Using a new definition on imaginary number ”i”, which I newly made in this article, I made a proof of Riemann Hypothesis

PROOF OF RIEMANN HYPOTHESIS

JEONG GEON KIM

Abstract. Using a new definition on imaginary number ”i”, which I newly made in this article, I made a proof of Riemann Hypothesis. My point of departure is on the possibility that: If any complex number is zero , then it can take a form ofe^ρ =p^ρ in complex space and exponent ρ is the root of Riemann zeta functionζ(ρ) = 0 and p is any prime number.

Problem : Prove the Riemann Hypothesis.

Riemann Hypothesis ( hereinafter called as RH ) is that ; If ζ(ρ) = 0, ρ= −2, −4, −6 · · · , then (ρ) = 1/2. (ρ = 1/2 + it, ρ ∈ C, t = 0 and t ∈ R).

(0.1) ζ(s) =

∞ n=1

1 n^s (n = 1, 2, 3· · · , p = 2, 3, 5 · · · , s ∈ C)

Notation :

(1) herein and hereinafterR, C mean the fields.

(2)

p mean product notation (subscript).

(3) i : imaginary number, (4) p or P : all primes

(5)−→: approach sign , (6) =⇒: if ∼, then ∼ sign . (7)⇐⇒: if and only if , (8)x^yz = x^yz , (9) log_e(x) = ln(x).

(10) Most of numbers and functions in this article are complex ones inC.

1. Preliminaries

Firstly, I would like to transform the problem statement to equivalent one as follows ;

(1.1) ζ(s) =

∞ n=1

1 n^s =

p

1

1− p^−s = 1

(1− 2^−s)(1− 3^−s)(1− 5^−s)· · ·, ((s) > 1).

Date: July 25, 2006.

2000Mathematics Subject Classification. NT, MSC-Class : 11M26.

Key words and phrases. Riemann Hypothesis, Prime Number Theory, Duality/Symmetry.

This small amateur article is due to my family’s full support and love and I wrote this for the pursuit of human knowledge, even though in a amateurism, with a joyful firm belief that doing mathematics has been one of the most beautiful hobbies of people since history began.

1

Since the Riemann’s zeta function was extended by its analytic continuation to whole C except s = 1 by Riemann in his paper in 1859, the equality between Riemann’s zeta function(ζ(s)) and Euler product had to be extended also in ana- lytic continuation of ζ(s). Therefore, the below equation deleting the condition of

(s) > 1 is possible in that context.

(1.2) 1

ζ(ρ) =

p

(1− p^−ρ) =

p

p^ρ− 1 p^ρ

So, equivalent problem statement is as follows ;

(1.3) { ∃ζ(ρ) ∀pj, ∀ρj | p^ρ_j^j = 0 ⇐⇒ ζ(ρ) = 0, (1/ζ(ρ)) = ±∞ }

(ρ = 1/2 + it∈ C, p : primes, j = 1, 2, 3, · · · , t = 0, t ∈ R).

Let’s define first the variables to be used in this article ;

Let z∈ C and y = πz²= ln(x)∈ C, x = exp(y) = e^πz2 ∈ C. And also let Wj∈ C

(1.4) W_j= p^ρ_j^j

Meanwhile, if all W_j is equal to 0 , then we can think that;

(1.5)

W_j = p^ρ_j^j = e^ρjln(pj)= e^ρj =|e^πz2|e^i(pj+2nπ)=|x|e^i(pj+2nπ)=√

ee^i(pj+2nπ)=√ ee^iPj.

Please note that I am using|e^πz2| = |x| =√

e as a modulus of a complex number W_j, which is actually total volume sum of every complex dimension (in real space, total volume sum of every even real dimension) when|y| = |lnx| = ln|x| = ln(√

e) = 1/2 =(ρ).

(1.6)

ρ_j= log_pjW_j= lnW_j= ln|x| + i(pj+ 2nπ) = ln(√

e) + i(p_j+ 2nπ) = (1/2) + it_j.

(n : integer, p_j : primes and also Arg of log function, j = 1, 2, 3,· · · ).

Please also note that: |y| = |πz²| = |V ol(C¹)| = |lnx| = ln|x| = ln(√

e) = 1/2 =

(ρ).

Three equations shown above are very critical and important points for my proof of RH and please see carefully these ones. While p_j are all primes, however, they

are also now Arg. of log function. From above, we can easily induce the value of |x| = |e^y| = √

e. However, as I will explain later, this value √

e is the ” Total Volume Sum of Every Complex Dimension ” when|y| = |lnx| = ln|x| = 1/2 = (ρ).

Meanwhile, we need to see the so called analytic continuation of ζ(s), its func- tional equation , its analytic continuation ξ(s) and its functional equation [1] [8] [6]

[7] [9] [10]. By quoting these equations here I would like to stress that volume for- mulas in differential geometry or analysis textbooks [4] are melted into these two functional equations.

(1.7) 2sin(πs)

(s− 1)ζ(s) = 2π

(−s)ζ(s), (sin(πs)/πs = 1/

(−s) (s)).

= i·

 _+∞

+∞

(−x)^s−1dx e^x− 1

= i·

±∞

n=±1

(−2nπi)^s−1(−2πi) = (2π)^s

∞ n=1

n^s−1((−i)^s−1+ i^s−1)

= (2π)^s

∞ n=1

n^s−1· 2sin(πs/2) = (2π)^s· 2sin(πs/2)

∞ n=1

n^s−1

(1.8) ζ(s) =

(−s)(2π)^s−12sin(πs/2)ζ(1− s)

(1.9) π^−s/2Γ(s/2)ζ(s) = π^−(1−s)/2Γ(1− s

2 )ζ(1− s), (s = 1).

By the below ξ(s), analytic continuation of ζ(s), and functional equation of ξ(s) we get the value of ζ(−2k) = 0, (k = 1, 2, 3, · · · ), so called trivial zeroes

−2, −4, −6, · · ·.

(1.10)

ξ(s) = ξ(1−s) = (1/2)s(s−1)π^−s/2Γ(s/2)ζ(s) = ζ(s)(s− 1)

U nit V ol (R^s) = ζ(s)s(s− 1) U nit Area (R^s−1).

(s= 0, −2, −4, −6, · · · , or 1 − s = 1, 3, 5, 7, · · · .).

