A Proof of Riemann Hypothesis

Tao LiuSchool of Science, Southwest University of Science and Technology, Mianyang, Sichuan 621010, China

State Key Laboratory of Environment-friendly Energy Materials,Southwest University of Science and Technology, 59 Qinglong Road, Mianyang, Sichuan 621010, China

Juhao WuStanford University, Stanford, California 94309, USA

(Dated: May 5, 2020)

The meromorphic function W (s) introduced in the Riemann-Zeta function ζ (s) =W (s)ζ (1−s) maps the lineof s = 1/2+ it onto the unit circle in W -space. |W (s)|= 0 gives the trivial zeroes of the Riemann-Zeta functionζ (s). In the range: 0 < |W (s)| 6= 1, ζ (s) does not have nontrivial zeroes. |W (s)|= 1 is the necessary conditionfor the nontrivial zeros of the Riemann-Zeta function. Writing s = σ + it, in the range: 0≤ σ ≤ 1, but σ 6= 1/2,even if |W (s)|= 1, the Riemann-Zeta function ζ (s) is non-zero. Based on these arguments, the nontrivial zerosof the Riemann-Zeta function ζ (s) can only be on the s= 1/2+ it critical line. Therefore a proof of the RiemannHypothesis is presented.

I. RIEMANN HYPOTHESIS

Let us briefly revisit the Riemann Hypothesis. Recall that the Riemann-Zeta function can be defined via Riemann’s functionalequation:

ζ (s) = 2sπ

s−1 sin(

πs2

)Γ(1− s)ζ (1− s), (1)

where Γ(1− s) is the analytical continuation of the factorial and s = σ + it, with σ ∈ R and t ∈ R both being real number. Noticethat s =−2n (n = 1,2, · · · ,∞) are the trivial zeroes of ζ (s) and s = 1 is its pole. The Riemann Hypothesis states on the necessarycondition of the nontrivial zeroes of the Riemann-Zeta function as: “all the nontrivial zeroes of ζ (s) is on the line: s = 1/2+ it”[1]. Up to now, it is proven that the nontrivial zeroes of ζ (s) can only be in the range: 0≤ σ ≤ 1 [2, 3]

II. THE MEROMORPHIC FUNCTION W (s)

A. Introduction of the meromorphic function W (s)

Based on Riemann’s functional equation (1), we can introduce a meromorphic function:

W (s) = 2sπ

s−1 sin(

πs2

)Γ(1− s), (2)

so that Eq. (1) is rewritten as:

ζ (s) =W (s)ζ (1− s). (3)

The distribution of the nontrivial zeros of the Riemann-Zeta function is closely related to the properties of the meromorphicfunction W (s). In the following, let us discuss the properties of W (s).

B. The properties of the meromorphic function W (s)

1. The reflection symmetry of W (s)

In the s-complex space, for arbitrary ε ∈ R, setting s0 = 1/2+ it, then the pair: s± = s0± ε is a mirror symmetric pair withrespect to s0 as shown in Fig. 1. The complex conjugates of s+ and s− are noted as s∗+ and s∗−. Under the W (s) map, s± andtheir mirror reflected complex conjugate s∗∓ are reciprocal pair.

W (s±)W(s∗∓)= 1. (4)

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FIG. 1. In the s-complex plane, s+ and s− are mirror symmetric with respect to s0; s∗+ and s∗−are mirror symmetric with respect to s0.

ProofBecause:

sin(

πs2

)=

π

Γ(1− s/2)Γ(s/2),

and

Γ(1− s)Γ(1− s/2)

=2−sΓ

( 1−s2

)√

π,

we can rewrite W (s) as:

W (s) = πs− 1

2Γ( 1−s

2

)Γ( s

2

) . (5)

On the complex plane of s, for arbitrary real number ε ∈ R, because:

W (s+) = eε+it Γ( 1

4 −ε

2 −it2

)Γ( 1

4 +ε

2 +it2

) ,and

W (s∗−) = e−ε−it Γ( 1

4 +ε

2 +it2

)Γ( 1

4 −ε

2 −it2

) = 1W (s+)

,

which is to say that in the W -space, W (s∗−) must be equal to 1/W (s+), which is the reciprocal of W (s+); we have:

W (s+)W (s∗−) =W (s+)1

W (s+)= 1.

Similarly, we can prove that in the W -space, W (s∗+) must be equal to 1/W (s−), which is the reciprocal of W (s−), i.e.,W (s∗+)W (s−) = 1. Therefore, property II B 1 is proven.

2. W (s) has trivial zeros at s = {−2n|n = 0,1,2, · · · ,∞} and pole s = {2n+1|n = 0,1,2, · · · ,∞}. W (s) at these trivial zeros and at thesepoles are reciprocal pairs: W (−2n)W (1+2n) = 1.

ProofFirst of all, it is obvious that:

W (−2n) = 2−2nπ−2n−1 sin(−nπ)Γ(1+2n) = 0

for (n = 0,1,2, · · · ), and

W (2n+1) = 22n+1π

2n sin(

nπ +π

2

)Γ(−2n) = ∞

3

for (n = 0,1,2, · · · ).Next, setting ε = 1

2 +2n, and t = 0, we have s+ = 2n+1, and s− =−2n. According to reflection symmetry relation as in Eq.(4), we have:

W (2n+1)W (−2n) = 1. (6)

This then serves as the proof for property II B 2.

3. W (s) is not zero at s = 2n (n = 1,2, · · · ).

ProofLet ε =− 1

2 +2n, and t = 0, then s+ = 2n, and s− = 1−2n. According to the reflection symmetry as in Eq. (4), we have:

W (2n)W (1−2n) = 1,

i.e.,

W (2n) =1

W (1−2n)=

121−2nπ−2n sin

(π

2 −nπ)

Γ(2n)=

(2π)2n

2(−1)n(2n−1)!6= 0.

This is a proof for property II B 3.

4. Monotonicity of the absolute value |W (s)| of W (s).Excluding the zeros and the poles, the absolute value |W (s)| of the map W (s) is a monotonic function of t except when σ = 1/2:

1. in the range 0 < t <+∞, when σ > 1/2, |W (s)| monotonically decreases with the increase of t; when σ < 1/2, |W (s)| monotonicallyincreases with the increase of t.

2. in the range −∞ < t < 0, when σ > 1/2, |W (s)| monotonically increases with the increase of t; when σ < 1/2, |W (s)| monotonicallydecreases with the increase of t.

