The is most likely not true

J.Sondow1 , C. Dumitrescu2, M. Wolf 3

1https://jonathansondow.github.io 2e-mail: [email protected] 3e-mail: [email protected]

Abstract We show that there is a possible contradiction between the Riemann’s Hypothesis and the on the strong universality of the zeta func- tion.

1 Introduction

In his only paper devoted to the published in 1859 [32] (it was also included as an appendix in [12]) Bernhard Riemann continued analytically the series

∞ X 1 , s = σ + it, σ > 1 (1) ns n=1 to the complex plane with exception of s = 1, where the above series is a harmonic divergent series. He has done it using the integral Γ(1 − s) Z (−z)s dz ζ(s) = , (2) 2πi ez − 1 z arXiv:2102.08313v2 [math.GM] 30 Apr 2021 C where the contour C is

6 C ' - -

& The definition of (−z)s is (−z)s = es log(−z), where the definition of log(−z) conforms to the usual definition of log(z) for z not on the negative real axis as the branch which is real for positive real z, thus (−z)s is not defined on the positive real axis,

1 see [12, p.10]. Appearing in (2) the gamma function Γ(z) has many representations, we present the Weierstrass product:

∞ e−Cz Y ez/k Γ(z) = . (3) z 1 + z  k=1 k Here C is the Euler–Mascheroni constant n ! X 1 C = lim − log(n) = 0.577216 .... (4) n→∞ k k=1 From (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 definition R ∞ −t z−1 of the gamma function given by the integral Γ(z) = 0 e t dt is valid only for Re[z] > 0. Recently perhaps over 100 representations of ζ(s) are known, for review of the integral and series representations see [27]. The function ζ(s) has trivial zeros at s = −2n, n = 1, 2, 3,... and nontrivial zeros in the critical strip 0 < Re(s) < 1. In [32] Riemann made the assumption, now called the Riemann Hypothesis (RH 1 for short in following), that all nontrivial zeros ρ lie on the critical line Re[s] = 2 : 1 ρ = 2 + iγ. Contemporary the above requirement is augmented by the demand that all nontrivial zeros are simple. The classical (from XX century) references on the RH are [39], [12], [17], [20]. In the 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 that it 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 not true, 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 found in [4]. When J. Derbyshire asked A. Odlyzko about his opinion on the validity of RH he replied “Either it’s true, or else it isn’t” [11, p. 357–358]. There were some attempts to prove RH using the physical methods, see [33] or [42]. We recall some linked to RH that were disproved in the past. Hasel- grove [15] disproved the P´olya’s stating that the “most” (i.e., 50% or more) of the natural numbers less than any given number have an odd number of prime factors, i.e. the function X L(x) := λ(n) (5) n≤x satisfies L(x) ≤ 0 for x ≥ 2, where λ(n) is Liouville’s function defined by λ(n) = (−1)Ω(n). Here Ω(n) is the number of, not necessarily distinct, prime factors in the decompo- r1 rn sition n = p1 ··· pα(n), with multiple factors counted multiply: r = r1 + ... + rn, i.e.

2 Ω(n) counts the total number of prime factors of an integer. From the truth of the P´olya Conjecture the RH follows, but not the other way around. The Haselgrove proof was indirect, and in 1960 Lehman [24] has found using the computer explicit counter–example: L(906180359) = 1. Many hopes to prove RH were linked to the Mertens conjecture. Let M(x) denote the defined by X M(x) = µ(n), (6) n≤x where µ(n) is the M¨obiusfunction   1 for n = 1 µ(n) = 0 when p2|n r  (−1) when n = p1p2 . . . pr.

From Mertens conjecture √ |M(x)| < x (7) again the RH would follow. However in 1985 A. Odlyzko and H. te Riele [29] disproved the Mertens conjecture, again not directly, but later it was shown by J. Pintz [30] that the first appears below exp(3.21 × 1064). The upper bound has since been lowered to exp(1.59 × 1040) [22]. Some analogs of the RH were proved and some other were disproved. Andr´e Weil proved the Riemann hypothesis to be true for field functions [41], while the Davenport-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 (also termed modern or enhanced) on the universality of the zeta function. First such theorem was proved in 1975 by S.M. Voronin [40]. He wanted to prove the RH but 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(1/2, 1) = 1 {s ∈ C : 2 < Re(s) < 1} ⊂ 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 K. Then for any  > 0 there exists a T (, f) such that    3 max f(s) − ζ s + + i T (, f) < . (8) |s|≤r 4 Put simply in words it means that the zeta function approximates locally any smooth function in a uniform way. We will use the strong (modern or enhanced) version of (8) (see [23], [26]):

