The Riemann Hypothesis is is most likely not true
TThe Riemann Hypothesis is not true
J.Sondow , C. Dumitrescu , M. Wolf https://jonathansondow.github.io e-mail: [email protected] e-mail: [email protected] Abstract
We show that there is a contradiction between the Riemann’s Hypoth-esis and the theorem on the strong universality of the zeta function.
In his only paper devoted to the number theory published in 1859 [30] (it was alsoincluded as an appendix in [12]) Bernhard Riemann continued analytically the series ∞ (cid:88) n =1 n s , s = σ + it, σ > s = 1, where the above series is a harmonicdivergent series. He has done it using the integral ζ ( s ) = Γ(1 − s )2 πi (cid:90) C ( − z ) s e z − dzz , (2)where the contour C is (cid:54) (cid:45) C (cid:39)(cid:38)(cid:24)(cid:25) (cid:45)(cid:27) Appearing in (2) the gamma function Γ( z ) has many representations, we presentthe Weierstrass product: Γ( z ) = e − γz z ∞ (cid:89) k =1 (cid:16) zk (cid:17) − e z/k . (3)1 a r X i v : . [ m a t h . G M ] F e b rom (3) it is seen that Γ( z ) is defined for all complex numbers z , except z = − n for integer n >
0, where are the simple poles of Γ( z ). The most popular definitionof the gamma function given by the integral Γ( z ) = (cid:82) ∞ e − t t z − dt is valid only for Re [ z ] >
0. Recently perhaps over 100 representations of ζ ( s ) are known, for reviewof the integral and series representations see [26]. The function ζ ( s ) has trivial zerosat s = − n, n = 1 , , , . . . and nontrivial zeros in the critical strip 0 < Re ( s ) < Riemann Hypothesis (RHfor short in following), that all nontrivial zeros ρ lie on the critical line Re [ s ] = : ρ = + iγ . Contemporary the above requirement is augmented by the demand thatall nontrivial zeros are simple.The classical (from XX century) references on the RH are [36], [12], [17], [19]. Inthe XXI century there appeared two monographs about the zeta function: [6] and [7].There was a lot of attempts to prove RH and the common opinion was thatit is true. However let us notice that there were famous mathematicians: J. E.Littlewood [1], P. Turan and A.M. Turing [5, p.1209] believing that the RH is nottrue, see the paper “On some reasons for doubting the Riemann hypothesis” [18](reprinted in [6, p.137]) written by A. Ivi´c. New arguments against RH can be foundin [4]. When J. Derbyshire asked A. Odlyzko about his opinion on the validity ofRH he replied “Either it’s true, or else it isn’t” [11, p. 357–358]. There were someattempts to prove RH using the physical methods, see [31] or [39].We recall some conjectures linked to RH that were disproved in the past. Hasel-grove [15] disproved the P´olya’s Conjecture stating that the “most” (i.e., 50% ormore) of the natural numbers less than any given number have an odd number ofprime factors, i.e. the function L ( x ) := (cid:88) n ≤ x λ ( n ) (4)satisfies L ( x ) ≤ x ≥
2, where λ ( n ) is Liouville’s function defined by λ ( n ) = ( − r ( n ) . Here r ( n ) is the number of, not necessarily distinct, prime factors in the decomposi-tion n = p r · · · p r n α ( n ) , with multiple factors counted multiply: r = r + . . . + r n . Fromthe truth of the P´olya Conjecture the RH follows, but not the other way around.The Haselgrove proof was indirect, and in 1960 Lehman [23] has found using thecomputer explicit counter–example: L (906180359) = 1.Many hopes to prove RH were linked to the Mertens conjecture. Let M ( x )denote the Mertens function defined by M ( x ) = (cid:88) n ≤ x µ ( n ) , (5)2here µ ( n ) is the M¨obius function µ ( n ) = n = 10 when p | n ( − r when n = p p . . . p r . From | M ( x ) | < x (6)again the RH would follow. However in 1985 A. Odlyzko and H. te Riele [27]disproved the Mertens conjecture, again not directly, but later it was shown by J.Pintz [28] that the first counterexample appears below exp(3 . × ). The upperbound has since been lowered to exp(1 . × ) [21].Some analogs of the RH were proved and some other were disproved. Andr´eWeil proved the Riemann hypothesis to be true for field functions [38], while theDavenport-Heilbronn zeta-function [10], which shares many properties with usual ζ ( s ), has zeros outside critical line and even to the right of Re ( s ) = 1, see [3].In this paper we are going to present incompatibility of RH with the strong (alsotermed modern or enhanced) theorems on the universality of the zeta function. Firstsuch theorem was proved in 1975 by S.M. Voronin [37]. He wanted to prove the RHbut instead he proved remarkable theorem on the universality of the Riemann ζ ( s )function: Voronin’s theorem : Let K be a compact subset of the strip D (1 / ,
1) = { s ∈ C : < Re ( s ) < } ⊂ C such that the complement of K is connected. Let f : K → C be a continuous function on K which is holomorphic on the interior of K and is not zero in U . Then for any (cid:15) > T ( (cid:15), f ) such thatmax | s |≤ r (cid:12)(cid:12)(cid:12)(cid:12) f ( s ) − ζ (cid:18) s + (cid:18)
34 + i T ( (cid:15), f ) (cid:19)(cid:19)(cid:12)(cid:12)(cid:12)(cid:12) < (cid:15). (7)Put simply in words it means that the zeta function approximates locally any smoothfunction in a uniform way. We will use the strong (modern or enhanced) version of(7) (see [22], [25]): Strong version of the universality theorem : Let K be a compact subset ofthe strip D (1 / ,
1) = { s ∈ C : < Re ( s ) < } ⊂ C with connected complement,and let f ( s ) be holomorphic function in the interior of K and non-vanishing on K .Then for any (cid:15) > T →∞ T µ (cid:26) τ ∈ [0 , T ] : sup s ∈ K | ζ ( s + iτ ) − f ( s ) | < (cid:15) (cid:27) > µ ( A ) is the Lebesgue measure of the set A . We stress that in these theoremsthe RH is not assumed .We give some heuristics behind our reasoning. The horizontal (i.e. at fixed t )plots of the | ζ ( s ) | in the critical strip show a local minimum at σ = (at nontrivialzeros) or close to it and to right of the critical line the modulus of the zeta increasesup to a local maxima. Sometimes this maximum is for σ ∈ ( ,
1) and sometimes itmoves to the right of σ = 1, see Fig.1. We will show, assuming RH , that at verylarge t there will be no a local maximum in the critical strip — the maximum willpermanently be outside critical strip, thus | ζ ( s ) | will be strictly increasing horizon-tally for < σ < t > T with some very large T . We introduce thenotation: D δ ( T ) = { s : 0 < − σ < δ and t > T } (9)i.e. D δ ( T ) denotes strip far away in the critical strip close to the line Re ( s ) = 1. Thebound T can be of the order of the estimate exp(1 . × ) for the violation of theMerten’s conjecture [21] or even much larger — let us recall here the first estimationof the Skewes number e e e ≈ [32]. This will lead to the contradictionwith well established strong universality theorem: namely the decreasing function f ( s ) = 1 /s cannot be approximated by infinite number of vertical shifts of increasinghorizontally ζ ( s ).In [16] on pp.60–61 we can read: “Long open conjectures in analysis tend to befalse ”. Figure 1: Plots of | ζ ( s ) | for t = 30982 .
39 and t = 101 . In this section we collect some known results we will need to prove our Theorem 1.
Proposition 1.
