THE DENSITY OF PRIMES OF THE FORM a + km

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1 THE DENSITY OF PRIMES OF THE FORM a + km HYUNG KYU JUN Abstract. The Dirichlet s theorem on arithmetic progressions states that the number of prime numbers less than of the form a + km is approimately when goes to infinity. In this paper, we will present a proof of the log theorem and discuss the notion of natural density and analytic density. Contents. Preliminary 2. The L-functions L(s, χ) 4 3. The Proof of the Dirichlet s theorem on Arithmetic Progressions 8 4. Natural Density and Analytic Density 3 5. The Proof of the Wiener-Ikehara Theorem 2 Acknowledgments 24 References 24 The Dirichlet s theorem on arithmetic progressions describes the density of the prime numbers. Specifically, denote π a () to be the number of primes p lesser than of the form a + km, then π a () log. In this paper, we will present a proof of Dirichlet s theorem on arithmetic progressions by analyzing the singularities of certain L-functions. Some preliminary backgrounds will be presented in the first section. In the second section some basic properties of the L-function L(s, χ) will be established. Section 3 contains the proof of Dirichlet s theorem using Wiener- Ikehara theorem, whose proof is postponed to section 5. In section 4 we will compare the natural density and the analytic density of a set.. Preliminary Let us recall some definitions that will be used throughout the paper in this section. Set a, m to be natural numbers such that m 2 and gcd(a, m) =. Denote ζ(s) to be the Riemann Zeta function, i.e. (.) ζ(s) = for Re(s) >. Facts.2. n= n s Date: AUGUST 3, 23.

2 2 HYUNG KYU JUN When Re(s) >, we have ζ(s) = p p s where the product is taken over all prime numbers. ζ(s) is holomorphic and is nonzero when Re(s) >. We have ζ(s) = s + φ(s) where φ(s) is holomorphic when Re(s) >. ζ(s) can be etended to a meromorphic function on C with a unique simple pole at s =, and the residue of ζ(s) at s = is. We omit the proofs of these facts. See [] or [7] for details. Definition.3. Let (.4) π a () = Definition.5. Let where Definition.6. Let Λ a (n) = ψ a () = p n p a(modm) p p a(modm) log p n = k Λ a (k) { log n, if n = p k, and p a (mod m), otherwise θ a () = p p a(modm) log p Let χ be a group homomorphism from the multiplicative group ((Z/mZ), ) to (C, ) such that χ(ab) = χ(a)χ(b). We etend the function to all of Z by putting χ(c) = if gcd(c, m) >. we call χ the Dirichlet character mod m. Lemma.7. (Orthogonality) Let G be the multiplicative group (Z/mZ) Then {, if χ = χ(g) =, otherwise g G Proof. Let χ. Choose h G such that χ(h). Then, χ(h) χ(g) = χ(hg) = χ(g) g G g G g G Hence, As χ(h), (χ(h) ) χ(g) = g G χ(g) = g G

3 THE DENSITY OF PRIMES OF THE FORM a + km 3 Corollary.8. Fi g G. Then χ(g) = χ Ĝ {, if g =, otherwise Proof. We apply lemma.5 to the dual group Ĝ. Net we will present Abel s summation formula which will be used etensively throughout the paper. Theorem.9. (Abel s summation formula) Let λ λ 2... be a sequence of real numbers such that λ n when n, and let A() = λ n a n where (a n ) is a sequence of comple numbers. If ρ : R C is a function(not necessarily continuous), we have (.) k n= k a n ρ(λ n ) = A(λ k )ρ(λ k ) A(λ n )(ρ(λ n+ ) ρ(λ n )) Furthermore, if ρ has a continuous derivative in (, ), and λ, then the above equation can be written as (.) a n ρ(λ n ) = A()ρ() A(t)ρ (t)dt λ λ n n= Proof. For convenience, let A(λ ) =. Then, we have k a n ρ(λ n ) = n= k (A(λ n ) A(λ n )ρ(λ n ) n= k = A(λ k )ρ(λ k ) A(λ n )(ρ(λ n+ ) ρ(λ n )) n= which proves (.). Now, assume that ρ has a continuous derivative ρ in (, ). Let k = ma{k λ k }. Then, k k λn+ a n ρ(λ n ) = A(λ n ) ρ (t)dt λ n n= n= And A(λ k )ρ(λ k ) = A()ρ() A(t)ρ (t)dt λ n as A(t) is a step function that remains constant in the interval [λ k, λ k+ ). Hence, (.) follows from (.). Corollary.2. With the same notations as in theorem.9, if A()ρ() when, we have a n ρ(λ n ) = A(t)ρ (t)dt λ n= Proof. Send in the equation (.).

