SIMPLE PROOF OF THE PRIME NUMBER THEOREM This writeup is drawn from a writeup by Paul Garrett for his complex analysis course, http://www-users.math.umn.edu/~garrett/m/complex/notes_2014-15/ 09_prime_number_theorem.pdf Especially, the bibliography of the source writeup contains relevant papers of Cheby- shev, Erd}os,Garrett, Hadamard, Newman, de la Vall´eePoussin, Riemann, Selberg, and Wiener. The prime-counting function, a function of a real variable, is π(x) = jfp : p ≤ xgj: That is, π(x) equals the number of prime numbers that are at most x. The Prime Number Theorem states that x π(x) ∼ log(x) meaning that limx!1 π(x)=(x= log(x)) = 1. The Chebyshev theta function, also a function of a real variable, is X #(x) = log p: p≤x A quick argument shows that #(x) = O(x), meaning that #(x) ≤ cx for some c and all large x; in fact, the argument produces such a c and the inequality holds for all x. A basic lemma of asymptotics specializes to show that if #(x) ∼ x, meaning that limx!1 #(x)=x = 1, then π(x) ∼ x= log x, giving the Prime Number Theorem. Thus the main work of this writeup is to go from #(x) = O(x) to #(x) ∼ x. With ζ(s) the Euler{Riemann zeta function, the dominant term of ζ0(s)/ζ(s) near s = 1 is a Dirichlet-like series closely related to #(x). This fact combines with the convergence theorem in section 4 below to finish the proof. Contents 1. Weak theta asymptotic 2 2. Lemma on asymptotics, beginning of the proof 2 2.1. Lemma 2 2.2. Beginning of the proof 3 3. Euler{Riemann zeta function 3 3.1. Zeta as a sum 3 3.2. Zeta as a product 4 3.3. Euler's proof 4 3.4. Continuation of zeta and its logarithmic derivative 5 3.5. Non-vanishing of zeta on Re(s) = 1 6 3.6. Improved continuation of the logarithmic derivative 7 3.7. Dominant term of the logarithmic derivative near s = 1 7 1 2 SIMPLE PROOF OF THE PRIME NUMBER THEOREM 4. Convergence theorem, corollary on asymptotics, end of the proof 7 4.1. Theorem 7 4.2. Corollary 10 4.3. End of the proof 11 1. Weak theta asymptotic A quick argument shows that #(x) = O(x) as follows. For any positive integer n, 2n Y 2n X 2n p ≤ < = 22n; n j n<p≤2n j=0 and so 0 1 X Y #(2n) − #(n) = log p = log @ pA < 2n log 2: n<p≤2n n<p≤2n It follows that m m+1 #(2 ) < 2 log 2; m 2 Z≥1; and now, because any x > 1 satisfies 2m−1 < x ≤ 2m for some such m, #(x) < 2m+1 log 2 < 4x log 2: So indeed #(x) = O(x). 2. Lemma on asymptotics, beginning of the proof 2.1. Lemma. The following lemma is elementary and ubiquitous in asymptotics. Lemma 2.1. Suppose that a sequence fcng satisfies X cn log n ∼ rx for some r: n≤x Then X rx c ∼ : n log x n≤x Proof. Name the two sums in the lemma, X X θ(x) = cn log n and '(x) = cn: n≤x n≤x Thus θ(x) ∼ rx, and we want to show that '(x) ∼ rx= log x. Because the step function θ(x) jumps by cn log n at each n, and the step function '(x) jumps by cn at each n, we have for t > 1 in the sense of Stieltjes integration, dθ(t) d'(t) = : log t SIMPLE PROOF OF THE PRIME NUMBER THEOREM 3 With \∗" denoting a fixed, large enough lower limit of integration, and with a Stieltjes integral and integration by parts, Z x Z x x Z x dθ(t) θ(t) θ(t) (1) '(x) ∼ d'(t) = = + 2 dt: t=∗ t=∗ log t log t t=∗ t=∗ t log t The boundary term is asymptotically rx= log x, as desired for '(x), so what needs to be shown is that the last integral in (1) is o(x= log x). Because θ(t)=t ∼ r for large t, estimate the integral of 1= log2 t, first breaking it into two pieces, p Z x 1 Z x 1 Z x 1 2 dt = 2 dt + p 2 dt: t=∗ log t t=∗ log t t= x log t For the first piece, p p p Z x Z x x 1 p 1 p 1 p 2 dt ≤ x 2 dt = − x ∼ x; t=∗ log t t=∗ t log t log t t=∗ while for the second, Z x 1 1 p 2x p 2 dt ≤ 2 p (x − x) ∼ 2 : t= x log t log x log x R x 2 Altogether t=∗ dt= log t is o(x= log x). Because θ(t)=t = O(1), the last integral in (1) is therefore o(x= log x) as well, and the argument is complete. 