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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).