Two classic theorems from number theory: The Prime Number Theorem and Dirichlet s Theorem

Transcription

1 Two classic theorems from number theory: The Prime Number Theorem and Dirichlet s Theorem Senior Exercise in Mathematics Lee Kennard 5 November, 2006

2 Contents 0 Notes and Notation 3 Introduction 4 2 Primes in the odd integers 5 2. The infinitude of rimes The Prime Number Theorem History of the Prime Number Theorem An analytic theorem Newman s roof of the Prime Number Theorem Primes in arithmetic rogressions Dirichlet s Theorem Euler s roof for q = Dirichlet characters and L-functions (rime modulus Dirichlet s roof (rime modulus Dirichlet characters and L-functions (general modulus Dirichlet s roof (general modulus A robabilistic interretation of Dirichlet s theorem The Generalized Prime Number Theorem A Aendix 49 A. Comlex Integration and Differentiation A.. Comlex Integration A..2 Comlex Differentiation A.2 The Leibniz Integral Rule A.3 Useful algebraic concets and facts A.3. The multilicative grou Z α is cyclic for all odd rime A.3.2 The multilicative grou Z 2α for n 3 is generated by 5 and A.3.3 The Legendre Symbol A.4 Carefully considered convergence issues A.4. Infinite roducts A.4.2 Combining absolutely convergent series A.4.3 The Dirichlet test for convergence A.4.4 The Taylor series for log( z converges for all z such that z

3 0 Notes and Notation Proving the Prime Number Theorem (PNT might have been sufficient for a senior exercise; it is a beautiful fact which is nontrivial to rove. I had thought about Dirichlet s theorem on rimes in arithmetic rogressions as well, but I thought that both would be too much, so my real lan was to do one or the other. Once I finished the roof of the PNT, however, I had some momentum, and I was having fun. Sure, I had learned about Abel s Lemma, the Dirichlet test for convergence, and the Leibniz integral rule, but all of these seemed to me disconnected tools, used together here only because they haened to be necessary for the roof. I longed for unity or, as least, something which I could add to make this exercise a whole. Since the Dirichlet theorem interested me, here is what I came u with. I asked the questions Do there exist infinitely many rimes (in the set of natural numbers N or in arithmetic rogressions in N and, if so, how are these rimes (in N or in these arithmetic rogressions distributed that is, how dense are they in N? Euclid and, later, Euler answered the first half of the first question; Dirichlet answered the second half; the Prime Number Theorem (PNT answered the first half of the second question; and the generalized PNT answered the second half. The answers to these four questions form the outline of this exercise. Along the way, we will do some analytic number theory: we will define the Riemann ζ-function and the Dirichlet L-functions; we will do some comlex analysis: we will use the notion of comactness several times (thank you Professor Schumacher, continuously deform contours of integration (thank you Professor Holdener, and analytically continue functions; we will do some abstract algebra: we will use Lagrange s theorem multile times (thank you Professor Aydin, rove that Z α is cyclic for odd rimes, and define Dirichlet characters on Z q for all ositive integers q; and, at one oint, we will even do some robability theory (thank you Professor Hartlaub. 2 The following notation will be used throughout the aer: N the set of ositive integers, 2, 3,... Z the set of integers...,, 0,,... R the set of real numbers C the set of comlex numbers Z q the ring Z/qZ Z q the multilicative grou of invertible elements of Z q R(s denotes the real art of s C I(s denotes the imaginary art of s C φ(n the Euler totient function ζ(s the Riemann zeta function (defined below In the words of John B. Fraleigh, author of A first course in abstract algebra: Never underestimate the ower of a theorem which counts something! 2 I also want to thank Professor Milnikel and Professor Brown; a large ortion of what I understand is understood because I understand linear algebra. 3

4 Introduction Prime numbers are the building blocks of the natural numbers, just as atoms are the building blocks of molecules. Though we cannot understand everything about the natural numbers via meditations on the rimes, we can surely learn a lot! A rime natural number is an irreducible number. That is, the only divisors of a rime are and itself. 3 Also true is the fact that if a rime number dividing a roduct ab of two integers a and b, then either divides a or divides b. This fact can be used to show that the factorization of natural numbers into rimes is unique. Since every natural number can be factored into rimes (using the fact that the quotient obtained by divided a number by a rime is strictly less than the number, we conclude that, for all natural numbers n >, there exists a unique set of rimes < 2 < < r and ositive integers a,..., a r such that n = r k= Here the nature of rimes as building blocks is clear. In this exercise, we shall take interest in the distribution of rimes in the natural numbers. We ask: How many rimes are there? About how many rimes are there less than a given x? How many rimes are there of the form a+bn where a and b are integers? And finally: About how many rimes of the form a + bn are there less than a given x? These four questions form the skeleton of this exercise. a k k. 3 We do not count as a rime for a few reasons, the most imortant of which is to reserve unique factorization. 4

5 2 Primes in the odd integers Since we do not count as a rime, and since we do not count 0 as a natural number, we might ask if 2 is rime. This is obvious, since and 2 divide 2, and there are no other natural numbers less than 2 which could divide it. Therefore 2 is the smallest rime. In fact, it is the only even rime. For if n is an even natural number besides two, then n = 2m for some natural number m >, and this imlies that (at least three natural numbers namely,, 2, and m all divide n. Two is the smallest rime; two is the only even rime; and many ways, two is the oddest of all rimes. Because two is so secial, we choose to ignore it throughout most of this aer. So we start by studying the distribution of rimes in the odd integers. 2. The infinitude of rimes From the definition, we see also that 3, 5, 7,, and 3 are all rime and that (by skiing to trile-digit numbers 0, 03, and 07 are all rime. If we work a little harder, we would find that 009, 03, 09, and 02 are rime. In fact, no matter what ower of 0 we go u to, we will always be able to find another rime. The reason: There exist infinitely many rimes. One incredible roof of this was given by Euclid. I might say that no senior exercise on the distribution of rimes would be comlete if it lacked this roof. It makes the non-mathematician smile; it makes us squeal. The roof goes as follows: Suose there are only finitely many rimes. Denote them by, 2,..., n. Define the natural number N by N = n k +. k= As we mentioned (but did not rove above, we can write N = n k= where the a k are nonnegative integers. 4 But then if a k > 0 for some k, then k divides both N and N, which means that k =, a contradiction. If a k = 0 for all k, then N =, contradicting the fact that N + >. Therefore, there must exist infinitely many rimes. Like all great mathematical theorems, this one has multile roofs. A second roof is in order for two reasons. First, it will rove a stronger fact from which the infinitude of rimes follows immediately. Second, it will introduce us to the Riemann zeta function, a fundamental function in analytic number theory. But before we get to this, we rove an imortant result which we will use throughout the aer: Theorem. Let χ : N C be a comletely multilicative 5 bounded function, and (formally define L(s, χ for all s C by χ(n L(s, χ = n s. 4 Before we secified that each a k > 0, but we allow some a k to be 0 here so we can more easily and clearly write N as a roduct of the rimes. 5 That is, χ(mn = χ(mχ(n for all m, n N (without restriction. n= a k k, 5

