Riemann’s Zeta Function and the Prime Number Theorem

18.785 Number theory I Lecture #16 16 Riemann’s zeta function and the prime number theorem We now divert our attention from algebraic number theory to talk about zeta functions and L-functions. As we shall see, every global field has a zeta function that is intimately related to the distribution of its primes. We begin with the zeta function of the rational field Q, which we will use to prove the prime number theorem. We will need some basic results from complex analysis, all of which can be found in any introductory textbook (such as [1, 2, 3, 7, 12]). A short glossary of terms and a list of the basic theorems we will use can be found at the end of these notes.1 16.1 The Riemann zeta function Definition 16.1. The Riemann zeta function is the complex function defined by the series X n−s , ζ(s) := n≥1 for Re(s) > 1, where n varies over positive integers. It is easy to verify that this series converges absolutely and locally uniformly on Re(s) > 1 (use the integral test on an open ball strictly to the right of the line Re(s) = 1). By Theorem 16.17, it defines a holomorphic function on Re(s) > 1, since each term n−s = e−s log n is holomorphic. Theorem 16.2 (Euler product). For Re(s) > 1 we have X Y ζ(s) = n−s = (1 − p−s )−1 , p n≥1 where the product converges absolutely. In particular, ζ(s) 6= 0 for Re(s) > 1. The product in the theorem above ranges over primes p. This is a standard practice in analytic number theory that we will follow: the symbol p always denotes a prime, and any sum or product over p is understood to be over primes, even if this is not explicitly stated. Proof. We have X n−s = XY p−vp (n)s = YX n≥1 p n≥1 p−es = Y p e≥0 (1 − p−s )−1 . p To justify the second equality, consider the partial zeta function ζm (s), which restricts the summation in ζ(s) to the set Sm of m-smooth integers (those with no prime factors p > m). If p1 , . . . , pk are the primes up to m, absolute convergence implies X X Y X Y ei ζm (s) := n−s = (pe11 · · · pekk )−s = (p−s ) = (1 − p−s )−1 . i e1 ,...,ek ≥0 n∈Sm 1≤i≤k ei ≥0 p≤m For any δ > 0 the sequence of functions ζm (s) converges uniformly on Re(s) > 1 + δ to ζ(s); indeed, for any  > 0 and any such s we have |ζm (s) − ζ(s)| ≤ X n≥m 1 n−s ≤ X n≥m |n−s | = X n≥m n− Re(s) ≤ Z ∞ m 1 x−1−δ dx ≤ m−δ < , δ Those familiar with this material should still glance at §16.3.2 which touches on some convergence issues that are particularly relevant to number theoretic applications. for all sufficiently large m. It follows that the sequence ζQ m (s) converges locally uniformly to ζ(s) on Re(s) > 1. The sequence of functions Pm (s) := p≤m (1 − p−s )−1 clearly converges Q locally uniformly to (1 − p−s )−1 on any region in which the latter function is absolutely convergent (or even just convergent). For any s in Re(s) > 1 we have X | log(1 − p−s )−1 | = p X X1 X XX |p−s |e = (|ps | − 1)−1 < ∞, p−es ≤ e p p p e≥1 e≥1 P where we have used the identity log(1 − z) = − n≥1 n1 z n , valid for |z| < 1. It follows that Q −s −1 is absolutely convergent (and in particular, nonzero) on Re(s) > 1. p (1 − p ) Theorem 16.3 (Analytic continuation I). For Re(s) > 1 we have ζ(s) = 1 + φ(s), s−1 where φ(s) is a holomorphic function on Re(s) > 0. Thus ζ(s) extends to a meromorphic function on Re(s) > 0 that has a simple pole at s = 1 with residue 1 and no other poles. Proof. For Re(s) > 1 we have X 1 ζ(s) − = n−s − s−1 n≥1 Z ∞ x −s Z X dx = n−s − 1 n≥1 n+1 x −s  dx n = XZ n+1