3 Strong version of the universality theorem: Let K be a compact subset of 1 the strip D(1/2, 1) = {s ∈ C : 2 < Re(s) < 1} ⊂ C with connected complement, and let f(s) be holomorphic function in the interior of K and non-vanishing on K. Then for any  > 0,

1   lim inf µ τ ∈ [0,T ] : sup |ζ(s + iτ) − f(s)| <  > 0 (9) T →∞ T s∈K holds.

Here µ(A) is the Lebesgue measure of the set A. We stress that in these theorems the RH is not assumed.

Our reasoning will be based on the following

Strong universality in modulus theorem: Consider the function

F (s) = eAsζ(s), (10) where A > 0 is a positive constant. Let K be a compact (non–empty) subset of the 1 strip D(1/2, 1) = {s ∈ C : 2 < Re(s) < 1} ⊂ C with connected complement and let f(s) be a holomorphic function in the interior of K and non – vanishing on K. Then for any  > 0   1 lim inf µ τ ∈ [0,T ] : sup |F (s + iτ)| − |f(s)| <  > 0 (11) T →∞ T s∈K Proof: We have F 0(s) ζ0(s) = A + (12) F (s) ζ(s) Next consider the function f(s) g(s) = . (13) eAs The function ζ(s) has the strong universality property so we have:   1 lim inf µ τ ∈ [0,T ] : sup ζ(s + iτ) − g(s) <  > 0 (14) T →∞ T s∈K We have the equivalence

As Ks ζ(s + iτ) − g(s) <  ⇔ e ζ(s + iτ) − f(s) <  e < M (15) where As σA A e = e < e =: M (16)

4 1 for 2 < σ < 1. So we have the implication:

As As e ζ(s + iτ) − g(s)| <  ⇒ |e ζ(s + iτ) − f(s)| < M (17) and finally: 0 |F (s + iτ)| − |f(s)| < M =  . (18) and it is what we wanted to prove.  We give some heuristics behind our reasoning. The horizontal (i.e. at fixed t) 1 plots of the |ζ(s)| in the critical strip show a local minimum at σ = 2 (at nontrivial zeros) or close to it and to right of the critical line the modulus of the zeta increases 1 up to a local maxima. Sometimes this maximum is for σ ∈ ( 2 , 1) and sometimes it moves to the right of σ = 1, see Fig.1. We will show, assuming RH, that at very large t there will be no a local maximum in the critical strip — the maximum will permanently be outside critical strip, thus |ζ(s)| will be strictly increasing horizon- 1 tally for 2 < σ < 1 and all t > T0 with some very large T0. We introduce the notation: Dδ(T0) = {s : 0 < 1 − σ < δ and t > T0} (19) 1 with 0 < δ < 2 . Thus Dδ(T0) denotes strip far away in the critical strip close to the line Re(s) = 1. The bound T0 can be of the order of the estimate exp(1.59 × 1040) for the violation of the Merten’s conjecture [22] or even much larger — let us 79 34 recall here the first estimation of the Skewes number eee ≈ 101010 [35]. This will lead to the contradiction with well established strong universality theorem: namely the decreasing function f(s) = 1/s cannot be approximated by infinite number of vertical shifts of increasing horizontally F (s). In [16] on pp.60–61 we can read: “Long open conjectures in analysis tend to be false ”.

Figure 1: Plots of |ζ(s)| for t = 30982.39 and t = 101.2.

5 2 Preliminary results

In this section we collect some known results we will need to sketch the proof of falsity of RH.

