VII.8. The Riemann Zeta Function.

Transcription

1 VII.8. The Riemann Zeta Function VII.8. The Riemann Zeta Function. Note. In this section, we define the Riemann zeta function and discuss its history. We relate this meromorphic function with a simple pole at z = (see Theorem VII.8.4) to, of all things, prime numbers. Note. Let z C and let n N. Then So n z = exp(z log n) = exp(re(z) log n) = n Re(z). n k z = k= So if Re(z) + ε then n exp( Re(z) log k) = k= n k z k= n k (+ε) = k= n k= n k Re(z). k= k +ε and the series n= n z converges uniformly and absolutely on G = {z Re(z) + ε} (absolutely by comparing to the p-series with p = + ε > and uniformly by the Weierstrass M-Test [Theorem II.6.2] with u n (z) = n z for Re(z) + ε and M n = /n +ε ). Now each u n is continuous on G and so each sum of u n s is continuous. Since the convergence is uniform, the limit n= n z is continuous on G. Also, each u n is analytic on G, so by Theorem VII.2., n= u n(z) is analytic on G. Definition VII.8.. The Riemann zeta function on {z Re(z) > } is defined as ζ(z) = n z. n=

2 VII.8. The Riemann Zeta Function 2 Note. We will eventually extend ζ(z) to a function analytic in whole complex plane except for z = where it will have a simple pole. The extension is not accomplished by analytic continuation (see Chapter IX), but by relating the zeta function to the gamma function. Note. In 859, Georg Bernhard Riemann published an 8 page paper, On the Number of Primes Less Than a Given Magnitude (Ueber die Anqahl der Primzahlen unter einer gegenen Grösse). A translation of this paper can be found in the appendix of H. M. Edwards Riemann s Zeta Function, Academic Press, Inc. (974) (this book has now been reprinted by Dover Publications). In the paper, Riemann comments that it is very likely that the complex zeros of the zeta function all have real part equal to /2, but that he has been unable to prove this [Edwards, page 6]. This conjecture which concerns the distribution of prime numbers is now the holy grail of open math problems and is known as the Riemann Hypothesis. Georg Friedrich Bernhard Riemann, (From the MacTutor History of Mathematics archive.)

3 VII.8. The Riemann Zeta Function 3 Note. We now relate ζ(z) to Γ(z). Recall that, by Theorem VII.7.5, for Re(z) > Γ(z) = e t t z a dt. By replacing the real variable t with nu where n > (t = nu and dt = n du) we get or Γ(z) = = n z Γ(z) = e nu (nu) z n du e nu u z n z du = n z e nt t z dt in terms of t) If Re(z) > and we sum over n N, we get ( ( ζ(z)γ(z) = n )Γ(z) z = n z Γ(z) ) = n= n= e nt t z dt for Re(z) >. ( n= ) e nt t z dt. (8.2) We want to take the sum inside the integral and simplify. We need some preliminary results first. Lemma VII.8.3. (a) Let S = {z Re(z) a} where a >. If ε > then there is a number δ, < δ <, such that for all z S we have β (e t ) t z dt < ε whenever δ > β > α >. α

4 VII.8. The Riemann Zeta Function 4 (b) Let S = {z Re(z) A} where A R. If ε > then there is a number κ > such that for all z S we have β (e t ) t z dt < ε whenever β > α > κ. α Corollary VII.8.4. (a) If S = {z a Re(z) A} where < a < A < then the integral converges uniformly on S. (e t ) t z dt (b) If S = {z Re(z) A} where < A <, then the integral converges uniformly on S. (e t ) t z dt Note. We now use the uniform convergence of Corollary 8.4 in Equation (8.2). Note. The following definition and result are from the exercises in Section VII.2. Definition. Let G be a region, let a R and suppose that f : [, ] G C is a continuous function. The integral F(z) = compact subsets on G if lim b b a subset of G. a f(t,z) dt is uniformly convergent on f(t,z) dt exists uniformly for z in any compact

