Analytic Number Theory

Home , Analytic number theory, Euler product

Andrew Kobin Fall 2013 Contents Contents

Contents

0 Introduction 1

1 Preliminaries 2 1.1 Basic Number Theory ...... 2 1.2 Basic Analysis ...... 3 1.3 Euler-Maclaurin Summation ...... 4 1.4 The Bernoulli Numbers ...... 13

2 Euler’s Work 15 2.1 On the Sums of Series of Reciprocals ...... 15 2.2 Newton’s Identities ...... 20 2.3 Euler’s Product Form ...... 22 2.4 The Prime Number Theorem ...... 31

3 Complex Analysis 35 3.1 Arithmetic ...... 35 3.2 Functions and Limits ...... 37 3.3 Line Integrals ...... 44 3.4 Diﬀerentiability ...... 46 3.5 Integration in the Complex Plane ...... 52 3.6 Singularities and the Residue Theorem ...... 57

4 The Zeta Function 62 4.1 The Functional Equation ...... 62 4.2 Finding the Zeros ...... 68 4.3 Sketch of the Prime Number Theorem ...... 69 4.4 Generalized Riemann Hypothesis ...... 71

i 0 Introduction

0 Introduction

These notes were compiled from a semester of lectures at Wake Forest University by Dr. ∞ X 1 John Webb. The primary focus is the Riemann Zeta Function: ζ(s) = ns n=1

∞ X 1 Example 0.0.1. ζ(1) = , the harmonic series, is a divergent series. n n=1

∞ X 1 π2 Example 0.0.2. We know ζ(2) = converges to , but how? n2 6 n=1 Euler’s Results π2 Proved that ζ(2) = 6 900+ papers over his lifetime

– Some 15 of them dealt with the Zeta Function

Riemann wrote one paper on the topic

– Revolutionized analytic number theory x Prime Number Theorem: π(x) ≈ , where π(x) is the number of primes less than log(x) or equal to x. Riemann gave a map of how to prove this theorem. Modern Research

Riemann Hypothesis

L-functions

Calculating zeroes of the Zeta Function

1 1 Preliminaries

1 Preliminaries

1.1 Basic Number Theory

Z is the integers {0, ±1, ±2,...}. Deﬁnition. For an integer n, we say d is a divisor of n, d | n, if there exists an integer c such that dc = n. Properties (1) n | n for all n ∈ Z (2) If d | n and n | m then d | m (3) If d | n and d | m then d | (xn + ym)

(4) If d | n then ad | an for all a ∈ Z (5) If ad | an then d | n if a 6= 0

(6)1 | n for all n ∈ Z (7) n | 0 for all n ∈ Z (8) If 0 | n then n = 0 (9) If d | n and n 6= 0 then |d| ≤ |n| (10) If d | n and n | d then |d| = |n| Deﬁnition. Given n, m ∈ Z, we say that d ≥ 1 is the greatest common divisor of n and m if d | n, d | m, and for all e that divides n and m, |e| ≤ d. Notes: Given another divisor of n, m, say e, then e | gcd(n, m) Given gcd(n, m), there exist integers x and y such that xn + ym = gcd(n, m)

{(xn + ym) | x, y ∈ Z} = {k · gcd(n, m) | k ∈ Z} Deﬁnition. An integer p ≥ 2 is prime if its only positive divisors are 1 and p. Deﬁnition. An integer n > 1 is composite if it is not prime. Note: 1 is not prime. Theorem 1.1.1 (Fundamental Theorem of Arithmetic). Any integer n, |n| > 1, can be uniquely written as the product of powers of distinct primes

e1 e2 er n = p1 p2 ··· pr , where pi is prime and ei ∈ Z for all i. Example 1.1.2. 2013 = 3 · 11 · 61.

2 1.2 Basic Analysis 1 Preliminaries

1.2 Basic Analysis

Deﬁnition. If f is a function, f(x) converges to L ∈ , denoted lim f(x) = L, if for all R x→∞ ε > 0 there exists N > 0 such that for all x > N, |f(x) − L| < ε.

Deﬁnition. A function f(x) diverges (to ∞), denoted lim f(x) = ∞, if for all M > 0 x→∞ there exists some N > 0 such that for all x > N, f(x) > M.

This can be adapted for −∞ as well.

Deﬁnition. Given functions f(x) and g(x) deﬁned on R (or Z) ≥ a, with g(x) > 0 and monotonic on [a, ∞), we say that f(x) = O(g(x)) if for all x ≥ a there exists some constant M > 0 such that |f(x)| ≤ Mg(x), also denoted f(x) << g(x).

Deﬁnition. Given functions f and g, f(x) >> g(x) if there exists some constant m > 0 such that |f(x)| ≥ mg(x) for all x > a.

Deﬁnition. If f(x) >> g(x) and f(x) << g(x) then f and g are said to have the same order, denoted f(x) g(x). X 1 Example 1.2.1. = O(log(x)) p p prime p≤x Proof. What if we sum over all integers?:

x X 1 n n=1

x x ∞ X 1 X 1 X Z ∞ We know ≥ . Recall the integral test: f(n) and f(x) dx both converge n p n=1 p=2 n=1 1 or both diverge (if f > 0 and f is monotone on [1, ∞)). So we have

x X 1 Z x 1 ≤ dt = log(x). n t n=2 1 In fact, x X 1 log(x) < < 2 log(x). n n=1 x x x X 1 X 1 X 1 Therefore ≤ < 2 log(x) for all x > 2. Hence = O(log(x)). p n p p=2 n=1 p=2

x x X 1 X 1 But is way bigger than so this is a bad approximation tool. n p n=1 p=2

3 1.3 Euler-Maclaurin Summation 1 Preliminaries

Example 1.2.2. sin(x) << 1

Let g(x) = 1, then | sin(x)| ≤ g(x) for all x ∈ R. So sin(x) << 1. But is sin(x) 1? No, since sin(x) = 0 at inﬁnitely many points. Example 1.2.3. f(x) = sin(x) + x First, f(x) << x because if M = 2, |f(x)| ≤ 2x for x > 0. And f(x) >> x because if m = 1/2, |f(x)| ≥ 1/2x for x > 0. Thus f(x) x. Deﬁnition. Two functions f and g are asymptotic to each other, denoted f(x) ∼ g(x), if f(x) lim = 1. x→∞ g(x) Example 1.2.4. f(x) = sin(x) + x, g(x) = x

f(x) sin(x) + x sin(x) lim = lim = lim + 1 = 0 + 1 = 1 x→∞ g(x) x→∞ x x→∞ x thus f(x) ∼ g(x). Proposition 1.2.5. If f(x) ∼ g(x) then f(x) g(x). Proof omitted. Note that the converse is not true in general. x X 1 Z x 1 Example 1.2.6. The integral test actually states that ∼ dt = log(x). n t n=1 1

1.3 Euler-Maclaurin Summation

n−1 X Z n Let f(x) > 0 and strictly decreasing on [1, ∞). Examine dn = f(k) − f(x) dx. k=1 1

f(x) 4

2 dn

1 2 4 6 n8

4 1.3 Euler-Maclaurin Summation 1 Preliminaries

Proposition 1.3.1. dn < f(1) for any n > 1.

n−1 X Z k+1 Proof. Rewrite dn = (f(k) − f(x)) dx. Then k=1 k

n−1 X dn < (f(k) − f(k + 1)) = f(1) − f(n) k=1

by telescoping series. And since f(n) > 0, f(1) − f(n) < f(1). Hence dn < f(1).

Let C(f) = lim dn. We know C(f) exists because dn is increasing but bounded. Then n→∞ we can write n X Z k+1 C(f) = lim [f(k) − f(x)] dx. n→∞ k=1 k ∞ X Z k+1 Let Ef (n) = f(n) + dn − C(f). Then Ef (n) > 0 since dn − C(f) = [f(k) − f(x)] dx. k=n k Together, this gives us

∞ X Z ∞ f(k) = f(x) dx + C(f) + Ef (n). k=1 1

∞ X 1 Goal: Approximate to at least 3 decimal places. n2 n=1

1 Deﬁnition. γ = C x is called the Euler constant. 1 Example 1.3.2. Let f(x) = x . Then

n X Z n f(k) = f(x) dx + γ + Ef (n) k=1 1 Z n 1 We know that dx = log(n). To approximate the remaining terms, we will ﬁrst prove 1 x the following theorem for the general case.

n X Z n Z n Theorem 1.3.3. f(k) = f(x) dx + (x − bxc) f 0(x) dx + f(1), where f has a con- k=1 1 1 tinuous ﬁrst derivative on [1, n].

Proof. We begin with n X Z n f(k) = f(x) dx + dn. k=1 1

5 1.3 Euler-Maclaurin Summation 1 Preliminaries

To ﬁnd dn,

n−1 X Z k+1 dn = [f(k) − f(x)] dx k=1 k which we will integrate by parts. Let

u = f(k) − f(x) dv = dx

du = −f 0(x) dx v = x − (k + 1) ← we get to choose a constant and integrate:

Z k+1 k+1 Z k+1 0 [f(k) − f(x)] dx = [f(k) − f(x)](x − (k + 1)) + (x − (k + 1))f (x) dx k k k Z k+1 = [(f(k) − f(k + 1)) · 0 − 0(−1)] + (x − (k + 1))f 0(x) dx k Z k+1 Z k+1 = 0 + (x − (k + 1))f 0(x) dx = (x − (k + 1))f 0(x) dx. k k Thus

n−1 n−1 X Z k+1 X Z k+1 [f(k) − f(x)] dx = (x − (k + 1))f 0(x) dx k=1 k k=1 k n−1 X Z k+1 = (x − bxc − 1)f 0(x) dx k=1 k n−1 n−1 X Z k+1 X Z k+1 = (x − bxc)f 0(x) dx − f 0(x) dx k=1 k k=1 k n−1 X Z k+1 Z n = (x − bxc)f 0(x) dx − f 0(x) dx k=1 k 1 n−1 X Z k+1 = (x − bxc)f 0(x) dx − (f(n) − f(1)). k=1 k This is a formula we can work with. Plugging it back into the series formula, we obtain

n X Z n Z n f(k) = f(x) dx + (x − bxc)f 0(x) dx + f(1). k=1 1 1

Note that the (x − bxc) part above bounds the integral, but we can do a little better. The function x − bxc is a 1-periodic function. By selecting x − bxc − 1/2 instead, we still have a 1-periodic function but one that will integrate to 0 over integer periods.

6 1.3 Euler-Maclaurin Summation 1 Preliminaries

2 x − bxc

1 2 3 4

1 x − bxc − 1/2

1 2 3 4

−1

( x − bxc − 1/2 x 6∈ Z So let P1(x) = 0 x ∈ Z. Now consider Z n Z n 0 0 (x − bxc)f (x) dx = (x − bxc − 1/2 + 1/2)f (x) dx 1 1 Z n Z n 0 1 0 = P1(x)f (x) dx + /2f (x) dx 1 1 Z n 0 1 = P1(x)f (x) dx + /2(f(n) − f(1)). 1 Putting this into the formula from Theorem 1.3.3 gives us the following theorem: Theorem 1.3.4 (First Derivative Form of Euler-Mclaurin Summation Formula). n Z n Z n X 0 1 f(k) = f(x) dx + P1(x)f (x) dx + /2(f(n) + f(1)) k=1 1 1 where f has a continuous ﬁrst derivative f 0 on [1, n]. n X 1 Example 1.3.5. Approximate k k=1

n 0 X 1 Z n 1 Z n 1 1 1 = dx + P (x) dx + + 1 k x 1 x 2 n k=1 1 1 1 Z n 1 1 1 ≤ log(x) + 2 dx + + 1 2 1 x 2 n ≤ log(x) + 1.

7 1.3 Euler-Maclaurin Summation 1 Preliminaries

Strategy: (1) Add up ﬁrst few terms by hand (2) Use Euler-Maclaurin Formula to estimate the tail Z n 0 (3) Bound P1(x)f (x) dx to within 3 decimals i ∞ X Z ∞ Recall: if f(x) > 0 and f(n) converges then f(x) dx converges and lim f(n) = 0. n→∞ n=1 1 Z ∞ Claim. |f 0(x)| dx converges. i Proof. Since f is a monotone decreasing function (from the integral test), f 0 is always negative. So Z ∞ Z ∞ |f 0(x)| dx = − f 0(x) dx i i Z n = − lim f 0(x) dx n→∞ i = lim [f(i) − f(n)] n→∞ = f(i) − 0 since f(n) → 0 = f(i). Z ∞ Thus |f 0(x)| dx converges to f(i). i A consequence of this is: Z ∞ 1 Z ∞ 0 0 1 |P1(x)f (x)| dx ≤ |f (x)| dx = /2f(i). i 2 i 1 Example 1.3.6. Approximate f(x) = to 3 decimals. x2 1 −2 Let f(x) = ; then f 0(x) = . We will ﬁnd an i such that x2 x3 Z ∞ 2P1(x) 3 dx ≤ 0.0005 i x so that the ± gap of the error is 0.001.

1 k3 k + 1/2 k + 1

k −1 (k+1)3

8 1.3 Euler-Maclaurin Summation 1 Preliminaries

0 2 For [k, k + 1/2] take the max value of |f (x)| = . k3

1 0 k3 P1(x) max |f (x)|

k k + 1/2

Z k+1/2 Z k+1/2 2P1(x) 2 Then − 3 dx ≤ P1(x) · 3 dx. k x k k area of triangle = 1 4k3 0 2 1 2 Likewise, for [k + / , k + 1] take min |f (x)| = (k+1)3 .

k + 1/2 k + 1

−1 (k+1)3

Z k+1 Z k+1 2P1(x) 2 Then − 3 dx ≥ P1(x) · 3 dx. k+1/2 x k+1/2 (k + 1)

area of triangle = 1 4(k+1)3 Z k+1 0 1 1 This gives us an estimate for the error term: P1(x)f (x) dx ≤ 3 − 3 . So we k 4k 4(k + 1) have ∞ Z ∞ X 1 1 1 P (x)f 0(x) dx ≤ − = 1 4k3 4(k + 1)3 4i3 i k=i 1 √ by telescoping sum. We want ≤ 0.0005 ⇒ i ≥ 3 500 = 7.9 ... So choose i = 8 and we 4i3 ∞ X 1 can estimate f(n) for f(x) = to within 3 decimals: x2 n=1 ∞ 7 X 1 X 1 Z ∞ 1 1 = + dx + f(1) n2 n2 x2 2 n=1 n=1 8 7 X 1 1 1 1 = + + n2 8 2 64 n=1 1 1 1 1 1 1 1 1 = 1 + + + + + + + + 4 9 16 25 36 49 8 128 ≈ 1.6446.

9 1.3 Euler-Maclaurin Summation 1 Preliminaries

Compare this to the real value, which is 1.64493 ...

Example 1.3.7. For 8-decimal accuracy, we want 1 ≤ 0.000000005 ⇒ i ≥ 368.4 so choose i = 369. 4i3 Given the First Derivative Form,

n X Z n 1 Z n f(k) = f(x) dx + (f(1) + f(n)) + P (x)f 0(x) dx, 2 1 k=1 1 1 error we want to reduce the error term further. Note the following

Z 1 P1(x) dx = 0 ← good cancellation. 0 We will integrate by parts on the error term. Let

0 u = f (x) dv = P1(x) dx

00 2 du = f (x) dx v = 1/2(x − x) + c

Z 1 2 where we get to pick c. If c = 1/12 then 1/2(x − x) + c dx = 0. So in order to make the 0 2 integral periodic and have good cancellation, let v = 1/2((x − bxc) − (x − bxc)) + 1/12. Note 1 2 1 that since P1(x) was piecewise continuous, /2((x − bxc) − (x − bxc)) + /12 is continuous as well (on [0, ∞)). Now to integrate,

Z n 0 0 2 n 1 2 1 12 P1(x)f (x) dx = f (x) / ((x − bxc) − (x − bxc)) + / 1 1 Z n 2 00 − 1/2((x − bxc) − (x − bxc)) + 1/12 f (x) dx 1 Z n 1 0 1 0 1 00 = /12f (n) − /12f (1) − /2P2(x)f (x) dx 1 Z n 2 1 1 00 where P2(x) = (x−bxc) −(x−bxc)+ /6. This gives us a new error term, /2P2(x)f (x) dx. 1 Theorem 1.3.8 (Second Derivative Form of Euler-Maclaurin Summation Formula).

n Z n Z n X 1 1 0 0 1 00 f(k) = f(x) dx + /2(f(1) + f(n)) + /12(f (n) − f (1)) − /2 P2(x)f (x) dx k=1 1 1

2 1 where P2(x) = (x − bxc) − (x − bxc) + /6 and f has continuous ﬁrst and second derivatives on [1, n].

