L-FUNCTIONS AND THE RIEMANN HYPOTHESIS. 1 n s

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1 L-FUNCTIONS AND THE RIEMANN HYPOTHESIS KEITH CONRAD (.). The zeta-function and Dirichlet L-functions For real s >, the infinite series n converges b the integral test. We want to use this series when s is a complex number. First we describe a simple convergence test for infinite series of complex numbers and then we explain what n s means when s C. Definition.. An infinite series of complex numbers n z n is defined, like an infinite series of real numbers, as the limit of its partial sums: n s z n lim n N z n. N n For an infinite series of real numbers n x n where the terms are not all positive, the most important convergence test is the absolute convergence test: if the nonnegative series n x n converges then the original series n x n converges. The absolute convergence test also works for an infinite series of complex numbers. Theorem.2. If n z n converges in R then n z n converges in C Proof. Let s N N n z n. To prove the numbers s N converge in C, the idea is to prove the form a Cauch sequence. For N > M, N N (.2) s N s M z N z N. nm+ nm+ Since the series of real numbers n z n is assumed to converge, the sequence of its partial sums N n z n is a Cauch sequence in R, so the numbers N nm+ z N become arbitraril small when M and N are large enough. That means b (.2) that s N s M is arbitraril small when M and N are large enough, so the numbers s N are a Cauch sequence in C and thus converge in C. Remark.3. Ever rearrangement of the terms in an absolutel convergent series in R also converges to the same value, and the same propert is true for absolutel convergent series in C. The converse is false: n ( )n /n converges but n ( )n /n n /n does not converge.

2 2 KEITH CONRAD We put these ideas to work to define the exponential function on C. From calculus, e x n for all x R. We use the right side to define the exponential function on complex numbers. Definition.4. For s C, e s : s n n!. n This series converges b the absolute convergence test (Theorem.2): s n n! s n n!, n n which is finite since it is e s. The important algebraic propert e x e e x+ for x, R holds for the complex exponential function: e z e w e z+w for all z, w C. In particular, e z e z e z z e, so e z : the complex exponential function is never zero. The next theorem sas the absolute value (modulus) of e s depends on s onl b its real part. Theorem.5. If s C then e s e Re(s). Proof. For ever complex number z x + i, z 2 x zz. Therefore x n n! e s 2 e s e s e s e s e s+s e 2 Re(s) (e Re(s) ) 2, so e s e Re(s) since e s and e Re(s) are both positive numbers. In particular, e s has a constant absolute value along ever vertical line in C: e x+i e x for all x, R. Definition.6. For a > in R, set a s : e s ln a. This function a s with positive base a has properties similar to the function e s : a z a w a z+w (so a z for all z C) and a s a Re(s). We will not define complex powers of general complex numbers, onl complex powers of positive numbers. For a positive integer n we have n s n Re(s) for s C, so the series (.) with s C is absolutel convergent when Re(s) >, and thus it is convergent b Theorem.2. Definition.7. For s C with Re(s) >, the Riemann zeta-function at s is ζ(s) n There is a connection between ζ(s) and the prime numbers, first discovered b Euler in the 7s for real s. Theorem.8. For s C with Re(s) >, ζ(s) p n s. /p s /2 s /3 s /5 s. Proof. The idea is to expand each factor /( /p s ) into a geometric series and multipl together all those geometric series. What is a geometric series in C? It is n zn for a

3 L-FUNCTIONS AND THE RIEMANN HYPOTHESIS 3 complex number z, and just like a geometric series of real numbers, this series converges if and onl if z <, in which case z n z. n Taking z /p s for a prime p, we have /p s < when /p Re(s) <, which means Re(s) >. Therefore (.3) Re(s) > /p s + p s + p 2s + p 3s + When we multipl together this series for p 2 and p 3 we get ( + 2 s + 4 s + 8 ) ( s s + 9 s + ) 27 s s + 3 s + 4 s + 6 s + 8 s + where the right side is the sum of all /n s where n has onl prime factors 2 or 3 (or both). If we multipl together the series (.3) as p runs over all prime numbers, we need Re(s) > rather than just Re(s) > to justif the calculations and we get the sum of /n s where n runs over all positive integers b the uniqueness of prime factorization. The sum of all /n s is ζ(s), so we are done. The product in Theorem.8 is called the Euler product representation of ζ(s). Here is how it appeared in Euler s paper, where he wrote /( /p s ) as p s /(p s ). Each factor in the Euler product is nonzero, and from this ζ(s) when Re(s) >. This propert is not obvious if we onl use the series that defines ζ(s) (how can ou tell when an infinite series is nonzero?). In the 83s, Dirichlet introduced a generalization of the Riemann zeta-function, where the coefficients in the series for ζ(s) are not all. Definition.9. For m, (Z/mZ) denotes the invertible numbers modulo m. A function χ: (Z/mZ) C is called a Dirichlet character, or a Dirichlet character mod m when we want to specif the modulus, if it is multiplicative 2 : χ(ab) χ(a)χ(b) for all a and b in (Z/mZ). Example.. The character χ 4 on (Z/4Z) is defined b the rule {, if a mod 4, χ 4 (a mod 4), if a 3 mod 4. This is multiplicative since 3 2 mod 4 and χ 4 (3) 2 ( ) 2 χ 4 (3 2 ). 2 In the language of abstract algebra, we call this a group homomorphism.