ProofThe fact that W (s∗) = 2s∗πs∗−1 sin

(πs∗2

)Γ(1− s∗) = W (s), excluding the zeros and the poles of W (s), because |W (s)|2 =

W (s)W (s∗), we only need to take the first-order derivative with respect to t on both sides of the equation:

2|W (s)| ddt|W (s)|= d

dt[W (s)W (s∗)] . (7)

The right hand side of Eq. (7) gives:

ddt

[W (s)W (s∗)] =i|W (s)|2

2

[Ψ

(σ − it

2

)−Ψ

(σ + it

2

)−Ψ

(1−σ − it

2

)+Ψ

(1−σ + it

2

)](8)

where Ψ(z) is the digamma function. Therefore, except for the zeros and the poles of W (s), according to Eq. (7) and Eq. (8),we have:

ddt|W (s)|= i|W (s)|

4

[Ψ

(σ − it

2

)−Ψ

(σ + it

2

)− Ψ

(1−σ − it

2

)+Ψ

(1−σ + it

2

)](9)

where the series expression of Ψ(z) is:

Ψ(σ + it) =−γ +∞

∑n=1

σ + it−1n(n+σ + it−1)

(10)

with γ being the Euler constant. Further noticing that:

Ψ

(σ − it

2

)−Ψ

(σ + it

2

)=

∞

∑n=1

−4it|it +2n+σ −2|2

,

4

and

−Ψ

(1−σ − it

2

)+Ψ

(1−σ + it

2

)=

∞

∑n=1

4it|it +2n−σ −1|2

,

so for a more explicit expression showing its being positive or negative, Eq. (9) can be rewritten as:

ddt|W (s)|=

(12−σ

)t|W (s)|

∞

∑n=1

8(n− 3

4

)|it +2n+σ −2|2|it +2n−σ −1|2

. (11)

With the summation being positively defined, and |W (s)|> 0, therefore:1) For arbitrary non-zero t, when and only when σ = 1/2, we have d

dt |W (s)|= 0. This is to say that σ = 1/2 is the only zeroof d

dt |W (s)|.2) When σ 6= 1/2, the value d

dt |W (s)| being positive or negative is uniquely determined by the coefficient (1/2−σ)t in Eq.(11).In the range: 0 < t < ∞,

when σ < 1/2, ddt |W (s)|> 0, therefore |W (s)| monotonically increases with the increase of t;

when σ > 1/2, ddt |W (s)|< 0, therefore |W (s)| monotonically decreases with the increase of t.

In the range: −∞ < t < 0,when σ < 1/2, d

dt |W (s)|< 0, therefore |W (s)| monotonically decreases with the increase of t;when σ > 1/2, d

dt |W (s)|> 0, therefore |W (s)| monotonically increases with the increase of t.So, property II B 4 is proven.

5. In the range: 0≤ σ ≤ 1, but σ 6= 1/2, the t satisfying |W (s)|= 1 is bounded: 2π < |t|< κ .

Proof1) Because |W (s)| = |W (s∗)|, the t satisfying |W (s)| = 1 is always symmetric with respect to ±t. So, let us only study t ≥ 0

case.2) Because W (s+)W (s∗−) = 1, we have |W (s+)||W (s−)|= 1, i.e., when s± = 1

2 ± ε + it, |W (s+)| and |W (s−)| have reflectionsymmetry.

3) In the range: 0 < σ < 1,when t = 0, |W (σ)| = W (σ), it is easy to prove dW (σ)

dσ> 0 (please refer to Appendix A), so W (σ) monotonically increases

with the increase of σ . Furthermore:

0≤W (σ)≤ 1, (0≤ σ ≤ 1/2);

and

1≤W (σ)≤ ∞, (1/2≤ σ ≤ 1).

When t increases from 0, based on the monotonicity of |W (s)|with respect to t and the reflection symmetry of |W (s+)| |W (s−)|=1, we know:

for a given σ being (0 ≤ σ < 1/2), |W (σ + it)| must monotonically increase to being larger than 1 from being less than 1;likewise, for a given σ being (1/2 < σ ≤ 1), |W (σ + it)| must monotonically decrease to being less than 1 from being largerthan 1. Therefore, there always exists t1 < t2, so that σ± = 1

2 ± ε with 0 < ε ≤ 1/2:

|W (σ−+ it1)|< 1 < |W (σ−+ it2)|, (0≤ σ− < 1/2).

Due to reflection symmetry, |W (σ−+ it1)||W (σ++ it1)|= 1, |W (σ−+ it2)||W (σ++ it2)|= 1, one must have:

|W (σ++ it2)|< 1 < |W (σ++ it1)|, (1/2 < σ+ ≤ 1).

4) t2 = 2.01π

When t2 = 2.01π , |W (0+ it2)|= 1.0025> 1,∣∣W ( 1

2 + it2)∣∣= 1, |W (1+ it2)|= 1

1.0025 < 1. It is easy to prove d|W (σ+it)|dσ

∣∣∣t=t2

< 0

(please refer to Appendix B), therefore |W (σ + it2)| monotonically decreases with the increase of σ . Therefore, we have:

|W (σ + it2)|> 1, (0≤ σ < 1/2);

5

and

|W (σ + it2)|< 1, (1/2 < σ ≤ 1).

This is to say that when t changes from 0 to t2, in the range 0≤σ < 1/2, the value |W (σ + it)|, starting from being smaller than1, always monotonically increases to be larger than 1; while in the range 1/2 < σ ≤ 1, the value |W (σ + it)|, starting from beinglarger than 1, always monotonically decreases to be smaller than 1. Therefore, when σ 6= 1/2, the t satisfying |W (σ + it)| = 1has an up limit, i.e., t won’t be larger than κ ≡ t2 = 2.01π . In the range 0≤ σ ≤ 1, for arbitrary σ 6= 1/2, due to the fact that thet satisfying |W (s)|= 1 is always reflection symmetric with respect to ±t, t has a low limit −κ , and an up limit κ .

5) t1 = 2π

When t1 = 2π , |W (0+ it1)|= 0.9999999991 < 1,∣∣W ( 1

2 + it1)∣∣= 1, |W (1+ it1)|= 1

0.9999999991 > 1. It is easy to prove (pleaserefer to Appendix C) that:

|W (σ + it1)|< 1 (0≤ σ < 1/2),

and

|W (σ + it1)|> 1 (1/2 < σ ≤ 1).

So in the range 0≤ σ ≤ 1, and σ 6= 1/2, the t satisfying |W (σ + it)|= 1 can only be in the range t1 < |t|< t2. That is to say, thet satisfying |W (s)|= 1 is bounded:

2π < |t|< κ. (12)

Just for illustration, with t1 = 2π and t2 = 2.01π , the continuous evolution of |W (σ + it)| with σ is shown in Fig. 2.

