About the Riemann Hypothesis

http://dx.doi.org/10.4236/jamp.2016.43061

About the Riemann Hypothesis

Jinhua Fei

Changling Company of Electronic Technology, Baoji, China

Received 1 February 2016; accepted 27 March 2016; published 30 March 2016

Copyright © 2016 by author and Scientific Research Publishing Inc.

This work is licensed under the Creative Commons Attribution International License (CC BY).

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

Abstract

The Riemann hypothesis is part of Hilbert’s eighth problem in David Hilbert’s list of 23 unsolved problems. It is also one of the Clay Mathematics Institute’s Millennium Prize Problems. Some thematicians consider it the most important unresolved problem in pure mathematics. Many ma-thematicians made a lot of efforts; they don’t have to prove the Riemann hypothesis. In this paper, I use the analytic methods to deny the Riemann Hypothesis; if there’s something wrong, please criticize and correct me.

Keywords

Riemann Hypothesis, Disavowal

1. Introduction

Riemann Hypothesis was posed by Riemann in early 50’s of the 19th century in his thesis titled “The Number of Primes Less than a Given Number”. It is one of the unsolved “super” problems of mathematics. The Riemann Hypothesis is closely related to the well-known Prime Number Theorem. The Riemann Hypothesis states

that all the nontrivial zeros of the zeta-function lie on the “critical line” : Re 1 2

s s

 ₌ 

 

 . In this paper, we use the

analytical methods, and refute the Riemann Hypothesis. For convenience, we will abbreviate the Riemann Hy-pothesis as RH.

2. Some Theorems in the Classic Theory

In this paper, Γ

( )

s is the Euler gamma function, ζ

( )

s is the Riemann zeta function.

Lemma 2.1. If Rew>0, then

( )

( )2

( )

1

d exp

2π

s

s w s w

i

−

Γ = −

where Re w is the real part of complex number w.

Let η>0 be given, when s≥η and args ≤ −π η, then

( )

1

log .

s s O s

  ′

Γ _{= +  }

 

Γ _{ }

If − ≤ ≤4 σ 4,t ≥1, then

(

)

1

(

)

3

2 π π 1 2 π

2π exp log 1 exp

2 2 2 2

it tσ t it t i O tσ t

σ −  λ σ   −  

Γ + = _− + − + _ − _+ _{ }− _

   

   

where λ=1 if t≥1, λ= −1 if t≤ −1. See [1] page 523, page 525.

Lemma 2.2. If Res>1, then

( )

( )

2 s n

n s

n

ζ ζ

∞

=

Λ ′

= −

∑

where Λ

( )

n is the Mangoldt function. Let s is any complex number, we have

( )

1

1 1 1 1 1

1

1 2 2

s c s

s ρ s

ζ

ζ ρ ρ

′ _{= − + +}  ₊ ₋ Γ ′ ₊ 

 

 

−

∑

 −  Γ  

where ρ be the nontrivial zeros of ζ

( )

s , c₁ be the positive constant.

We write s= +σ it. If − ≤ ≤1 σ 2,− <π Im log

{

(

s−1

)

}

≤π, − <π Im log

{

(

s−ρ

)

}

≤π, then

( )

(

)

(

)

(

(

)

)

1

log log 1 log log 2

t

s s s O t

γ

ζ ρ

− ≤

= − − +

∑

− + +

where Im s is the imaginary part of complex number s. See [2] page 4, page 31, page 218.

Lemma 2.3. Let N T

( )

is the number of zeros of ζ

( )

s in the rectangle 0< <σ 1, 0< <t T. then

( )

7

( )

1

log

2π 2π 2π 8

T T T

N T S T O

T

 

= − + + _{+  }

 

where

( )

1arg 1 .

π 2

S T = ζ _ +iT_

 

See [3] page 98.

Lemma 2.4. Assume that RH, If x≥2, then

( )

( )

( )

2n x

x n x R x

ψ

≤ ≤

=

∑

Λ = +

where

( )

1 2 2_log

R x x x.

See [3] page 113.

3. Some Preparation Work

Lemma 3.1. Assume that RH, and 0 1 50

δ

< ≤ , then

(

)

(

)

1

2 2

0 0

1 1

2

log i d 1 and log i d 1

δ

δ

ζ σ γ σ ζ σ γ σ

−

− +

+ +

∫



∫



where γ₀ is the ordinate of nontrivial first zero of ζ

( )

s , γ ≈₀ 14.134725.