4 4 KEITH CONRAD It does not make sense to define a character χ on (Z/4Z) b {, if a mod 4, χ(a mod 4) i, if a 3 mod 4. since this is not multiplicative: 3 2 mod 4 but χ(3) 2 i 2 while χ(). Example.. In (Z/5Z), ever number is a power of 2: 2 mod 5, 2 2 mod 5, mod 5, and mod 5. A power 2 k mod 5 depends on k modulo 4 since 2 k+4l 2 k 6 l 2 k mod 5 for all l Z. Therefore we can define a character χ 5 on (Z/5Z) having values in the 4th roots of unit in C b the rule χ 5 (2 k mod 5) i k. Here is an explicit formula for the values of this character:, if a mod 5, i, if a 2 mod 5, χ 5 (a) i, if a 3 mod 5,, if a 4 mod 5. A Dirichlet character mod m is a multiplicative function defined on the integers that are relativel prime to m. If χ is on all of (Z/mZ), we call χ the trivial Dirichlet character mod m and write χ m. There is a trivial Dirichlet character for each modulus. Each element of (Z/mZ) has finite order: a ϕ(m) mod m for all a in (Z/mZ). Therefore the values of a Dirichlet character χ on (Z/mZ) have to be roots of unit in C: a ϕ(m) mod m χ(a) ϕ(m) in C. We can consider χ as a function of period m on all integers, not just on integers relativel prime to m, b defining χ(n) if gcd(n, m) >. As a function on Z, χ remains multiplicative: χ(ab) χ(a)χ(b) for all integers a and b. Example.2. The character χ 4 from Example. is defined on all integers b, if a mod 4, χ 4 (a), if a 3 mod 4,, if a is even. Example.3. The character χ 5 from Example. is defined on all integers b, if a mod 5, i, if a 2 mod 5, χ 5 (a) i, if a 3 mod 5,, if a 4 mod 5,, if a is a multiple of 5. Example.4. The trivial character mod m is defined on all integers b {, if (n, m), m (n), if (n, m) >,

5 L-FUNCTIONS AND THE RIEMANN HYPOTHESIS 5 Definition.5. For a Dirichlet character χ, the Dirichlet L-function 3 of χ for Re(s) > is (.4) L(s, χ) χ(n) n s. n Wh do we sa Re(s) > in the definition of L(s, χ)? Since χ(n) is either a root of unit or, χ(n) or. Therefore χ(n) for all positive integers n, so the series (.4) converges for Re(s) > because it is absolutel convergent: n χ(n) n s n Re(s) < n n Example.6. The L-function of χ 4 is L(s, χ 4 ) n χ(n) n s converges n χ(n) n s converges. χ 4 (n) n s 3 s + 5 s 7 s + 9 s s + 3 s 5 s + when Re(s) >, with alternating signs in the numerators and powers of odd numbers in the denominators. Example.7. The L-function of χ 5 is L(s, χ 5 ) n when Re(s) >. χ 5 (n) n s + i 2 s i 3 s 4 s + 6 s + i 7 s i 8 s 9 s + s + Example.8. The L-function of the trivial character mod m is L(s, m ) m (n) n s n s, n (n,m) which looks like the zeta-function without terms at integers that have a factor in common with m. Since a Dirichlet character χ is multiplicative, L(s, χ) has an Euler product: if Re(s) >, L(s, χ) p χ(p)/p s χ(2)/2 s χ(3)/3 s χ(5)/5 s χ(7)/7 s. Proving this infinite product equals the series (.4) is similar to the proof that ζ(s) has an Euler product, and is left to the reader. Like the zeta-function, from the Euler product we have L(s, χ) when Re(s) >. Example.9. For Re(s) >, L(s, χ 4 ) p χ 4 (p)/p s + /3 s /5 s + /7 s + / s /3 s. The Euler factor at p 2 is since χ 4 (2). Although χ 4 (n) has alternating values,,,,... on odd numbers, it does not have alternating values on odd prime numbers! 3 It is not known wh Dirichlet denoted his functions with an L.

6 6 KEITH CONRAD Example.2. For Re(s) >, L(s, χ 5 ) p χ 5 (p)/p s i/2 s + i/3 s i/7 s / s + i/3 s. The Euler factor at p 5 is since χ 5 (5). Example.2. For m >, L(s, m ) p m (p)/p s (p,m) /p s. This is the Euler product for the zeta-function with the factors at primes dividing m removed. The importance of ζ(s) and the functions L(s, χ) in number theor is that important theorems about prime numbers depend on properties of these functions, but these properties involve the functions outside the region Re(s) > where the are initiall defined. Our next goal is to explain how to extend ζ(s) and L(s, χ) to the whole complex plane, except at s in the case of the zeta-function. 2. The Γ-function To define ζ(s) and L(s, χ) beond Re(s) > we will use an idea of Riemann, which involves the Γ-function, which is defined b an improper integral depending on a parameter. In 729, Euler essentiall discovered that n! x n e x dx for integers n. The right side makes sense even if n is not an integer: check as an exercise that the improper integral x t e x dx converges for all real t >. Since x s e x x Re(s) e x, the complex-valued integral x s e x dx makes sense for all s with Re(s) > (this is analogous to the absolute convergence test for infinite series of complex numbers). Definition 2.. For s C with Re(s) >, we define Γ(s) x s e x dx x. Notice the /x at the end of the integrand. That is wh the integral converges when x s /x x Re(s) has exponent greater than, which means Re(s) >. We have Γ(n + ) n! for all integers n. Remark 2.2. When we integrate a complex-valued function, there is no geometric interpretation of its value as an area, volume, and so on. Integrals of complex-valued functions could be defined b integrating the real and imaginar parts: if f(x) u(x) + iv(x) where u(x) and v(x) are real-valued functions then b a f(x) dx b a u(x) + i b a v(x) dx. However, it is better to define these integrals as limits of Riemann sums, just like in calculus, but using limits of complex numbers instead of limits of real numbers.