4 4 HYUNG KYU JUN Corollary.3. (Abel s lemma) Let (a n ) (b n ) be two sequences. Define. Then, l n A k,l = a i and S k,n = a i b i i=k i=k n (.4) S k,n = A k,i (b i b i+ ) + A k,n b n i=k Proof. Put a i = A n,i A n,i and the result is immediate. 2. The L-functions L(s, χ) In this section, we will present some facts about L-functions associated to Dirichlet characters. Definition 2.. Let χ be a Dirichlet character mod m and s C. Define χ(n) L(s, χ) = n s for Re(s) >. n= Immediately we have the following lemma: Lemma 2.2. L(s, ) = ζ(s) p m( p s ) We want to understand the behavior of L(s, χ) when s tends to. It turns out that if χ =, then L(s, χ) diverges when s and has a simple pole there. On the other hand, when χ, we have that L(, χ). Corollary 2.3. L(s, ) etends to a meromorphic function for Re(s) > It has a unique simple pole at s = with res(l(s, ), ) =. Proof. This follows from the fact that ζ(s) etends to a meromorphic function for Re(s) > with a unique simple pole at s =. We also know that res(ζ, ) =. Applying Lemma 2.2, we get res(l(s, ), ) = ( p ) res(ζ, ) = p m Let G(s) denote the ordinary Dirichlet series, which means that a n G(s) = n s. Note that L-functions are ordinary Dirichlet series with positive coefficients. We could use Landau s theorem as a very powerful tool to determine the radius of convergence for an ordinary Dirichlet series with positive coefficients. n=

5 THE DENSITY OF PRIMES OF THE FORM a + km 5 Theorem 2.4. (Landau) Let G(s) be an ordinary Dirichlet series with positive coefficients. Then, the domain of convergence of G(s) is bounded by a singularity of G(s) located on the real ais. In other words, if σ > is the abscissa of convergence of G(s), then G(s) has a singularity at σ. Proof. Let G(s) = where a n for all n N. Suppose that G(s) converges for Re(s) > ρ with ρ R. Assume that G(s) can be etended analytically to a function holomorphic in a neighborhood of the point z = ρ in the comple plane. We will show that there eists ɛ > such that G(s) converges for Re(s) > ρ ɛ. Replace s by s ρ in the equation. Now assume ρ =. We see that G(s) is holomorphic in Re(s) > plus a neighborhood of s =, so it is holomorphic in a disk s + ɛ for some ɛ >. Moreover, the Taylor series of G(s) converges in the disk. From comple analysis, we know that the k th derivative of G(s) can be epressed as G (k) (s) = n N n= a n n s ( log n) k a n n s when Re(s) >. Hence, G (k) () = ( ) k (log n) k a n n n N The Taylor series epansion of G is : G(s) = For s = ɛ, we have k= k! (s )k G (k) (), where s + ɛ. We know that G( ɛ) = k= ( ) k G (k) () = n k! ( + ɛ)k ( )G (k) (). (log n) k a n n is a convergent series with positive terms. Thus, G( ɛ) = k! ( + ɛ)k (log n) k a n n converges. We get k, n N

6 6 HYUNG KYU JUN G( ɛ) = n = n a n n k= a n n n+ɛ = n k! ( + ɛ)k (log n) k a n n ɛ Hence, the series converges for s = ɛ, and it also converges for Re(s) > ɛ. Now, let us consider L(s, χ) when χ. Lemma 2.5. Let f(s) = where a n. If A k,l = l a n are bounded, then f converges for Re(s) >. k n= Proof. Assume that A k,l < M R +. From (.4), we see that a n n s and thus, Hence f converges. m a ( m i M i s i s ) + (i + ) s m s i=l l m i=l a i i s M l s. Proposition 2.6. If χ, L(s, χ) converges when Re(s) > and converges absolutely when Re(s) >. If Re(s) >, we have L(s, χ) = p χ(p) p s Proof. For Re(s) >, the absolute convergence follows from the fact that converges absolutely when k >. In order to show the convergence of the series for Re(s) >, we use the above lemma 2.5 Using the lemma, it is enough to show A l,n = n χ(i) are bounded for all l, n. From Lemma.7 (orthogonality), we know that l+m χ(i) =. Hence, it is enough to consider l n such that n l < m. As χ(i) = if gcd(i, m) = and χ(i) = otherwise, we see that n n A l,n = χ(i) χ(i) l l l i=l n= n k