(1.11)

ξ(2s) = ξ(1−2s) = s(2s−1)π^−sΓ(s)ζ(2s) = ζ(2s)(2s− 1)

U nit V ol (R^2s) = ζ(2s)2s(2s− 1) U nit Area (R^2s−1).

(2s= 0, −2, −4, −6, · · · or 1 − 2s = 1, 3, 5, 7, · · · .).

Unit Volume and Unit Area in this equation are the ones made from unit radius r = 1. By this equation, we now know that: ξ(2) = ξ(−1), ξ(3) = ξ(−2), ξ(4) = ξ(−3), ξ(5) = ξ(−4), etc., simply meaning that ξ(s) is symmetrical function of the symmetry axis of (s) = 1/2. Meanwhile, I would like to present here to you the Volume Formulas, which are now being widely used in Mathematics, for the purpose of showing you what kind of problems are involved in them and what are their implications in connection with the problem of RH.

(1.12) V ol (Rⁿ) = 2π^n/2rⁿ

nΓ(n/2) = (πr²)^(n/2)

(n/2)Γ(n/2) =(πr²)^(n/2) (n/2)! .

(n : integer, but n= 0, −2, −4, −6, · · · r : radius)

(1.13) V ol (R⁰) = (π^0/2)r⁰

(0/2)Γ(0/2)= 1

0· (−1)! = 1

0· (±∞) = 0/0

Please note that ; 0· Γ(0) = Γ(1). Because ,if it is, we should accept the quite absurd result of ; V ol(R⁰) = (0/n)Γ(0/n)^π0r0 = 1 = _0·Γ(0/n)ⁿ = n = (n·0)Γ(n·0)¹ = 1/n.

And in the case of 2sin(πs)

(s− 1) = 2sin(πs)Γ(s) = 2π/

(−s) = 2π/Γ(1− s) shown above in the analytic continuation of ζ(s), if s = 0, then 0· Γ(0) = 2π = 2π · 0 · Γ(0), which is quite absurd result taken when we accept 0· Γ(0) = Γ(1) = 1.

However, this is the source of complexity in the quest for the proof of RH.

The recursion formula for Gamma function [5] : Γ(z+1) = zΓ(z) can be extended to the left half of the complex plane, however, in this case z= 0, −1, −2, −3, · · ·.

(1.14) V ol (Cⁿ) = V ol (R²ⁿ) = (πr²)ⁿ

n Γ(n) = (πr²)ⁿ n!

(n= 0, −1, −2, −3, · · ·).

(1.15) V ol (C⁰) = V ol (R^2·0) = V ol (R⁰) = 0/0 = indeterminate.

So, we come to know that the value of volume forR⁰andC⁰cannot be calculated whatever the volume formula was adopted for calculation.

(1.16)

V ol (C⁻ⁿ) = V ol(R⁻²ⁿ) = 1

(−n) Γ(−n)(πr²)ⁿ = 1

Γ(1− n)(πr²)ⁿ = 1

(−n)!(πr²)ⁿ = 0

(n = 1, 2, 3,· · · , r = 0).

If r = 0, V ol(C⁻ⁿ) = V ol(R⁻²ⁿ) is 0/0(indeterminate).

(1.17) Area (Rⁿ⁻¹) = 2π^n/2rⁿ⁻¹

Γ(n/2) = d V ol(Rⁿ) dr

(1.18) Area (R²ⁿ⁻¹) =(2n)πⁿr²ⁿ⁻¹

n! =d V ol(R²ⁿ)

dr =d V ol(Cⁿ) dr .

Let’s return to the main issue and please note that the variable ”x” in this article has the following meaning :

(1.19) x =

n

(πz²)ⁿ

n! =

n

V ol(Cⁿ) =

n

V ol(R²ⁿ) = exp(πz²) = exp(y).

”x” is total volume sum of every complex dimension. ( n = 0, 1, 2, 3,· · · , radius z ∈ C , y = V ol(C¹) ∈ C, x ∈ C ). Especially, √

e = |x| is just the total volume sum of every complex dimension when (ρ) = 1/2.

In ”x”, however, please note that V ol(C⁰ =R⁰) is included as one(1), not as (0/0). That means we implicitly accept 0· Γ(0) = Γ(1) = 1.

Seeing the above mentioned analytic continuation of ζ(s),which is deeply in- volved with the Volume Formulas, we can think that the complex variable ”s” of ζ(s) and ξ(s) may mean the dimension number expressed in complex num- ber for various complex spaces.

Meanwhile, our new problem statement is now being changed concisely as follows :

Prove this statement ;∀Wj, W_j = 0. (j = 1, 2, 3,· · · ).

From a normal viewpoint of complex analysis, W_j = e^ρj = P_j^ρj =√

ee^i(Pj+2nπ) can never be 0. In this regard, we need to think the imaginary number ”i” in dif- ferent perspectives in order to tackle and to prove the RH.

2. New definition of imaginary number ”i”

What is meant in that all W_j = p^ρ_j^j = e^ρj = 0∈ x ? In other words, what is the meaning of : ρ_j= 1/2 + it_j = 1/2 + i(p_j+ 2nπ)∈ y(= V ol(C¹) = πz²)−→ (−)∞ ? If we interpret ρ ∈ y as a volume of complex one dimension, what are the meanings of minus infinity V ol(C¹) value and zero value of Volume Sum of Every Complex Dimension, W = e^ρ= p^ρ∈ x, in this case ?

While, if each and every ρ ,which also may mean the various dimension number in a sense that ρ is in ”s” and ”s” of ζ(s) can be dimension number as I mentioned before in analytic continuation of ζ(s) , approaches to minus infinity, then what kind of implications are involved in ρ and ”i” in this context ?

Can each of complex number ”i” and ρ ,regardless of their possible implications as a volume or as a dimension number, approach to or equal to minus infinity in

”normal” mathematical justifications ?

I think this is the reason why RH has been such a difficult task to tackle.

If we calculate the unit volume of±∞ dimensions in real and complex spaces as shown below, we come to know that each unit volume of−∞ dimension in real and complex space is zero or indeterminate, while each unit volume of +∞ di- mension in real and complex space is easily calculated as zero(0).

(2.1)

U nit V ol (R^−∞) = 2· π^−∞

−∞ · Γ(−∞) = 0

Γ(−∞ + 1) = 0

(−∞)! = 0 = Unit V ol (R^∞).