Proposition 1. The Riemann–von Mangoldt formula gives the number N(T ) of zeros of the zeta function ζ(s) with imaginary part in the interval (0,T )

T T T 7  1  N(T ) = log − + + O + S(T ). (20) 2π 2π 2π 8 T

Here S(t) denote the value of 1 S(t) = π−1 arg ζ( + it) (21) 2

1 obtained by continuous variation along the straight lines joining 2, 2+iT , and 2 +iT , starting with the value 0. For proof see any book on the . The formula (20) was already stated by Riemann in his notable paper [32]. We will use [38, Lemma 2.1], but we found this fact appeared already in 1973 in [37, not labelled formula after eq.(20) on p. 151]

Proposition 2. : For any holomorphic function f(s) with f(s) 6= 0 for all s = σ+it in some open domain D we have (here f 0(s) denotes the complex derivative with respect to s) f 0(s) 1 ∂|f(s)| Re = . (22) f(s) |f(s)| ∂σ Proof: First we have: ∂|f(s)| ∂u ∂v |f(s)| = u(s) + v(s) . (23) ∂σ ∂σ ∂σ The Cauchy–Riemann equations for real and imaginary part of f(s) = u(s) + iv(s) asserts that ∂u(s) ∂v(s) ∂v(s) ∂u(s) f 0(s) = + i = − i . (24) ∂σ ∂σ ∂t ∂t From these equations we have:

f 0(s) u + iv uu + vv + i(uv − vu ) = σ σ = σ σ σ σ (25) f(s) u + iv |f(s)|2 and taking the real parts and using (23) gives (22). 

It follows that to show |ζ(s)| is increasing in some domain Dδ(T0) we need to 0 show that Re(ζ (s)/ζ(s)) > 0 in this set Dδ(T0). Thus we will use the following formula for the logarithmic derivative of the ζ(s) function (see e.g. [9, Chapt.12]):

6 Proposition 3. If s 6= ρ and ρ 6= −2n and s 6= 1 then

ζ0(s) 1 1 1 Γ0( s + 1) X  1 1 = B + + log(π) − 2 + + . (26) ζ(s) 1 − s 2 2 Γ( s + 1) s − ρ ρ 2 ρ

The last summation extends over all nontrivial zeros of ζ(s), i.e. assuming the RH 1 over ρ = 2 ± iγ. The constant B is the following sum [12, pp. 67 and 159], [9, pp.80–82]:

X 1 1 1 B = − = − 1 − C + log(4π) = −0.0230957 .... (27) ρ 2 2 ρ

The sum (27) is real and convergent when zeros ρ and complex conjugate ρ are paired together and summed according to increasing absolute values of the imaginary parts of ρ. However when we consider the real parts all the series that will emerge below (see e.g. (30)) are absolutely convergent so the summation order doesn’t matter. Taking real parts makes denominators larger because of terms γ2. Also we remark, that Γ0(s) ψ(s) := Γ(s) where ψ(s) is the digamma function. In [2, eq. (6.3.18)] we found: 1 1 ψ(s) = log(s) − − + ... (28) 2s 12s2 In our case it reads:  σ + it   t   1  Re ψ + 1 = log + O . (29) 2 2 t2

We are interested in the real part of (26) (confront (22)) thus we write: ! ζ0(s)  1  1 1 s  X 1 Re = Re + log(π)−Re ψ + 1 +Re . (30) ζ(s) 1 − s 2 2 2 s − ρ ρ

The last two terms have opposite signs and, as we will see later, their dependence on t will cancel exactly. The existing in the literature bounds on last term in (30) are of the form, see e.g. [39, p.9.6]: ! X 1 Re = O(log(t)) (31) s − ρ ρ and in view of (28) it is not sufficient for our purposes. In the next Section we will obtain more accurate estimation of this sum over nontrivial zeros of ζ(s).

7 3 Ancillary lemmas

In this section we will prove some lemmas needed to compute terms in (30). The 1 1 term s−1 vanishes for large t. We assume everywhere the RH: ρ = 2 ± iγ, γ ∈ R+, thus in the real part of the sum over ρ in (26) we write :

 1  σ − 1 σ − 1 Re = 2 = 2 (32) 2 1 2 2 s − ρ |s − ρ| (σ − 2 ) + (t − γ) and for s ∈ Dδ(T0) the sum over ρ is positive. We start with calculating the last term in (30):

Lemma 4. Assume RH. For all s with ζ(s) 6= 0 we have:

X 1 Z ∞ u − t = 2 N(u) du (33) |s − ρ|2 |s − ( 1 + iu)|4 Im(ρ)>0 2π 2

Here N(u) counts nontrivial zeros ρ of ζ(s) in the critical strip with 0 < Im(ρ) < u.