7 THE DENSITY OF PRIMES OF THE FORM a + km 7 The partial sums are bounded, so L(s, χ) converges. Net we will prove that L(s, χ) when χ. In order to show this, we need to define a function ζ m and inspect how the function behaves near s =. Definition 2.7. Let (2.8) ζ m (s) = χ L(s, χ) Let p be a prime not dividing m. I will simply write p as the image of p in the group G = (Z/mZ). Denote f(p) to be the order of p in G. Let g(p) = f(p). We can rewrite ζ m with respect to prime numbers that do not divide m. Proposition 2.9. ζ m (s) = ( ) g(p) p m p f(p)s In order to prove the proposition, we need a lemma. Lemma 2.. If p is a prime not dividing m, we have ( χ(p)t ) = ( T f(p) ) g(p). χ Proof. Let W be the set of all f(p) th roots of unity. We have ( wt ) = T f(p). w W For all w W, we have g(p) characters χ in Ĝ such that χ(p) = w, as g(p) is th order of the quotient of G by the cyclic subgroup < p >. Hence, the lemma follows. Proof. (of Proposition 2.9) ζ m (s) = χ L(s, χ). Replace L(s, χ) using Proposition 2.6, we have ζ m (s) = ( χ(p) χ Ĝ p ) = s p m ( χ(p) p ). s By Lemma 2., we get ( χ(p) ( p s ) = Thus the proposition follows. Proposition 2.. L(, χ) if χ. χ p χ p f(p)s )g(p).

8 8 HYUNG KYU JUN Proof. We will prove this result by contradiction. We know that L(s, χ ) is holomorphic in a neighborhood of s = (Proposition 2.6), and L(s, ) etends to a meromorphic function for Re(s) > with a unique simple pole at s = (Corollary 2.3). Now, suppose that the proposition is not true. There eists χ such that L(, χ) =, and we conclude that ζ m (s) = L(s, χ) is holomorphic at s =. χ As ζ m is an ordinary Dirichlet series with positive coefficients, Landau s theorem (2.4) shows that the abscissa of convergence σ. We will show that this is impossible. The p th factor of ζ m is ( p f(p)s ) g(p) = ( + p f(p)s + p 2f(p)s +...) g(p) and it dominates the series + p s + p 2s +... Hence, all the coefficients of ζ m are greater than those of the series F (s) = n s (n,m)= A problem occurs because the series F diverges at s =. We see that ζ m (s) F (s) for all s σ. Hence, ζ m diverges at s =. We have shown that ζ m cannot be holomorphic for the entire half plane Re(s) >. Therefore L(, χ) if χ. Now we are ready to prove the theorem of arithmetic progressions. 3. The Proof of the Dirichlet s theorem on Arithmetic Progressions Recall that a, m are natural numbers with m 2 and gcd(a, m) =. The aim of this section is to prove the following theorem. Theorem 3.. There eist infinitely many prime numbers p such that p a (mod p). Moreover π a () log as. Showing an asymptotic behavior of π a is quite strenuous, so we instead try to observe the behavior of ψ a when. Lemma 3.2. When, the following are equivalent: π a () log ψ a () θ a () Proof. By Abel s summation formula (Corollary.8), we have and π a () = log p log p = θ a() log + θ a (t) 2 t log 2 t dt

9 THE DENSITY OF PRIMES OF THE FORM a + km 9 Also, we see that 2 θ a (t) t log 2 dt = O( t log 2 ). Hence π a () log ψ a() ψ a () = θ a () + θ a ( /2 ) + θ a ( /3 ) +... θ a() when. Now we are reduced to showing that (3.3) ψ a (). Let (3.4) F a (s) =. For Re(s) > we define log L(s, χ) as log L(s, χ) = p n= Λ a (n) n s log χ(p)p s = n,p χ(p) n np ns. The series n,p χ(p) n np ns is obviously convergent. Then we can set ( ) log L(s, χ) = L (s, χ) L(s, χ) = d ds n= Lemma 3.5. For s C and Re(s) >, (3.6) F a (s) = χ(a) L (s, χ) L(s, χ). χ χ(n)λ(n) n s. Proof. In essence, this is Fourier analysis on the finite abelian group G = (Z/mZ). We see that χ χ(a) L (s, χ) L(s, χ) = χ ( χ(a) n= χ(n)λ(n) ) n s = ( n= Using Corollary.8, we see that the above equation equals to χ χ(a) L (s, χ) L(s, χ) = n= Λ a (n) n s. χ ) Λ(n) χ(na ) n s We will relate F a and ψ a by a specific formula, and prove (3.3) from this formula using the Wiener-Ikehara theorem (Theorem 3.3). Let us first recall a simple lemma from comple analysis: Lemma 3.7. Let f be a meromorphic function. If f has at most a pole at z, then res(f /f, z ) = ord(f, z ).

