AA Proof of Riemann Hypothesis
Tao Liu
School of Science, Southwest University of Science and Technology, Mianyang, Sichuan 621010, ChinaState Key Laboratory of Environment-friendly Energy Materials,Southwest University of Science and Technology, 59 Qinglong Road, Mianyang, Sichuan 621010, China
Juhao Wu
Stanford 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 ( − s ) maps the lineof s = / + it onto the unit circle in W -space. | W ( s ) | = ζ ( s ) . In the range: 0 < | W ( s ) | (cid:54) = ζ ( s ) does not have nontrivial zeroes. | W ( s ) | = s = σ + it , in the range: 0 ≤ σ ≤
1, but σ (cid:54) = / | 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 = / + 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 ) = s π s − sin (cid:16) π s (cid:17) Γ ( − s ) ζ ( − s ) , (1)where Γ ( − s ) is the analytical continuation of the factorial and s = σ + it , with σ ∈ R and t ∈ R both being real number. Noticethat s = − n ( n = , , · · · , ∞ ) are the trivial zeroes of ζ ( s ) and s = ζ ( s ) is on the line: s = / + it ”[1]. Up to now, it is proven that the nontrivial zeroes of ζ ( s ) can only be in the range: 0 ≤ σ ≤ 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 ) = s π s − sin (cid:16) π s (cid:17) Γ ( − s ) , (2)so that Eq. (1) is rewritten as: ζ ( s ) = W ( s ) ζ ( − 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 s = / + it , then the pair: s ± = s ± ε is a mirror symmetric pair withrespect to s 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 (cid:0) s ∗∓ (cid:1) = . (4) a r X i v : . [ m a t h . G M ] M a y FIG. 1. In the s -complex plane, s + and s − are mirror symmetric with respect to s ; s ∗ + and s ∗− are mirror symmetric with respect to s . Proof
Because: sin (cid:16) π s (cid:17) = π Γ ( − s / ) Γ ( s / ) , and Γ ( − s ) Γ ( − s / ) = − s Γ (cid:0) − s (cid:1) √ π , we can rewrite W ( s ) as: W ( s ) = π s − Γ (cid:0) − s (cid:1) Γ (cid:0) s (cid:1) . (5)On the complex plane of s , for arbitrary real number ε ∈ R , because: W ( s + ) = e ε + it Γ (cid:0) − ε − it (cid:1) Γ (cid:0) + ε + it (cid:1) , and W ( s ∗− ) = e − ε − it Γ (cid:0) + ε + it (cid:1) Γ (cid:0) − ε − it (cid:1) = W ( 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 + ) W ( s + ) = . 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 = {− n | n = , , , ··· , ∞ } and pole s = { n + | n = , , , ··· , ∞ } . W ( s ) at these trivial zeros and at thesepoles are reciprocal pairs: W ( − n ) W ( + n ) = . Proof
First of all, it is obvious that: W ( − n ) = − n π − n − sin ( − n π ) Γ ( + n ) = n = , , , · · · ), and W ( n + ) = n + π n sin (cid:16) n π + π (cid:17) Γ ( − n ) = ∞ for ( n = , , , · · · ).Next, setting ε = + n , and t =
0, we have s + = n +
1, and s − = − n . According to reflection symmetry relation as in Eq.(4), we have: W ( n + ) W ( − n ) = . (6)This then serves as the proof for property II B 2.
3. W ( s ) is not zero at s = n (n = , , ··· ). Proof
Let ε = − + n , and t =
0, then s + = n , and s − = − n . According to the reflection symmetry as in Eq. (4), we have: W ( n ) W ( − n ) = , i.e. , W ( n ) = W ( − n ) = − n π − n sin (cid:0) π − n π (cid:1) Γ ( n ) = ( π ) n ( − ) n ( n − ) ! (cid:54) = . 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. in the range < t < + ∞ , when σ > / , | W ( s ) | monotonically decreases with the increase of t; when σ < / , | W ( s ) | monotonicallyincreases with the increase of t.2. in the range − ∞ < t < , when σ > / , | W ( s ) | monotonically increases with the increase of t; when σ < / , | W ( s ) | monotonicallydecreases with the increase of t. Proof
The fact that W ( s ∗ ) = s ∗ π s ∗ − sin (cid:16) π s ∗ (cid:17) Γ ( − s ∗ ) = W ( s ) , excluding the zeros and the poles of W ( s ) , because | W ( s ) | = 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 ) | = ddt [ W ( s ) W ( s ∗ )] . (7)The right hand side of Eq. (7) gives: ddt [ W ( s ) W ( s ∗ )] = i | W ( s ) | (cid:20) Ψ (cid:18) σ − it (cid:19) − Ψ (cid:18) σ + it (cid:19) − Ψ (cid:18) − σ − it (cid:19) + Ψ (cid:18) − σ + it (cid:19)(cid:21) (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 ) | (cid:20) Ψ (cid:18) σ − it (cid:19) − Ψ (cid:18) σ + it (cid:19) − Ψ (cid:18) − σ − it (cid:19) + Ψ (cid:18) − σ + it (cid:19)(cid:21) (9)where the series expression of Ψ ( z ) is: Ψ ( σ + it ) = − γ + ∞ ∑ n = σ + it − n ( n + σ + it − ) (10)with γ being the Euler constant. Further noticing that: Ψ (cid:18) σ − it (cid:19) − Ψ (cid:18) σ + it (cid:19) = ∞ ∑ n = − it | it + n + σ − | , and − Ψ (cid:18) − σ − it (cid:19) + Ψ (cid:18) − σ + it (cid:19) = ∞ ∑ n = it | it + n − σ − | , so for a more explicit expression showing its being positive or negative, Eq. (9) can be rewritten as: ddt | W ( s ) | = (cid:18) − σ (cid:19) t | W ( s ) | ∞ ∑ n = (cid:0) n − (cid:1) | it + n + σ − | | it + n − σ − | . (11)With the summation being positively defined, and | W ( s ) | >
0, therefore:1) For arbitrary non-zero t , when and only when σ = /
2, we have ddt | W ( s ) | =
0. This is to say that σ = / ddt | W ( s ) | .2) When σ (cid:54) = /
2, the value ddt | W ( s ) | being positive or negative is uniquely determined by the coefficient ( / − σ ) t in Eq.(11).In the range: 0 < t < ∞ ,when σ < / ddt | W ( s ) | >
0, therefore | W ( s ) | monotonically increases with the increase of t ;when σ > / ddt | W ( s ) | <
0, therefore | W ( s ) | monotonically decreases with the increase of t .In the range: − ∞ < t < σ < / ddt | W ( s ) | <
0, therefore | W ( s ) | monotonically decreases with the increase of t ;when σ > / ddt | W ( s ) | >
0, therefore | W ( s ) | monotonically increases with the increase of t .So, property II B 4 is proven.
