Riemann Zeta Function: The Riemann Hypothesis Origin, the Factorization Error, and. the Count of the Primes

Transcription

1 Gauge Insiue Journal, Volume 5, No 4, November 009 Riemann Zea Funcion: The Riemann Hypohesis Origin, he Facoriaion Error, and he Coun of he Primes Augus 008 Keywords: Zea Funcion, Gamma funcion, Ellipic Funcions, Analyic Funcions, Meromorphic funcions 000 mahemaics subjec Classificaion 33E0; 30A0, 6D5, 30D0

2 Gauge Insiue Journal, Volume 5, No 4, November 009 Conens Inroducion 0. The facoriaion error..8. The Coun of Prime Numbers. Gauss approximaion.... Gauss Approximaion and he prime number heorem. Zea Series in Re >. The Euler produc Zea Inegral in Re > 3. Gamma Inegral in Re > Gamma for any Zea Inegral in x > Zea for any 4. The pah Inegral formula for Zea 4. The Inegraion pah The firs inegral The second inegral The hird inegral he pah inegral formula A second pah inegral formula for Zea..9

3 Gauge Insiue Journal, Volume 5, No 4, November Zea is finie for all Zea has eros a = n The Zea funcional equaion 5. sin π ()() ζ ( π)[( i) i ]{... } Γ = Zea Funcional equaion Funcional equaion wih respec o = Poisson Ellipic Funcion ψ () 6. Poisson Ellipic funcion Poisson funcional equaion Riemann s Formula for Zea 7. Riemann s formula for Zea The funcion ξ () 8. Definiion of ξ () 5 8. Formulas for ξ () Riemann s firs formula for ξ () riemann s second formula for ξ ().5 9. The Funcion ξ ( + iw) 9. Formula for ξ ( + iw) Taylor Series for ξ ( + iw)

4 Gauge Insiue Journal, Volume 5, No 4, November The Zeros of ξ () 0. The eros of ξ () All he eros of ξ () are in 0 < x < For R large enough, log ξ( ) Rlog R in w = R Hadamard Esimae Riemann Esimae The Hypohesis.65. Facoriaion of ξ. Weiersrass facoriaion of ξ ().67 h ( ). ξ() = e ρ ρ ( )( ) n= n n 68.3 Facoriaion of ξ ( + iw), and ξ ().70.4 Riemann s Facoriaion error..73. The Number of Primes<. π (), and F () Zea in erms of f () 3. Definiion of f () ζ () in erms of f () f () in erms of Zea 4. f () in erms of ζ () 8 4

5 Gauge Insiue Journal, Volume 5, No 4, November f () in erms of ξ () 5. f () in erms of ξ () The log ξ(0) erm y= y= d = d = πi log πi..89 y= y= y= log ξ(0) d log ξ(0) log πi = =.9 log y= 7. Terms wih log( ) 7. d( log Γ ( + ) ) = d( log( + ) ) n n= β u= β u du + cons, if σ < 0 y= logu u= d log πi = log β y = u= β u + > u= 0 du$ cons, if σ 0 logu If σ < 0, 7.4 If σ > 0, y= u= β u d log = du πi log β logu...98 y= u= y= u= β u d log = du πi log logu...00 β y= u= If σ > 0, hen u= β u β Li( ) i, for upper-half-plane pah du π = 05 β logu Li( ) + πi, for lower-half-plane pah u= 0 5

6 Gauge Insiue Journal, Volume 5, No 4, November 009 y= β Li( ) πi, for upper-half-plane pah 7.6 If σ > 0, d log β πi = 08 log β Li( ) + πi, for lower-half-plane pah y= 8. The log( ) erm y= d log( ) = Li( ).09 πi log 8. ( ) y= 9. The log + α α erm The assumpion of he Hypohesis 9. Implici claim. 9.3 If he eros of ξ are on x = and are sequenced by sie y= + iα iα d log Li( ) Li( ) πi + = + log y α α = 4 α 9.4 The Hypohesis Origin.4 0. The log ( ) Γ + erm y= u= πi u u u 0. d log ( ) Γ + = du...6 log log y= u=. The Coun and Densiy of he Primes. If he eros of ξ, α 's, are posiive and sequenced by sie, u= f () = Li() + log ξ(0) Li( ) + Li( ) + du ulogu u + iα iα.8 α. The Formula for he coun of he primes 8 u= 6

