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1. Introduction : One of the Most Difficult Problems Today to Mathematicians and Physicists Is Riemann Hypothesis

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1. Introduction : One of the Most Difficult Problems Today to Mathematicians and Physicists Is Riemann Hypothesis Analytical expression of Complex Riemann Xi function ξ(s) and proof of Riemann Hypothesis. Debasis Biswas Chakdaha W.B – 741 235. India Email – [email protected]. Abstract : [In this paper explicit analytical expression for Riemann Xi function ξ(s) is worked out for complex values of s. From this expression Riemann Hypothesis is proved. Analytic Expression for non-trivial Zeros of Riemann Zeta function ζ(s) is also found. A second proof of Riemann Hypothesis is also given.] Keywords : Riemann Zeta and Riemann Xi function, Riemann Hypothesis, Critical line, non-trivial zeros. 1. Introduction : One of the most difficult problems today to mathematicians and physicists is Riemann Hypothesis. It is a conjecture proposed by Bernhard Riemann which says that all the complex zeros of Riemann Zeta function ζ(s) has real part σ = ½ . Till now this conjecture is neither proved nor disproved. In this paper an explicit analytical expression of Riemann Xi function ξ(s) is found. From this expression of ξ(s) expression for zeros of ξ(s) is derived. As it is known that all the zeros of ξ(s) are identical with nontrivial zeros of Riemann zeta function, the zeros of ξ(s) are identified as nontrivial zeros of Riemann zeta function ζ(s). The paper is organized as follows. In section 2, we summarize the definition of Riemann Xi and Riemann zeta function, mention the connection between ξ(s) and ζ(s) . We also mention some important results related to zeros of ζ(s). Next in section 3, the expression for ξ(s) in terms of an arbitrary function is derived. In section 4, the analytical expression for ξ(s) is worked out using a theorem of analysis. In section 5, Riemann Hypothesis is proved and equation for nontrivial zeros of ζ(s) is derived. In Section 6, a second proof of Riemann Hypothesis is given. Section 7 contains conclusion and comments. An appendix is added after section 7. 2 2. Riemann zeta and Xi function and Riemann Hypothesis For real s > 1, Riemann zeta function ζ(s) is defined as [1, 2, 3] ∞ −푠 ζ(s) = ∑휈=1 휈 … (2.1) The definite integral equivalent of (2.1) is ∞ ts−1 Г(s) ζ(s) = ∫ dt ; real s > 1 …(2.2) 0 et − 1 and Г(s) ≡ Gamma function ζ(s) can be analytically continued from (2.1) and can be defined for 0 < real s <1 ∞ 1 1 Г(s) ζ(s) = ∫ ( − ) ts−1 dt ; 0 < real s <1 … (2.3) 0 et−1 t Riemann zeta function satisfies a well-known functional equation [3] : s 1−s − − ( ) 1−s π 2Г(s/2)ζ(s) = π 2 Г( ) ζ(1 − s) … (2.4) 2 This functional equation (2.4) is also a definition of ζ(s) over whole complex s-plane except a singularity at s = 1. A consequence of symmetry of (2.4) suggests that if there is a complex zero of ζ(s) for Re s = ½ + δ, then there must be another zero of ζ(s) for Re s = ½ − δ. Riemann introduced another function