7 L-FUNCTIONS AND THE RIEMANN HYPOTHESIS 7 Using integration b parts, which is valid for complex-valued functions, check that (2.) Γ(s + ) sγ(s) when Re(s) >. Rewriting this formula as Γ(s + ) (2.2) Γ(s), s the right side makes sense when Re(s + ) >, meaning when Re(s) >, except at s. Therefore we use (2.2) to define Γ(s) for < Re(s) except when s. At s there is definitel a problem, since the numerator Γ(s + ) is Γ()! while the denominator is : we have to set Γ(). Now (2.2) is true for all s with Re(s) >. If Re(s) >, so Re(s + ) >, then (2.2) implies Γ(s + ) Γ(s + 2)/(s + ), so Re(s) > Γ(s) Γ(s + ) s Γ(s + 2)/(s + ) s Γ(s + 2) s(s + ). The expression at the end makes sense when Re(s) > 2 except at s and s. Therefore we can define Γ(s) for 2 < Re(s) b the formula Γ(s) Γ(s + 2) s(s + ), and that makes (2.2) true for all s with Re(s) > 2 except at s and s, where Γ() Γ( ). We can use (2.2) to extend Γ(s) consistentl to Re(s) > 3 b Γ(s) Γ(s + 3) s(s + )(s + 3) except at s,, and 2, where Γ(s), and more generall Γ(s) extends to Re(s) > k for each k Z + b (2.3) Γ(s) Γ(s + k) s(s + ) (s + k ). except at s,,..., (k ), where Γ(s). In this wa, Γ(s) can be defined on all of C except at and the negative integers, where Γ(s). Equation (2.) is now true for all s C. Although it is not obvious, it turns out that Γ(s) for all s C. Knowing the Γ- function never vanishes and that it becomes infinite at,, 2,... and nowhere else will be important later when we extend ζ(s) and L(s, χ) to s C and ask where these functions are. Remark 2.3. The original definition of Γ(s) for Re(s) > as an integral uses as the lower bound of integration. If the lower bound of integration were a positive number, the integral would make sense from the beginning on the whole complex plane: check as an exercise that for c >, the integral c x t e x dx/x converges for all t R, so c x s e x dx/x converges for all s C (analogue of absolute convergence test for complex series). 3. Extending the zeta-function to C Riemann used Γ(s) to extend the definition of ζ(s) beond the region Re(s) >. Here is part of what he showed in his onl paper on number theor. (Riemann s primar interests were in geometr, analsis, and mathematical phsics.)

8 8 KEITH CONRAD Theorem 3. (Riemann, 859). The function Z(s) π s/2 Γ ( s 2) ζ(s) can be extended from Re(s) > to the whole complex plane, except at s and s, and it satisfies the functional equation Z(s) Z( s). The function Z(s) is called the completed Riemann zeta-function. It is definitel not obvious wh π s/2 Γ(s/2) is a reasonable factor to use here! Nearl ears later, Tate s thesis (95) explained this, but it requires ideas beond the scope of these lectures (p-adic numbers and adeles). Proof. For Re(s) >, we rewrite π s/2 Γ(s/2) using a change of variables: Γ(s) x s e x dx ( s x x π s/2 Γ 2) s/2 dx π s/2 e x x t s/2 πt dt e t, where t x/π. (Observe for c > that d(cx)/(cx) dx/x. We will use this again below.) Now we multipl this b ζ(s). For Re(s) >, ( s π s/2 Γ ζ(s) 2) ( s ) π s/2 Γ 2 n s n n n n n t s/2 dt e πt ns t t s/2 dt (n 2 e πt ) s/2 t s/2 e πn2 d, where t/n 2. Interchanging the sum and integral (this can be justified), we get ( s (3.) π s/2 Γ ζ(s) 2) s/2 e πn2 d e πn2 s/2 d. n Set h() n e πn2 e π + e 4π + e 9π + for >. For large we have h() e π. The series h() runs over positive integers. The related series over all integers θ() n Z e πn2 + 2e π + 2e 4π + 2e 9π + + 2h() is a famous function in analsis, and Riemann knew a remarkable connection (found b Jacobi) between its values at and /: ( ) (3.2) θ θ().

9 L-FUNCTIONS AND THE RIEMANN HYPOTHESIS 9 A proof of this formula needs Fourier analsis and the Poisson summation formula; we do not discuss it here, but see numerical data in the table below. For large, θ() is nearl and θ(/) is nearl θ() θ(/) In terms of h(), (3.2) sas ( ) (3.3) h 2 ( ( + 2h()) ) + h(). 2 Returning to (3.), break up the integral over (, ) into integrals over (, ) and (, ), and then replace with / to write the integral over (, ) as an integral over (, ), noting d(/)/(/) d/: π s/2 Γ ( s 2) ζ(s) s/2 d h() s/2 d h() h ( + ) s/2 d + s/2 d h() s/2 d h(). Since the second integral has a positive lower bound of integration and h() e π for large, the second integral converges for all s C. Recall we said earlier that the integral defining the Γ-function would converge for all s if the lower bound of integration were positive (Remark 2.3); the same thing is happening here when the lower bound of integration is. B (3.3), the first integral above is ( + ) h() 2 s/2 d 2 ( ) and the second integral above converges for all s. Check as an exercise that d Re(a) > a a, so for Re(s) >, 2 ( ) s/2 d a s/2 d ( 2 (s )/2 ( 2 (s )/2 ) s/2 s/2 + ) d s s. We therefore have obtained the following formula for π s/2 Γ(s/2)ζ(s): ( s π s/2 s/2 d Γ ζ(s) h() 2) + (3.4) ( s)/2 d h() ( s)/2 d h() + s s h()( s/2 + ( s)/2 ) d s s.

