On Hilbert s 8th Problem

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1 On Hilbert s 8th Problem Nick Polson University of Chicago August, 8th 17

2 Riemann ζ-function Hadamard factorisation of Xi-function to characterize the zeros of the zeta function Hilbert s 8th Problem Riemann defines the zeta function, ζ(s), as the analytic continuation of n=1 n s on Re(s) > 1 and the ξ-function by ζ(s) = n=1 n s, ξ(s) = 1 s(s 1 1)π s Γ ( 1 s) ζ(s). (1) The Riemann Hypothesis (RH) states that all the non-trivial zeroes of ζ(s) lie on the critical line Re(s) = 1, or equivalently, those of ξ( 1 + is) = ξ( 1 is) lie on the real axis.

3 Riemann ξ-function The ξ-function is an entire function of order one. Titchmarsh (.1.5) provides a Hadamard s factorization, for all values of s, with ξ() = ζ() = 1, ξ(s) = ξ()e ( b s 1 s ) e ρ s () ρ ρ where b = 1 log(4π) 1 1 γ. The zeros, ρ, of ξ correspond to the non-trivial zeros of ζ. If RH is true, the zeroes of ξ are of the form 1 ± iτ, as ξ(s) = ξ(1 s), and ξ(s) = ξ() ( 1 s ) ( 1 + iτ 1 s ) 1 iτ. (3) τ>

4 Outline The argument to prove RH proceeds in three steps. (a) RH is equivalent to the existence of a generalised gamma convolution (GGC) r.v. (Bondesson, p.14, Roynette and Yor), denoted by H ξ 1, whose Laplace transform expresses the reciprocal ξ-function as ξ ( 1 ) ξ ( 1 + s ) = E(exp( shξ 1 )). (4) (b) Theorem 1 constructs a GGC random variable H ξ α for α > 1, such that ξ(α) ξ(α + s s) e bα = E(exp( shα)) ξ where b α = ξ ξ (α) + 1 α 1. (5) (c) Theorem then extends (b) to the case α = 1. Lemma 1 helps identify the GGC random variable H ξ α. Propositions 1 and for GGC representations of ζ and (α 1)/(s + (α 1))

5 Reciprocal ξ-function To prove (a). Under RH, the Hadamard factorisation is ξ(s) = ξ() ( 1 s ) ( 1 + iτ 1 s ) 1 iτ τ> Now ξ( 1 ) = ξ() τ> τ /( τ ) and Hence, the reciprocal ξ-function satisfies ξ(s) = ξ() τ> (s 1 ) + τ (6) τ. (7) ( ) ξ 1 ( = ξ 1 + s) τ τ> τ + s. (8) Frullani s identity, log(z/z + s ) = (1 e s t )e tz dt/t, yields ( ) ξ 1 ( = ξ 1 + s) τ { ( ) } z τ> τ + s = exp log z + s U(dz) (9) ( = exp (1 e s t ξ )g 1 (t) dt ) = E(exp( s H ξ 1 )). t (1) Here g ξ 1 (t) = e tz U 1 (dz) and U 1 (dz) = τ> δ τ (dz) is the Thorin measure with δ a Dirac The GGC H ξ D 1 = τ> Yτ, where Yτ Exp(τ ), directly satisfies ( ) τ τ> τ +s = E exp( s H ξ 1 ).

6 Reciprocal ξ-function A Hadamard representation for ξ(α + s), with α > 1 Theorem The reciprocal ξ-function satisfies for α > 1 and s >, where ( ξ(α) ξ(α + s) = ebαs exp and g ξ α is completely monotone b α = ξ ξ (α) + 1 α 1 g ξ α(t) = (1 e 1 st )g ξ α(t) dt t ) (11) (1) e tz U ξ α(dz). (13)

7 Reciprocal ξ-function To derive ξ(α)/ξ(α + s), use ξ (α) = (α 1)π 1 α Γ (1 + 1 α) ζ (α) (14) ξ(α + s) = (α 1 + s)π 1 (α+s) Γ (1 + 1 (α + s)) ζ(α + s). (15) Now, ξ() = ξ(1) = 1 and ξ( 1 ) and ξ ( 1 ) = so ξ ( 1 )/ξ( 1 ) =. Taking derivatives of log ξ(s) = log(s 1) 1 s log π + log Γ (1 + 1 s) + log ζ(s), ξ ξ (α) = 1 α 1 1 ζ log π + ζ (α) + 1 ψ (1 + 1 α). (16) with ψ(s) = Γ (s)/γ(s). Hence, = ξ ξ ( 1 ) = 1 log π + ζ ζ ( 1 ) + 1 ψ ( 5 4 ). (17)