known as Riemann Xi function ξ(s) which satisfies a functional equation : ξ(s) = ξ(1 − s) … (2.5) The Riemann Xi and Riemann zeta function are connected through the equation [1] : s 1 − ξ(s) = s(s – 1 ) π 2Г(s/2)ζ(s) … (2.6) 2 The Riemann zeta function has real and complex zeros [2]. The real zeros of ζ(s) are known as trivial zeros and are given by ζ(– 2m) = 0 ; m = 1, 2, 3, . … (2.7) The complex zeros of ζ(s) lie [1] within the strip 0 <Re s < 1. This strip is known as critical strip and zeros lying in critical strip are known as nontrivial zeros. And the straight line Re s = ½ is called critical line . Now a consequence of (2.6) is that the nontrivial zeros of Riemann zeta function ζ(s) are identical with zeros of Riemann Xi function ξ(s) [3]. Riemann Hypothesis is a conjecture made by Riemann which says that the nontrivial zeros of Riemann zeta function ζ(s) has real part σ = ½ . This implies that the zeros of Riemann Xi function ξ(s) has also real part of σ = ½ . This Hypothesis has not yet been proved or disproved. 3 The literatures on ξ(s) and ζ(s) are extremely large. Only a few important results related to zeros of ζ(s) are mentioned. G. H. Hardy [4] has proved that an infinite number of zeros lie on the critical line σ = ½. J. B. Conrey [5] has shown that more than 40% zeros of ζ(s) lie on critical line. N. Levinson [6] has shown that more than one third zeros of ζ(s) lie on critical line. 3. Expression of Riemann Xi function ξ(s) in terms of an arbitrary function. As the nontrivial zeros of ζ(s) are identical with zeros of ξ(s) , we will consider ξ(s) and find its analytical expression in terms of an arbitrary function. This is more convenient which will be clear later. Firstly, we will consider solution of (2.5) which is satisfied by ξ(s) . The solution of (2.5) in terms of an arbitrary function was given by Hymer [7]. Following Hymer [7] the general solution of the equation (2.5) can be written as in terms of an arbitrary function. Hymer considers the functional equation of more general form φ (a + bs) = nφ(s) ; n, a, b are constants …(3.1) and gives the solution of (3.1) as log n Coa log b 2π Coa φ(s) = (C s − ) θ [Cos { . log (C s − )}] …(3.2) o 1−b o log b o 1−b Co ≡ arbitrary constant θo ≡ arbitrary function Now for a = 1, b = – 1 and n = 1 equation (3.1) reduces to φ (1 – s) = φ(s) …(3.3) Equation (2.5) and (3.3) are of same form. Hence solution of (2.5) and (3.3) will have identical form. The solution of (3.3) in terms of an arbitrary function θ0 follows from (3.2) : log 1 1 2π 1 φ(s) = (s − )log(−1) θ [Cos { . log (s − )}] 2 o log(−1) 2 taking Co = 1 in (3.2) 2nπi 1 ( ) 2π 1 i.e., φ(s) = (s − ) 2n+1 πi θ [Cos { . log (s − )}] 2 o log(−1) 2 2n 1 ( ) 2π 1 = (s − ) 2n+1 θ [Cos { . log (s − )}] 2 o (2n+1)πi 2 2n 1 ( ) −2i 1 = (s − ) 2n+1 θ [Cos { . log (s − )}] 2 o (2n+1) 2 2n 1 ( ) 2 1 = (s − ) 2n+1 θ [Cosh { . log (s − )}] …(3.4) 2 o (2n+1) 2 4 A glance at (3.4) suggests that φ(s) can be written more conveniently in terms of another arbitrary function. Therefore φ(s) = ψ0(s − ½) …(3.5) ψ0(s − ½) is another arbitrary function. Now comparing (2.5), (3.3) and (3.5) we can write the solution of (2.5) in terms of an arbitrary function : ξ(s) = ψ (s – ½) …(3.6) Equation (3.6) is the solution of (2.5) and represents the Riemann Xi function ξ(s) in terms of an arbitrary function. 