7 Gauge Insiue Journal, Volume 5, No 4, November The approximae densiy of f ()...4 Riemann s Approximaion for F ()..3.5 Comparison wih Gauss approximaion.3.6 The effec of he Hypohesis series..4.7 Flucuaions of f ().6 References.7 7

8 Gauge Insiue Journal, Volume 5, No 4, November 009 Inroducion Riemann's 859 Zea paper defines he Zea funcion and uses is properies o approximae he coun of prime numbers up o a number, and he densiy of he primes a he number The few pages paper oulines a book ha was never wrien by Riemann. The paper sums up Riemann s resuls on he Zea funcion, and on he coun of he prime numbers, wih a few connecing words, and no proofs/explanaions. Aemps o wrie he book were made by Tichmarsh, and by Edwards, bu none followed hrough Riemann s wriing. Only by saying faihful o Riemann s developmen of he Zea funcion we can Ge o he origin and meaning of he Riemann Hypohesis, Correc he facoriaion error, Apply Riemann s Formula for he Coun of he Primes, Acquire he ools needed for he Hypohesis Proof In 006, I aemped o wrie ha book, [Dan], and his is an exensive revision of ha aemp. 0. The Facoriaion Error Riemann defines he auxiliary funcion 8

9 Gauge Insiue Journal, Volume 5, No 4, November 009 / ξ() ( ) π Γ ( /+ )() ζ ha has he same eros as ζ () in 0 < x < We show ha i has he facoriaion ξ() = ξ(0) / iα /+ iα n = n n = ξ(0) + α + /4 n = where he /± are he eros of ξ (). αn n We furher show ha if he eros are all on he line x = /, his facoriaion for ξ () produces he erm n= + iα Li( n iα ) Li( n + ) in he formula for he coun of he primes, where Li( ) is he Logarihmic inegral. Riemann obained he erroneous facoriaion ( /) ξ() = ξ + = / n= αn ha does no produce he Logarihmic inegral series erm. Already in 860, Genocchi poined ou ha he formula should use ξ. = 0 9

10 Gauge Insiue Journal, Volume 5, No 4, November 009 Bu ha only divered aenion from he error in he facors of he produc, and he main problem seemed o be ha a derivaion was missing. In 893, Hadamard supplied he derivaion and obained ξ() = ξ(0), ρ where he ρ 's are he eros of ξ (). Hadamard formula does no exhibi he connecion o he α n 's, and so far as I can ell, he connecion o he Logarihmic inegral series in he formula for he coun of he primes, was never made. Indeed, o make ha connecion, one has o follow hrough he whole paper. Thus, producing he correc derivaions and resuls of he Zea paper amouns o execuion of he book ha is oulined in he Zea paper. ρ I use common noaions and erms such as = x + iy no s = σ + iτ w(), no s () s Γ () no Π( s ). "eros of ξ " no "roos of he equaion ξ () = 0." Oherwise, Riemann's noaions are kep unchanged. 0

11 Gauge Insiue Journal, Volume 5, No 4, November 009 The Coun of Prime Numbers. Gauss approximaion In his 859 Zea Paper, Riemann's aim was...o repor on a sudy of he frequency wih which prime numbers occur. A opic ha seems worhy of such reporing, because of he ineres shown in i by Gauss and Dirichle over many years. Gauss (849) compued u= u= du log u o approximae π (), he Number of Primes <. His resuls are lised in he following Gauss able [Ed, p.3] u= du π() Error logu u= 500, 000 4, 556 4, ,000,000 78,50 78, ,500,000 4, 4,63. 5.,000,000 48,883 49, ,500,000 83,06 83, ,000,000 6,745 6,

12 Gauge Insiue Journal, Volume 5, No 4, November 009. Gauss Approximaion, and he Prime Number Theorem We have where u= du = Li() Li(), logu u= is he Logarihmic Inegral. du Li( ) = lim + ε 0 logu u= ε u= u= 0 u= + ε du logu By he Prime Number Theorem (Hadamard-896), π() Li(), as. The Prime Number Theorem subsaniaes he Gauss approximaion