10 HYUNG KYU JUN Proof. Translate if necessary, we may take z =. Assume that it has at most a pole at. Then f has only a finite number of negative terms. Hence, let us write Then, f(z) = a m z m + (higher terms) a m, m Z. (3.8) f(z) = a m z m ( + h(z)) where h(z) is a power series with no constant term. Differentiating both sides, we get (3.9) f (z) = ma m z m + a m z m h (z) Divide (3.9) by (3.8), we see that As h(z) has no constant term, Hence, res(f /f, ) = ord(f, ). f f = m z + h (z) + h(z). h (z) +h(z) is holomorphic at. Lemma 3.. The function F a (s) can be etended to a meromorphic function on an open set O containing {s C Re(s), s }. Moreover, F a has a simple pole at s = with residue. Proof. From Proposition 2., if χ, we see that L(s, χ) is holomorphic and is nonzero on an open set O containing {s C Re(s), s }. Hence L (s, χ)/l(s, χ) is holomorphic. On the other hand, L(s, ) has a unique simple pole at s =. From Lemma 3.7, we see that L (s, )/L(s, ) has a simple pole at s = with residue. Hence, using the relation (3.6) from Lemma 3.5, we see that F a has a simple pole with res(f a, ) = Using Corollary.2 (Abel s summation formula), we see that If we put t = e, we get (3.) F a (s) = n= F a (s) s = Λ a (n) n s = s ψ a ()e s d. Lemma 3. tells that ( Fa (s) ) (3.2) res, = s. ψ a (t) dt. ts+

11 THE DENSITY OF PRIMES OF THE FORM a + km Theorem 3.3. (Wiener-Ikehara) Let A() be a non-negative, non-decreasing function in an interval [, ). Assume that for σ >, the integral A()e s d, s = σ + it converges to the function F(s), where F is holomorphic for σ ecept for a simple pole at s = with residue γ. Then, (3.4) lim e A() = γ The proof of the theorem is long and requires techniques from functional analysis, so we will present it later in Section 5.. Proof. (of Theorem 3.) We will show that ψ a () Set A() = ψ a (e ). We know that ψ a is non-decreasing and non-negative. From (3.), we see that F (s) = F a(s) = A()e s d s and (3.2) tells us that the residue of F at s = is. Hence, by Theorem 3.3, we conclude that e ψ a (e ) as. In other words, we have ψ a () as. log From Lemma 3.2, we conclude that π a () as. This completes the proof of the Dirichlet s theorem on Arithmetic Progressions. Corollary 3.5. (Dirichlet) There are infinitey many primes of the form a + km where gcd(a, m) =. Proof. The number of the primes of the form a + km is given as π a () and it is not bounded above. log, Corollary 3.6. Primes are evenly distributed among the congruence classes modulo m. Proof. π a () log, and log does not depend on a. We will finish this section by presenting another proof showing ψ a () contour integral instead of Theorem 3.. Proof. Fi R and define Claim: ψ a (). s+ by g(s) = F a (s) s(s + ). From Lemma 3., we see that g(s) is meromorphic on an open set O containing {s C Re(s), s } with a simple pole at s = and