Proof: The lower limit of integration is 2π as the first nontrivial zero is larger 1 t  ρ1 = 2 + i14.134725 ... and it will assure positivity of log 2π . We assume RH 1 ρn = 2 + iγn. We will use the Abel summation in the form: Z y X 0 anφ(n) = A(y)φ(y) − A(x)φ(x) − A(u)φ (u) du. (34) x

In our case we take a(n) = N(γn+1) − N(γn) thus A(u) = N(u). For x = 2π and y → ∞ two first boundary terms above are zero and we are left with:

X 1 Z ∞ u − t = 2 N(u) du. (35) |s − ρ|2 ((σ − 1 )2 + (u − t)2)2 Im(ρ)>0 2π 2

 Lemma 5. Assume RH. We have  X 1  1  t  1 Z ∞ u − t t Re = log +2(σ− ) S(u) du+O( ). s − ρ 2 2π 2 ((σ − 1 )2 + (u − t)2)2 log(t) ρ 2π 2 (36)

8 7 Proof: First we notice that the term 8 as well as O(1/t) does not contribute under the integral:

7 Z ∞ u − t Z ∞  1  u − t du + O du = 8 ((σ − 1 )2 + (u − t)2)2 u ((σ − 1 )2 + (u − t)2)2 π 2 π 2 (37)  1  O −−−−−−→t→∞ 0 . t2

From (20) we have:

X 1 Z ∞  u  u  u  u − t = 2 log − +S(u) du. (38) |s − ρ|2 2π 2π 2π ((σ − 1 )2 + (u − t)2)2 Im(ρ)>0 2π 2

We have following primitives:   arctan u−t Z u u − t 1  σ− 1 u  du = 2 − (39) 1 2 2 2 1 1 2 2 π ((σ − 2 ) + (u − t) ) 2π (σ − 2 ) (σ − 2 ) + (u − t) and in the limit t → ∞ we have Z ∞ u u − t 1 du −−−−−−→t→∞ (40) 1 2 2 2 1 π π ((σ − 2 ) + (u − t) ) 2(σ − 2 ) Integrating by parts we have: Z ∞ u  u  u − t log du = 1 2 2 2 2π π 2π ((σ − 2 ) + (u − t) ) ∞ 1  u   1  u log · − − π 2π 1 2 2 2((σ − 2 ) + (u − t) ) 2π ! (41) Z ∞  u  1 log + 1 · − du = 1 2 2 2π 2π 2(σ − 2 ) + (u − t) 1 Z ∞ 1 1 Z ∞  u  1 du + log du 1 2 2 1 2 2 2π 2π (σ − 2 ) + (u − t) 2π 2π 2π (σ − 2 ) + (u − t) For the first integral we obtain

∞  u−t  ∞ arctan 1 Z 1 1 σ− 1 1 du = 2 −−−−−−→t→∞ . (42) 1 2 2 1 1 2π 2π (σ − 2 ) + (u − t) 2π (σ − 2 ) 2(σ − 2 ) 2π

9 For the second integral we have:

1 Z ∞ log u  I(σ, t) = 2π du = 1 2 2 2π 2π (σ − 2 ) + (t − u) ( 1 1   1   −4  2 log(π) + log + log × 2π 4i(σ − 1 ) i(σ − 1 ) − t i(σ − 1 ) + t 2 2 2 (43)   1   1  × log 1 − log − 1 − i(σ − 2 ) − t i(σ − 2 ) + t ) 1   2π   2π  1 Li2 1 − Li2 − 1 2i(σ − 2 ) i(σ − 2 ) + t i(σ − 2 ) − t where Li2(z) is the dilogarithm function:

∞ X zn Li (z) ≡ , |z| < 1. (44) 2 n2 n=1 Using the twice integrated geometrical series

∞ X zn z + (1 − z) ln(1 − z) = , |z| < 1. (45) n(n + 1) z n=1 we obtain ∞ ∞ X zn X zn z z2 z3 Li (z) = = + + + + ... (46) 2 n2 n(n + 1) 2 12 36 n=1 n=1 ∞ z + (1 − z) ln(1 − z) X zn = + (47) z n2(n + 1) n=1 Using the de l’Hospitale rule for the first term above, it follows that in the limit t → ∞ the dilogarithms in (43) disappear. The same conclusion can be reached more rigorously (but in much longer way) using the Kummer’s theorem on the integral (explicit) form of the Li2(z) for complex z, see e.g. [21, p. 67]. The final result is: 1  t  1 I(σ, t) = 1 log + o(t), (48) 2(σ − 2 ) 2π 2π where o(t) → 0 when t → ∞. 1 In (30) the sum is over all zeros 2 ± iγ and the sum over negative imaginary nontrivial zeros is a positive constant because (32) is still valid for these zeros and 1 for 2 < σ < 1 all these terms are positive. This sum is finite as terms are smaller than in the series X 1 X 1 C log 4π = Re = 1 + − = 0.0230957 ..., (49) γ2 + 1 ρ 2 2 γ>0 4 ρ