(2.2)

U nit V ol (C^−∞) = 1

π^∞· (−∞)! = 0

Γ(−∞ + 1) = 0

(−∞)! = 0 = Unit V ol (C^∞).

(2.3) U nit V ol (R^∞) = 2· π^∞

∞ · Γ(∞)= 2· π^∞

(∞)! = U nit V ol (C^∞) = π^∞ (∞)! = 0.

(2.4) U nit V ol (R^−∞) = U nit V ol(R^−2·∞) = U nit V ol (C^−∞) = 0 or (0/0).

(2.5) U nit V ol (R⁻ⁿ) = U nit V ol (C⁻ⁿ) = U nit V ol (R⁻²ⁿ) = 0.

( n: plus integer, n=0 , n−→ ∞ ).

(2.6)

U nit V ol (R^s) (−∞ < s < −∞ + 1) = 0/0, Unit V ol (R⁰) = U nit V ol (C⁰) = 0/0.

( s: s is not integer, s is in-between unit interval of integers ).

THESE ARE the only two cases of the indeterminate Unit Volume quantities in real and complex spaces. While, V ol(C⁻ⁿ) = V ol(R⁻²ⁿ) is indeterminate(0/0) if r = 0.

(2.7) ξ(−∞) = ζ(−∞)(−∞ − 1)

U nit V ol (R^−∞) = ξ(1 +∞) = ζ(1 +∞)(∞) U nit V ol (R^1+∞).

So, ζ(−∞) is indeterminate when ”s” is in-between very big minus integers.

And, ζ(−∞) is zero(0) when ”s” is on that very big minus integer as well as the ζ(s) values on trivial zeroes.

While, the reason why the volume of−∞ dimension in real and complex space is indeterminate is due to the involvement of Γ(s) function in denominator of each volume formula. Γ(s) function approaches ±∞ in the domain of zero and minus integer. However, in each interval between very big minus integers Γ(s) function approaches to zero(0) as ”s” approaches to minus infinity. So, volume can be zero(0) or indeterminate (0/0) when ”s” is minus infinity .

Now, we come to know that ζ(−∞) = 0 when complex variable ”s” is located just on the very big minus integer. Does this imply any message to the RH that ζ(ρ) = 0 for all ρ ?

For the reminder, I will show also the case of s = 0, s = 1 as below, which is fault. And this is the result when we simply apply s = 0, s = 1 to the Riemann’s functional equation of ξ(s).

(2.8) ξ(0) = ζ(0)(0− 1)

U nit V ol (R⁰) = ξ(1) = ζ(1)(1− 1) U nit V ol (R¹)= 0

0 ∈ R.

(2.9)

ζ(0)(−1)Unit V ol(R¹) = (−1/2)(−1)(2) = 1 = ζ(1)(0)Unit V ol(R⁰) =∞·0·(0/0).

THEREFORE, THERE IS NO EQUALITY LIKE ξ(1) = ξ(0) IF WE SIMPLY SEE THE FUNCTIONAL EQUATION OF ξ(s). BECAUSE, ζ(s) DOES HAVE A SIMPLE POLE AT s = 1,WITH RESIDUE 1.

HOWEVER, USING THE RIEMANN-SIEGEL FORMULA [1], WE CAN CAL- CULATE ζ(0) =−1/2 AND INSERTING UNIT VOL(R⁰ =C⁰) = 1 TO FUNC- TIONAL EQUATION OF ξ(s) WE CAN INDUCE THE EQUALITY OF ξ(0) = ξ(1) = 1/2.

Considering the arguments made so far, we can now make new additional con- dition to the analytic continuation of ζ(s), i.e. ξ(s), as follows :

ξ(s) is entire function in C , except s = 0, −2, −4, −6, · · · , or 1 − s = 1, 3, 5, 7,· · · , (−∞) < s < (−∞ + 1) or ∞ < (1 − s) < ∞ + 1. In inequality cases, each of ”s” and ”(1-s)” is located in the open unit integer interval (n, n + 1).

To see clearly the problems involved in relation to this newly added condition for analytic continuation of ζ(s), we need to see the so called Riemann-Siegel Formula here [1] :

(2.10) ζ(ρ) = (−ρ)!

2πi

 _+∞

+∞

(−x)^ρ e^x− 1

dx

x = 0 ∈ C, (ρ = 1 2 + it).

(2.11) ζ(1− ρ) = (ρ− 1)!

2πi

 _+∞

+∞

(−x)^−ρ+1 e^x− 1

dx

x = 0 ∈ C.

(2.12) ζ(ρ)Γ(ρ) = ζ(ρ)· (ρ − 1)! =

 _∞

0

x^ρ e^x− 1

dx x = 0.

(2.13)

ζ(−n) = n!

2πi

 _+∞

+∞

(−x)⁻ⁿ e^x− 1

dx

x = (−1)ⁿB_n+1

n + 1= ξ(−n) Unit V ol (R⁻ⁿ)

−(n + 1) ∈ R.

(n = 0, 1, 2, 3,· · · , Bn : Bernoulli N umbers, Odd B_2n+1are all zero(0) except B₁=

−1/2).

(2.14) (−1)ⁿ⁺¹B_n+1= ξ(−n) Unit V ol (R⁻ⁿ) ∈ R.

(n = 0, 1, 2, 3,· · · ) (2.15)

(−1)²ⁿ⁺¹B_2n+1= ξ(−2n) Unit V ol (R⁻²ⁿ) = ξ(−2n) Unit V ol (C⁻ⁿ) = 0 ∈ C.

(n = 0, 1, 2, 3,· · · ).

(2.16) ζ(0) = 1 2πi

 _+∞

+∞

dx

(e^x− 1)x= B₁= −1

2 = ξ(0) U nit V ol (R⁰)

−1 .

PLEASE NOTE THAT right hand side of above equation can only be justifiable when U nit V ol(R⁰) = U nit V ol(C⁰) = V ol(R⁰) = V ol(C⁰) = 1, which implicitly means 0· Γ(0) = Γ(1) = 1 , and ξ(0) = 1/2 = ξ(1). So it means Riemann [7] showed us this result in his 1859 paper through the calculation of contour integral in complex space in extending ζ(s) toC as a meromorphic function with only a simple pole at s = 1 with residue 1, which cannot easily be imagined if we simply see his functional equation of ξ(s).