The Riemann–von Mangoldt formula gives the number N ( T ) ofzeros of the zeta function ζ ( s ) with imaginary part in the interval (0 , T ) N ( T ) = T π log T π − T π + O (log T ) . (10)4or proof see any book on the Riemann zeta function. The formula (10) wasalready stated by Riemann in his notable paper [30].We will use [35, Lemma 2.1], but we found this fact appeared already in 1973 in[34, not labelled formula after eq.(20) on p. 151] Proposition 2. : For any holomorphic function f ( s ) with f ( s ) (cid:54) = 0 for all s = σ + it in some open domain D we have (here f (cid:48) ( s ) denotes the complex derivative withrespect to s ) Re (cid:16) f (cid:48) ( s ) f ( s ) (cid:17) = 1 | f ( s ) | ∂ | f ( s ) | ∂σ . (11) Proof:
First we have: | f ( s ) | ∂ | f ( s ) | ∂σ = u ( s ) ∂u∂σ + v ( s ) ∂v∂σ . (12)The Cauchy–Riemann equations for real and imaginary part of f ( s ) = u ( s ) + iv ( s )asserts that f (cid:48) ( s ) = ∂u ( s ) ∂σ + i ∂v ( s ) ∂σ = ∂v ( s ) ∂t − i ∂u ( s ) ∂t . (13)From these equations we have: f (cid:48) ( s ) f ( s ) = u σ + iv σ u + iv = uu σ + vv σ + i ( uv σ − vu σ ) | f ( s ) | (14)and taking the real parts and using (12) gives (11). (cid:3) It follows that to show | ζ ( s ) | is increasing in some domain D δ ( T ) we need toshow that Re ( ζ (cid:48) ( s ) /ζ ( s )) > D δ ( T ). Thus we will use the followingformula for the logarithmic derivative of the ζ ( s ) function (see e.g. [9, Chapt.12]): Proposition 3. If s (cid:54) = ρ and ρ (cid:54) = − n and s (cid:54) = 1 then ζ (cid:48) ( s ) ζ ( s ) = B + 11 − s + 12 log( π ) −
12 Γ (cid:48) ( s + 1)Γ( s + 1) + (cid:88) ρ (cid:16) s − ρ + 1 ρ (cid:17) . (15) The last summation extends over all nontrivial zeros of ζ ( s ) , i.e. assuming the RHover ρ = ± iγ . The constant B in (15) has a few representations, see e.g. [9, Chapt.12]: B = −
12 + C − π ) = − . . . . (16)where C is the Euler–Mascheroni constant C = lim n →∞ (cid:0)(cid:80) nk =1 1 k − log( n ) (cid:1) = 0 . . . . .This B is the following sum [12, pp. 67 and 159], [9, pp.80–82]: B = − (cid:88) ρ ρ = − − C + 12 log(4 π ) . (17)5he sum (17) is real and convergent when zeros ρ and complex conjugate ρ are pairedtogether and summed according to increasing absolute values of the imaginary partsof ρ . However when we consider the real parts all the series that will emerge below(see e.g. (19)) are absolutely convergent so the summation order doesn’t matter.Taking real parts makes denominators larger because of terms γ .Also we remark, that Γ (cid:48) ( s )Γ( s ) = ψ ( s ) , where ψ ( s ) is the digamma function. In [2, eq. (6.3.18)] we found: ψ ( s ) = log( s ) − s − s + . . . (18)We are interested in the real part of (15) (confront (11)) thus we write: Re (cid:18) ζ (cid:48) ( s ) ζ ( s ) (cid:19) = Re (cid:18) − s (cid:19) + 12 log( π ) − Re (cid:18) ψ (cid:16) s (cid:17)(cid:19) + Re (cid:32)(cid:88) ρ s − ρ (cid:33) (19)The last two terms have opposite signs and, as we will see later, their dependenceon t will cancel exactly. The existing in the literature bounds on last term in (19)are of the form, see e.g. [36, p.9.6]: Re (cid:32)(cid:88) ρ s − ρ (cid:33) = O (log( t )) (20)and in view of (18) it is not sufficient for our purposes. In the next Section we willobtain more accurate estimation of this sum over nontrivial zeros of ζ ( s ). In this section we will prove some lemmas needed to compute terms in (19). Theterm s − vanishes for large t . We assume everywhere the RH: ρ = ± iγ, γ ∈ R + ,thus in the real part of the sum over ρ in (15) we write : Re (cid:18) s − ρ (cid:19) = σ − | s − ρ | = σ − ( σ − ) + ( t − γ ) (21)and for s ∈ D δ ( T ) the sum over ρ is positive. We start with calculating the lastterm in (19): Lemma 4.