FIG. 2. When t = 2π , and 2.01π , the continuous evolution of |W (σ + it)| in the range of 0≤ σ ≤ 1. In the plot: ↑ stands for the monotonicincrease direction of |W (s)| with t; while ↓ stands for the monotonic decrease direction of |W (s)| with t.

So, property II B 5 is proven.

III. THE DISTRIBUTION OF THE NONTRIVIAL ZEROS OF THE RIEMANN-ZETA FUNCTION

Let us take the absolute value of the Riemann Equation in (3) to have:

|ζ (s)|= |W (s)||ζ (1− s)|. (13)

In the following, we will first prove a few lemmas of the Riemann-Zeta function ζ (s) based on the properties of the meromorphicfunction W (s).

Lemma 1: the set of the zeroes of W (s) = 0: s = {−2n|n = 0,1,2, · · · ,∞} contains all the trivial zeroes of Riemann-Zetafunction ζ (s). There is no nontrivial zero in this set.Proof

Notice that the set of all the trivial zeroes of ζ (s) is: s = {−2n|n = 1,2, · · · ,∞}, which contains all the elements of the set ofthe zeroes of W (s) except n = 0. However, when n = 0, we have ζ (0) =−1/2 6= 0. Therefore, Lemma 1 is proven.

Lemma 2: Riemann-Zeta function ζ (s) and ζ (1− s) do not have nontrivial zeroes in the range: 0 < |W (s)| 6= 1.Proof

1) If |W (s)| 6= 1, then it must be true that |ζ (s)| 6= |ζ (1− s)|.

6

Assuming that |ζ (s)| = |ζ (1− s)|, then the necessary condition for the equation |ζ (s)| = |ζ (1− s)| being equivalent to theRiemann equation |ζ (s)| = |W (s)||ζ (1− s)| is |W (s)| = 1 (please refer to Appendix D). Then this is in conflict of |W (s)| 6= 1.Therefore, the assumption could not hold and it has to be true that |ζ (s)| 6= |ζ (1− s)|.

2) When |W (s)|> 0, for the Riemann-Zeta function satisfying |ζ (s)| 6= |ζ (1− s)|, it is impossible to have |ζ (s)|= 0.Assuming that |ζ (s)|= 0, then due to |ζ (s)|= |W (s)||ζ (1− s)|, it must be true that |ζ (1− s)|= 0, so that |ζ (s)|= |ζ (1− s)|.

Then this is in conflict with the statement that |ζ (s)| 6= |ζ (1− s)|. Therefore, the assumption ζ (s) = 0 can not hold.3) When |W (s)|> 0, for the Riemann-Zeta function satisfying |ζ (s)| 6= |ζ (1− s)|, |ζ (1− s)|= 0 can only be valid on the

trivial zeroes.Assuming that |ζ (1− s)| = 0, then due to |ζ (s)| = |W (s)||ζ (1− s)|, one would have |ζ (s)| = 0 with the only exception of

|W (s)|= ∞, i.e., s is a pole of |W (s)|. However, |ζ (s)|= 0 is in conflict with |ζ (s)| 6= |ζ (1− s)|. So, for |ζ (1− s)|= 0 and also|ζ (s)| 6= |ζ (1− s)|, the only possibility is |W (s)|= ∞, i.e., s is the pole of |W (s)|. Now, the poles of |W (s)| are s = 2n+1, where|ζ (1− s)|= |ζ (−2n)|= 0, i.e., these are just the trivial zeroes.

Based on the above 1), 2), and 3), we know that the Riemann-Zeta function does not have nontrivial zeroes for 0 < |W (s)| 6= 1.So Lemma 2 is proven.

Lemma 3: |W (s)|= 1 is the necessary condition for the nontrivial zeroes of the Riemann-Zeta function ζ (s).Proof

Based on Lemma 1: the set of zeroes from W (s) = 0, i.e., s = {−2n|n = 0,1,2, · · · ,∞} does not contain nontrivial zeroes ofthe Riemann-Zeta function ζ (s).

Based on Lemma 2: In the range: 0 < |W (s)| 6= 1, there is no nontrivial zeroes of the Riemann-Zeta function ζ (s).Therefore, the nontrivial zeroes of the Riemann-Zeta function can only be on |W (s)|= 1. Indeed, when |W (s)|= 1, we must

have:

|ζ (s)|= |ζ (1− s)|. (14)

If |ζ (s)|= 0, then |ζ (1− s)|= 0. However, |ζ (s)|= |ζ (1− s)| does not guarantee ζ (s) to be zero. Therefore, |W (s)|= 1 is onlythe necessary condition of the nontrivial zeroes of the Riemann-Zeta function ζ (s).

So Lemma 3 is proven.Corollary 1: for s = 1/2+ it being the nontrivial zeroes of ζ (s), the necessary condition for W (s) is that |W (s)|= 1.Inserting s = 1

2 + it and s∗ = 12 − it into Eq. (5), we have:

W (s)W (s) = πs+s∗−1

Γ( 1−s

2

)Γ

(1−s∗

2

)Γ( s

2

)Γ( s∗

2

)∣∣∣∣∣∣s= 1

2+it,s∗= 12−it

= 1. (15)

So we have: |W (s)|= 1. The geometric illustration is that W (s) maps the straight line s = 1/2+ it in s-space to the unit circle inthe W -space as in Fig. 3 (a) and (b).

FIG. 3. Writing W (s) = u(σ , t)+ iv(σ , t), the meromorphic function W (s) maps the straight line s = 1/2+ it in s-space (a) onto the unit circlein W -space (b). In the Figure, there are 8,000 points in the range: t =−100 · · ·100.

Lemma 4: Riemann-Zeta function does not have nontrivial zeroes in the range 0 ≤ σ < 1/2 and 1/2 < σ ≤ 1; and|t|> κ and |t|< 2π .Proof

According to Lemma 3, |W (s)| = 1 is the necessary condition for the Riemann-Zeta function to have nontrivial zeroes. Onthe other hand, according to the property II B 5 of the meromorphic function: for |W (s)| = 1 but σ 6= 1/2, t is bounded, i.e.,2π < |t| < κ with κ = 2.01π . Therefore, in the range |t| > κ and |t| < 2π , we have |W (s)| 6= 1. Hence, except σ = 1/2,Riemann-Zeta function does not have nontrivial zeros in the range 0≤ σ ≤ 1 and |t|< 2π and |t|> κ . So Lemma 4 is proven.

