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Advances in Pure Mathematics, 2020, 10, 86-99

https://www.scirp.org/journal/apm

ISSN Online: 2160-0384

ISSN Print: 2160-0368

DOI:

10.4236/apm.2020.102006 Feb. 28, 2020 86

Advances in Pure Mathematics

A Standard Method to Prove That the Riemann

Zeta Function Equation Has No Non-Trivial

Zeros

Xiaochun Mei

Department of Theoretical Physics and Pure Mathematics, Institute of Innovative Physics in Fuzhou, Fuzhou, China

Abstract

A standard method is proposed to prove strictly that the Riemann Zeta func-

tion equation has no non-

trivial zeros. The real part and imaginary part of the

Riemann Zeta function equation are separated completely. Suppose

( ) ( ) ( )

12

, ,0s ab i ab

ξξ ξ

=+=

but

( ) ( ) ( )

12

, ,0s ab i ab

ζζ ζ

=+≠

with

s a ib

= +

at first. By comparing the real part and the imaginary part of Zeta

function equation individually, a set of equation about

a

and

b

is ob-

tained. It is proved that this equation set only has the solutions of trivial ze-

ros. In order to obtain possible non-trivial zeros, the

only way is to suppose

that

( )

1

,0ab

ζ

=

and

()

2,0ab

ζ

=

. However, by using the compassion me-

thod of infinite series, it is proved that

( )

1

,0ab

ζ

≠

and

( )

2,0ab

ζ

≠

. So the

Riemann Zeta function equation has no non-trivial zeros. The Riemann hy-

pothesis does not hold.

Keywords

Riemann Hypothesis, Riemann Zeta Function, Riemann Zeta Function

Equation, Jacobi’s Function, Residue Theorem, Cauchy-Riemann Equation

1. Introduction

In the author’s previous paper titled “The inconsistency problem of Riemann

Zeta function equation” [1], it was proved that after complex continuation was

considered, on the real axis, the Riemann Zeta function equation had serious

inconsistency. The Riemann hypothesis was meaningless [1].

In the present discussions of Riemann hypothesis and the calculations of Zeta

function’s zeros, approximate methods are commonly used. The real part and

How to cite this paper:

Mei, X.C. (2020)

A

Standard Method to Prove Tha t the Ri

e-

mann Zeta Function Equation Has No

Non

-Trivial Zeros.

Advances in Pure M

a-

thematics

,

10

, 86-99.

https://doi.org/10.4236/apm.2020.102006

Received:

January 11, 2020

Accepted:

February 25, 2020

Published:

February 28, 2020

Copyright © 20

20 by author(s) and

Scientific

Research Publishing Inc.

This work i s licensed under the Crea tive

Commons Attribution International

License (CC BY

4.0).

http://creativecommons.org/licenses/by/4.0/

Open Access

X. C. Mei

DOI:

10.4236/apm.2020.102006 87

Advances in Pure Mathematics

imaginary part of Zeta function equation are mixed together so that problems

become complicated. Because the Cauchy-Riemann Equation which an analyzed

function should satisfy was violated, the obtained zeros are not the real ones of

strict Zeta functions equation [1].

In this paper, a standard method is proposed to separate the real part and the

imaginary part of Zeta function equation completely. Then by comparing the

real part and the imaginary part individually, it is proved strictly that on whole

complex plane, the Zeta function equation has no non-trivial zeros. All trivial

zeros are located on the real axis. The Riemann hypothesis is proved untenable

again.

The Riemann Zeta function has two forms. One is the form of series summa-

tion and another is the form of integral. The form of series summation is more

fundamental with

( ) ( )

1, Re 1

s

n

sn s

ζ

∞−

=

= >

∑

(1)

Her

s a ib C=+∈

is a complex number. Based on Equation (1), by using the

Gama function

( )

sΓ

and introducing the counter integral on the complex

plane, Riemann obtained the integral form of Zeta function , as well as the alge-

braic relation of Zeta function, called as the Riemann Zeta function Equation [2]

[3].

( ) ( ) ( )

( )

11

11

2 2 sin 2 1

ss

s

nn

n s sn

ππ

∞∞

−−−

−

= =

= Γ−

∑∑

(2)

According the definition of Equation (1), we also have

( )

( )

1

1

1

s

n

sn

ζ

∞−−

=

−=

∑

(3)

Substituting Equation (1) and Equation (3) in Equation (2), Equation (2) can

be written as

( ) ( ) ( ) ( ) ( ) ( )

1

2 2 sin 2 1 1 , Re 1

s

s s sss

ζ ππ ζ

−

= Γ− − ≠

(4)

According to the common understanding at present, on the right side of Equ-

ation (4), the definition domain of function is extended from

( )

Re 1s>

to

whole complex plane except the point

( )

Re 1s=

. So Equation (4) is considered

as the new definition of Zeta function after complex continuation. However, by

examining Riemann’s deduction carefully in his original paper proposed in 1859,

we can see that Equation (4) is only a simplified symbol form. The original form

of Equation (4) should be Equation (2) [1].