10 KEITH CONRAD Up to this point, Re(s) >. The formula (3.4) makes sense for all s C other than and : the integral converges for all s and the terms /s and /( s) are meaningful when s {, }. Therefore we use (3.4) to extend the meaning of Z(s) : π s/2 Γ(s/2)ζ(s) to all of C {, }. The formula in (3.4) is unchanged when we replace s with s, so Z(s) Z( s). The most important step in Riemann s proof is using the formula (3.2). In a certain sense, (3.2) is equivalent to the functional equation for Z(s). Remark 3.2. We onl proved the functional equation Z(s) Z( s) after we found the formula (3.4) that let us extend Z(s) from Re(s) > to C {, }. It would be absurd to tr to prove Z(s) Z( s) if the domain for Z(s) does not include both s and s! Corollar 3.3. For Re(s) > and Re(s) <, Z(s). Proof. First we will treat the case Re(s) > and then we will use the functional equation Z(s) Z( s) to treat the case Re(s) <. If Re(s) > then Z(s) π s/2 Γ(s/2)ζ(s) and the three factors are each nonzero when Re(s) > : ζ(s) b the Euler product, Γ(s/2) because the Γ-function is finite and nonzero on Re(s) >, and π s/2 because π s/2 for all s C. If Re(s) < then Re( s) > so Z(s) Z( s) b what we first showed when the real part is greater than. Corollar 3.4. The Riemann zeta-function extends from Re(s) > to all s C except at s, where ζ(). We have ζ() /2 and ζ(s) when s is a negative even integer. Proof. For s C, the definition Z(s) π s/2 Γ(s/2)ζ(s) motivates us to define (3.5) ζ(s) πs/2 Z(s) Γ(s/2). This formula is consistent with the original definition of ζ(s) when Re(s) >. The function Z(s) makes sense everwhere except at and, while Γ(s/2) for s {, 2, 4,... } and Γ(s/2) for all s, so the above definition of ζ(s) makes sense everwhere except perhaps at s, s, and s { 2, 4, 6,...}. We now look more closel at these possibilities. If s { 2, 4, 6,...} then Z(s) b Corollar 3.3 while Γ(s/2), so it is natural to interpret (3.5) as saing ζ(s). What happens to (3.5) when s and s? When s, π s/2 /Γ(s/2) π /2 /Γ(/2) is finite and nonzero 4, while Z(), so ζ(). The case s is more subtle, since Z(s) and Γ(s/2) are both infinite at s. Recall sγ(s) Γ(s + ) for s C. This tells us s 2 Γ ( s 2 ) Γ ( s 2 + ), and when s we have Γ(s/2 + ) Γ(). Therefore we write ζ(s) πs/2 Z(s) Γ(s/2) πs/2 (s/2)z(s) (s/2)γ(s/2) πs/2 sz(s)/2 Γ(s/2 + ). 4 In fact this value is : Γ(/2) π.

11 B (3.4), sz(s) s Therefore as s, so we set ζ() /2. L-FUNCTIONS AND THE RIEMANN HYPOTHESIS h()( s/2 + ( s)/2 ) d s lim sz(s). s s ζ(s) πs/2 sz(s)/2 Γ(s/2 + ) π ( )/2 Γ() 2, From now on, we consider ζ(s) as a function on C that is finite everwhere except at s. Remember that its original definition as a series onl makes sense when Re(s) >. Remark 3.5. It is possible to write the functional equation Z(s) Z( s) as a functional equation relating ζ(s) and ζ( s), but it looks awful: ζ( s) ( πs ) π (2π)s sin Γ( s)ζ( s). 2 When ou think about s and s, have the picture below in mind: s and s are smmetric around the point /2, which is the midpoint of the line between them. We know ζ(s) for Re(s) > b the Euler product, and the formula ζ(s) πs/2 Z(s) Γ(s/2) implies ζ(s) for Re(s) < except at negative even integers. Negative even integers are called trivial zeros of ζ(s). Other zeros are called nontrivial and satisf Re(s). It is not that hard to show ζ(s) < when < s <, so there are no real nontrivial zeros. It is much harder to show ζ(s) along the whole line Re(s) : this is equivalent to the prime number theorem! Using the functional equation, it follows that ζ(s) along the whole line Re(s). One of the most famous problems in mathematics is about the location of nontrivial zeros of ζ(s). Riemann Hpothesis: Ever nontrivial zero of the zeta-function has real part 2. The first few nontrivial zeros of ζ(s) with positive imaginar part have the form /2 + it for the following approximate values of t: 4.347, 2.22, There are infinitel man nontrivial zeros of ζ(s), and unlike the trivial zeros, there is no simple formula for an of them.