n−s − x−s dx. n≥1 n R n+1 For each n ≥ 1 the function φn (s) := n (n−s − x−s )dx is holomorphic on Re(s) > 0. For each fixed s in Re(s) > 0 and x ∈ [n, n + 1] we have Z x Z x Z x |s| |s| |s| −s −s −s−1 |n − x | = st dt ≤ dt = dt ≤ 1+Re(s) , s+1 1+Re(s) | n n n |t n t and therefore Z n+1 |φn (s)| ≤ n−s − x−s dx ≤ n |s| n1+Re(s) . For any s0 with Re(s0 ) > 0, if we put  := Re(s0 )/2 and U := B< (s0 ), then for each n ≥ 1, |s0 | +  =: Mn , n1+ s∈U P P and n Mn = (|s0 | + )ζ(1 + ) converges. The series n φn thus P converges locally normally on Re(s) > 0. By the Weierstrass M -test (Theorem 16.19), n φn converges to a function 1 φ(s) = ζ(s) − s−1 that is holomorphic on Re(s) > 0. sup |φn (s)| ≤ We now show that ζ(s) has no zeros on Re(s) = 1; this fact is crucial to the prime number theorem. For this we use the following ingenious lemma, attributed to Mertens.2 Lemma 16.4 (Mertens). For x, y ∈ R with x > 1 we have |ζ(x)3 ζ(x + iy)4 ζ(x + 2iy)| ≥ 1. 2 If this lemma strikes you as pulling a rabbit out of a hat, well, it is. For a slight variation, see [15, IV], which uses an alternative approach due to Hadamard. Proof. From the Euler product ζ(s) = log |ζ(s)| = − X Q − p−s )−1 , we see that for Re(s) > 1 we have p (1 log |1 − p−s | = − X p Re log(1 − p−s ) = X X Re(p−ns ) p since log |z| = Re log z and log(1 − z) = − P log |ζ(x + iy)| = p n≥1 zn n≥1 n n , for |z| < 1. Plugging in s = x + iy yields X X cos(ny log p) p n≥1 npnx , since Re(p−ns ) = p−nx Re(e−iny log p ) = p−nx cos(−ny log p) = p−nx cos(ny log p). Thus log |ζ(x)3 ζ(x + iy)4 ζ(x + 2iy)| = X X 3 + 4 cos(ny log p) + cos(2ny log p) p n≥1 npnx . We now note that the trigonometric identity cos(2θ) = 2 cos2 θ − 1 implies 3 + 4 cos θ + cos(2θ) = 2(1 + cos θ)2 ≥ 0. Taking θ = ny log p yields log |ζ(x)3 ζ(x + iy)4 ζ(x + 2iy)| ≥ 0, which proves the lemma. Corollary 16.5. ζ(s) has no zeros on Re(s) ≥ 1. Proof. We know from Theorem 16.2 that ζ(s) has no zeros on Re(s) > 1, so suppose ζ(1 + iy) = 0 for some y ∈ R. Then y 6= 0, since ζ(s) has a pole at s = 1, and we know that ζ(s) does not have a pole at 1 + 2iy 6= 1, by Theorem 16.3. We therefore must have (1) lim |ζ(x)3 ζ(x + iy)4 ζ(x + 2iy)| = 0, x→1 since ζ(s) has a simple pole at s = 1, a zero at 1 + iy, and no pole at 1 + 2iy. But this contradicts Lemma 16.4. 16.2 The Prime Number Theorem The prime counting function π : R → Z≥0 is defined by X π(x) := 1; p≤x it counts the number of primes up to x. The prime number theorem (PNT) states that π(x) ∼ x . log x The notation f (x) ∼ g(x) means limx→∞ f (x)/g(x) = 1; one says that f is asymptotic to g. This conjectured growth rate for π(x) dates back to Gauss and Legendre in the late 18th century. In fact Gauss believed the asymptotically equivalent but more accurate statement3 Z x dt π(x) ∼ Li(x) := . 2 log t 3 More accurate in the sense that |π(x) − Li(x)| grows more slowly than |π(x) − x | log x as x → ∞. However it was not until a century later that the prime number theorem was independently proved by Hadamard [5] and de la Vallée Poussin [9] in 1896. Their proofs are both based on the work of Riemann [10], who in 1860 showed that there is a precise connection between the zeros of ζ(s) and the distribution of primes (we shall say more about this later), but was unable to prove the prime number theorem. The proof we will give is more recent and due to Newman [8], but it relies on the same properties of the Riemann zeta function that were exploited by both Hadamard and de la Vallée, the most essential of which is the fact that ζ(s) has no zeros on Re(s) ≥ 1 (Corollary 16.5). A concise version of Newman’s proof by Zagier can be found in [15]; we will follow Zagier’s outline but be slightly more expansive in our presentation. We