(2.17) ζ(−1) = 1 2πi

 _+∞

+∞

−1 e^x− 1

dx x² =−1

12 = −B2

2 = ξ(−1)

−2

π^−1/2

(−1/2)! = ξ(−1)

−2π .

(2.18) ζ(−3) = 6 2πi

 _+∞

+∞

−1 e^x− 1

dx x⁴ = 1

120=−B4

4 = ξ(−3)

−4

π^−3/2

(−3/2)! = ξ(−3) 8π² .

(2.19) ξ(0) = 1

2 = ξ(1), ξ(−1) = π

6, ξ(−2) = 0

0, ξ(−3) = π²

15, ξ(−4) = 0 0,· · ·

(2.20) ζ(0) =−1

2 , ζ(−1) = −1

12, ζ(−2) = 0, ζ(−3) = 1

120, ζ(−4) = 0, · · ·

(2.21) ζ(2) = π²

6 , ζ(4) = π⁴

90,· · · , ζ(2n) = (2π)²ⁿ· (−1)ⁿ⁺¹· B2n

2· (2n)!

(2.22) 0 = ζ(ρ) = ξ(ρ)· Unit V ol(R^ρ)

ρ− 1 = ξ(ρ)· π^ρ/2

(ρ− 1) · (ρ/2)! ∈ C.

(2.23) 0 = ξ(ρ) = (ρ− 1)ζ(ρ)

U nit V ol (R^ρ) = ξ(1− ρ) = (−ρ)ζ(1 − ρ)

U nit V ol (R^1−ρ) ∈ C.

Since ξ(ρ) is analytic, Unit Vol (R^ρ) and Unit Vol (R^1−ρ) cannot be 0 or±∞ . So, in order for ζ(ρ) equals to be 0 for all ρ ; (1) ξ(ρ) must be 0 , (2) (ρ/2)!= 0 or

±∞ , (3)(^1−ρ₂ )!= 0 or ± ∞.

Therefore, ξ(ρ) = (1/2)ρ(ρ− 1)π^−ρ/2Γ(ρ/2)ζ(ρ) = (ρ− 1)π^−ρ/2(^ρ₂)!· ζ(ρ) = 0, so (^ρ₂)! = 0 or ± ∞ . And (^1−ρ₂ )!= 0 or ± ∞. However, what’s the meaning of (^ρ₂)!= 0 or ± ∞, (^1−ρ₂ )!= 0 or ± ∞ ? In other words, when the values of Unit Vol (R^ρ) and Unit Vol (R^1−ρ) do not have 0 or±∞, what do these facts have any implication to(ρ) = 1/2 , ρ −→ (−)∞, and to ∀Wj, W_j = e^ρj = 0?

In real world, s!−→ 0 only when ”s” is in the open integer interval (−∞, −∞+1) or when ”1-s” is in the open integer interval (∞, ∞ + 1) , as I explained before.

In this context, how can we find out the method to prove (^ρ₂)! = 0 or ± ∞, (^1−ρ₂ )!= 0 or ± ∞ in complex space for the advancement to the proof of RH?

In this environment, we should invent new methods or concepts for the proof of RH. As I think, one of those things should be commenced in the way of rethinking imaginary number itself. Then, how do we rethink about the ”i”, basic building block for modern sciences and mathematics ? And, after all, how can we incor- porate new idea into our new problem statement of: ∀Wj, W_j =√

ee^iPj = 0 (j = 1, 2, 3,· · · ) ?

Now therefore, it is time for us to cast a serious question to ourselves:

What is ”i” ? [11]

i = e^ln(i) = e^iπ2 · 1 = e^iπ₂ · e^2nπi = e^iπ2(1+4n) = √

e^iπ(1+4n) = √

e^2ln(i) =

√e^ln(−1)=√

−1, 1 = e^2nπi, ln(i) = ^iπ₂ +2nπi = ^iπ₂(1+4n), (−2i)ln(i) = ln(i⁻²ⁱ) = π(1 + 4n), e^π(1+4n)= i⁽⁻²ⁱ⁾= 1⁻ⁱ= ₁¹i =_e−2nπ¹ = e^2nπ, etc.(n: integer).

These are the simple examples of the equations which use the imaginary num- ber ”i”. Especially, ln(i) = log_e(i) is multi-valued, so, 4ln(i) = 2πi(1 + 4n) = 2nπi(_n¹ + 4)= ln(i⁴) = ln(1) = ln(e^2nπi) = 2nπi.

Meanwhile, let’s see the below reasonings adding new concepts to ”i” [2] :

(2.24) G(z) =

∞ n=1

(1 + z

n)e^−z/n= 1

Γ(1 + z) = 1 zΓ(z) (Γ(z) : Euler Gamma F unction)

(2.25)

G(z)G(−z) = 1

Γ(1 + z)· Γ(1 − z) =sin(πz)

πz = sinc(z) = 1

z!(−z)! = 1 B(1 + z, 1− z) (B(1 + z, 1− z) : Euler Beta F unction, sinc(z) : sinc(z) function) [13] [14]

(2.26) 1

G(z− 1) = Γ(z) = 1 e^γ(z)·z· z

∞ n=1

(1 + z

n)⁻¹e^z/n

(γ : Euler Gamma Constant, Above f orm is reciprocal of W eierstrass f orm)

(2.27) G(z− 1) = ze^γ(z)G(z) = 1 Γ(z)

(2.28) G(z− 1)

G(z) =Γ(1 + z)

Γ(z) = ze^γ(z)= z, exp(γ(z)) = e^γ(z)= 1(z) = 1 (2.29) e^γ= 1 = e^2nπi= e^−γ= e^−2nπi, γ = 2nπi, n = γ

2πi (n : integer, let^s call this number as the Rotation Number ”n”).

Other authors use this number as a ”winding number” or as a ”index”.

(2.30)

4ln(i) = γ(1

n+ 4) = 2πi(1 + 4n) = 2nπi(1

n+ 4)= 2nπi = γ = ln(1) = ln(e^2nπi)

(2.31) ln(i) = γ(1

4n+ 1) = πi

2 (1 + 4n) =πi

2 + 2nπi = πi 2 + γ

(2.32) 1 = G(0) = e^γG(1) = G(1), e^−γ= G(1) = 1 =

∞ n=1

(1 + 1 n)e^−1/n

(2.33) γ = (

∞ n=1

1

n)− ln∞ = 0.577215664901532 · · · = 2nπi = ln(1) = 0.