8 Reciprocal ξ-representation Theorem 1 follows from the decomposition { ( ξ(α) ξ ξ(α + s) exp s ξ (α) 1 )} ) (α 1) Γ (1 + 1 = α 1 s + (α 1) α e s 1 ψ(1+ 1 α) ( ) ζ(α)e s ζ ζ (α) Γ (α + s) ζ (α + s) (18) where Each term will be treated separately. ψ(s) = Γ ( ) e t Γ (s) = e st t 1 e t dt. (19)

9 Zeta representation Euler s product formula, for α > 1, gives ζ (α + s) = ( 1 p α s ) 1 = ζ p(α+s) where ζ p(s) := p s /(p s 1) p prime and ζ(α) = p ζp(α), yields log ζ(α + s) ζ(α) = p = log p prime 1 p α 1 p α s = p (e sx 1)e αx µ ζ (dx) x where µ ζ (dx) = p r=1 (log p)δ r log p(dx). r=1 () 1 r p αr (e sr log p 1) (1) ()

10 GGC Identity Given the identities, valid for s > and x >, e sx + sx 1 = 1 e x /t t = (1 e 1 st )(1 e x t ) x dt (3) πt 3 e tz sin (x z/) πz dz (4) Lemma Suppose that the function f satisfies ( ) f (α + s) log s f (α) f (α) f (α) = for some arbitrary µ(dx). (e sx 1 + sx)e αx µ(dx) x (5)

11 GGC Identity Then this can be re-expressed as ( ) f (α) log + s f (α) f (α + s) f (α) = where ν α(t) is the completely monotone function ν α(t) = 1 ( e tz π When f (s) = Γ(1 + 1 s) then (1 e 1 st ) να(t) t sin (x z/)e αx µ(dx) x dt (6) ) dz πz (7) When f (s) = ζ p(s) then µ Γ (dx) = dx e x 1 (8) µ ζ (dx) = k 1(log p)δ k log p (dx). (9)

12 ζ-representation Lemma 1 for the ζ-function yields Proposition For α > 1 and s >, with ζ(α + s) ζ s e ζ (α) = exp ζ(α) µ ζ (dx) x = p ( µ ζ p(dx) x (e sx + sx 1)e αx µ ζ (dx) x = n ) (3) Λ(n) log n δ log n(dx) (31) where Λ(n) is the von Mangoldt function, Λ(n) = log p if n = p r, with p prime.

13 ξ-representation Moreover, with ν ζ α(t) = p prime νζ p,α(t), ( ζ(α) ζ ζ(α + s) es ζ (α) = exp where ν ζ α(t)dt < for α > 1, and ν ζ α(t) = 1 πt = 1 πt n p prime (1 e 1 st ) νζ α(t) t ) dt, (3) Λ(n) ( ) 1 e (log n)/t (33) n α log p r 1 1 ( p αr ) 1 e (r log p)/t. (34)

14 (α 1)/(s + (α 1)) representation The term (α 1)/(s + (α 1)) in the Hadamard product follows from Lemma The function a/(s + a), with a = α 1, has representation a s + a = e sx ae ax dx = e 1 s t xe x t πt 3 ae ax dx dt = exp { with completely monotone function (1 e 1 st ) ν α(t) t } dt ν α(t) = 1 e 1 at erfc(a 1 t) = E(e tz a ) (35) where Z a has density a/π x(a + x), for x >.

15 Hadamard factorisation The Hadamard factorization for the reciprocal ξ-function, for α > 1, is ( ξ(α) ξ(α + s) = ebαs exp (1 e 1 st ) g ) α(t) ξ dt t (36) where b α = ξ (α) 1, with completely monotone function ξ α 1 g ξ α(t) = ν α(t) + ν Γ α(t) + ν ζ α(t). (37) Equivalently, there exists H ξ α, with GGC density, such that ξ(α) ξ(α + s s) e bα = E(exp( shα)) ξ. (38) This completes the derivation of Theorem 1.

