Erratum to “The Riemann Hypothesis-Millennium Prize Problem” [Advances in Pure Mathematics 6 (2016) 915-920] ()

A. A. Durmagambetov^{}

L. N. Gumilyov Eurasian National University.

**DOI: **10.4236/apm.2016.613077
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L. N. Gumilyov Eurasian National University.

The original online version of this article (Durmagambetov, A.A. (2016) The Riemann Hypothesis-Millennium Prize Problem. Advances in Pure Mathematics, 6, 915-920. 10.4236/apm.2016.612069) unfortunately contains a mistake. The author wishes to correct the errors in Theorem 2 of the result part.

Keywords

Euler, Chebyshev, Dirichlet, Riemann, Hypothesis, Zeta Function, Muntz, Function, Complex Numbers, Regular Function, Integral Function Representation, Millennium Prize Problem

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Durmagambetov, A. (2016) Erratum to “The Riemann Hypothesis-Millennium Prize Problem” [Advances in Pure Mathematics 6 (2016) 915-920]. *Advances in Pure Mathematics*, **6**, 1053-1056. doi: 10.4236/apm.2016.613077.

Theorem 1. Let the function be limited on every finite interval, and (x) is

continuous and limited on every finite interval then

(1)

Corollary 1. Let the function, , then

(2)

(3)

Our goal is to use this theorem on the analogs of zeta functions. We are interested in the analytical properties of the following generalizations of zeta functions:

(4)

(5)

(6)

(7)

Let N be the set of all natural numbers and ―the set of all natural numbers without

Below we will always let, this limitation is introduced only to simplify the calculations. Considering all the information above let us rewrite

For the function let us apply the results obtained by Muntz for the zeta function representation. With the help of the given definitions we formulate the analog of Muntz theorem.

Lemma 1. Let the function

then (8)

(9)

PROOF: According to the theorem conditions we have

(10)

Lemma 2. Let the function

(11)

then

(12)

(13)

(14)

(15)

PROOF: Follows from computing of integrals.

Lemma 3. Let the function

,

, then

(16)

(17)

PROOF: Computing the sums , we have

(18)

Theorem 2. Let the function

,

, then

(19)

PROOF: Using Corollary 1. we have

(20)

(21)

(22)

(23)

(24)

(25)

From the last equation we obtain the regularity of the function as s satisfied

Theorem 3. The Riemann’s function has nontrivial zeros only on the line;

PROOF: For, we have

(26)

Applying the formula from the theorem 2

(27)

estimating by the module

(28)

Estimating the zeta function, potentiating, we obtain

(29)

According to the theorem 1 limited for z from the following multitude

(30)

similarly, applying the theorem 2 for we obtain its limitation in the same multitude. For the function we have a limitation for all z, belonging to the half-plane Re(s) > 1/2 + 1/R. similarly, applying the theorem 2 for we obtain its limitation in the same multitude and finally we obtain:

(31)

These estimations for prove that zate function does not have zeros on the half-plane due to the integral representation (3) these results are projected on the half-plane for the case of nontrivial zeros. The Riemann’s hypothesis is proved.

Conflicts of Interest

The authors declare no conflicts of interest.

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[5] |
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[7] |
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[9] |
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