6 Then L(s, χ converges absolutely for all R(s > and has the following roduct reresentation which converges absolutely for all R(s > : L(s, χ = χ( s where the roduct is over all rimes. Proof. For a rime, consider the series k=0 k=0 ( χ( k = k=0 s k=0 k=0 χ( k ks for a variable s C. Since χ is bounded, there exists a real number M such that χ(n M for all n. Therefore, χ ( k ks M ks = M ( k M R(s = R(s < since R(s <, so this series absolutely converges for all rime. If we enumerate the rimes = 2, 2 = 3, 3 = 5,..., then for all N N, ( ( kj N N χ( j s = χ( j N χ k j j j s =. j j= j= k j =0 Since we can multily convergent series in the natural way (see Aendix A.4.2, this roduct is equal to ( N χ k ( j j χ k k 2 2 k N N χ(n k = k N = j= k = ( js s = j k = k N = k k 2 2 k N n s, N n S N where S N is the set of ositive integers whose greatest rime divisor is less than or equal to N. (That is, S N is the set of n N such that N for all rime which divide n. The justification for this last ste is what makes this roof beautiful. We first note that the multilication of absolutely convergent series yields another absolutely convergent series. So rearranging the terms is justified. To show that the sums are equal, we show that each term in each sum shows u in the other, and that no term shows u twice in either sum. Starting on the left side of the equality, to each (k,..., k N -term corresonds a unique n S N, namely, n = k k N N, such that the (k,..., k N - term on the left equals the n-term on the right. On the other hand, given the n-th term on the right, since n S N, n has a factorization into rimes less than or equal to N. Since this factorization is unique, there exists a unique N-tule (k,..., k N such that the n-term on the right equals the (k,..., k N -term on the left. Since the terms in each series show u exactly once in the other series, the series are equal. Finally, since every n N can be factored into rimes, j= k j =0 k js j χ(n lim n s = n S N n= N χ(n n s = L(s, χ, 6

7 and since χ(n n s M <, nr(s n=0 n=0 the series L(s, χ converges (absolutely for all R(s >, giving us L(s, χ = lim N χ(n n s = lim n S N N N j= χ( j s j = χ( s, as claimed. Finally, to show that the roduct converges absolutely, it will be sufficient (see Aendix A.4. to show that the series ( χ( s converges absolutely. But this is straightforward. We have that χ(n M for all n N. For a technical reason, redefine M if necessary such that χ(n M and R(s M for all rimes. (This is always ossible because the set of all R(s is a discrete subset of R. Then χ( s = χ( s χ( M n R(s M, and this finite because R(s >, so the roduct converges absolutely for R(s >. With this fact, we make a definition and derive an amazing corollary. Definition 2. Define the Riemann zeta function ζ(s for all s C by ζ(s = (Note: we will often omit the bounds on the sum; when we do so, it should be assumed the sum over n is over n N. n= Clearly, if χ(n = for all n N, then L(s, χ = ζ(s (in the notation of the revious theorem. Since the constant function χ(n = is comletely multilicative and bounded, we obtain the following corollary: Corollary 3 (Euler s roduct formula. The Riemann zeta function converges absolutely for all R(s >, and it has the roduct reresentation which also converges absolutely for R(s >. n n s = ζ(s = n s. s, Meditations on the equality of this infinite sum and infinite roduct are good for the soul, I believe, if and only if one understands the equality. Euler sure did, and his soul was well. The equality in Corollary 3 has been called the all-imortant formula and the golden key by Havil 6 and Derbyshire, 7 resectively. 6 Havil, J. The All-Imortant Formula. Princeton, NJ: Princeton University Press, Derbyshire, J. Prime Obsession: Bernhard Riemann and the Greatest Unsolved Problem in Mathematics. New York, NY: Penguin, n= 7

8 Furthermore, this equality is an analytic one, yet it is equivalent to the Fundamental Theorem of Arithmetic. For we used the Fundamental Theorem in our roof of the equality, and, conversely, if the equality holds, then every n N, the n-term in the series must corresond to one and only one term in the infinite roduct, once the multilication is erformed. Using Corollary 3, we can easily rovide our second roof of the infinitude of the rimes. Considering the series reresentation of ζ(s, we see that ζ( is the harmonic series which diverges to infinity. Suose, there existed finitely many rimes. Then by Corollary 3, lim ζ(s = lim s + s + which contradicts ζ(s as s The Prime Number Theorem s = <, With the infinitude of the rimes roven, one might ask how dense the rimes are in N. For examle, the sequences 2, 4, 6, 8,... and 4, 6, 64, 256,... each contain an infinite number of terms, but the first sequence, in a sense, contains many more terms. To make this notion rigorous, we could define ψ (x = #{2m m N, 2m x} and ψ 2 (x = #{4 m m N, 4 m x}. Then ψ (x/x is aroximately 2 while ψ 2(x/x is aroximately log 4 log x x, which goes to 0 as x and is therefore less than 2. In a similar way, we give one (famous answer to the density of rimes question by defining the function π(x = #{ x is rime}. Then, for examle, π(6 = #{2, 3, 5} = 3 and π(7 = #{2, 3, 5, 7} = 4. We will use this definition of π(x throughout the aer. The Prime Number Theorem (or PNT gives us an asymtotic formula for counting the number of rimes less than x. The PNT states that π(x x log x where the logarithm is taken base e (here and throughout the aer and the notation f(x g(x means that f(x lim x g(x =. That is, to count the number of rimes less than, say, 000, we could calculate 000/ log The actual value is π(000 = 68. The ratio of the actual value to the aroximate is about.6. If we do the same for a million, we have π(0 6 = 78, 498 and 0 6 / log(0 6 72, 382. The ratio between these values is.084. As we will rove, this ratio converges to. As an interesting aside, one might notice the resence of the transcendental number e in the formula given by the PNT. Given how uncomfortable the Greeks were when they discovered the irrational numbers, imagine how distraught they would be uon discovering that e shows u in a theorem describing the density in N of the multilicative building blocks of N. This is an elementary question, yet the answer requires, not just irrational numbers, but transcendental numbers! Perhas this is art of the reason why algebraic rodigy Niels Abel described the PNT as erhas the most remarkable theorem in mathematics. 8