16 Extension to ξ( 1 )/ξ( 1 + s) for s > Pólya finds X ξ, with a symmetric density p(x), such that where, p(x) 4π e 9 x πex ξ( 1 + s) ξ( 1 ) = E(e sx ξ ), s C as x and p(x) = 1 ξ( 1 ) p n(x) and p n(x) := n π(πn e x 3)e 5 x n πe x. n=1

17 Calculating ξ( 1 )/ξ( 1 + s) Calculate ξ( 1 )/ξ( 1 + s) via G(x) and its transform, m G (s), G(x) = k=1 ( 1) k+1 Γ(k) ξ( 1 )ξ( 3 + k) ξ( 1 + k) x k and m G (s) = Existence of m G (s) from Grosswald. Hayman-Grosswald bound x s 1 G(x)dx. (39) ( ) k log ξ(k + 1 ) = 1 k log πe log k log( 1 π) + o(1) as k ξ( 1 ) ( ( ) ) k ξ( 1 = exp 1 + k) k log 74 πe log k 14 log( 1 π) + log ξ( 1 ) + o(1) as k. (4)

18 Transform m G (x) Calculate m G (s), where Y ξ = e X ξ, with density q(y), and z = xy, ξ( 1 s) ξ( 1 ) m G (s) = E(Y s ξ ) x s 1 G(x)dx (41) = y s q(y)x s 1 G(x)dxdy (4) = q(y)z s 1 G (z/y) dzdy (43) = z s 1 ( 1) k+1 ξ( 3 Γ(k) + k) ξ( 1 ) y k q(y)dy ξ( k=1 1 + k) z k dz (44) = z s 1 ( 1) k+1 ξ( 3 Γ(k) + k)zk dz (45) k=1 = Γ(1 + s)ξ( 3 s) (46) by Ramanujan s master theorem. Fubini follows from tail of Pólya s distribution.

19 Transform m G (x) Using (38) with 1 < α 3 and cα = ξ( 1 1 )e bα(α ) /ξ(α), yields ξ( 1 ) ξ( 1 + k) = ξ( 1 ) ( ) ξ(α + k α + 1 ) = cαe e bk (k α+ 1 ) H for k = 1,,... (47) where H = H ξ α and b = b α. Apply with α = 3. This allows us to directly calculate m G (s).

20 Hardy-Ramanujan Figure : Master Theorem b 3 =

21 GGC Construction Calculate m G (s) = x s 1 G(x)dx using (47) with α = 3 { } ( 1) k+1 G(x) = c αe e kx (k 1)H x k. (48) Γ(k) k=1 Here ξ( 3 + k)/ξ( 3 ) = e bk E(e kx ) from (36). Ramanujan s master theorem yields m G (s) = cγ(1 + s)e ( e sx (s+1) H ) (49) From (46) and (49), with b = b 5 4, c = ce 1 4 b /ξ( 5 4 ), H = H 5 4, for < s < 1 16, ξ( 1 ) ξ( 1 + s) = ξ( 1 )ξ( 3 s) ξ( 1 s) = ce 1 ξ( 3 s) (5) ( e sx ( s+1) H ) e b s E(e ( 1 4 s) H). (51)

22 GGC Construction Theorem The reciprocal Riemann ξ-function satisfies a HCM condition ξ( 1 ) ( ) ξ( 1 + s) = E exp( sh ξ 1 ) (5) where H ξ 1 is GGC. The GGC property is also preserved under exponential tilting. Then (5) is the LT of a GGC from the composition theorem (Theorem in [1, p.41]), as are functions φ H ( N k=1 c ks a k ) when c k > and < a k 1 (Theorem 3.3. [1, p.41]).

23 Riemann ζ-zeros The LT, E(exp( sh ξ 1 )), of a GGC distribution, is analytic in the whole complex plane cut along the negative real axis, and, in particular, it cannot have any singularities in that cut plane. As ξ ( 1 ) /ξ ( 1 + s ) = E(exp( sh ξ 1 )) for < s < 1, then, by analytic 4 continuation, this equality must hold for all values of s in the cut plane, namely C \ (, ). Hence, the denominator, ξ( 1 + s) cannot have any zeros there, and ξ( 1 + s) has no zeros for Re(s) >. Then ξ(s) has no zeroes for Re(s) > 1 Re(s) < 1 either. and, as ξ(s) = ξ(1 s), no zeroes for Therefore, all non-trivial zeros of the ζ function lie on the critical line 1 + is. arxiv