Riemann s Functional Equation is Not Valid and its Implication on the Riemann Hypothesis. Armando M. Evangelista Jr.

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1 Riemann Functional Equation i Not Valid and it Implication on the Riemann Hypothei By Armando M. Evangelita Jr. On November 4, 28 ABSTRACT Riemann functional equation wa formulated by Riemann that uppoedly extended the domain of the Riemann zeta function from the right half-plane into the entire complex plane except at =. It alo led him to obtained a real function that contain zero for ome. Now, the real function wa alo related to the zeta function which in turn ha omething to do with the ditribution of prime number. Thi led him to obtained a formula for relating the zero of the zeta function to the number of prime given a certain number. Riemann then conjectured that all the zero of the zeta function are at = ½ + ωi, i, which i now known a the Riemann Hypothei. Hence Riemann functional equation i the foundation upon which the Riemann Hypothei i baed. But the functional equation, a hall be hown here, i not valid uch that the Riemann Hypothei crumble on it claim.

2 Riemann Functional Equation The Riemann zeta function (or zeta function) i hown below () ζ () = n = n= n = σ + ω i, where i a complex variable with real part σ and imaginary part ω, and i i the imaginary unit, i =. It i known that the infinite erie in () i undefined if σ, conditionally convergent if < σ, and abolutely converge if σ >. ζ () i related to the ditribution of prime number for one obtain from () the infinite product, (2) ζ() = ( 2 )( 3 )( 5 )( 7 ) ( p ) = p p. Thi had driven Riemann to obtained a formula for relating the uppoedly non-trivial zero of ζ (½+ωi, i) with the number of prime given a certain number. A imple inpection of (2) and one can eaily conclude that uch zero are nowhere to be found. Analytic Continuation If a function f () i analytic in domain D and a econd function f 2 () i analytic in domain D 2 : (a) f 2 () f () for each in the interection D D 2. (b) f 2 () = f () for each in the interection D D 2. (c) If f 2 () = f () for each in the interection D D 2 and if D D 2 or D 2 i the union D D 2, then, f 2 () i the analytic continuation of f () into the econd domain D 2. (a) imply point to the fact that the two function though eemingly equal, are not; while (b) how that they are only equal at their common point. Analytic continuation i only recommended in (c) ince f () domain will be extended from D into domain D 2. For example, conider the three function hown below f () = f 2 () = = ( n) = n=, >, n = n = n, <, n=

3 and f 3 () =,, that have imilar cloed form,. But f 2 () f () ince they are on different domain while f () = f 3 () only at their interection, that i, when their modulu are greater than one. The function f 3 () occupie the entire domain except at =, hence f 3 () i the analytic continuation of either f () or f 2 (). Table Some Value for f (), f 2 (), and f 3 () f () f 2 () f 3 () -2 /3 undefined /3 - undefined undefined /2 undefined undefined undefined undefined 2 - undefined undefined -2 A one can ee from ome of the value given on Table, at their interection, f 3 () = f (), a expected. Now, conider the widely known Riemann functional equation (3) ζ () = 2 π in ( π 2 ) Γ( )ζ ( ), where Γ( ) and ζ ( ) are the reflection of the gamma and the zeta function,. Formula (3) i conidered a the analytic continuation of () into the entire complex plane except at =. The ocalled trivial zero of (3) are at each even integer = -2n, n. Table 2 Some Value of ζ () for () and (3) ζ () Formula () Formula (3) ζ ( 4) undefined ζ ( 3) undefined /2 ζ ( 2) undefined ζ ( ) undefined -/2 ζ () + / 2 + /3+ /4 + = undefined

4 ζ (2) π 2 / 6 I m not ure ζ (4) π 4 / 9 I m not ure A can be een from Table 2, the two formula have no common point and are unequal for all. Hence (3) i not the analytic continuation of ζ () into the entire complex plane. Simple analyi will reveal that (3) i fale: ζ () a defined by () need σ >/2 to converge while the right-ide function in (3) need σ </2 due to the preence of ζ ( ) [3]. The two function, therefore, have no common point and are not equal. Simple Guide for Analytic Continuation. Let a function f () be analytic in domain D, check if f () for value of that are not in D. If there i a function f 2 () that i analytic in domain D 2 equal to f () at D and D D 2, then, f 2 () i the analytical continuation of f (). 2. If f () = for all value of that are not in D, then, it i completely defined by it domain D. 3. If f () i an integral, perform and 2 after the integral ha been evaluated, becaue analytic continuation can not be perform if the integral doen t exit! For example, conider the function below f () = σ >, for value of <, f (), hence one can analytically continue f () into f 2 (), if there i one and maybe valid for all except at =. That i, f 2 () =. Conider the integral f () = e ( )t dt. The integral exit if σ > and it value being /( ) f () = σ >.