(2.34) i⁻²ⁱ= e^−2iln(i) = e^(1+4n)π = e^π−2γi= 1⁻ⁱ= e^2nπ

(2.35)

π = 2γi−2iln(i) = 2i(γ−ln(i)) = 2iπ 2i = 2i(

∞ n=1

1

n−ln∞−ln(i)) = 2i(

∞ n=1

1

n−ln(∞i))

(2.36) i⁻²ⁱ= e^(1+4n)π=

⎧⎨

⎩

e^π : if n−→ 0₋

∞ : if n −→ +∞

0 : if n−→ −∞

Please refer to the special remarks made at the end of the proof of RH.

(2.37)

i = e^ln(i)= e^iπ2(1+4n)= e^γ(1+4n1)= γ 2nπ =

⎧⎨

⎩

0 : if n−→ ±∞

±∞ : if n−→ 0±

i = _2kπ^γ : if n = k (k : other integer)

Please be reminded here that γ = 2π·n·i and 0·Γ(0) = 2π·0·Γ(0) = Γ(1) = 1, which is the quite absurd result we implicitly accepted in the analytic continuation of ζ(s) at s = 0 , i.e., in the calculation of ζ(0) =−1/2 and V ol(R⁰=C⁰) = 1.

I already mentioned before that accepting 0· Γ(0) = Γ(1) = 1 in the process of analytic continuation of ζ(s) is the source of complexity of RH.

And now in my new definition of imaginary number ”i”, the ”γ = 2π·n·i”

and ”2π· 0 · Γ(0) = 0 · Γ(0)” are both showing some similarities.

(2.38) i ={0, ±∞, γ

2nπ}, 1 = e^γ = e^2nπi= e^−γ= e^−2nπi (∞, 1 = e^2nπi= e^γ, 0, i, γ = 2nπi∈ C)

Let’s call the set of all possible combinations of (n· i), A , then γ = 2πA.

(2.39) A = {n · i| (n · i), (0 · ∞)}, γ = 2πA = ln(1) = ln(e^γ) = 2nπi, n : integer.

Therefore, as far as we consider the Euler Constant γ as a real value of 0.57721· · ·, imaginary number ”i” can have real value and ”i” is not a fixed value in this case.

It is multi-valued and closely interconnected with the Rotation Number ”n = _2πi^γ ”.

What a surprising result it is! And what’s the real meaning of ”i”? In this argument, we can conclude that ”i” has no definite size in real space, however, it can be projected/converted to various real values including 0 and ± ∞ through the newly defined Euler Constant γ = 2nπi (n ∈ Z). In that sense, I think that idθ = d(iθ) [12] can be justified ,which just means ”i” is not affected by the total dif- ferential. Imaginary number ”i” is dimensionless and gauge free in physicist’s sense.

In summary, we can conclude that ”i = _2nπ^γ ” has no definite size in R , but, can be projected/converted to real space through γ = 2nπi =

0.57721· · · as various discrete real values 0, ±∞,_2nπ^γ according to the Ro- tation Number ”n = _2πi^γ ”. In that case, it is not a continuum entity but can exist as multi-valued discrete numbers inR.

The reason why I am showing here that the imaginary number ”i” has so unique character is : If ”i” has minus infinity as one of its values, then RH has a pos- sibility of its proof. So to speak, all W_j = √

ee^iPj = √

ee^i(Pj+2nπ) = √ ee^iPje^γ can be 0 in case we apply 1 = e^γ = e^2nπi and i = −∞ to Wj. Because,

∀ W_j, W_j = 0 (j = 1, 2, 3,· · · ) in this case, thus making the proof of RH nat- urally. Therefore, the imperative requirement for us to prove RH is to prove i = −∞ for our problem statement of ∀Wj, W_j = 0. Therefore, i = −∞ re- quirement is only possible when the Rotation Number ”n” −→ 0₋ by the equation (2.37).

Meanwhile, another examples of interesting results arising out of the reasonings made above in connection with the prime number theorem function π(x) are as follows :

(2.40)

∞ n=1

1 n−

p

1

p = ln∞ + γ − lnln∞ + γ

= ln( ∞

ln∞) + 2γ = ln(π(∞)) + 2γ = ln∞ + 2γ, (lnln∞ = 0) ∈ C

(π(x) = number of primes less than x, π(∞) =_ln(∞)^∞ , p : primes).

(2.41) 

p

1 p =

 _x=ln∞

x=1

d(lnx) =

 _x=ln∞

x=1

dx x =

 _t=∞

t=e

dt lnt· t

= [lnx]^x=ln∞_x=1 = lnln∞ − ln(1) = lnln∞ − γ = −γ ∈ C

( t = e^x, x = e^y, x = lnt, y = lnx, t = e^ey)

Please note that ”t” will have the same meaning hereinafter continu- ously.

Euler’s formula for the sum of reciprocal primes is

p1

p = lnln(∞), so there is just a difference of γ between his and mine when we apply 1 = e^γ = e^2nπi, ln(1) = γ = 2nπi ∈ C in the above calculations. The reason of difference is : γ = 0 =

(γ) ∈ R and e^γ = e^(γ)= e⁰= 1∈ R. While, let’s analyze differently as below :

(2.42) 

p

1

p = lnln∞ − γ =

 _t=∞

t=e

dt lnt· t =

 _t=∞

t=e

d(lnt) lnt =

 _t=∞

t=e

dln(lnt)

=

 _t=∞

t=e

d{d(lnt)} = [d(lnt)]^t=∞_t=e = [dx]^x=ln∞_x=1=eγ = [y]^y=lnln∞_y=γ = d(ln∞) − d(1)

= d²(∞) − d(1) = d(∞)

∞ − γ = ln∞

∞ − γ = 1

π(∞)− γ = −γ,

(d²(∞) = d(ln∞) = lnln∞ = 0, d(∞) = ln(∞), γ = d(1) = ln(e^γ) = ln(1))∈ C

Seeing the above calculation, we can induce that the total differential (d(x)) and complex log function (ln(x)) can be interchanged without any limitation at x =∞, x = ln∞, x = e and x = 1 = e^γ ∈ C when we properly change the variable of the function ”x” .