(

)

(

)

0

0 0 0

1

1

log log log

2

i i i O

γ γ

ζ σ γ σ γ γ γ

− ≤

 

+ _ − + − _+

 

∑



because

(

)

2

2

0 0 0

1 1 1 1

log log

2 i i 2 2 i Arg 2 i i

σ γ γ  σ γ γ  σ γ γ

 _{− + −} _{= } ₋  _{+ − +}  _{− + −} 

  _  _  

  _  _  

(

)

2

2

0 0

1 1

log log 1

2 i i 2

σ γ γ  σ γ γ 

 _{− + −}  _ ₋  _{+ − +}

  _  _

  _  _

and

(

)

2 2

2 0

1 1 9

log log log 1

2 2 4

σ  σ γ γ 

 ₋  _{≤ } ₋  _{+ − ≤}  ₊ 

  _  _  

  _  _  

(

)

2 2

2 0

1 1

log log 1

2 2

σ γ γ σ

_{ }  _{ }

− + − − +

_{ }  _{ }

_{ }  _{ }

  

therefore

2

0

1 1

log log 1.

2 i i 2

σ γ γ σ

 _{− + −}   ₋  ₊

   

    

And because

( )

( )

3 2

3 2 3

2

3 2

2 2 2

2 2

1 1

2 2

2

1 3

2 2

1

1 1

1 1 1

log d log d log d

2 2 2

1 1

2 log d 2 log d

2 2

2 log d 2 log d

2 log 2 d O 1 O 1

δ δ

δ

δ

δ

σ σ σ σ σ σ

σ σ σ σ

σ σ σ σ

δ δ σ

+ +

+

 ₋  ₌  ₋  ₊  ₋ 

     

     

   

= − _ − _ + _ − _

   

= − +

= + + =

∫

∫

∫

∫

∫

∫

∫

∫

therefore

(

)

0

2 2

0 0

1

1 1

2 2

2 2

1 2

1

log d log d 1

2

1

log d 1 1.

2

i i i

γ γ

δ δ

δ

ζ σ γ σ σ γ γ σ

σ σ

− ≤

+ +

+

 

+ _ − + − _ +

 

 ₋  ₊

 

 

∑

∫

∫

∫



 

Similarly, we have

(

)

1 2

0 1

log i d 1.

δ

ζ σ γ σ

−

−

+

∫



This completes the proof of Lemma 3.1. Throughout the paper, we write

1

i , , 50, 2π.

z a b a T b

T

It is easy to see that

( )

₍

2

₎

12 1

0

π 1 1

arctg , 1 , .

2 2 1 4π π

k k

k k

b a

h h h

a k b T T

+ ∞

+ =

= − = − ≤ ≤

+

∑

Lemma 3.2. We calculate the three complex numbers. Because

(

)

1

(

)

1

2 2 2_{exp arctg} 2 2 2_exp π

2

b

a ib a b i a b i ih

a

   

+ = + _{ }= + _ − _

   

therefore when t is the real number, we have

(

)

3 4

3

2 2 8 2_exp 3π 3 π _exp π

8 4 2 2

t i it

z − = a +b − _i −i h+ t−th_ _ t−th_

   

(

)

1 2

1

2 2 4 2_exp π π _exp π

4 2 2 2

t i

it h

z− − = a +b − − _−i +i + t−th_ _ t−th_

   

(

)

(

)

1 2

1 1

2 2 4 2

2 _exp π π _exp π

4 2 2 2

t i

it it h

z− + = a+ib − + = a +b − + _−i +i − t+th_{ }− t+th_

   

the three complex numbers required below.

Lemma 3.3.

( ) ( )(

)

( )

3 4

d 1

s

s ζ s a ib s

ζ

− −

′

Γ +

∫



Proof. By Lemma 2.1 and Lemma 3.2, we have

( ) ( )(

)

( )

(

)

(

)

(

)

(

)

( )

(

)

(

)

3 4

3 4

3 4

5 4

5 4

3 3

d d

4 4

3 3

d

4 4

2 log 2 exp d

2 log 2 d 1.

s it

it

s s a ib s i it it a ib t

it it a ib t

t t th t

t t t

ζ ζ

ζ ζ

ζ ζ

+∞

− −

−∞ −

+∞

−

−∞ +∞

−

−∞

+∞ ₋

−∞

′   ′ 

Γ + = Γ − +_{  }− + _ +

   

′

   

Γ − +_{  }− + _ +

   

+ + −

+ +

∫

∫

∫

∫

∫





 

This completes the proof of Lemma 3.3.