10 1.3 Euler-Maclaurin Summation 1 Preliminaries

If we want to reﬁne further, we want Z x P3(x) = 3 P2(t) dt + c 0 Z 1 where the coeﬃcient 3 is chosen so P3(x) is monic, and c is chosen such that P3(x) dx = 0. 0 3 3 2 1 This give us P3(x) = x − /2x + /2x + c, so

Z 1 4 3 2 1 3 2 x x x (x − 3/2x + 1/2 + c) dx = − + + cx 0 4 2 4 0 1 1 1 = − + + c = 0 4 2 4

⇒ c = 0! Z x 3 3 2 1 Thus we set P3(x) = 3 P2(t) dt = x − /2x + /2x. To integrate by parts again, let 0 00 u = f (x) dv = P2(x) dx

000 1 du = f (x) dx v = /3P3(x). Then 1 Z n 1 Z n 00 1 00 000 P2(x)f (x) dx = /6P3(x)f (x) − P3(x)f (x) dx. 2 1 6 1

Note that from 1 to n, P3(x) = 0. Thus we have the following theorem: Theorem 1.3.9 (Third Derivative Form of Euler-Maclaurin Summation Formula). n Z n Z n X 1 1 0 0 1 000 f(k) = f(x) dx + /2(f(1) + f(n)) + /12(f (n) − f (1)) + /6 P3(x)f (x) dx k=1 1 1

3 3 2 1 where P3(x) = x − /2x + /2x and f has continuous ﬁrst, second and third derivatives on [1, n]. 1 Example 1.3.10. Let’s apply this to f(x) = x2 Z k+1 0 2 00 6 000 24 000 Note that f (x) = − 3 , f (x) = 4 , f (x) = − 5 . Let’s look at P3(x)f (x) dx . x x x k

k k + 1

11 1.3 Euler-Maclaurin Summation 1 Preliminaries

Then we have

Z k+1 24 Z k+1/2 24 Z k+1 P (x)f 000(x) dx ≤ P (x) dx − P (x) dx 3 5 3 5 3 k x k (k + 1) k+1/2 and 1 Z 1/2 x4 x3 x2 /2 1 1 1 1

P3(x) dx = − + = − + = . 0 4 2 4 0 64 16 16 64 Thus

k+1 Z 24 1 24 1 1 P (x)f 000(x) dx ≤ − since the integral from 0 to /2 3 5 5 is the same as from 1/2 to 1 k k 64 (k + 1) 64 3 1 1 = − . 8 k5 (k + 1)5

This becomes

Z ∞ ∞ 000 3 X 1 1 3 1 P3(x)f (x) dx ≤ − = 8 k5 (k + 1)5 8 i5 i k=i by telescoping sum. So let’s estimate:

∞ i−1 Z ∞ Z ∞ X 1 X 1 1 1 1 1 2 1 24P3(x) = + dx + + − dx. n2 n2 x2 2 i2 12 i3 6 x5 n=1 n=1 i i

1 1 1 Our error ≤ so let’s get within .5 × 10−6. Then ≤ 8 × 10−6 ⇒ i5 ≥ 1.25 × 105 16 i5 i5 which gives us i ≈ 12.

Let’s look at our error terms so far:

1 P1(x) = x − bxc − /2 2 1 P2(x) = (x − bxc) − (x − bxc) + /6 3 3 2 1 P3(x) = x − /2x + /2x.

Z x Z 1 In general, Pk(x) = k Pk−1(t) dt + bk, where bk is chosen such that Pk(x) dx = 0. 0 0

12 1.4 The Bernoulli Numbers 1 Preliminaries

1.4 The Bernoulli Numbers

For k ≥ 1, let Z x Bk(x) = k Bk−1(t) dt + bk 0 Z 1 where bk is chosen such that Bk(x) dx = 0. Then we have 0

B0(x) = 1 1 B1(x) = x − /2 2 1 B2(x) = x − x + /6 3 3 2 1 B3(x) = x − /2x + /2.

Bk(x) is known as the kth Bernoulli polynomial, and the sequence of bk terms are called the Bernoulli numbers.

k Proposition 1.4.1. Bk(x) = (−1) Bk(1 − x). Proof. If k = 1, then

1 B1(1 − x) = (1 − x) − /2 = 1/2 − x

= −B1(x)

k−1 so the base case holds. Now assume Bk−1(x) = (−1) Bk−1(1 − x). We have that Z x Bk(x) = k Bk−1(t) dt + bk 0 Z x k−1 = k (−1) Bk−1(1 − t) dt + bk. 0 Let u = 1 − t, so that du = −dt. Then

Z 1−x k Bk(x) = k (−1) Bk−1(u) du + bk 1 Z 1 k−1 = k (−1) Bk−1(u) du + bk. 1−x

Z 1 Z 1−x Z 1 Note that 0 = Bk−1(t) dt = Bk−1(t) dt + Bk−1(t) dt. Then we can substitute: 0 0 1−x

Z 1−x k k Bk(x) = k (−1) Bk−1(u) du + bk = (−1) Bk(1 − x). 0

13 1.4 The Bernoulli Numbers 1 Preliminaries

Note that Z 1 Bk(1) = k Bk−1(t) dt + bk = k(0) + bk = bk = Bk(0) 0

so Bk(1) = Bk(0). And if k is odd, then Bk(1) = −Bk(0) = −bk, but these also equal bk, hence bk = 0 if k is odd.

1 Proposition 1.4.2. For k ≥ 2, if k is even then Bk(x) = 0 for exactly one value in [0, /2]. 1 And if k is odd, Bk(x) = 0 iﬀ x = 0, /2 or 1. Proof. Let k = 2, then see graph. Now suppose k is odd and the above holds for k −1, which 1 1 is even. We know that Bk(0) = Bk ( /2) = 0. Suppose that Bk(c) = 0 for some c ∈ (0, /2). 1 By Rolle’s Theorem, since Bk(0) = Bk(c) = Bk ( /2) there must be an a and b such that 1 0 0 0 0 < a < c < b < /2 and Bk(a) = Bk(b) = 0. But Bk(x) = kBk−1(x) which is even. By 1 inductive hypothesis, there’s only value in [0, /2] such that Bk−1(x) = 0, a contradiction. 1 Thus for k odd, Bk(x) = 0 iﬀ x = 0, /2 (or 1 by extension). Now suppose k is even and the 1 hypothesis holds for k − 1. Suppose Bk(t1) = Bk(t2) = 0 for t1, t2 ∈ [0, /2] with t1 6= t2. By 0 Rolle’s Theorem, Bk(x) = kBk−1(x) has a zero in (t1, t2). But since k−1 is odd, Bk−1(x) 6= 0 1 on the interval (0, /2), contradiction our choice of t1, t2. Hence if k is even, Bk(x) = 0 for exactly one value between 0 and 1/2. Properties of Bernoulli Numbers

(1) bk = 0 if k is odd

1 (2) The critical points of Bk(x) are x = 0, /2, 1 if k is even, so bk is either a max or in on 1 [0, 1], and Bk ( /2) is the opposite

k k X k X k (3) b = b — in fact, B (x) = xrb k r k−r k r k−r r=0 r=0

1 1−k (4) |bk| ≥ |Bk(x)| on the interval [0, 1] if k is even, and Bk ( /2) = −(1 − 2 )bk for k 1 even, so |bk| − |Bk ( /2) | is very small

(5) |P2m+1(x)| ≤ (2m + 1)|b2m+1|

∞ x X xm (6) = b ex − 1 m m! m=0 Theorem 1.4.3 (General Form for Euler-Maclaurin Summation). If f has 2m+1 derivatives on [i, n],

n Z n m X X b2r (2r−1) (2r−1) f(k) = f(x) dx + 1/2(f(1) + f(n)) − f (n) − f (1) (2r)! k=i i r=1 Z n 1 (2m+1) + P2m+1(x)f (x) dx (2m + 1)! i

where P2m+1(x) = B2m+1(x − bxc).

14 2 Euler’s Work

2 Euler’s Work

2.1 On the Sums of Series of Reciprocals

In this ﬁrst section, we will follow the work of Leonhard Euler in his seminal paper On the Sums of Series of Reciprocals, published in 1735. The main result, which we prove in detail ∞ X 1 π2 twice, is the now-famous identity = . n2 6 n=1 We begin with some notation. Let s represent an arbitrary angle of the unit circle. Then y = sin(s) and x = cos(s). It is known that s3 s5 s7 y = s − + − + ... 3! 5! 7! which corresponds to the Maclarin series for sine. Note that since sin(s) is periodic, the above equation holds for inﬁnite values of s. We can transform the equation into the following: s s3 s5 s7 0 = 1 − + − + − ... y 3!y 5!y 7!y Pretend this is a polynomial; then it can be written two ways:

n n−1 n−2 P1(x) = x + an−1x + an−2x + ... + a1x + a0

= (x − b1)(x − b2)(x − b3) ··· (x − bn)

where b1, b2, . . . , bn are all roots of P1(x). Suppose instead we have

2 n P2(x) = 1 + a1x + a2x + ... + anx

with roots b1, b2, . . . , bn.

Claim. Then we can write P2 as x x x P2(x) = 1 − 1 − ··· 1 − . b1 b2 bn

Proof. If we evaluate P2 at one of its roots, bi, we have bi bi bi P2(bi) = 1 − 1 − ··· 1 − ··· = 0. b1 b2 bi =0

So the bi’s indeed satisfy their deﬁnition as roots. Thus our two expressions are degree n polynomials with the same roots, so they can only diﬀer by a factor of k. Plugging in x = 0, we can solve for k = 1. Thus our two expressions are equivalent:

n P2(x) = 1 + a1x + ... + anx x x x = 1 − 1 − ··· 1 − . b1 b2 bn

15 2.1 On the Sums of Series of Reciprocals 2 Euler’s Work

As a sidenote, if f(z) and g(z) are analytic (inﬁnitely diﬀerentiable) on a domain D ⊂ C, f(z) = 0 ⇔ g(z) = 0 for all z ∈ D, and if this holds, f(z) = kg(z). This is a result from complex analysis, which was not available to Euler at the time. However, his conclusion was correct.

s s3 Now consider f(s) = 1 − + − ... Then the roots of f(s) are all the angles y 3!y A, B, C, D, . . . such that

y = sin(A) = sin(B) = sin(C) = sin(D) = ...

and we can write s s s f(s) = 1 − 1 − 1 − ··· A B C Returning to our polynomial again, x x P2(x) = 1 − 1 − ··· = 1 + a1x + ... b1 b2

0 −x x then if we want a1, we must find all possible ways of getting x , e.g. a1 = − − ... b1 b2 s s3 s s s So for f(s) = 1 − + − ... = 1 − 1 − 1 − ··· , y 3!y A B C 1 1 1 1 = + + + ... y A B C For the second coefficient, we have 1 1 1 1 0 = + + + + ... AB AC BC AD where the denominators are all possible products of pairs of roots of f(s). For the third coefficient, 1 1 1 1 = − + + + ... 3!y ABC ABD ACD Let A be the smallest arc such that sin(A) = y. Then sin(A + 2πk) = y for all k ∈ Z. Thus we can replace each of the roots of f(s) with A + 2πk for some k ∈ Z: s s s s s3 1 − 1 − 1 − ··· = 1 − + − ... A π − A A + 2π y 3!y Then by the above, 1 1 1 1 = + + + ... y A π − A A + 2π 1 1 1 0 = + + + ... A(π − A) A(A + 2π) (π − A)(A + 2π) 1 −1 −1 −1 = + + + ... 3!y A(π − A)(A + 2π) A(π − A)(−A − π) (π − A)(A + 2π)(−A − π)

16 2.1 On the Sums of Series of Reciprocals 2 Euler’s Work

Now deﬁne

α = a + b + c + d + e + ... β = ab + ac + ad + bc + bd + ... γ = abc + abd + acd + bcd + ....

So alpha is the sum of single terms, β is the sum of all possible products of two terms, and γ is the sum of products of three terms.

Claim. a2 + b2 + c2 + ... = α2 − 2β.

Proof. For a + b = α, ab = β and

a2 + b2 = a2 + b2 + 2ab − 2ab = (a + b)2 − 2ab = α2 − 2β.

The rest of the proof can be shown by induction. Here Euler is creating symmetric polynomials.

Deﬁnition. A symmetric polynomial is a polynomial that is ﬁxed by all possible permu- tations on its variables.

There is only one degree-1 symmetric polynomial of n variables: x1 + x2 + x3 + ... + xn. Claim. a3 + b3 + c3 + ... = α3 − 3αβ + 3γ

Proof omitted.

Claim. a4 + b4 + c4 + ... = α4 − 4α2β + 4αγ + 2β2 − 4δ.

Proof omitted.

Let P = a + b + c + ... = α then Q = a2 + b2 + c2 + ... = α2 − 2β = P α − 2β R = a3 + b3 + c3 + ... = α3 − 3αβ + 3γ = Qα − P β + 3γ S,T, etc. follow from here.

Returning to our series in question, we have 1 1 1 1 = + + + ... y A π − A A + 2π

17 2.1 On the Sums of Series of Reciprocals 2 Euler’s Work

where A is the least angle such that y = sin(A). But this just gives us 1 = α y 0 = β −1 = γ 3!y 0 = δ etc.

Q 1 Since β = 0, Q = P α − 2β = P α and R = Qα + 3γ = − , and this holds for all values y 2y of y = sin(A). We will now choose y = 1, so A = π/2. All of our roots now come in equal 1 1 1 1 pairs: , , , ,... Then π/2 π/2 5π/2 5π/2

1 1 1 1 1 α = = + + + + ... 1 A π − A A − 2π −π − A 2 2 2 2 2 2 = + − − + + − ... π π 3π 3π 5π 5π 4 1 1 1 1 = 1 − + − + − ... . π 3 5 7 9 1 1 1 1 π So 1 − + − + − ... = . Note that this looks like a case of the Taylor series for 3 5 7 9 4 tan−1(x):

∞ X (−1)nx2n+1 tan−1(x) = 2n + 1 n=0 ∞ X (−1)n π tan−1(1) = = . 2n + 1 4 n=0 We can then write Q as

Q = a2 + b2 + c2 + d2 + ... 2 2 2 2 −22 −22 2 2 2 2 = + + + + + + ... π π 3π 3π 5π 5π 8 1 1 1 = 1 + + + + ... . π2 9 25 49

This gives us our ﬁrst important result:

∞ π2 1 1 1 X 1 = 1 + + + + ... = . 8 9 25 49 (2n + 1)2 n=0

18 2.1 On the Sums of Series of Reciprocals 2 Euler’s Work

∞ X 1 1 1 1 Now let z = = 1 + + + + ... Then to produce all the even terms, divide by 4: n2 4 9 16 n=1 z 1 1 1 1 1 = + + + + + ... 4 4 16 36 64 100

π2 z 4 So z − / just gives us back the odd terms, which we have shown equal 8 : z π2 ⇒ z − = 4 8 π2 ⇒ z = . 6 For an alternate proof, set y = 0 at the beginning. Then the roots of our equation will be ±π, ±2π, ±3π, . . ., giving us α = 0 and β = − 1/6. Thus Q = −2β = 1/3, and we can proceed to solve for ζ(2) as before:

1 1 2 1 2 −12 1 2 Q = = + + + + ... 3 π π 2π 2π 2 1 1 1 = 1 + + + + ... π2 4 9 16 ∞ π2 X 1 ⇒ = . 6 n2 n=1

∞ X 1 We can solve for other identities in the same way. For example, we ﬁnd that if y = n4 n=1 ∞ y X 1 then = , implying 16 (2n)4 n=1

∞ y X 1 π4 y − = = 16 (2n − 1)4 32 · 3 n=1 π4 ⇒ y = . 90 Although Euler did not provide a general formula for ζ(2n) in this paper, his methods here can be extended to show that for all n,

(−1)n+1b (2π)2n ζ(2n) = 2n 2(2n)!

where b2n is the Bernoulli number for k = 2n.