13 Gauge Insiue Journal, Volume 5, No 4, November 009 Zea Series in Re >. The Euler Produc For Re >, Riemann wroe: ζ( ) n = p...my saring poin was he observaion of Euler ha he produc equals / p= prime p, n= nauraln where p ranges over all he prime numbers, andn over all he naural numbers. I denoe by ζ (), 3

14 Gauge Insiue Journal, Volume 5, No 4, November 009 he funcion of he complex variable, defined by hese wo expressions when hey converge. Proof: Euler's produc formula, defined for k =, 3,... by = , k k k k 3 n k k k k 3 5 p can be exended o complex numbers wih Re >, because Thus, in Re >. x n n n= naural n= naural ζ() = n / p. n= naural p= prime For a proof ha he equaliy beween he sum and he produc in Euler Produc Formula holds for complex numbers, see [Kara], or [Ed] 4

15 Gauge Insiue Journal, Volume 5, No 4, November Zea Inegral in Re > 3. Gamma Inegral in Re > 0 Proof: For x > 0, Euler s inegral, can be wrien as For he firs inegral, Γ () = = = 0 = = 0+ e e d d = = d + d. e e = 0+ = = = = x x lim x d d = = e x = 0+ x x 0+ = 0+ = 0+. Therefore he firs inegral converges uniformly for any x δ > 0. For he second inegral, 5

16 Gauge Insiue Journal, Volume 5, No 4, November 009 = = x d = e = = Inegraing by pars, e d = x = = / e = x = ( e ) + ( x ) d. e = = x = = / e = x 3 = + ( x ) ( e ) + ( x ) ( x ) d e e =.. Evenually, afer [ x ] such inegraions, he power of in he inegrand becomes x [ x], and we have = = ( x )...( x [ x]) d < ( x )...( x [ x]) d < [ x] + x e e = = Therefore, he Gamma inegral is defined, and converges uniformly for x > Gamma for any Γ () is exended for any by n! n Γ () = lim n ( + )( + )...( + n ) 6

17 Gauge Insiue Journal, Volume 5, No 4, November 009 = n lim ( + )( + / )...( + / n ) n This is analyic for any excep for he poles a = 0,,, Zea Inegral in x > For x >, ζ() Riemann wroe: and = = e = 0 d = = 0 = () e = Γ n= naural n / p= prime p = 0 e d d converge only when Re >. However, i is easy o find for ζ () an expression ha holds for any. Firs of all, by applying he equaion = = 0 n e d = Γ() n 7

18 Gauge Insiue Journal, Volume 5, No 4, November 009 Proof: we ge = Γ ()() ζ = d. e = 0 By 3., for x > 0, Euler's Gamma funcion is By he change of variable u Tha is, for n =,, 3,... N, Γ () = u= u= 0 = n, Γ () = n u e u du = = 0 = 0 e n d = Γ () = d, e = Γ () = d, e = 0..., = Γ () = d. N N e = 0 Therefore, summing his equaliies, for x > 0, =.. N N e e e = 0. Γ () = + +. d 8

19 Gauge Insiue Journal, Volume 5, No 4, November 009 Since converges only for x >, hen, leing N, we n n= obain for x >, = N N e e e = 0. Γ ( ) = lim d Uniform convergence of... e + e + = e for δ > 0, allows bringing he limi in, and summing under he inegral sign. Thus, for x >, The inegral, can be wrien as For he second inegral = Γ ( ) = d. e = 0 = e = 0+ d = = d + d e. e = 0+ = 9

20 Gauge Insiue Journal, Volume 5, No 4, November 009 = = x = x d = d d / / / e ( e )( e + ) e = = = The las inegral converges similarly o = d, in 3.. e = For he firs inegral, = = x d = d e e = 0+ = 0+ = = x d e e. = 0+ The inegral diverges in 0 < x, and converges only in x >. Therefore, for x >, = ζ() = d Γ () e = 0. 0