12 2 HYUNG KYU JUN Then, consider res(g, ) = 2 2 res(f a, ) = 2 2. (3.7) 2πi c+ i c i g(s)ds. We want to epress (3.7) in two different ways. First, let us use the fact that F a (s) = Λ a (n)/n s. Hence, (3.7) becomes (3.8) Λ a (n) c+ i s+ ds 2πi c i n s s(s + ). Note that we need to show the termwise integration is valid so that (3.7) is equivalent to (3.8). We can rewrite g(s) as g(s) = ( Λ a (n) ( ) ( ) s n s Λ a (n) ( ) ) s+ n. n s + Von Mangoldt s method from [3] shows that each of the two terms can be integrated termwise. Therefore, the termwise integration is valid. Now, using the fact that we have s+ n s s(s + ) = s ( ) s n n s + ( ) s+, n c+ i 2πi c i s+ ds n s s(s + ) = 2πi c+ i c i ( ) s ds n s n 2πi c++ i c+ i (3.9) { c+ i s+ ds 2πi c i n s s(s + ) = n if n, if n Thus (3.8) becomes (3.2) (3.8) = n Λ a (n)( n) = ψ a (t)dt ( ) v dv n v. The second way of epressing (3.7) is to do contour integral of a rectangle defined by Re(s) = c and Re(s) = with a small protuberance near s =, and two edges in the infinity. We know that g(s) is meromorphic on an open set O with a unique simple pole at s =. Hence, by the residue theorem, Hence, γ g(s)ds = res(g, ) = 2 2.

13 THE DENSITY OF PRIMES OF THE FORM a + km 3 (3.2) 2πi c+ i c i g(s)ds γ g(s)ds = 2 2 as we see that integral over other edges of the protuberated rectangle is o( 2 ). Hence, putting (3.2) and (3.8) together, we get (3.22). ψ a (t)dt 2 2 We are reduced to showing that (3.22) implies ψ a (). The reader is encouraged to look at [3] for a rigorous proof. Although in [3] the author shows that 2 ψ(t)dt 2 implies ψ(), we can almost eactly follow the argument to prove our result. Here, we will provide a simpler argument that (3.22) ψ a () from the fundamental theorem of calculus: ψ a (t)dt ψ a (t)dt = o(2 ). Hence, from the fundamental theorem of calculus we get ψ a () = + o() by differentiating both sides. The above equation is equivalent to ψ a (). 4. Natural Density and Analytic Density In this section, we will discuss two different notions of density. We will present another proof of Dirichlet s theorem on arithmetic progressions (i.e. Corollary 3.5) using analytic density. Let P be the set of all prime numbers. Let A be any subset of P. Denote A n = {a A a n} and P n = {p P p n}. Definition 4.. The natural density N(A) of a subset A P is defined to be if the limit eists. A n N(A) = lim n P n

14 4 HYUNG KYU JUN Definition 4.2. The analytic density, or Dirichlet density of A P is defined to be if the limit eists. D(A) = lim s + a s a A p s p P We can etend the definitions of two kinds of density for two sets A B N. Theorem 4.3. Let A be a subset of B N. Assume that χ B (m) m s when s. m= If N(A) (with respect to B) eists and is equal to k R, then D(A) also eists and equals k. Proof. Let χ A and χ B be characteristic functions of the sets A and B. Set s C with Re(s). Then we have that By Corollary.3, we know where n m= Hence, when n, we have k = lim n n χ A (m) n. χ B (m) m= m= n χ A (m) m s = ( m s (m + ) s )A m + A n n s, m= m= A m = χ A (m) m s = m χ A (i) < m. i= m= ( m s (m + ) s ) A m, as An n s n n s = n s when n. By assumption, for any ɛ >, there is an N N such that if n N, k ɛ < n χ A (m) n < k + ɛ χ B (m) m= m= k ɛ < A n B n < k + ɛ where B n is defined similar to A n. Hence, (4.4) (k ɛ)b n < A n < (k + ɛ)b n for all n N.

15 THE DENSITY OF PRIMES OF THE FORM a + km 5 Choose suitably large l N such that (4.5) (k ɛ)b n l < A n < (k + ɛ)b n + l for all n N. This is possible because there are at most finitely many terms that do not satisfy the inequality (4.4). We see that.. m= m= χ A (m) m s = m= ( m s (m + ) s ) A m. Using the former inequality (4.5), we get χ A (m) ( m s < (k + ɛ) m s ) (m + ) s B m + l We see that Moreover, We get (k + ɛ) m= m= ( m s (m + ) s ) B m = (k + ɛ) l m= ( m s (m + ) s ) = l m= m= ( m s (m + ) s ). χ B (m) m s (k ɛ) m= (4.6) k ɛ χ B (m) m s l < l χb (m) m s < m= χ A (m) m s < (k + ɛ) χa (m) m s χb (m) m s < k + ɛ + From the assumption we have χ B (m) m s when s, so m= m= l χ B (m) m s whens. m= χ B (m) m s l χb (m) m s Since ɛ > is arbitrary, we can send ɛ in (4.6), and send s. Consequently, we get the desired result. Therefore, (Analytic Density) = (Natural Density) = k. Remark 4.7. Note that the converse is not always true. Eample 4.8. Let A N be the set of natural numbers which have first digit. + l

16 6 HYUNG KYU JUN Proof. For simplicity denote N(n) = A n N n, where N n = {k N k n}. We see that A has an analytic density, but it does not have a natural density, as lim sup N(n) lim inf N(n). n n For simplicity denote N(m) = An N. n A simple counting argument shows that Hence, we see that A n = k 9 if n = k. lim inf N(n) n 9 If n = 2 k, we see that (in fact equality holds here). A n = k 9 + k = k+, which implies 9 lim sup N(n) 5 n 9 Therefore, natural density does not eist. Now, let us show that the analytic density eists nevertheless. I claim that a s a A n s log 2 when s. n N From facts.2, we know that ζ(s) = n N n s s claim is equivalent to lim s a s a A s = lim(s ) s a A For s >, we see from the figure below that a s = log 2. when s. Hence, the ( 2 k ) k= s d a s ( 2 k ) k a A k= s d + k since (Blue areas) : shift all the blue regions above [ k, 2 k ] to the rectangle above [, 2]. Obviously the sum of all blue areas above [ k, 2 k ] is less than that of the rectangle above [, 2], which has area. We can calculate ( 2 k ) s d k k= using basic integration and facts about geometric progression.

17 THE DENSITY OF PRIMES OF THE FORM a + km 7 Figure. y = / s and a s a A ( 2 k ) s d = k k= Hence, we have = s k= k= [ s (2 k ) s ( k ) s 2 2 s 2 s ( k ) s = 2 2 s s s 2 s s 2 2 s s s 2 s s a s 2 2 s s s 2 s s + a A multiplying by s > and sending s + gives which shows that log 2 log 2 D(A) log log, D(A) = log 2. Similarly, if we define A k = {n N n has first digit k}, we can show D(A k ) = log ( + k ). Note that ] 9 k= ( ) 9 ( ) log + = log + = log =. k k k=

18 8 HYUNG KYU JUN Surprisingly, the same holds for the set of prime numbers which have first digit : it also has analytic density log 2 but it does not have natural density. For more information, see [] page 76, [5], and [6]. We also note that the conclusion aligns with Benford s Law, or the first-digit law in statistics which claims ( that ) numbers with the leading digit k {,, 9} occur with probability log + k. We should ( note ) that the probability of finding the numbers with first digit k is not log + k (As we just noted in the eample, natural ( density) of A k does not eist). Nonetheless, the analytic density of each A k is log + k. We will show that D(P a ) = of primes congruent to a modulo m is infinite. 4.3 and the fact that N(P a ) = following [], which shows that the number Of course, we can use theorem (one could use the prime number theorem π() = log together with theorem 3.). However, there is a direct way to see that D(P a ) =. The proof is mainly based on the facts that L(s, χ) when χ (proposition 2.) and L(, s) has a simple pole at s = (corollary 2.3). Lemma 4.9. p s log s p P Proof. From the facts.2, we can epress ζ as ζ(s) = p Hence, if we take the logarithm, we get where µ(s) = p P k 2 follows. log ζ(s) = p P k when s. = s + φ(s). p s kp ks = p P p s + µ(s), is bounded. As ζ has a simple pole in s =, the lemma kp ks Proposition 4.. If D(A) eists, then p P D(A) = lim s + Proof. Lemma 4.9 indicates that p s log s a s a A log s when s. Let P a = {p P p a (mod m)}. The goal now is to show that P a has analytic density. In order to prove this, we will first establish three lemmas.

19 THE DENSITY OF PRIMES OF THE FORM a + km 9 Let us define f χ (s) = p m χ(p) p s. Note that f χ obviously converges for Re(s) >. Lemma 4.. If χ =, then f χ log s when s. Proof. Observe that f χ differs from the series p s at finite number of terms (i.e. prime numbers p such that p m. The result immediately follows from Lemma 4.9. Lemma 4.2. If χ, then f χ remains bounded when s. Proof. Recall that we defined log L(s, χ) in the proof of lemma 3.2 as follow: log L(s, χ) = p log χ(p)p s = n,p χ(p) n np ns. Now, we can rewrite the above equation as We see that log L(s, χ) = n,p χ(p) n np ns p = p χ(p) p s χ(p) p s = f χ. + n,p 2 χ(p) n np ns. Proposition 2. shows that log L(s, χ) remains bounded when s. n,p 2 χ(p) n np ns also remains bounded when s. Therefore, f χ remains bounded when s. Define g a (s) = p P a p s. We see that the analytic density of P a is g a (s) (4.3) D(P a ) = lim s log. s Let us eamine the behavior of g a when s. Lemma 4.4. We have g a (s) = χ(a)f χ (s). χ