So, I will use the equation , total differential d(·) = ln(·) ∈ C, continu- ously hereinafter in this article.

d(∞) = ln∞ = ∞ − 1, d(1) = ln(1) = ln(e^γ) = γ = 2nπi= 0 ∈ C (2.43)

d(ln∞) = d(∞)

∞ = ln∞

∞ = lnln∞ = d(d(∞)) = d²(∞) = 0 ∈ C (2.44)

d(∞) = ∞ · (

p

1

p+ γ) =∞ · lnln∞ = ∞ · d²(∞) = ∞ · 0 = ln(∞) ∈ C (2.45)

π(∞) = ∞

ln∞ = 1

p1

p + γ = 1

lnln∞ = 1 d²(∞) =1

0 =∞ = 1 + ln(∞) ∈ C (2.46)

∞ = 1

d= 1 + d(∞) = 1 + ln(∞), d · ∞ = 1 = e^γ= d(∞) = ∞ · 0 = ln∞ ∈ C (2.47)

d = 1

∞= e^γ

∞ = 0, (d : total differential, d = 0) ∈ C (2.48)



p

1

p+ γ = 0∈ C (2.49)

γ = 2nπi = ln(1) = d(1) = ln(e^γ) = d(e^γ)∈ C (2.50)

PLEASE NOTE HERE THAT d(1) = d(e^2nπi) = d(e^γ) = ln(e^γ) = γ = ln(1) = 0 and d(∞) = ln(∞) = ∞ · 0 = ∞ · (

p1

p + γ) = 0 mean complex numbers ∞ and 1 = e^2nπi = e^γ are not pure Constants in this sense. And Euler Constant γ is also not a pure Constant in a sense that it is connected with rotation number ”n” and multi-valued ”i” in complex space. In that con- text, e^γ = e^−γ= 1(= the value of e^γ = e^0.577···∈ R) is justified in C.

(2.51)

∞ n=1

1 n+

p

1

p = lnln∞ + ln∞ = ln∞ = d(∞) = ∞ · 0

= ln(∞ · ln∞) = lnln(∞^∞)= lnln(∞) = d²(∞) = 0 ∈ C

(2.52)

∞ n=1

1

n= ln(∞) + γ = ln(∞) + ln(1) = ln(∞) + γ

= d(∞) + d(1) = d(∞) + γ = ln(∞) = d(∞) = ∞ · 0 = 0 ∈ C

(2.53) ∞ = 1 d =1

0 = π(∞) = ∞

d(∞) = ∞ ln∞

=∞ · 0 + 1 =

 _x=1

x=0

dx x =

 _x=1

x=0

d(lnx) = ln(1)− ln(0) = γ + ∞ , (d : total dif f erential, d= 0) ∈ C

(2.54) d(∞) = ln∞ = ∞ − 1 = −ln(0) − 1 ∈ R

 ₁

0

dx

x = [lnx]¹₀= ln(1)− ln(0) = 0 − (−∞) = ∞

= (

∞ k=1

1

k) + 1− γ(Euler Constant) = ln∞ + 1, (ln(1) = 0) ∈ R

 _∞

0

dx x =

 _∞

0 d(lnx) = [lnx]^∞₀ = ln∞ − (−∞) = ln∞ + ∞ = 2ln∞ + 1 ∈ R

(2.55) d(∞) = ln(∞) = ∞ − 1 ∈ C,

 _∞

0

dx x =

 ₁

0

dx x +

 _∞

1

dx

x = [lnx]^∞₀ = (γ+∞)+(ln∞−γ) = ∞+ln∞ = 2ln∞+1 ∈ C

3. The Proof of Riemann Hypothesis

Summarizing the key results of the reasonings made so far, those are as follows :

lnln∞ = ∞, ln(∞) = d(∞), ln(1) = d(1) = 0 ∈ R (3.1)

∞ = ln∞ + 1, γ = (

∞ n=1

1

n)− ln∞ = 0.57721 · · · ∈ R (3.2)

∞ = ln∞ + 1 ∈ R, C (3.3)

1 = e^γ = e^−γ∈ C (3.4)

lnln∞ = d²(∞) = 0, ln∞ = d(∞) ∈ C (3.5)

ln(1) = ln(e^γ) = d(1) = γ = 2nπi∈ C, (n ∈ Z) (3.6)

∞ n=1

1 n+

p

1

p = ln∞ ∈ C, (p : primes) (3.7)

∞ n=1

1 n−

p

1

p = ln∞ + 2γ ∈ C (3.8)



p

1

p=−γ ∈ C (3.9)

∞ n=1

1

n = ln∞ + γ ∈ C (3.10)

γ = 2nπi = ln(1) = ln(e^γ) =−

p

1 p∈ C (3.11)

Rotation N umber(n) = γ

2πi, Imaginary N umber ”i” = γ (3.12) 2nπ

if ”n”−→ 0−, then i−→ −∞ ∈ C (3.13)

(3.14) if ”i” −→ −∞, then ∀Wj ∈ C, Wj = 0 ∈ C , (j = 1, 2, 3, · · · )

[PROOF] :

We will continuously use the variable x = exp(πz²) , y = lnx = πz² and t = e^x= e^ey ∈ C as defined previously. And before we go into the conclusion, I quote here the meromorphic counting theorem [3]:

[MEROMORPHIC COUNTING THEOREM]

Suppose that C is a simple closed positively oriented path, Ω is the region inside C, and f is meromorphic on Ω , analytic and non-vanishing on C. Let N(f) denote the number of zeros of f inside Ω and P(f) denote the number of poles of f inside Ω, counted according to multiplicity. Then N(f) and P(f) are finite and

(3.15) N (f )− P (f) = 1

2πi



C

f^(z) f (z)dz

According to the above theorem, let’s define γ and Rotation Number(n) as below and path C, Ω will be used same as the theorem above.

(3.16) y = γ = γ(z) =



C

dx(z) x(z) =



C

d(ln(x(z))) = ln(1(z)) = d(1(z)) = 2nπi

= ln(e^γ(z)) = d(e^γ(z)) = d(e^2nπi) = y(z) = πz²= V ol(C¹) = V ol(R²), x = x(z) = e^πz2= e^y = 1(z) = e^2nπi= 1

x = e^y(z)= x(z) =

∞ n=−∞

(πz²)ⁿ n!