Lemma 3.4.

(

)

0

1

2 2

1

log d log

2 2π

it t

it a ib t T

γ +∞

− −

 

Γ_ + _ +

 

∫



Proof. By Lemma 2.1 and Lemma 3.2, we have

(

)

(

)

(

(

)

)

(

)

(

)

( )

(

)

0

0

0 1 2

1

2 2 4 2 2 2

1 2 2 4 1

1 2 2 4

1 2

1

log d

2 2π

π

2π exp exp log 1 log d

4 2 2π

exp log d

π

2π exp

4 2

it

t i

t it a ib t

h t

a b i i th it t a b t

O a b t th t t

h I a b i i I

γ

γ

γ +∞

− −

+∞

− −

∞

− ₋

−

 

Γ_ + _ +

 

   

= + _− + _ − + − + _{ }

   

 

+ _ + − _

 

  

= _ + _− + _+

 

 

∫

we write

(

)

1

2 2 2_, 2π 2π 1

r= a +b ≤ ≤r +

(

)

(

)

( )

₍

₍

₎

₎

( ) (

)

( )

( ) (

)

0

0

0

1 exp log log 1 log d

2π

exp

log d exp log log 1

log log 2π

exp

log log 2π d exp log

log log

exp

exp log log 2π d exp log

log log

t

I th it t r t

th t

it t r

i t i r

th t

i t it

t r re

th t

i th t it

t r re

γ

γ

γ

+∞

+∞

+∞

 

= − + − − _{ }

 

−  

= _{ } − −

−  

−  

= − − _{ }

−  

−

  

= − _ − + − _

− 

 

∫

∫

∫

( )

( )

( )

( )

(

)

( )

( )

( )

( )

(

)

0

0

0

2

2

exp exp

1 exp log exp log d

log log log log 2π

exp exp

1 exp log exp log d

log log log log 2π

th th r t

O i h th h it t

t r t t r re

th th r t

O i h th h it t

t r t t r re

γ

γ

γ

+∞

+∞

+∞



 



  _{− −}   _{ }

   

= + − − + −_ − _{  }

 _ − − _   

 

  _{− −}   _{ }

   

= + − − + − − _{ }

 

 _ − − _   

 

∫

∫

∫



( )

(

)

0

2

1

exp d 1

log log

h th t

t t r

γ

+∞ 

 − + 

 ₋ 

 

∫



( )

( )

( )

(

)

2

2 0

2

2 0

1 1

2

2

1 2 1 2

exp log d exp log d

log d log exp d log log

h

h h

h

I t th t t t th t t

t t t h h th t h T

γ

γ −

− −

−

∞

− −

∞

− −

− + −

+ −

∫

∫

∫

∫



  

This completes the proof of Lemma 3.4.

Lemma 3.5.

0

1 1

1 2

2 2

1 1

log d log

2 2

it it

it z it z z t t T

γ

+∞ _{− − − −}

−

   

′

Γ_ + _ − Γ_ + _

   

∫



Proof. When t≥γ₀, by Lemma 2.1, we have

1

1 1 1 1 1

log

2 2

2 2 2

π π π

exp log exp exp log .

2 2 2

it it it it it

t t t− t t t

         

′

Γ _ + _ Γ_ + _{ } + _+ Γ_ + _{ } + _

   

     

₋  ₊ ₋  ₋ 

     

     



 

By Lemma 2.1 and Lemma 3.2, we have

( )

( )

( )

( )

1 1

2 2

0

2

2

0 0

2

2 0

1

1 1 1

1 2 2 2

1 1

log d

2 2

exp log d exp log d exp log d

log d log exp d log .

it it

h

h

h

h

it z it z z t t

t th t t t th t t t th t t

t t t h h th t T

γ

γ γ

γ

−

− −

− +∞

− − − − −

+∞ +∞

− − −

+∞

− −

   

′

Γ _ + _ − Γ_ + _

   

− − + −

+ −

∫

∫

∫

∫

∫

∫

 

This completes the proof of Lemma 3.5.