5 The function f 2 () =, i the analytic continuation of f () into the domain D 2, that i, analytic for all except at =. But if one look at the zeta function ζ () = n = n= n, one can eaily verify that ζ () = if σ : thu, a valid function equal to ζ () at σ doe not exit. Therefore, ζ () could not be extended on the left half-plane. The zeta function i completely defined by () on the right half-plane. Now, conider the integral hown below (4) I () = ( z) e z dz R( z) > and σ >, C formula (4) have to complex quantitie z and, where z i the complex variable, and i a contant relative to z. The integral will converge if the real part of z and are greater than zero. Since i a contant in (4), one can move (-) outide the integral ign, that i I () = ( ) C z e z dz, ( ) i a multivalued complex quantity with principal value ( ) ( ) = e ±(π i). Integrating (4) over the Hankel contour : it tart from + toward the circle with a very mall radiu ρ then goe around the circle counterclockwie once and goe to back +. I () = ( ) ( 2π x e x dx + (ρe iθ ) e θi ρ ei ρe iθ dθ + (x e 2π i ) e xe2 π idx), o the econd integral above approache zero a ρ I () = ( ) ( x e x dx + + (x e 2 π i ) e xe2π i dx ) = ( ) ( x e x dx ) (e 2 π i ),

6 and then chooing the principal value of ( ) = e i π, (5) I () = 2i in(π )Γ(). The integral in (4) i only valid if σ >, but (5) i now valid for negative value of σ except for the negative integer due to analytic continuation and ha zero at = n for integer n. Now, conider the contour integral ( x) (6) I () = C e x dx = ( ) x dx x > and σ >, e x and ince I () = ( ) ( (7) thu, x e x dx + 2π (ρ e iθ ) e ρ e iθ ρ ieiθ dθ + ( xe 2π i ) e xe 2π i ) dx, ( ) ( x e x dx ) (e2π i ), x dx = ζ ()Γ(), e x (8) I () = 2iin(π ) ζ ()Γ(). Due to the preence of ζ () on (8), I () i only valid on the right half-plane. By uing the reflection formula and formula (6), (7), and (8), one obtain Γ()Γ( ) = π in(π ), (9) ζ () = Γ( ) 2π i ( x) e x dx.

7 Riemann argued that ζ () defined by (9) i now valid over the entire -plane except at =, by imply rearranging formula (6), (7), and (8)! One mut bear in mind that the integral mut firt be evaluated in order for analytical continuation to be applied. Hence (9) i imply an identity, that i, the zeta function i equal to itelf ζ () = Γ( ) 2π i ( x) e x Γ( ) dx = { 2i in(π )ζ ()Γ()} = ζ (). 2π i Moreover, ( x) e x dx involve the integral x e x dx, which will only be valid for poitive value of σ and undefined for negative value of σ and σ =. PROOF : x e x dx = x e nx dx = ( n= n= n )( x e x dx ) and the integral x e x dx i only defined if σ >, that i, ( x e x dx = x x! + x2 2! x3 3! + x4 4! + ( )n x n n! ) dx = ( x x + (+)! + x +2 (+2)2! x +3 (+3)3! + x +4 (+4)4! + ( )n x +n (+n)n! ) ince if one ubtitute negative value of σ or σ = on the lower limit of the lat expreion above, the integral won t exit! Remember that analytic continuation hould be perform after the integral ha been obtained and not before. Thu, the integral in (9) i not valid if σ and (8) i the valid integral.,