5 VII.8. The Riemann Zeta Function 5 Theorem VII.8.A/Exercise VII.2.2. Let G be a region, a R, f : [a, ) G C is a continuous function, and suppose the integral F(z) = a f(t,z) dt is uniformly convergent on compact subsets of G. Suppose for each t (a, ), f(t,z) is analytic on G. Then F is analytic on G and F (k) (z) = a k f(t,z) z k dt. Corollary VII.8.B. Let G be a region, a R, f : (,] G C is a continuous function, and suppose the integral F(z) = f(t,z) dt is uniformly convergent on compact subsets of G. Suppose for each t (,], f(t,z) is analytic on G. Then F is analytic on G and F (k) (z) = a k f(t,z) z k dt. Note. If K is a compact subset of G = {z Re(z) > } then K is closed and bounded (by the Heine-Borel Theorem) and so K {z a Re(z) A} for some a,a R with a < a A. With f(t,z) = (e t ) t z = (e t ) e (z )log t, we have that for each t (a, ), f(t,z) = (e t ) (t ) z is analytic on G. For compact set K G where K {z a Re(z) A} where < a < A <, so by Corollary VII.8.4(a), F(z) = f(t,z) dt = (et ) t z dt converges uniformly on K and so F(z) is uniformly convergent on compact subsets of G. Therefore, by Theorem VII.8.A/Exercise VII.2.2, f(z) = (et ) t z dt is analytic on G.

6 VII.8. The Riemann Zeta Function 6 Proposition VII.8.5. For Re(z) > ζ(z)γ(z) = (e t ) t z dt. Note. We now extend ζ(z) from {z Re(z) > } to all of C (except z = ). We use the relationship between ζ(z) and Γ(z) given in Proposition VII.8.5 to make this extension. Note. We now explore the Laurent series for (e z ). Notice that lim z z(ez ) = lim z z e z = lim z e z = (L Hôpital s Rule holds here), so (e z ) has a pole of order at z = (see Definition V..6). By Corollary V..8(b), the coefficients a n in the Laurent series are for n 2. By Proposition V.2.4, a = Res((e z ) ;) = z(e z ) z= = (by the limit argument above). so z(e z ) (with the removable singularity at z = removed) is entire. The coefficient of z in the power series representation of x(e z ) is [z(e z ) ] z= = (ez ) ze z z= e z e z ze z = lim (e z ) 2 z 2(e z )z z z = lim z 2(e z ) = lim z 2e = z 2. So in the Laurent series for (e z ), the constant term is /2. Therefore, the Laurent series of (e z ) is of the form e z = z 2 + a n z n n=

7 VII.8. The Riemann Zeta Function 7 for some a,a 2,... So e t t has a limit as t (its /2) and e t t remains bounded in a neighborhood of t =. But this implies that the integral ( e t ) t z dt t Note. We now extend the definition of ζ from {z Re(z) > } to C \ {}. We do so stepwise. STEP. Extend the definition of ζ from {z Re(z) > } to {z Re(z) > }. STEP 2. Extend the definition of ζ from {z Re(z) > } to {z Re(z) > }. STEP 3. Extend the definition of ζ from {z Re(z) > } to all of C \ {}. This will give ζ as a meromorphic function with a simple pole at z = (see Theorem VII.8.4). At each step, we extend ζ to a new set which overlaps with a previous set; we must confirm that the resulting ζ is well-defined. That is, we must confirm that the definitions are consistent on the overlaps of the sets. Note. We will explore analytic continuation in Chapter IX. One could use the definition of ζ on {z Re(z) > } as ζ(z) = n= n z and analytic continuation to extend it to C \ {}. However, we follow the steps given above. In fact, this is the approach taken by Riemann himself. The following quote is from H. M. Edwards Riemann s Zeta Function, Academic Press (974), page 9: It is interesting to note that Riemann does not speak of the analytic continuation of the function n s beyond the half plane Re s >, but speaks rather of finding a formula for it which remains valid for all s. This indicates that he viewed the problem in terms more analogous