Lemma 3.6. Assume that RH, then

( )

( )

0 0

1 1

2 2

1 1

d 1 and d 1

2 2

it it

it z S t t it z S t t

γ γ

+∞ _{  − −} +∞ _{  − −}

′

Γ_ + _ Γ_ + _

   

∫



∫



where

( )

1arg 1 .

π 2

S t = ζ_ +it_

 

Proof. By Lemma 3.2, it is easy to see that

(

)

(

)

1 1

2 2 4 2 2

1 2 2 4 2

1 1 π π

exp

2 2 4 2 2

1 π π π

exp cos sin .

2 2 4 2 4 2

t i it

t i

h

it z it a b i i t th

h h

it a b t th i

− − − −

− −

     

′ ′

Γ_ + _ = Γ_ + _ + _− + + − _

     

 

       

′

= Γ_ + _ + _ − _{ }− + _+ _− + _

       

We write

( )

( )

(

2 2

)

2

s

H s = Γ′ s a +b −

( )

( )

(

)

1

π π

exp exp 1

2 2

G s =H s _ __−i +ih s_ _+ __−i +ih_ s− _

   

   

 

( )

( )

(

)

2

π π

exp exp 1

2 2

G s =H s _{ }_−i +ih s_ _− __−i +ih_ s− _

   

   

 

( )

(

)

(

)

3

π π

1 exp exp 1

2 2

G s =H −s _ __−i +ih s_ _+ __−i +ih_ s− _

   

   

 

( )

(

)

(

)

4

π π

1 exp exp 1 .

2 2

G s =H −s _ __−i +ih s_ _− __−i +ih_ s− __

   

   

 

It is easy to see that

(

)

1

2 2 4 2

1

1 1 π π

2 exp cos

2 2 2 4 2

t

i _h

G _ +it_= Γ′_ +it_ a +b − − _ t−ht_ _− + _

       

(

)

1

2 2 4 2

2

1 1 π π

2 exp sin

2 2 2 4 2

t

i _h

G _ +it_= Γi ′_ +it_ a +b − − _ t−ht_ _− + _

       

(

)

1

2 2 4 2

3

1 1 π π

2 exp cos

2 2 2 4 2

t

i _h

G _ +it_= Γ′_ −it_ a +b − + _ t−ht_ _− + _

       

(

)

1

2 2 4 2

4

1 1 π π

2 exp sin .

2 2 2 4 2

t

i _h

G _ +it_= Γi ′_ −it_ a +b − + _ t−ht_ _− + _

       

Assume that RH and 0 1

50

δ

< ≤ , by the contour integration method, we have

( )

( )

0

( )

( )

0

( )

( )

0 0

1 1

1

2 2

1 1 1

1 1 1

2

log d log d log d 0

i _i i

i i

i

G s s s G s s s G s s s

δ _γ δ γ

γ δ γ

ζ ζ ζ

− + ∞ _{− +} − +

− + ∞ − +

− +

+ + =

∫

∫

∫

( )

( )

0

( )

( )

0

( )

( )

0 0

1 1

1

2 2

1 1 1 1 2

1 1 1

2

log d log d log d .

i _i i

i i

i

G s s s G s s s G s s s J J

δ _γ δ γ

γ δ γ

ζ

ζ

ζ

− + ∞ _{− +} − +

− + ∞ − +

− +

= − − = +

By Lemma 2.1 and Lemma 3.2,

( )

( )

(

)

(

)

(

)

(

)

(

)

( )

(

)

0

0

0 0

0 1

1 1 1

1

3 2 2

3 2 2

log d 1 log 1 d

π

1 exp log d log exp d

2

log d 1.

i

i

J G s s s G it it t

it t th t t t t th t

t t t

γ

γ

γ γ

γ

ζ ζ

− + ∞

− + ∞

∞ ∞ ₋

∞ ₋

= − − + − +

 

′

Γ − + _ − _ −

 

∫

∫

∫

∫

∫



 

 

By Lemma 2.1, Lemma 3.1 and Lemma 3.2, we have

( )

( )

(

)

(

)

(

)

0 0

0

0

1 1

2 2

2 1 1 0 0

1 1

1 2

0 1

log d log d

log d 1.

i

i

J G s s s G i i

i

δ γ δ

γ δ

ζ σ γ ζ σ γ σ

ζ σ γ σ

− + −

− + −

−

−

= − + +

+

∫

∫

∫



 

When δ →0, we have

0 1

1 1

log d 1.