12 2 KEITH CONRAD 4. Extending Dirichlet L-functions to C Following the ideas from the previous section, we will extend each L(s, χ), for nontrivial χ, from the region Re(s) > to the whole complex plane. A propert of a Dirichlet character, called its parit (being even or odd) will be important. Since (χ( )) 2 χ(( ) 2 ) χ(), we have χ( ) or χ( ). Definition 4.. A Dirichlet character χ is called even if χ( ) and it is called odd if χ( ). Since χ( a) χ( )χ(a), whether χ is even or odd as a character is the same as whether χ is even or odd as a function on Z. Example 4.2. Ever trivial character is even. Example 4.3. The characters χ 4 and χ 5 from Section are odd. Example 4.4. For an odd prime p, the Legendre smbol ( a p ), where ( ) { a, if a mod p, p, if a mod p for a mod p, is a character on (Z/pZ). Here is a table of its values when p 7. a ) ( a 7 For example, ( 2 7 ) since 2 9 mod 7. The multiplicativit of the Legendre smbol, ( ab p ) ( a p )( b p ), is not obvious. It sas in particular that a product of two nonsquares mod p is a square mod p, and this propert is often false when the modulus is composite. For example, the onl squares in (Z/5Z) {, 2, 4, 7, 8,, 3, 4 mod 5} are and 4, so the product of two nonsquares such as 2 and 7 is 4, which is also not a square. A basic theorem in number theor sas for odd primes p, mod p if and onl if p mod 4, so the Legendre smbol mod p is even if p mod 4 and odd if p 3 mod 4. Definition 4.5. For a nontrivial character χ mod m, its completed L-function is ( s Λ(s, χ) m s/2 π s/2 Γ L(s, χ) 2) if χ is even and ( ) s + Λ(s, χ) m (s+)/2 π (s+)/2 Γ L(s, χ) 2 if χ is odd. To write these in a unified notation, let δ {, } be the integer such that χ( ) ( ) δ, so δ for even χ and δ for odd χ. Then both formulas above sa ( ) s + δ Λ(s, χ) m (s+δ)/2 π (s+δ)/2 Γ L(s, χ). 2 Note the modulus m of χ plas a direct role in the definition of Λ(s, χ). Theorem 4.6. For a nontrivial Dirichlet character χ, Λ(s, χ) can be extended from Re(s) > to all of C, with finite values everwhere, and satisfies the functional equation Λ(s, χ) w χ Λ( s, χ). for a complex number w χ such that w χ.

13 L-FUNCTIONS AND THE RIEMANN HYPOTHESIS 3 There are few things to sa about Theorem 4.6 before we discuss its proof. While Z(s) is infinite at s and s, Λ(s, χ) for nontrivial χ is finite everwhere. This is an important difference between the case of trivial characters (whose L- functions are the zeta-function with at most finitel man Euler factors removed) and nontrivial characters. Unless χ χ, which means χ takes values in {±}, the functional equation does not relate Λ(s, χ) with Λ( s, χ), but with the completed L-function of the conjugate character χ. When χ takes values in {±} it turns out that w χ, so the functional equation sas Λ(s, χ) Λ( s, χ), which looks like the functional equation Z(s) Z( s). We are avoiding a technical issue in our statement of Theorem 4.6: the functional equation is not true for some nontrivial characters! It is onl true when χ has an additional propert called being primitive. Primitivit is not a propert we want to discuss here (ou can read about it in analtic number theor books), and in practice it is a mild condition: ever Dirichlet character that is not primitive can be associated to a primitive Dirichlet character, and their L-functions are closel related (the differ in a finite number of Euler factors), so properties of L-functions for nonprimitive characters often can be reduced to the case of L-functions of primitive characters. We will indicate in the proof of Theorem 4.6 where the missing propert of primitivit is needed. The characters χ 4, χ 5, and ( p ) are all primitive. ( s Proof. In the proof of Theorem 3., we saw for Re(s) > that π s/2 Γ 2) B similar reasoning, for Re(s) > ( ) s + (4.) π (s+)/2 Γ 2 x (s+)/2 π (s+)/2 e x dx x t (s+)/2 πt dt e t, t s/2 πt dt e t. where t x/π. We will use these formulas to write Λ(s, χ) for Re(s) > as an integral. When χ mod m is even and Re(s) >, ( s Λ(s, χ) m s/2 π s/2 Γ L(s, χ) 2) ( s ) χ(n) m s/2 π s/2 Γ 2 n s n n χ(n) χ(n) n n χ(n) where mt/n 2. Interchanging the sum and integral, (4.2) Λ(s, χ) n m χ(n) s/2 e πn2 /m d s/2 ts/2 dt e πt s n t m s/2 t s/2 dt (n 2 e πt ) s/2 t s/2 e πn2 /m d, χ(n)e πn2 /m n s/2 d.

14 4 KEITH CONRAD Looking at the series in (4.2), set for >, so h(, χ) n χ(n)e πn2 /m e π/m + χ(2)e 4π/m + χ(3)e 9π/m + (4.3) Λ(s, χ) s/2 d h(, χ) and for large, h(, χ) e π/m. Define a related series over all integers: θ(, χ) n Z χ(n)e πn2 /m n (χ(n) + χ( n))e πn2 /m n 2χ(n)e πn2 /m, where the last equation uses the fact that χ is even (if χ were odd then χ(n) + χ( n), so this definition of θ(, χ) would be ). Unlike θ(), the constant term of θ(, χ) is since χ(). We have (4.4) θ(, χ) 2e π/m + 2χ(2)e 4π/m + 2χ(3)e 9π/m + 2h(, χ). When χ mod m is odd and Re(s) >, ( ) s + Λ(s, χ) m (s+)/2 π (s+)/2 Γ L(s, χ) 2 ( ) s + χ(n) m (s+)/2 π (s+)/2 Γ 2 n s n n χ(n) m nχ(n) n nχ(n) n n nχ(n) (s+)/2 t(s+)/2 n s m (s+)/2 t (s+)/2 n s+ πt dt e t πt dt e t m (s+)/2 t (s+)/2 πt dt (n 2 ) (s+)/2 e t (s+)/2 e πn2 /m d, b (4.) where mt/n 2. Interchanging the sum and integral, (4.5) Λ(s, χ) n Looking at the series in (4.5), set nχ(n) s+ 2 e πn 2 /m d nχ(n)e πn2 /m s+ 2 n h(, χ) n nχ(n)e πn2 /m e π/m + 2χ(2)e 4π/m + 3χ(3)e 9π/m + d. for >, so (4.6) Λ(s, χ) h(, χ) (s+)/2 d

