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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 “Supper” 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” . In this paper, we use Nevanlinna’s Second Main Theorem in the value distribution theory, refute the Riemann Hypothesis. In reference [7], we have already given a proof of refute the Riemann Hypothesis. In this paper, we gave out the second proof, please read the reference.

In the 19th century, the famous mathematician E. Picard obtained the pathbreaking result: Any non-constant entire function f ( z ) must take every finite complex value infinitely many times, with at most one exception. Later, E. Borel, by introducing the concept of the order of an entire function, gave the above result a more precise formulation.

This result, generally known as the Picard-Borel theorem, lays the foundation for the theory of value distribution and since then it has been the source of many research papers on this subject. R. Nevanlinna made the decisive contribution to the development of the theory of value distribution. The Picard-Borel Theorem is a direct consequence of Nevanlinna theory.

In this paper, we use Nevanlinna’s Second Main Theorem in the value distribution theory; we got an important the conclusion by Riemann hypothesis. This conclusion contradicts the References [

We give some notations, definitions and theorems in the theory of value distribution, its contents are in the references [

We write

log + x = { log x 1 ≤ x 0 0 ≤ x < 1

It is easy to see that log x ≤ log + x .

Let f ( z ) be a non-constant meromorphic function in the circle | z | < R , 0 < R < + ∞ . we denote by n ( r , f ) the number of poles of f ( z ) on | z | ≤ r ( 0 < r < R ) , each pole being counted with its proper multiplicity. Denote by n ( 0 , f ) the multiplicity of the pole of f ( z ) at the origin. For arbitrary

complex number a ≠ ∞ , we denote by n ( r , 1 f − a ) the number of zeros of f ( z ) − a on | z | ≤ r ( 0 < r < R ) , each zero being counted with its proper multiplicity. Denote by n ( 0 , 1 f − a ) the multiplicity of the zero of f ( z ) − a at the origin.

We write

m ( r , f ) = 1 2 π ∫ 0 2 π log + | f ( r e i φ ) | d φ

N ( r , f ) = ∫ 0 r n ( t , f ) − n ( 0 , f ) t d t + n ( 0 , f ) log r .

When a ≠ ∞ ,

N ( r , 1 f − a ) = ∫ 0 r n ( t , 1 f − a ) − n ( 0 , 1 f − a ) t d t + n ( 0 , 1 f − a ) log r

and T ( r , f ) = m ( r , f ) + N ( r , f ) , T ( r , f ) is called the characteristic function of f ( z ) .

Lemma 2.1. If f ( z ) is a analytical function in the circle | z | < R ( 0 < R < ∞ ) . we have

T ( r , f ) ≤ log + M ( r , f ) ≤ ρ + r ρ − r T ( ρ , f ) ( 0 < r < ρ < R )

where M ( r , f ) = max | z | = r | f ( z ) |

Lemma 2.1 follows from the References [

Lemma 2.2. Let f ( z ) be a non-constant meromorphic function in the circle | z | < R ( 0 < R < ∞ ) . a λ ( λ = 1 , 2 , ⋯ , h ) and b μ ( μ = 1 , 2 , ⋯ , k ) are the zeros and poles of f ( z ) in the circle | z | < ρ ( 0 < ρ < R ) respectively, each zero or pole appears as its multiplicity indicates, and z = 0 is neither zero nor pole of the function f ( z ) , then, in the circle | z | < ρ , we have the following formula

log | f ( 0 ) | = 1 2 π ∫ 0 2 π log | f ( ρ e i φ ) | d φ − ∑ λ = 1 h log ρ | a λ | + ∑ μ = 1 k log ρ | b μ | .

This formula is called Jensen formula.

Lemma 2.2 follows from the References [

Lemma 2.3. Let f ( z ) be the meromorphic function in the circle | z | ≤ R , and f ( 0 ) ≠ 0 , ∞ , 1 , f ′ ( 0 ) ≠ 0. when 0 < r < R , we have

T ( r , f ) < 2 { N ( R , 1 f ) + N ( R , 1 f − 1 ) + N ( R , f ) } + 4 log + | f ( 0 ) | + 2 log + 1 R | f ′ ( 0 ) | + 36 log R R − r + 5220

This is a form of Nevanlinna’s Second Main Theorem.

Lemma 2.3 follows from the References [

Lemma 2.4. Let f ( x ) be decreasing and non-negative for x ≥ a . Then the limit

lim N → ∞ ( ∑ n = a N f ( n ) − ∫ a N f ( x ) d x ) = η

exists, and that 0 ≤ η ≤ f ( a ) . Moreover, if f ( x ) → 0 as x → ∞ , then for ξ ≥ a + 1 , we have

| ∑ a ≤ n ≤ ξ f ( n ) − ∫ a ξ f ( x ) d x − η | ≤ f ( ξ − 1 )

The lemma 2.4 follows from the References [

Lemma 2.5. When σ ≥ 1 2 , | t | ≥ 2 , we have

| ζ ( σ + i t ) | ≤ c 1 | t | 1 2

Where ζ ( s ) is Riemann zeta function.