2 2

G it it t

γ

ζ

∞ _{   }

+ +

   

   

∫



Similarly,

0 2

1 1

log d 1.

2 2

G it it t

γ

ζ

∞ _{   }

+ +

   

   

∫



Assume that RH and 0 1

50

δ

< ≤ , by the contour integration method, we have

( )

( )

0

( )

( )

0

( )

( )

0 0

1

2

2 2

3 3 3

1 1

2

2 2

log d log d log d 0

i _i

i

i i i

G s s s G s s s G s s s

δ γ _γ

γ δ δ γ

ζ ζ ζ

+ + ₊

+ ∞

+ + + ∞ + +

+ + =

∫

∫

∫

( )

( )

( )

( )

( )

( )

0

0

0 ₀

1 2

3 1 2

2 2

3 3

1 2

2

log d

log d log d

i

i

i i

i i

G s s s

G s s s G s s s

δ γ

δ

γ

γ _{δ γ}

ζ

ζ ζ

+ +

+ + ∞

+ + ∞

+ + +

= − −

∫

∫

∫

same as above

( )

( )

0 1 2

3 1 2

log d 1.

i

i

G s s s

δ

δ γ

ζ

+ + ∞

+ +

∫



When δ →0, we have

0 3

1 1

log d 1.

2 2

G it it t

γ

ζ

∞ _{   }

+ +

   

   

∫



Similarly,

0 4

1 1

log d 1.

2 2

G it it t

γ

ζ

∞ _{   }

+ +

   

   

Synthesize the above conclusion, we have

0

0

1 3

1 1 1

log d

2 2 2

1 1 π π 1

2 exp cos log d

2 2 2 4 2 2

1 π π

4 Re exp cos

2 2 4 2

G it G it it t

h

H it H it t ht it t

h

H it t ht

γ

γ

ζ

ζ

∞

∞

  ₊ ₊  ₊   ₊ 

     

 _{   } _{ }

 

          

= _{ } + _+ _ − _{ } − _{ }− + _{ } + _

         

  

      

= _{ } + _{ } − _{ }− + _

     

 

∫

∫

0

1 1

log arg d

2 it i 2 it t

γ

ζ ζ

∞ _{    }

+ + +

    

 _{   }

 

∫

therefore

( )

0

1 π π

Re exp cos d 1.

2 2 4 2

h

H it t ht S t t

γ

∞_{      }

+ − − +

 _ _ _ _ _ _

  

∫



Similarly,

0

0

1 3

1 1 1

log d

2 2 2

1 π π 1 1

4 Im exp cos log arg d

2 2 4 2 2 2

G it G it it t

h

i H it t ht it i it t

γ

γ

ζ

ζ ζ

∞

∞

  ₊ ₋  ₊   ₊ 

     

 _{   } _{ }

 

 

          

= _{ } + _{ } − _{ }− + _{ } + _+ _ + _

         

   

∫

∫

therefore

( )

0

1 π π

Im exp cos d 1.

2 2 4 2

h

H it t ht S t t

γ

∞_{      }

+ − − +

 _ _ _ _ _ _

  

∫



Similarly,

( )

0

1 π π

Re exp sin d 1

2 2 4 2

h

H it t ht S t t

γ

∞_{  }

 ₊   ₋  _{− +} 

     

 _{ } _{   }

  

∫



( )

0

1 π π

Im exp sin d 1.

2 2 4 2

h

H it t ht S t t

γ

∞_{  }

 ₊   ₋  _{− +} 

 _ _ _ _ _ _

  

∫



Therefore

( )

0

1 2

1

d 1.

2

it

it z S t t

γ

+∞ _{  − −}

′ Γ_ + _

 

∫



We use the same process, we can get

( )

0

1 2

1

d 1.

2

it

it z S t t

γ

+∞ _{  − −}

Γ_ + _

 

∫



This completes the proof of Lemma 3.6.

Lemma 3.7. Assume that RH, we have

(

)

1

2 2

1

log 2

i

i a ib γ T

γ

γ − −

−∞< <+∞

 

Γ_ + _ +

 

∑



where γ be the ordinates of the nontrivial zeros of ζ

( )

s .

Proof.