=

∞ n=−∞

yⁿ n! =

∞ n=−∞

πⁿ

n!z²ⁿ= 1

= Laurent Series in y(= πz²) = V olume Sum of Every Cⁿ

= V ol(C⁰=R⁰) used in analytic continuation of ζ(s),

t = e^ey = e^eγ= e¹= e = exp(V olume Sum of Every Cⁿ) = Constant.

(3.17) n = n(z) = γ(z) 2πi

= 1 2πi



C

dx(z) x(z) = 1

2πi



C

d(lnx(z))

= ln(x(z)) 2πi =y(z)

2πi =πz² 2πi = z²

2i

So, we come to know that y = γ, x = 1, t = e. And also know that :

(3.18) n(z) = N (x(z))− P (x(z)) = γ(z) 2πi =y(z)

2πi =z²

2i = 2nπi 2πi = πz²

2πi

Now therefore, IF THERE ARE values of z∈ C (= the radius of V ol(C¹)), which satisfy the imperative requirements of RH, i.e.:

(3.19) {Rotation Number (n(z)) −→ 0₋} =⇒ {i −→ −∞}

THEN, the existence of the required values of z∈ C which are satisfying the equations (3.16), (3.17), (3.18), and (3.19) proves the RH.

THERE ARE those required values of z. Let’s call the set of those values of z

∈ C, B, which satisfy the requirements of RH,i.e., (3.16), (3.17), (3.18), and (3.19) :

(3.20) B = {z ∈ C| 2ni = z², z= 0, n −→ 0₋, n= 0}

IN SUMMARY :

{∃z ∈ B| (n −→ 0−) =⇒ (i −→ −∞) =⇒ (ρ −→ −∞) =⇒ (∀Wj, W_j = e^ρj = 0)}, (j = 1, 2, 3, · · · ).

ρ = 1/2 + it = 1/2 + ie =−∞ + e/2 = 1/2 − ∞ · e = −∞ − ∞ · e = −∞.

1/2 =−∞ = d(0) = ln(0) = −ζ(0) is from (2.16) and i = −∞ is from (2.37), (3.18).

SPECIAL REMARKS :

If radius ”z” of V ol(C¹) = πz² is ”n” , i.e., ”z=n” , then γ = 2πni = 2πzi = πz², z = 2i = 2(−∞) = 2(1/2) = 1 = x. Surprisingly, γ = π = y, z = n = 1 = x.

So, in this case, the setB is as follows ;

(3.21) B = {z = n = x = 1 ∈ C| n = z −→ 0₋, n = z= 0, γ = 2πni = 2πzi

= πz²= π = y, n = z = 2i = 2(−∞) = 2(1/2) = 1 = x = e^γ = e^π}

Therefore, in this special case, x = 1 = e^γ = e^π(= i⁻²ⁱwhen n−→ 0− : please refer to the equation (2.36)), which was mentioned by David Hilbert in 1900 2nd ICM at Paris under the sub-heading No.7, Irrationality and Transcendence of Cer- tain Numbers. So, I think I also proved the transcendence of the Euler Constant gamma : γ is a transcendental number.

This completes the proof of RH, which is also the subject of sub-heading No. 8 of Hilbert’s address on 23 problems at 1900 Paris 2nd ICM..

While, as to the D(x) = |π(x) − Li(x)| < C√

x· lnx, (x > 0), the proof is as below , and , herein x = 1 = e^γ, y = γ = lnx = 2nπi and t=e=Constant will be continuously used:

Now we let Li(x) =_x

e dt lnt :

Here we define∞ as ∞ = ζ(0) = _n=∞

n=1 1

n⁰ = 1 + 1 +· · · = ln∞ + 1 = −1/2 =

−d(0) = −ln(0) = −(ρ).

(3.22) Li(x) =

 _t=x

t=e

dt lnt =

 _t=x

t=e

t· dt lnt· t =

 _t=x

t=e

t· dln(lnt) = t ·

 _x

e

d²lnt

= t

 _d(x)

d(e)

d(lnt) (By Stokes^s) =

 _d(x)=ln(x)

d(e)=ln(e)=1=e^γ

t· dt t

=

 _lnx

1=e^γdt = [t]^lnx₁ = lnx− 1 = y − x = lnx − e^γ = lnx− e^2nπi

= γ− 1 = (−

p

1 p)− 1 =

 _x

0 lntdt =

 _x

e

lntdt = xy− x = 1 · γ − 1.

Duality : 1/lnt = lnt = 1/ln(e) = ln(e) = d(e) = 1 = Id. of M ultiplicative Group.

(ln(∞) = d(∞) = ∞ − 1 ∈ C, R), (x = 1 = e^γ, y = lnx = γ = 2nπi∈ C) .

(3.23)

|x|−→∞lim D(x) =|∞ − ln(∞) + 1| = |ln(∞) + 1 − ln(∞) + 1| = |2| = 2 · 

|γ| ·1 γ 

≤ |γ| +1 γ

 < ln∞ + ∞ = 2ln(∞) + 1 = 2 · ∞ − 1 = ∞ − 1 = ln(∞) = Max. |γ|

< C· (Max. |γ|) < C ·

M ax.|x = 1 = e^γ| · (ln∞)

= C·√

∞ · (ln∞) = lim

|x|−→∞C·√

x· lnx, (x > 0).

(C > 1 : Constant), (∞ = ln∞ + 1 ∈ C, R), (√

AB≤ ^A+B₂ , A, B > 0).