15 L-FUNCTIONS AND THE RIEMANN HYPOTHESIS 5 and for large, h(, χ) e π/m. Define a related series over all integers: θ(, χ) n Z nχ(n)e πn2 /m n (nχ(n) + ( n)χ( n))e πn2 /m n 2nχ(n)e πn2 /m, where the last equation uses the fact that χ is odd (if χ were even then nχ(n)+( n)χ( n) so this definition of θ(, χ) would be ). We have (4.7) θ(, χ) 2e π/m + 4χ(2)e 4π/m + 6χ(3)e 9π/m + 2h(, χ) for >. As in the case of even characters, θ(, χ) has constant term. In (3.2) we gave an important formula that connects θ() and θ(/). This formula has a generalization to θ(, χ), depending on if χ is even or odd: there is a complex number w χ with absolute value such that ( ) (4.8) θ, χ w χ θ(, χ) if χ is even and (4.9) θ ( ), χ w χ 3/2 θ(, χ) if χ is odd. The proof of these formulas involves Fourier analsis, like (3.2) does, and we omit the details. 5 Notice these formulas use χ on the right side. Without giving an exact formula for w χ, 6 we can read off one propert of these numbers of absolute value : (4.) w χ w χ. For example, b the definition of θ(, χ) whether χ is even or odd, we have θ(, χ) θ(, χ), so b appling complex conjugation to both sides of (4.8) or b replacing χ with χ everwhere in (4.8) we get two formulas when χ is even: θ(/, χ) w χ θ(, χ) and θ(/, χ) wχ θ(, χ). From this we get (4.) since all terms besides w χ and w χ on both sides of each formula are equal and θ(, χ) is not identicall zero. The proof of (4.) for odd χ is similar, using (4.9) instead of (4.8). Since h(, χ) 2θ(, χ) whether χ is even or odd (see (4.4) and (4.7)), formulas (4.8) and (4.9) turn into formulas using h(, χ) b dividing both sides of (4.8) and (4.9) b 2: ( ) (4.) h, χ w χ h(, χ) if χ is even and (4.2) h ( ), χ w χ 3/2 h(, χ) if χ is odd. We will use these to get a formula for Λ(s, χ) that makes sense at all s C. 5 This is exactl the place in the proof where we need χ to be primitive. For nonprimitive characters, (4.8) and (4.9) are actuall not true. 6 For primitive χ, wχ a mod m χ(a)e2πia/m / m if χ is even, and w χ a mod m χ(a)e2πia/m /(i m) if χ is odd.

16 6 KEITH CONRAD Case : χ is even. Start from (4.3) and break up the integral as + : Λ(s, χ) s/2 d h(, χ) s/2 d h(, χ) + s/2 d h(, χ). The second integral makes sense for all s C since the lower bound of integration is positive and since h(, χ) e π/m for large. Make the change of variables / in the first integral: d(/)/(/) d/, so Λ(s, χ) s/2 d h (/, χ) + w χ h(, χ) s/2 d + ( s)/2 d w χ h(, χ) + s/2 d h(, χ) s/2 d h(, χ) s/2 d h(, χ). b (4.) Now the first integral makes sense for all s C, so this provides a method of extending Λ(s, χ) to all s C (finite values everwhere). Combining the integrals into a single integral over (, ), (4.3) Λ(s, χ) ( h(, χ) s/2 + w χ h(, χ) ( s)/2) d. To prove the functional equation Λ(s, χ) w χ Λ( s, χ), replace χ with χ and s with s in (4.3) to get Λ( s, χ) B (4.), w χ w χ w χ w χ w χ 2, so w χ Λ( s, χ) ( h(, χ) ( s)/2 + w χ h(, χ) s/2) d ( w χ h(, χ) ( s)/2 + h(, χ) s/2) d b (4.3). This completes the proof of Theorem 4.6 when χ is even. 7 Case 2: χ is odd. B (4.6), Λ(s, χ) (s+)/2 d h(, χ) (s+)/2 d h(, χ) + Λ(s, χ) (s+)/2 d h(, χ). The second integral makes sense for all s C since the lower bound of integration is positive and since h(, χ) e π/m for large. B the change of variables / in the first 7 Strictl speaking, this proves the theorem when χ is even and primitive.