5. In the range: ≤ σ ≤ , but σ (cid:54) = / , the t satisfying | W ( s ) | = is bounded: π < | t | < κ . Proof
1) Because | W ( s ) | = | W ( s ∗ ) | , the t satisfying | W ( s ) | = ± t . So, let us only study t ≥ W ( s + ) W ( s ∗− ) =
1, we have | W ( s + ) || W ( s − ) | = i.e. , when s ± = ± ε + it , | W ( s + ) | and | W ( s − ) | have reflectionsymmetry.3) In the range: 0 < σ < t = | W ( σ ) | = W ( σ ) , it is easy to prove dW ( σ ) d σ > W ( σ ) monotonically increaseswith the increase of σ . Furthermore: 0 ≤ W ( σ ) ≤ , ( ≤ σ ≤ / ) ;and 1 ≤ W ( σ ) ≤ ∞ , ( / ≤ σ ≤ ) . 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 < / | W ( σ + it ) | must monotonically increase to being larger than 1 from being less than 1;likewise, for a given σ being (1 / < σ ≤ | W ( σ + it ) | must monotonically decrease to being less than 1 from being largerthan 1. Therefore, there always exists t < t , so that σ ± = ± ε with 0 < ε ≤ / | W ( σ − + it ) | < < | W ( σ − + it ) | , ( ≤ σ − < / ) . Due to reflection symmetry, | W ( σ − + it ) || W ( σ + + it ) | = | W ( σ − + it ) || W ( σ + + it ) | =
1, one must have: | W ( σ + + it ) | < < | W ( σ + + it ) | , ( / < σ + ≤ ) . t = . π When t = . π , | W ( + it ) | = . > (cid:12)(cid:12) W (cid:0) + it (cid:1)(cid:12)(cid:12) = | W ( + it ) | = . <
1. It is easy to prove d | W ( σ + it ) | d σ (cid:12)(cid:12)(cid:12) t = t < | W ( σ + it ) | monotonically decreases with the increase of σ . Therefore, we have: | W ( σ + it ) | > , ( ≤ σ < / ) ;and | W ( σ + it ) | < , ( / < σ ≤ ) . This is to say that when t changes from 0 to t , in the range 0 ≤ σ < /
2, the value | W ( σ + it ) | , starting from being smaller than1, always monotonically increases to be larger than 1; while in the range 1 / < σ ≤
1, the value | W ( σ + it ) | , starting from beinglarger than 1, always monotonically decreases to be smaller than 1. Therefore, when σ (cid:54) = /
2, the t satisfying | W ( σ + it ) | = t won’t be larger than κ ≡ t = . π . In the range 0 ≤ σ ≤
1, for arbitrary σ (cid:54) = /
2, due to the fact that the t satisfying | W ( s ) | = ± t , t has a low limit − κ , and an up limit κ .5) t = π When t = π , | W ( + it ) | = . < (cid:12)(cid:12) W (cid:0) + it (cid:1)(cid:12)(cid:12) = | W ( + it ) | = . >
1. It is easy to prove (pleaserefer to Appendix C) that: | W ( σ + it ) | < ( ≤ σ < / ) , and | W ( σ + it ) | > ( / < σ ≤ ) . So in the range 0 ≤ σ ≤
1, and σ (cid:54) = /
2, the t satisfying | W ( σ + it ) | = t < | t | < t . That is to say, the t satisfying | W ( s ) | = π < | t | < κ . (12)Just for illustration, with t = π and t = . π , the continuous evolution of | W ( σ + it ) | with σ is shown in Fig. 2. FIG. 2. When t = π , and 2 . π , 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 ( − 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 ) = : s = {− n | n = , , , · · · , ∞ } 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 = {− n | n = , , · · · , ∞ } , which contains all the elements of the set ofthe zeroes of W ( s ) except n =
0. However, when n =
0, we have ζ ( ) = − / (cid:54) =
0. Therefore, Lemma 1 is proven.
Lemma 2: Riemann-Zeta function ζ ( s ) and ζ ( − s ) do not have nontrivial zeroes in the range: < | W ( s ) | (cid:54) = . Proof
1) If | W ( s ) | (cid:54) = , then it must be true that | ζ ( s ) | (cid:54) = | ζ ( − s ) | . Assuming that | ζ ( s ) | = | ζ ( − s ) | , then the necessary condition for the equation | ζ ( s ) | = | ζ ( − s ) | being equivalent to theRiemann equation | ζ ( s ) | = | W ( s ) || ζ ( − s ) | is | W ( s ) | = | W ( s ) | (cid:54) = | ζ ( s ) | (cid:54) = | ζ ( − s ) | .