19 2.2 Newton’s Identities 2 Euler’s Work

2.2 Newton’s Identities

This section provides a brief review of Newton’s identities, which were available to Euler at the time he wrote On the Sums of Series of Reciprocals. In this paper, Euler used the notation seen in Section 3.1; here we will instead adopt a more modern notation for Newton’s identities. Let k, l ≥ 1 and m, r ≥ 0 be integers. Deﬁne

∞ X k tk = xn n=1 X sl = xi1 xi2 ··· xil

i1,...,il distinct

X m u(m, r) = xj0 xj1 xj2 ··· xjr j0,j1,...,jr distinct with indices ic and jd positive integers.

Lemma 2.2.1. Let k, l ≥ 2. Then tksl = u(k + 1, l − 1) + u(k, l). Proof. Let k, l ≥ 2. Then

∞ ! ! X k X tksl = xn xi1 xi2 ··· xil

n=1 i1,...,il distinct

k k k k = x1 + x2 + x3 + ... + xl + ... (x1x2 ··· xl + x1x2 ··· xl−1xl+1 + x1x2 ··· xl−1xl+2 + ...)

k+1 k+1 k+1 k+1 = x1 x2 ··· xl + x1x2 ··· xl + ... + x1x2 ··· xl + ... + x1 x2 ··· xl−1xl+1

k+1 k + x1x2 ··· xl−1xl+1 + ... + x1x2 ··· xl−1xl xl+1 + ...

k+1 k+1 k = x1 x2 ··· xl + x1x2 ··· xl + ... + x1x2 ··· xlxl+1 + ...

  ! X X = xk+1x ··· x + xk x ··· x  j0 j1 jl−1  i1 i2 il j0,...,jl−1 distinct i1,...il distinct

= u(k + 1, l − 1) + u(k, l).

Lemma 2.2.2. Let l ≥ 1. Then t1sl = u(2, l − 1) + (l + 1)sl+1.

20 2.2 Newton’s Identities 2 Euler’s Work

Proof. Consider

∞ ! ! X X t1sl = xn xil xi2 ··· xil

n=1 i1,...,il

= (x1 + x2 + x3 + ...)(x1x2x3 ··· xl + x1x2x3 ··· + x2x3 ··· xl+1 + ...)

2 2 2 = x1x2x3 ··· xl + x2x1x3 ··· xl + x3x1x2 ··· xl + ...   X 2 =  xj0 xj1 ··· xjl−1  + (x1x2x3 ··· xl+1 + x1x2x3 ··· xlxl+2 + ...) j0,j1,...,jl−1

= u(2, l − 1) + ((l + 1)(x1x2x3 ··· xl) + (l + 1)(x1x2x3 ··· xl−1xl+1) + ...)

! X = u(2, l − 1) + (l + 1) xj1 xj2 ··· xjl

j1,j2,...,jl

= u(2, l − 1) + (l + 1)sl+1.

These lemmas are used to prove the main theorem in this section, Newton’s Identities.

Theorem 2.2.3 (Newton’s Identities). Let k ≥ 1. Then

k−1 k tk − tk−1s1 + tk−2s2 − ... + (−1) t1sk−1 + (−1) ksk = 0.

Proof. First consider

∞ X k tk − tk−1s1 = xn − (u(k, 0) + u(k − 1, 1)) n=1 ∞ X X X = xk − xk − xk−1x n j0 j0 j1 n=1 distinct j0 j0,j1 distinct ∞ ∞ X X X = xk − xk − xk−1x n n j0 j1 n=1 n=1 j0,j1 distinct X = − xk−1x j0 j1 j0,j1 distinct = −u(k − 1, 1).

Next,

tk − tk−1s1 + tk−2s2 = −u(k − 1, 1) + (u(k − 1, 1) + u(k − 2, 2)) = u(k − 2, 2).

21 2.3 Euler’s Product Form 2 Euler’s Work

And

tk − tk−1s1 + tk−2s2 − tk−3s3 = u(k − 2, 2) − (u(k − 2, 2) + u(k − 3, 3)) = −u(k − 3, 3)

and so forth. Eventually we will obtain

k−2 k−2 tk − tk−1s1 + ... + (−1) t2sk−2 = (−1) u(2, k − 2) k−1 k−1 +(−1) t1sk−1 + (−1) t1sk−1 k−2 = (−1) [u(2, k − 1) − u(2, k − 1) − ksk] k−1 = (−1) ksk.

Finally, putting the last term in, we obtain

k−1 k k−1 k tk − ... + (−1) t1sk−1 + (−1) ksk = (−1) ksk + (−1) ksk k−1 = (−1) (ksk − ksk) = 0.

2.3 Euler’s Product Form

In this section we study the important Euler’s Product Form, which is usually written

∞ −1 X 1 Y 1 = 1 − . ms ps m=1 p prime

In Euler’s paper Various Observations about Inﬁnite Series (1737), he made use of the following notation 1 1 1 2n 3n 5n 7n 1 + + + + ... = · · · ··· , 2n 3n 4n 2n − 1 3n − 1 5n − 1 7n − 1 which of course is equivalent to our more modern notation for Euler’s Product Form. We will follow Euler’s proof below. Theorem 2.3.1 (Euler’s Product Form). 1 1 1 2n 3n 5n 7n 1 + + + + ... = · · · ··· 2n 3n 4n 2n − 1 3n − 1 5n − 1 7n − 1 In other words, ∞ −1 X 1 Y 1 = 1 − . ms ps m=1 p prime Proof. We will show that

n n n (2 − 1)(3 − 1) ··· (pi − 1) 1 1 1 n n n x = 1 + n + n + n + ... 2 3 ··· pi pi+1 pi+2 (pi+1pi+2)

22 2.3 Euler’s Product Form 2 Euler’s Work

The base case is easy. Now assume the property holds for all primes up to pi. Then

n n n (2 − 1)(3 − 1) ··· (pi − 1) 1 1 n n n x = 1 + n + n + ... 2 3 ··· pi pi+1 pi+2

n n n (2 − 1)(3 − 1) ··· (pi − 1) 1 1 1 1 n n n n x = n + 2n + n n + ... 2 3 ··· pi pi+1 pi+1 pi+1 pi+1pi+2 Thus n n n (2 − 1)(3 − 1) ··· (pi − 1) 1 1 1 n n n 1 − n x = 1 + n + n + ... 2 3 ··· pi pi+1 pi+1 pi+2

1 1 1 − n + 2n + n n + ... pi+1 pi+1 pi+1pi+2

n n n n (2 − 1)(3 − 1) ··· (pi − 1)(pi+1 − 1) 1 1 n n n n x = 1 + n + n + ... 2 3 ··· pi pi+1 pi+1 pi+2 By induction, the property holds for all p and the desired result follows. An alternate proof of Euler’s Product Form is given here. The proof utilizes the Fundamental Theorem of Arithmetic (1.1.1), which states that every natural number factors uniquely into the product of some primes. Proof.

−1 ! Y 1 Y 1 1 − = ps 1 − 1 p prime p prime ps

Y 1 1 1 = 1 + + + + ... by geometric sum ps p2s p3s p prime

∞ X 1 = by Fundamental Theorem of Arithmetic (1.1.1) ms m=1

= ζ(s).

X 1 X 1 In the next sequence, we will prove that diverges by showing >> log log(x). p p p prime p prime p≤x p≤x

Steps: (a) f(x) ∼ g(x) ⇒ log(f(x)) ∼ log(g(x))

23 2.3 Euler’s Product Form 2 Euler’s Work

−1 X 1 Y 1 (b) < 1 − n p n≤x p prime

(c) − log(1 − t) ≤ 2t for t ∈ [0, 1/2] X 1 X 1 (d) Recall that ∼ log(x) to show >> log log(x) n p n≤x p prime p≤x Proof of (a): Suppose f(x) ∼ g(x). Then

f(x) lim = 1 x→∞ g(x) f(x) lim log = log(1) = 0. x→∞ g(x)

So lim [log(f(x)) − log(g(x))] = 0 ⇒ lim log(f(x)) = lim log(g(x)) and we conclude x→∞ x→∞ x→∞ log(f(x)) lim = 1. x→∞ log(g(x))

Hence log(f(x)) ∼ log(g(x)).

Proof of (b): Consider

−1 ! ∞ n Y 1 Y 1 Y X 1 1 − = = . p 1 − 1 p p≤x p≤x p p≤x n=0

∞ n X 1 1 1 Y X 1 Since = 1 + + ... + , then contains all of the terms of the former, n 2 x p n≤x p≤x n=0 plus the product of the reciprocals of all primes less than p. Therefore it must be that

−1 X 1 Y 1 < 1 − . n p n≤x p≤x

Proof of (c): See graph of functions.

24 2.3 Euler’s Product Form 2 Euler’s Work

−1 X 1 Y 1 Proof of (d): By (b), < 1 − . Thus n p n≤x p≤x

! −1! X 1 Y 1 log < log 1 − n p n≤x p≤x

X 1 = − log 1 − p p≤x

X 2 1 ≤ by (c) with t = p p p≤x

X 1 = 2 . p p≤x ! X 1 X 1 X 1 So log << . And since ∼ log(x), (a) gives us n p n n≤x p≤x n≤x ! X 1 log ∼ log log(x). n n≤x

X 1 X 1 Hence log log(x) << which is suﬃcient to prove that diverges. p p p≤x p prime Next, we provide another proof of the divergence of the sum of reciprocals of primes. The terminology loosely follows another paper by Euler, but we also employ series and product notation. X 1 Theorem 2.3.2. diverges. p p prime

∞ X 1 X 1 Proof. Since < which converges, then ps ns p prime n=1

1 1 1 X 1 A = + + + ... = 2s 3s 5s ps p prime 1 1 1 X 1 B = + + + ... = 22s 32s 52s p2s p prime

25 2.3 Euler’s Product Form 2 Euler’s Work

etc. all converge for s > 1. Next, consider

∞ ! −1! X 1 Y 1 log = log 1 − by Euler’s Product Form ns ps n=1 p prime

X = − log 1 − p−s p prime

∞ X X p−sn = by power series for − log(1 − x) n p prime n=1

∞ X X p−sn = we can switch order of summation since n series is absolutely convergent n=1 p prime

∞ ! X 1 X = p−sn n n=1 p prime

= A + 1/2B + 1/3C + 1/4D + ...

Also note that ! Y ps A + 1/2B + 1/3C + ... = log ps − 1 p prime

X ps = log . ps − 1 p prime

Claim. 1/2B + 1/3C + 1/4D + ... converges. Proof. Consider the B term ﬁrst:

∞ 1 X −2s 1 X −2s 1/2B = p ≤ n 2 2 p prime n=2

1 Z ∞ 1 ≤ 2s dx 2 1 x

1 1 −1 ∞ = 2s−1 · 2 x 2s − 1 1

1 1 = . 2 2s − 1

26 2.3 Euler’s Product Form 2 Euler’s Work

1 X −ks 1 So 1/2B converges. Likewise, p ≤ by replacing 2 with k above. Also k k(ks − 1) p prime 1 1 note that ≤ . Then we have k(ks − 1) (k − 1)2

∞ X X −ns 1/2B + 1/3C + 1/4D + ... = p n=2 p prime ∞ X 1 ≤ n(ns − 1) n=2 ∞ X 1 ≤ (n − 1)2 n=2 ∞ X 1 π2 = = . n2 6 n=1

Hence 1/2B + 1/3C + ... converges as claimed.

+ Now, we showed that as s → 1 , A + 1/2B + 1/3C + ... diverges. And since we proved that X 1 for any s, 1/2B + 1/3C +... converges, it must be that A diverges. Thus diverges. p p prime

X 1 Our next goal is to show that − log log(x) < 15 for x suﬃciently large. Recall p p≤x that for s > 1, ∞ ! X 1 X 1 π2 0 < log − < < 2. n2 ps 6 n=1 p prime

∞ 1 X 1 1 Lemma 2.3.3. For s > 1, < < + 1. s − 1 ns s − 1 n=1 1 Proof. Let f(x) = . Then since f is monotone decreasing, the integral test gives us xs

∞ Z ∞ dx X 1 < . xs ns 1 n=1 We know Z ∞ −s+1 ∞ dx x −1 1 s = = = . 1 x −s + 1 1 −s + 1 s − 1 Likewise, we can split oﬀ the ﬁrst term of the series (since we can’t integrate → 0) to obtain

∞ X 1 Z ∞ dx 1 < + 1 = + 1. ns xs s − 1 n=1 1

27 2.3 Euler’s Product Form 2 Euler’s Work

Thus we have the desired bound: ∞ 1 X 1 1 < < + 1. s − 1 ns s − 1 n=1

Corollary 2.3.4. For s > 1, 1 < (s − 1)ζ(s) < s. Proof. Follows from Lemma 2.3.3.

X 1 1 Lemma 2.3.5. − log < 2 when s ∈ (0, 1/2). ps+1 s p prime Proof. As noted on the previous page,

∞ ! X 1 X 1 −2 < − log < 0. ps+1 ns+1 p prime n=1 Furthermore, Lemma 2.3.3 gives us ∞ X 1 3 1 < s < s + 1 < ns+1 2 n=1 ∞ ! X 1 3 ⇒ 0 = log(1) < log(s) + log < log < log(e) = 1. ns+1 2 n=1 And by adding the two inequalities together, we get

∞ ! ∞ ! X 1 X 1 X 1 −2 < + log(s) − log + log < 1 ps+1 ns ns p prime n=1 n=1 X 1 1 ⇒ −2 < − log < 1 < 2. ps+1 s p prime

X 1 1 Thus − log < 2. ps+1 s p prime The proof of the following theorem results from Euler and the Prime Harmonic Function, by Paul Pollack at UGA.

X 1 Theorem 2.3.6. − log log(x) < 15 for suﬃciently large x. p p≤x Proof. Let λ(t) be a bounded function on [0, 1], x > e4 ≈ 80. Deﬁne ! 1 1 −1 X log(x) F (λ; x) = 1 λ p . p log(x) p prime p Our road map is as follows:

28 2.3 Euler’s Product Form 2 Euler’s Work

X 1 (1) Pick λ such that F (λ ; x) = 0 0 p p≤x

(2) Find better λ’s that bound λ0 above and below

(3) Obtain bounds on F (λ0; x) First, deﬁne ( 1 1 t e ≤ t ≤ 1 λ0(t) = 1 0 0 ≤ t < e .

−1 1 We calculate where p log(x) = by e

−1 1 p log(x) = e 1 p log(x) = e 1 log(p) = 1 log(x) log(p) = log(x) p = x.