should note that there are also “elementary" proofs of the prime number theorem independently obtained by Erdös [4] and Selberg [11] in the 1940s that do not use the Riemann zeta function, but they are elementary only in the sense that they do not use complex analysis; the details of these proofs are considerably more complicated than the one we will give. Rather than work directly with π(x), it is more convenient to work with the log-weighted prime-counting function defined by Chebyshev4 X ϑ(x) := log p, p≤x whose growth rate differs from that of π(x) by a logarithmic factor. Theorem 16.6 (Chebyshev). π(x) ∼ x log x if and only if ϑ(x) ∼ x. Proof. We clearly have 0 ≤ ϑ(x) ≤ π(x) log x, thus ϑ(x) π(x) log x ≤ . x x For every  ∈ (0, 1) we have ϑ(x) ≥ X   log p ≥ (1 − )(log x) π(x) − π(x1− ) x1− 1 and all  > 0 we have |F (λx) − F (x)| <  for all sufficiently large x. Fix λ > 1 and suppose there is an unbounded sequence (xn ) such that f (xn ) ≥ λxn for all n ≥ 1. For each xn we have Z λxn Z λxn Z λ λ−t f (t) − t λxn − t F (λxn ) − F (xn ) = dt ≥ dt = dt = c, 2 2 t t t2 xn xn 1 for some c > 0, where we used the fact that f is non-decreasing to get the middle inequality. Taking  < c, we have |F (λxn ) − F (xn )| = c >  for arbitrarily large xn , a contradiction. Thus f (x) < λx for all sufficiently large x. A similar argument shows that f (x) > λ1 x for all sufficiently large x. These inequalities hold for all λ > 1, so limx→∞ f (x)/x = 1. Equivalently, f (x) ∼ x. 5 The equality sign in the big-O notation f (x) = O(g(x)) is a standard abuse of notation; it simply means lim supx→∞ |f (x)|/|g(x)| < ∞ (and nothing more). In more complicated equalities a big-O expression should P be interpreted as a set of functions, one of which makes the equality true, for example, n≥1 n1 = log n+O(1). In order to show that the hypothesis of Lemma 16.8 is satisfied for f = ϑ, we will work with the function H(t) = ϑ(et )e−t − 1; the change of variables t = eu shows that Z ∞ Z ∞ ϑ(t) − t H(u)du converges . dt converges ⇐⇒ t2 1 0 We now recall the Laplace transform. Definition 16.9. Let h : R>0 → R be a piecewise continuous function. The Laplace transform Lh of h is the complex function defined by Z ∞ Lh(s) := e−st h(t)dt, 0 which is holomorphic on Re(s) > c for any c ∈ R for which h(t) = O(ect ). The following properties of the Laplace transform are easily verified. • L(g + h) = Lg + Lh, and for any a ∈ R we have L(ah) = aLh. • If h(t) = a ∈ R is constant then Lh(s) = as . • L(eat h(t))(s) = L(h)(s − a) for all a ∈ R. We now define the auxiliary function Φ(s) := X p−s log p, p which is related to ϑ(x) by the following lemma. Lemma 16.10. L(ϑ(et ))(s) = Φ(s) s is holomorphic on Re(s) > 1. Proof. By Lemma 16.7, ϑ(et ) = O(et ), so L(ϑ(et )) is holomorphic on Re(s) > 1. Let pn be the nth prime, and put p0 := 0. The function ϑ(et ) is constant on t ∈ (log pn , log pn+1 ), so Z log pn+1 Z log pn+1   1 −s e−st ϑ(et )dt = ϑ(pn ) e−st dt = ϑ(pn ) p−s − p n n+1 . s log pn log pn We then have (Lϑ(et ))(s) = Z ∞ e−st ϑ(et )dt = 0 = = = ∞   1X −s ϑ(pn ) p−s − p n n+1 s 1 s 1 s 1 s n=1 ∞ X ∞ ϑ(pn )p−s n − n=1 ∞  X n=1 ∞ X 1X ϑ(pn−1 )p−s n s n=1  ϑ(pn ) − ϑ(pn−1 ) p−s n p−s n log pn = n=1 Φ(s) . s Let us now consider the function H(t) := ϑ(et )e−t − 1. It follows from the lemma and standard properties of the Laplace transform that on Re(s) > 0 we have LH(s) = L(ϑ(et )e−t )(s) − (L1)(s) = L(ϑ(et ))(s + 1) − 1 Φ(s + 1) 1 = − . s s+1 s Lemma 16.11. The function Φ(s) − that is holomorphic on Re(s) ≥ 1. 1 s−1 extends to a meromorphic function on Re(s) > 1 2 Proof. By Theorem 16.3, ζ(s) extends to a meromorphic function on Re(s) > 0, which we also denote ζ(s), that has only a simple pole at s = 1 and no zeros on Re(s) ≥ 1, by Corollary 16.5. It follows that the logarithmic derivative ζ 0 (s)/ζ(s) of ζ(s) is meromorphic on Re(s) > 0, with no zeros on Re(s) ≥ 1 and only a simple pole at s = 1 with residue −1 (see §16.3.1 for standard facts about the logarithmic derivative of a meromorphic function). In terms of the Euler product, for Re(s) > 1 we have6 !0 !0 Y X