2) When | W ( s ) | > , for the Riemann-Zeta function satisfying | ζ ( s ) | (cid:54) = | ζ ( − s ) | , it is impossible to have | ζ ( s ) | = . Assuming that | ζ ( s ) | =
0, then due to | ζ ( s ) | = | W ( s ) || ζ ( − s ) | , it must be true that | ζ ( − s ) | =
0, so that | ζ ( s ) | = | ζ ( − s ) | .Then this is in conflict with the statement that | ζ ( s ) | (cid:54) = | ζ ( − s ) | . Therefore, the assumption ζ ( s ) =
3) When | W ( s ) | > , for the Riemann-Zeta function satisfying | ζ ( s ) | (cid:54) = | ζ ( − s ) | , | ζ ( − s ) | = can only be valid on thetrivial zeroes. Assuming that | ζ ( − s ) | =
0, then due to | ζ ( s ) | = | W ( s ) || ζ ( − s ) | , one would have | ζ ( s ) | = | W ( s ) | = ∞ , i.e. , s is a pole of | W ( s ) | . However, | ζ ( s ) | = | ζ ( s ) | (cid:54) = | ζ ( − s ) | . So, for | ζ ( − s ) | = | ζ ( s ) | (cid:54) = | ζ ( − s ) | , the only possibility is | W ( s ) | = ∞ , i.e. , s is the pole of | W ( s ) | . Now, the poles of | W ( s ) | are s = n +
1, where | ζ ( − s ) | = | ζ ( − n ) | = 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 ) | (cid:54) = Lemma 3: | W ( s ) | = 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 ) = i.e. , s = {− n | n = , , , · · · , ∞ } does not contain nontrivial zeroes ofthe Riemann-Zeta function ζ ( s ) .Based on Lemma 2: In the range: 0 < | W ( s ) | (cid:54) =
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 musthave: | ζ ( s ) | = | ζ ( − s ) | . (14)If | ζ ( s ) | =
0, then | ζ ( − s ) | =
0. However, | ζ ( s ) | = | ζ ( − s ) | does not guarantee ζ ( s ) to be zero. Therefore, | W ( s ) | = ζ ( s ) .So Lemma 3 is proven. Corollary 1: for s = / + it being the nontrivial zeroes of ζ ( s ) , the necessary condition for W ( s ) is that | W ( s ) | = . Inserting s = + it and s ∗ = − it into Eq. (5), we have: W ( s ) W ( s ) = π s + s ∗ − Γ (cid:0) − s (cid:1) Γ (cid:16) − s ∗ (cid:17) Γ (cid:0) s (cid:1) Γ (cid:0) s ∗ (cid:1) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) s = + it , s ∗ = − it = . (15)So we have: | W ( s ) | =
1. The geometric illustration is that W ( s ) maps the straight line s = / + 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 = / + it in s -space (a) onto the unit circlein W -space (b). In the Figure, there are 8 ,
000 points in the range: t = − ··· Lemma 4: Riemann-Zeta function does not have nontrivial zeroes in the range ≤ σ < / and / < σ ≤ ; and | t | > κ and | t | < π . Proof
According to Lemma 3, | W ( s ) | = | W ( s ) | = σ (cid:54) = / t is bounded, i.e. ,2 π < | t | < κ with κ = . π . Therefore, in the range | t | > κ and | t | < π , we have | W ( s ) | (cid:54) =
1. Hence, except σ = / ≤ σ ≤ | t | < π and | t | > κ . So Lemma 4 is proven. Lemma 5: Riemann-Zeta function does not have nontrivial zeroes in the range ≤ σ < / and / < σ ≤ ; and π < | t | < κ . Proof
1) In the range 0 ≤ σ < /
2, in order to prove that the Riemann-Zeta function ζ ( s ) (cid:54) = ζ ( − s ) (cid:54) = | W ( s ) | monotonically increases from 0 < | W ( s ) | < | W ( s ) | =
1, we only need to prove that when | W ( s ) | =
1, it is not in the form of . According to Eq. (3) and Eq. (5), we have: | W ( s ) | = | ζ ( s ) || ζ ( − s ) | = π σ − (cid:12)(cid:12) Γ (cid:0) − s (cid:1)(cid:12)(cid:12)(cid:12)(cid:12) Γ (cid:0) s (cid:1)(cid:12)(cid:12) . (16)Notice that when t > ddt (cid:12)(cid:12)(cid:12)(cid:12) Γ (cid:18) − s (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) = − t ∞ ∑ n = (cid:12)(cid:12) Γ (cid:0) − s (cid:1)(cid:12)(cid:12) | it + n − σ − | < , (17) ddt (cid:12)(cid:12)(cid:12) Γ (cid:16) s (cid:17)(cid:12)(cid:12)(cid:12) = − t ∞ ∑ n = (cid:12)(cid:12) Γ (cid:0) s (cid:1)(cid:12)(cid:12) | it + n + σ − | < . (18)So, both (cid:12)(cid:12) Γ (cid:0) − s (cid:1)(cid:12)(cid:12) and (cid:12)(cid:12) Γ (cid:0) s (cid:1)(cid:12)(cid:12) are continuous functions which monotonically decrease with increasing t . For arbitrary 0 ≤ σ < , π σ − / is bounded and not equal to zero. Therefore, when and only when t → ∞ , we have (cid:12)(cid:12) Γ (cid:0) − s (cid:1)(cid:12)(cid:12) →
0, and (cid:12)(cid:12) Γ (cid:0) s (cid:1)(cid:12)(cid:12) → t → ∞ | ζ ( s ) || ζ ( − s ) | = lim t → ∞ π σ − (cid:12)(cid:12) Γ (cid:0) − s (cid:1)(cid:12)(cid:12)(cid:12)(cid:12) Γ (cid:0) s (cid:1)(cid:12)(cid:12) → . (19)Because when t >
0, both (cid:12)(cid:12) Γ (cid:0) − s (cid:1)(cid:12)(cid:12) and (cid:12)(cid:12) Γ (cid:0) s (cid:1)(cid:12)(cid:12) are continuous function with no singular point, when σ (cid:54) = , the ratio of theabsolute values of the Riemann-Zeta functions ζ ( s ) and ζ ( − s ) : | ζ ( s ) || ζ ( − s ) | continuously monotonically increase with increasing t , when and only when t → ∞ , can it be in the form of . Therefore, logically for a certain arbitrary finite t > | ζ ( s ) || ζ ( − s ) | can notbe in the form of (please refer to Appendix F). In particular, when t continuously changes from 2 π to 2 . π , 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 ( − s ) | = π σ − (cid:12)(cid:12) Γ (cid:0) − s (cid:1)(cid:12)(cid:12)(cid:12)(cid:12) Γ (cid:0) s (cid:1)(cid:12)(cid:12) (cid:57) . (20)Therefore in the range: 0 ≤ σ < / π < | t | < κ , the Riemann-Zeta function satisfying | W ( s ) | = | W ( s ) | is reflection symmetric with respect to σ = /
2, in the range: 1 / < σ ≤
1, the Riemann-Zetafunction satisfying | W ( s ) | = ≤ σ ≤
1, but σ (cid:54) = , numerical calculations as shown in Fig. 4 of the absolute value of the Riemann-Zetafunctions satisfying | W ( s ) | = s = / + 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 ( − 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 ≤ σ ≥ | W ( s ) | .1) The zeros of | W ( s ) | correspond to the singular points of (cid:12)(cid:12) Γ (cid:0) s (cid:1)(cid:12)(cid:12) = | Γ ( − n ) | . While (cid:12)(cid:12) Γ (cid:0) − s (cid:1)(cid:12)(cid:12) = | Γ (cid:0) n + (cid:1) | are not singular. FIG. 4. A numerical example of the absolute values of the Riemann-Zeta functions | ζ ( s ) | and | ζ ( − s ) | in the range 0 ≤ σ ≤
1, but σ (cid:54) = ;while satisfying | W ( s ) | =
1. Notice that we do not intentionally exclude the point of σ = on the curve, since here we just want to give anillustration.