21 Gauge Insiue Journal, Volume 5, No 4, November Zea for any 4. The Pah Inegral Formula for Zea λ= + i0 λ= i0 ( λ) ζ() = i dλ sin πγ( ) λ e he pah sars a i0, encircles = 0, and reurns o + i0. Riemann wroe...consider he inegral + i0 i0 ( λ) dλ e λ along a closed pah from λ = i0 o λ = + i0

22 Gauge Insiue Journal, Volume 5, No 4, November 009 clockwise around a domain ha conains he singulariy a = 0, bu none of he singulariies a = πin, for n =,,3... To define he muli-valued funcion we choose he branch of ha is real for The inegral equals ( )log( ) ( ) e λ λ =, log( λ) λ < 0. = = 0 iπ iπ ( ) ( e e ) d, e where he inegraion pah is along he real axis. Thus, we obain + i0 i0 ( λ) sin πγ ()() ζ = i dλ. e λ PROOF:

23 Gauge Insiue Journal, Volume 5, No 4, November The Inegraion Pah To evaluae he inegral we choose a pah ha λ= + iδ ( λ) dλ, λ e λ= iδ runs from λ = iδ o λ = δ iδ, along λ = iδ. runs from λ = δ iδ o λ = δ + iδ, encircling = 0 along iθ iθ λ = δ e εe. runs from λ = δ + iδ o λ = + iδ, along λ = + iδ. Then, he inegral equals λ = δ iδ λ = δ + iδ λ = + iδ ( λ) ( λ) ( λ) dλ+ dλ+ dλ λ λ λ e e e. λ= iδ λ= δ iδ λ= δ+ iδ To say in he same branch of log( λ), we should no cross he cu along he posiive x axis. 3

24 Gauge Insiue Journal, Volume 5, No 4, November For δ 0, he Firs Inegral e = iπ e = 0+ d Proof: For he firs inegral, we roae [ iδ] clockwise, muliplying i by o obain Tha is, Therefore, e i π [ iδ]. i λ = [ iδ] = [ iδ] e π. iπ iπ ( λ) = ([ iδ] e ) = e [ iδ] and we have λ = δ iδ ( λ) = iπ ( iδ) λ = λ iδ e e λ= iδ = δ d e d. Since cosδ, we have, e e iδ e + i δ + iδ iδ + ( e )( e ) ( e ) e iδ e 4

25 Gauge Insiue Journal, Volume 5, No 4, November 009 δ ( i ) iδ e e Also, for δ 0, δ ( i ) iδ e e. Therefore, by Lebesgue Dominan Convergence, as δ 0, firs inegral e = iπ e = 0+ d. 4.4 For δ 0, and x >, he Second Inegral 0 Proof: In he second inegral, and λ = δ ε λ= δ+ iδ θ= π x ( λ) ε dλ iθ exp( εe ) εdθ. λ e λ= δ iδ θ= 0 To apply Lebesgue Dominan Convergence, we need o confirm ha as ε 0, x ε i exp( εe θ ) 0. We will avoid differeniaing square roos by showing ha 5

26 Gauge Insiue Journal, Volume 5, No 4, November 009 Now, x ε i exp( εe θ ) ( i θ iθ )( ) iθ εe εe exp( εe ) = e e 0. ( ) ε cos θ cos i sin i sin ε θ ε θ ε θ = e + e e + e cos cos ε θ ε θ = e + e cos( εsin θ) 0, as ε 0. d exp ( e iθ ε ) = dε d = e + e dε εcosθ cos ( ε θ cos( ε sin θ ) ) εcosθ εcosθ ( ) = cos θe e cos θ cos( εsin θ) sin θ sin( εsin θ) εcosθ εcosθ ( ) = cos θe e cos( θ + εsin θ) cos ( e ) εcos θ ε θ θ θ ε θ = e cos cos( + sin ) 0, as ε 0. d dε iθ exp( εe ) = d e cos e cos ( ) ε θ ε θ = cos θ cos( θ + εsin θ) dε εcos θ ε θ cos ( e ) = e cos θ cos θ cos( θ + εsin θ) + 6