0 ζ 0 (s) − = − log ζ(s) = − log (1 − p−s )−1 = log(1 − p−s ) ζ(s) p p  X log p X 1 X p−s log p 1 = + log p = = 1 − p−s ps − 1 ps ps (ps − 1) p p p X log p = Φ(s) + . ps (ps − 1) p The sum on the RHS converges absolutely and locally uniformly to a holomorphic function on Re(s) > 1/2. The LHS is meromorphic on Re(s) > 0, and on Re(s) ≥ 1 it has only a 1 simple pole at s = 1 with residue 1. It follows that Φ(s) − s−1 extends to a meromorphic 1 function on Re(s) > 2 that is holomorphic on Re(s) ≥ 1. 1 Corollary 16.12. The functions Φ(s + 1) − 1s and (LH)(s) = Φ(s+1) s+1 − s both extend to meromorphic functions on Re(s) > − 12 that are holomorphic on Re(s) ≥ 0. Proof. The first statement follows immediately from the lemma. For the second, note that   Φ(s + 1) 1 1 1 1 − = Φ(s + 1) − − s+1 s s+1 s s+1 is meromorphic on Re(s) > − 21 and holomorphic on Re(s) ≥ 0, since it is a sum of products of such functions. The final step of the proof relies on the following analytic result due to Newman [8]. Theorem 16.13. Let f : R≥0 → R be a bounded piecewise continuous function, and suppose its R ∞Laplace transform extends to a holomorphic function g(s) on Re(s) ≥ 0. Then the integral 0 f (t)dt converges and is equal to g(0). Proof. Without loss of generality weR assume f (t) ≤ 1 for all t ≥ 0. For τ ∈ R>0 , define Rτ ∞ gτ (s) := 0 f (t)e−st dt, By definition 0 f (t)dt = limτ →∞ gτ (0), thus it suffices to prove lim gτ (0) = g(0). τ →∞ For r > 0, let γr be the boundary of the region {s : |s| ≤ r and Re(s) ≥ −δr } with δr > 0 chosen so that g is holomorphic on γr ; such a δr exists because g is holomorphic on Re(s) ≥ 0, hence on some open ball B≤2δ(y) (iy) for each y ∈ [−r, r], and we may take 6 As is standard when computing logarithmic derivatives, we are taking the principal branch of the complex logarithm and can safely ignore the negative real axis where it is not defined since we are assuming Re(s) > 1. δr := inf{δ(y) : y ∈ [r, −r]}, which is positive because [−r, r] is compact. Each γr is a 2 simple closed curve, and for each τ > 0 the function h(s) := (g(s) − gτ (s))esτ (1 + rs2 ) is holomorphic on a region containing γr . Using Cauchy’s integral formula (Theorem 16.26) to evaluate h(0) yields  Z   1 1 s (2) g(0) − gτ (0) = h(0) = g(s) − gτ (s) esτ + 2 ds. 2πi γr s r We will show the LHS tends to 0 as τ → ∞ by showing that for any  > 0 we can set r = 3/ > 0 so that the absolute value of the RHS is less than  for all sufficiently large τ . Let γr+ denote the part of γr in Re(s) > 0, a semicircle of radius r. The integrand is absolutely bounded by 1/r on γr+ , since for |s| = r and Re(s) > 0 we have sτ g(s) − gτ (s) · e  1 s + 2 s r  Z ∞ f (t)e−st dt · = τ ∞ eRe(s)τ r s · + r s r eRe(s)τ 2 Re(s) · r r τ Re(s)τ − Re(s)τ e e 2 Re(s) = · · Re(s) r r 2 = 2/r . Z ≤ Therefore 1 2πi Z  γr+ e− Re(s)t dt ·   1 s 2 1 1 sτ g(s) − gτ (s) e + 2 ds ≤ · πr · 2 = s r 2π r r (3) Now let γr− be the part of γr in Re(s) < 0, a truncated semi-circle. For any fixed r, the first term g(s)esτ (s−1 + sr−2 ) in the integrand of (2) tends to 0 as τ → ∞ for Re(s) < 0 and |s| ≤ r. For the second term we note that since gτ (s) is holomorphic on C, it makes no difference if we instead integrate over the semicircle of radius r in Re(s) < 0. For |s| = r and Re(s) < 0 we then have gτ (s)e sτ  1 s + 2 s r  Z τ f (t)e−st dt · = 0 τ eRe(s)τ r s · + r s r eRe(s)τ (−2 Re(s)) r r 0 ! − Re(s)τ Re(s)τ e e (−2 Re(s)) = 1− Re(s) r r Z ≤ e− Re(s)t dt · = 2/r2 · (1 − eRe(s)τ Re(s)), where the factor (1 − eRe(s)τ Re(s)) on the RHS tends to 1 as τ → ∞ since Re(s) < 0. We thus obtain the bound 1/r + o(1) when we replace γr+ with γr− in (3), and the RHS of (2) is bounded by 2/r + o(1) as τ → ∞. It follows that for any  > 0, for r = 3/ > 0 we have |g(0) − gτ (0)| < 3/r =  for all sufficiently large τ . Therefore limτ →∞ gτ (0) = g(0) as desired. Remark 16.14. Theorem 16.13 is an example of what is known as a Tauberian theorem. For a piecewise continuous function f : R≥0 → R, its Laplace transform Z ∞ Lf (s) := e−st f (t)dt, 0 is typically not defined on Re(s) ≤ c, where c is the least c for which f (t) = O(ect ). Now it may happen that the function Lf has an analytic continuation to a larger domain; for 1 example, if f (t) = et then (Lf )(s) = s−1 extends to a holomorphic function on C−{1}. But plugging values of s with Re(s) ≤ c into the integral usually does not work; in our f (t) = et example, the integral diverges on Re(s) ≤ 1. The theorem says that when Lf extends to a holomorphic function on the entire half-plane Re(s) ≥ 0, its value at s = 0 is exactly what we would get by simply plugging 0 into the integral defining Lf . More generally, Tauberian theorems refer to results related toRtransforms f → T (f ) that ∞ allow us to deduce properties of f (such as the convergence of 0 f (t)dt) from properties of T (f ) (such as analytic continuation to Re(s) ≥ 0). The term “Tauberian" was coined by Hardy and Littlewood and refers to Alfred Tauber, who proved a theorem of this type as a partial converse to a theorem of Abel. Theorem 16.15 (Prime Number Theorem). π(x) ∼ x log x . Proof. H(t) = ϑ(et )e−t − 1 is piecewise continuous and bounded, by Lemma 16.7, and its Laplace transform extends to a holomorphic function on Re(s) ≥ 0, by Corollary 16.12. Theorem 16.13 then implies that the integral Z ∞ Z ∞  H(t)dt = ϑ(et )e−t − 1 dt 0 0 converges. Replacing t with log x, we see that  Z ∞ Z ∞ 1 dx ϑ(x) − x ϑ(x) − 1 = dx x x x2 1 1 converges. Lemma 16.8 implies ϑ(x) ∼ x, equivalently, π(x) ∼ x log x , by Theorem 16.6. One disadvantage of our proof is that it does not give us an error term. Using more sophisticated methods, Korobov [6] and Vinogradov [14] independently obtained the bound ! x
, π(x) = Li(x) + O exp (log x)3/5+o(1) in which we note that the error term is bounded by O(x/(log x)n ) for all n but not by O(x1− ) for any  > 0. Assuming the Riemann Hypothesis, which states that the zeros of ζ(s) in the critical strip 0 < Re(s) < 1 all lie on the line Re(s) = 12 , one can prove π(x) = Li(x) + O(x1/2+o(1) ). More generally, if we knew that ζ(s) has no zeros in the critical strip with real part greater than c, for some c ≥ 1/2 strictly less than 1, we could prove π(x) = Li(x) + O(xc+o(1) ). There thus remains a large gap between what we can prove about the distribution of prime numbers and what we believe to be true. Remarkably, other than refinements to the o(1) term appearing in the Korobov-Vinogradov bound, essentially no progress has been made on this problem in the last 60 years. 16.3 A quick recap of some basic complex analysis The complex numbers C are a topological field under the distance metric d(x, y) = |x − y| √ induced by the standard absolute value |z| := z z̄, which is also a norm on C as an Rvector space; all references to the topology on C (open, compact, convergence, limits, etc.) are made with this understanding. 