In this original paper, Riemann introduced another form of Zeta function eq-

uation [4]

( ) ( ) ( ) ( )

2

112

2

s

s ss s s

ξπ ζ

−

= −Γ

(5)

By using the formula

( ) ( )

1x xx

θθ

=

of Jacobi’s function, Riemann proved

that the function (5) had the symmetry

( ) ( )

1ss

ξξ

= −

. It was considered that

X. C. Mei

DOI:

10.4236/apm.2020.102006 88 Advances in Pu

re Mathematics

( )

s

ζ

described by Equation (4) and

( )

s

ξ

described by Equation (5) had the

same non-trivial zeros, though they had different forms. In practical discussion,

Equation (5) was used to calculate the zeros of Riemann hypothesis in general.

Riemann guessed that all non-trivial zeros were located on the critical line

()

Re 1 2

s=

of complex plane, but had not provided any concrete zero. Over

one hundred years, mathematicians had done a lot of research on the Riemann

hypothesis, trying to prove or falsify it, but nothing worked. The Riemann hy-

pothesis becomes the world mathematics problem.

However, there is a third possibility,

i.e

., there is something wrong with the

Riemann Zeta function equation itself, so that the Riemann hypothesis can not

be proved. It was proved in the author’s previous paper that there were four ba-

sic mistakes in the Riemann’s original paper in 1859, the Riemann Zeta function

did not hold, the Riemann hypothesis becomes meaningless [1].

1) An integral item around the original point of coordinate system was neg-

lected in Riemann’s original paper. The item was convergent when

( )

Re 1s>

,

but infinite when

( )

Re 1s≤

. That is to say, the integral form of Riemann Zeta

function has not changed its divergence of series summation form. The Riemann

Zeta function Equations (2) and (4) do not hold.

2) The Riemann Zeta function equation has serious inconsistency. The so-called

continuation of function indicates that a function which has no meaning in a

certain domain is re-defined so that it becomes meaningful in this domain. But

there is a basic requirement for the function’s continuation,

i

.

e

., this new defined

function should have the same form with original function in the original do-

main. Otherwise, the extended function can not be regarded as the continuation

of original function [5]. According to this basic principle, in the domain of

( )

Re 1s>

, the left side of Equation (4) should be in the form of Equation (1).

On the other hand, Equation (4) has definition on whole complex plane ex-

cept the point

( )

Re 1s=

i.e

.,

1

a≠

but

b

can be arbitrary. Taking

3.5a=

and

0b=

, Equation (4) should be effective. Because the original form of Equa-

tion (4) is Equation (2), it means that Equation (2) should be effective on

3.5a=

and

0b=

. However, it is proved that the left side of Equation (2) is a

limited value but the right side of Equation (2) is infinite when

3.5a=

and

0b=

, so the two sides of Riemann Zeta function equation are inconsistent [1].

In fact, it is proved that on the real axis, the Riemann Zeta function equation

only holds at the point

12

sa= =

. However, at this point, the Zeta function is

infinite, rather than zero. At the other points of real axis, if the left side of Equa-

tion (2) is convergent, the right side of Equation (2) is divergent, and vice versa.

So the two sides of the Riemann Zeta function equation are incompatible.

3) A summation formula was used in the deduction of the integral form of

Riemann zeta function. The applicable condition of this formula is

0x>

. At

point

0x=

, the formula becomes meaningless. However, the lower limit of

Zeta function integral is

0x=

, so this formula can not be used. The integral

form of Riemann zeta function does not hold.

X. C. Mei

DOI:

10.4236/apm.2020.102006 89

Advances in Pure Mathematics

4) The formula

( ) ( )

1x xx

θθ

=

of Jacobi function was used to prove the

symmetry of Zeta function. The applicable condition of this formula is

0x>

[4]. But the lower limit of integral involved in the deduction is

0x=

. Therefore,

the formula can not be used too, the symmetry

( ) ( )

1ss

ξξ

= −

does not hold.

The zeros calculation of Riemann Zeta function were discussed in the paper

[1]. At present, it has been proved by manual and computer numerical methods

that there are lot of zeros on the critical line of

12a=

. The number has ex-

ceeded 10 trillion [6]. The paper pointed out that all methods used in the calcu-

lations were approximate ones. For example, Equation (5) was developed into

the infinite series called as the Riemann-Siegal formula, then the zero of each

polynomial formula was calculated. The result violated the symmetry of the

Cauchy-Riemann formula that any analytic functions should satisfy, so they

were not the true zeros of strict Zeta functions.

In this paper, regardless of these problems mentioned above, we suppose that

the Riemann Zeta function equation still holds and discuss the zero problem of

Zeta function strictly. A simple and standard method is proposed to prove that

the Riemann Zeta function equation has no non-trivial zeros on whole complex

plane.

Let

12

i

ξξ ξ

= +

and

12

i

ζζ ζ

= +

, by separating each item of Equation (5)

into real and imaginary parts, Equation (5) is written as the forms that real part

and imaginary part are separated completely. Then we discuss the zeros of real

part and imaginary part individually.