2) Both the non-singular (cid:12)(cid:12) Γ (cid:0) s (cid:1)(cid:12)(cid:12) and the non-singular (cid:12)(cid:12) Γ (cid:0) − s (cid:1)(cid:12)(cid:12) monotonically decrease, when t monotonically increases nearthe zeros of | W ( s ) | . When and only when t → ∞ , we have (cid:12)(cid:12) Γ (cid:0) − s (cid:1)(cid:12)(cid:12) →
0, and (cid:12)(cid:12) Γ (cid:0) s (cid:1)(cid:12)(cid:12) → i.e. ,lim t → ∞ | ζ ( s ) || ζ ( − s ) | = lim t → ∞ π σ − (cid:12)(cid:12) Γ (cid:0) − s (cid:1)(cid:12)(cid:12)(cid:12)(cid:12) Γ (cid:0) s (cid:1)(cid:12)(cid:12) → .
3) Near its zeros: ( − n ), (0 < | W ( s ) | < t , | W ( s ) | will complete all the { s = − n + ε + it } points satisfying | W ( s ) | = σ < complex plane. In this case:1 = | W ( s ) | = | ζ ( s ) || ζ ( − s ) | = π − n + ε − (cid:12)(cid:12) Γ (cid:0) n + − ε − it (cid:1)(cid:12)(cid:12)(cid:12)(cid:12) Γ (cid:0) − n + ε + it (cid:1)(cid:12)(cid:12) (cid:57) , and t is bounded.This is to say that the | ζ ( s ) || ζ ( − s ) | = | W ( s ) | = . Therefore, | ζ ( s ) | and | ζ ( − s ) | both do nothave nontrivial zeros in the σ < complex plane. Furthermore, based on the reflection symmetry of W ( s ) with respect to σ = ,we readily know that the Riemann-Zeta function does not have nontrivial zero in the σ > 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 ≤ s = / + it critical line [1].In summary, a proof of the Riemann Hypothesis is presented in this paper. Appendix A: Proof of d | W ( s ) | d σ (cid:12)(cid:12)(cid:12) t = > Because: d | W ( s ) | d σ = | W ( σ ) | (cid:20) ( π ) − Ψ (cid:16) s (cid:17) − Ψ (cid:18) s ∗ (cid:19) − Ψ (cid:18) − s (cid:19) − Ψ (cid:18) − s ∗ (cid:19)(cid:21) (A1)and making use of: Ψ ( s ) = − γ + ∞ ∑ n = s − n ( n + s − ) , when t =
0, we have: Ψ (cid:16) s (cid:17) + Ψ (cid:18) s ∗ (cid:19) = − γ − ∞ ∑ n = ( − σ ) n ( n + σ − ) , (A2)and Ψ (cid:18) − s (cid:19) + Ψ (cid:18) − s ∗ (cid:19) = − γ − ∞ ∑ n = ( + σ ) n ( n − σ − ) . (A3)Now plugging Eq. (A2) and Eq. (A3) into Eq. (A1), we have: d | W ( s ) | d σ (cid:12)(cid:12)(cid:12)(cid:12) t = = | W ( σ ) | (cid:40) ( π ) + γ + ∞ ∑ n = (cid:20) ( − σ ) n ( n + σ − ) + ( + σ ) n ( n − σ − ) (cid:21)(cid:41) . (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 that d | W ( s ) | d σ (cid:12)(cid:12)(cid:12) t = > Appendix B: Proof of d | W ( s ) | d σ (cid:12)(cid:12)(cid:12) t = . π < Setting: G ( σ , t ) = ( π ) − Ψ (cid:16) s (cid:17) − Ψ (cid:18) s ∗ (cid:19) − Ψ (cid:18) − s (cid:19) − Ψ (cid:18) − s ∗ (cid:19) (B1)then Eq. (A1) can be rewritten as: d | W ( s ) | d σ = | W ( s ) | G ( σ , t ) . (B2)In order to prove d | W ( s ) | d σ (cid:12)(cid:12)(cid:12) t = . π <
0, one needs only to prove the maximum value of G ( σ , . π ) satisfies: max [ G ( σ , . π )] <
1) when σ = / and t = . π , G ( / , . π ) is the maximum value The first order and second order derivatives of G ( σ , t ) with respect to σ are: dG ( σ , t ) d σ = (cid:20) − Ψ (cid:16) , s (cid:17) − Ψ (cid:18) , s ∗ (cid:19) + Ψ (cid:18) , − s (cid:19) + Ψ (cid:18) , − s ∗ (cid:19)(cid:21) , (B3)and d G ( σ , t ) d σ = (cid:20) − Ψ (cid:16) , s (cid:17) − Ψ (cid:18) , s ∗ (cid:19) − Ψ (cid:18) , − s (cid:19) − Ψ (cid:18) , − s ∗ (cid:19)(cid:21) , (B4)where Ψ ( , z ) = d Ψ ( z ) dz , Ψ ( , z ) = d Ψ ( z ) dz . Because when σ = , 1 − s = s ∗ : Ψ (cid:18) , − s (cid:19) = Ψ (cid:18) , s ∗ (cid:19) , and Ψ (cid:18) , − s ∗ (cid:19) = Ψ (cid:16) , s (cid:17) . Hence, we have: dG ( σ , t ) d σ (cid:12)(cid:12)(cid:12)(cid:12) σ = = , while d G ( σ , t ) d σ (cid:12)(cid:12)(cid:12)(cid:12) σ = , t = . π = − . < . G (cid:0) , . π (cid:1) is the maximum value, and G (cid:18) , . π (cid:19) = − . < . (B5)
2) when t = . π , G (cid:0) , . π (cid:1) is the maximum value in the range ≤ σ ≤ F ( σ , t ) = (cid:20) Ψ (cid:16) s (cid:17) + Ψ (cid:18) s ∗ (cid:19) + Ψ (cid:18) − s (cid:19) + Ψ (cid:18) − s ∗ (cid:19)(cid:21) , (B6)then G ( σ , t ) = ( π ) − F ( σ , t ) . (B7)Because in the range 0 ≤ σ ≤