10 see e.g.[28, Corollary 10.14, eq.(10.30)]. Finally we have that  X 1  1  t  1 Z ∞ u − t log(t) Re = log +2(σ− ) S(u) du+O . s − ρ 2 2π 2 ((σ − 1 )2 + (u − t)2)2 t ρ 2π 2 (50) for large t, what we wanted to prove.  Then it follows that the terms log(t) and all constants in (30) cancel on rhs and we are left with: ζ0(s) 1 Z ∞ u − t log(t) Re = 2(σ − ) S(u) du + O . (51) 1 2 2 2 ζ(s) 2 2π ((σ − 2 ) + (u − t) ) t To get full disproof of the RH we need some good estimation of the last integral of S(u). The function S(t) is very fast oscillating, see Fig.3, and we expect the integral Z ∞ u − t S(u) du (52) 1 2 2 2 2π ((σ − 2 ) + (u − t) ) to be bounded for t ∈ R, see [19]. In the case of our integral in (52) only what happens in a small neighborhood of t matters, far from t the integrand decays very fast. When t grows the oscillations of S(t) become denser and denser in a neighborhood of t , thus making the cancelations upon integration more effective. In [11, p. 358] we can read the opinion of Odlyzko that RH can be violated when S(t) reaches value near 100. In fact Selberg [34] proved that S(t) is unbounded.

4 Possible disproof of the Riemann Hypothesis

If the RH is true, then it is known (see e.g. [39]) that !  t  S(t) = O log (53) log log(t)

The outcome of the analysis performed in the previous sections is that for t > T0 1 and 1 − δ < σ < 1 with 0 < δ < 2 the real part of the logarithmic derivative of ζ(s) is bounded ζ0  Re (s) = O(1). (54) ζ Then we have: F 0  ζ0(s) Re (s) = A + Re > 0 (55) F ζ(s) for large enough A to overcome bound in (54). Hence for the function F (s) from (22) we have that for all t > T0 and 1 − δ < σ < 1 modulus |F (s)| is strictly horizontally increasing for A > 0 in the Dδ(T0), i.e. ∂|F (s)| > 0 for s = σ + it ∈ D (T ). (56) ∂σ δ 0

11 Figure 2: The plot of S(t) for t ∈ (106, 106 + 20). All values are contained in the interval (−1, 1).

At this point we have a contradiction of RH with strong theorem on the universality in modulus, as the function F (s) can not be universal because it can not approximate e.g. the function f(s) = 1/s which has negative horizontal derivative for Re(s) = 1 σ ∈ ( 2 , 1). 1 How zeros outside Re(s) = 2 can restore compatibility with universality theo- rems? First of all the inequality (54) cannot be proved: suppose there are zeros 1 ρ = β + it with 2 < β < 1 (then there are also 3 additional symmetrical ze- 1 ros β − it, 1 − β ± it). Then instead of the positive term σ − 2 there will be 1 σ − β, 2 < β < 1 which can have any sign. The mechanism is following: the zeros 1 off critical line will pull local maxima of |ζ(s)| to the right of Re(s) = 2 . Around 1 every zero β + it, β 6= 2 the decreasing and increasing of |ζ(s)|, and hence of F (s), will be present on both sides of the zeros ρ = β + it restoring ability of ζ(s) and F (s) to approximate functions defined in strong version of the universality theorem.

12 Also, as the density of “approximation” regions in universality theorems is positive, there has to be infinity of zeros 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, is dense 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, 1). This means that we suspect the RH can be maximally violated. Let us remark, that the Lerch zeta-functions are, in a certain sense, universal and it was proved, that the real parts of zeros of this zeta-functions 1 form a dense subset of the interval ( 2 , 1), see [14].