At first, suppose that

1

0

ξ

=

and

2

0

ξ

=

, but

1

0

ζ

≠

and

2

0

ζ

≠

, we obtain

a set of equation about

a

and

b

. It is proved that the only solution to this eq-

uation set is

1

a=

and

0b=

. But they are the trivial zeros located on the real

axis, not non-trivial zeros. So Equation (5) has no non-trivial zeros. By the same

method, it also is proved that Equation (4) only has no trivial zeros which are

located at the points

2an= −

(

0,1, 2,n=

) and

0b=

.

At last, in order to obtain possible non-trivial zeros, we take

1

0

ζ

=

and

2

0

ζ

=

,

i.e

., the summation form of Zeta function itself is equal to zero. Howev-

er, by using the compassion method of infinite series, it is proved that

1

ζ

and

2

ζ

can not be zeros simultaneously.

Therefore, we prove that the Riemann Zeta function equation has no non-trivial

zeros again, the Riemann hypothesis does not hold.

2. The Proof That the Zeta Function Equation (5) Has No

Non-Trivial Zeros

We discuss the zeros of Equation (5) in this section. Then discuss the zeros of

Equation (4) in next section.

Theorem 1. On the complex plane, if the real part and imaginary part of zeta

function

( )

s

ζ

are not equal to zeros, the Zeta function equation (5) has no

non-trivial zeros. The trivial zero is located on the real axis at the point

1a=

and

0b=

.

X. C. Mei

DOI:

10.4236/apm.2020.102006 90 Advances in Pu

re Mathematics

Proof: We separate the real part and imaginary part of

)(s

ξ

and

( )

s

ζ

,

write them as

( ) ( ) ( )

12

,,s ab i ab

ξξ ξ

= +

(6)

( ) ( ) ( )

12

,,s ab i ab

ζζ ζ

= +

(7)

Here

1

ξ

,

2

ξ

,

1

ζ

and

2

ζ

are real functions. By using formula

ln

et

t=

, we

have

( )

( )

( )

( ) ( )

2

2 22

2ln 2

2 ln 2

2

ee

cos ln 2 sin ln 2

a ib

s a ib

ib ib

aa

a

b ib

π

π

π π ππ

ππ

ππ π

−+

− −−

−−

−−

−

= =

= =

= −

(8)

( ) ( )( ) ( ) ( )

2

1 1 12s s a ib a ib a a b i ab b− = + −+ = − − + −

(9)

( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

2

22

22

11

211 cos ln 2 2 sin ln 2

212 cos ln 2 1 sin ln 2

2

s

a

a

ss

a a b b ab b b

i ab b b a a b b

π

π ππ

ππ π

−

−

−

−

= −− + −

+ − − −−

(10)

Let

( ) ( ) ( ) ( )

22

1

11 cos ln 2 2 sin ln 2

2

a

G a a b b ab b b

π ππ

−

= −− + −

( ) ( ) ( ) ( )

22

2

12 cos ln 2 1 sin ln 2

2

a

G ab b b a a b b

ππ π

−

= − − −−

(11)

We get

( )

212

11

2ss s G iG

π

−−= +

(12)

Here

1

G

and

2

G

are real functions. On the other hand, the definition of

real Gama function is [6]

( )

1

0

e d 0, 0

ta

a tt a

∞−−

Γ= > >

∫

(13)

Let

a s a ib→=+

, we obtain the complex continuation of Gama function.

We have

( )

( )

( ) ( )

( )

( ) ( )

ln 2

21 21 2 21

00 0

21

0

12

2e de de e d

e cos ln 2 sin ln 2 d

,,

ib t

ts ta ib ta

ta

s tttttt t

t bt i bt t

ab i ab

∞∞ ∞

−− −− −−

∞−−

Γ= = =

= +

=Γ +Γ

∫∫ ∫

∫

(14)

( ) ( )

21

10

, e cos ln 2 d

ta

ab t b t t

∞−−

Γ=

∫

X. C. Mei

DOI:

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Advances in Pure Mathematics

( ) ( )

21

20

, e sin ln 2 d

ta

ab t b t t

∞−−

Γ=

∫

(15)

1

Γ

and

2

Γ

are also real functions. Therefore, according to the equations

above, Equation (5) can be written as

( )( )( )

( ) ( )

( ) ( )

1 2 1 2 1 21 2

11 2 2 1 12 21 2

12 21 1 11 2 2 2

i G iG i i

GG GG

iG G G G

ξξ ζζ

ζζ

ζζ

+ = + Γ+Γ +

= Γ− Γ − Γ+ Γ

+ Γ+ Γ + Γ− Γ

(16)

The real part and imaginary part of Equation (16) are separated with

( ) ( )

1 11 22 1 1 2 21 2

GG GG

ξζζ

= Γ− Γ − Γ+ Γ

(17)

( ) ( )

2 12 21 1 11 2 2 2

GG GG

ξζζ

= Γ + Γ + Γ− Γ

(18)