(

)

(

)

(

)

0 0

1 2

1 1

2 2

1 2

1 2

1 1

2 2

i

i i

i a ib

i a ib i a ib A A

γ γ

γ γ

γ γ γ γ

γ

γ γ

− − −∞< <+∞

− − − +

< <+∞ < <+∞

 

Γ_ + _ +

 

   

= Γ_ + _ + + Γ_ − _ + = +

   

∑

(

)

( )

0 0

1 1

2 2

1

1 1

d

2 2

it i

A i a ib γ it z N t

γ γ γ

γ

− − +∞ − −

≤ <+∞

   

= Γ_ + _ + = Γ_ + _

   

∑

∫

by Lemma 2.3, the above formula

( )

( )

1 2

0

0 0

1 1

2 2

1 7 1

d log

2 2π 2π 2π 8

1 1 1 1

log d d .

2π 2 2π 2

it

it it

t t t

it z S t O t t

it z t it z S t O t

γ

γ γ

+∞

− −

+∞ _{− −} +∞ _{− −}

 

   

= Γ_ + _{ } − + + + _{ }

    

 

       

= Γ_ + _{  } + Γ_ + _{ } + _{ }

        

∫

∫

∫

By Lemma 3.4, the above formula

( )

(

)

0

0

1 1

2 2

1 1

2

2 2

1 1

log d

2 2

1 1 1

log d log

2 2

it it

it it

i it z i it z z S t t

O it z it z z t O T

t

γ

γ

+∞ _{− − − −}

+∞

− − − −

 _′    

= − _ Γ _ + _ − Γ_ + _{ }

   

 

 _{     } 

′

+ _ Γ_ + _ + Γ_ + _{  } _+

     

 

∫

∫

by Lemma 3.5 and Lemma 3.6, above formulas log2T . By Lemma 2.1 and Lemma 3.2, we have

(

)

(

)

0 0

1 2 2

1

exp π 1.

2

i

A i a ib γ h

γ γ γ γ

γ − + γ γ

≤ <+∞ ≤ <+∞

 

= Γ_ − _ + − +

 

∑



∑



This completes the proof of Lemma 3.7.

Lemma 3.8. Assume that RH, if T ≥2, then

( )

1

2 2

2

exp log .

n

n

n T O T T

T ∞

=

 

 

Λ _− _= + _{ }

   

∑

Proof. By Lemma 2.4, we have

( )

( )

(

( )

)

( )

( )

(

)

( )

(

)

( )

( )

2 2 2

2 2

1 2 2

2

1 1

2

2 2

0 1 2

exp exp d exp d

1

exp d exp d 1

2 1

exp log exp d 1

log log exp d 1

n

n x x

n x x R x

T T T

x x

x R x x O

T T T

x

T O x x x O

T T T

T O T x x T x x O

T O T

ψ

∞ ∞

∞

=

∞ ∞

∞

∞

     

Λ _− _= _− _ = _− _ +

     

   

= _− _ + _− _ +

   

 

   

= _− _+ _{ }− _{ }+

     

 

= + _ + − _+

 

= +

∑

∫

∫

∫

∫

∫

∫

2

log T .

 

 

 

This completes the proof of Lemma 3.8.

4. Conclusions

When a 1,T 50,b 2π

T

= ≥ = , n is the positive integer; by Lemma 2.1, we have

( )(

)

( )2

(

)

1

d exp exp .

2π

s s n

s a ib n s an ibn

i T

− −  

Γ + = − − = _− _

 

∫

( )

( ) ( )(

)

( )

2 2

1

exp d .

2π

s n

n

n s s a ib s

T i

ζ ζ

∞ ₋

=

′

 

− Λ _− _= Γ +

 

∑

∫

By Lemma 2.2 and RH, the above formula is

(

)

(

)

( )

( ) ( )(

)

( )

1

1

2

3 4

1 1

0 d .

2 2π

i s

a ib i a ib s s a ib s

i

γ γ

ζ ζ

γ

ζ ζ

− − − −

−∞< <+∞ ₋

′ ′

 

= − + + Γ_ + _ + + + Γ +

 

∑

∫

By Lemma 3.3 and Lemma 3.7, the above formula is _log2_T.

By Lemma 3.8, we get a contradiction; therefore the RH is incorrect.

References

[1] Montgomery, H.L. and Vaughan, R.C. (2006) Multiplicative Number Theory I. Classical Theory. Cambridge Univer-sity Press, Cambridge.http://dx.doi.org/10.1017/CBO9780511618314

[2] Titchmarsh, E.C. (1988) The Theory of the Riemann Zeta Function. Oxford University Press, Oxford.