7

Lemma 5: Riemann-Zeta function does not have nontrivial zeroes in the range 0 ≤ σ < 1/2 and 1/2 < σ ≤ 1; and2π < |t|< κ .Proof

1) In the range 0 ≤ σ < 1/2, in order to prove that the Riemann-Zeta function ζ (s) 6= 0 and ζ (1− s) 6= 0 when |W (s)|monotonically increases from 0 < |W (s)|< 1 to |W (s)|= 1, we only need to prove that when |W (s)|= 1, it is not in the form of00 .

According to Eq. (3) and Eq. (5), we have:

|W (s)|= |ζ (s)||ζ (1− s)|

= πσ− 1

2

∣∣Γ( 1−s2

)∣∣∣∣Γ( s2

)∣∣ . (16)

Notice that when t > 0 (please refer to Appendix E):

ddt

∣∣∣∣Γ(1− s2

)∣∣∣∣=−t∞

∑n=1

∣∣Γ( 1−s2

)∣∣|it +2n−σ −1|2

< 0, (17)

ddt

∣∣∣Γ( s2

)∣∣∣=−t∞

∑n=1

∣∣Γ( s2

)∣∣|it +2n+σ −2|2

< 0. (18)

So, both∣∣Γ( 1−s

2

)∣∣ and∣∣Γ( s

2

)∣∣ are continuous functions which monotonically decrease with increasing t. For arbitrary 0≤σ < 12 ,

πσ−1/2 is bounded and not equal to zero. Therefore, when and only when t → ∞, we have∣∣Γ( 1−s

2

)∣∣→ 0, and∣∣Γ( s

2

)∣∣→ 0.According to Eq. (16), we have:

limt→∞

|ζ (s)||ζ (1− s)|

= limt→∞

πσ− 1

2

∣∣Γ( 1−s2

)∣∣∣∣Γ( s2

)∣∣ → 00. (19)

Because when t > 0, both∣∣Γ( 1−s

2

)∣∣ and∣∣Γ( s

2

)∣∣ are continuous function with no singular point, when σ 6= 12 , the ratio of the

absolute values of the Riemann-Zeta functions ζ (s) and ζ (1− s): |ζ (s)||ζ (1−s)| continuously monotonically increase with increasing

t, when and only when t→ ∞, can it be in the form of 00 . Therefore, logically for a certain arbitrary finite t > 0, |ζ (s)|

|ζ (1−s)| can not

be in the form of 00 (please refer to Appendix F). In particular, when t continuously changes from 2π to 2.01π , the t satisfying

|W (s)|= 1, according to the property II B 5, has an up limit, t < κ . In this case, according to Eq. (16) and Eq. (19), we have:

1 = |W (s)|= |ζ (s)||ζ (1− s)|

= πσ− 1

2

∣∣Γ( 1−s2

)∣∣∣∣Γ( s2

)∣∣ 900. (20)

Therefore in the range: 0≤ σ < 1/2 and 2π < |t|< κ , the Riemann-Zeta function satisfying |W (s)|= 1 does not have nontrivialzero.

2) Due to the fact that |W (s)| is reflection symmetric with respect to σ = 1/2, in the range: 1/2 < σ ≤ 1, the Riemann-Zetafunction satisfying |W (s)|= 1 does not have nontrivial zero, either.

3) In the range 0 ≤ σ ≤ 1, but σ 6= 12 , numerical calculations as shown in Fig. 4 of the absolute value of the Riemann-Zeta

functions satisfying |W (s)|= 1 agree with the theoretical predictions in 1) and 2).So Lemma 5 is proven.Now based on Lemma 1, Lemma 2, Lemma 3, Lemma 4, Lemma 5, and Corollary 1, we finally reach the conclusion that the

nontrivial zeroes of the Riemann-Zeta function can only locate on the s = 1/2+ it critical line. The Riemann Hypothesis is thenproven.

IV. DISCUSSION

In this paper, we introduce a meromorphic function W (s) via the Riemann functional equation: ζ (s) ≡W (s)ζ (1− s). Bystudying the properties of the absolute value of this meromorphic function |W (s)|, we find that the necessary condition forRiemann-Zeta function to have nontrivial zero is |W (s)| = 1. With this necessary condition, we not only prove the RiemannHypothesis in the range 0 ≤ σ ≤ 1, but also we give a direct explanation why Riemann-Zeta function does not have nontrivialzero in the range σ ≤ 0 and σ ≥ 1 on the complex plane based on the zeros and singular points of |W (s)|.

1) The zeros of |W (s)| correspond to the singular points of∣∣Γ( s

2

)∣∣= |Γ(−n)|. While∣∣Γ( 1−s

2

)∣∣= |Γ( 2n+12

)| are not singular.

8

FIG. 4. A numerical example of the absolute values of the Riemann-Zeta functions |ζ (s)| and |ζ (1− s)| in the range 0≤ σ ≤ 1, but σ 6= 12 ;

while satisfying |W (s)| = 1. Notice that we do not intentionally exclude the point of σ = 12 on the curve, since here we just want to give an

illustration.

2) Both the non-singular∣∣Γ( s

2

)∣∣ and the non-singular∣∣Γ( 1−s

2

)∣∣monotonically decrease, when t monotonically increases nearthe zeros of |W (s)|. When and only when t→ ∞, we have

∣∣Γ( 1−s2

)∣∣→ 0, and∣∣Γ( s

2

)∣∣→ 0, i.e.,

limt→∞

|ζ (s)||ζ (1− s)|

= limt→∞

πσ− 1

2

∣∣Γ( 1−s2

)∣∣∣∣Γ( s2

)∣∣ → 00.

3) Near its zeros: (−2n), (0 < |W (s)|< 1), during the monotonic increase of t, |W (s)| will complete all the {s =−2n+ε + it}points satisfying |W (s)|= 1 in the σ < 1

2 complex plane. In this case:

1 = |W (s)|= |ζ (s)||ζ (1− s)|

= π−2n+ε− 1

2

∣∣Γ(n+ 1−ε

2 −it2

)∣∣∣∣Γ(−n+ ε

2 +it2

)∣∣ 9 00,

and t is bounded.This is to say that the |ζ (s)|

|ζ (1−s)| = 1 satisfying |W (s)| = 1 is not in the form of 00 . Therefore, |ζ (s)| and |ζ (1− s)| both do not

have nontrivial zeros in the σ < 12 complex plane. Furthermore, based on the reflection symmetry of W (s) with respect to σ = 1

2 ,we readily know that the Riemann-Zeta function does not have nontrivial zero in the σ > 1

2 complex plane.The above statement agrees with the known conclusion that the nontrivial zeroes of ζ (s) can only be in the range: 0≤ σ ≤ 1

[2, 3]; and the nontrivial zeroes of the Riemann-Zeta function can only locate on the s = 1/2+ it critical line [1].In summary, a proof of the Riemann Hypothesis is presented in this paper.