If the Zeta function equation has zeros, its real part and imaginary part should

be equal to zero simultaneously. Let

1

0

ξ

=

and

2

0

ξ

=

, we obtain

( ) ( )

11 2 2 1 12 2 1 2

0GG GG

ζζ

Γ− Γ − Γ+ Γ =

(19)

( ) ( )

12 21 1 11 2 2 2

0GG GG

ζζ

Γ+ Γ + Γ− Γ =

(20)

If

1

0

ζ

≠

and

2

0

ζ

≠

, we can obtain from Equation (19)

12 21

12

11 22

GG

GG

ζζ

Γ+ Γ

=Γ− Γ

(21)

Substitute Equation (21) in Equation (20), we get

( ) ( )

2

12 21 2 11 22 2

11 22

0

GG GG

GG

ζζ

Γ+ Γ + Γ− Γ =

Γ− Γ

(22)

or

( ) ( )

22

12 21 11 2 2

0GG GGΓ + Γ + Γ− Γ =

(23)

Because it is the square summation of two items, each one in Equation (23)

should be zero simultaneously

12 21

0GGΓ+ Γ=

(24)

11 22

0GGΓ− Γ=

(25)

From Equation (25), we have

1 22 1

GGΓ= Γ

. Substitute it in Equation (24),

we get

22

12

0GG+=

. Because

2

1

G

and

2

2

G

can not be negative, we can only

have

1

0G=

and

2

0G=

. According to Equation (11), the results are

( ) ( ) ( ) ( )

2

1 cos ln 2 2 sin ln 2 0a a b b ab b b

ππ

−− + − =

(26)

( ) ( ) ( ) ( )

2

1 sin ln 2 2 cos ln 2 0a a b b ab b b

ππ

− −− + − =

(27)

To square them and add them together, we get

( ) ( )

22

2

120a a b ab b

−− + − =

(28)

Equation (28) indicates

( )

2

10aa b−− =

and

20ab b−=

. Due to that

a

and

b

are real numbers, the solutions of these two formulas are

0,1a=

and

0b=

. Obviously, they are trivial zeros located on the real axis. In fact, because

X. C. Mei

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re Mathematics

of

2

0

s

π

−

≠

this solution is the result of

( )

10ss−=

in Equation (9). Let its

real part and imaginary part be equal to zeros simultaneously, we have

( )

2

10aa b−− =

and

20ab b−=

. Obviously, the value

12a=

of Riemann’s

hypothesis is not the solution of Equation (28).

The result above has nothing to do with the

( )

sΓ

function. We should con-

sider the zeros of

( )

sΓ

function. It is obvious that when

0b≠

,

( )

1

,abΓ

and

( )

2

,abΓ

described by Equation (15) are not equal to each other. So they can not be

equal to zeros simultaneously (if they have zeros). Similar to the real

( )

aΓ

func-

tion, it indicates that the

( )

sΓ

function of complex continuation has no zeros too.

As we known that when

0b=

and

20

a>

or

0a>

,

( )

2aΓ

is limited

but not equal to zero. When

2 0, 1, 2,a= −−

, or

0, 2, 4,a= −−

,

( )

2aΓ

is

infinite [5]. Therefore, after

( )

2sΓ

function is considered, the zero

0

a=

in

Equation (28) is canceled. The trivial zero of Zeta function Equation (5) is lo-

cated at

1

a=

and

0b=

. Thus, the proof of Theorem 1 is finished.

Besides, if we want to look for the non-trivial zeros of

( )

s

ξ

, the last way is to

let

1

0

ζ

=

and

2

0

ζ

=

in which the non-trivial zeros may be contained. In

this case, the problem whether or not the series summation form of Zeta

function can be equal to zero is involved. We will discuss this problem in Sec-

tion 4.

3. The Proof That the Zeta Function Equation (4) Has No

Non-Trivial Zeros

Theorem 2. On the complex plane, if the real part and imaginary part of Zeta

function

()

1s

ζ

−

are not equal to zeros, the Zeta function equation Equation

(4) has no non-trivial zeros. The trivial zeros are located on the real axis at the

points

2an= −

(

0,1, 2,n=

) and

0b=

.

Proof: Let

( ) ( )

ss

ζζ

′

→

and write Equation (4) as

( ) ( ) ( ) ( ) ( )

1

2 2 sin 2 1 1

s

s s ss

ζ ππ ζ

−

′= Γ− −

(29)

According to current understanding,

( )

s

ζ

′

is considered as the new defini-

tion of Zeta function after complex continuation was carried out. But

( )

1s

ζ

−

still has the same form of Equation (1), because Equation (3) was used in the last

steep of Riemann’s deduction. By using the formula

2

ei

i

π

=

, we have

( )

( ) ( )

( ) ( )

222 22

12 12

22

ee e

sin 2 e e

22

1e e ee

2

is is i i a ib i a ib

ia ia

bb

si

πππ ππ

ππ

ππ

π

−−+ −+

− −+

−

−

= = −

= −

(30)

() ( ) ( ) ( ) ( )