9 2.2. History of the Prime Number Theorem A timeline of the history of the PNT is as follows: 793 Gauss begins thinking the function π(x. 798 Gauss and Legendre conjecture something close to the PNT; Legendre secifically conjectures that π(x = where A = and B = x A log x+b 848 Chebyshev roved that if π(x x/ log x N for some ositive integer N, then N =. 852 Chebyshev uses elementary roerties of the factorial and logarithmic functions to show that there existed x 0 such that for all x > x 0, where B 0.92 and 6 5 B.. B < π(x x/ log x < 6 5 B, 892 Sylvester imroved on Chebyshev s bound, increasing the value of B to about 0.956, but he closes his aer with a essimistic comment about the generalizability of Chebyshev and his (elementary methods. 896 Hadamard and de la Vallée Poussin alied Hadamard s theory of integral functions to the Riemann zeta function, as well as a simle trigonometric identity to rove the PNT. 92 Hardy asks whether he should believe that there exists an elementary roof (i.e., one which does not involve comlex analysis to what seems like an elementary fact about the natural numbers. 932 Landau and Wiener rovide simlified roofs of the PNT. 948 Paul Erdös and Atle Selberg find a truly elementary roof, to the surrise of Hardy and many others. (The develoment of the roof and of the controversy over who got the credit is an interesting story, which is summarized in [G]. 980 Donald Newman, following in the footstes of Wiener and Ikehara s simlifications, simlifies contour integral methods used in older roofs, to both avoid estimates at and the use of Fourier transforms. D. Zagier s summarized roof sans only three ages. Many great theorems have multile roofs. As this brief history hoefully shows, the PNT is not lacking in roofs. The roof we give is essentially Newman s, but we follow Zagier s aer, which ulls results from Euler, Riemann, Chebyshev, Hadamard, de la Vallée Poussin, Mertens, and Newman An analytic theorem Before we get to Newman s roof of the PNT, we first state and rove what we, following Newman, will call the Analytic Theorem. It is, for his uroses, a lemma, so we state and rove it here. The result is the most difficult to rove of the sequence of facts Newman roves on his way to roving the PNT. It might seem unmotivated, but rest assured: it is just what we need! But first, for our uroses, we rove the following lemma (the roof of which must have been so obvious to Zagier that he felt no need to mention it in his aer: 9

10 Lemma 4. Fix R > 0. Suose g : C C is holomorhic on R(s 0. Then there exists η > 0 such that g is holomorhic at s for all s C such that s R and R(s η. Proof. Since g is holomorhic for R(s 0, it is (in articular holomorhic on A 0 = {s C : R(s = 0, s R}. By definition, for each s A 0, there exists r s > 0 such that g is holomorhic at each oint in the disk B rs (s = {w C : w s < r s }. Because this disk is oen, it makes sense to say that g is holomorhic on the disk or, more loosely, at each oint the disk. Since A 0 can be written A 0 = {iy : R y R}, A 0 is clearly closed and bounded, which means that it is comact. Note that s A 0 B rs (s A 0, so by the definition of comactness, there exist s,..., s n A 0 such that n B rk (s k A 0, k= where we have set r k = r sk for each k to simlify the notation. If needed, relabel the n oints s,..., s n such that y < y 2 < < y n, where y k = I(s k for each k. Also, if B rk (s k B rl (s l for some k l, then throw out s k from the list and relabel. Next, define w 0 = Ri, w n = Ri, and w j = (s j+ ir j+ + (s j + ir j 2 for all j =,..., n. As is clear from Figure, w 0 B r (s, w n B rn (s n, and w j B rj (s j B rj+ (s j+ for each j =,..., n. Therefore, since each B rk (s k is oen, which also imlies that the intersection of any two is also oen, we can choose ositive numbers η 0,..., η n such that B η0 (w 0 B r (s, B ηn (w n B rn (s n, and B ηj (w j B rj (s j B rj+ (s j+ for all j =,..., n. Set η = 2 min{η 0,..., η n }. Then for each j =,..., n, both w j η B rj (s j and w j η B rj (s j. (The 2 in the definition of η was so that we could claim that the w j η are in the balls. Since balls are convex, this imlies that, for each j =,..., n, every oint on the segment connecting w j and w j is in B rj (s j and is therefore in n k= B r k (s k. Since the union of these segments is the segment connecting w 0 = Ri and w n = Ri, we have that where A η n B rk (s k, k= A η = { η + iy : R y R}. Furthermore, given any s = σ + it C such that s R and η < σ < 0, we have it A 0 and η +iy A η, which imlies that iy, η +iy n k= B r k (s k. In fact, since η +iy is further from each s k than iy, there exists k such that iy, η + iy B rk (s k. Thus by the convexity of B rk (s k, 0

11 Figure : This figure demonstrates how we choose the oints w j ; here, the balls centered at s, s 2, and s 3 cover A 0. η + iy B rk (s k for all η (0, η. Since s = σ + iy was an arbitrary comlex number such that s R and η < σ < 0, we have that A η n B rk (s k k= for all η < η 0 (where A η is defined in the same way as A η. So, in fact, the union of all such A η is contained in n k= B r k (s k, which (remember contained only oints at which g(s is holomorhic. Therefore, g(s is holomorhic on A η = {σ + iy : η σ 0, R y R}. η η 0 In articular, g(s is holomorhic for all s C such that s R and R(s η. Now we move on to the theorem. Theorem 5 (Analytic Theorem. Let f(x be a bounded and locally integrable function 8 and suose that the function g(s = 0 f(xe sx dx, defined for R(s > 0, extends holomorhically to R(s 0. Then 0 f(xdx exists and is equal to g(0. Proof. For all T > 0, define g T (s = T 0 f(xe sx dx. 8 A locally integrable function is one whose integral is finite over any comact subset of the function s domain.

12 We wish to show that lim T g T (0 = g(0, as this would rove the theorem. Our strategy will be to estimate the difference between g T (0 and g(0 in terms of a contour integral in C, and to rove that it goes to 0. To this end, let ɛ > 0. Our first ste is choosing real numbers M, R = R(M, ɛ, η = η(r, M = M (η, R, δ = δ(η, M, and T 0 = T 0 (δ, ɛ, M, R. We will then let T > T 0, and show that this imlies g T (0 g(0 < ɛ by means of an estimate of a contour integral. To begin, we know that f(x is bounded. Choose M > 0 such that f(x M for all x 0. Next, choose R > 3M/ɛ. Given R, use Lemma 4 and choose η > 0 such that g(s is holomorhic at all s C such that s R and η σ. Since g(s is holomorhic on the closed and bounded that is, comact subset containing all s C such that s R and σ > η, g(s is bounded on this set. Choose M > 0 such that g(s M for all s in this region. Next, choose δ > 0 such that δ < η and sin ( δ R < πɛ 9M. (Note that such a δ exists, because sin (δ/r 0 continuously as δ 0. Finally, choose T 0 > 0 such that e δt 0 < πδɛ 8RM. (Again, δ > 0, so e δt 0 0 continuously as T 0, so this is ossible. Let T > T 0. We claim that g T (0 g(0 < ɛ. To comute this estimate, we rewrite this difference as the value of a contour integral. Define the contour C in C as the boundary of the region {s C : s R, R(s δ}, ositively oriented. Around this contour, we will integrate ( h(s = (g T (s g(se st + s2 R 2 s. Note the following g(s is holomorhic on C (by our choice of δ, which is less that η, so g(s is bounded on C. By the Leibniz Integral Rule (see Aendix A.2, since 0 and T do not deend on s, d T ds g T (s = 0 ( f(xe sx T dx = f(x( xe sx dx. s 0 Since f(x is locally integrable, the integrand is locally integrable. Therefore, this integral, which is the derivative of g T (s, exists for all s C. So g T (s is entire and, as a result, bounded on any bounded subset of C. e st ( + s 2 /R 2 is entire and, as a result, bounded on any bounded subset of C. s has a simle ole at s = 0. h(s is holomorhic excet for a simle ole at 0, since it is the roduct of s and functions which are holomorhic inside C. 2