Appendix A: Proof of d|W (s)|dσ

∣∣∣t=0

> 0

Because:

d|W (s)|dσ

=|W (σ)|

4

[4ln(π)−Ψ

( s2

)−Ψ

(s∗

2

)−Ψ

(1− s

2

)−Ψ

(1− s∗

2

)](A1)

and making use of:

Ψ(s) =−γ +∞

∑n=1

s−1n(n+ s−1)

,

when t = 0, we have:

Ψ

( s2

)+Ψ

(s∗

2

)=−2γ−

∞

∑n=1

2(2−σ)

n(2n+σ −2), (A2)

9

and

Ψ

(1− s

2

)+Ψ

(1− s∗

2

)=−2γ−

∞

∑n=1

2(1+σ)

n(2n−σ −1). (A3)

Now plugging Eq. (A2) and Eq. (A3) into Eq. (A1), we have:

d|W (s)|dσ

∣∣∣∣t=0

=|W (σ)|

4

{4ln(π)+4γ +

∞

∑n=1

[2(2−σ)

n(2n+σ −2)+

2(1+σ)

n(2n−σ −1)

]}. (A4)

Now, it is clear that each term in the infinite summation in Eq. (A4) is positive when 0 < σ < 1, therefore it is true thatd|W (s)|

dσ

∣∣∣t=0

> 0.

Appendix B: Proof of d|W (s)|dσ

∣∣∣t=2.01π

< 0

Setting:

G(σ , t) = 4ln(π)−Ψ

( s2

)−Ψ

(s∗

2

)−Ψ

(1− s

2

)−Ψ

(1− s∗

2

)(B1)

then Eq. (A1) can be rewritten as:

d|W (s)|dσ

=|W (s)|

4G(σ , t). (B2)

In order to prove d|W (s)|dσ

∣∣∣t=2.01π

< 0, one needs only to prove the maximum value of G(σ ,2.01π) satisfies: max[G(σ ,2.01π)]<

0.1) when σ = 1/2 and t = 2.01π , G(1/2,2.01π) is the maximum valueThe first order and second order derivatives of G(σ , t) with respect to σ are:

dG(σ , t)dσ

=12

[−Ψ

(1,

s2

)−Ψ

(1,

s∗

2

)+Ψ

(1,

1− s2

)+Ψ

(1,

1− s∗

2

)], (B3)

and

d2G(σ , t)dσ2 =

14

[−Ψ

(2,

s2

)−Ψ

(2,

s∗

2

)−Ψ

(2,

1− s2

)−Ψ

(2,

1− s∗

2

)], (B4)

where Ψ(1,z) = dΨ(z)dz , Ψ(2,z) = d2Ψ(z)

dz2 . Because when σ = 12 , 1− s = s∗:

Ψ

(1,

1− s2

)= Ψ

(1,

s∗

2

),

and

Ψ

(1,

1− s∗

2

)= Ψ

(1,

s2

).

Hence, we have:

dG(σ , t)dσ

∣∣∣∣σ= 1

2

= 0,

while

d2G(σ , t)dσ2

∣∣∣∣σ= 1

2 ,t=2.01π

=−1.009542407 < 0.

10

Therefore, G( 1

2 ,2.01π)

is the maximum value, and

G(

12,2.01π

)=−0.015751728 < 0. (B5)

2) when t = 2.01π , G( 1

2 ,2.01π)

is the maximum value in the range 0≤ σ ≤ 1Setting

F(σ , t) =[

Ψ

( s2

)+Ψ

(s∗

2

)+Ψ

(1− s

2

)+Ψ

(1− s∗

2

)], (B6)

then

G(σ , t) = 4ln(π)−F(σ , t). (B7)

Because in the range 0≤ σ ≤ 1, and t 6= 0, F(σ , t) is an analytical function, there exists any order of derivative of F(σ , t) withrespect to σ :

dn

dσn F(σ , t) =12n

[Ψ

(n,

s2

)+(−1)n

Ψ

(n,

1− s∗

2

)+Ψ

(n,

s∗

2

)+(−1)n

Ψ

(n,

1− s2

)]. (B8)

When σ = 12 , we have: s = 1− s∗ = 1

2 + it, and s∗ = 1− s = 12 − it, plugging into Eq. (B8), we have:

dn

dσn F(σ , t)∣∣∣∣σ=1/2

=

{0, (n = 2k+1)

122k−1

[Ψ(2k, 1

4 +it2

)+Ψ

(2k, 1

4 −it2

)], (n = 2k) . (B9)

Let us Taylor expand F(σ , t) at σ = 12 , we have:

F(σ , t) =∞

∑n=0

1n!

dn

dσn F(σ , t)

∣∣∣∣∣σ= 1

2

(σ − 1

2

)n

=∞

∑k=0

Ψ(2k, 1

4 +it2

)+Ψ

(2k, 1

4 −it2

)(2k)!22k−1

(σ − 1

2

)2k

. (B10)

Therefore,

ddσ

F(σ , t) =∞

∑k=1

2k[Ψ(2k, 1

4 +it2

)+Ψ

(2k, 1

4 −it2

)](2k)!22k−1

(σ − 1

2

)2k−1

=∞

∑k=1

Ψ(2k, 1

4 +it2

)+Ψ

(2k, 1

4 −it2

)(2k−1)!22k−1

(σ − 1

2

)2k−1

.

(B11)Setting σ − 1

2 = δ , according to Eq. (B7) and Eq. (B10), we get the Taylor expansion of G(σ , t):

G(σ , t) = 4ln(π)−∞

∑k=0

Ψ(2k, 1

4 +it2

)+Ψ

(2k, 1

4 −it2

)22k−1(2k)!

δ2k. (B12)

Notice that:

ddσ

G(σ , t) =− ddσ

F(σ , t),

and plugging into Eq. (B11), we readily get the Taylor expansion of ddσ

G(σ , t):

ddσ

G(σ , t) =−∞

∑k=1

Ψ(2k, 1

4 +it2

)+Ψ

(2k, 1

4 −it2

)22k−1(2k−1)!

δ2k−1. (B13)

In particular, setting t = 2.01π , we have:

dG(σ ,2.01π)

dσ= c1δ + c3δ

3 + c5δ5 +O(δ 7), (B14)

where c2k−1 =−Ψ(2k, 1

4+2.01πt

2 )+Ψ(2k, 14−

2.01πt2 )

(2k−1)!22k−1 ; in particular,

c1 =−1.0095424×10−1,

11

c3 =+2.5705715×10−3,

and

c5 =−6.6264219×10−5.