20 2 HYUNG KYU JUN Proof. The proof is similar to that of Lemma 3.5. If we substitute f χ in the given formula, we get χ(a)f χ (s) = ( ) χ(pa ) /p s. χ p m χ From Corollary.8, we see that χ(pa ) = χ {, if pa (mod m), if not Therefore, pa (mod m) p a (mod m). χ(a)f χ (s) = g a (s). χ Now we are ready to prove that the analytic density of P a = {p P a (mod m)} is. Theorem 4.5. Proof. From lemma 4., f log D(P a ) = s when s Lemma 4.2 tells that all other f χ are bounded. Hence, we conclude that p from lemma 4.4. From the equation (4.3), we see that g a (s) log s D(P a ) =. Note that we cannot directly tell that the natural density of P a is from this proof. The proof only tells that D(P a ) = >. 5. The Proof of the Wiener-Ikehara Theorem This section is entirely devoted to prove the Wiener-Ikehara theorem (theorem 3.3) we used in section 3. We will follow [4] with some additional eplanation and clarification if applicable. Let us first recall the Wiener-Ikehara Theorem.

21 THE DENSITY OF PRIMES OF THE FORM a + km 2 Theorem 5.. (Wiener-Ikehara) Let A() be a non-negative, non-decreasing function in an interval [, ). Assume that for σ >, the integral A()e s d, s = σ + it converges to the function F(s), where F is holomorphic for σ ecept for a simple pole at s = with residue γ. Then, (5.2) lim e A() = γ Eample 5.3. An obvious (but uninteresting) eample would be A() = e. It is non-negative, non-decreasing function of [, ). Also, we see that A()e s d = s. and s is holomorphic for σ ecept for s = where it has a simple pole with a residue. Evidently, lim e e = = res( s, ). Let us normalize A() to make γ =. (Replace A() by A() γ B() = e A(). We will first show that for all λ >, if necessary). Set λy lim B(y v v λ )sin2 v 2 dv = π, and then deduce lim B() = by showing that Lemma 5.4. For all λ >, Proof. For σ >, we have Hence, Set f(s) = lim sup λy lim f(s) s = B() lim inf B(). B(y v λ )sin2 v v 2 dv = π. A()e s d, and s = e (s )s d (B() )e (s ) d, σ >. g(s) = f(s) s, and g ɛ(t) = g( + ɛ + it) for ɛ >. Then g(s) is analytic for Re(s) = σ as f is a meromorphic function with a unique simple pole at s = with residue. For λ >, we have

22 22 HYUNG KYU JUN (5.5) = 2 2λ 2λ 2λ ( g ɛ (t) 2 2λ t 2λ ) e iyt dt ( t )e iyt( ) (B() )e (ɛ+it) d dt. 2λ The order of integration in the above equation can be interchanged by Fubini s theorem. Since A() is nonnegative and nondecreasing, if s R and >, f(s) = A()e s d A() e us du = A()e s. s Hence A() sf(s)e s. As f(s) is holomorphic for σ >, we see that sf(s) is a constant number for σ >. Consequently A() = O(e s ) for any s >, and A() e as, or equivalently A() = o(e s ) (if not, we will have that s f(s) = A()e s d diverges). Therefore, for any δ >, we have B()e δ = A()e (+δ) = o(). As a result, the integral (B() )e (ɛ+it) d converges uniformly in t [ 2λ, 2λ]. Fubini s theorem tells that the order of integration in (5.5) is interchangeable. We now have (5.6) 2 2λ 2λ ( g ɛ (t) t 2λ ) e iyt dt = = (B() )e ɛ ( 2λ 2λ ( 2 (B() )e ɛ sin2 λ(y ) λ(y ) 2 t 2λ d. ) ) e i(y )t dt d Because g is analytic if σ, and g ɛ (t) g( + it) uniformly in an interval t [ 2λ, 2λ] when ɛ. Hence, lim ɛ Therefore, the limit e ɛ sin2 λ(y ) λ(y ) 2 d = lim ɛ B()e ɛ sin2 λ(y ) λ(y ) 2 d sin 2 λ(y ) λ(y ) 2 d. eists. As the integrand is nonnegative and monotonically increasing as ɛ, we can apply the monotone convergence theorem and get lim ɛ B()e ɛ sin2 λ(y ) λ(y ) 2 d = This implies that B() sin2 λ(y ) λ(y ) 2 d.