11 1 1

ln2

22 22 22 2 22 e

s a ib a ib a ib

π

π π ππ π

− −+ − −

= = =

(31)

( ) ( )

( )

( ) ( )

( )

( ) ( )

1

112 12

ln2 2 2

11 2 ln 2 1 2 ln 2

22

2 2 sin 2

2 e e e ee

2 e e ee

s

aia ia

ib b b

aia b ia b

bb

s

ππ

ππ π

ππ ππ

ππ

ππ

π

π

−

−− −+

−

−− + −+ −

−

= −

= −

(32)

Let

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Advances in Pure Mathematics

( ) ( )

1 2 ln2 , 1 2 ln2AabBab

ππ ππ

=−+ =+−

(33)

Equation (32) can be written as

( ) ( )

( )

( )

112

12 2 22

2 2 sin 2

2 e cos e cos e sin e sin

s

ab b bb

s G iG

A Bi A B

π π ππ

ππ

π

−

−−−

= +

= −+ +

(34)

( )

( )

12 /2

1

2 e cos e cos

abb

G AB

ππ

π

−−

= −

( )

( )

122

2

2 e sin e sin

abb

G AB

ππ

π

−−

= +

(35)

Here

A

,

B

,

1

G

and

2

G

are real functions. We also have [6]

( )

( ) ( )

( )

( ) ( )

1 1 ln

00 0

1

0

12

1 e de deed

e cos ln sin ln d

,,

t s t a ib t a ib t

ta

s tttttt t

t bt i bt t

ab i ab

∞∞ ∞

− −− −− − −− −

∞−−

Γ− = = =

= −

=Γ +Γ

∫∫ ∫

∫

(36)

() ( ) ( ) ( )

12

00

, e cos ln d , , e sin ln d

ta ta

ab t b t t ab t b t t

∞∞

−− −−

Γ= Γ=−

∫∫

(37)

Here

1

Γ

and

2

Γ

are also real functions. Thus, Equation (4) can be written

as

( )( )( )

( ) ( )

( ) ( )

12 1 21 212

11 2 2 1 12 21 2

12 21 1 11 2 2 2

i G iG i i

GG GG

iG G G G

ζζζ ζζ

ζζ

ζζ

′′ ′

= + = + Γ+Γ +

= Γ− Γ − Γ+ Γ

+ Γ+ Γ + Γ− Γ

(38)

By separating real part and imaginary part, we get

( ) ( )

1 11 22 1 1 2 21 2

GG GG

ζζζ

′= Γ− Γ − Γ+ Γ

(39)

( ) ( )

2 12 21 1 11 2 2 2

GG GG

ζ ζζ

′= Γ + Γ + Γ− Γ

(40)

If the Zeta function equation has zeros, its real part and imaginary part should

be zeros simultaneously. Suppose that

1

0

ζ

≠

and

2

0

ζ

≠

, according to the

same method as shown in Section 2, we get

12 21

0GGΓ+ Γ=

(41)

11 22

0GGΓ− Γ=

(42)

Form Equation (42), we have

1 22 1

GGΓ= Γ

. Substituting it in Equation (41),

we get

22

12

0GG+=

. The only solutions are

1

0G=

and

2

0G=

. According to

Equation (35), we obtain

22

e cos e cos 0

bb

AB

ππ

−

−=

(43)

22

e sin e sin 0

bb

AB

ππ

−

+=

(44)

To square Equation (43) and Equation (44), then add them together, we get

( )

e e 2cos

bb

AB

ππ

−

+= +

(45)

To square Equation (43) and Equation (44) and subtract them, we get

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( )

e cos2 e cos2 2cos

bb

A B AB

ππ

−+=−

(46)

Substituting Equation (33) in Equation (45) and Equation (46), we obtain

( )

e e 2cos

bb

a

ππ

π

−

+=

(47)

( ) ( ) ( )

e cos 2 ln2 e cos 2 ln2 2cos 2 ln 2

bb

ab ab b

ππ

ππ ππ π

−

− +− −=−

(48)

If

0b≠

, by developing

e

b

π

−

and

e

b

π

into the series, we can prove

e e2

bb

ππ

−+>

. However, because of

( )

2cos 2a

π

≤

, Equation (47) does not hold.

So the only solution of Equation (47) is

0b=

and

( )

2 0,1,2,a nn=±=

representing the trivial zeros located on the real axis. In this case, Equations (47)

and (48) become the same with the same solution.

If taking

12a=

, we have

( )

2cos 2 0

π

=

. Because of

e e0

bb

ππ

−+≠

, Equa-

tion (47) can not hold too. Therefore,

12

a=

is not the solution of Equation

(4) too. In fact, because of

is

20

π

≠

, the solutions

0b=

and

2an= ±

are the

result of

( )

sin 2 0s

π

=

in Equation (29), having nothing to do with

Γ

func-

tion too.

But if let

0b=

and

1

aa

′

−= −

in Equation (37), we have

1aa

′= −

. When

0a′≤

or

1

a≥

, we have

1

Γ →∞

. Therefore, the corresponding zeros

2,4, 6,a=

in (47) can be removed. The trivial zeros of Equation (4) only ap-

pear at the points

2an= −

(

0,1, 2,n=

) and

0b=

. Thus, the proof of Theo-

rem 2 is finished.