For a sufficiently large N, |c5| > ∑Nk=4 |c2k−1| → 1.824122120× 10−6, i.e., neglecting the terms with order higher than O(δ 7)

does not change the conclusion on dG(σ ,2.01π)dσ

being positive or negative.When δ < 0: c1δ > 0, c3δ 3 < 0, c5δ 5 > 0, and c1δ +c3δ 3 > 0, so at the left side of the zero

(− 1

2 ≤ δ < 0), d

dσG(σ , t)

∣∣t=2.01π

>

0, so G(σ ,2.01π) monotonically increases with σ .When δ > 0: c1δ < 0, c3δ 3 > 0, c5δ 5 < 0, and c1δ +c3δ 3 < 0, so at the right side of the zero

(0 < δ ≤ 1

2

), d

dσG(σ , t)

∣∣t=2.01π

<

0, so G(σ ,2.01π) monotonically decreases with σ .So, G(1/2,2.01π) must be the maximum in the range 0≤ σ ≤ 1, i.e., max[G(σ ,2.01π)] = G(1/2,2.01π).

Appendix C: Proof of |W (σ−+2πi)|< 1; |W (σ++2πi)|> 1 with(σ± = 1

2 ± ε)

ProofBased on the reflection symmetry, we have |W (σ−+2πi)||W (σ++2πi)|= 1. To prove |W (σ++2πi)|> 1, we only need to

prove |W (σ−+2πi)|< 1.1) G(σ ,2π) monotonically increases in the range 0≤ σ < 1

2 , and monotonically decreases in the range 12 < σ ≤ 1.

Setting σ = 12 +δ , with − 1

2 ≤ δ ≤ 12 , when t = 2π , according to Eq. (B12), we have:

G(σ ,2π) = 4ln(π)−∞

∑n=0

Ψ(2n, 1

4 +πi)+Ψ

(2n, 1

4 −πi)

22n−1(2n)!δ

2n. (C1)

Neglecting terms with order higher than O(δ 6), we have:

G(σ ,2π)≈ 4ln(π)+g0 +g2δ2 +g4δ

4, (C2)

with g0 =−4.5746788, g2 =−5.0986449×10−2, g4 = 6.5574618×10−4.Using Eq. (C1), we have:

dG(σ ,2π)

dσ=−

∞

∑n=1

Ψ(2n, 1

4 +πi)+Ψ

(2n, 1

4 −πi)

22n−1(2n−1)!δ

2n−1. (C3)

i.e.,

dG(σ ,2π)

dσ= g′1δ +g′3δ

3 +g′5δ5 +O(δ 7), (C4)

where g′2k−1 =−Ψ(2n, 1

4+πi)+Ψ(2n, 14−πi)

22n−1(2n−1)! , in particular:

g′1 =−1.0197290×10−1,

g′3 =+2.6229847×10−3,

and

g′5 =−6.8327078×10−5.

For a sufficiently large N, |g′5| > ∑Nk=4 |g′2k−1| → 1.904573728× 10−6, i.e., neglecting the terms with order higher than O(δ 7)

does not change the conclusion of dG(σ ,2π)dσ

being positive or negative.According to Eq. (C4), we have when δ < 0: g′1δ > 0, g′3δ 3 < 0, g′5δ 5 > 0, and g′1δ +g′3δ 3 > 0, so

dG(σ ,2π)

dσ≈ g′1δ +g′3δ

3 +g′5δ5 > 0

(−1

2≤ δ < 0

).

12

When δ > 0: g′1δ < 0, g′3δ 3 > 0, g′5δ 5 < 0, and g′1δ +g′3δ 3 < 0, so

dG(σ ,2π)

dσ≈ g′1δ +g′3δ

3 +g′5δ5 < 0

(0 < δ ≤ 1

2

).

So G(σ ,2π) monotonically increases in the range 0≤ σ < 12 , and G(σ ,2π) monotonically decreases in the range 1

2 < σ ≤ 1.2) There are two symmetric zero points: σ0± = 1

2 +δ± of G(σ ,2π) in the range 0≤ σ < 1According to Eq. (C2), setting G(σ ,2π) = 0, i.e.,

4 ln(π)+g0 +g2δ2 +g4δ

4 = 0,

we have δ± =±0.2885526325.3) in the range 0 ≤ σ < 1/2, |W (σ +2πi)| monotonically decreases at the left side of σ0− = 1

2 + δ− and monotonicallyincreases at the right side.

According to the above item 1), G(σ ,2π) monotonically increases; according to the above item 2), G( 1

2 +δ−,2π)= 0,

noticing that G(0,2π) = −0.008465084 < 0, G( 1

2 ,2π)= 0.004240720 > 0, we find that during the continuous change of σ

from 0 to 12 , G(σ ,2π) monotonically evolves from being less than zero to being larger than zero. Therefore, we have:

d|W (σ +2πi)|dσ

∣∣∣∣σ< 1

2+δ−

=|W (σ +2πi)|

4G(σ ,2π)

∣∣∣∣σ< 1

2+δ−

< 0,

and

d|W (σ +2πi)|dσ

∣∣∣∣σ= 1

2+δ−

=|W (σ +2πi)|

4G(σ ,2π)

∣∣∣∣σ= 1

2+δ−

= 0,

and

d|W (σ +2πi)|dσ

∣∣∣∣σ> 1

2+δ−

=|W (σ +2πi)|

4G(σ ,2π)

∣∣∣∣σ> 1

2+δ−

> 0.

That is to say, |W (σ +2πi)| monotonically decreases at the left side of σ0− = 12 + δ− and monotonically increases at the right

side.4) in the range 0≤ σ < 1/2, |W (σ +2πi)|< 1.According to the above item 3), |W (σ + 2πi)| monotonically decreases at the left side of σ0− = 1

2 + δ− and monotonicallyincreases at the right side; therefore, the values of |W (σ +2πi)| at the two ends: σ = 0 and σ = 1

2 must be larger than its value atpoints between the two ends. Because |W (0+2πi)|= 0.9999999991 < 1, and |W (1/2+2πi)|= 1, we have |W (σ−+2πi)|< 1for(0≤ σ < 1

2

).