( 1 −1 p log(x) 0 < p ≤ x So λ0 p log(x) = Then we have 0 x < p. ! 1 1 −1 X log(x) F (λ0; x) = 1 λ0 p p log(x) p prime p

! ! 1 1 1 1 1 X log(x) X = 1 p + 1 · 0 p log(x) p log(x) p≤x p p>x p

X 1 = . p p≤x

Consider λ0(t); we will pick linear λU and λL to bound λ0 on the interval.

λL

1 λ0 λU

t 1 e 1

29 2.3 Euler’s Product Form 2 Euler’s Work

But ﬁrst we need to bound F (λ; x) when λ is linear. Suppose λ(t) = a + bt. Then ! 1 1 −1 X log(x) F (λ; x) = 1 λ p p log(x) p prime p

! ! X 1 b = 1 a + 1 1+ log(x) log(x) p prime p p

X 1 X 1 = a 1 + b 2 . 1+ log(x) 1+ log(x) p prime p p prime p

4 1 1 2 1 Notice that since x > e , log(x) < 4 and log(x) < 2 . So Lemma 2.3.5 gives us

1 1 X 1 − log log(x) < 2 letting s = 1+ log(x) log(x) p prime p

and 1 log(x) 2 X 2 − log < 2 letting s = . 1+ log(x) 2 log(x) p prime p Now we can multiply through by a and b to get X 1 −2|a| < a 1 − a log log(x) < 2|a| 1+ log(x) p prime p

X 1 log(x) −2|b| < b 2 − b log < 2|b|. 1+ log(x) 2 p prime p And we can add the inequalities: X 1 X 1 log(x) −2(|a| + |b|) < a 1 + b 2 − a log log(x) − b log < 2(|a| + |b|) 1+ log(x) 1+ log(x) 2 p prime p p prime p

−2(|a| + |b|) < F (λ; x) − (a + b) log log(x) + b log(2) < 2(|a| + |b|). Thus |F (λ; x) − (a + b) log log(x)| < 2(|a| + |b|) + |b| log(2) if λ is linear. Now we will pick e 1 λ (t) = −et + (e + 1) and λ (t) = t − by our graph on the previous page. Notice U L e − 1 e − 1 that both lines pass through (1, 1), so λU (1) = aU + bU = 1 and λL(1) = aL + bL = 1. Thus by the bound on linear λ’s,

|F (λU ; x) − (a + b) log log(x)| < 2(|a| + |b|) + |b| log(2)

⇒ |F (λU ; x) − log log(x)| < 2(|e + 1| + | − e|) + | − e| log(2) < 2(2e + 1) + e log(2) < 12.88 + 1.885 < 15

⇒ F (λU ; x) < log log(x) + 15.

30 2.4 The Prime Number Theorem 2 Euler’s Work

Likewise, the lower bound yields

|F (λL; x) − (a + b) log log(x)| < 2(|a| + |b|) + |b| log(2) −1 e e ⇒ |F (λL; x) − log log(x)| < 2 + + log(2) e − 1 e − 1 e − 1 e + 1 e < 2 + log(2) e − 1 e − 1 < 4.33 + 1.097 < 15.

Hence F (λL; x) > log log(x) − 15. Putting this together, we have

log log(x) − 15 < F (λL; x) ≤ F (λ0; x) ≤ F (λU ; x) < log log(x) + 15,

X 1 so |F (λ ; x) − log log(x)| < 15. But we determined that F (λ ; x) = so we can conclude 0 0 p p≤x that X 1 − log log(x) < 15. p p≤x

The main result of this theorem is X 1 ∼ log log(x). p p≤x

2.4 The Prime Number Theorem

X 1 In the Section 2.3, we proved that ∼ log log(x). Now deﬁne the “indicator function” p p≤x γ by ( 1 n prime γ(n) = 0 n composite for natural numbers n. Then X γ(n) X 1 = n p n≤x p≤x X γ(n) ⇒ ∼ log log(x). n n≤x

A natural question we can ask is: How often is γ(n) = 1? (i.e. How often is n prime?) Consider Z x log log(x) = f(t) dt 2

31 2.4 The Prime Number Theorem 2 Euler’s Work

for some function f(t). Then

d d Z x log log(x) = f(t) dt dx dx 2 1 = f(x). x log(x)

Thus we have X γ(n) Z x 1 ∼ dt n t log(t) n≤x 2 X γ(n) X 1 ⇒ ∼ . n n log(n) n≤x n≤x

1 γ(n) 1 This suggests that EV (γ(n)) = , and the ratio will converge to as n log(n) n n log(n) gets big. X Deﬁne π(x) to be number of primes p ≤ x. Then π(x) = γ(n). Furthermore, deﬁne n≤x Z x dt the logarithmic integral by Li(x) = . Gauss conjectured that π(x) = Li(x), and 2 log(t) while this is not stricly true, his intuition was correct that π(x) grows at about the same rate as the logarithmic integral (i.e. they are asymptotic). x Lemma 2.4.1. Li(x) ∼ . log(x) Proof. We will integrate by parts. Let 1 u = v = t log(t)

−1 du = dv = dt. t(log(t))2

Then Z x dt t x Z x t x 2 Z x dt Li(x) = = + 2 dt = − + 2 . 2 log(t) log(t) 2 2 t(log(t)) log(x) log(2) 2 (log(t)) To integrate by parts again, let 1 u = v = t (log(t))2

−2 du = dv = dt. t(log(t))3

32 2.4 The Prime Number Theorem 2 Euler’s Work

Then x 2 t x Z x 2t Li(x) = − + 2 + 3 dt log(x) log(2) (log(t)) 2 2 t(log(t))

x x Z x dt = + 2 + c + 2 3 log(x) (log(x)) 2 (log(t))

x 3x ≤ + + c. log(x) (log(x))2 x And clearly ≤ Li(x) so we have log(x) x x 3x ≤ Li(x) ≤ + + c log(x) log(x) (log(x))2

Li(x) 3 c log(x) ⇒ 1 ≤ x ≤ 1 + + . log(x) log(x) x

Thus as x → ∞ this becomes Li(x) x 1 ≤ x ≤ 1 + 0 + 0 ⇒ Li(x) ∼ . log(x) log(x)

The Prime Number Theorem is perhaps the most important results from analytic number theory. The theorem was proven separately by Hadamard and de LaValle-Poussin using Riemann’s work on the zeta function. x Theorem 2.4.2 (Prime Number Theorem). π(x) ∼ . log(x) On the previous page, we showed that x Li(x) ∼ log(x)

so one route to proving the Prime Number Theorem would be to show that π(x) ∼ Li(x), which is of course much harder. Simple calculus shows that x ≤ π(x) ≤ Li(x) log(x) ↑ closer and it turns out that

x log π(x) − ∼ 2 log |Li(x) − π(x)| . log(x)

33 2.4 The Prime Number Theorem 2 Euler’s Work

x √ Modern calculations also indicate that π(x) − ≈ x. log(x) Deﬁne the functions

π1(x) = #p ≤ x such that p ≡ 1 (mod 4)

π3(x) = #p ≤ x such that p ≡ 3 (mod 4).

Notice that 1+π1(x)+π3(x) = π(x), so at least one of these functions must diverge as x → ∞.

Questions:

Are π1(x) and π3(x) both inﬁnite?

Calculations show that π1(x) < π3(x) for many x. Does this hold for all x?

Dirichlet’s Theorem: Take any n, a ∈ Z with gcd(a, n) = 1. Then there are an inﬁnite number of primes p ≡ a (mod n).

Corollary 2.4.3. For any relatively prime integers a, n, πa,n(x) equals the number of primes p ≤ x such that p ≡ a (mod n) diverges to inﬁnity as x gets big.

A special case of this is that π1(x) and π3(x) both diverge.

Corollary 2.4.4. If (a, n) = 1 and (b, n) = 1, then πa,n(x) ∼ πb,n(x).

34 3 Complex Analysis

3 Complex Analysis

In this chapter we survey the basic results in complex analysis that will√ be useful in number theory applications. Recall the deﬁnition of the imaginary number i = −1.

Deﬁnition. A complex number is a number of the form z = x + iy for x, y ∈ R. The set of all complex numbers is denoted C.

These numbers lie on what is known as the complex plane, denoted C. y (x, y)

In this way we can view the real part x and the imaginary part y of x + iy separately. The set of all complex numbers is denoted C, and they form an algebraic ﬁeld under the operations

Addition: (x1, y1) + (x2, y2) = (x1 + x2, y1 + y2). Scaling: k(x, y) = (kx, ky) where k is a real scalar.

Multiplication: (x1, y1)(x2, y2) = (x1x2 − y1y2, x1y2 + x2y1). Note that this multiplication diﬀers from the usual multiplication on R, as in Euclidean geometry.

3.1 Arithmetic

For a complex number z = x + iy we will denote the real and imaginary parts by x = Re(z) and y = Im(z). As a vector space, C has the following special attributes for each vector (complex number).

Deﬁnition. For a complex number z = x + iy, the modulus or absolute value of z is |z| = px2 + y2 and the complex conjugate of z is z¯ = x − iy.

Note that |z| and |z¯| are always equal. Geometrically, the modulus represents the distance in the complex plane from the origin (0, 0) to (x, y).

Proposition 3.1.1. For z, w ∈ C, (i) |zw| = |z| |w|.

(ii) zw =z ¯w¯.

35 3.1 Arithmetic 3 Complex Analysis

Since C is a ﬁeld, there is also a notion of divisibility for complex numbers. In particular if x + iy, u + iv ∈ C and u + iv 6= 0, we deﬁne x + iy xu + yv + i(yu − xv) = . u + iv u2 + v2

x+iy One can check that this is the appropriate formula by multiplying and dividing u+iv by the conjugate u − iv. As in the xy-plane, there is a polar coordinate system for complex numbers: if z = x + iy −1 y then we set r = |z|, x = r cos θ and y = r sin θ where θ = tan x . This gives us z = |z|(cos θ + i sin θ).

Multiplication is compatible with polar representations, for if z = |z|(cos θ + i sin θ) and w = |w|(cos ψ + i sin ψ) we have

zw = |z| |w|(cos θ + i sin θ)(cos ψ + i sin ψ) = |z| |w|(cos θ cos ψ − sin θ sin ψ) + i(cos θ sin ψ + sin θ cos ψ) = |z| |w|(cos(θ + ψ) + i sin(θ + ψ)).

z |z| Likewise, w = |w| (cos(θ − ψ) + i sin(θ − ψ)). Taking powers of complex numbers, e.g. zn, is sometimes diﬃcult to compute, since multiplication isn’t quite as straightforward in the complex plane. However, there is a result which utilizes the polar representation of a complex number to simplify the expression. Theorem 3.1.2 (De Moivre’s Theorem). For all integers n, (cos θ + i sin θ)n = cos(nθ) + i sin(nθ). Proof. We prove this using induction on n. For the base case n = 1, we simply have

(cos θ + i sin θ)1 = cos θ + i sin θ.

Now assume De Moivre’s Theorem holds for n. Then we have

(cos θ + i sin θ)n+1 = (cos θ + i sin θ)n(cos θ + i sin θ) = (cos(nθ) + i sin(nθ))(cos θ + i sin θ) = (cos(nθ) cos θ − sin(nθ) sin θ) + i(sin θ cos(nθ) + cos θ sin(nθ)) = cos((n + 1)θ) + i sin((n + 1)θ).

Definition. When we write z = |z|(cos θ + i sin θ), the angle θ is called the argument of z, denoted arg z. We often want to restrict our attention to a single, canonical value of θ for any z. Thus we define the principal argument θ = Arg z, where −π ≤ θ ≤ π. Proposition 3.1.3. Arg(zw) = Arg z + Arg w, where these may differ by a multiple of 2π.

36 3.2 Functions and Limits 3 Complex Analysis

3π Example 3.1.4. Let z = −1 + i and w = i. Then zw = −1 − i, Arg(zw) = − 4 and 3π π 5π 3π Arg z + Arg w = + = ≡ − mod 2π. 4 2 4 4 Continuing with the geometric parallels between Euclidean space and the complex plane, we have the important triangle inequality for complex numbers:

|z + w| ≤ |z| + |w|.

There is also a related inequality, sometimes called the reverse triangle inequality:

|z| − |w| ≤ |z − w|. The original purpose of complex numbers was to compute roots of all polynomials, so it will be desirable to be able to compute roots of complex numbers. In other words, if w = |w|(cos ψ + i sin ψ), what is w1/n? Let z = w1/n, so that zn = w. Then using De Moivre’s Theorem (3.1.2) we have

|w|(cos ψ + i sin ψ) = (|z|(cos θ + i sin θ))n = |z|n(cos(nθ) + i sin(nθ)).

Solving for θ, we see that ψ + 2πk cos ψ = cos(nθ) =⇒ nθ = ψ + 2πk =⇒ θ = n for some integer k. Hence our expression for w1/n is ψ + 2πk ψ + 2πk z = w1/n = |w|1/n cos + i sin . n n

For the nth root of w, that is w1/n, this formula gives all possible roots. In fact there are n distinct roots; all others are repeated values. 2 p 2 2 Recall that the equation of a circle in R is (x − x0) + (y − y0) = r for r > 0. In the complex plane, this is expressed by |z − z0| = r.

3.2 Functions and Limits

In this section we introduce functions that have values in the complex plane.

Deﬁnition. A function of a complex variable z is a map f : D → C for some subset D ⊆ C, i.e. f assigns a complex number to each z ∈ D. Deﬁnition. The domain of a complex-valued function f is the set of all values z for which the function operates; this is usually denoted D. The range is all possible values of the function, denoted Im f or f(D).

Example 3.2.1. Let f(z) = z2. The domain of f is all of C, while the range of f is the closed upper half plane {z ∈ C | Im(z) ≥ 0}.

37 3.2 Functions and Limits 3 Complex Analysis

y y

f x x

1 Example 3.2.2. f(z) = z−1 has domain D = {z ∈ C | z 6= 1} and range f(D) = {z ∈ C | z 6= 0}.

Deﬁnition. A sequence is a complex-valued function whose domain is the set of positive integers, written (zn) = (z1, z2, z3,...) where each zi is a complex number.

Definition. A sequence (zn) is said to have a limit L if, given any ε > 0 there is some N ∈ such that |zn − L| < ε for all n ≥ N. In this case we write lim zn = L and say that N n→∞ (zn) converges to L. If no such L exists, then (zn) is said to diverge. The definitions of sequence and limit are nearly identical to their counterparts in real analysis. However, in the complex plane every number has a real and an imaginary part. The following proposition helps us relate the definition of a complex limit to its real and imaginary parts.

Proposition 3.2.3. Let zn = xn + iyn and z = x + iy. Then lim zn = z ⇐⇒ lim xn = x n→∞ n→∞ and lim yn = y. n→∞

Proof. ( =⇒ ) If lim zn = z then the inequalities |xn − x| ≤ |zn − z| and |yn − y| ≤ |zn − z| n→∞ directly imply that (xn) and (yn) converge to x and y, respectively. ( ⇒ = ) On the other hand, suppose (xn) → x and (yn) → y. If ε > 0 is given, we may ε ε choose N1 and N2 such that |xn − x| < 2 for all n ≥ N1 and |yn − y| < 2 for all n ≥ N2. Let N = max{N1,N2}. Then for all n ≥ N the triangle inequality gives us ε ε |z − z| ≤ |x − x| + |y − y| < + = ε. n n n 2 2

Hence (zn) converges to z = x + iy. As a result, we have

Corollary 3.2.4. If zn → z then |zn| → |z|. The converse to this is generally false. For example, the sequence |in| converges to 1 since |in| = |i|n = 1n = 1 for all n; however, in = (i, −1, −i, 1, i, −1,...) and this ﬂuctuates inﬁnitely often between these four values, so the sequence diverges.