So, looking for possible non-trivial zeros of Equation (4), the only way for us

is to consider

( )

1

,0ab

ζ

=

and

()

2,0ab

ζ

=

. The problem is involved whether

or not the series summation form of Zeta function can be equal to zero. Ac-

cording to Equation (1), we have

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )

( )

12

11 1

1 ln 2 1 ln 3 1 ln

11 1

11 1

1 ,,

11 1

123

12 e 3 e e

1 2 cos ln 2 3 cos ln3 cos ln

2 sin ln 2 3 sin ln3 sin ln

ss s

a ib a ib a ib n

aa a

aa a

s ab i ab

nn

b b n bn

i b b n bn

ζζζ

−− −

−− −

−− −

−− −

−= +

= + + +⋅⋅⋅+ +⋅⋅⋅

= + + +⋅⋅⋅+ +

=+ + ++ +

+ + ++ +

(49)

So we obtain

( ) ( ) ( ) ( )

11 1

1

, 1 2 cos ln 2 3 cos ln3 cos ln

aa a

ab b b n b n

ζ

−− −

=+ + ++ +

(50)

( ) ( ) ( ) ( )

11 1

2

, 2 sin ln 2 3 sin ln3 sin ln

aa a

ab b b n b n

ζ

−− −

= + ++ +

(51)

We discuss the zeros of Equations (50) and (51) in next section.

4. The Proof That the Series Summation Formula of

Riemann Zeta Function Has No Zeros

4.1. The Convergence of Summation Form of Zeta Function

In order to discuss the zeros of the summation form of Zeta function, we should

discuss its convergence. If

sa=

is a real number, Equation (1) is divergent

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when

1a<

without zeros. When

1

a>

, the series is convergent and great than

zero, so (1) has no zeros. The proof is as below [7].

Suppose

1a>

,

p

is prime number, by considering the Euler prime number

product formula

( )

( )

11

1 1 exp ln 1

aa a

pp p

ap p p

ζ

−−

−− −

=− ≥+ =− +

∏∏ ∑

(52)

For any

0x>

, we have

( )

ln 1 xx+<

. Because

a

p

−

<∞

∑

is convergent,

we have

()

exp 0

a

p

ap

ζ

−

≥− >

∑

(53)

If

s a ib= +

is a complex number, the situation is more complicated. It does

not seems to have strict proof to conform

( )

0s

ζ

≠

at present. Let’s discuss

this problem below. According to the judgment formula of series convergence of

complex function showing in Equation (1), we have

()

11

lim lim lim 1

1 1/

1

a ib

a ib

na ib

nn n

n

un

un

n

+

+

+

+

→∞ →∞ →∞

= = =

+

+

(54)

Because the radius of convergence is 1, we can not judge the convergence of

Equation (1). By using the Euler formula, we write Equation (1) as

( )

( ) ( ) ( )

( ) ( ) ( )

( )

11 1

123

1 2 cos ln2 3 cos ln3 cos ln

2 sin ln2 3 sin ln3 sin ln

a ib a ib a ib

aa a

aa a

sn

b b n bn

i b b n bn

ζ

++ +

−− −

−− −

=+ + ++ +

=+ + ++ +

− + ++ +

(55)

By separating the real part and imaginary part, let

( )

s u iv

ζ

= +

, we get

( ) ( ) ( ) ( )

, 1 2 cos ln2 3 cos ln3 cos ln

aa a

uab b b n b n

− −⋅ −

=+ + ++ +

(56)

() ( ) ( ) ( )

( )

, 2 sin ln2 3 sin ln3 sin ln

aa a

vab b b n b n

−− −

=− + ++ +

(57)

By using the formula

( ) ( )

ln 1 ln ln 1 1

nn n+= + +

and considering the con-

vergences of Equations (56) and (57), we have:

( )

( )

( ) ( )

( )

( )

( ) ( )

( )

( )

1cos ln 1

lim lim 1 cos ln

cos ln ln 1 1

lim 1 1 cos ln

cos ln

lim 1

cos ln

a

na

nn

n

a

n

n

n bn

uun bn

bnb n

n bn

bn

bn

+

→∞ →∞

→∞

→∞

+

=+

++

=+

= =

(58)

( )

( )

( ) ( )

( )

( )

1

sin ln 1 sin ln

lim lim lim 1

sin ln

1 sin ln

a

na

nn n

n

n bn bn

vv bn

n bn

+

→∞ →∞ →∞

+

= = =

+

(59)

The radius of convergence is still equal to 1. The convergences of Equation

(56) and Equation (57) can not be determined. Because the formulas contain

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Trigonometric functions, the items in the formula can be positive or negative,

we can not ensure that the result of summation is always great than zero. This is

different from the situation when s is a real number.