8 VII.8. The Riemann Zeta Function 8 to the extension [by formulas]... than to a piece-by-piece extension of the function in the manner that analytic continuation is customarily taught today. The view of analytic continuation in terms of chains of disks and power series convergent in each disk descends from Weierstrass and is quite antithetical to Riemann s basic philosophy that analytic functions should be dealt with globally, not locally in terms of power series. Lemma VII.8.C. The function on G = {z Re(z) > }. ( e t ) t z dt is an analytic function t Note. For Re(z) > we have ( e t t ) t z dt + (z ) + t z e t dt = = = t z e t dt t z t z 2 dt + (z ) e t dt t= z tz + (z ) t= t z+ e t dt z + (z ) = t z e t dt t z+ e t dt. So the first quantity equals ζ(z)γ(z) for Re(z) > by Proposition VII.8.5. This motivates us to extend the definition of ζ from Re(z) > to Re(z) >, thus completing STEP in our extension of ζ.

9 VII.8. The Riemann Zeta Function 9 Definition. For Re(z) > define the Riemann zeta function as ζ(z) = ( ( Γ(z) e t ) t z dt + (z ) t z ) + t e t dt. (8.7) Note. On Re(z) >, ζ is a meromorphic function with a simple pole at z =. We ( know that e t ) t z dt is an analytic function on Re(z) > as argued t t z above and dt is uniformly convergent on compact subsets if Re(z) > e t by Corollary VII.8.4(b) and so is analytic by Theorem VII.8.A/Exercise VII.2.2. By Proposition V.2.4, Res(ζ;) = (z )ζ(z) = lim(z )ζ(z) = z= z (since the analytic functions given by the integrals produce in the limit and only z lim = remains; recall that Γ() = ). That is, the residue of ζ at z = is z z. Note. We now give an alternative representation of ζ for < Re(z) <. For < Re < then So by (8.7) ζ(z)γ(z) = t z 2 dt tz t= z = t= z. ( e t ) t z dt t z 2 dt + t = t z e t dt t z dt for < Re(z) <. (8.8) e t

10 VII.8. The Riemann Zeta Function Lemma VII.8.D. The function ( e t t + ) t z dt is an analytic func- 2 tion on G = {z Re(z) > }. Lemma VII.8.E. The function on G = {z Re(z) < }. ( e t ) t z dt is an analytic function t Note. From (8.8) we have for < Re(z) < that ( e t t + ) t z dt 2 2z + ( e t ) t z dt t = = = ( e t ) t z dt + t 2 ( ) e t t ( e t t t z dt + 2z t2 t= t z dt 2z t= 2z ) t z dt = ζ(z)γ(z). However, by Lemmas VII.8.D and VII.8.E, the integrals in the first part of this equation are also valid on { < Re(z) < }. This motivates our next definition which extends ζ to {z Re(z) > } and accomplishes STEP 2 of the extension of ζ. Definition. For < Re(z) < define the Riemann zeta function as ζ(z) = ( ( Γ(z) e t t + ) t z dt ( 2 2z + e t ) ) t z dt. (8.9) t

11 VII.8. The Riemann Zeta Function Note. The term /(2z) in equation (8.9) makes it appear that ζ may have a pole at z =. Now each of the integrals in (8.9) represents analytic functions throughout G = {z < Re(z) < }. Also,.Γ(z) is analytic on G. Notice that /(2zΓ(z)) = /(2Γ(z + )) by Theorem VII.7.7 (the Functional Equation for Γ), so /(2zΓ(z)) is analytic at z = ; its value is /(2Γ()) = /2. So ζ is analytic throughout {z < Re(z) < }. Theorem VII.8.3. Riemann s Functional Equation. For < Re(z) < we have ζ(z) = 2(2π) z Γ( z)ζ( z) sin πz/2. Note. Conway claims that The same type of reasoning given that (8.3) holds for < Re(z) < (page 92). Since the right hand side of Riemann s Functional Equation is analytic in all of Re(z) <, we use this as STEP 3 in the extension of ζ to all of C \ {}. Definition. For Re(z) < define the Riemann zeta function as ( ) ζ(z) = 2(2π) z Γ( z)ζ( z) sin 2 πz. Note. Notice that, by definition, ζ satisfies Riemann s Functional Equation for Re(z) < by definition and satisfies it for < re(z) < by Conway s claim mentioned above. Treating both sides of Riemann s Functional Equation as analytic