Proposition 3.2.5. Suppose lim zn = z. Then n→∞

(i) For any complex scalar k 6= 0, lim kzn = kz. n→∞

38 3.2 Functions and Limits 3 Complex Analysis

1 1 (ii) If zn 6= 0 for any n and z 6= 0, then lim = . n→∞ zn z

Proof. (i) Let ε > 0 be given. By convergence of (zn) there exists a positive integer N such ε that |zn − z| < |k| . Then for all n ≥ N, ε |kz − kz| = |k| |z − z| < |k| = ε. n n |k|

Hence (kzn) → kz. |z| (ii) First we can choose an N1 such that |zn − z| < 2 for all n ≥ N1. Note that by the reverse triangle inequality, |z| |z| |z | ≥ |z| − |z − z| > |z| − = . n n 2 2

We use this to control the |zn| term in the calculations below. Next for any ε > 0 there is an |z|2ε N2 such that for all n ≥ N2, |zn − z| < 2 . Let N = max{N1,N2}. Then for any n ≥ N, 2 1 1 z − zn |zn − z| 2 1 2 |z| ε − = = ≤ |zn − z| < 2 = ε. zn z znz |zn| |z| |z| |z| |z| 2 Hence 1 → 1 . zn z This shows that limits of complex sequences behave as expected (by which we mean they behave as their counterparts do in the real case). We also have

Theorem 3.2.6. If (zn) converges to z and (wn) converges to w, then the sequence (znwn) converges to zw.

Deﬁnition. Given a function f(z) with domain D and a point z0 either in D or in the boundary ∂D of D, we say f has a limit at z0 if lim f(z) = L z→z0 for some L ∈ C. Explicitly, f(z) has limit L at z0 if for every ε > 0 there exists a δ > 0 such that 0 < |z − z0| < δ implies |f(z) − L| < ε.

Deﬁnition. f(z) is continuous at a point z0 in its domain if lim f(z) exists and it equals z→z0 f(z0). In particular, f(z) is continuous if for every ε > 0 there exists a δ > 0 such that if |z − z0| < δ then |f(z) − f(z0)| < ε. Example 3.2.7. The function f(z) = |z|2 is continuous on its domain C. For example, f(z) ε has limit 4 at z0 = 2i. To see this, let ε > 0 and deﬁne δ1 = 1, δ2 = 5 and δ = min{δ1, δ2}. Note that by the reverse triangle inequality, |z| ≤ |z − 2i| + |2i| < 1 + 2 = 3; we will use this below. Then if 0 < |z − 2i| < δ we have |f(z) − f(2i)| = ||z|2 − 4| = ||z| + 2| · ||z| − 2| = (|z| + 2)|z − 2i| ε < (3 + 2) = ε. 5 Hence lim f(z) = 4 as claimed. z→2i

39 3.2 Functions and Limits 3 Complex Analysis

z Example 3.2.8. Consider the function f(z) = where z = x + iy 6= 0 andz ¯ = x − iy, its z¯ complex conjugate. Does lim f(z) exist? Well consider this limit along two different paths z→0 in the complex plane: 0 + iy lim f(z) = = −1 (x,y)→(0,y) 0 − iy x + i0 lim f(z) = = 1. (x,y)→(x,0) x − i0 z Since these limits are different, the limit of the function must not exist. Hence is not z¯ continuous at z0 = 0. Definition. A function f(z) has a limit at infinity, denoted lim f(z) = L, if for any z→∞ ε > 0 there is a (large) number M such that |f(z) − L| < ε whenever |z| ≥ M. Note that there is no restriction on arg z; only |z| is required to be large.

1 Example 3.2.9. The family of functions f(z) = zm has a limit L = 0 as z → ∞ for all 1 m = 1, 2, 3,.... To see this, let ε > 0 and choose M = ε1/m . Then if |z| ≥ M, m m 1 1 1 1/m m = ≥ = (ε ) = ε. zm |z| M By properties of limits, we have

Proposition 3.2.10.

n 1) Every polynomial p(z) = a0 + a1z + ... + anz is continuous on the complex plane.

p(z) 2) If p(z) and q(z) are polynomials, then their quotient q(z) is continuous at all points such that q(z) 6= 0.

Every complex-valued function f(z) can be written as f(z) = u(z) + iv(z), where u and v are each real-valued functions. This allows us to view every complex function by its real and imaginary parts. It is easy to see that all of the results on continuity for functions of the real numbers now apply for complex-valued functions. In particular,

Proposition 3.2.11. Let f = u + iv be a complex-valued function. Then f is continuous at z0 if and only if u and v are both continuous at z0.

n X Definition. For complex numbers z1, z2,... their nth partial sum is zj = z1 + ... + zn. j=1 Definition. An infinite series of complex numbers is a limit of partial sums

∞ n X X zj = lim zj. n→∞ j=1 j=1

40 3.2 Functions and Limits 3 Complex Analysis

n X Deﬁnition. We say an inﬁnite series of partial sums sn = zj converges if s = lim sn n→∞ j=1 exists. Otherwise, the series diverges.

In the complex case, we can write each zj = xj + iyj so every inﬁnite series may be written as the sum of a real and imaginary series:

∞ ∞ ∞ X X X zj = xj + i yj. j=1 j=1 j=1 P P P As with functions, the series zj converges if and only if xj and yj converge. In other words, lim sn only converges when lim xn and lim yn both exist. n→∞ n→∞ n→∞

∞ ∞ ∞ X X X Deﬁnition. A series zj has absolute convergence if |zj| converges. If zj con- j=1 j=1 j=1 verges but the absolute series does not converge, we say the series converges conditionally.

∞ ∞ ∞ X P X Notice that if zj converges (absolutely) then both xj and yj converge (abso- j=1 j=1 j=1 lutely) as well. The triangle inequality for series looks like

∞ ∞ X X zj ≤ |zj|. j=1 j=1

Recall from single-variable calculus the exponential function ex. This function has many deﬁnitions, with the two most important being

xt ex = lim 1 + t→∞ t ∞ X xn and ex = . n! n=1 In complex analysis, we deﬁne

Deﬁnition. For z = x + iy, the complex exponential function ez is deﬁned by

ez = ex(cos y + i sin y).

The special case eit = cos t + i sin t is called Euler’s formula. Euler was the ﬁrst to realize the connection between the exponential function and sine and cosine. This amazing identity, called “the most remarkable formula in mathematics” by Feynman, has been around since 1748 and has far-reaching implications in many branches of mathematics and physics. The following proposition shows that this deﬁnition captures all of the nice properties of ex from the real case. We will see in a moment that in the complex plane, the exponential function has even deeper properties and an essential connection to the geometry of C.

41 3.2 Functions and Limits 3 Complex Analysis

Proposition 3.2.12. For complex numbers z and w,

(a) ez+w = ezew.

1 −z (b) ez = e . (c) ez+2πi = ez, that is, the complex exponential function is periodic with period 2πi.

(d) If z = x + iy, |ez| = ex and therefore |eiy| = 1.

(e) ez 6= 0 for any z ∈ C. Proof. (a) Let z = x + iy and w = x0 + iy0. Then

ez+w = e(x+x0)+i(y+y0) = ex+x0 (cos(y + y0) + i sin(y + y0)) = exex0 (cos y + i sin y)(cos y0 + i sin y0) = ezew

(the last part uses a trick similar to the one used in the proof of De Moivre’s Theorem (3.1.2)). (b) follows from (a) and trig properties. (c) follows directly from the definition of ez. (d) follows from the fact that for any θ, | cos θ + i sin θ| = 1. (e) By part (d), |ex+iy| = ex, and x is real so ex is always nonzero. Therefore |ez| 6= 0 which implies ez 6= 0. Note that part (c) of Proposition 3.2.12 implies that f(z) = ez is not a one-to-one function on the complex plane. This is unfortunate, since that was one of the nice attributes of ex in the real case, as it allowed us to define an inverse, the logarithm log x. We next show how to construct a partial solution to this problem. Let w = ex+iy. We seek a function F such that F (w) = x + iy and eF (x+iy) = x + iy. Note that since |w| = ex and these are real numbers, we have x = ln |w|. This allows us to define

Deﬁnition. The formal logarithm is written log z = ln |z| + i arg z.

This is not a function (meaning it is not well-defined), since arg z represents a set of values which differ by 2kπ for integers k. We remedy this by making branch cuts of the complex plane. This is done by taking a ray from the origin, say with angle θ and defining the branch (θ, θ + 2π] so that log z is well-defined on this domain. The most important branch is

Deﬁnition. Let Arg z denote the argument of z in the branch (−π, π]; this is called the principal branch. Then we deﬁne the principal logarithm by

Log z = ln |z| + i Arg z.

Proposition 3.2.13. On the principal branch, Log ez = eLog z = z.

42 3.2 Functions and Limits 3 Complex Analysis

Proof. Let z = x + iy with Arg z = θ ∈ (−π, π]. Then on one hand,

Log ez = ln |ez| + i Arg ez = ln ex + iy = x + iy = z

and on the other hand,

eLog z = eln |z|+i Arg z = eln |z|(cos θ + i sin θ) = |z|(cos θ + i sin θ) = z.

Note that these require that we restrict our attention to a single branch (it may not even be the principal branch) for the expressions to be well-defined. Recall that f(z) = u(z) + iv(z) is continuous if and only if u and v are continuous. Well Arg z has no limit at values along the negative real axis. Therefore Log z is not continuous at any point Re(z) ≤ 0. However, making a different branch cut allows us to define a function with different continuity. As in the real case, exponentials for bases other than e are permitted. They relate to the logarithm by az = ez log a where log a is defined on a fixed branch of the logarithm. The complex trigonometric functions are defined in terms of ez.

Deﬁnition. The complex cosine and complex sine functions are deﬁned by

1 iz −iz 1 iz −iz cos z = 2 (e + e ) and sin z = 2i (e − e ).

Note that the complex trig functions coincide with their real counterparts, for if x ∈ R we have

1 ix −ix 1 2 (e + e ) = 2 (cos x + i sin x + cos(−x) + i sin(−x)) 1 = 2 (cos x + i sin x + cos x − i sin x) = cos x 1 ix −ix 1 and 2i (e − e ) = 2i (cos x + i sin x − (cos(−x) + i sin(−x))) 1 = 2i (cos x + i sin x − cos x + i sin x) = sin x. The complex cosine and sine functions are also periodic, with period 2π like the real-valued cosine and sine. Using the fact that ez is periodic, we can write

1 i(z+2π) −i(z+2π) cos(z + 2π) = 2 (e + e ) 1 iz 2πi −iz −2πi = 2 (e e + e e ) 1 iz −iz = 2 (e + e ) = cos z 1 i(z+2π) −i(z+2π) and sin(z + 2π) = 2i (e − e ) 1 iz 2πi −iz −2πi = 2i )(e e − e e ) 1 iz −iz = 2i (e − e ) = sin z. Many other properties of the real trig functions carry over the complex case. Just to name a few,

43 3.3 Line Integrals 3 Complex Analysis

(a) cos(−z) = cos z and sin(−z) = − sin z

π π (b) sin z + 2 = cos z and cos z + 2 = − sin z (c) sin(z + w) = sin z cos w + cos z sin w (d) cos(z + w) = cos z cos w − sin z sin w (e) cos2 z + sin2 z = 1 (f) cos2 z − sin2 z = cos(2z) (g) When we deﬁne the derivative of a complex-valued function in Section 3.4, we will see that the derivatives of cos z and sin z are similar to the real case.

3.3 Line Integrals

If f :[a, b] → C is a complex-valued function which is continuous on some interval [a, b] where a, b ∈ R, then the integral of f over [a, b] is simply Z b Z b Z b f(t) dt = Re(f(t)) dt + i Im(f(t)) dt. a a a For functions that take on values over some region in the complex plane, we integrate over curves. Deﬁnition. Let f(z) be a complex-valued function which is continuous on some region D ⊆ C and let γ be a smooth curve contained in D that is parametrized by γ(t), a ≤ t ≤ b. Then the line integral of f over γ is Z Z b f(z) dz = f(γ(t))γ0(t) dt. γ a

a γ(t)

Remember that a curve is smooth if its ﬁrst derivative γ0(t) exists and is continuous on [a, b]. Since the curves are all functions on a real interval [a, b], we need not worry about complex derivatives yet; γ0(t) is just the ﬁrst derivative in the normal sense. Some important examples of parametrizations in the complex plane are

44 3.3 Line Integrals 3 Complex Analysis

Example 3.3.1. A curve γ is simple if γ(t1) 6= γ(t2) whenever a < t1 < t2 < b. In plain language, a simple curve does not intersect itself; it is an embedding of the interval [a, b] into C. The easiest simple curve to parametrize is a line:

z0 γ

If γ is the line between z0 and z1, then we parametrize it by γ(t) = z0 + t(z1 − z0) for 0 ≤ t ≤ 1. Example 3.3.2. A curve γ is closed if γ(a) = γ(b), i.e. it starts and ends in the same location. The canonical example of a simple closed curve is a circle: γ

it This is parametrized by γ(t) = z0 + re for 0 ≤ t ≤ 2π. Z Example 3.3.3. Let’s compute the line integral z2 dz over the line from (0, 0) to (2, 3) in γ the complex plane.

z1 = 2 + 3i γ

z0 = 0 + 0i

We parametrize the curve by γ(t) = 2t + 3it, 0 ≤ t ≤ 1. Then using the formula above, we compute Z Z 1 Z 1 z2 dz = γ(t)2γ0(t) dt = (2t + 3it)2(2 + 3i) dt γ 0 0 Z 1 Z 1 = (4t2 − 9t2 + 12it2)(2 + 3i) dt = (−5t2 + 12it2)(2 + 3i) dt 0 0 Z 1 1 2 2 46 3 3 1 46 = (−46t + 9it ) dt = − t + 3it 0 = − + 3i. 0 3 0 3

45 3.4 Diﬀerentiability 3 Complex Analysis

Example 3.3.4. Just as reversing the order of a and b in a real integral changes the integral by −1, one can reverse the orientation of a smooth curve γ to switch the sign of the line integral along γ. Let −γ denote the curve γ with orientation reversed. Then Z Z f(z) dz = − f(z) dz. −γ γ Deﬁnition. The length of a curve γ is given by the integral

Z b Z b |γ0(t)| dt = px0(t)2 + y0(t)2 dt a a where γ(t) = x(t) + iy(t), a ≤ t ≤ b is a parametrization of γ.

Example 3.3.5. Let γ be the unit circle, which has the parametrization γ(t) = eit, 0 ≤ t ≤ 2π. Let’s verify the circumference of the circle with the formula for the length of γ:

Z 2π Z 2π Z 2π |γ0(t)| dt = |ieit| dt = dt = 2π. 0 0 0 The next proposition contains some useful properties of the line integral.

Proposition 3.3.6. Suppose γ is a smooth curve and f and g are continuous, complex-valued functions on a domain containing γ. Z Z Z (a) (f(z) + g(z)) dz = f(z) dz + g(z) dz. γ γ γ Z Z (b) For any c ∈ C, cf(z) dz = c f(z) dz. γ γ (c) If τ is a curve whose initial point is the terminal point of γ, then γτ is deﬁned to be the curve obtained by following γ and then τ. The integral over γτ is given by Z Z Z f(z) dz = f(z) dz + f(z) dz. γτ γ τ

(d) f(z) dz ≤ max |f(z)| · length(γ). γ z∈γ

3.4 Diﬀerentiability z Recall that the function f(z) = is not continuous at z = 0. This points to the fact z¯ 0 that complex functions are somehow diﬀerent than their real brethren, and in particular the convergence of a function in C is much stronger than convergence in R.