4.2. The Zeros of Common Analytic Functions

In the theory of complex functions, the analytic nature of functions is very im-

portant. Many theorems cannot be used for non-analytic function. For example,

the residue theorem is effective only for analytic functions. On the other hand, a

complex function can always be written as

( ) ( ) ( )

,,f z uxy ivxy= +

(60)

Here

z x iy

= +

. If

( )

fz

is analytic one, its real part and imaginary part are

related. The Cauchy-Riemann formula should be satisfied with [6]

,

uvu v

xyy x

∂∂∂ ∂

= = −

∂∂∂ ∂

(61)

In the current calculation of the zero point of Riemann Zeta function, some

approximate methods are adopted. Because Equation (61) is ignored, what ob-

tained are not real zeros of Zeta function [1].

It should be emphasized that when we calculate the zero points of analytic

functions, we need to separate the real part and the imaginary part. Because it's

possible to have a situation where the real part or the imaginary part is equal to

zero, but they are not equal to zero simultaneously. However, in the current zero

calculation of Riemann hypothesis, the real part and imaginary part are often

mixed together, making the problem ambiguous.

4.3. The Proof That the Series Summation Formula of Riemann

Zeta Function Has No Zeros on the Complex Plane

Theorem 3. The series summation formula of Riemann Zeta function has no

zeros on whole complex plane.

Proof: By using the formula

ln

e

a an

n=

, we write Equations (50) and (51) as

( ) ( )

( )

( )

( )

( )

1 ln2 1 ln3

1, , 1 e cos ln2 e cos ln3

aa

ab uab b b

ζ

−−

==+++

(62)

( ) ( )

( )

( )

( )

( )

1 ln 2 1 ln 3

2

, , e sin ln 2 e sin ln 3

aa

ab vab b b

ζ

−−

== ++

(63)

It is easy to prove that Equations (62) and (63) satisfy Equation (61). So the

summation form of Zeta function is an analytic one. We prove below that Equa-

tions (62) and (63) can not be equal to zero simultaneously.

Let’s discuss the simplest situation to take the first two items in

( )

10

s

ζ

−=

as shown in Equation (49). According to Equations (62) and (63), we have

( )

( )

1 ln2

1 e cos ln2 0

a

b

−

+=

(64)

( )

( )

1

e sin ln 2 0

a

b

−

=

(65)

Suppose that

a

is a limited number, the solution of Equation (65) is

ln2bn

π

=

. If

n

is an even number, we have

cos 1n

π

=

. Substituting it in Equa-

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tion (64), we get

( )

1 ln2

1e 0

a−

+=

or

( ) ( )

1 ln2 ln 1a−=−

(66)

Because

a

is a real number,

( )

ln 1 lne

i

i

π

π

−= =

is not a real number, so

Equation (64) has no solution of real number. When

n

is an odd number, we

have

cos 1n

π

= −

. Substituting it in Equation (64), we get

( )

1 ln2

1e 0

a−

−=

or

( )

1 ln2 ln1 0a−==

(67)

Therefore, the solutions of Equations (64) and (65) are

1a=

and

( )

2 1 ln 2bn

π

= +

(

0,1, 2,n=

). If

1a≠

, Equations (64) and (65) can not be

equal to zeros simultaneously.

Then, let’s discuss the first three items of Equation (49). By considering

ln

e

a an

n=

and

ln

e

ib ib n

n=

, multiplying the first three items with

( ) ( )

( )

( ) ( )

1

1 1 ln 2 3

2 3 23 e

a

s s ib

−

−− − ⋅

= ⋅

, and let it equal to zero, we get

( )

( ) ( ) ( ) ( )

1ln 2 3 1 1

ln 2 ln 3

23 e 2 e 3 e 0

aib a a

ib ib

−−⋅ − −

−−

⋅ ++=

(68)

By considering the Euler’s formula and separating real part and imaginary

part, we obtain

( )

( )

( )

( )

( )

( )

( )

( )

111

2 3 cos ln 2 3 2 cos ln2 3 cos ln 3 0

aaa

b bb

−−−

⋅ ⋅+ + =

(69)

()

()

()

( )

()

()

( )

( )

111

2 3 sin ln 2 3 2 sin ln 2 3 sin ln3 0

aaa

b bb

−−−

⋅ ⋅+ + =

(70)

We write Equation (70) as

( )

( )

( )

() ( )

111

2 3 cos ln 2 3 2 cos ln 2 3 cos ln 3 0

2 22

aaa

b bb

π ππ

−−−

⋅ − ⋅+ − + − =

(71)

It can be seen that Equations (69) and (71) are completely symmetric with the

same parameters before cosine function. Because

a

and

b

are arbitrary, in or-

der to make Equations (69) and (71) tenable for arbitrary

a

and

b

, the only

way is to let

( ) ( )

ln 2 3 ln 2 3 , ln 2 ln 2, ln3 ln3

2 22

b b b bb b

π ππ

⋅=− ⋅ =− =−

(72)

or

( )

,,

4ln 2 3 4ln2 4ln3

b bb

π ππ

= = =

⋅

(73)

However, these three relations are contradictory. So Equations (69) and (70)

can not be equal to zero simultaneously.