1, and t (cid:54) = F ( σ , t ) is an analytical function, there exists any order of derivative of F ( σ , t ) withrespect to σ : d n d σ n F ( σ , t ) = n (cid:20) Ψ (cid:16) n , s (cid:17) + ( − ) n Ψ (cid:18) n , − s ∗ (cid:19) + Ψ (cid:18) n , s ∗ (cid:19) + ( − ) n Ψ (cid:18) n , − s (cid:19)(cid:21) . (B8)When σ = , we have: s = − s ∗ = + it , and s ∗ = − s = − it , plugging into Eq. (B8), we have: d n d σ n F ( σ , t ) (cid:12)(cid:12)(cid:12)(cid:12) σ = / = (cid:26) , ( n = k + ) k − (cid:2) Ψ (cid:0) k , + it (cid:1) + Ψ (cid:0) k , − it (cid:1)(cid:3) , ( n = k ) . (B9)Let us Taylor expand F ( σ , t ) at σ = , we have: F ( σ , t ) = ∞ ∑ n = n ! d n d σ n F ( σ , t ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) σ = (cid:18) σ − (cid:19) n = ∞ ∑ k = Ψ (cid:0) k , + it (cid:1) + Ψ (cid:0) k , − it (cid:1) ( k ) !2 k − (cid:18) σ − (cid:19) k . (B10)Therefore, dd σ F ( σ , t ) = ∞ ∑ k = k (cid:2) Ψ (cid:0) k , + it (cid:1) + Ψ (cid:0) k , − it (cid:1)(cid:3) ( k ) !2 k − (cid:18) σ − (cid:19) k − = ∞ ∑ k = Ψ (cid:0) k , + it (cid:1) + Ψ (cid:0) k , − it (cid:1) ( k − ) !2 k − (cid:18) σ − (cid:19) k − . (B11)Setting σ − = δ , according to Eq. (B7) and Eq. (B10), we get the Taylor expansion of G ( σ , t ) : G ( σ , t ) = ( π ) − ∞ ∑ k = Ψ (cid:0) k , + it (cid:1) + Ψ (cid:0) k , − it (cid:1) k − ( k ) ! δ k . (B12)Notice that: dd σ G ( σ , t ) = − dd σ F ( σ , t ) , and plugging into Eq. (B11), we readily get the Taylor expansion of dd σ G ( σ , t ) : dd σ G ( σ , t ) = − ∞ ∑ k = Ψ (cid:0) k , + it (cid:1) + Ψ (cid:0) k , − it (cid:1) k − ( k − ) ! δ k − . (B13)In particular, setting t = . π , we have: dG ( σ , . π ) d σ = c δ + c δ + c δ + O ( δ ) , (B14)where c k − = − Ψ ( k , + . π t ) + Ψ ( k , − . π t ) ( k − ) !2 k − ; in particular, c = − . × − , c = + . × − , and c = − . × − . For a sufficiently large N , | c | > ∑ Nk = | c k − | → . × − , i.e., neglecting the terms with order higher than O ( δ ) does not change the conclusion on dG ( σ , . π ) d σ being positive or negative.When δ < c δ > c δ < c δ >
0, and c δ + c δ >
0, so at the left side of the zero (cid:0) − ≤ δ < (cid:1) , dd σ G ( σ , t ) (cid:12)(cid:12) t = . π >
0, so G ( σ , . π ) monotonically increases with σ .When δ > c δ < c δ > c δ <
0, and c δ + c δ <
0, so at the right side of the zero (cid:0) < δ ≤ (cid:1) , dd σ G ( σ , t ) (cid:12)(cid:12) t = . π <
0, so G ( σ , . π ) monotonically decreases with σ .So, G ( / , . π ) must be the maximum in the range 0 ≤ σ ≤ i.e. , max [ G ( σ , . π )] = G ( / , . π ) . Appendix C: Proof of | W ( σ − + π i ) | < ; | W ( σ + + π i ) | > with (cid:0) σ ± = ± ε (cid:1) Proof
Based on the reflection symmetry, we have | W ( σ − + π i ) || W ( σ + + π i ) | =
1. To prove | W ( σ + + π i ) | >
1, we only need toprove | W ( σ − + π i ) | < G ( σ , π ) monotonically increases in the range ≤ σ < , and monotonically decreases in the range < σ ≤ . Setting σ = + δ , with − ≤ δ ≤ , when t = π , according to Eq. (B12), we have: G ( σ , π ) = ( π ) − ∞ ∑ n = Ψ (cid:0) n , + π i (cid:1) + Ψ (cid:0) n , − π i (cid:1) n − ( n ) ! δ n . (C1)Neglecting terms with order higher than O ( δ ) , we have: G ( σ , π ) ≈ ( π ) + g + g δ + g δ , (C2)with g = − . g = − . × − , g = . × − .Using Eq. (C1), we have: dG ( σ , π ) d σ = − ∞ ∑ n = Ψ (cid:0) n , + π i (cid:1) + Ψ (cid:0) n , − π i (cid:1) n − ( n − ) ! δ n − . (C3) i.e. , dG ( σ , π ) d σ = g (cid:48) δ + g (cid:48) δ + g (cid:48) δ + O ( δ ) , (C4)where g (cid:48) k − = − Ψ ( n , + π i ) + Ψ ( n , − π i ) n − ( n − ) ! , in particular: g (cid:48) = − . × − , g (cid:48) = + . × − , and g (cid:48) = − . × − . For a sufficiently large N , | g (cid:48) | > ∑ Nk = | g (cid:48) k − | → . × − , i.e. , neglecting the terms with order higher than O ( δ ) does not change the conclusion of dG ( σ , π ) d σ being positive or negative.According to Eq. (C4), we have when δ < g (cid:48) δ > g (cid:48) δ < g (cid:48) δ >
0, and g (cid:48) δ + g (cid:48) δ >
0, so dG ( σ , π ) d σ ≈ g (cid:48) δ + g (cid:48) δ + g (cid:48) δ > (cid:18) − ≤ δ < (cid:19) . δ > g (cid:48) δ < g (cid:48) δ > g (cid:48) δ <
0, and g (cid:48) δ + g (cid:48) δ <
0, so dG ( σ , π ) d σ ≈ g (cid:48) δ + g (cid:48) δ + g (cid:48) δ < (cid:18) < δ ≤ (cid:19) . So G ( σ , π ) monotonically increases in the range 0 ≤ σ < , and G ( σ , π ) monotonically decreases in the range < σ ≤
2) There are two symmetric zero points: σ ± = + δ ± of G ( σ , π ) in the range ≤ σ < G ( σ , π ) = i.e. ,4 ln ( π ) + g + g δ + g δ = , we have δ ± = ± .