8 The Miapplication of the Reidue Theorem Riemann applied the Reidue Theorem on (6) by auming the quantitie 2πni ni for n = ±, ±2, ±3, and o on, a the pole of (6). The function q(z) = /(e z ) in (6) i a periodic function of z, q(z +2πi ) = e ( z + 2π i) = e z R(z) >, ince the exponential function e Z i a periodic function with period 2πni i. Hence the term 2πni ni are the multiple of the fundamental period 2πni i. It i, therefore, not valid to treat them a the pole of (6). In fact, the only pole of q(z) i at z = with reidue, q(z) = e z = e z + e 2z + e 3 z + e nz = e nz, R(z)>, n= o that it contour integral on a imple cloed path i 2πni i, that i, dz e z = 2π i. Unfortunately, one can not ue z = on (6) ince it will be undefined. Thu, Riemann determination led him to apply the Reidue Theorem on (6), that i, ( z) C e z dz = 2π i ( n= ( 2πni) + (2π ni) ) = 2π iin ( π 2 ) 2 π ζ ( ), and by equating the lat expreion to formula (8), (9) 2iin(π )ζ ()Γ() = 2πi in ( π 2 ) 2 π ζ ( ), one obtain the Riemann functional equation (3), And, by uing equation ζ () = 2 π in ( π 2 ) Γ( )ζ ( ). Γ ()Γ ( ) = π in(π) and πγ() = 2 Γ ( 2) Γ ( + 2 ),

9 one arrive at the equality, () π 2 Γ ( ( 2 ) ζ () = π 2 ) Γ( 2 ) ζ ( ). The equality in (), if it i valid, will mean that the function to the left, ay Φ(), will be equal to it reflection Φ( ) ; which mean that at = ½ + ωi, i, Φ(/2 + ωi ) i equal to Φ(/2 ωi), and, in accordance with the Reflection Principle Φ() = Φ(), Φ(/2 + ωi ) will be a real function. And o, if Φ(/2 + ωi ) i real and ha zero for ome value of ωi,, then ζ(/2 + ω i) ha alo zero from (). Since, it i known, that Γ ( ha no 2) zero for all, Riemann conjectured that all the zero of ζ () are at = ½ + ωi,i, which i the famou Riemann Hypothei. The function Φ() ha no zero if it ain t real, becaue it will be jut like ζ () that ha no zero ince it modulu i alway greater than zero on the right half-plane. The zero that many are looking for are mot likely the zero of of the real part of Φ(/2 + ωi ), that i, R {Φ(/2 + ω i)} = for ome value of ω, but Φ(/2 + ωi ) doe not converge due to ζ(/2 + ω i) [3]. ( z) Thu, the Riemann Hypothei relie upon the validity of C e z dz being equal to 2π iin ( π 2 ) 2 π ζ ( ) in oder for () to be true o that Φ(/2 + ωi ) will be a real function that ha zero which ha the implication of ζ(/2 + ω i) having zero. I ve hown that not the cae, and Riemann functional equation could not be valid and the Riemann hypothei ret on a fale premie.

10 Concluion (a) The Riemann zeta function i completely defined by () on the right half-plane. (b) The analytic continuation of an integral function mut be perform after the integral ha been evaluated and not before. (c) The function 2iin(π )ζ ()Γ() i a valid contour integral of C 2π iin ( π 2 ) 2 π ζ ( ) i not. ( x) e x dx while ( x) (d) I have noticed recently that Riemann applied the Reidue theorem on dx in the C e x clockwie direction; it doen t matter, ince the Reidue theorem can not be applied to it. (e) From (c), the functional equation π 2 Γ ( ( 2 ) ζ () = π 2 ) Γ( 2 ) ζ ( ) i not true. (f) From (d), Φ(/2 + ωi ) i not a real function and ha no zero jut like the zeta function on the right half-plane. (g) From (e), the Riemann hypothei i, therefore, fale. REFERENCES: [] Riemann, Bernhard (859). On the Number of Prime Number le than a Given Quantity. [2] Brown, Jame Ward; Churchill, Ruel V. (996). Complex Variable and Application (6 th ed.). Singapore: McGraw Hill International Edition. [3] Evangelita, Armando M. (28). The Domain of the Riemann Zeta Function on the Complex Plane. LINKS:

Riemann s Functional Equation is Not a Valid Function and Its Implication on the Riemann Hypothesis. Armando M. Evangelista Jr.

Riemann s Functional Equation is Not a Valid Function and Its Implication on the Riemann Hypothesis. Armando M. Evangelista Jr. Riemann Functional Equation i Not a Valid Function and It Implication on the Riemann Hypothei By Armando M. Evangelita Jr. armando78973@gmail.com On Augut 28, 28 ABSTRACT Riemann functional equation wa