Off cause, to the series of which item’s number is limited, the proof above is

not strict. But for the series with infinite items, this method is standard one. Let

( )

11 1

11 1

11 0

23

ss s

n

sp

ζ

−− −

−=+ + ++ +=

(74)

by multiplying two sides of Equation (74) with

( )

( )

( )

( )

( )

23 1

11

ln

23 1

23 e

nn

sa

ib p p p p

n nn

p pp p p

−

−−

−

−

⋅ ⋅⋅⋅

, we get

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( )

( )

( )

( )

( )

( )

( )

( )

( )

( )

( )

( )

23 1 23 2

23 1 23 1

11

ln ln

23 1 23 2

11

ln ln

31 231

ee

e e0

n nn

n nn

aa

ib p p p ib p p p p

n nn

aa

ib p p p ib p p p p

nn nn

pp p pp p p

p pp pp pp

−−

−−

−−

−−

−−

−−

−−

−−

+ +⋅⋅⋅

++ =

(75)

For infinite series, we have

n

p→∞

. By dividing real part and imaginary

part, similar to Equations (69) and (71), we obtain

( )

( )

( )

( )

( )

( )

( )

( )

( )

( )

( )

( )

( )

( )

( )

( )

1

23 1 23 1

1

23 2 23 2

1

3 1 31

1

23 1 23 1

cos ln

cos ln

cos ln

cos ln 0

a

nn

a

nn nn

a

nn nn

a

nn nn

pp p b pp p

pp p p b pp p p

p pp b p pp

pp p p b pp p p

−

−−

−

−−

−

−−

−

−−

+ ⋅⋅⋅

+ + ⋅⋅⋅

+ ⋅⋅⋅ =

(76)

( )

( )

( )

( )

( )

( )

( )

( )

( )

( )

( )

( )

1

23 1 23 1

1

23 2 23 2

1

31 31

1

23 1 23 1

cos ln

2

cos ln

2

cos ln

2

cos ln 0

2

a

nn

a

nn nn

a

nn nn

a

nn nn

pp p b pp p

pp p p b pp p p

p pp b p pp

pp p p b pp p p

π

π

π

π

−

−−

−

−−

−

−−

−

−−

−

+−

++ −

+ −=

(77)

According to the theory of infinite series, when the number of items tends to in-

finite, to make Equations (76) and (77) be tenable simultaneously, besides

a→∞

,

the only way is to let the corresponding terms are equal to each other, we have

( ) ( )

23 1 23 2

,

4ln 4ln

n nn

bb

pp p pp p p

ππ

−−

= =

( ) ()

31 231

,

4ln 4ln

nn nn

bb

p pp pp pp

ππ

−−

= =

(78)

And so on. However, Equation (78) are impossible, so Equations (76) and (77)

can not be equal to zeros simultaneously. The summation form of Zeta function

( )

1s

ζ

−

can not be zero, and so do for

( )

s

ζ

′

to let

1

ss

′→−

. Then the

proof of Theorem 3 is finished.

The result above has nothing to do with the value of

a

, as long as

a<∞

, no

matter

1

a>

or

1a≤

. If take

0b=

, i.e., on the real axis, Equation (51) be-

comes zero but Equation (50)is not equal to zero. We have

( ) ( )

11 1

1

1 1 12 3 0

aa a

aa n

ζζ

−− −

−= −=+ + ++ +≠

(79)

If taking

12a=

, we have

1 12a−=

. It is proved in [1] that only in this

case, the two sides of Equation (2) can be equal to each other. But the result is

infinite, rather than zero, so it is meaningless.

( )

11 1

12 1 23 n

ζ

= + + + + + →∞

(80)

According to the discussion above, the Zeta function equations Equation (4)

and Equation (5) has no non-trivial zeros.

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Up to now, we have fully explained the problem of Riemann hypothesis, and

obtained the following Theorem 4.

Theorem 4: The Riemann Zeta function Equations (4) and (5) have no

non-trivial zeros on whole complex plane. The Riemann hypothesis does not

hold.

5. Conclusions

The Riemann Zeta function equation has two forms. They are considered with

the same non-trivial zeros. The Riemann hypothesis claims that all non-trivial

zeros were located on the critical line

( )

Re 1 2s=

of complex plane, but it can

not be proved up to now.

In the author previous paper “The inconsistency problem of Riemann Zeta

function equation”, it was revealed that there were four basic mistakes in the

Riemann’s original paper proposed in 1895. The Riemann Zeta function equa-

tion did not hold and the Riemann hypothesis was meaningless.

In this paper, we suppose that the Riemann Zeta function equations still hold.

By separating the Zeta function equation into real part and imaginary part com-

pletely, it is proved that the Riemann Zeta function equations have no non-trivial

zeros. The summation form of Zeta function itself also has no zeros. The Rie-

mann hypothesis is proved untenable again from another angle.

Conflicts of Interest

The author declares no conflicts of interest regarding the publication of this pa-

per.

References

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Grosse.

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