46 3.4 Diﬀerentiability 3 Complex Analysis

Deﬁnition. The derivative of a complex function f(z) at a point z0 ∈ C is deﬁned by

0 f(z0 + h) − f(z0) f(z) − f(z0) f (z0) = lim = lim . h→0 h z→z0 z − z0

If these limits exist, we say f(z) is differentiable at z0. This definition is the same as in the real case, although as discussed above the notion of a limit is much stronger in C. In the complex world, we have a further notion of differentia- bility:

Definition. A complex function f(z) is holomorphic at z0 ∈ C if f(z) is differentiable on some open disk centered at z0. Functions which are holomorphic on the whole complex plane C are called entire. Example 3.4.1. Many familiar functions from real analysis have the same derivative in the complex plane. For example, f(z) = z2 has derivative 2z which may be confirmed by computing either of the above limits. In fact this holds for all z ∈ C so z2 is an entire function.

Example 3.4.2. Complex conjugation is not diﬀerentiable at any z0 ∈ C since z¯ − z¯ z − z z¯ lim 0 = lim 0 = lim z→z0 z − z0 z→z0 z − z0 z→0 z does not exist as we have seen.

Most of the nice properties of real derivatives carry over to the complex place.

Proposition 3.4.3. Let f and g be diﬀerentiable at z ∈ C. (a) (f(z) + g(z))0 = f 0(z) + g0(z).

(b) For any c ∈ C, (cf)0(z) = cf 0(z). (c) (fg)0(z) = f 0(z)g(z) + f(z)g0(z).

f(z)0 f 0(z)g(z) − f(z)g0(z) (d) If g(z) 6= 0 then = . g(z) g(z)2 (e) (zn)0 = nzn−1. In particular this means that polynomials are entire.

(f) If g is diﬀerentiable at f(z) then (g(f(z)))0 = g0(f(z))f 0(z).

The fundamental property in this section is a pair of equations called the Cauchy- Riemann Equations, which relate the derivative f 0(z) to the partial derivatives with respect to the real and imaginary parts of z.

47 3.4 Diﬀerentiability 3 Complex Analysis

Theorem 3.4.4 (Cauchy-Riemann Equations). Let f(z) = u(x, y) + iv(x, y) be a complex function which is continuous at z0 = x0 + iy0. Then f(z) is diﬀerentiable at z0 if and only ∂u ∂u ∂v ∂v if the partial derivatives ∂x , ∂y , ∂x and ∂y exist, are continuous and satisfy ∂u ∂v ∂u ∂v = and = − ∂x ∂y ∂y ∂x

on some neighborhood of z0.

Proof. ( =⇒ ) If f(z) is diﬀerentiable at z0 = x0 + iy0 then

0 f(z0 + h) − f(z0) f (z0) = lim . h→0 h

First consider approaching z along the line (x0 + h) + iy0: f((x + h) + iy ) − f(x + iy ) u(x + h, y ) + iv(x + h, y ) − u(x , y ) − iv(x , y ) lim 0 0 0 0 = lim 0 0 0 0 0 0 0 0 h→0 h h→0 h u(x + h, y ) − u(x , y ) v(x + h, y ) − v(x , y ) = lim 0 0 0 0 + i 0 0 0 0 h→0 h h ∂u ∂v = + i = f 0(z ). ∂x ∂x 0

Next, approach along x0 + i(y0 + h): f(x + i(y + h)) − f(x + iy ) u(x , y + h) + iv(x , y + h) − u(x , y ) − iv(x , y ) lim 0 0 0 0 = lim 0 0 0 0 0 0 0 0 ih→0 ih ih→0 ih u(x , y + h) − u(x , y ) v(x , y + h) − v(x , y ) = lim 0 0 0 0 + i 0 0 0 0 h→0 ih ih 1 ∂u ∂v ∂v ∂u = + = − i = f 0(z ). i ∂y ∂y ∂y ∂y 0

0 Setting these two expressions for f (z0) equal gives the result, since the real and imaginary parts of the resulting expression must be equal. ( ⇒ = ) The converse requires a little more care. We will show that f(z) is differentiable 0 ∂f ∂u ∂v at z0 with derivative f (z0) = ∂x (z0) = ∂x (z0) + i ∂x (z0). We first break up the difference quotient, using h = hx + ihy:

f(z + h) − f(z ) f(z + h) − f(z + h ) + f(z + h ) − f(z ) 0 0 = 0 0 x 0 x 0 h h f(z + h + ih ) − f(z + h ) f(z + h ) − f(z ) = 0 x y 0 x + 0 x 0 h h h f(z + h + ih ) − f(z + h ) h f(z + h ) − f(z ) = y · 0 x y 0 x + x · 0 x 0 . h hy h hx Elsewhere, we have ∂f h ∂f h ∂f (z ) = y · (z ) + x · (z ). ∂x 0 h ∂y 0 h ∂x 0

48 3.4 Diﬀerentiability 3 Complex Analysis

Now we subtract these two expressions and take a limit, which gives f(z0 + h) − f(z0) ∂f hy f(z0 + hx + ihy) − f(z0 + hx) ∂f lim − (z0) = lim − (z0) h→0 h ∂x h→0 h hy ∂y hx f(z0 + hx) − f(z0) ∂f + lim − (z0) . h→0 h hx ∂x

hx If we can show that the limits on the right are both 0, then we’re done. The ratios h and hy h are both bounded by the triangle inequality, so it suﬃces to prove the the expressions in parentheses tend to 0. The second term goes to 0 since by deﬁnition,

∂f f(z0 + hx) − f(z0) (z0) = lim . ∂x hx→0 hx

The other expression is more problematic, since it involves both hx and hy. However, the Mean Value Theorem from real analysis gives us real numbers 0 < a, b < 1 such that

u(x0 + hx, y0 + hy) − u(x0 + hx, y0) = uy(x0 + hx, y0 + ahy) hy

v(x0 + hx, y0 + hy) − v(x0 + hx, y0) and = vy(x0 + hx, y0 + bhy). hy Substituting these expressions into the ﬁrst term above gives us

f(z0 + hx + ihy) − f(z0 + hx) ∂f − (z0) = uy(x0 + hx, y0 + ahy) + ivy(x0 + hx, y0 + bhy) hy ∂y

− uy(x0, y0) − ivy(x0, y0)

= (uy(x0 + hx, y0 + ahy) − uy(x0, y0))

+ i(vy(x0 + hx, y0 + bhy) − vy(x0, y0)).

Finally, these two pieces each tend to 0 since uy and vy are assumed to be continuous at z0 = x0 + iy0. This ﬁnishes the proof. Example 3.4.5. Consider f(z) = Log z using the principal branch D as its domain. We may write this as

1 2 2 y f(z) = ln |z| + i Arg z = 2 ln(x + y ) + i arctan x . 1 2 2 y So one sees that u(x, y) = 2 ln(x + y ) and v(x, y) = arctan x . We calculate the partials: x y 1 −y u = v = − = x x2 + y2 x x2 y 2 x2 + y2 1 + x y 1 1 x u = v = = . y x2 + y2 y x y 2 x2 + y2 1 + x

Hence ux = vy and uy = −vx so f(z) satisﬁes the Cauchy-Riemann equations on D, meaning it is diﬀerentiable. Moreover, we can write its derivative as x y x − iy z¯ 1 f 0(z) = u + iv = − i = = = . x x x2 + y2 x2 + y2 x2 + y2 |z|2 |z|

49 3.4 Diﬀerentiability 3 Complex Analysis

∞ X 1 Example 3.4.6. The zeta function converges absolutely for Re(s) > 1. In fact, it is ns n=1 holomorphic when Re(s) > 1. There is another function that is an analytic continuation of ζ(s) on all of C r {1}. Note: sometimes functions have “functional equations”, e.g. − sin(−z) = sin(z). Suppose f(z) is analytic for Re(z) > a and f ∗(z) is an analytic continuation of f(z) on C, and f ∗(z) has a functional equation that relates values of f ∗(z) on Re(z) < a to values on Re(z) > a. Then this functional equation can give us information for f(z) on “bad” domains. Example 3.4.7. For the zeta function, we will prove that there is a function ξ(s) = g(s)ζ(s) that is analytic on Re(s) > 1, which has an analytic continuation

ξ∗(s) = ξ∗(1 − s).

evaluate ξ∗(s) using ξ∗(1−s) ξ(s) well-deﬁned no info

Re(s) = 1/2

Deﬁnition. A power series is an inﬁnite series of the form

∞ X n an(z − z0) . n=0

Such a series is said to be centered about z0. Example 3.4.8. Power series are really a generalization of a geometric series

∞ X zn n=0 1 centered about z = 0, where all the coeﬃcients are 1. This series converges to exactly 0 1 − r when |z| < 1. We will see that power series behave in similar ways, and when they converge, they converge to complex functions that we may be interested in.

∞ X n For a power series an(z − z0) we have three cases for convergence: n=0

50 3.4 Diﬀerentiability 3 Complex Analysis

(1) The series only converges at z = z0. In this case, the radius of convergence of the series is 0.

(2) The series converges for all z in a disc of ﬁnite radius R centered at z0.

(3) The series converges for all z ∈ C, in which case we say the series has an inﬁnite radius of convergence.

A power series with positive or inﬁnite radius of convergence represents a function that is holomorphic within the disc of convergence of the series. This is one of the most important facts in complex analysis, so we take a moment to formalize it here.

∞ X n Theorem 3.4.9. Suppose an(z − z0) has a positive or inﬁnite radius of convergence R. n=0 Then it represents a function f(z) which is holomorphic on D = {z ∈ C : |z − z0| < R}. Now that we know that power series are holomorphic (diﬀerentiable) on their discs of convergence, we can take derivatives.

∞ X n Theorem 3.4.10. Suppose an(z − z0) has a positive or inﬁnite radius of convergence n=0 R. Then its derivative is also a power series:

∞ 0 X n−1 f (z) = nan(z − z0) n=1 which has radius of convergence R.

This can be applied repeatedly to obtain the Taylor series expansion of f(z) about z0:

∞ (n) X f (z0) f(z) = (z − z )n. n! 0 n=0 Example 3.4.11. The Taylor series for the exponential function is

∞ X zn ez = . n! n=0 Using the formulas for cos z and sin z from Section 3.2, we can derive their Taylor series as well:

∞ X (−1)n cos z = (z − z )2n (2n)! 0 n=0 ∞ X (−1)n sin z = (z − z )2n+1. (2n + 1)! 0 n=0

51 3.5 Integration in the Complex Plane 3 Complex Analysis

3.5 Integration in the Complex Plane

We now arrive at a theorem of central importance in complex analysis. The statement of the theorem is simple, but as we will see, this result has far-reaching implications in the complex world. Theorem 3.5.1 (Cauchy’s Theorem). Let f(z) be a complex function that is holomorphic on domain D, and suppose γ is any piecewise smooth, simple, closed curve in D. Then Z f(z) dz = 0. γ Proof. By assumption f 0(z) is continuous on D and γ has interior Ω within D. We compute Z Z Z f(z) dz = (u + iv)(dx + i dy) = (u dx − v dy + i(v dx + u dy)) γ γ γ Z Z = (u dx − v dx) + i (v dx + u dy) γ γ ZZ ZZ = (−vx − uy) dxdy + i (ux − vy) dxdy by Green’s Theorem Ω Ω ZZ ZZ = (−vx + vx) dxdy + i (ux − ux) dxdy by Cauchy-Riemann equations Ω Ω = 0 + i0 = 0.

Some immediate consequences of Cauchy’s Theorem are

Corollary 3.5.2 (Independence of Path). If γ1 and γ2 are curves with the same initial and terminal points lying in a domain on which f(z) is holomorphic, then Z Z f(z) dz = f(z) dz. γ1 γ2

Corollary 3.5.3 (Deformation of Path). Suppose γ1 and γ2 are two simple, closed curves with the same orientation, with γ2 lying on the interior of γ1.

γ2

γ1

If f(z) is holomorphic on the region between γ1 and γ2 then Z Z f(z) dz = f(z) dz. γ1 γ2

52 3.5 Integration in the Complex Plane 3 Complex Analysis

Corollary 3.5.4 (Fundamental Theorem of Calculus). If f(z) is holomorphic on a simply- connected domain D, then there is a holomorphic function F satisfying Z F (z) = f(z) dz γ for any γ lying in D. Equivalently, F satisﬁes F 0(z) = f(z) on all of D.

Theorem 3.5.5 (Cauchy’s Integral Formula). Suppose f is holomorphic on a domain D and γ is a simple closed curve on D, with positive orientation and interior Ω. Then for all z ∈ Ω, 1 Z f(ζ) f(z) = dζ. 2πi γ ζ − z

C z0 D

Ω γ

Proof. Fix z ∈ Ω and let C be a circle with center z contained in Ω. Note that for any f(ζ) z ∈ D, is holomorphic on D {z}. By deformation of path, ζ − z r

1 Z f(ζ) 1 Z f(ζ) dζ = dζ. 2πi γ ζ − z 2πi C ζ − z

We parametrize C by z + reit for 0 ≤ t ≤ 2π and write

Z Z 2π it 1 f(ζ) 1 f(z + re ) it dζ = it ire dt 2πi C ζ − z 2πi 0 re 1 Z 2π = f(z + reit) dt. 2π 0 Now take the limit as r → 0. Since f(z) is continuous, we can bring the limit inside the integral: 1 Z 2π 1 Z 2π lim f(z + reit) dt = f(z) dt. r→0 2π 0 2π 0 Notice that f(z) doesn’t depend on t, so we can integrate this easily and see that it equals f(z). This proves the theorem.

53 3.5 Integration in the Complex Plane 3 Complex Analysis

The next theorem shows that Cauchy’s Integral Formula is intimately related to complex power series.

Theorem 3.5.6. Let f be holomorphic on a domain D and suppose z0 is a point in D such that the circle |z − z0| < R for some real R lies in D. Let γ be a simple closed curve lying within this circle and containing z0 on its interior. Then ∞ Z X k 1 f(ζ) f(z) = ak(z − z0) where ak = k+1 dζ 2πi (ζ − z0) k=0 γ

Proof. Let ∆ = {z : |z − z0| < R}. By deformation of path, it suﬃces to consider when γ is a circle. For a ﬁxed r < R, we take γ to be the positively-oriented circle γ : |z − z0| = r. By Cauchy’s Integral Formula (3.5.5), 1 Z f(ζ) f(z) = dζ 2πi γ ζ − z

for any z on the interior of γ. For any one of these z’s, let s = |z −z0| so that s < r. Consider 1 1 1 1 = = · . z−z0 ζ − z (ζ − z0) − (z − z0) ζ − z0 1 − ζ−z0 |z − z | s Note that 0 = < 1. This allows us to introduce the series as a convergent geometric |ζ − z0| r series: ∞ k 1 1 X z − z0 = . ζ − z ζ − z0 ζ − z0 k=0 Using this and the expression given by Cauchy’s integral formula above, we are able to write 1 Z f(ζ) f(z) = dζ 2πi γ ζ − z Z ∞ k 1 f(ζ) X z − z0 = dζ 2πi ζ − z0 ζ − z0 γ k=0 ∞ Z 1 X k f(ζ) = (z − z0) k+1 dζ. 2πi (ζ − z0) k=0 γ

Corollary 3.5.7. If f(z) is holomorphic on D, f has derivatives of all orders on D and each derivative is holomorphic on D. Proof. By Theorem 3.5.6, f(z) can be written as a power series with positive radius of convergence, ∞ Z X k 1 f(ζ) f(z) = ak(z − z0) with ak = k+1 dζ, 2πi (ζ − z0) k=0 γ

for some γ about z0. We will see below that we can differentiate (and antidifferentiate) power series, so f(z) is infinitely differentiable on the region of convergence of the power series.