5 Summary and final remarks

The main ingredients of our reasoning are: the strong universality theorem, the re- lation (22) and Lemma 5. It seems that the idea presented above can be applied to other 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 Dedekind zeta-functions, for a review see e.g. [26]. From (20) it follows that at large t between t and t + 1 there are approximately 1 t  1000000 2π log 2π zeros. For t of the order say 2π × 10 it gives almost 160000 zeros packed tightly on the interval of the length 1. On the other hand there should be occasionally such t that the large values of |ζ(s)| should be attained on the critical strip of the order t, see e.g. [36]. Hence the plot of ζ(s) at very large Im(s) should be very wild and zeros off the critical line can save the situation. Let us cite here the opinion of Ivi´c: “... the values for which ζ(s) will exhibit its true asymptotic behaviour must be really very large.”, see [18, p.12]. Let us remark that the function

k(x) = ex sin(ex) (57) is even worse than ζ(s): k(x) has exponentially increasing amplitude and up to x there are approximately ex/π zeros hence the interval between two consecutive zeros decreases 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 T0. There are effective versions of the universality theorems e.g. [31], and obtained numbers are enormously large. We can expect that T0 is also very large, maybe of the order of the Skewes number. Thus we can say that RH is practically true. Levinson [25] proved that more than one-third of zeros of Riemann’s ζ(s) are on critical line by relating the zeros of the zeta function to those of its derivative, and 1 Conrey [8] improved this further to two-fifths (precisely 40.77 % have Re(ρ) = 2 ). The present record seems to belong to S. Feng, who proved that at least 41.28% of the zeros of the Riemann zeta function are on the critical line [13]. In view of our

13 result we can expect theorems like: no more than say 99% or even no more than 70% of nontrivial zeros of ζ(s) are on critical line.

Acknowledgment: We are grateful to Professor Philippe Blanc for pointing to us error in the previous version of this paper and e-mail exchange.

References

[1] http://empslocal.ex.ac.uk/people/staff/mrwatkin/zeta/littlewood.htm.

[2] M. Abramowitz and I. A. Stegun. Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables. Dover, New York, ninth Dover printing, tenth GPO printing edition, 1964.

[3] E. P. Balanzario and J. S´anchez-Ortiz. Zeros of the Davenport-Heilbronn coun- terexample. Mathematics of Computation, 76(260):2045–2050, Oct 2007.

[4] P. Blanc. A New Reason for Doubting the Riemann Hypothesis. Experimental Mathematics, pages 1–5, Apr 2019.

[5] A. R. Booker. Notices of the AMS., 54(3):1208–1211, Nov 2006.

[6] P. Borwein, S. Choi, B. Rooney, and A. Weirathmueller. The Riemann Hypoth- esis: A Resource For The Afficionado And Virtuoso Alike. Springer Verlag, Berlin, Heidelberg, New York, 2007.

[7] K. Broughan. Equivalents of the Riemann Hypothesis, volume 1 and 2. Cam- bridge University Press, 2017.

[8] J. B. Conrey. More than two-fifths of the zeros of the Riemann zeta function are on the critical line. J. Reine Angew. Math., 399:1–26, 1989.

[9] H. Davenport. Multiplicative Number Theory, volume 74 of Graduate Texts in Mathematics. Springer-Verlag, New York, second edition, 1980. Revised by Hugh L. Montgomery.

[10] H. Davenport and H. Heilbronn. On the Zeros of Certain II. Journal of the London Mathematical Society, s1-11(4):307–312, Oct 1936.

[11] J. Derbyshire. Prime Obsession. Bernhard Riemann and the greatest unsolved problem in mathematics. Joseph Henry Press, Washington, 2003.

[12] H. M. Edwards. Riemann’s zeta function. Academic Press, 1974. Pure and Applied Mathematics, Vol. 58.

14 [13] S. Feng. Zeros of the Riemann zeta function on the critical line. Journal of Number Theory, 132:511–542, 2012.

[14] R. Garunkstis and A. Grigutis. The size of the Lerch zeta-function at places symmetric with respect to the line Re(s) = 1/2. Czechoslovak Mathematical Journal, 69(1):25–37, Apr 2018.

[15] C. B. Haselgrove. A disproof of a conjecture of Polya. Mathematika, 5:141–145, 1958.

[16] J. Holt. When Einstein Walked with Goedel : Excursions to the Edge of Thought. Farrar Straus And Giroux, New York, 2018.

[17] A. Ivi´c. The Riemann zeta-function: the theory of the Riemann zeta-function with applications. Wiley, New York, 1985.