16.3.1 Glossary of terms and standard theorems Let f and g denote complex functions defined on an open subset of C. • f is differentiable at z0 if limz→z0 f (z)−f (z0 ) z−z0 exists. • f is holomorphic at z0 if it is differentiable on an open neighborhood of z0 . • f is analytic at z0 if there of z0 in which f can be defined by P is an open neighborhood n a power series f (z) = n=0 an (z − z0 ) ; equivalently, f is infinitely differentiable and has a convergent Taylor series on an open neighborhood of z0 . • Theorem: f is holomorphic at z0 if and only if it is analytic at z0 . • Theorem: If C is a connected set containing a nonempty open set U and f and g are holomorphic on C with f|U = g|U , then f|C = g|C . • With U and C as above, if f is holomorphic on U and g is holomorphic on C with f|U = g|U , then g is the (unique) analytic continuation of f to C and f extends to g. • If f is holomorphic on a punctured open neighborhood of z0 and |f (z)| → ∞ as z → z0 then z0 is a pole of f ; note that the set of poles of f is necessarily a discrete set. • f is meromorphic at z0 if it is holomorphic at z0 or has z0 as a pole. • Theorem: at z0 then it can be defined by a Laurent series P If f is meromorphic n f (z) = n≥n0 an (z − z0 ) that converges on an open punctured neighborhood of z0 . • The order of vanishing ordz0 (f ) of a nonzero function f that is meromorphic at z0 is the least index n of the nonzero coefficients an in its Laurent series expansion at z0 . Thus z0 is a pole of f iff ordz0 (f ) < 0 and z0 is a zero of f iff ordz0 (f ) > 0. • If ordz0 (f ) = 1 then z0 is a simple zero of f , and if ordz0 (f ) = −1 it is a simple pole. • The residue resz0 (f ) of a function P f meromorphic at z0 is the coefficient a−1 in its Laurent series expansion f (z) = n≥n0 an (z − z0 )n at z0 . • Theorem: If z0 is a simple pole of f then resz0 (f ) = limz→z0 (z − z0 )f (z). • Theorem: If f is meromorphic on a set S then so is its logarithmic derivative f 0 /f , and f 0 /f has only simple poles in S and resz0 (f 0 /f ) = ordz0 (f ) for all z0 ∈ S. In particular the poles of f 0 /f are precisely the zeros and poles of f . 16.3.2 Convergence P P Recall that a series ∞ n=1 an of complex numbers converges absolutely if the series n |an | of nonnegative real numbers converges. An equivalent definition is that the function a(n) := an is integrable with respect to the counting measure µ on the set of positive integers N. Indeed, if the series is absolutely convergent then Z ∞ X an = a(n)µ, n=1 N and if the series is not absolutely convergent, the integral is not defined. Absolute convergence is effectively built-in to the definition of the Lebesgue integral, which requires that in order for the function a(n) = x(n) + iy(n) to be integrable, the positive real functions |x(n)| and |y(n)| must both be integrable (summable), and separately computes sums of the positive and negative subsequences of (x(n)) and (y(n)) as suprema over finite subsets. The measure-theoretic perspective has some distinct advantages. It makes it immediately clear that we may replace the index set N with any set of the same cardinality, since the counting measure depends only on the cardinality of N, not its ordering. We are thus free to sum over any countable index set, including Z, Q, any finite product of countable sets, and any countable coproduct of countable sets (such as countable direct sums of Z); such sums are ubiquitous in number theory and many cannot