54 3.5 Integration in the Complex Plane 3 Complex Analysis

Theorem 3.5.6 suggests a powerful connection between power series and holomorphic functions in the complex plane. In this section we prove that every power series represents a holomorphic function on its region of convergence and every holomorphic function has a power series representation on its domain. First, we need a converse to Cauchy’s Theorem (3.5.1). Theorem 3.5.8 (Morera’s Theorem). Suppose f(z) is continuous on a domain D and Z f(z) dz = 0 γ for all smooth, closed curves γ in D. Then f is holomorphic on D. Proof. We may assume D is connected; otherwise the proof can be repeated on each con- Z nected component of D. Fix z0 ∈ D and define F (z) = f(ζ) dζ where γ is any smooth γ curve connecting z0 and z. By independence of path, F (z) is well-defined for all z ∈ D. Since all closed curves γ give F = 0 and f(z) is continuous, it follows that F 0(z) = f(z), that is, F is an antiderivative of f. Then F (z) is holomorphic on D, which by Corollary 3.5.7 implies that f(z) is also holomorphic on D. We prove the first direction of the power series-holomorphic function connection below. ∞ X k Theorem 3.5.9. Suppose f(z) = ak(z − z0) has a positive radius of convergence R. k=0 Then f is a holomorphic function on the domain D = {z ∈ C : |z − z0| < R}. Proof. Given any closed curve γ in D, Z ∞ X k ak(z − z0) dz = 0 γ k=0 by continuity of the power series on its region of convergence. Then Morera’s Theorem says that f(z) is holomorphic on D. Now we know that power series are differentiable on their region of convergence. The next result says that we can differentiate power series term-by-term, just as in the real case. ∞ X k Theorem 3.5.10. Suppose f(z) = ak(z − z0) has positive radius of convergence R. k=0 Then f(z) is differentiable with

∞ 0 X k−1 f (z) = kak(z − z0) k=1 which also has radius of convergence R. We can repeatedly apply Theorem 3.5.10 to subsequent derivatives of f to obtain a statement of Taylor’s Theorem for complex functions:

55 3.5 Integration in the Complex Plane 3 Complex Analysis

∞ X k Theorem 3.5.11. Suppose f(z) = ak(z−z0) has a positive radius of convergence. Then k=0

f (k)(z ) a = 0 . k k! We now turn to the other connection between holomorphic functions and power series. Well actually, we have already proven (Corollary 3.5.7) that holomorphic functions have power series representations, which we recall here.

Theorem 3.5.12. Let f be holomorphic on a domain D. Then ∞ Z X k 1 f(ζ) f(z) = ak(z − z0) for ak = k+1 dζ 2πi (ζ − z0) k=0 γ

where z0 ∈ D and γ is a simple closed curve lying in D and containing z0 on its interior. We immediately obtain the following generalization of Cauchy’s integral formula (3.5.5).

Corollary 3.5.13. Suppose f is holomorphic on a domain D and γ is a simple closed curve in D, positively oriented and with interior Ω. Then for all z ∈ Ω and n ∈ N, Z (n) n! f(ζ) f (z) = n+1 dζ. 2πi γ (ζ − z) We now deﬁne what it means for a function to be analytic on a certain region in the complex plane.

Deﬁnition. A function f(z) that is continuous on a region D ⊆ C is analytic at z0 ∈ D if f equals its Taylor series expansion about z0 and f is analytic on D if it is analytic at every point in D.

The following theorem summarizes everything we have learned so far about holomorphic functions in the complex plane.

Theorem 3.5.14. For a complex function f(z) which is continuous on a domain D, the following are equivalent:

(1) f(z) is diﬀerentiable on some open disk centered at z0 ∈ D, that is, f is holomorphic at z0.

(2) The Taylor series expansion of f(z) about z0 converges to f(z) with positive radius of convergence, i.e. f is analytic.

(3) f(z) satisﬁes the Cauchy-Riemann equations on some neighborhood of z0. Z (4) f(z) dz = 0 for every simple closed curve γ inside D with z0 on its interior (Cauchy’s γ Theorem and Morera’s Theorem).

56 3.6 Singularities and the Residue Theorem 3 Complex Analysis

We conclude with a consequence of the generalized Cauchy’s integral formula to entire functions that are bounded. Theorem 3.5.15 (Liouville’s Theorem). If f(z) is entire and there exists a constant M such that |f(z)| ≤ M for all z ∈ C, then f is a constant function.

Proof. Let z0 ∈ C and take Cr to be the circle centered at z0 with radius r > 0. By Corollary 3.5.13, Z 0 1 f(ζ) f (z0) = 2 dζ. 2πi Cr (ζ − z0) it Parametrize the circle by Cr : z0 + re , 0 ≤ t ≤ 2π. Then

Z 2π it 0 1 f(z0 + re ) it f (z0) = 2 2it ire dt 2πi 0 r e Z 2π it 1 f(z0 + re ) = it dt. 2πr 0 e Taking the modulus of both sides and applying the triangle inequality for integrals, we have

0 As we take r → 0, this expression tends to 0 as well, showing |f (z0)| = 0. Since z0 was arbitrary, we have shown that f(z) is constant.

3.6 Singularities and the Residue Theorem

With Theorem 3.5.12, we saw that an analytic function can be written ∞ Z X k 1 f(ζ) f(z) = ak(z − z0) where ak = k+1 dζ 2πi (ζ − z0) k=0 γ for all z in its domain D. This is highly useful, but when f(z) is not analytic on a domain D we still want a way of representing f as a series. This motivates the introduction and application of Laurent series:

Deﬁnition. A Laurent series is a series expansion of a function f(z) about a point z0 not in the domain of f in terms of two inﬁnite power series, a positive and negative one:

∞ ∞ X k X −k X k f(z) = ak(z − z0) + bk(z − z0) = ck(z − z0) . k=0 k=1 k∈Z

57 3.6 Singularities and the Residue Theorem 3 Complex Analysis

Remark. A Laurent series converges if and only if both the positive and negative series converge. Absolute and uniform convergence are deﬁned analagously. Notice that any Taylor series is a Laurent series whose negative part vanishes.

We should take a moment to explicitly describe the region of convergence of a Laurent series. Suppose ∞ ∞ X k X k X −k ck(z − z0) = ak(z − z0) + bk(z − z0) . k∈Z k=0 k=1

The positive series has some radius convergence R1, that is, the series converges on the region 1 {z ∈ : |z − z0| < R1}. Similarly, the negative series is just a power series in so it C z−z0 has radius of convergence 1 , i.e. it converges when 1 < 1 . This can be written as the R2 |z−z0| R2 complement of a closed disk, {z ∈ C : |z − z0| > R2}. Thus we see that the Laurent series is convergent on an annular region {z ∈ C : R2 < |z − z0| < R1} (as long as R2 < R1). By Theorem 3.5.9, the Laurent series represents an analytic function f(z) on the region D = {z ∈ C : R2 < |z − z0| < R1}. This is made explicit in the next theorem.

Theorem 3.6.1. Suppose f is a holomorphic function on D = {z ∈ C : R1 < |z −z0| < R2}. Then f is equal to its Laurent series expansion about z0 which can be written

∞ ∞ X k X −k f(z) = ak(z − z0) + bk(z − z0) k=0 k=1 1 Z f(ζ) 1 Z f(ζ) where ak = k+1 dζ and bk = −k+1 dζ 2πi C2 (ζ − z0) 2πi C1 (ζ − z0)

for circles C1 and C2 centered at z0 with radii R1 and R2, respectively. Proof. Apply Cauchy’s Theorem (3.5.1) and related results to both series.

Remark. By the deﬁnition of their coeﬃcients in terms of the integrals above, Laurent series expansions are unique.

Laurent series give us a way to deal with ‘holes’ in the domain of a function which is otherwise holomorphic on the region. Such functions have a special name:

Definition. A complex function f(z) is meromorphic on a domain D if it is holomorphic on D r {z1, z2, . . . , zr} where r is finite. A singularity is the name we give to a ‘hole’ in the domain of a complex function. Below we describe the three different types of singularities a function may have.

Deﬁnition. If f(z) is holomorphic on the punctured disk D = {z ∈ C : 0 < |z − z0| < R} for some R > 0 (R may be inﬁnite) but not at z0 then z0 is called an isolated singularity of f. The three types of isolated singularities are

(a) z0 is a removable singularity if there is a function g which is holomorphic on the disk D ∪ {z0} = {z ∈ C : |z − z0| < R} such that f(z) = g(z) for all z ∈ D.

58 3.6 Singularities and the Residue Theorem 3 Complex Analysis

(b) z0 is a pole if lim |f(z)| = ∞. In particular, z0 is a pole of order m if z0 is z→z0 1 a root of f(z) with multiplicity m. Equivalently, m is the smallest integer such that m+1 lim (z − z0) f(z) = 0. z→z0

(c) z0 is an essential singularity if it is neither removable nor a pole. The isolated singularities of a function may be characterized in terms of Laurent series expansions of the function.

Proposition 3.6.2. Let z0 be an isolated singularity of f(z) and suppose f(z) has a Laurent series expansion ∞ ∞ X n X −n f(z) = an(z − z0) + bn(z − z0) n=0 n=1 in the region 0 < |z − z0| < R.

(a) z0 is a removable singularity if and only if bn = 0 for all n and there is a function g, ( f(z) z 6= z g(z) = 0 a0 z = z0,

which is analytic in |z − z0| < R.

(b) z0 is a pole of f(z) if and only if all but a ﬁnite number of the bn vanish. Speciﬁcally, if bn = 0 for all n > m then z0 is a pole of order m and f can be written

∞ bm bm−1 b1 X f(z) = + + ... + + a (z − z )n. (z − z )m (z − z )m−1 z − z n 0 0 0 0 n=0

(c) z0 is an essential singularity if and only if infinitely many of the bn are nonzero. We saw there is a connection between the coefficients of the negative part of the Lau- rent series of a function and contour integrals of the function about its singularities. The coefficient b1 in a Laurent series is of particular importance, so much so that it has a special name.

Deﬁnition. Let z0 be an isolated singularity of f(z). The residue of f at z0 is 1 Z Res(f; z0) := f(z) dz 2πi C where C : |z − z0| = r for some 0 < r < R, the radius of convergence of the Laurent series for f. This is in turn equal to the b1 coeﬃcient of the Laurent series. There is a nice formula for the residues of removable singularities and poles.

Proposition 3.6.3. Suppose z0 is a nonessential singularity of f(z).

59 3.6 Singularities and the Residue Theorem 3 Complex Analysis

(a) If z0 is a removable singularity, Res(f; z0) = 0.

(b) If z0 is a pole of order m, then

m−1 1 d m Res(f; z0) = lim m−1 (z − z0) f(z). (m − 1)! z→z0 dz

Proof. (a) follows from Cauchy’s Theorem (3.5.1), and (b) is a simple application of Taylor’s Theorem to the series ∞ m X n+m (z − z0) f(z) = cn(z − z0) . n=−m

The formula for Res(f; z0) follows from the identiﬁcation of the residue and b1.

Proposition 3.6.4. Suppose f and g are analytic on |z − z0| < r for some z0 ∈ C and 0 r > 0, and suppose g(z0) = 0 but g (z0) 6= 0. Then f f(z0) Res ; z0 = 0 . g g (z0)

Proof. Let g(z) have the following power series centered at z0 (by assumption the series has no c0 coeﬃcient):

∞ ∞ X k X k g(z) = ck(z − z0) = (z − z0) ak(z − z0) k=1 k=0

where ak = ck−1; call the analytic function represented by this new series h(z). Note that h(z0) = c1 6= 0, so f(z) f(z) = g(z) (z − z0)h(z) f and h is analytic at z0. Using the definition of residue in terms of the Laurent series coeffi- f f cients, the residue of g is equal to the constant term of the series for h (the n = −1 term of f f(z0) 0 the series for ). This is computed to be , but by the way we defined h, h(z0) = g (z0). g h(z0) Hence f f(z0) Res ; z0 = 0 . g g (z0)

We ﬁnally arrive at the central theorem in basic complex analysis: the Residue Theorem.

Theorem 3.6.5 (The Residue Theorem). Suppose f(z) is meromorphic on a region D; let z1, . . . , zn be the isolated singularties of f inside D. If γ is a piecewise smooth, positively oriented, simple closed curve lying in D that does not pass through any of the zi then

n Z X f(z) dz = 2πi Res(f; zi). γ i=1

60 3.6 Singularities and the Residue Theorem 3 Complex Analysis

Proof. Draw a positively-oriented circle Ci around each singularity zi such that zi is the only singularity of f on its interior. The case where n = 3 is illustrated below.

z2 z1

0 Then γ is contractible to a curve γ which connects the Ci together and otherwise contains no singularities on its interior. Such a contraction is shown in the next ﬁgure.

z z2 γ0 1

n Z Z X Z Then f(z) dz = f(z) dz + f(z) dz but by construction, f(z) is holomorphic on 0 γ γ i=1 Ci the interior of γ0, so by Cauchy’s Theorem (3.5.1) this part equals 0. Evaluate the remaining terms using the deﬁnition of residue to produce the main summation formula:

n n Z X Z X f(z) dz = f(z) dz = 2πi Res(f; zi). γ i=1 Ci i=1

61 4 The Zeta Function

4 The Zeta Function

4.1 The Functional Equation

Proposition 4.1.1. If ζ(s) has an analytic continuation to all of C with an isolated singularity at s = 1, then the pole at s = 1 has order 1.

∞ X 1 Proof. In the half-plane Re(s) > 1, we can deﬁne ζ(s) = in the usual way. Then we ns n=1 have ∞ Z ∞ 1 X 1 Z ∞ 1 dx ≤ ≤ 1 + dx xs ns xs 1 n=1 1

1 1 ≤ ζ(s) ≤ 1 + s − 1 s − 1

1 ≤ (s − 1)ζ(s) ≤ s, and by the Squeeze Theorem, lim (s − 1)ζ(s) = 1, so s→1+ s∈R lim(s − 1)ζ(s) = lim (s − 1)ζ(s) = 1. s→1 s→1+ s∈C s∈R Thus we can conclude that s = 1 is a pole of order 1. As a result, we can calculate the residue of ζ(s) at s = 1 by 1 Z g(z) Res(ζ; 1) = dz = g(1) = 1 2πi γ z − 1 by the characterization of simple poles, evaluated using the formula above. So the residue of the zeta function at s = 1 is 1.

To ﬁnd an analytic continuation of ζ(s), recall the function (from homework) Z ∞ I(s) = e−tts dt 0 which converges absolutely for Re(s) > −1. And I(s) = P(s), which is analytic everywhere on C except negative integers. Deﬁne the Gamma function Γ(s) = I(s − 1). Then we substitute t = nx to obtain Z ∞ Γ(s) = e−nx(nx)s−1n dx 0 Z ∞ Γ(s) −nx s−1 s = e x dx. n 0

62 4.1 The Functional Equation 4 The Zeta Function

Note here that ns appears, so we want to sum over all n to get our hands on the zeta function. Doing so yields ∞ ∞ X Γ(s) X Z ∞ = e−nxxs−1 dx. ns n=1 n=1 0 By Fubini’s Theorem, we can switch the summation and integral if the absolute value of the right side is ﬁnite. So consider

N X Z ∞ Z ∞ |e−nxxs−1| dx = |e−nxxs−1| dx since ﬁnite sums swap order n=1 0 0

∞ X = xRe(s)−1 e−nx n=1

e−x = xRe(s)−1 by geometric series 1 − e−x

xRe(s)−1 = . ex − 1 Finally,

∞ Z ∞ X Z ∞ xRe(s)−1 |e−nxxs−1| dx = dx ex − 1 0 n=1 0

Z 1 xRe(s)−1 Z ∞ xRe(s)−1 = x dx + x dx 0 e − 1 1 e − 1

Z 1 xRe(s)−1 Z ∞ 1 ≤ dx + xRe(s)−1e−x dx. 0 x 1 2

63 4.1 The Functional Equation 4 The Zeta Function

ex (The ﬁrst integral is due to ex − 1 ≥ 1 for 0 ≤ x ≤ 1, and the second is because ex − 1 > 2 for x ≥ 1.) Note that this is integrable, so we can swap the integral and summation above, giving: ∞ Z ∞ X Z ∞ xs−1 ζ(s)Γ(s) = e−nxxs−1 dx = dx ex − 1 0 n=1 0 by geometric series.