Lemma 2.5 follows from the References [

Lemma 2.6. Let f ( z ) be the analytic function in the circle | z | ≤ R , let M ( r ) and A ( r ) denote the maxima of | f ( z ) | and Re f ( z ) on | z | = r respectively. Then for 0 < r < R , we have

M ( r ) ≤ 2 r R − r A ( R ) + R + r R − r | f ( 0 ) |

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

Lemma 2.6 follows from the References [

Let s = σ + i t is the complex number, when σ > 1 , Riemann zeta function is

ζ ( s ) = ∑ n = 1 ∞ 1 n s

When σ > 1 , we have

log ζ ( s ) = ∑ n = 2 ∞ Λ ( n ) n s log n

where Λ ( n ) is Mangoldt function.

Lemma 3.1. If t is any real number, we have

1) 0.0426 ≤ | log ζ ( 4 + i t ) | ≤ 0.0824

2) | ζ ( 4 + i t ) − 1 | ≥ 0.0426

3) 0.917 ≤ | ζ ( 4 + i t ) | ≤ 1.0824

4) | ζ ′ ( 4 + i t ) | ≥ 0.012

Proof.

1) | log ζ ( 4 + i t ) | ≤ ∑ n = 2 ∞ Λ ( n ) n 4 log n ≤ ∑ n = 2 ∞ 1 n 4 = π 4 90 − 1 ≤ 0.0824 | log ζ ( 4 + i t ) | ≥ 1 2 4 − ∑ n = 3 ∞ 1 n 4 = 1 + 2 2 4 − ∑ n = 1 ∞ 1 n 4 = 9 8 − π 4 90 ≥ 0.0426.

2) | ζ ( 4 + i t ) − 1 | = | ∑ n = 2 ∞ 1 n 4 + i t | ≥ 1 2 4 − ∑ n = 3 ∞ 1 n 4 = 1 + 2 2 4 − ∑ n = 1 ∞ 1 n 4 = 9 8 − π 4 90 ≥ 0.0426.

3) | ζ ( 4 + i t ) | = | ∑ n = 1 ∞ 1 n 4 + i t | ≤ ∑ n = 1 ∞ 1 n 4 = π 4 90 ≤ 1.0824 | ζ ( 4 + i t ) | = | ∑ n = 1 ∞ 1 n 4 + i t | ≥ 1 − ∑ n = 2 ∞ 1 n 4 = 2 − ∑ n = 1 ∞ 1 n 4 = 2 − π 4 90 ≥ 0.917.

4) | ζ ′ ( 4 + i t ) | = | ∑ n = 2 ∞ log n n 4 + i t | ≥ log 2 2 4 − ∑ n = 3 ∞ log n n 4 .

by Lemma 2.4, we have

∑ n = 3 ∞ log n n 4 = ∫ 3 ∞ log x x 4 d x + η

where 0 ≤ η ≤ log 3 3 4

∫ 3 ∞ log x x 4 d x = − 1 3 ∫ 3 ∞ log x d x − 3 = log 3 3 4 + 1 3 ∫ 3 ∞ x − 4 d x = log 3 3 4 − 1 3 2 ∫ 3 ∞ d x − 3 = log 3 3 4 + 1 3 5 .

Therefore

∑ n = 3 ∞ log n n 4 ≤ log 3 3 4 + 1 3 5 + log 3 3 4 | ζ ′ ( 4 + i t ) | ≥ log 2 2 4 − 2 log 3 3 4 − 1 3 5 ≥ 0.012.

This completes the proof of Lemma 3.1.

Now, we assume that Riemann hypothesis is correct, and abbreviation as RH. In other words, when σ > 1 2 , the function ζ ( σ + i t ) has no zeros. The function log ζ ( σ + i t ) is a multi-valued analytic function in the region σ > 1 2 , t ≥ 1. we choose the principal branch of the function log ζ ( σ + i t ) , therefore, if ζ ( σ + i t ) = 1 , then log ζ ( σ + i t ) = 0 .

Let δ = 1 100 , c 1 , c 2 , ⋯ , is the positive constant.

Lemma 3.2. If RH is correct, when δ = 1 100 , σ ≥ 1 2 + 2 δ , t ≥ 16 , we have

| log ζ ( σ + i t ) | ≤ c 2 log t + c 3

Proof. In Lemma 2.6, we choose f ( z ) = log ζ ( z + 4 + i t ) , R = 7 2 − δ , r = 7 2 − 2 δ , t ≥ 16.