3) in the range ≤ σ < / , | W ( σ + π i ) | monotonically decreases at the left side of σ − = + δ − and monotonicallyincreases at the right side. According to the above item 1), G ( σ , π ) monotonically increases; according to the above item 2), G (cid:0) + δ − , π (cid:1) = G ( , π ) = − . < G (cid:0) , π (cid:1) = . >
0, we find that during the continuous change of σ from 0 to , G ( σ , π ) monotonically evolves from being less than zero to being larger than zero. Therefore, we have: d | W ( σ + π i ) | d σ (cid:12)(cid:12)(cid:12)(cid:12) σ < + δ − = | W ( σ + π i ) | G ( σ , π ) (cid:12)(cid:12)(cid:12)(cid:12) σ < + δ − < , and d | W ( σ + π i ) | d σ (cid:12)(cid:12)(cid:12)(cid:12) σ = + δ − = | W ( σ + π i ) | G ( σ , π ) (cid:12)(cid:12)(cid:12)(cid:12) σ = + δ − = , and d | W ( σ + π i ) | d σ (cid:12)(cid:12)(cid:12)(cid:12) σ > + δ − = | W ( σ + π i ) | G ( σ , π ) (cid:12)(cid:12)(cid:12)(cid:12) σ > + δ − > . That is to say, | W ( σ + π i ) | monotonically decreases at the left side of σ − = + δ − and monotonically increases at the rightside.
4) in the range ≤ σ < / , | W ( σ + π i ) | < . According to the above item 3), | W ( σ + π i ) | monotonically decreases at the left side of σ − = + δ − and monotonicallyincreases at the right side; therefore, the values of | W ( σ + π i ) | at the two ends: σ = σ = must be larger than its value atpoints between the two ends. Because | W ( + π i ) | = . <
1, and | W ( / + π i ) | =
1, we have | W ( σ − + π i ) | < (cid:0) ≤ σ < (cid:1) . Appendix D: When | ζ ( s ) | = | ζ ( − s ) | , it must be true that | W ( s ) | = Proof
According to the Riemann-Zeta equation ζ ( s ) = W ( s ) ζ ( − s ) , we have: W ( s ) = ζ ( s ) ζ ( − s ) . Because W ( s ) = W ( s ∗ ) , we have ζ ( s ) ζ ( − s ) = ζ ( s ∗ ) ζ ( − s ∗ ) . Therefore: | W ( s ) | = W ( s ) W ( s ) = ζ ( s ) ζ ( s ) ζ ( − s ) ζ ( − s ) = ζ ( s ) ζ ( s ∗ ) ζ ( − s ) ζ ( − s ∗ ) . (D1)Here, s = σ + it , and s ∗ = σ − it .Assuming | ζ ( s ) | = | ζ ( − s ) | , (D2)we have: | ζ ( s ) | = | ζ ( − s ) | , (D3)3where | ζ ( s ) | = ζ ( s ) ζ ( s ∗ ) and | ζ ( − s ) | = ζ ( − s ) ζ ( − 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 ( − s ) ζ ( − s ∗ ) .