Assume RH. For all s with ζ ( s ) (cid:54) = 0 we have: (cid:88) Im ( ρ ) > | s − ρ | = 2 (cid:90) ∞ π N ( u ) u − t | s − ( + iu ) | du (22) Here N ( u ) counts nontrivial zeros ρ of ζ ( s ) in the critical strip with < Im ( ρ ) < u . roof: The lower limit of integration is 2 π as the first nontrivial zero is larger ρ = + i . . . . and it will assure positivity of log (cid:0) t π (cid:1) . We assume RH ρ n = + iγ n . We will use the Abel summation in the form: (cid:88) x
Assume RH. We have with positive
Const > and for t = Re ( s ) → ∞ : Re (cid:16) (cid:88) ρ s − ρ (cid:17) →
12 log( t ) + Const. (25)
Proof:
First we notice that the term log( t ) under big– O in (10) does not con-tribute to the sum over ρ in the limit of large t : (cid:90) ∞ π log( u ) u − t (( σ − ) + ( u − t ) ) du = 14 (cid:16) t arctan (cid:16) u − tσ − (cid:17) ( σ − ) + ( σ − ) t +1( σ − ) + t log (cid:18) u ( σ − ) + ( u − t ) (cid:19) − u )( σ − ) + ( u − t ) (cid:17)(cid:12)(cid:12)(cid:12)(cid:12) ∞ π t →∞ −−−−−−−→ . (26)From (10) we have: (cid:88) Im ( ρ ) > | s − ρ | = 2 (cid:90) ∞ π (cid:16) u π log (cid:16) u π (cid:17) − u π (cid:17) u − t (( σ − ) + ( u − t ) ) du. (27)We have following primitives: (cid:90) uπ u − t (( σ − ) + ( u − t ) ) du = 12 π (cid:16) arctan (cid:16) u − tσ − (cid:17) ( σ − ) − u ( σ − ) + ( u − t ) (cid:17) (28)7nd in the limit t → ∞ we have (cid:90) ∞ π uπ u − t (( σ − ) + ( u − t ) ) du t →∞ −−−−−−→ σ − ) (29)Integrating by parts we have: (cid:90) ∞ π uπ log (cid:16) u π (cid:17) u − t (( σ − ) + ( u − t ) ) du =1 π (cid:32) u log (cid:16) u π (cid:17) · (cid:16) − σ − ) + ( u − t ) ) (cid:17)(cid:12)(cid:12)(cid:12)(cid:12) ∞ π − (cid:90) ∞ π (cid:0) log (cid:16) u π (cid:17) + 1 (cid:1) · (cid:0) − σ − ) + ( u − t ) (cid:1) du (cid:33) =12 π (cid:90) ∞ π σ − ) + ( u − t ) du + 12 π (cid:90) ∞ π log (cid:16) u π (cid:17) σ − ) + ( u − t ) du (30)For the first integral we obtain12 π (cid:90) ∞ π σ − ) + ( u − t ) du = 12 π arctan (cid:16) u − tσ − (cid:17) ( σ − ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ∞ π t →∞ −−−−−−→ σ − ) . (31)For the second integral we have: I ( σ, t ) = 12 π (cid:90) ∞ π log (cid:0) u π (cid:1) ( σ − ) + ( t − u ) du =12 π (cid:40) i ( σ − ) (cid:18) π ) + log (cid:18) i ( σ − ) − t (cid:19) + log (cid:18) − i ( σ − ) + t (cid:19)(cid:19) ×× (cid:18) log (cid:18) i ( σ − ) − t (cid:19) − log (cid:18) − i ( σ − ) + t (cid:19)(cid:19) − i ( σ − ) (cid:18) Li (cid:18) πi ( σ − ) + t (cid:19) − Li (cid:18) − πi ( σ − ) − t (cid:19)(cid:19) (cid:41) (32)where Li ( z ) is the dilogarithm function:Li ( z ) ≡ ∞ (cid:88) n =1 z n n , | z | < . (33)Using the twice integrated geometrical series ∞ (cid:88) n =1 z n n ( n + 1) = z + (1 − z ) ln(1 − z ) z , | z | < . (34)8e obtain Li ( z ) = ∞ (cid:88) n =1 z n n = ∞ (cid:88) n =1 z n n ( n + 1) + z z
12 + z
36 + . . . (35)= z + (1 − z ) ln(1 − z ) z + ∞ (cid:88) n =1 z n n ( n + 1) (36)Using the de l’Hospitale rule for the first term above, it follows that in the limit t → ∞ the dilogarithms in (32) disappear. The same conclusion can be reached morerigorously (but in much longer way) using the Kummer’s theorem on the integral(explicit) form of the Li ( z ) for complex z , see e.g. [20, p. 67]. The final result is: I ( σ, t ) = 12( σ − ) log( t ) + 12 π o ( t ) , (37)where o ( t ) → t → ∞ .In (19) the sum is over all zeros ± iγ and the sum over negative imaginarynontrivial zeros is a positive constant because (21) is still valid for these zeros andfor < σ < γ n ≈ πn/ log( n ) thus there is a majorization with 1 < α < < Re (cid:88) Im ( ρ ) < s − ρ = (cid:18) σ − (cid:19) (cid:88) γ n (cid:18) σ − ) + ( t − γ n ) (cid:19) < (cid:18) σ − (cid:19) (cid:88) n n α < ∞ . (38)Finally we have12 log( t ) < Re (cid:16) (cid:88) ρ s − ρ (cid:17) <
12 log( t ) + Const for large t (39)what we wanted to prove. (cid:3) Lemma 6.
Assuming RH and for s ∈ D δ ( T ) with < δ < and with some verylarge T and C > we have: < Re (cid:18) ζ (cid:48) ( s ) ζ ( s ) (cid:19) <
12 log( π ) + C . (40) In particular we can write: Re (cid:18) ζ (cid:48) ( s ) ζ ( s ) (cid:19) t →∞ −−−−−−→
12 log( π ) + C . (41) Proof:
The proof of the this inequalities follows by collecting results of theabove Lemmas. (cid:3) Disproof of the Riemann Hypothesis
The outcome of the analysis performed in the previous sections is that for t > T and 1 − δ < σ < < δ < the inequality Re (cid:18) ζ (cid:48) ζ ( s ) (cid:19) > t > T and 1 − δ < σ < | ζ ( s ) | is strictly horizontally increasing in the D δ ( T ), i.e. ∂ | ζ ( s ) | ∂σ > s = σ + it ∈ D δ ( T ) . (43)Now we will show that there is a contradiction of RH with strong version of theuniversality theorem. Theorem 1:
The Riemann Hypothesis in incompatible with the strong univer-sality theorem.