Z (−z)s dz Next, deﬁne F(s) = lim , where γε,δ is the given contour: ε,δ→0 z γε,δ e − 1 z

II I δ ε

III

We can rewrite (−z)s in pieces:  (e−πiz)s on part I  (−z)s = smooth,continuous on part II  (eπiz)s on part III.

For part I, the parametrization z = x + iε, δ0 ≤ x < ∞, x → ∞ makes the integral become

Z δ0 −πi s Z ∞ s−1 Z ∞ s−1 (e (x + iε)) dx −πis (x + iε) ε→0 −πis x x+iε = −e x+iε dx −−→ −e x dx. ∞ e − 1 x + iε δ0 e − 1 δ0 e − 1 Similarly, the parametrization z = x − iε, δ0 ≤ x < ∞, x → ∞ makes part III look like Z ∞ πi s Z ∞ s−1 Z ∞ s−1 (e (x − iε)) dx πis (x − iε) ε→0 πis x x−iε = e x−iε dx −−→ e x dx. δ0 e − 1 x − iε δ0 e − 1 δ0 e − 1 Now for part II, we want the integral to vanishe as δ → 0. The parametrization z = δeiθ, τ ≤ θ ≤ 2π − τ, where τ is arbitrarily small, gives us Z (−z)s dz Z 2π−τ (−δeiθ)s iδeiθ dθ = z δeiθ iθ II e − 1 z τ e − 1 δe

Z 2π−τ (−δeiθ)s = i dθ. δeiθ τ e − 1 On the whole path, we have the following bounds:

64 4.1 The Functional Equation 4 The Zeta Function

| − δeiθ|s = | − 1|s|δ|s|eiθ|s = δs

iθ |eδe − 1| = |eσ+iκ − 1| where δeiθ = σ + iκ ≥ |e−δ − 1| since eσ+iκ is closest to 1 when θ = π, giving σ = −δ, κ = 0 δ ≥ since δ is small. 2

Then the part II integral becomes Z 2π−τ (−δeiθ)s Z 2π−τ δs i dθ ≤ i dθ δeiθ δ τ e − 1 τ 2

Z 2π−τ = 2iδs−1 dθ τ

< 4πiδs−1 since τ > 0.

If Re(s) > 1, 4πiδs−1 → 0 as δ → 0. Putting everything together, we have Z (−z)s dz Z ∞ xs−1 Z ∞ xs−1 lim = lim −e−πis dx + eπis dx + 4πiδs−1 ε,δ→0 z δ→0 x x γε,δ e − 1 z δ e − 1 δ e − 1

Z ∞ s−1 πis −πis x = (e − e ) x dx + 0 0 e − 1

= 2i sin(πs)Γ(s − 1)ζ(s).

This holds for all Re(s) > 1. Next, solve for the zeta function: 1 Z (−z)s dz ζ(s) = z . 2i sin(πs)Γ(s) γ e − 1 z πs We proved for homework that sin(πs) = , so P(s)P(−s) 1 P(s)P(−s) P(−s) = = 2i sin(πs)P(s − 1) 2iπsP(s − 1) 2πi since P(s) = sP(s − 1). Thus the functional equation for the zeta function is: Γ(1 − s) Z (−z)s dz ζ(s) = z Re(s) > 1 2πi γ e − 1 z which is an analytic continuation to the entire complex plane minus s = 1. Note that P(−s) is deﬁned everywhere except positive integers. But ζ(s) is deﬁned at these points. More- over, the functional equation for ζ(s) covers the rest of the complex plane, namely Re(s) ≤ 1,

65 4.1 The Functional Equation 4 The Zeta Function

s 6= 1, so we have values for ζ(s) everywhere except s = 1. Since the functional equation is analytic around s = 1, we see that s = 1 is a simple pole.

What happens to the functional equation for Re(s) < 0? P(−s) (and Γ(1 − s)) are well-deﬁned, so we will examine the integral part. Consider the contour Dn:

2π (n + 1/2) poles

By the Residue Theorem,

Z (−z)s dz X (−z)s = 2πi Res ; α . ez − 1 z (ez − 1)z Dn poles α α∈Dn Since we cut out z = 0, the only poles occur when ez − 1 = 0 ⇒ ez = 1 ⇒ z = 2πik for integers 0 < |k| ≤ n. We calculate the residue at z = 2πik by

(−z)s Res ; 2πik = g(2πik) (ez − 1)z

(−z)s g(z) where = . Then apply L’Hˆopital’sRule: (ez − 1)z z − 2πik z − 2πik 1 lim = lim = 1. z→2πik ez − 1 z→2πik ez Thus we obtain  (−z)s (z − 2πik) z 6= 2πik  z g(z) = (e − 1)z (−2πik)s  = −(−2πik)s−1 z = 2πik. −2πik

66 4.1 The Functional Equation 4 The Zeta Function

Hence the residue at z = 2πik is −(−2πik)s−1. We can plug this into the integration formula, which gives us

n X (−z)s X 2πi Res ; α = −2πi (2πik)s−1 + (−2πik)s−1 (ez − 1)z α∈Dn k=1 n X = −2πi(2π)s−1 is−1 + (−i)s−1 ks−1 k=1 n π X 1 = −2πi(2π)s−1 2 sin s . 2 k1−s k=1 Consider as n → ∞,

Z (−z)s dz Z (−z)s dz Z (−z)s dz z = − z + z . Dn e − 1 z γn e − 1 z |z|=2π(n+1/2) e − 1 z

Z (−z)s dz Claim. As n → ∞, z −→ 0. |z|=2π(n+1/2) e − 1 z

z z 1 1 Proof. Consider e − 1 on |z| = 2π(n + 1/2). By work in class, |e − 1| ≥ ⇒ ≤ 2. 2 ez − 1 Also, s (−z) Re(s−1) = (2π(n + 1/2)) −→ 0 as n → ∞. z This gives us

Z s (−z) dz Re(s−1) 1 2 1 2 z ≤ 2π(n + / )2π ·2 · (2π(n + / )) |z|=2π(n+1/2) e − 1 z length of path Re(s) = 4π (2π(n + 1/2)) .

Re(s) And since Re(s) < 0, lim 4π (2π(n + 1/2)) = 0. n→∞ Z s s−1 π (−z) dz Hence as n → ∞, 2πi(2π) · 2 sin s ζ(1 − s) = z . Then the functional 2 γ e − 1 z equation for Re(s) < 0 looks like π ζ(s) = P(−s)(2π)s−1 · 2 sin s ζ(1 − s). 2 Both ζ(s) and its functional equation are analytic everywhere except s = 1. Since they are analytic continuations of each other, the functional equation will continue to match ζ(s) everywhere (except s = 1).

67 4.2 Finding the Zeros 4 The Zeta Function

4.2 Finding the Zeros

The formula (−1)n+1b (2π)2n ζ(2n) = 2n , 2(2n)! where b2n is a Bernoulli number, gives values for ζ(s) at positive, even integers. It also turns out that (−1)nb ζ(−n) = n+1 n + 1 holds for all negative integers (which was proven for homework). The functional equation gives us (−1)nb π ζ(−n) = n+1 = P(n)(2π)−(n+1) · 2 sin (−n) ζ(1 + n). n + 1 2 And we have  −1 n ≡ 1 (mod 4) π π  sin (−n) = − sin n = 0 n is even 2 2 1 n ≡ 3 (mod 4). When n = 0, both sides have zero factors so we can’t get any information about ζ(1 + n) (which is good). On the other hand, we can write

n+1 n+1 (−1) 2 (2π) b ζ(1 + n) = n+1 2(n + 1)!

n+1 π 2 since P(n) = n! and (−1) will give us the correct values of sin 2 (−n) by above.

What have we accomplished so far? P(−s) Z (−z)s dz (1) We showed that ζ(s) = z is analytic on C except for a simple pole 2πi γ e − 1 z at s = 1. π (2) For Re(s) < 0, the functional equation ζ(s) = P(−s)(2π)s−1 ·2 sin s ζ(1−s) holds 2 on the entire complex plane, except the pole at s = 1. (3) By Euler’s Product Form, ζ(s) 6= 0 when Re(s) > 1. (4) The functional equation tells us there are no nontrivial zeros when Re(s) < 0. (There are zeros at the negative even integers since P(−s) is deﬁned using Bernoulli numbers; these are called the trivial zeros of the zeta function.) (5) Any nontrivial zeros are found on the critical strip 0 < Re(s) < 1. Why is knowing about the zeros of the zeta function important? Well a major implication of Riemann’s paper is that x π(x) ∼ ⇔ ζ(s) 6= 0 when Re(s) = 1. log(x) In addition, they provide a route to proving the Prime Number Theorem (2.4.2).

68 4.3 Sketch of the Prime Number Theorem 4 The Zeta Function

1 The Riemann Hypothesis. Every nontrivial zero of ζ(s) lies on the critical line Re(s) = 2 . Implications of the Riemann Hypothesis: (1) This would conﬁrm the Prime Number Theorem (2.4.2). x √ (2) Moreover, π(x) = + O( x log(x)). log(x) (3) This essentially describes the possible sizes of the gap between two successive primes.

(4) Consider

1 Y = 1 − p−s ζ(s) p prime X (−1)k = where k = # primes dividing n ns n square-free ∞ X µ(n) = ns n=1 where µ(n) is the M¨obiusfunction deﬁned by ( (−1)k n is square-free and k = # primes dividing n µ(n) = 0 otherwise.

4.3 Sketch of the Prime Number Theorem

x π(x) log(x) Recall π(x) ∼ ⇔ lim = 1. Chebyshev introduced the function ψ(x) log(x) x→∞ x deﬁned by X ψ(x) = log(p). pm≤x Note that we can rewrite this as X log(x) X ψ(x) = log(p) = Λ(n) log(p) p≤x n≤x where Λ(n) is the Von Mangoldt function given by ( log(p) if n = pm Λ(n) = 0 otherwise.

Then we have X log(x) X log(x) X ψ(x) = log(p) ≤ log(p) = log(x) = π(x) log(x). log(p) log(p) p≤x p≤x p≤x

69 4.3 Sketch of the Prime Number Theorem 4 The Zeta Function

π(x) log(x) ψ(x) Theorem 4.3.1 (Chebyshev). lim sup = lim sup . x→∞ x x→∞ x Proof omitted.

ψ(x) Therefore if lim exists and is equal to 1, we have x→∞ x ψ(x) π(x) log(x) ≤ x x π(x) log(x) from above, so lim would equal 1 as well. This is as far as Chebyshev got. x→∞ x Recall Euler’s Product Form: −1 Y 1 X 1 ζ(s) = 1 − ⇒ log(ζ(s)) = − log 1 − ps ps p prime p prime ∞ X X 1 = + by Taylor series for log(1 − x). mpms p prime m=1 Deriving both sides gives

∞ ∞ ζ0(s) X X 1 −m log(p) X X log(p) X Λ(n) = = − = − . ζ(s) m pms pms ns p prime m=1 p prime m=1 n≤x −ζ0(s) Z ∞ ψ(x) Claim. = s s+1 dx. ζ(s) 1 x Proof. Consider Λ(n) = ψ(n) − ψ(n − 1). Then

N N X Λ(n) X ψ(n) − ψ(n − 1) ψ(1) = + ns ns 1 n=1 n=2 goes to 0

N−1 ψ(N) X 1 1 = − − ψ(n) N s (n + 1)s ns n=2

N−1 ψ(N) X Z n+1 s = + ψ(n) dx N s xs+1 n=2 n

ψ(N) Z N ψ(x) = s + s s+1 dx. N 2 x ψ(N) Chebyshev showed that ψ(x) = O(x) so −→ 0 as N → ∞ (if Re(s) > 1), and ψ(x) = 0 N s for 1 ≤ x ≤ 2. Thus we have proven the claim.

70 4.4 Generalized Riemann Hypothesis 4 The Zeta Function

By Mellin Inversion,

1 Z a+i∞ −ζ0(s) xs ψ(x) = ds for some a ∈ R, a > 1 2πi a−i∞ ζ(s) s which Von Mangoldt evaluated to be

X xρ ζ0(0) ψ(x) = x − − ρ ζ(0) ρ where ρ are all the zeros of ζ(s). This is as far as Von Mangoldt got.

Now consider ρ for the trivial zeros of ζ(s):

∞ ∞ X xρ X x−2n −1 X 1 1 1 = − = = log 1 − −→ 0 as x → ∞. ρ 2n 2 nx2n 2 x2 ρ n=1 n=1

It turns out that for nontrivial ρ,

1 X xρ lim = 0 ⇔ Re(ρ) < 1 for all ρ. x→∞ x ρ ρ

Thus if the Riemann Hypothesis holds, it would give the smallest possible error term for our ψ(x) approximation above. Once we have ψ(x) ∼ x, the PNT follows.

4.4 Generalized Riemann Hypothesis

Deﬁnition. A character χ is a map G → C∗ where G is a group and χ is a group homo- morphism such that

χ(1) = 1

χ(ab) = χ(a)χ(b).

× For our purposes, let G = Zn , the multiplicative group of integers modulo n. We extend these characters to functions on the integers in the following way.

Deﬁnition. The Dirichlet character is given by χ : Z → C where for a ﬁxed n ∈ Z+, χ(a) = χ(a+n) for all a ∈ Z. Then n is called the modulus of χ, and we have the following properties:

χ(1) = 1

χ(ab) = χ(a)χ(b)

χ(c) = 0 if gcd(c, n) > 1

If χ(a) 6= 0 then |χ(a)| = 1, by which a is an nth root of unity.

71 4.4 Generalized Riemann Hypothesis 4 The Zeta Function

Example 4.4.1. The trivial character χ0 maps every integer to 1. Example 4.4.2. A principal character mod n maps x 7→ 1 iff gcd(x, n) = 1 and x 7→ 0 otherwise. For instance, a principal character mod 3 maps 1 7→ 1 4 7→ 1 2 7→ 1 5 7→ 1 3 7→ 0 6 7→ 0 Definition. For a Dirichlet character χ, the Dirichlet L-function is defined as

∞ X χ(n) L(s, χ) = . ns n=1

Note the special case ζ(s) = L(s, χ0). Moreover, since |χ(n)| ≤ 1 for all n, L(s, χ) converges absolutely for Re(s) > 1.

The product form: Since χ functions are multiplicative, we can write L-functions in Euler’s product form: −1 Y χ(p) L(s, χ) = 1 − . ps p prime

Analytic continuation: Similar to ζ(s), L(s, χ) can be extended to the entire complex plane with at most a simple pole at s = 1.

Functional equations: L-functions have functional equations of the form

L(1 − s, χ) = (∗∗) L(s, χ).

Dirichlet’s Theorem (Primes in Arithmetic Progression): Given integers a and b with gcd(a, b) = 1, the sequence (a + bn) represents inﬁnitely many primes.

Dirichlet’s Theorem (Strong Version): If gcd(a, p) = 1 then x π (x) ∼ a (p − 1) log(x) where πa(x) = # of primes q ≡ a (mod p) less than x.

These are proven using L-functions on the line Re(s) = 1. This brings us to what many would call the central topic in modern analytic number theory.

Generalized Riemann Hypothesis. For any Dirichlet L-function, L(s, χ) = 0 if and only 1 + if Re(s) = 2 or s = −2n for n ∈ Z .

72 4.4 Generalized Riemann Hypothesis 4 The Zeta Function

Several implications of GRH are:

∗ Tells us a lot about the deep inner structure of Zn. ∗ 2 Shows that Zn can be generated by less than 2(log(n)) elements. ∗ 6 Shows that Zp has a primitive root of size c(log(p)) for some uniform constant c.