, and uniformly converge. So we can interchange the summation and integration symbols. There- 44

50 fore, ψ () 2πi 2πi 2πi 2πi 2πi 2πi c+ i n c i c+ i c i n c+ i c i c+ i c i c+ i c i c+ i c i Λ(n)(/n) s ds s(s + ) Λ(n)(/n) s ds s(s + ) s ( Λ(n) s(s + ) n s s s(s + ) s s(s + ) s s(s + ) n p prime p prime ( ( ζ (s) ζ(s) α ) ds ) ln(p) p αs ds /p s ln(p) /p s ds ) ds. (35) Dividing by yields the desired result. Theorem 8. If c > and we have ψ () 2 ( ) 2 2 2πi c+ i c i s h(s)ds, where h(s) ( ζ (s) s(s + ) ζ(s) ). s 45

51 Proof. ψ () 2 ( ) 2 c+ i 2 2πi 2πi 2πi 2πi 2πi c i c+ i c i c+ i c i c+ i c i c+ i c i s s(s + ) s s(s + ) s s(s + ) s s(s + ) s h(s)ds. ( ζ (s) ζ(s) ( ζ (s) ζ(s) ( ζ (s) ζ(s) ) ds ( 2 ds 2 ) ) ds 2πi ) ( ζ (s) ζ(s) s c+ i c i s (s )(s)(s + ) ds ) ds (/) s s(s + )(s + 2) ds (36) Note that we get from line to line 2 by using a previous lemma for k 2 and u /, and from lines 2 to 3 by replacing s with s. So we need to show that as the integrand 0. Having the integrand tend to 0 as will then imply that ψ () 2 /2 as. Letting c, we have + i i h( + it)e it ln() dt. Thus we need to eamine ζ(s) and ζ (s) near σ. However, our current denition of the Riemann Zeta Function isn't valid for σ. We will epand on this in the net section. 46

52 4.2 The Riemann Zeta Function Near s This subsection will be devoted to nding bounds, estimates, and properties of ζ(s) and ζ (s) near the line σ. We rst prove the following lemma: Lemma 9. Proof. ζ(s) N n s N s+ d + N s s. Euler's summation formula states that if f is smooth (continuous derivative) on [a, b], 0 < a < b, then f(n) a n b b a f(t)dt + b Taking f(n) n s, a N, and b, we get a ( t t ) f (t)dt + f(b) ( b b ) f(a) ( a a ). ζ(s) n N n N n N n N n N n N n N n n s n s + n s + nn+ N n s + N [ n s + n s t s dt + N t s dt s N t s (s ) ] N ( )( t t st s ) dt + lim s N N N s t t t s+ s N n s N s (s ) s N n s + N (s ) s s N n s + N s s s N t t t s+ t t t s+ dt + lim t t t s+ t t t s+ s dt + lim dt + lim dt + lim dt + lim s s. s s (37) 47

53 Since s s and lim s 0, the result holds. Using this representation of ζ(s), we can see that ζ (s) N n ln(n) n s ( ) ln() + s N s+ d N s+ d N s ln(n) s N s (s ) 2. We can now use these results, along with discussion from section 3.3 to nd approimations and results for the Riemann zeta function near the line Re(s). This representation can be used to show that for all α > 0, ζ(s) O(ln(t)) and ζ (s) O(ln 2 (t)) for σ /2 and t e. The following results come from []. To show that ζ( + it) 0, we use the following: Theorem 20. For σ >, ζ 3 (σ) ζ(σ + it) 4 ζ(σ + 2it). To prove this, we make use of a result from [] which states that for Re(s) >, ζ(s) e F (s), where F (s) Λ(n) n2 ln(n) n. s Some manipulation can be done with this representation, and also the fact that Λ(n) n2 ln(n) Then, n s p prime m ( ζ 3 (σ) ep Also, ζ(s) ep ( p m So, mp ms, for σ >. p m ) 3 ( mp mσ ep cos(mt ln(p)) mp mσ ). ζ(σ + it) ( 4 ep p m p m 4 cos ( mt ln(p) ) ). mp mσ ) 3 mp mσ. 48