be meaningfully interpreted as limits of partial sums in the usual sense, since this assumes that the index set is well ordered (not the case with Q, for example). The measure-theoretic view makes P also makes it clear that we may convert any absolutely convergent sum• of the form X×Y into an iterated sum P P theorem. X Y (or vice versa), via Fubini’s Q We say that an infinite product is absolutely conn an of nonzero P Q complex numbers P vergent when the sum n log an is, in which case n an := exp( n log an ).7 This implies that an absolutely convergent product cannot converge to zero, and the sequence (an ) must converge to 1 (no matter how we order the an ). All of our remarks above about absolutely convergent series apply to absolutely convergent products as well. A series or product of complex functions fn (z) is absolutely convergent on S if the series or product of complex numbers fn (z0 ) is absolutely convergent for all z0 ∈ S. Definition 16.16. A sequence of complex functions (fn ) converges uniformly on S if there is a function f such that for every  > 0 there is an integer N for which supz∈S |fn (z)−f (z)| <  for all n ≥ N . The sequence (fn ) converges locally uniformly on S if every z0 ∈ S has an open neighborhood U for which (fn ) converges uniformly on U ∩S. When applied to a series of functions these terms refer to the sequence of partial sums. Because C is locally compact, locally uniform convergence is the same thing as compact convergence: a sequence of functions converges locally uniformly on S if and only if it converges uniformly on every compact subset of S. Theorem 16.17. A sequence or series of holomorphic functions fn that converges locally uniformly on an open set U converges to a holomorphic function f on U , and the sequence or series of derivatives fn0 then converges locally uniformly to f 0 (and if none of the fn has a zero in U and f 6= 0, then f has no zeros in U ). Proof. See [3, Thm. III.1.3] and [3, Thm. III.7.2]. P Definition 16.18. n (z) converges normally on a set S P P A series of complex functions n fP if n kfn k := n supz∈S |fn (z)| converges. The series n fnP (z) converges locally normally on S if every z0 ∈ S has an open neighborhood U on which n fn (z) converges normally. Theorem 16.19 (Weierstrass M-test). Every locally normally convergent series of P functions converges absolutely and locally uniformly. Moreover, a series n fn of holomorphic functions on converges locally normally converges to a holomorphic function f PS that 0 on S, and then n fn converges locally normally to f 0 . 7 In this definition we use the principal branch of log z := log |z| + i Arg z with Arg z ∈ (−π, π). Proof. See [3, Thm. III.1.6]. P Remark 16.20. To show a series n fn is locally normally convergent on a set S amounts to proving that for every z0 ∈ S there is an open neighborhood P U of z0 and a sequence of real numbers (Mn ) such that |fn (z)| ≤ Mn for z ∈ U ∩ S and n Mn < ∞, whence the term “M -test". 16.3.3 Contour integration We shall restrict our attention to integrals along contours defined by piecewise-smooth parameterized curves; this covers all the cases we shall need. Definition 16.21. A parameterized curve is a continuous function γ : [a, b] → C whose domain is a compact interval [a, b] ⊆ R. We say that γ is smooth if it has a continuous nonzero derivative on [a, b], and piecewise-smooth if [a, b] can be partitioned into finitely many subintervals on which the restriction of γ is smooth. We say that γ is closed if γ(a) = γ(b), and simple if it is injective on [a, b) and (a, b]. Henceforth we will use the term curve