13 From these considerations, since 0 is in the region of which C is a boundary, Cauchy s Residue Theorem imlies that h(sds = 2πiRes(h, 0 = 2πi lim sh(s = 2πi(g T (0 g(0, s 0 C where the last equality follows because g(s and g T (s are holomorhic and therefore continuous at 0. From this, we reevaluate our goal: We must show that 2π h(s = g T (0 g(0 < ɛ. C We slit the estimate into three ieces. Define C ± = C {s C : ±R(s > 0}. Then h(sds h(sds C C + + g T (se ( st + s2 ds C R 2 s + g(se ( st + s2 ds C R 2 s ( For the first integral, we note first that, since R(s > 0 for s C +, g(s g T (s = f(xe sx dx f(x e R(sx dx M T T e R(sx dx = T M R(s e R(sT for all s C +. Also, since s = R for s = σ + it C +, ( est + s2 R 2 s = e R(sT σ 2 + t 2 + (σ + it 2 R 3 = e R(sT 2σ σ + it R 3 = 2e R(sT R(s R 2. Therefore, if we denote the arc length of C + by l(c +, h(sds l(c + max h(s C + s C + ( 2πR 2 = 2πM R < 2π ɛ 3 ( M R(s e R(sT ( e R(sT 2R(s R 2 (since R > 3M/ɛ. So far, so good. Now we estimate the second iece. Since g T (s is entire, we can continuously deform 9 C into C = {s C : s = R, R(s < 0} and relace the integral over 9 See Aendix A... 3

14 C with the integral over C. Now for all s C, and, as in the first integral, So C g T (s = T 0 M R(s f(x e R(sx dx M ( est + s2 R 2 s ( g T (se st + s2 ds R 2 s ( e R(sT T 0 = 2eR(sT R(s R 2. e R(sx dx M R(s e R(sT ( l(c max g s C T (se st + s2 R 2 s 2πR ( ( M 2 R(s e R(sT R(sT R(s 2e R 2 = 2πM R < 2π ɛ 3. Finally, we estimate the third integral in three arts. Let C = {s C : R(s = δ}, These three contours clearly artition C, so g(se st C C 2 = {s C : R(s > δ, I(s > 0}, C 3 = {s C : R(s > δ, I(s < 0}. ( + s2 R 2 ds s 3 g(se st C j j= For the integral over C, we have l(c 2R, g(s M, (by our choice of T 0, and So g(se st C e st = e δt e δt 0 < πδɛ 8RM ( + s2 R 2 s ( + δ = 2 δ. ( + s2 R 2 ds s ( + s2 R 2 ds s. ( l(c max s C g(sest + s2 R 2 s (2RM e δt 2 δ < 4RM ( πδɛ δ 8M = 2π ɛ 9. 4

15 For the integral over C 2, we have l(c 2 = R sin ( δ R < πɛr 9M (by our choice of δ, g(s M, e st = e R(sT, and ( + s2 R 2 s ( + R = 2 R. So g(se st C 2 ( + s2 R 2 ds s ( l(c 2 max s C 2 g(sest + s2 R 2 s ( ( ( δ 2 R sin M R R < ( πɛr 9M = 2π ɛ 9. 2M A similar no, an identical calculation yields the same bound for the integral over C 3. Therefore, the third integral in Equation is bounded by R 2π ɛ 9 + 2π ɛ 9 + 2π ɛ 9 = 2π ɛ 3, which means the the integral over C is bounded by So 2π ɛ 3 + 2π ɛ 3 + 2π ɛ 3 = 2πɛ. 2π g T (0 g(0 = which, after dividing both sides by 2π, gives us our result. C h(sds < 2πɛ, Newman s roof of the Prime Number Theorem The roof follows Zagier s sequence of stes, u to some reordering of the lemmas (including the analytic theorem. Zagier was kind enough to leave lenty of details for the reader to work out. For examle, his roofs of Lemmas 0, 2, and 3 are two sentences long, and his roof of Lemma is a mere three. I have greatly exanded on his roofs with the intention of making it understandable to the undergraduate who has had some analysis. My hoe is that one could read this and understand it in one (ossibly long sitting, as oosed to the multile (infinitely long sittings which I have required. We begin with a definition: Definition 6. For all x R and s C, define Φ(s = log s and ϑ(x = x log, where the sums over run over the rime natural numbers. 5

16 With these definitions in hand, we rove the Prime Number Theorem (following [Z]. The roof follows easily from a sequence of stes, which we rove here as lemmas. The first two lemmas build u to the third: Lemma 7. For R(s >, ζ(s and Φ(s are absolutely convergent. Proof. Corollary 3 roves that ζ(s is absolutely convergent, so we must show that Φ(s is too. Suose σ = R(s >. Then consider that log s n=2 log n n σ log 2 2 σ + 2 log n n σ dn. A comutation shows that the value of the integral is (σ 2 when σ >, from which we conclude that Φ(s absolutely converges for all σ >. Lemma 8. For all δ >, ζ(s and Φ(s are uniformly convergent on σ = R(s δ. (When this is the case, we say these series are locally uniformly convergent. Proof. We use the Weierstrass M-test. Fix δ >. For all n N, set M n = n δ and M n = n δ log n. Then because δ >, both n= M n and n= M n converge. Also, n s = n σ M n and n s log n = n σ log n M n for all σ >. So by the Weierstrass M-test, ζ(s and Φ(s converge uniformly for σ δ. From these two lemmas we can rove the following lemma: Lemma 9. For R(s >, ζ(s and Φ(s are holomorhic. Proof. Let σ = R(s >. By Lemma 8, ζ(s and Φ(s converge uniformly at s. Then if we can show that the termwise derivatives of these series also converge uniformly at s, then we could conclude that ζ (s and Φ (s exist and are equal to their resective termwise derivatives (see Aendix A..2. To do this, we comute the series of termwise derivatives, obtaining d ds n s = log n n s and n= n= d log dx s = log2 s These are clearly uniformly convergent by the Weierstrass M-test; the argument is similar to that which roved Lemma 8. In the next two lemmas, we modify ζ(s and Φ(s slightly so as to enlarge the region over which it is holomorhic. More secifically, we subtract (s from both functions and rove that the modified functions are holomorhic over some larger region. Lemma 0. For σ = R(s > 0, ζ(s /(s is holomorhic. (That is, ζ(s s extends holomorhically to R(s > 0. Proof. For σ >, we have s = dx x s = n+ dx n= n x s. 6