1) If ζ ( s ) (cid:54) =
0, then it must be true that ζ ( − s ) (cid:54) =
0. According to Eq. (D1) and Eq. (D2), it must be true that: | W ( s ) | = | ζ ( s ) || ζ ( − s ) | = . (D4)2) If ζ ( s ) =
0, then it must be true that ζ ( − s ) =
0. In this case, we have | W ( s ) | = . In this following, we prove the limitof this is 1.2.1) In the range 0 ≤ σ ≤
1, and t (cid:54) =
0, due to the fact that ζ ( s ) , ζ ( s ∗ ) , ζ ( − s ) , and ζ ( − s ∗ ) are all analytical functions.Furthermore, there exists their arbitrary order of partial derivative with respect to σ and t : ∂ m ζ ( s ) ∂σ m , ∂ m ζ ( s ) ∂ t m , ∂ n ζ ( s ∗ ) ∂σ n , ∂ n ζ ( s ∗ ) ∂ t n ; ∂ m ζ ( − s ) ∂σ m , ∂ m ζ ( − s ) ∂ t m , ∂ n ζ ( − s ∗ ) ∂σ n , ∂ n ζ ( − s ∗ ) ∂ t n . 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: s n = σ n + it n with ( n = , , , · · · ), so that ζ ( s n ) =
0. Then the complex conjugate ζ ( s n ) = ζ ( s ∗ n ) = σ n and t n , they are the zeros of the real function P ( σ n , t ) . Then there exists the Taylor expansion of P ( σ n , t ) around the zeros, t n : P ( σ n , t ) = ∞ ∑ j = j ! d j P ( σ n , t ) dt j (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) t = t n ( t − t n ) j (D5)where the derivatives in the expansion coefficients: (cid:26) d j P ( σ n , t ) dt j (cid:12)(cid:12)(cid:12) t = t n , j = , , , · · · , ∞ (cid:27) won’t be all equal to zero according tothe series expansion theorem. Notice that, the zeroth order expansion coefficient of P ( σ n , t ) is just P ( σ n , t ) ≡ ζ ( s n ) ζ ( s ∗ n ) = dP ( σ n , t ) dt (cid:12)(cid:12)(cid:12)(cid:12) t = t n = (cid:18) d ζ ( σ n + it ) dt ζ ( σ n − it ) + ζ ( σ n + it ) d ζ ( σ n − it ) dt (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) t = t n = . There must exit a certain k ≥
2, so that:lim t → t n (cid:18) d k P ( σ n , t ) dt k (cid:19) = lim t → t n d k dt k ∞ ∑ j = j ! d j P ( σ n , t ) dt j (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) t = t n ( t − t n ) j = d k P ( σ n , t ) dt k (cid:12)(cid:12)(cid:12)(cid:12) t = t n (cid:54) = . (D6)2.3) With the assumption in Eq. (D2), following Eq. (D3), it must be true that: | ζ ( − s n ) | = i.e. , t n is the zeros of the realfunction Q ( σ n , t ) . Similarly, there exists the Taylor expansion around the zeros t n : Q ( σ n , t ) = ∞ ∑ j = j ! d j Q ( σ n , t ) dt j (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) t = t n ( t − t n ) j . (D7)where the derivatives in the expansion coefficients: (cid:26) d j Q ( σ n , t ) dt j (cid:12)(cid:12)(cid:12) t = t n , j = , , , · · · , ∞ (cid:27) won’t be all equal to zero. Because boththe zeroth order expansion coefficient and the first order expansion coefficient are equal to zero; there must exit a certain order k ≥
2, so that: lim t → t n (cid:18) d k Q ( σ n , t ) dt k (cid:19) = lim t → t n d k dt k ∞ ∑ j = j ! d j Q ( σ n , t ) dt j (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) t = t n ( t − t n ) j = d k Q ( σ n , t ) dt k (cid:12)(cid:12)(cid:12)(cid:12) t = t n (cid:54) = . (D8)According to Eq. (D3), Eq. (D6), and Eq. (D8), there must exist a certain k ≥
2, so that: d k P ( σ n , t ) dt k (cid:12)(cid:12)(cid:12)(cid:12) t = t n = d k Q ( σ n , t ) dt k (cid:12)(cid:12)(cid:12)(cid:12) t = t n (cid:54) = . (D9)4For | W ( σ n + it n ) | = P ( σ n , t ) Q ( σ n , t ) = , because:lim t → t n | W ( σ n + it ) | = lim t → t n P ( σ n , t ) Q ( σ n , t ) = lim t → t n d k dt k P ( σ n , t ) d k dt k Q ( σ n , t ) = d k P ( σ n , t ) dt k (cid:12)(cid:12)(cid:12) t = t n d k Q ( σ n , t ) dt k (cid:12)(cid:12)(cid:12) t = t n . (D10)According to Eq. (D9) and Eq. (D10), we have: lim t → t n | W ( σ n + it ) | = . (D11)Based on the Taylor expansion of the real functions P ( σ n , t ) and Q ( σ n , t ) around their zeros, we can prove by the same procedurethat there exists a certain k ≥
2, so that: d k P ( σ , t n ) d σ k (cid:12)(cid:12)(cid:12)(cid:12) σ = σ n = d k Q ( σ , t n ) d σ k (cid:12)(cid:12)(cid:12)(cid:12) σ = σ n (cid:54) = . (D12)Therefore, we have: lim σ → σ n | W ( σ + it n ) | = lim σ → σ n P ( σ , t n ) Q ( σ , t n ) = lim σ → σ n d k d σ k P ( σ , t n ) d k d σ k Q ( σ , t n ) = d k P ( σ , t n ) d σ k (cid:12)(cid:12)(cid:12) σ = σ n d k Q ( σ , t n ) d σ k (cid:12)(cid:12)(cid:12) σ = σ n = . (D13)According to Eq. (D11) and Eq. (D13), we have: lim t → t n | W ( σ n + it ) | = , (D14)and lim σ → σ n | W ( σ + it n ) | = . (D15)The above serves as a proof. Appendix E: Proof of ddt (cid:12)(cid:12) Γ (cid:0) − s (cid:1)(cid:12)(cid:12) = ∑ ∞ n = − t | Γ ( − s ) | | it + n − σ − | , ddt (cid:12)(cid:12) Γ (cid:0) s (cid:1)(cid:12)(cid:12) = ∑ ∞ n = − t | Γ ( s ) | | it + n + σ − | Because: (cid:12)(cid:12)(cid:12) Γ (cid:16) s (cid:17)(cid:12)(cid:12)(cid:12) = Γ (cid:16) s (cid:17) Γ (cid:18) s ∗ (cid:19) . (E1) (cid:12)(cid:12)(cid:12)(cid:12) Γ (cid:18) − s (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) = Γ (cid:18) − s (cid:19) Γ (cid:18) − s ∗ (cid:19) . (E2)Furthermore, in the range 0 ≤ σ ≤ t (cid:54) =
