Riemann s explicit formula

Transcription

1 Garrett Continuing to review the simple case (haha!) of number theor over Z: Another example of the possibl-suprising application of othe things to number theor. Riemann s explicit formula More interesting than a Prime Number Theorem is the precise relationship between primes and zeros of zeta found b Riemann. The idea applies to an zeta or L-function for which we know an analtic continuation and other reasonable properties. It took 40 ears for [Hadamard 893], [vonmangoldt 895], and others to complete Riemann s sketch of the Explicit Formula relating primes to zeros of the Euler-Riemann zeta function. Even then, lacking a zero-free strip inside the critical strip, the Explicit Formula does not ield a Prime Number Theorem, despite giving a precise relationship between primes and zeros of zeta.

2 Garrett Riemann s explicit formula Riemann s dramatic relation between primes and zeros of the zeta function depends on man ideas undeveloped in Riemann s time. Thus, the following sketch, roughl following Riemann, is not a proof. Rather, the sketch tells which supporting ideas need development to produce a proof. that ζ(s) has both its Euler product expansion in a half-plane ζ(s) = n n s = p prime p s (Res > ) Riemann alread knew ζ(s) has a meromorphic continuation throughout C (see below).

3 Garrett If we believe, as Riemann did, and as Hadamard and others later proved, that is also has a Riemann-Hadamard product expansion (s ) ζ(s) = e a+bs ( s ) e s/ n= ( + s ) e s/2n 2n product over non-trivial zero of ζ, for all s C. Then, following Riemann, extract tangible information from the equalit of the two products (s ) p p s = e a+bs ( s ) e s/ n= ( + s ) e s/2n (Res > ) 2n

4 Garrett First, take logarithmic derivatives of both sides, using log( x) = x + x 2 /2 + x 3 / on the left-hand side: = b + s m, p ( s + ) log p p ms + n ( s + 2n ) 2n A slight rearrangement: m, p log p p ms = s b ( s + ) n ( s + 2n ) 2n The left-hand side needs Res > for convergence, while the righthand side converges for all s C apart from the visible poles at, the non-trivial zeros, and the trivial zeros 2, 4, 6,....

5 Garrett Next, appl the Perron identit 2πi σ+i σ i Y s s ds = (for Y > ) 0 (for 0 < Y < ) Reall, we have to be slightl careful: lim T 2πi σ+it σ it Y s s ds = (for Y > ) 0 (for 0 < Y < ) (for σ > 0) (for σ > 0) If we can appl this to entire expressions, b f lim T 2πi σ+it σ it f(s) Xs s ds (with σ > ) term-wise to the left-hand side, and use residues term-wise to evaluate the right-hand side, we would have p m <X log p = (X ) b ( X + + ) n ( X 2n 2n + 2n ) 2n

6 Garrett which simplifies to von Mangoldt s reformulation of Riemann s Explicit Formula: p m <X log p = X (b + ) X + n X 2n 2n Slightl more precisel, because of the wa the Perron integral transform is applied, and the fragilit of the convergence, p m <X log p = X (b + ) lim T Im () <T X + n X 2n 2n The Riemann-Hadamard product needs both generalities about Weierstraß-Hadamard product expressions for entire functions of prescribed growth, and specifics about the growth of the analtic continuation of ζ(s).

7 Garrett For future reference: The two sides of the equalit of logarithmic derivatives are ver different. The logarithmic derivative of the Euler product converges well in right half-planes, and converges all the better farther to the right. The logarithmic derivative of the Riemann-Hadamard product does not converge powerfull, but is not restricted to a half-plane, and its poles are exhibited explicitl b the expression.

8 Garrett Analtic continuation and functional equation of ζ(s) The following ideas gained publicit and importance from Riemann s paper, but were apparentl known before Riemann s time. The ke is that the completed zeta function has an integral representation in terms of an automorphic form, namel, the simplest theta function. Both the analtic continuation and the functional equation of zeta follow from this integral representation using a parallel functional equation of the theta function, the latter demonstrated b Poisson summation.