17 integral, Λ(s, χ) L-FUNCTIONS AND THE RIEMANN HYPOTHESIS 7 (s+)/2 d h (/, χ) + w χ 3/2 (s+)/2 d h(, χ) ( s+)/2 d w χ h(, χ) + + (s+)/2 d h(, χ) (s+)/2 d h(, χ) (s+)/2 d h(, χ). b (4.2) These integrals each make sense for all s C, so this provides a method of extending Λ(s, χ) to all s C (finite values everwhere, as before). Combining the integrals into a single integral over (, ), ( (4.4) Λ(s, χ) h(, χ) (s+)/2 + w χ h(, χ) ( s+)/2) d. To prove the functional equation Λ(s, χ) w χ Λ( s, χ), replace χ with χ and s with s in (4.4) to get ( Λ( s, χ) h(, χ) ( s+)/2 + w χ h(, χ) (s+)/2) d B (4.), w χ w χ w χ w χ w χ 2, so ( w χ Λ( s, χ) w χ h(, χ) ( s+)/2 + h(, χ) (s+)/2) d Λ(s, χ) b (4.4). This completes the proof of Theorem 4.6 when χ is odd. 8 Remark 4.7. While θ() has a nonzero constant term (at n ), the other functions θ(, χ) for nontrivial χ have constant term, and this is the reason wh Z(s) has value at s and while Λ(s, χ) is finite everwhere. Corollar 4.8. For Re(s) > and Re(s) <, Λ(s, χ). Proof. First we will treat the case Re(s) > and then we will use the functional equation for Λ(s, χ) to treat the case Re(s) <. If Re(s) >, Λ(s, χ) is m s/2 π s/2 Γ(s/2)L(s, χ) or m (s+)/2 π (s+)/2 Γ((s + )/2)L(s, χ) depending on whether χ is even or odd, and all the factors are nonzero: L(s, χ) b the Euler product, Γ(s/2) and Γ((s + )/2) are not because the Γ-function is finite and nonzero on Re(s) >, and exponential functions are nonzero everwhere. The exact same reasoning shows Λ(s, χ) when Re(s) >. If Re(s) < then Re( s) > and Λ(s, χ) w χ Λ( s, χ) b the functional equation, so the right side is not b what we showed about the completed L-function of χ. Corollar 4.9. For nontrivial χ, L(s, χ) extends from Re(s) > to all s C. If χ is even then L(s, χ) when s is or a negative even integer, and if χ is odd then L(s, χ) when s is a negative odd integer. Notice the contrast with ζ(s): ζ() /2 while the corollar is saing L(, χ) if χ is an even nontrivial character. 9 8 Strictl speaking, this proves the theorem when χ is odd and primitive. 9 When a character is not primitive, Corollar 4.9 is still true for its L-function.

18 8 KEITH CONRAD Proof. For all s C, the definition of Λ(s, χ) motivates us to define (4.5) L(s, χ) m s/2 π s/2 Λ(s, χ) Γ(s/2) if χ is even and (4.6) L(s, χ) m (s+)/2 π (s+)/2 Λ(s, χ) Γ((s + )/2) if χ is odd. These are consistent with the original definition of L(s, χ) when Re(s) >. The function Λ(s, χ) makes sense everwhere (finite values) on C, while Γ(s/2) for s {, 2, 4,... } and Γ((s + )/2) for s {, 3, 5,... }, so the above definition of L(s, χ) makes sense everwhere except perhaps at and negative even integers if χ is even, and except for negative odd integers if χ is odd. What happens in these cases? If χ is even and s { 2, 4, 6,...} then Λ(s, χ) b Corollar 4.8 while Γ(s/2), so it is natural to interpret (4.5) as saing L(s, χ). If χ is odd and s {, 3, 5,...} then Λ(s, χ) b Corollar 4.8 while Γ((s+)/2), so it is natural to interpret (4.6) as saing L(s, χ). What happens to (4.5) when s? The numerator is Λ(, χ) and the denominator is Γ(). Using the functional equation, Λ(, χ) w χ Λ(, χ), and a hard theorem in analtic number theor sas the completed L-function of a nontrivial Dirichlet character is in C at s. Therefore Λ(, χ) C, so L(, χ) using (4.5). From now on, we consider L(s, χ) for nontrivial χ as a function on C that is finite everwhere. Remark 4.. It is possible to write the functional equation Λ(s, χ) w χ Λ( s, χ) as an ugl functional equation for L(s, χ): ( /2 s (2π)s πs ) L(s, χ) w χ m sin Γ( s)l( s, χ) π 2 for even χ and for odd χ. ( /2 s (2π)s πs ) L(s, χ) w χ m cos Γ( s)l( s, χ) π 2 The zeros of L(s, χ) at integers are called trivial and all other zeros of L(s, χ) are called nontrivial. 2 Generalized Riemann Hpothesis: Ever nontrivial zero of L(s, χ) whose real part is strictl between and has real part 2. Example 4.. The first few nontrivial zeros of L(s, χ 4 ) with positive imaginar part have the form /2 + it for the following approximate values of t: 6.29,.2437, This is a contrast to the case of the completed zeta-function, where Z(). Such functional equations are onl true for primitive χ. 2 There ma be additional zeros on the imaginar axis if χ is not primitive and these are considered trivial zeros also.

19 L-FUNCTIONS AND THE RIEMANN HYPOTHESIS 9 Example 4.2. The first few nontrivial zeros of L(s, χ 5 ) with positive imaginar part have the form /2 + it for the following approximate values of t: 6.835, , For ever nontrivial Dirichlet character χ, there are infinitel man nontrivial zeros of L(s, χ). While it is not hard to show ζ(s) on the interval (, ), without the Generalized Riemann Hpothesis there is no proof that L(s, χ) on the interval (, ) for all χ, except possibl at s /2. It is conjectured that L(/2, χ) for all Dirichlet characters χ. 5. Analtic functions We created extensions of Z(s) and Λ(s, χ) from Re(s) > to C (allowing that Z() and Z() ), from which we got extensions of ζ(s) and L(s, χ) from Re(s) > to C (allowing that ζ() ). In what sense can we sa these extensions are unique without referring to explicit formulas? To answer this we will use the language of analtic functions. Definition 5.. A domain in C is a connected open subset. Intuitivel, this means a domain is in one piece (connected) and when an point in the domain is moved b a small amount it remains in the domain. Examples of domains in C include open discs and open half-planes such as {s : Re(s) > }. Domains do not include an point on their boundar. Definition 5.2. A function f : Ω C on a domain Ω is called analtic if it can be written as a power series near each point of Ω: for ever a Ω we can write for all s close to a. f(s) n c n (s a) n Example 5.3. The function e s is defined on C and is analtic there: for each a C and s C, e s e a e s a e a (s a) n a (s a)n e. n! n! n n Example 5.4. Ever rational function is analtic on the domain that is C without the finitel man points where the denominator is. For instance, /(s 2 + ) is analtic on C {±i}. Example 5.5. For s C, /n s is analtic since /n s n s e s ln n, and this can be written as a power series in s a for all a C b the same ideas as in Example 5.3. Example 5.6. The infinite series ζ(s) n /ns and L(s, χ) n χ(n)/ns are both analtic on Re(s) >. Example 5.7. The function Γ(s) is analtic on C {,, 2, 3,...}. Example 5.8. The absolute value function f(s) s is not analtic on an domain in C. Analtic functions have a unique extension propert, as described in the following theorem that is proved in courses on complex analsis. Theorem 5.9. If f : Ω C is analtic on a domain Ω and Ω is a larger domain, then there is at most one extension of f to an analtic function on Ω.