[18] A. Ivi`c. On some reasons for doubting the Riemann hypothesis. arXiv:math.NT/0311162, Nov. 2003.

[19] A. Karatsuba and M. A. Korolov. The argument of the Riemann zeta function. Russian Math. Surveys, 60(3):433–488, 2005.

[20] A. Karatsuba and S. M. Voronin. The Riemann zeta-function. Walter de Gruyter, Berlin New York, 1992.

[21] A. N. Kirillov. Dilogarithm identities. Progress of Theoretical Physics Supple- ment, 118:61–142, 1995.

[22] T. Kotnik and H. te Riele. The Mertens conjecture revisited. In 7-th ANTS, volume 4076 of Lecture Notes in Computer Science, pages 156–167, 2006.

[23] A. Lauriniˇckas. Limit Theorems for the Riemann Zeta-Function. Mathematics and Its Applications. Kluwer Academic Publishers, Dordrecht, 1996.

[24] S. Lehman, R. On Liouville’s Function. Math. Comput., 4:311–320, 1960.

[25] N. Levinson. More than one third of zeros of Riemann’s zeta-function are on σ = 1/2. Advances in Math., 13:383–436, 1974.

[26] K. Matsumoto. A Survey On The Theory Of Universality For Zeta And L- Functions. In Number Theory: Plowing and Starring Through High Wave Forms, pages 95–144. World Scientific, Feb 2015.

[27] M. Milgram. Integral and Series Representations of Riemann’s Zeta Function and Dirichlet’s Eta Function and a Medley of Related Results. Journal of Mathematics, 2013:Article ID 181724, 2013.

15 [28] H. L. Montgomery and R. C. Vaughan. Multiplicative Number Theory I: Clas- sical Theory. Cambridge Studies in Advanced Mathematics. Cambridge Uni- versity Press, 2006.

[29] A. M. Odlyzko and H. J. J. te Riele. Disproof of the Mertens Conjecture. J. Reine Angew. Math., 357:138–160, 1985.

[30] J. Pintz. An effective disproof of the Mertens conjecture. Asterisque, 147- 148:325–333, 1987.

[31] R. R. Garunkˇstis, A. Laurincikas, K. Matsumoto, J. Steuding, and R. Steuding. Effective uniform approximation by the Riemann zeta-function. Publicacions Matematiques, 54:209–219, Jan 2010.

[32] B. Riemann. Ueber¨ die Anzahl der Primzahlen unter einer gegebenen Gr¨osse. Monatsberichte der K¨oniglichPreußischen Akademie der Wissenschaften zu Berlin., pages 671–680, November 1859. english translation available at http://www.maths.tcd.ie/pub/HistMath/People/Riemann.

[33] D. Schumayer and D. A. W. Hutchinson. Physics of the Riemann hypothesis. Rev. Mod. Phys., 83(2):307–330, Apr 2011.

[34] A. Selberg. On the remainder in the formula for N(T ), the number of zeros of ζ(s) in the strip 0 < t < T . Avh. Norske Vid. Akad. Oslo. I., pages 1–27, 1944.

[35] S. Skewes. On the difference π(x) − Li(x). J. London Math. Soc., 8:277–283, 1934.

[36] K. Soundararajan. Extreme values of zeta and L-functions. Mathematische Annalen, 342(2):467–486, May 2008.

[37] R. Spira. Zeros of ζ0(s) and the Riemann hypothesis. Illinois J. Math., 17(1):147–152, 03 1973.

[38] G. K. Srinivasan and P. Zvengrowski. On the Horizontal Monotonicity of |Γ(s)|. Canadian Mathematical Bulletin, 54(3):538–543, Sep 2011.

[39] E. C. Titchmarsh. The Theory of the Riemann Zeta-function. The Clarendon Press Oxford University Press, New York, sec. ed. edition, 1986. Edited and with a preface by D. R. Heath-Brown.

[40] S. M. Voronin. Theorem on the Universality of the Riemann Zeta Function. Izv. Akad. Nauk SSSR, Ser. Matem., 39:475–486, 1975. Reprinted in Math. USSR Izv. 9, 443-445, 1975.

[41] A. Weil. Sur les courbes algebriques et les varietes qui s’en deduisent. Hermann, Paris, 1948.

16 [42] M. Wolf. Will a physicist prove the Riemann hypothesis? Reports on Progress in Physics, 83(3):36001, Mar 2020.

17