23 THE DENSITY OF PRIMES OF THE FORM a + km 23 2λ 2λ ( g(t) t 2λ ) e iyt dt = B() sin2 λ(y ) λ(y ) 2 d sin 2 λ(y ) λ(y ) 2 d by (5.6). If we let y, then (LHS) due to the Riemann-Lesbegue lemma. On (RHS), second term gives Hence, lim sin 2 λ(y ) λ(y ) 2 d = lim λy sin 2 v v 2 dv = π. λy (5.7) lim and we finish the proof of the lemma. B(y v v λ )sin2 v 2 dv = π Now, let us show that lim B() = using Lemma 5.4. Lemma 5.8. lim B() = Note that this is the Wiener-Ikehara theorem. Proof. We will first show that (5.9) lim sup B() and then show (5.) lim inf B(). For (5.9), choose a and λ R +. Let y > a λ. Then, from (5.7), we have a lim sup a B(y v v λ )sin2 v 2 dv π since the intergrand is nonnegative. Because A(u) = B(u)e u is nondecreasing, we have ( e y a/λ B y a ) ( e y v/λ B y v ) for v [ a, a]. λ λ Then, Hence, ( B y v ) ( B y a ) ( e (v a)/λ B y a ) e 2a/λ. λ λ λ i.e., a lim sup a B(y a λ )e 2a/λ sin2 v v 2 dv π lim sup B(y a a sin 2 v λ )e 2a/λ a v 2 dv π.

24 24 HYUNG KYU JUN As we fied a and λ, we have lim sup conclude B(y a/λ) = lim sup B(y). We then a e 2a/λ lim sup B(y) a sin 2 v v 2 dv π for all a > and λ >. Let a and λ while a/λ. Then, as the above inequality holds for all a, λ > That is, lim sup B(y) sin 2 v v 2 dv π π lim sup so the inequality (5.9) holds. B(y) π lim sup B(y), We will now show that (5.) also holds, which completes the proof. Inequality (5.9) implies that B(s) c, for suitably large c. Let s fi a, λ > as before. If y is large enough, we have (5.) λy B(y v ( v a λ )sin2 v 2 dv c As before, if v [ a, a] we have which implies that (5.2) a a B ( y v λ + a a sin 2 v v 2 dv + ( B y v λ ) ( a ) B y + e 2a/λ, λ B(y v v λ )sin2 v 2 dv B ( y + a ) a e 2a/λ λ a From (5.7), (5.), and (5.2) we conclude ( a sin 2 v π c v 2 dv + a i.e, ( a sin 2 v π c v 2 dv + a sin 2 ) v v 2 dv sin 2 ) v v 2 dv Again, send a, λ while a λ. Then, π π lim inf a )sin 2 v v 2 dv. sin 2 v v 2 dv. sin 2 ) v v 2 dv + lim inf B( y + a ) a e 2a/λ λ a a + lim inf B(y)e2a/λ a B(y). lim inf B(y). sin 2 v v 2 dv. sin 2 v v 2 dv.

25 THE DENSITY OF PRIMES OF THE FORM a + km 25 Hence, (5.) is verified. We have proved lim B(y) =, which is the Wiener Ikehara theorem. Acknowledgments. It is my pleasure to thank my mentor, Tianqi Fan for her help and helpful advice throughout this project. I also thank for Professor Emerton and Professor Narasimhan for their helpful comments for the project. Without their help, the project must have gone astray. I also thank Professor May for organizing REU this year. References [] J. P. Serre. A course in Arithmetic 5th. NY: Springer, 996. [2] S. Lang. Comple Analysis 4th. NY: Springer, 999. [3] H. M. Edwards. Riemann s Zeta Function NY: Dover Publications, Inc, 2. [4] K. Chandrasekharan Introduction to Analytic Number Theory NY: Springer, 968. [5] D. I. A. Cohen, T. M. Katz Prime Numbers and the First Digit Phenomenon J. Number Theory 8(984), [6] D. I. A. Cohen, An eplanation of the first digit phenomenon, J. Combin. Theory. Ser. A 2. (976), [7] M. Ram Murty Problems in Analytic Number Theory NY: Springer, 2.