0 <|y = γ = lnx = 2nπi| ≤ ln∞, |y| = 0, (n ∈ Z, but n −→ 0₋, n= 0) (3.24)

0 <1 γ

 < ∞ = ln∞ + 1 (3.25)

0 <|x = 1 = e^γ| ≤ ∞ = ln∞ + 1, |x| = 0, (d = 1/∞ = 0, d −→ 0) (3.26)

0 < e^−∞≤ |t = e| ≤ e^∞=∞, |t| = 0 (3.27)

∞ = ζ(0) = 1 + 1 + · · · = ζ(1) + 1 − γ = ln∞ + 1 = −1/2 = −d(0) = −ln(0) (3.28)

x = d(e) = ln(e) = 1 = e^γ = Id. of M ultiplicative Group (3.29)

y = d(1) = ln(e^γ) = γ = 2nπi (3.30)

t = e^x= e = Constant (3.31)

d(·) = ln(·) (3.32)

ζ(1)− ζ(0) = γ − 1 =

 _x

e

dt lnt =

 _x

e

lntdt (3.33)

(3.34)

Density of primes = Li(∞)

∞ =

p

1

p =−γ =

 _t=∞

t=e

dt lnt· t =

 _∞

e

dln(lnt)

=

 _t=∞

t=e

d(d(lnt)) =

 _dt=lnt=ln∞

dt=lnt=ln(e)=1=e^γ

d(lnt) (By Stokes^s)

=

 _ln∞

1=e^γ

dt

t = [lnt]^ln∞_1=eγ = lnln(∞) − γ = d²(∞) − γ = −γ

(3.35) Li(x) =

 _x

e

dt lnt =

 _x

e

t· dt

t· lnt = γ− 1 = y − x = ζ(1) − ζ(0) =

 _x

e

lntdt

This completes the proof. .

[Post Script after the Proof of RH]

I want to add more equations as below. Anyhow, it seems to me that below equa- tions are very different from the Einstein’s Field Equation of General Relativity(GR) which is non-linear partial differential equation :

If we define d(0) = ln(0) =−∞ = −ζ(0) ∈ C, then below equation is possible.

(3.36) d(0) + d(1) + d(∞) = ln(0) + ln(1) + ln(∞) = −ζ(0) + γ + ln(∞)

=−ζ(1) + γ − 1 + γ + ln(∞) = γ − 1 = ζ(1) − ζ(0) =

 _x

e

dt lnt =

 _x

e

lntdt∈ C.

(3.37) ζ(1) = d(1) + d(∞) = ln(1 = e^γ) + ln(∞) = γ + ln(∞) ∈ C

(3.38) ζ(0) =−d(0) = −ln(0) = ∞ = ln(∞) + 1 ∈ C

(3.39) γ =−

p

1

p= d(1) = ln(1) = ln(e^γ) = lnln(e) = d²(e), (p : primes)∈ C (3.40)

d(0)+d(1)+d(e)+d(∞) = ln(0)+ln(1)+ln(e)+ln(∞) = γ = 2nπi = d(1) = d²(e)∈ C

(1 = e^γ = e^2nπi = e^−γ = e^−2nπi, γ = 2nπi ∈ C, γ = 0.57721 · · · ∈ R, n ∈ Z but n = 0 and n −→ 0−, d : total dif f erential).

Please compare the the below two equations, there are similarities. But, both are definitely different. :

(3.41) R_μν−1

2Rg_μν+ Λg_μν=−8πG c⁴ T_μν

(3.42) d²(e)− d(1) + (−d(0) − d(∞)) = d²(e^e)

(γ− γ + 1 = 1, d²(e) = d(1) = γ, d(e) =−d(0) − d(∞) = d²(e^e) = 1).

(3.43) 1· 1 · 1 · · · = e^γ· e^−γ· e^γ· · · = e^{γ−γ+γ···}= 1

(d(1) = γ : Id. of Additive Group, d(e) = 1 : Id. of M ultiplicative Group) Before closing the article, I would like to leave a short comment, which I feel in the pursuit for the proof of RH ;

My proof of RH, as I think, was made through the countless trial and error and with dauntless childlike imagination. It seems to me that those thoughts for proof are still being astray somewhere in space-time like the lost arrows shot to the skies aiming at the heart of a myth, the Riemann Hypothesis(RH).

The following is the famous Chapter 1 of ” The Way and its Virtue ” made by Chinese Philosopher Lao-Tzu ( B.C. 600 , China ). Chapter 1 was handwritten by me on a sunny day of 2005 in my room at Seoul. My own translation of the Chapter 1 is introduced here, since I feel that RH does have similarities with this Chapter 1.

Attached is my photo taken in 1974 at the age of 20 on the summit of In-Soo Cliff near Seoul after 7 hour of rock climbing. This photo was posted on Vol. 58 (June 1974) of Korean Mountaineering Monthly Magazine ” San (Mountain) ”.

Once the way was told, the way is not the Way : Once named, the name is not the Name.

State of no name, it’s the beginning of Universe : Once named, it’s the mother of Everything.

Therefore, with empty humble mind you shall see the Wonders, With greedy wanton mind you will see the Boundaries.

Both of these, however, are the same.

Emerged from different names.

We call this the Deep.

Deeper and deeper, that’s the way to the door of Wonders.

References

1. Harold M. Edwards : Riemann’s Zeta Function, pp.12,13,137,etc. Dover Publications, 2001.

2. Lars V. Ahlfors : Complex Analysis, pp.198-200., McGraw-Hill, 1979.

3. Nakhle H. Asmar : Applied Complex Analysis with Partial Differential Equa- tions, pp.371, Prentice Hall, 2002.

4. William R. Wade : An Introduction to Analysis, pp.431, Prentice Hall, 2000.

5. Daniel Zwillinger : CRC Standard Mathemathical Tables and Formulae, pp.540, Chapman and Hall/CRC, 2003.

6. J. Brian Conrey : The Riemann Hypothesis, pp.341-353, Notices of AMS , March 2003.

7. David R. Wilkins : Translations of Riemann’s 1859 Paper On the Number of Prime Numbers less than a Given Quantity, Dec. 1998.

8. Enrico Bombieri : Problems of the Millennium: The Riemann Hypothesis-CMI’s Official Problem Description, CMI, 2000.

9. Peter Sarnak : Problems of the Millennium: The Riemann Hypothesis (2004), CMI, 2004.

10. Barry Mazur : Are there still unsolved problems about the numbers 1,2,3,4,...?, CMI Lecture at MIT, May 3, 2005.

11. William Timothy Gowers : Is there such a thing as infinity?, CMI Lecture at Har- vard, March 22, 2004.

12. Lynn H. Loomis, Shlomo Sternberg : Advanced Calculus, pp. 400, Addison Wesley, 1968.

13. George B. Arfken, Hans J. Weber : Mathematical Methods for Physicists, pp.614, Academic Press, 1995.

14. D.W. Jordan, P. Smith : Mathematical Techniques, pp.529, Oxford University Press, 2002.

68-896, Jangwee-Dong, Seongbuk-Ku, Seoul, Korea. 136-829 E-mail address: jeogkim @ kornet.net