12 VII.8. The Riemann Zeta Function 2 functions on C\{} and the fact that both sides are equal on a subset of C\{} with a limit point, then by Corollary IV.3.8, the functional equation holds throughout C \ {}. We summarize our knowledge of ζ as follows. Theorem VII.8.4. The zeta function is meromorphic in C with only a simple pole at z = and Res(ζ;) =. For z, ζ satisfies Riemann s Functional Equation. Note. Since Γ has simple poles at z =,, 2,... then Γ( z) has a simple pole at z =,2,3,... Now ζ is analytic at z = 2,3,4,..., so from Riemann s Functional Equation (Theorem VII.8.3) we have for z = 2,3,4,... that ζ( z) sin(πz/2) = (otherwise ζ would not be analytic at z = 2,3,4,...). Now sin(πz/2) = for z = 2,4,6,... so we must have ζ( z) = for z = 3,5,7,... (since sin(πz/2) = ± for these values of z). That is, ζ(z) = for z = 2, 4, 6,... Note. Conway comments that: Similar reasoning gives that ζ has no other zeros outside the closed strip {z Re(z) } (page 93). Definition VII.8.6. The points z = 2, 4, 6,... are the trivial zeros of ζ and the strip {z Re(z) } is the critical strip.

13 VII.8. The Riemann Zeta Function 3 Note. We are now motivated and have the background to state the Riemann Hypothesis. The Riemann Hypothesis. If z is a zero of the zeta function in the critical strip then Re(z) = /2. Theorem VII.8.7. Euler s Theorem. If Re(z) > then ζ(z) = n= where {p n } is the sequence of prime numbers. p z n Note. The exercises in the section give several number theoretic properties of ζ(z). In particular, Exercises VII.8.2 VII.8.5. Exercise VII.8.2. Use Euler s Theorem to prove that n= p n =. Notice that this implies that there are an infinite number of primes. Exercise VII.8.3. Prove that ζ 2 (z) = n= d(n)/nz for Re(z) >, where d(n) is the number of divisors of n. Exercise VII.8.4. Prove that ζ(z)ζ(z ) = n= σ(n)/nz for Re(z) >, where σ(n) is the sum of the divisors of n. Exercise VII.8.5. Prove that ζ(z )/ζ(z) = n= ϕ(n)/nz for Re(z) >, where ϕ(n) is the number of integers less than n and which are relatively prime to n.

14 VII.8. The Riemann Zeta Function 4 Note. In the 25 BBC and the Open University presented the three part documentary series The Music of the Primes, written by Marcus de Sautoy. In it, the modular surface of the Riemann zeta function is given. The modular surface as viewed in the quadrant where both real and imaginary parts of z are positive. The nontrivial zeros of ζ indicated by white circles.

15 VII.8. The Riemann Zeta Function 5 A view out the line Re(z) = /2 with the pole at z = visible on the left. The nontrivial zeros indicated as beams lining up along the line Re(z) = /2.

16 VII.8. The Riemann Zeta Function 6 Note. Another view of the modular surface is given at the Michigan State University website (accessed 8/27/27): In this view of the modular surface ζ(x + iy), the pole is on the left and the trivial zeros are off in the valley on the left behind the pole. Several of the nontrivial zeros are visible running off to the right. The nontrivial zeros of ζ with imaginary parts between and 5 are /2 + iy where (to two decimal places) y is: 4.3, 2.2, 25., 3.42, 32.94, 37.59, 4.92, 43.33, 48., and (from accessed 8/27/27). Revised: 8/27/27