to refer to any piecewise-smooth parameterized curve γ, or to its oriented image of in the complex plane (directed from γ(a) to γ(b)), which we may also denote γ. Definition 16.22. Let f : Ω → C be a continuous function and let γ be a curve in Ω. We define the contour integral Z Z f (z)dz := γ b f (γ(t))γ 0 (t)dt, a whenever the integralR on the RHS (which is defined as a Riemann sum in the usual way) converges. Whether γ f (z)dz converges, and if so, to what value, does not depend on the parameterization of γ: ifR γ 0 is another parameterized curve with the same (oriented) image R as γ, then γ 0 f (z)dz = γ f (z)dz. We have the following analog of the fundamental theorem of calculus. Theorem 16.23. Let γ : [a, b] → C be a curve in an open set Ω and let f : Ω → C be a holomorphic function Then Z f 0 (z)dz = f (γ(b)) − f (γ(a)). γ Proof. See [2, Prop. 4.12]. Recall that the Jordan curve theorem implies that every simple closed curve γ partitions C into two components, one of which we may unambiguously designate as the interior (the one on the left as we travel along our oriented curve). We say that γ is contained in an open set U if both γ and its interior lie in U . The interior of γ is a simply connected set, and if an open set U contains γ then it contains a simply connected open set that contains γ. Theorem 16.24 (Cauchy’s Theorem). Let U be an open set containing a simple closed curve γ. For any function f that is holomorphic on U we have Z f (z)dz = 0. γ Proof. See [2, Thm. 8.6] (we can restrict U to a simply connected set). Cauchy’s theorem generalizes to meromorphic functions. Theorem 16.25 (Cauchy Residue Formula). Let U be an open set containing a simple closed curve γ. Let f be a function that is meromorphic on U , let z1 , . . . , zn be the poles of f that lie in the interior of γ, and suppose that no pole of f lies on γ. Then Z f (z)dz = 2πi γ n X reszi (f ). i=1 Proof. See [2, Thm. 10.5] (we can restrict U to a simply connected set). R it To see where the 2πi comes from, consider γ dz z with γ(t) = e for t ∈ [0, 2π]. In general one weights residues by a corresponding winding number, but the winding number of a simple closed curve about a point in its interior is always 1. Theorem 16.26 (Cauchy’s Integral Formula). Let U be an open set containing a simple closed curve γ. For any function f holomorphic on U and a in the interior of γ, Z 1 f (z) f (a) = dz. 2πi γ z − a Proof. Apply Cauchy’s residue formula to g(z) = f (z)/(z − a); the only poles of g in the interior of γ are a simple pole at z = a with resa (g) = f (a). Cauchy’s residue formula can also be used to recover the coefficients f (n) (a)/n! appearing in the Laurent series expansion of a meromorphic function at a (apply it to f (z)/(z −a)n+1 ). One of many useful consequences of this is Liouville’s theorem, which can be proved by showing that the Laurent series expansion of a bounded holomorphic function on C about any point has only one nonzero coefficient (the constant coefficient). Theorem 16.27 (Liouville’s theorem). Bounded entire functions are constant. Proof. See [2, Thm. 5.10]. We also have the following converse of Cauchy’s theorem. Theorem 16.28 (Morera’s Theorem). Let f be a continuous function and on an open set U , and suppose that for every simple closed curve γ contained in U we have Z f (z)dz = 0. γ Then f is holomorphic on U . Proof. See [3, Thm. II.3.5]. References [1] Lars V. Ahlfors, Complex analysis: an introduction to the theory of analytic functions of one complex variable, 3rd edition, McGraw-Hill, 1979. 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