9 Garrett Elementar-but-doomed argument It is worthwhile to see that simple calculus can extend the domain of ζ(s) a little. The idea is to pa attention to quantitative aspects of the integral test. That is, = n ζ(s) s = ζ(s) ( n+ n s dx ) x s dx x s = n n ( n s [ s n s ]) (n + ) s Even for complex s, we have a Talor-Maclaurin expansion with error term (n + ) s = ( n ( + n )) s = n s ( + s n + O( n 2 ) ) = n s s n s + O( s n s+ ) The constant in the big-o term is uniform in n for fixed s. Thus, n s [ s n s ] (n + ) s = O ( n s+ )

10 Garrett That is, for fixed Re(s) > 0, we have absolute convergence of n in the larger region Re(s) > 0. ( n s [ s n s ]) (n + ) s A similar but increasingl complicated device produces a meromorphic continuation to half-planes Re(s) > l. However, this approach is under-powered...

11 Garrett A more serious argument Euler s integral for the gamma function is Γ(s) = e t t s dt t Among other roles, the gamma function Γ(s) interpolates the factorial function: integration b parts ields Γ(n) = (n )! for positive integer n. Theorem The completed zeta function 0 ξ(s) = π s 2 Γ( s 2 ) ζ(s) has an analtic continuation to s C, except for simple poles at s = 0,, and has the functional equation ξ( s) = ξ(s)... and (anticipating the Riemann-Hadamard product issues) s(s )ξ(s) is entire and bounded in vertical strips.

12 Garrett The following proof-sketch is itself an archetpe. The simplest theta function is θ(z) = n Z e πin2 z with z in the complex upper half-plane H. B Riemann s time, Jacobi s functional equation of θ(z) was well-known, as the simplest example of a larger technical phenomenon: θ(z) = iz θ( /z) (Proven below.) The modified version θ(i) 2 = n= e πn2 appears just below.

13 The connection to ζ(s) is the integral presentation: Claim: For Re(s) > Garrett π s/2 Γ( s 2 ) ζ(s) = 0 θ(i) 2 s/2 d Meaning? An integral against t s with dt/t, a Mellin transform, is just a Fourier transform in different coordinates. Starting from the integral, for Re(s) >, compute directl 0 θ(i) 2 s/2 d = 0 n e πn2 s/2 d = n 0 e πn2 s/2 d = π s/2 n n 2s 0 e s/2 d b replacing b /(πn 2 ), and this is = π s/2 Γ( s 2 ) n = ξ(s) (for Re(s) > ) ns

14 Garrett Bibliograph [Edwards 974] H.M. Edwards, Riemann s zeta function, Academic Press, New York, 974. [Hadamard 893] J. Hadamard, Étude sur les Propriétés des Fonctions Entières et en Particulier d une Fonction Considérée par Riemann, J. Math. Pures Appl. p, [Ivić 985] A. Ivić, The Riemann zeta function, J. Wile, New York, 985. [Iwaniec 2002] H. Iwaniec, Spectral Methods of Automorphic Forms, 2nd edition, AMS, Providence, [First edition, Revisto Mathematica Iberoamericana, 995.] [vonmangoldt 895] H. von Mangoldt, Zu Riemann s Abhandlung Über die Anzahl der Primzahlen unter einer gegebenen Grösse, J. Reine Angew. Math. 4, [Patterson 988] S. J. Patterson, An introduction to the theor of the Riemann zeta function, Cambridge Universit Press, 988. [Titchmarsh 95] E.C. Titchmarsh, The theor of the Riemann zeta function, Oxford Universit Press, 95. [Weil 952] A. Weil, Sur les formules explicites de la théorie des nombres premiers, Comm. Lund (vol. dedié à Marcel Riesz) (Oeuvres sci. [952b], Vol. II, Springer, New York, 979.) [Weil 972] A. Weil, Sur les formules explicites de la théorie des nombres, Izv. Akad. Nauk. (ser. Math.) 36 (972), 3-8. (Oeuvres sci. [972], Vol. III, Springer, New York, 979.)