54 Using this we also get ζ(σ + 2it) ( p cos ( 2mt ln(p) ) ). m mp mσ Multiplying these three results means we are multiplying eponentials, hence are adding the fractions in the eponents. Our fraction looks like cos ( mt ln(p) ) + cos ( 2mt ln(p) ) mp mσ. Letting θ mt ln(p), we use the fact that, cos θ + cos 2θ cos θ + 2 cos 2 θ 2 cos 2 θ + 4 cos θ ( cos 2 θ + 2 cos θ + ) (38) 2 ( cos θ + ) 2 0. So, ζ 3 (σ) ζ(σ + it) 4 ζ(σ + 2it) ( ep p m ) 0 mp mσ e 0. Therefore the result will hold. We now have everything we need to prove our goal result. Theorem 2. ζ( + it) 0 for any t R. Proof. We can divide the above result by σ for σ >, to get ζ 3 (σ) ζ(σ + it) 4 ζ(σ + 2it) σ σ (σ ) 3 ζ 3 (σ) ζ(σ + it) 4 ζ(σ + 2it) (σ ) 4 σ ( (σ )ζ(σ) ) 3 ζ( + it) σ 49 4 ζ( + 2it) σ.

55 We eamine rst lim σ +((σ )(ζ(σ)) 3. Note that since ζ(s) has a residue at a pole s, then, lim )ζ(s). σ +(σ Also, lim ζ(σ + 2it) ζ( + 2it). σ + In order to show that ζ( + it) 0, assume for a contradiction that ζ( + it) 0 for some t R. Then, ζ(σ + it) σ 4 ζ(σ + it) ζ( + it) σ 4, by assumption. Taking the limit of the above as σ + gives us the derivative of ζ( + it). So, lim ζ(σ + it) ζ( + it) σ σ + 4 ζ ( + it) 4. Using the above results, we see that ζ 3 (σ) ζ(σ + it) 4 ζ(σ + 2it) lim ζ ( + it) 4 ζ( + 2it) ; σ + σ however lim σ + σ. These two results contradict one another. Therefore ζ( + it) 0, for any t R. This is the main result we need in order to prove the Prime Number Theorem, but one more result is necessary. Theorem 22. For σ and t e, there eists M > 0, such that ζ (s) ζ(s) < M ln9 (t). We will not present the proof of this here, but it is the result of previous results, and the reader is referred to [] for further clarication. 50

56 4.3 The Final Step We now want to show that as c+ i s ( ζ (s) 2πi c i s(s + ) ζ(s) ) 0. s We can use information from sections 3.3, 4. and 4.2 to show this. Theorem 23. [] For, ψ () 2 ( ) 2 h( + it)e it ln() dt, 2 2π where h( + it) dt converges. Thus, by the Riemann-Lebesgue Lemma, ψ () 2 /2, and thus ψ() as. Proof. Recall that 4. was devoted to proving that ψ () 2 ( ) 2 2 2πi c+ i c i s h(s)ds, ( ) where h(s) s(s+) ζ (s) ζ(s) s. This result is true for c > and. We want to show this also holds for c. Consider the region R below. 5

57 Note that this is a simple, closed, positively oriented curve. Since s h(s) is analytic inside and on R, we can use Cauchy's Theorem to see that s h(s)ds 0. R We can use this fact to help us evaluate parts of this integral. Our goal is to see what happens as T. 52

58 Consider the upper line on R. Note that s(s + ) (σ + it)(σ + + it) (σ + it )(σ + it ) σ + it σ + it T 2. (39) Similarly, s(s + )(s ) T 3 T 2, since T. Also, if T e, then by section 4.2, there eists a constant M > 0, such that ζ (s)/ζ(s) M ln 9 (T ), if σ. Thus, h(s) M ln9 (T ) T 2. Using this result, we have that c+it +it s c h(s)ds c M ln9 (T ) T 2 dσ, since here we are only considering the real part of the integrand. Integrating with respect to σ yields, c+it +it s h(s)ds Mc ln9 (T ) T 2 (c ). Now letting T, we get the following: ln 9 (T ) lim T T 2 0. (40) So as T, c+it +it s h(s)ds 0, 53