17 Furthermore, this series absolutely converges for σ > since n+ dx x s n+ dx x σ = ( σ n + = n σ <. n= n n= n n= Therefore, since the series for ζ(s also converges absolutely for σ > by Corollary 3, we can subtract the series termwise, obtaining ζ(s s = n= n+ n dx n s n= n+ n dx x s = n= n+ n ( n s x s dx, (2 which is an absolutely convergent, holomorhic function for σ > (because ζ(s and (s are for σ >. Now we show that this series converges absolutely for all σ > 0. We evaluate the following in two different ways. On one hand, On the other hand, n+ s so n x n n+ s du u n s+ dx x n du u s+ dx n+ = ((n + n(x n max n= n+ n n ( n s x s n x n+ n u x ( n s x s dx s n σ+, n= dx. s s (( u s+ n σ+, which converges because σ + >. Hence the series in Equation 2 converges absolutely for σ > 0. Now consider the termwise derivative of the series in Equation 2: ( d ζ(s = ds s n= d ds n+ n ( n s x s dx = n= n+ n ( log n n s + log x x s dx. (Note the use of the Leibniz integral rule again! If this series converges uniformly for σ > 0, we will conclude that ζ(s (s is holomorhic for σ > 0. To rove this, we use the same trick as before: n+ ( log n n s + log x n+ x ( x s dx = s log u u s+ dudx n which imlies n= n+ n n n ((n + n(x n max + σ log(n + n σ+, ( log n n s + log x x s dx n= n x n+ n u x + σ log(n + n σ+. s log u u s+ Since σ + >, the series for d ds (ζ(s (s converges absolutely and therefore exists. 7

18 Lemma. For σ = R(s, ζ(s 0 and Φ(s /(s is holomorhic. Proof. First, for σ >, the roduct reresenting ζ(s converges absolutely by Corollary 3, hence it converges to a nonzero finite number (see Aendix A.4.. Later in the roof, we show that ζ(s 0 for σ = on our way to showing the second half of the result. By Lemma 9, Φ(s is holomorhic for R(s >. So since the same is true for (s, the result clearly follows for R(s >. It is left to show that Φ(s (s is holomorhic for R(s =. As a first ste, note that Φ(s and the series s ( s log converge absolutely for σ >. Hence Φ(s + log s ( s = ( log s + s ( s = log s. (3 Now consider that ζ (s ζ(s = d ζ(s dz s = ( d ζ(s dz s q s q = ( s log ( s 2 ( s = log s. (4 Equating Equations 3 and 4, solving for Φ(s, and subtracting (s, we obtain Φ(s ( ζ s = (s ζ(s + s Considering that log s ( s + log ( σ 2 log 4 2σ, log s ( s. (5 we could rove that the sum over on the right-hand side of Equation 5 is uniformly convergent for σ > 2 by using the Weierstrass M-test with M = log 4 2σ. We would then rove the same about the sequence of derivatives of the artial sums and conclude that it is holomorhic for σ > 2. Also, as we now show, the other term on the right-hand side of Equation 5 is holomorhic on σ > 0, excet where it is undefined. This would rove that the left-hand side of Equation 5, Φ(s (s, is holomorhic (excet where it is undefined for σ > 0. By Lemma 0, the function ζ(s (s ζ(s = s s is holomorhic for σ(s > 0. Hence the numerator of this function is holomorhic for σ > 0, which imlies that (s ζ(s is holomorhic for σ > 0. Since comlex-differentiability imlies that derivatives of all orders exist (see Aendix A..2, we conclude that d ds ((s ζ(s is also holomorhic. Hence ζ (s ζ(s d s = ds ((s ζ(s (s ζ(s 8

19 is holomorhic for σ > 0 excet where it is undefined, which could only be where (s ζ(s = 0. We first check that (s ζ(s 0 at s =. Since ζ(s (s is holomorhic on σ > 0 by Lemma 0, it is holomorhic at s =. In articular, the (finite value at exists and the limit of ζ(s (s is that value. Then we have lim s ( ζ(s s = lim s ( (s ζ(s s. (6 Because s converges to 0 as s, it must be the case that (s ζ(s converges to 0 as well. So (s ζ(s is not 0 at s =. Now we check that (s ζ(s 0 for σ and t = I(s 0. That is, we want to show that ζ( + it 0 for all real t 0. (This will also comlete the roof of the first art of this theorem. Suose on the contrary that ζ(s has a zero of order µ at s = + it. Because ζ(s s is holomorhic at s = + it and since t 0, ζ(s is holomorhic at s = + it. Therefore, by Lemma 0, µ 0. Write ζ(s = (s it µ g(s. Showing that µ = 0 would show that ζ(s has no zeros on the line σ =, so we roceed to this now. We need to get our hands on µ. To do this, note that ζ (s = µ(s it µ g(s+(s it µ g (s. This imlies, for all ɛ > 0, ζ ( ± it + ɛ ζ( ± it + ɛ = µ ( ± it + ɛ it + g ( ± it + ɛ g( ± it + ɛ. Multilying both sides by ɛ and taking ɛ 0 yields ( ± it + ɛ lim ɛ 0 ɛζ + ζ( ± it + ɛ = µ + lim ( ± it + ɛ ɛ 0 ɛg + g( ± it + ɛ. (7 But ζ(s, being once differentiable for σ >, is infinitely differentiable for σ >, so g (s is differentiable and therefore bounded for σ >. Also g( + it 0 because otherwise + it would have order strictly greater than µ. Finally, since (s it µ g(s = ζ(s = n= n s = ζ(s = (s itµ g(s, it follows that g( + it = g( it 0 = 0. Thus the limit involving g(s and g (s in Equation 7 is 0, and the right-hand side is equal to µ. Great! Now to interret the left-hand side, we use Equation 5. Set s = ± it in this equation, multily both sides by a ositive ɛ and take ɛ 0 +. Then the ɛ(s terms vanish, as does the term involving the sum over because it is bounded. This leaves us with ( ± it + ɛ lim ɛφ( ± it + ɛ = lim ɛ 0 + ɛ 0 ɛζ = µ. (8 + ζ( ± it + ɛ Now it will be convenient to consider the oint + 2it as a zero of order ν. As we argued for µ, we have that ν 0. (The actual value of ν has no bearing on our roof; it is merely a convenience to introduce it. Now by analogous reasoning, we could rove an analogue of Equation 8 for ν: lim ɛφ( ± 2it + ɛ = ν. (9 ɛ 0 + Finally, if we substitute + ɛ into Equation 5, multily by ɛ > 0, and take ɛ 0 +, we obtain lim ɛφ( + ɛ =. (0 ɛ 0 + 9