Because log ζ ( z + 4 + i t ) is the analytic function in the circle | z | ≤ R , by Lemma 2.6, in the circle | z | ≤ r , we have

| log ζ ( z + 4 + i t ) | ≤ 7 δ ( A ( R ) + | log ζ ( 4 + i t ) | )

by Lemma 2.5, we have

A ( R ) = max | z − z 0 | = R log | ζ ( z + 4 + i t ) | ≤ 1 2 log t + log c 1

by Lemma 3.1, we have

| log ζ ( z + 4 + i t ) | ≤ c 2 log t + c 3

therefore, when σ ≥ 1 2 + 2 δ , we have

| log ζ ( σ + i t ) | ≤ c 2 log t + c 3

This completes the proof of Lemma 3.2.

Lemma 3.3. If RH is correct, when δ = 1 100 , t ≥ 16 , ρ = 7 2 − 2 δ , in the circle | z | ≤ ρ , we have

N ( ρ , 1 ζ ( z + 4 + i t ) − 1 ) ≤ log log t + c 4

Proof. In the Lemma 2.2, we choose

f ( z ) = log ζ ( z + 4 + i t ) , R = 7 2 − δ , ρ = 7 2 − 2 δ . a λ ( λ = 1 , 2 , ⋯ , h ) are the zeros

of the function log ζ ( z + 4 + i t ) in the circle | z | < ρ , each zero appears as its multiplicity indicates. Because the function log ζ ( z + 4 + i t ) has no poles in the circle | z | < ρ , and log ζ ( 4 + i t ) is not equal to zero, we have

log | log ζ ( 4 + i t ) | = 1 2 π ∫ 0 2 π log | l o g ζ ( 4 + i t + ρ e i φ ) | d φ − ∑ λ = 1 h log ρ | a λ |

by Lemma 3.1 and Lemma 3.2, we have

∑ λ = 1 h log ρ | a λ | ≤ log log t + c 4 .

Because z = 0 is neither zero nor pole of the function log ζ ( z + 4 + i t ) , we have

∑ λ = 1 h log ρ | a λ | = ∫ 0 ρ ( l o g ρ t ) d n ( t , 1 f ) = [ ( l o g ρ t ) n ( t , 1 f ) ] 0 ρ + ∫ 0 ρ n ( t , 1 f ) t d t = ∫ 0 ρ n ( t , 1 f ) t d t = N ( ρ , 1 f ) = N ( ρ , 1 log ζ ( z + 4 + i t ) ) ≥ N ( ρ , 1 ζ ( z + 4 + i t ) − 1 ) .

This completes the proof of Lemma 3.3.

Theorem. If RH is correct, when σ ≥ 1 2 + 4 δ , δ = 1 100 , t ≥ 16 , we have

| ζ ( σ + i t ) | ≤ c 8 ( log t ) c 6

Proof. In Lemma 2.3, we choose f ( z ) = ζ ( z + 4 + i t ) , t ≥ 16 , R = 7 2 − 2 δ , r = 7 2 − 3 δ . by Lemma3.1, we have f ( 0 ) = ζ ( 4 + i t ) ≠ 0 , ∞ , 1 , and | f ′ ( 0 ) | = | ζ ′ ( 4 + i t ) | ≥ 0.012 , | f ( 0 ) | = | ζ ( 4 + i t ) | ≤ 1.0824. Because ζ ( z + 4 + i t ) is the analytic function, and it have neither zeros nor poles in the circle | z | ≤ R , we have

N ( R , 1 f ) = 0 , N ( R , f ) = 0

therefore, by Lemma 3.3, we have

T ( r , ζ ( z + 4 + i t ) ) ≤ 2 log log t + c 5

In Lemma 2.1, we choose R = 7 2 − 2 δ , ρ = 7 2 − 3 δ , r = 7 2 − 4 δ . by the maximal principle, in the circle | z | ≤ r , we have

log + | ζ ( z + 4 + i t ) | ≤ c 6 log log t + c 7

Therefore, when σ ≥ 1 2 + 4 δ , we have

log + | ζ ( σ + i t ) | ≤ c 6 log log t + c 7

log | ζ ( σ + i t ) | ≤ c 6 log log t + c 7

| ζ ( σ + i t ) | ≤ c 8 ( log t ) c 6

This completes the proof of Theorem.

The conclusion of Theorem contradicts the References [

Fei, C.L. (2017) Riemann Hypothesis and Value Distribution Theory. Journal of Applied Mathematics and Physics, 5, 734-740. https://doi.org/10.4236/jamp.2017.53062