Suppose that the Riemann hypothesis for the Riemann $\zeta$-function is false. There have yet been no workable results that the falsity of the Riemann hypothesis implies. The result of this article may be considered as more workable for the sake of deducing other new results on the Riemann hypothesis.
Deep Dive into A new result under the negation of the Riemann hypothesis.
Suppose that the Riemann hypothesis for the Riemann $\zeta$-function is false. There have yet been no workable results that the falsity of the Riemann hypothesis implies. The result of this article may be considered as more workable for the sake of deducing other new results on the Riemann hypothesis.
In this article, we treat the distribution of poles of the function
where Λ(n) is the arithmetical Mangoldt Λ-function and p ∈ Q ∩ (0, 1).
To express the main result of our discussion, the following theorem will become the starting point.
Theorem 1. Suppose that M (s, 1/2) has a pole at s = ρ, 1/2 < Re(ρ) < 1 and Im(ρ) some large positive real number. Then to each p = 1/2+a/b (a/b small) sufficiently close to p = 1/2, there corresponds a pole (or several of those) of M (s, a/b). The formula for residues and poles of M (s, a/b) is expressed by Dirichlet L-functions; specifically, as s approaches a pole of M (s, 1/2 + a/b), we have
where
and χ denotes Dirichlet character.
We note that the poles of M (s, 1/2) are exactly those of (ζ ′ /ζ)(s), where ζ denotes the Riemann zeta-function.
Besides, e -2πi(1/2+a/b)p = e -πip-2πiap/b = -e -2πiap/b for p odd primes. The main result is described as follows. That such residues exist is clear by Theorem 1.
The following is used in the proof of Theorem 1
where Φ(p, r, s) ≡ j≥1 e -2πipj (j + r) -s .
We will present a supporting argument for Theorem 1 in Section 3.
The following formula for the Γ-function [3, pp.405]
as |t| → ∞, will be often used.
2 The proof of Theorem 2
We define [3]
and
where 0 < q < 1 with 1 -q < c < 1. Shifting the path in I 2 to the left, we find out that
express (1 + x) -q as a power series in x at x = 0 and use Γ(s)s = Γ(s + 1). We use these two formulas (3) and (4) to evaluate
Here, we recall that [4]
where 1 2πi (c) F(s)x -s ds = f (x) and 1 2πi (c) G(s)x -s ds = g(x). We associate F and f with Γ(s) and e -x (by (3)), and G and g with Γ(s + q -1)Γ(1 -s) and Γ(q)x q-1 (1 + x) -q (by ( 4)), respectively. This gives
Thus choosing x → xpn with -π/2 < arg(p) ≤ 0, multiplying by
and summing over n ≥ 2, we have for 1 < Re(κ) < 2 and δ > 0 (so that c + δ + κ > 1),
From here on, we keep 1 < Re(κ) < 2 unless otherwise mentioned. We use the the change of variables z = xw, let x = 2πe πi/2 , and obtain
We choose q = 1 + δ by analytic continuation; the restriction 0 < q < 1 at the beginning is no longer relevant here.
Next, we rewrite the integral ∞ 0 on the right side of (8) as
where in obtaining the first equality, we used Cauchy’s theorem on contour integrals with the paths {0 ≤ w ≤ R}, {w = Re it : arg(p) ≤ t ≤ 0}, and {w = qe i arg(p) : 0 ≤ q ≤ R}, letting R → ∞, and then made the change of variables q = |p|w ′ .
By
and so (8) is rewritten as
Here, it is plain that with the dominated convergence theorem,
e -2πip(j+r) (j + r) -1-δ dr.
In order to analyze Y 2 , we use a variation of the following relation [5, pp.60]
where x is not an integer, N is the integer nearest to x, c > 0, σ + c > 1, a n ≪ ψ(n) for some non-decreasing ψ, and the series
converges absolutely for σ > 1 with
In the proof of (11) available in [5], we replace “n” and “x” by j + r and n, respectively, and obtain
where r ∈ (0, 1) and
Choosing s → 1 + δ and a j → e -2πipj in (12), we have
Furthermore, multiplying both sides by Λ(n)n -κ and summing all over the positive integers n ≥ 2, we get
where
By (13), we see that
Hence, the integral in Y 2 is rewritten as lim
Φ(p, r, 1 + δ + w)e -2πipr dr dw w ;
(15) the first equality is by pointwise convergence of ( 14), and the second by uniform convergence of ( 14) for each N . But if c and κ are sufficiently large so that κ -c > 1 and 1 + δ + c > 2, then with integration by parts, it is easy to show that
and so the last integral (c) in (15) converges absolutely. This in turn enables us to put the limit N → ∞ inside the integral symbol (c) (use the dominated convergence theorem); we get
By ( 17), (10) becomes
This completes the proof of Theorem 2.
We need the following lemma.
Lemma 1. M (s, q) is meromorphic in s for fixed q in the region Re(s) > 1/2
The lemma follows easily by applying the formula
which can be easily shown with the well-known identity concerning Dirichlet characters [2] χ∈
here, for each k, there exists only one 1 ≤ k ′ ≤ b which is equal to k modulo b so that the sum of ( 19) is equal to e -2πiak ′ /b = e -2πiak/b . We put q = a/b, a/b < 1, and get with (19),
This completes the proof of the lemma. Now, suppose that ρ n 1 ,p = 1/2+η n 1 +iγ n 1 is a pole of M (s, p) with η n 1 > 0.
As the major purpose of this discussion is to present a general idea of how we apply our result to the study of the zeros of the Riemann zeta-function, we assume that there exists only one such pole in the region Re(s) > 1/2.
Using the residue theorem in (1), we shift the path of the integral on the left side to σ = 1/2; then it becomes 1 2πi (c) M (s + δ + κ, p)Γ(s)Γ(s + δ)Γ(1 -s)π -s e -πis/2 ds = π -ρn 1 ,p+δ+κ e -πi(ρn 1 ,p-δ-κ)/2
where c ρn 1 ,p is the residue of M (s, p) at s = ρ n 1 ,p , E is the collection of all the terms associated with poles other than ones related to zeros of the zetafunction and
with β j = 1/2 + η j and δ, κ > 0 small. Let N (σ, T ) be the number of the nontrivial zeros o
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