20 To comlete the roof which, recall, required us to show that µ = 0 we come u with an equality in terms of µ and ν, from which we will conclude that µ = 0. To do this, note that 0 log +ɛ (2R(it/2 4 = log +ɛ (it/2 + it/2 4 = log +ɛ+2it + 4 log +ɛ+it + 6 log +ɛ +4 log +ɛ it + log +ɛ 2it = Φ( + 2it + ɛ + 4Φ( + it + ɛ + 6Φ( + ɛ +4Φ( it + ɛ + Φ( 2it + ɛ. ( Now multily both sides of Equation by ɛ > 0, take ɛ 0 +, and use Equations 8, 9, and 0 to arrive at 0 ν 4µ + 6 4µ ν = 6 8µ 2ν. Since ν 0, this inequality imlies 8µ 6, from which we (finally! conclude that µ = 0. The next ste uses some new notation, which we introduce now before continuing. Given functions defined on R, we say that f(x = O(g(x if there exist C R and x 0 R such that f(x C g(x for all x > x 0. For examle, the function f(x = log x+log 2 x+x is O(x because, in the limit as x, the x term dominates, from which we could rove that f(x = O(x using the definition. Now we go back to the roof of the Prime Number Theorem. Lemma 2. ϑ(x = O(x. Proof. First let n N. Then consider that e ϑ(2n ϑ(n = ex n< 2n log = n< 2n where k ranges over the integers (in the secified range and over the rimes. Now let a = n<k 2n k, b = k n k, and c = n< 2n Then ex(ϑ(2n ϑ(n = c, and a b = ( ( 2n n, where 2n n is the binomial coefficient. Now note that b a since ( 2n n is always an integer a cool fact in itself! Also note that c a since c is a roduct over a subset of the set of terms over which a is a roduct. Furthermore, note that if is a rime dividing c, then > n, which imlies k for all terms k in the roduct a. Hence b and c are relatively rime, from which we conclude that bc divides a. In articular, bc a, so e ϑ(2n ϑ(n = c a ( 2n b = n 2n k=0. ( 2n = ( + 2n = 2 2n. k 20

21 Taking logarithms, we have ϑ(2n ϑ(n 2n log 2. Next, we use this last equality to rove that ϑ(x ϑ( x 2 2x log 2 for all x (0,. If x (0, 2], the inequality holds trivially because ϑ(x = ϑ( x 2 = 0 unless x = 2, in which case the inequality reduces to log 2 4 log 2. So let x (2,, and choose m N such that 2m < x 2m +. Then ϑ(x = log log ϑ(2m + log(2m + x 2m+ (where the last inequality is strict if and only if 2m + is rime. Also, ( x ϑ = log log = ϑ(m. 2 x m 2 Hence ( x ϑ(x ϑ 2 ϑ(2m ϑ(m + log(2m + 2m log 2 + log(2m + x log 2 + log(x + 2x log 2, where the last inequality follows since log(x + x log 2 for all x > x 0 = 2. Finally, we use this estimate to rove the lemma! Choose C = 2 log 2 and x 0 = 2, and let x > x 0. Next, choose r N such that 2 r x < 2 r+. Then ϑ(x/2 r = 0, and hence which roves the lemma. Lemma 3. The imroer integral r ( ( x ( x r ( x ϑ(x = ϑ 2 k ϑ 2 k log 2 2 k k= k= ( 2 r = x log 2 2 2x log 2 = Cx, ϑ(x x x 2 dx converges. Proof. Let I be the value of the integral, be it finite or infinite. We are going to aly the analytic theorem to rove that this integral convergence, so, in order to do this, we must check that the hyotheses of Theorem 5. We change variables x = e t in the integral to get ( ϑ(x dx I = x x = ( ( ϑ e t e t dt = f(tdt, where we have defined f(t = ϑ(e t e t. Note that, by Lemma 2, for some constant C, 0 f(t ϑ(e t e t + (Ce t e t + = C + for all t ast some t 0. Since f(t has only jum discontinuities at the rime t, it is clearly bounded on (0, t 0 ] and hence bounded on (0,. Furthermore, for any comact K (0,, f(t has only finitely many discontinuities, so the integral of f(t over K is finite. Hence f is locally integrable. Now define g(z = 0 f(te zt dt for all R(z 0. (As we will show, g(z is holomorhic on R(z 0, so the convergence of the integral is not a roblem. As we will also show, 0 ϑ(e t e (z+t dt = 2 0 Φ(z + z +, (2

22 which imlies or, equivalently, g(z = 0 ϑ(e t e (z+t dt 0 e zt dt = g(z = (Φ(z + z z +. Φ(z + z + If this holds for R(z > 0, then R(z + >, then Lemma would then imly that Φ(z + z is holomorhic for R(z 0. The assumtions of the analytic theorem would therefore be satisfied, from which we could conclude that I, the integral in question which equals 0 f(tdt, exists and is equal to g(0 (a bounded quantity since g(z is holomorhic at z = 0, thus roving the theorem. All we need to do is rove is Equation 2. To do this, we let J be the value of the integral in Equation 2, set s = z +, and make the variable substitution x = e t to get sj = sϑ(x x s+ dx = k=0 k+ k sϑ(x dx, xs+ where, 2,... is an enumeration of the rimes and 0 = for convenience. The reason for slitting u the integral in this way is because, for x ( k, k+, ϑ(x = x log = k log = ϑ( k. z This allows us to ull the ϑ(x out of the integral in order to evaluate it: k+ s ( sj = ϑ( k k x s+ dx = ϑ( k s k s. k+ k=0 This almost looks like a telescoing series how we would love to see one of those right now! In fact, if we note that ϑ( k = ϑ( k+ log k+ and recall that the series ϑ( s and Φ(s converge absolutely, we arrive at exactly this: sj = k=0 ( ϑ( k s ϑ( k+ k s + log k+ k+ s k+ k=0 = k=0 ( ϑ( k s ϑ( k+ k s + Φ(s. k+ Since the value of the telescoing series is ϑ(/ = 0 for all R(s >, we have J = Φ(z + /(z + for R(z > 0, from which Equation 2 follows immediately. Lemma 4. ϑ(x x. Proof. We first show the following: For all ɛ > 0, there exists x 0 > 0 such that x > x 0 imlies ϑ(x/x < ɛ. A arallel roof (which we omit will show the corresonding inequality ϑ(x/x < ɛ, which will rove (straight from the definition of convergence that ϑ(x/x as x. We roceed by contradiction. Suose there exists an ɛ > 0 such that for all x 0 > 0, there exists x > x 0 for which ϑ(x/x ɛ or, equivalently, ϑ(x λx where λ = + ɛ >. Fix such an ɛ and let x 0 =. Choose x > x 0 such that ϑ(x λx. Then, since ϑ(x is an increasing function, λx x ϑ(t t λx t 2 dt x ϑ(x t λx t 2 dt x λx t t 2 dt. This integral s value is C(λ = λ log λ, which is a strictly ositive quantity since λ >. 22