0, both Γ (cid:0) s (cid:1) and Γ (cid:0) − s (cid:1) are analytical functions.We take derivative of Eq. (E1) and Eq. (E2) with respect to t : ddt (cid:12)(cid:12)(cid:12) Γ (cid:16) s (cid:17)(cid:12)(cid:12)(cid:12) = i (cid:12)(cid:12)(cid:12) Γ (cid:16) s (cid:17)(cid:12)(cid:12)(cid:12) (cid:20) Ψ (cid:16) s (cid:17) − Ψ (cid:18) s ∗ (cid:19)(cid:21) , (E3) ddt (cid:12)(cid:12)(cid:12)(cid:12) Γ (cid:18) − s (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) = i (cid:12)(cid:12)(cid:12)(cid:12) Γ (cid:18) − s (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) (cid:20) Ψ (cid:18) − s ∗ (cid:19) − Ψ (cid:18) − s (cid:19)(cid:21) . (E4)According to Eq. (10), Ψ (cid:18) σ − it (cid:19) − Ψ (cid:18) σ + it (cid:19) = ∞ ∑ n = − it | it + n + σ − | , (E5)5 − Ψ (cid:18) − σ − it (cid:19) + Ψ (cid:18) − σ + it (cid:19) = ∞ ∑ n = it | it + n − σ − | , (E6)Plugging Eq. (E5) into Eq. (E3), and plugging Eq. (E6) into Eq. (E4), we have: ddt (cid:12)(cid:12)(cid:12) Γ (cid:16) s (cid:17)(cid:12)(cid:12)(cid:12) = i (cid:12)(cid:12)(cid:12) Γ (cid:16) s (cid:17)(cid:12)(cid:12)(cid:12) ∞ ∑ n = it | it + n + σ − | . (E7) ddt (cid:12)(cid:12)(cid:12)(cid:12) Γ (cid:18) − s (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) = i (cid:12)(cid:12)(cid:12)(cid:12) Γ (cid:18) − s (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) ∞ ∑ n = it | it + n − σ − | . (E8)Notice that: ddt (cid:12)(cid:12) Γ (cid:0) s (cid:1)(cid:12)(cid:12) = (cid:12)(cid:12) Γ (cid:0) s (cid:1)(cid:12)(cid:12) ddt (cid:12)(cid:12) Γ (cid:0) s (cid:1)(cid:12)(cid:12) , ddt (cid:12)(cid:12) Γ (cid:0) − s (cid:1)(cid:12)(cid:12) = (cid:12)(cid:12) Γ (cid:0) − s (cid:1)(cid:12)(cid:12) ddt (cid:12)(cid:12) Γ (cid:0) − s (cid:1)(cid:12)(cid:12) , therefore, Eq. (E7) and Eq. (E8) canfinally simplified as: ddt (cid:12)(cid:12)(cid:12) Γ (cid:16) s (cid:17)(cid:12)(cid:12)(cid:12) = − t ∞ ∑ n = (cid:12)(cid:12) Γ (cid:0) s (cid:1)(cid:12)(cid:12) | it + n + σ − | , ddt (cid:12)(cid:12)(cid:12)(cid:12) Γ (cid:18) − s (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) = − t ∞ ∑ n = (cid:12)(cid:12) Γ (cid:0) − s (cid:1)(cid:12)(cid:12) | it + n − σ − | . This serves as a proof.
Appendix F: For an arbitrary finite t > , the | ζ ( s ) || ζ ( − s ) | satisfying ddt (cid:16) | ζ ( s ) || ζ ( − s ) | (cid:17) = ddt | W ( s ) | (cid:54) = , can not be in the form of . Proof
According to Eq. (11), when | W ( s ) | >
0, and σ < , we always have: ddt (cid:18) | ζ ( s ) || ζ ( − s ) | (cid:19) = | ζ ( − s ) | d | ζ ( s ) | dt − | ζ ( s ) | d | ζ ( − s ) | dt | ζ ( − s ) | = ddt | W ( s ) | > | ζ ( − s ) | d | ζ ( s ) | dt − | ζ ( s ) | d | ζ ( − s ) | dt > . (F2)Notice that: d | ζ ( s ) | dt = | ζ ( s ) | d | ζ ( s ) | dt = d ζ ( s ) dt ζ ( s ∗ ) + d ζ ( s ∗ ) dt ζ ( s ) , we have: d | ζ ( s ) | dt = | ζ ( s ) | (cid:20) d ζ ( s ) dt ζ ( s ∗ ) + d ζ ( s ∗ ) dt ζ ( s ) (cid:21) . Similarly, we have: d | ζ ( − s ) | dt = | ζ ( − s ) | (cid:20) d ζ ( − s ) dt ζ ( − s ∗ ) + d ζ ( − s ∗ ) dt ζ ( − s ) (cid:21) . Therefore, Eq. (F2) can be rewritten as:12 (cid:40) d ζ ( s ) dt ζ ( s ∗ ) + d ζ ( s ∗ ) dt ζ ( s ) | W ( s ) | − | W ( s ) | (cid:20) d ζ ( − s ) dt ζ ( − s ∗ ) + d ζ ( − s ∗ ) dt ζ ( − s ) (cid:21)(cid:41) > . (F3)Notice that, for an arbitrary finite t > | W ( s ) | > ζ ( s ) , ζ ( s ∗ ) , ζ ( − s ) , ζ ( − s ∗ ) are allanalytical functions. Furthermore, their derivatives d ζ ( s ) dt , d ζ ( s ∗ ) dt , d ζ ( − s ) dt , d ζ ( − s ∗ ) dt do not have singular points. If there exists a6certain s n , so that | ζ ( s n ) | = | ζ ( − s n ) | =
0, then it must be true that ζ ( s n ) = ζ ( s ∗ n ) = ζ ( − s n ) = ζ ( − s ∗ n ) =
0. Pluggingthese into Eq. (F3), we have:12 · d ζ ( s ) dt (cid:12)(cid:12)(cid:12) s = s n + · d ζ ( s ∗ ) dt (cid:12)(cid:12)(cid:12) s ∗ = s ∗ n | W ( s n ) | − | W ( s n ) | (cid:34) · d ζ ( − s ) dt (cid:12)(cid:12)(cid:12)(cid:12) s = s n + · d ζ ( − s ∗ ) dt (cid:12)(cid:12)(cid:12)(cid:12) s ∗ = s ∗ n (cid:35) = > . This is to say that | ζ ( s n ) | = | ζ ( − s n ) | = σ < , and an arbitrary finite t >
0, the | ζ ( s ) || ζ ( − s ) | satisfying ddt (cid:16) | ζ ( s ) || ζ ( − s ) | (cid:17) >
0, can not be in the form of .Similarly, for σ > , and an arbitrary finite t >
0, the | ζ ( s ) || ζ ( − s ) | satisfying ddt (cid:16) | ζ ( s ) || ζ ( − s ) | (cid:17) <
0, can not be in the form of .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 , 199 (1896).[3] Ch.J. de la Vallee-Poussin, “ Recherches analytiques sur la theorie des nombers premiers ”, Ann. Soc. Sci. Bruxelles20