Appendix D: When |ζ (s)|= |ζ (1− s)|, it must be true that |W (s)|= 1

ProofAccording to the Riemann-Zeta equation ζ (s) = W (s)ζ (1− s), we have: W (s) = ζ (s)

ζ (1−s) . Because W (s) = W (s∗), we haveζ (s)

ζ (1−s)= ζ (s∗)

ζ (1−s∗) . Therefore:

|W (s)|2 =W (s)W (s) =ζ (s)ζ (s)

ζ (1− s)ζ (1− s)=

ζ (s)ζ (s∗)ζ (1− s)ζ (1− s∗)

. (D1)

Here, s = σ + it, and s∗ = σ − it.Assuming

|ζ (s)|= |ζ (1− s)|, (D2)

we have:

|ζ (s)|2 = |ζ (1− s)|2, (D3)

13

where |ζ (s)|2 = ζ (s)ζ (s∗) and |ζ (1− s)|2 = ζ (1− s)ζ (1− s∗) are both real function of σ and t. For the convenience ofdiscussion, let us denote:

P(σ , t)≡ ζ (s)ζ (s∗),

and

Q(σ , t)≡ ζ (1− s)ζ (1− s∗).

1) If ζ (s) 6= 0, then it must be true that ζ (1− s) 6= 0. According to Eq. (D1) and Eq. (D2), it must be true that:

|W (s)|= |ζ (s)||ζ (1− s)|

= 1. (D4)

2) If ζ (s) = 0, then it must be true that ζ (1− s) = 0. In this case, we have |W (s)| = 00 . In this following, we prove the limit

of this 00 is 1.

2.1) In the range 0 ≤ σ ≤ 1, and t 6= 0, due to the fact that ζ (s), ζ (s∗), ζ (1− s), and ζ (1− s∗) are all analytical functions.Furthermore, there exists their arbitrary order of partial derivative with respect to σ and t: ∂ mζ (s)

∂σm , ∂ mζ (s)∂ tm , ∂ nζ (s∗)

∂σn , ∂ nζ (s∗)∂ tn ;

∂ mζ (1−s)∂σm , ∂ mζ (1−s)

∂ tm , ∂ nζ (1−s∗)∂σn , ∂ nζ (1−s∗)

∂ tn . Therefore the real functions P(σ , t) and Q(σ , t) are partial differentiable to arbitraryorder with respect to σ and t.

2.2) If ζ (s) has zeros: sn = σn + itn with (n = 1,2,3, · · · ), so that ζ (sn) = 0. Then the complex conjugate ζ (sn) = ζ (s∗n) = 0.Therefore, for these σn and tn, they are the zeros of the real function P(σn, t). Then there exists the Taylor expansion of P(σn, t)around the zeros, tn:

P(σn, t) =∞

∑j=0

1j!

d jP(σn, t)dt j

∣∣∣∣∣t=tn

(t− tn) j (D5)

where the derivatives in the expansion coefficients:{

d jP(σn,t)dt j

∣∣∣t=tn

, j = 0,1,2, · · · ,∞}

won’t be all equal to zero according to

the series expansion theorem. Notice that, the zeroth order expansion coefficient of P(σn, t) is just P(σn, t) ≡ ζ (sn)ζ (s∗n) = 0.The first order expansion coefficient is then:

dP(σn, t)dt

∣∣∣∣t=tn

=

(dζ (σn + it)

dtζ (σn− it)+ζ (σn + it)

dζ (σn− it)dt

)∣∣∣∣t=tn

= 0.

There must exit a certain k ≥ 2, so that:

limt→tn

(dkP(σn, t)

dtk

)= lim

t→tn

dk

dtk

∞

∑j=0

1j!

d jP(σn, t)dt j

∣∣∣∣∣t=tn

(t− tn) j

=dkP(σn, t)

dtk

∣∣∣∣t=tn6= 0. (D6)

2.3) With the assumption in Eq. (D2), following Eq. (D3), it must be true that: |ζ (1− sn)|2 = 0, i.e., tn is the zeros of the realfunction Q(σn, t). Similarly, there exists the Taylor expansion around the zeros tn:

Q(σn, t) =∞

∑j=0

1j!

d jQ(σn, t)dt j

∣∣∣∣∣t=tn

(t− tn) j. (D7)

where the derivatives in the expansion coefficients:{

d jQ(σn,t)dt j

∣∣∣t=tn

, j = 0,1,2, · · · ,∞}

won’t be all equal to zero. Because both

the zeroth order expansion coefficient and the first order expansion coefficient are equal to zero; there must exit a certain orderk ≥ 2, so that:

limt→tn

(dkQ(σn, t)

dtk

)= lim

t→tn

dk

dtk

∞

∑j=0

1j!

d jQ(σn, t)dt j

∣∣∣∣∣t=tn

(t− tn) j

=dkQ(σn, t)

dtk

∣∣∣∣t=tn6= 0. (D8)

According to Eq. (D3), Eq. (D6), and Eq. (D8), there must exist a certain k ≥ 2, so that:

dkP(σn, t)dtk

∣∣∣∣t=tn

=dkQ(σn, t)

dtk

∣∣∣∣t=tn6= 0. (D9)

14

For |W (σn + itn)|2 = P(σn,t)Q(σn,t)

= 00 , because:

limt→tn|W (σn + it)|2 = lim

t→tn

P(σn, t)Q(σn, t)

= limt→tn

dk

dtk P(σn, t)dk

dtk Q(σn, t)=

dkP(σn,t)dtk

∣∣∣t=tn

dkQ(σn,t)dtk

∣∣∣t=tn

. (D10)

According to Eq. (D9) and Eq. (D10), we have:

limt→tn|W (σn + it)|2 = 1. (D11)

Based on the Taylor expansion of the real functions P(σn, t) and Q(σn, t) around their zeros, we can prove by the same procedurethat there exists a certain k ≥ 2, so that:

dkP(σ , tn)dσ k

∣∣∣∣σ=σn

=dkQ(σ , tn)

dσ k

∣∣∣∣σ=σn

6= 0. (D12)

Therefore, we have:

limσ→σn

|W (σ + itn)|2 = limσ→σn

P(σ , tn)Q(σ , tn)

= limσ→σn

dk

dσk P(σ , tn)dk

dσ k Q(σ , tn)=

dkP(σ ,tn)dσ k

∣∣∣σ=σn

dkQ(σ ,tn)dσ k

∣∣∣σ=σn

= 1. (D13)

According to Eq. (D11) and Eq. (D13), we have:

limt→tn|W (σn + it)|= 1, (D14)

and

limσ→σn

|W (σ + itn)|= 1. (D15)

The above serves as a proof.