23 Now we reeat this rocess. Set x = x, x 0 = λx, and find an x > x 0 such that ϑ(x λx. Set x 2 = x and continue inductively to define a sequence x, x 2,... where each x n+ > λx n and λxn+ x n+ ϑ(t t t 2 dt C(λ for each n. Then ϑ(t t t 2 dt n= λxn x n ϑ(t t t 2 dt C(λ, which clearly diverges since C(λ > 0. This contradicts Lemma 3, which concludes the roof. With all of the hard work done, we can rove the Prime Number Theorem: Theorem 5 (The Prime Number Theorem. Let π(x denote the number of rimes less than or equal to x. Then π(x x/ log x. Proof. Let ɛ > 0. Without loss of generality, assume that ɛ <. To hel with the estimates, we Choose η (0, ɛ such that η < n= ɛ +η 4+ɛ. (This inequality imlies η < + ɛ 2. Choose x > 0 such that x > x imlies ( ηx < ϑ(x < ( + ηx. (This is ossible by Lemma 4. Choose x 2 > 0 such that x η log x < + ɛ 3 for all x > x 2. (This is ossible because x η log x 0 as x, as can be seen by alying l Hôital s rule. Set x 0 = max{x, x 2 }, and let x > x 0. Then from ( ηx < ϑ(x = x log x log x = π(x log x, we deduce π(x log x/x > η > ɛ. Now we obtain the other estimate. log > ( η log x. Hence ( + ηx > ϑ(x x η < x Dividing both sides by ( ηx yields from which we conclude Since log x is an increasing function, > x η imlies log x η < x log x = ( η log x ( π(x π(x η ( η log x ( π(x x η. π(x log x x + η η > π(x log x x log x x η, < + η η + log x x η < ɛ 2 + ɛ 2 = ɛ. Thus π(x log x/x is within ɛ of for all x > x 0, as required! Sit. Meditate. Enjoy the beauty which lies before us. 0 0 What comes next is sure to shock those who do not know and leasantly remind those who do. Either way, the task before us is a great one, so if a break is desired, now is the time. 23

24 3 Primes in arithmetic rogressions We move on now to a toic which motivated the birth of analytic number theory: rimes in arithmetic rogressions. So far, we have roven that there exist infinitely many rimes, and we have given a formula for the density of these rimes in the set of natural numbers. Ignoring the rime 2, we could have equivalently considered these questions about the odd numbers that is, the arithmetic rogression { + 2n} n=0 and come to the same conclusions. In this section, we generalize to arbitrary arithmetic rogressions {a + qn} n=0 where a, q N, and we ask the same two questions: Are there infinitely many rimes, and what is their density in N? 3. Dirichlet s Theorem Dirichlet s theorem, or, more roerly, Dirichlet s theorem on rimes in arithmetic rogressions, states that there exist infinitely many rimes in any arithmetic rogression a, a + q, a + 2q, a + 3q,... so long as gcd(a, q =. (If a and q have a common factor, then a + qn for all n N is comosite. We will rove this result in a sequence of stes. Our roof will follow Davenort s, which rimarily follows Dirichlet s original. We give Euler s roof for q = 2. We already have seen the result for q = 2, both from Euclid s argument and by considering the Euler roduct formula for ζ(s, but we give a third roof which, for one thing, roves a stronger fact, and, for another thing, ultimately insired Dirichlet who generalized the argument to comose a roof of his theorem. (Note: This is the way Dirichlet did it, and it is how Davenort resents it. 3.. Euler s roof for q = 2 In this section, we rovide a third roof of the fact that there exist infinitely many rimes. Recall that ζ(s = n n s = ( s for all s (,. Taking logarithms and exanding the logarithms into ower series (which will be allowed since s <, imlies that s is within the radius of convergence of, we obtain log ζ(s = log ( s = log( s = ( m m ( s m = m= m sm m= = s + δ(s (3 where δ(s = = m sm < sm = m=2 m=2 s ( s < 2 < ζ(2. ( 2s s Technically, we are about to turn the logarithm of an infinite roduct into an infinite sum, which may cause roblems that do not arise with finite sums and roducts. We excuse ourselves in Aendix A.4. by roving that this works as it does in the finite case. 24

25 (Note that we have sneakily rearranged the order of summation, but this is okay since the series converges absolutely, as can be seen since m sm < sm and sm <. Since log ζ(s as s +, and since δ(s is bounded in the same limit, we conclude from Equation 3 that diverges. This fact clearly imlies the infinitude of the rimes, and it is interesting in its own right the rimes are not too sarsely scattered in N. In order to rove Dirichlet s theorem, we will show the corresonding (generalized fact: qa diverges, where the notation q a means that is congruent to a modulo q, and the sum over q a is over all rimes congruent to a modulo q. This is quite a task, however, so we build our bridge to the theorem in stes. First, we let q > 2 be rime, define Dirichlet characters and Dirichlet L-functions for q, and rove the theorem for q. We then let q be comosite, define general Dirichlet characters and Dirichlet L-functions, and rove the theorem in general. Each ste for the second case is a generalization of the corresonding ste in the first case. Let s begin! 3..2 Dirichlet characters and L-functions (rime modulus In this section, we construct Dirichlet characters of a rime modulus and rove a key roerty about them. Dirichlet characters for the modulus q are arithmetic functions χ : N C, which are q- eriodic (that is, eriodic with eriod q and comletely multilicative (that is, χ(mn = χ(mχ(n for all m, n N. We also require that there exists a linear combination whose value is if n q a and 0 otherwise. (Note: Dirichlet characters satisfy the hyotheses of Theorem. To construct the Dirichlet characters for the rime modulus q, consider the multilicative grou Z q = Z q \{0}. As is well known, this grou is cyclic (see Aendix A.3., so there exists an element g Z q such that g generates Z q. For all n Z q, define ν(n to be the index of n relative to g; that is, define ν(n such that g ν(n q n. Then, for each of the q rimitive (q -th roots of unity ω, define the Dirichlet character on Z q corresonding to ω by ω ν(n. Finally, eriodically extend the definition of ω ν(n to define the character for all n N which are not divisible by q, and for those n for which q n, define ω ν(n = 0. The character ω ν(n is clearly well-defined, for n = m imlies n q m, which imlies g ν(n n m g ν(m modulo q, which imlies ν(n = ν(m + k(q for some integer k. Therefore, since ω q =, ω ν(n = ω ν(m. Also, the function is eriodic by construction. To show comlete multilicativity, note that k = mn, for m 0 n, imlies k mn modulo q, which imlies g ν(k k mn g ν(m g ν(n = g ν(m+ν(n. Since g q modulo q, ν(k = ν(m + ν(n + l(q for some l Z, which imlies ω ν(k = ω ν(m ω ν(n ( ω q l = ω ν(m ω ν(n. Finally, if m or n is 0, then so is k, which imlies that ω ν(k = 0 = ω ν(m ω ν(n. 25