2.2. Beginning of the proof. Consider the prime-indicator sequence, fcng = fc1; c2;::: g where ( 1 if n is prime cn = 0 otherwise: The Chebyshev theta function and the prime-counting function function are natu- rally re-expressed using this sequence, X X #(x) = cn log n and π(x) = cn: n≤x n≤x Consequently the lemma reduces the Prime Number Theorem to showing that #(x) ∼ x Already #(x) = O(x) is established, so the work is to go from this to the boxed result. 3. Euler{Riemann zeta function 3.1. Zeta as a sum. The Euler{Riemann zeta function is initially defined as a sum on an open right half plane, 1 X ζ(s) = n−s; Re(s) > 1: n=1 This sum converges absolutely on Re(s) > 1 because jn−sj = n−Re(s), and hence PN −s it indeed converges on Re(s) > 1. Each truncation n=1 n of the sum is entire. 4 SIMPLE PROOF OF THE PRIME NUMBER THEOREM Let K denote a compact subset of Re(s) > 1. There exists some σ > 1 such that Re(s) ≥ σ on K, and so 1 1 X −s X −σ n ≤ n ; s 2 K: n=N n=N This shows that the sum ζ(s) converges uniformly on K. Altogether, ζ(s) is holo- morphic on Re(s) > 1. 3.2. Zeta as a product. The Euler{Riemann zeta function has a second expres- sion as a product of so-called Euler factors over the prime numbers, Y ζ(s) = (1 − p−s)−1; Re(s) > 1: p The equality of the product and sum expressions of ζ(s) for Re(s) > 1 is a matter of the geometric series formula and the Fundamental Theorem of Arithmetic, as follows. Consider any positive integer k, let p1; : : : ; pk denote the first k primes, compute k k Mi k Mi Y −s −1 Y X −mis Y X −mis (1 − pi ) = lim pi = lim pi Mi!1 M1;:::;Mk!1 i=1 i=1 mi=0 i=1 mi=0 X X = lim n−s = n−s; M1;:::;Mk!1 Qk mi Qk mi n= i=1 pi n= i=1 pi mi≤Mi each i Q −s −1 P1 −s and take the limit as k goes to 1 to get the result, p(1 − p ) = n=1 n . Now the product form of ζ(s) inherits the holomorphy of the sum form. Also we can show that the product is a holomorphic function on Re(s) > 1 with no reference to its matching the sum. Recall a general result for a product Q1 n=1(1 + 'n(s)) with each 'n holomorphic on a domain Ω, as follows. Suppose that: For every compact K in Ω there exists a summable sequence fxng = fxn(K)g of nonnegative real numbers such that j'n(s)j ≤ xn for all n, uniformly over s 2 K. Q1 Then n=1(1 + 'n(s)) is holomorphic on Ω. In our case, Ω is Re(s) > 1, and −s −1 −s −1 −s 'n(s) is (1 − p ) − 1 = (1 − p ) p if n is a prime p, while 'n = 0 if n is composite. Let K be a compact subset of Re(s) > 1. There exists some σ > 1 such −σ that Re(s) ≥ σ on K. Let fxng = f2n g. For any prime p, for all s 2 K, −s −1 −s −σ j'p(s)j = j(1 − p ) p j ≤ 2p = xp; Q −s −1 and j'n(s)j = 0 ≤ xn for composite n and s 2 K. Thus the product p(1 − p ) is holomorphic on Re(s) > 1, as claimed. 3.3. Euler's proof. Using the product form of ζ(s), consider the logarithm of the zeta function for s approaching 1 from the right, X X X 1 (2) log ζ(s) = log((1 − p−s)−1) = : mpms p p m≥1 SIMPLE PROOF OF THE PRIME NUMBER THEOREM 5 This decomposes into two terms, X 1 X X 1 log ζ(s) = + : ps mpms p p m≥2 The sum form of ζ(s) shows that ζ diverges at 1, and hence so does log ζ although more slowly. The second sum is readily bounded by 1, 1 X X 1 X 1 X 1 X 1 < = < = 1: mpms p2s(1 − p−s) ps(ps − 1) k(k − 1) p m≥2 p p k=2 P −s So the first sum p p is asymptotic to log ζ(s) as s goes to 1, and consequently the prime numbers are dense enough to make the sum diverge at s = 1.