A Standard Method to Prove That the Riemann Zeta Function Equation Has No Non-Trivial Zeros

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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
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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
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( )
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:
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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:
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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
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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

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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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( ) ( )
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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