26 Hence ω ν(n is comletely multilicative. Finally, we must show that some linear combination of the ω ν(n gives us or 0, according to whether n q a or n q a, resectively. But first, how many ω ν(n are there? When we defined the functions, we secified that ω q =, so each ω is one of q rimitive (q -th roots of unity. On the other hand, if ω ν(n = ω ν(n 2, then, in articular, they are equal for n = g, for which ν(n =, which imlies ω = ω 2. So there are q functions, each of them distinct. Before we construct such a linear combination, we give an examle of a set of characters. Examle 6. Let q = 5. We choose 2 as our generator. (This choice is arbitrary, but this will not resent a roblem to us later on. We tyically choose some generator at the beginning and it is assumed that we kee the same generator. Then ν( = 0, ν(2 =, ν(3 = 3, and ν(4 = 2. The (q -th roots of unity are, i,, and i, and the corresonding characters are, i ν(n, ( ν(n, and ( i ν(n. A table of these values is given here: i ν(n i i ( ν(n ( i ν(n i i Moving on, we rove the key roerty about Dirichlet characters. Theorem 7. For a rime q > 2, q ω ν(a ω ν(n = ω {, n q a; 0, otherwise (4 where the sum is over all (q -th roots of unity. Proof. First, note that if q k, for some k Z, then ω k = for all (q -th roots of unity ω, and hence ω ωk = q. On the other hand, if q k, then ω k, and owers of ω k generate all of the -th roots of unity, where g = gcd(k, q. From this, we have that ( q g {ω kn } l n=0 = {ωkn } 2l n=l = = {ω kn } gl n=(g l. So ω k = ω g l ω kn = j= n=0 g j= ( ω kl ω k = 0 since ω kl = ( ω k l =. Therefore, if q does not divide k, then ω ωk = 0. We have roven ω k = ω { q, q divides k; 0, q does not divide k. If we divide both sides by q, we set k = ν(a + ν(n, and note that q k if and only if k q 0, which holds if and only if ν(a q ν(n, which in turn holds if and only if n q a, we see that our result follows from the above equality. 26

27 Hence the q arithmetic functions ω ν(n form a set of Dirichlet characters for the rime modulus q. With these characters in hand, Dirichlet defined what are now called Dirichlet L- functions. Following Dirichlet, we now roceed with the definition of these functions. Definition 8. For each Dirichlet character ω ν(n, define the Dirichlet L-function by for all s C. L ω (s = ω ν(n n s n= By Theorem, L ω (s absolutely converges for R(s > and has the following roduct reresentation L ω (s = ω ν( s, which also converges absolutely for R(s >. Now recall that ω ν(n = 0 for all n which are divisible by q, so, in fact, L ω (s = n q0 ω ν(n n s. (5 Also, since ω ν(q = 0, the = q term in the roduct reresentation is, so we have Now note that since L ω (s = q ω ν(. (6 s ω ν( s = R(s < < /2, no term in the roduct is 0. So since L ω (s converges absolutely by Theorem, we have that L ω (s 0 for all s C with R(s >. (This is a general fact about infinite roducts. See Aendix A.4. for an exlanation of why this is true. Similar to our calculation of log ζ(s in Section 3.., since L ω (s 0, we have log L ω (s = q ( m ω ν( s m = m= q m= m ω ν(m ms by the multilicativity of ω ν(n. Now we multily both sides by ω ν(a, sum over all (q -th roots of unity ω, rearrange infinite sums at will (because they are all absolutely convergent!, and obtain ω ν(a log L ω (s = ω ω Because of Equation 4, this imlies that ω ν(a q m ω ν(m ms m= = m ms q m= ω ω ν(a log L ω (s = (q ω q 27 m= m qa ω ν(a ω ν(m. m ms. (7

28 On the right-hand side of Equation 7, we have (q s + (q qa q m=2 m qa m ms. Since the double sum over and m has all ositive terms, it is bounded below by 0 and above by δ(s < ζ(2, as we saw in Section 3... Therefore, we have roven the following lemma to Dirichlet s theorem: Lemma 9. If ω ν(a log L ω (s diverges to infinity as s +, then the series s diverges to infinity. ω The conclusion of this theorem, if true, would rove Dirichlet s theorem. Since we want to rove the conclusion, let us consider this lemma s hyothesis. One of the (q roots of unity is itself, so the sum on the left-hand side of Equation 7 is ω ν(a log L ω (s = log L (s + ω ν(a log L ω (s. (8 ω ω If we consider the ω = term, we see from Equation 6 and Corollary 3 that L (s = q s = ( q s ζ(s, (9 so since ζ(s as s + yet s remains bounded, L ω (s as s +. This is a good start! If we can show that each of the remaining terms is bounded as s +, we will aly Lemma 9 to conclude Dirichlet s theorem. Before embarking on this venture, however, we can simlify our task a bit. First, note that the ω ν(n have magnitude, so they remain bounded as s +. This means that what we have to show is that log L ω (s is bounded as s +. Since the logarithm function is continuous (for R(s > 0 we have chosen the rincile branch of the logarithm without mentioning it, we have that lim log L ω(s = log lim L ω(s. s + s + Hence our task reduces to roving that L ω (s remains finite but nonzero as s +. But we can simlify our task once more. We can show that L ω (s is continuous and therefore finite at s =, so lim s + L ω (s = L ω (. Once we do so, we will have the following recursor to Dirichlet s theorem: Lemma 20. If L ω (s 0 for all ω, then Dirichlet s theorem holds. Proof. We first rove that L ω (s is continuous using the Dirichlet test for convergence and the series reresentation of L ω (s given in Equation 5. To do this, first note that n s decreases as n increases and converges to 0. Second, note that for all N N, N n= ων(n is bounded above by q 2. The reason for this is that as n runs through q consecutive integers, ν(n runs through each ossible index 0,,..., q 2, so q ω ν(n = ω ν = 0 n= 28 q 2 ν=0