Appendix E: Proof of ddt

∣∣Γ( 1−s2)∣∣= ∑

∞n=1

−t|Γ( 1−s2 )|

|it+2n−σ−1|2 , ddt

∣∣Γ( s2)∣∣= ∑

∞n=1

−t|Γ( s2 )|

|it+2n+σ−2|2

Because: ∣∣∣Γ( s2

)∣∣∣2 = Γ

( s2

)Γ

(s∗

2

). (E1)

∣∣∣∣Γ(1− s2

)∣∣∣∣2 = Γ

(1− s

2

)Γ

(1− s∗

2

). (E2)

Furthermore, in the range 0≤ σ ≤ 1 and t 6= 0, both Γ( s

2

)and Γ

( 1−s2

)are analytical functions.

We take derivative of Eq. (E1) and Eq. (E2) with respect to t:

ddt

∣∣∣Γ( s2

)∣∣∣2 = i2

∣∣∣Γ( s2

)∣∣∣2 [Ψ

( s2

)−Ψ

(s∗

2

)], (E3)

ddt

∣∣∣∣Γ(1− s2

)∣∣∣∣2 = i2

∣∣∣∣Γ(1− s2

)∣∣∣∣2 [Ψ

(1− s∗

2

)−Ψ

(1− s

2

)]. (E4)

According to Eq. (10),

Ψ

(σ − it

2

)−Ψ

(σ + it

2

)=

∞

∑n=1

−4it|it +2n+σ −2|2

, (E5)

15

−Ψ

(1−σ − it

2

)+Ψ

(1−σ + it

2

)=

∞

∑n=1

4it|it +2n−σ −1|2

, (E6)

Plugging Eq. (E5) into Eq. (E3), and plugging Eq. (E6) into Eq. (E4), we have:

ddt

∣∣∣Γ( s2

)∣∣∣2 = i2

∣∣∣Γ( s2

)∣∣∣2 ∞

∑n=1

4it|it +2n+σ −2|2

. (E7)

ddt

∣∣∣∣Γ(1− s2

)∣∣∣∣2 = i2

∣∣∣∣Γ(1− s2

)∣∣∣∣2 ∞

∑n=1

4it|it +2n−σ −1|2

. (E8)

Notice that: ddt

∣∣Γ( s2

)∣∣2 = 2∣∣Γ( s

2

)∣∣ ddt

∣∣Γ( s2

)∣∣, ddt

∣∣Γ( 1−s2

)∣∣2 = 2∣∣Γ( 1−s

2

)∣∣ ddt

∣∣Γ( 1−s2

)∣∣, therefore, Eq. (E7) and Eq. (E8) canfinally simplified as:

ddt

∣∣∣Γ( s2

)∣∣∣=−t∞

∑n=1

∣∣Γ( s2

)∣∣|it +2n+σ −2|2

,

ddt

∣∣∣∣Γ(1− s2

)∣∣∣∣=−t∞

∑n=1

∣∣Γ( 1−s2

)∣∣|it +2n−σ −1|2

.

This serves as a proof.

Appendix F: For an arbitrary finite t > 0, the |ζ (s)||ζ (1−s)| satisfying d

dt

(|ζ (s)||ζ (1−s)|

)= d

dt |W (s)| 6= 0, can not be in the form of 00 .

ProofAccording to Eq. (11), when |W (s)|> 0, and σ < 1

2 , we always have:

ddt

(|ζ (s)||ζ (1− s)|

)=|ζ (1− s)| d|ζ (s)|dt −|ζ (s)|

d|ζ (1−s)|dt

|ζ (1− s)|2=

ddt|W (s)|> 0 (F1)

therefore:

|ζ (1− s)|d|ζ (s)|dt

−|ζ (s)|d|ζ (1− s)|dt

> 0. (F2)

Notice that:

d|ζ (s)|2

dt= 2|ζ (s)|d|ζ (s)|

dt=

dζ (s)dt

ζ (s∗)+dζ (s∗)

dtζ (s),

we have:

d|ζ (s)|dt

=1

2|ζ (s)|

[dζ (s)

dtζ (s∗)+

dζ (s∗)dt

ζ (s)].

Similarly, we have:

d|ζ (1− s)|dt

=1

2|ζ (1− s)|

[dζ (1− s)

dtζ (1− s∗)+

dζ (1− s∗)dt

ζ (1− s)].

Therefore, Eq. (F2) can be rewritten as:

12

{dζ (s)

dt ζ (s∗)+ dζ (s∗)dt ζ (s)

|W (s)|− |W (s)|

[dζ (1− s)

dtζ (1− s∗)+

dζ (1− s∗)dt

ζ (1− s)]}

> 0. (F3)

Notice that, for an arbitrary finite t > 0, |W (s)| > 0 and have no singular points, ζ (s), ζ (s∗), ζ (1− s), ζ (1− s∗) are allanalytical functions. Furthermore, their derivatives dζ (s)

dt , dζ (s∗)dt , dζ (1−s)

dt , dζ (1−s∗)dt do not have singular points. If there exists a

16

certain sn, so that |ζ (sn) |= 0, |ζ (1− sn) |= 0, then it must be true that ζ (sn) = ζ (s∗n) = 0, ζ (1− sn) = ζ (1− s∗n) = 0. Pluggingthese into Eq. (F3), we have:

12

0 · dζ (s)

dt

∣∣∣s=sn

+0 · dζ (s∗)dt

∣∣∣s∗=s∗n

|W (sn)|− |W (sn)|

[0 · dζ (1− s)

dt

∣∣∣∣s=sn

+0 · dζ (1− s∗)dt

∣∣∣∣s∗=s∗n

]= 0 > 0.

This is to say that |ζ (sn)|= 0, |ζ (1− sn)|= 0 must lead to the invalidation of inequality relation in Eq. (F3). Therefore, forσ < 1

2 , and an arbitrary finite t > 0, the |ζ (s)||ζ (1−s)| satisfying d

dt

(|ζ (s)||ζ (1−s)|

)> 0, can not be in the form of 0

0 .

Similarly, for σ > 12 , and an arbitrary finite t > 0, the |ζ (s)|

|ζ (1−s)| satisfying ddt

(|ζ (s)||ζ (1−s)|

)< 0, can not be in the form of 0

0 .This serves as a proof.

REFERENCES

[1] Bernhard Rieman, “Ueber die Anzahl der Primzahlen unter einer gegebenen Grosse”, Monatsberichte der Berliner Akademie. In Gesam-melte Werke, Teubner, Leipzig (1859).

[2] Jacques Hadamard, “Sur la distribution des zeros de la fonction ζ (s)et ses consequences arithmetiques”, Bulletin de la Societe Mathema-tique de France 14, 199 (1896).

[3] Ch.J. de la Vallee-Poussin, “Recherches analytiques sur la theorie des nombers premiers”, Ann. Soc. Sci. Bruxelles 20, 183 (1896).