27 Gauge Insiue Journal, Volume 5, No 4, November 009 cos ( e ) εcos θ ε θ θ θ θ ε θ θ + e cos cos + sin( + sin )sin as ε 0. Therefore, by L'Hospial, for x >, and for ε 0, x x ε Dεε lim = lim ε 0 0 iθ ε iθ exp( εe ) Dε exp( εe ) = lim ε 0 Dε D x εε iθ exp( εe ) = lim ε 0 Dε ( x x ) ε iθ x exp( εe ) ( 0) =, as ε 0. ( ) Hence, as ε 0, and for x >, x ε i exp( εe θ ) 0. Therefore, by Lebesgue Dominan Convergence, for δ 0, and for x >, Second Inegral For δ 0 Proof:, he Third Inegral e = iπ e = 0+ d 7

28 Gauge Insiue Journal, Volume 5, No 4, November 009 For he hird inegral we roae [ + iδ] couner-clockwise, muliplying i by o obain Hence, i e π [ + iδ]. λ = [ + iδ] = [ + iδ] e iπ and we have iπ iπ ( λ) = ([ + iδ] e ) = e [ + iδ] λ = + iδ ( λ) = iπ ( + iδ) λ = λ + iδ e e λ= δ+ iδ = δ d e d Thus, by Lebesgue Dominan Convergence, as δ 0, Third Inegral e = iπ e = 0+ d. 4.6 The Pah-Inegral Formula 4. Proof: For δ 0, and for x >, we obain + i0 ( λ) = iπ iπ λ = ( ) λ e e i0 = 0 d e e d = = sin( i π) d e = 0 8

29 Gauge Insiue Journal, Volume 5, No 4, November 009 By 3., for x >, = = 0 d =Γ()() ζ e Hence, + i0 i0 ( λ) e λ dλ = sin( i π)()() Γ ζ Therefore, λ= + i0 λ= i0 ( λ) ζ() = i dλ sin πγ( ). λ e The righ hand side is defined for any. Since i equals Zea in he half plane x >, i is he analyic coninuaion of Zea o he complex plane for any. 4.7 A Second pah Inegral Formula for Zea Proof: + i0 i 0 Γ( ) ( λ) ζ() = dλ. πi e λ By 4., + i0 i0 ( λ) ζ() = dλ sin( i π)() Γ. e λ 9

30 Gauge Insiue Journal, Volume 5, No 4, November 009 Subsiuing π Γ( ) Γ ( ) =, sin( π) we obain + i0 i0 Γ( ) ( λ) ζ() = dλ. πi e λ 4.8 Zea is finie for all Riemann wroe Proof: For any, ζ() is a single-valued, and finie funcion. Γ( ) has simple poles a =,, 3..., bu only = is a pole of , or 3 + i0 i0 ( λ) i dλ sin( π) Γ( ). e λ I follows ha he inegral + i0 i0 ( λ) dλ e λ has eros ha cancel he poles a =, 3..., Zea has only one pole a =, and is finie for all. 30

31 Gauge Insiue Journal, Volume 5, No 4, November Zea has eros a = n for n =,,3... Proof: The Bernoulli numbers are defined by B n e = n= 0 B n n! n ha converges in < π. We have and ( ) ( ) =, n + ζ n B n + n B n + = 0 Thus, Zea has eros a for n =,,3... = n 3

32 Gauge Insiue Journal, Volume 5, No 4, November The Zea Funcional Equaion 5. sin πγ ()() ζ = ( π)[( i) + i ]{ } Riemann wroe 3 If x < 0, we inegrae along a clockwise-oriened pah around he complemenary domain in he complex plane. Then, he inegral is infiniesimal, because i is over values ha have infiniely large modulus. In he complemenary domain, he inegrand has singulariies only a λ =± πin, for n =,,3... Therefore, he inegral equals he sum of he clockwiseoriened pah- inegrals around hese singulariies. Since he clockwise-oriened pah-inegral around a singulariy = πin is we have ( πi)( πin), sin ()() ( ) [( ) π ζ π n i i Γ = + ]. n 3