20 2 KEITH CONRAD This theorem is not saing an analtic function on a domain can be extended to an analtic function on ever larger domain (tr to extend /s from C {} to C), but onl that if this can be done then an two was of doing this must lead to the same result. There is nothing like Theorem 5.9 for continuous functions being extended to continuous functions on a larger domain. It can be shown from the integral formulas for Z(s) and Λ(s, χ) that these functions are analtic on C {, } and on C, and then that ζ(s) and L(s, χ) are analtic on C {} and on C, so b Theorem 5.9 the extensions we have found of these functions from the half-plane Re(s) > are the onl possible analtic extensions. In this sense we have found the right extensions of ζ(s) and L(s, χ) to C. 6. Contour integrals We now turn to a ver important tool for studing analtic functions: complex contour integration. It will not at first look like this has anthing to do with describing ζ(s) or L(s, χ), or using the Riemann Hpothesis, so some patience will be needed. Definition 6.. If γ : [a, b] C is a differentiable path 3 and f is a continuous complexvalued function on the image of γ, then we define the complex contour integral of f along γ to be b (6.) f(s) ds : f(γ(t))γ (t) dt. γ a If γ is onl piecewise differentiable then we define γ f(s) ds to be the sum of the contour integrals of f on each piecewise differentiable piece of γ. The right side of (6.) can be calculated from an antiderivative (indefinite integral) of the integrand: if g (t) f(γ(t))γ (t) then γ f(s) ds g(b) g(a). This will be used below in Example 6.2. Our definition of contour integrals looks similar to line integrals of real-valued functions in R 2, but there is a crucial difference: the multiplication f(γ(t))γ (t) is a product of complex numbers, not an inner product of vectors in R 2. The value of γ f(s) ds is a complex number, not (usuall) a real number. Its value does not have a direct geometric interpretation as an area, volume, densit, etc., but we will see that the machiner of contour integration is ver powerful. The integral γ f(s) ds can be defined as a limit of Riemann sums, but for the sake of being efficient we have given the more direct definition above instead of the more conceptual definition using limits. Example 6.2. Let s compute s n ds where C + R is a circle centered at the origin of C + R radius R >, going once around the origin counterclockwise, The + in C + R refers to this positive orientation on the circle. This path around the circle can be described b γ : [, 2π] C where γ(t) Re it, 3 This means that if we write γ in terms of its real and imaginar parts, sa γ(t) u(t) + iv(t), then the component functions u(t) and v(t) are differentiable.

21 so γ (t) ire it and then s n ds C + R L-FUNCTIONS AND THE RIEMANN HYPOTHESIS 2 γ s n ds 2π 2π (Re it ) n (ire it ) dt ir n+ e i(n+)t dt. This last integral can be computed b finding an antiderivative of e i(n+)t. Separate cases are needed when n (so n + ) and when n (so n + ). Case : n. An antiderivative of e i(n+)t is e i(n+)t /(i(n + )), so 2π e i(n+)t dt ei(n+)t i(n + ) 2π e2πi(n+) e i(n + ) so C + s n ds. R Case 2: n. Here e i(n+)t, so an antiderivative is t and 2π e i(n+)t 2π dt t 2π 2π, so C + R Putting these two cases together, s n ds C + R 2π ds ir + dt 2πi. s {, if n, 2πi, if n. i(n + ), It is intriguing that the answer does not depend on R: for all circles centered at the origin, with a fixed choice of n, we get the same result! In light of the values of this integral, it is convenient to divide b 2πi to make the two values and : s n ds 2πi C + R {, if n, if n. Remark 6.3. If the path around the circle goes once around in the clockwise direction (negative orientation), using path γ(t) Re it for t 2π, then the contour integral would be { 2π s n ds ir n+ e i(n+)t, if n, dt C R 2π( i), if n, so { s n, if n, ds 2πi C, if n, R This illustrates how a contour integral depends on how γ traces out a path: if the image of γ traces out a path in the reverse direction or multiple times then γ f(s) ds changes. We will not focus on this aspect, which would require using the concept of a winding number. We can now integrate a power series n c ns n that converges on the circle C + R, assuming we can interchange the order of summation and integration: (6.2) c n s n ds. C + R n c n s n ds n C + R

22 22 KEITH CONRAD That does not look ver interesting. Add to the power series a finite number of negative powers of s (which all make sense on C + R, since is not on the circle) and then we find c n s n ds c C + R C + s ds + c n s n ds 2πic R n C + R and C + R n 2 n c n s n ds c 2 C + R and more generall for an N <, (6.3) s 2 ds + c C + R c n s n ds 2πic. C + R n N s ds + c n s n ds 2πic, n C + R This last formula is also valid if N, since in that case the coefficient c of /s is and the integral is also b (6.2). Dividing b 2πi in (6.3), c n s n ds c. 2πi C + R n N This calculation shows that the coefficient c of /s, unlike ever other coefficient in n N c ns n, can be detected b contour integration. This will lead to a ver powerful result in complex analsis called the residue theorem.