4. Explicit analytical expression of Riemann Xi function ξ(s). To derive the exact analytical expression of ξ(s) we state a result of analysis in the form of a theorem. The theorem is due to J. Brill [8] Theorem 1. (Due to J. Brill) The theorem states that if α is a root of Aα2 + Bα + C = 0 A, B, C are constants …(4.1) Then φ (y + αx) can be expressed as φ (y + αx) = η + αθ …(4.2) where α is independent of x, y and η = η(x, y), θ = θ(x, y) satisfy 1 ∂θ 1 휕η ∂θ 1 휕η = ( − ) = – …(4.3) A ∂y B ∂y ∂x C ∂x We will use the results from (4.1) to (4.3) of above theorem to find analytic expression of Riemann Xi function ξ(s), where s is complex. Conventionally s is written as s = σ + it ; i = √− 1 …(4.4) σ, t being real and imaginary parts of s respectively. Now from (4.3) we find 1 ∂θ 1 휕η + = 0 …(4.5) A ∂y C ∂x 1 ∂θ 1 휕η 1 휕θ − + = 0 …(4.6) A ∂y B ∂y B ∂x 1 휕η 1 휕η 1 휕θ + − = 0 …(4.7) C ∂x B ∂y B ∂x 5 From (4.7) one finds 1 ∂2η 1 ∂2η 1 ∂2θ + − = 0 …(4.8) C ∂y ∂x B ∂y2 B ∂y ∂x 1 ∂2θ 1 ∂2η and from (4.5) + = 0 …(4.9) A ∂x ∂y C ∂x2 ∂2θ Now from (4.8) and (4.9) eliminating we get assuming partial derivatives are continuous ∂x ∂y ∂2η B ∂2η C ∂2η + + = 0 …(4.10) ∂x2 A ∂x ∂y A ∂y2 Likewise from (4.5) ∂2θ A ∂2η + = 0 …(4.11) ∂y2 C ∂x ∂y And from (4.6) 1 ∂2θ 1 ∂2η 1 ∂2θ − + = 0 …(4.12) A ∂x ∂y B ∂x ∂y B ∂x2 휕2η Again assuming partial derivatives are continuous, we eliminate from (4.11) and (4.12). 휕x ∂y The result is : 휕2θ B ∂2θ C 휕2θ + + = 0 …(4.13) 휕x2 A ∂x ∂y A 휕y2 A comparison of (4.10) and (4.13) shows that we can choose η (x, y) = θ (x, y) …(4.14) Now in (4.1) we take α = 1 … (4.15) And in (4.2) we take 1 y = (σ − ) 2 x = i t … (4.16) Then in view of (4.15) we find from (4.1) A + B + C = 0 … (4.17) and from (4.2), (4.14), (4.15) and (4.16) 1 1 1 φ(σ − + i t) = η(σ − , i t) + θ(σ − , i t) … (4.18) 2 2 2 Using (4.16), (4.10) can be rewritten as 2 ∂2η iB ∂2η C 휕 η − − 1 + 2 = 0 ∂t2 A ∂t.∂(σ − ) A 1 2 휕(σ − ) 2 6 ∂2η iB ∂2η C 휕2η i.e., + 1 − 2 = 0 ∂t2 A 휕t.휕(σ − ) A 1 2 휕(σ − ) 2 2 ∂2η iB ∂2η A + B 휕 η i.e., + 1 + 2 = 0, using (4.17) ∂t2 A 휕t.휕(σ − ) A 1 2 휕(σ − ) 2 2 2 ∂ η ∂ η 휕2η B Therefore, + ik 1 + (1 + k) = 0 ; k = = constant …(4.19) ∂t2 휕t.휕(σ − ) 1 2 A 2 휕(σ − 2) Likewise (4.13) reduces to 2 2 ∂ θ ∂ θ 휕2θ + ik 1 + (1 + k) = 0 …(4.20) ∂t2 ∂t.휕(σ − ) 1 2 2 휕(σ − 2) Equation (4.2), (4.14), (4.18), (4.19) and (4.20) imply that (3.6) can be written as 1 ξ(s) = ψ(s ‒ ) 2 1 = ψ(σ ‒ + i t) 2 1 1 = η (σ ‒ , i t) + θ (σ ‒ , i t) …(4.21) 0 2 0 2 Where η0 and θ0 satisfy (4.19) and (4.20) and η0 = θ0 …(4.22) Hence from (4.21) and (4.22) the expression for ξ(s) can be written as 1 ξ(s) = 2η (σ ‒ , i t) …(4.23) 0 2 where η0 satisfies ∂2η ∂2η 2 0 0 휕 η0 2 + ik 1 + (1 + k) 2 = 0 …(4.24) ∂t 휕t.휕(σ − ) 1 2 휕(σ − 2) There exist methods for solution of (4.24).
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