Proof:
Choose f ( s ) = s . This function fulfills assumptions of the strong versionof the universality theorem (8) and its modulus is strictly decreasing thus can notbe approximated by infinite number of shifts of ζ ( s ) as beginning with some (large) T modulus | ζ ( s ) | is strictly increasing horizontally. More formally let us consider f ( s ) = 1 /s in some set K in the strip D (1 / ,
1) and take values of this function onsome horizontal level Im ( s ) = τ at two points σ + iτ and σ + iτ with σ < σ .According to the strong universality theorem there should be infinity of verticaltranslates of the ζ ( s ) approximating f ( s ) with arbitrary accuracy (cid:15) . These shifts of ζ ( s ) finally have to fall into D δ ( T ), where | ζ ( s ) | is increasing horizontally: | ζ ( σ + it ) | < | ζ ( σ + it ) | (44)Thus we write: (cid:12)(cid:12)(cid:12)(cid:12) ζ ( σ , + it ) − σ , + iτ (cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12) ζ ( σ , + it ) − σ , − iτσ , + τ (cid:12)(cid:12)(cid:12)(cid:12) < (cid:15). (45)For complex numbers a, b the inequality | a − b | ≥ (cid:12)(cid:12) | a | − | b | (cid:12)(cid:12) holds thus we have: (cid:15) > (cid:12)(cid:12)(cid:12)(cid:12) ζ ( σ , + it ) − σ , − iτσ , + τ (cid:12)(cid:12)(cid:12)(cid:12) ≥ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ζ ( σ , + it ) (cid:12)(cid:12) − (cid:113) σ , + τ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) . (46)After elementary manipulations we obtain | ζ ( σ + it ) | < | ζ ( σ + it ) | + 2 (cid:15) + 1 (cid:112) σ + τ − (cid:112) σ + τ . (47)10ecause for σ < σ the last difference above is negative thus for sufficiently small (cid:15) : 2 (cid:15) < (cid:112) σ + τ − (cid:112) σ + τ (48)the inequality follows: | ζ ( σ + it ) | < | ζ ( σ + it ) | . (49)Obtained inequality contradicts (44) (and (43)), thus the RH cannot be true. (cid:3) How zeros outside Re ( s ) = can restore compatibility with universality theo-rems? First of all the inequality (42) cannot be proved: suppose there are zeros ρ = β + it with < β < β − it, − β ± it ). Then instead of the positive term σ − there will be σ − β, < β < | ζ ( s ) | to the right of Re ( s ) = . Aroundevery zero β + it, β (cid:54) = the decreasing and increasing of | ζ ( s ) | will be present onboth sides of the zeros ρ = β + it restoring ability of ζ ( s ) to approximate functionsdefined in strong version of the universality theorem. Also, as the density of “ap-proximation” regions in universality theorems is positive, there has to be infinity ofzeros off critical line and very close to the left of Re ( s ) = 1. Thus we make the Conjecture:
The set of σ such that there exists t such that ζ ( σ + it ) = 0, isdense in (0 ,
1) (in usual topology).In other words: the real parts of the nontrivial zeros of the zeta function ζ ( s )form a dense subset of the interval (0 , , The main ingredients of our reasoning are: the strong universality theorem, the re-lation (11) and Lemma 6 It seems that the idea presented above can be applied toother zeta functions, for which the strong universality theorems holds, like the Hur-witz zeta functions ζ ( s, a ) or Dirichlet L -functions with Euler products or Dedekindzeta-functions, for a review see e.g. [25].From (10) it follows that at large t between t and t + 1 there are approximately π log (cid:0) t π (cid:1) zeros. For t of the order say 2 π × it gives almost 160000 zerospacked tightly on the interval of the length 1. On the other hand there should beoccasionally such t that the large values of | ζ ( s ) | should be attained on the criticalstrip of the order t , see e.g. [33]. Hence the plot of ζ ( s ) at very large Im ( s ) shouldbe very wild and zeros off the critical line can save the situation. Let us remark that11he function k ( x ) = e x sin( e x ) (50)is even worse than ζ ( s ): k ( x ) has exponentially increasing amplitude and up to x there are approximately e x /π zeros hence the interval between two consecutive zerosdecreases like e − x . The set of zeros of k ( x ) has an accumulation point in the infinity.We are at present unable to give the estimate for T . There are effective versionsof the universality theorems e.g. [29], and obtained numbers are enormously large.We can expect that T is also very large, maybe of the order of the Skewes number.Thus we can say that RH is practically true .Levinson [24] proved that more than one-third of zeros of Riemann’s ζ ( s ) are oncritical line by relating the zeros of the zeta function to those of its derivative, andConrey [8] improved this further to two-fifths (precisely 40.77 % have Re ( ρ ) = ).The present record seems to belong to S. Feng, who proved that at least 41.28% ofthe zeros of the Riemann zeta function are on the critical line [13]. In view of ourresult we can expect theorems like: no more than say 99% or even no more than70% of nontrivial zeros of ζ ( s ) are on critical line. References [1] http://empslocal.ex.ac.uk/people/staff/mrwatkin/zeta/littlewood.htm.[2] M. Abramowitz and I. A. Stegun.
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