33 Gauge Insiue Journal, Volume 5, No 4, November 009 Proof: By 4., we have + i0 i0 ( λ) sin πγ ()() ζ = i dλ e λ We will evaluae he inegral along a clockwise-oriened pah around he complemenary domain. By he Residue Theorem, he clockwise-oriened inegral equals e λ λ πin = ε ( λ) dλ ( λ) ( λ) ( πi)res = πi lim ( λ πin) λ e By L'Hospial, λ πin λ λ= πin e ( λ πin) lim = lim =. λ e e λ πin λ λ πin Hence, around he singulariy he inegral equals λ = πin, λ πin ( πi) lim ( λ) = ( πi)( πin). Similarly, around he singulariy 33

34 Gauge Insiue Journal, Volume 5, No 4, November 009 λ = πin, he inegral equals ( πi)( πin). The inegral around he complemenary domain is he sum of he inegrals around all he singulariies and i equals Thus, n= = ( πi) [( πin) + ( πin) ] = i( π) [( i) + ( i) ] n n = { } 3 sin πγ ( ) ζ( ) = ( π) [( i) + i ] Zea Funcional Equaion / Define η() π Γ ( /)() ζ. Then η() = η( ) for any. The equaion η() = η( ) is solved also by ( ). Riemann wroe 34

35 Gauge Insiue Journal, Volume 5, No 4, November 009 Proof: By using known properies of Γ () we obain he following relaion beween ζ () and ζ( ) By secion 5., Pu and π / Γ ( /) ζ( ) is unchanged if replaces. { } 3 sin πγ ( ) ζ( ) = ( π) [( i) + i ] ζ = ( ), 3 [( i) + i ] = i( i) i( i) log( i ) log i = ie [ e ] ( i π/) ( i π/) π = ie [ e ] = sin Then, Subsiue and sin πγ ()() ζ = π sin( π /)( ζ ). π sin( π / ) =, Γ ( /) Γ ( /) π sin( π) Γ ( ) =. Γ ( ) Then, 35

36 Gauge Insiue Journal, Volume 5, No 4, November 009 ζ() = π ζ( ). Γ( ) Γ( / ) Γ( / ) Tha is, / ( )/ / π Γ ( /) ζ( ) = π π Γ( ) ζ( ). Γ( / ) By [Mag, p. 3] Hence, / () π + Γ = Γ()( Γ ). / π Γ( ) = Γ([ ] / ) Γ( / ). Namely, π Γ ( ) ([ ]/) Γ( / ) = Γ. / Therefore, / ( )/ π Γ ( /) ζ( ) = π Γ([ ]/) ζ( ). Tha is, η() = η( ). 5.3 Funcional equaion wih respec o =. For x = α, η( α + iy) = η( + α iy) 36

37 Gauge Insiue Journal, Volume 5, No 4, November 009 Proof: Make he x -ranslaion Then, x = α. = x + iy = α + iy, = x iy = + α iy. The funcional equaion for ζ () is The poins η( α + iy) = η( + α iy). and are symmeric wih respec o α + iy, + α iy, =. 37

38 Gauge Insiue Journal, Volume 5, No 4, November Poisson Ellipic Funcion ψ (). 6. Poisson Ellipic Funcion For δ > 0, he funcions ψn ( ) π , e π 3 π N π e e e converge uniformly o he Poisson Ellipic Funcion ψ( ) π e π 3 π e e Riemann cies Jacobi s reaise on ellipic funcions [5, p. 84], as his source for he funcional equaion for ψ (). Bu in [Jaco, p. 60], Jacobi aribues he formula o Poisson. 6. Poisson Funcional Equaion / / ψ() = ψ(/ ) + Proof: Fix u, and inegrae he funcion 38

39 Gauge Insiue Journal, Volume 5, No 4, November 009 πθ ( iu) πθ πθ i u π u e = e e e in he ani-clockwise direcion, along he recangle wih verices (,0), (, iu), (, iu), (,0). Two of hese inegrals vanish: θ= + iu θ= + iu πθ πiu θ πu πθ πiu θ πu e e e dθ e e e dθ θ= θ= and π πiu θ πu = e e e u = 0 0. θ= θ= + iu θ= πθ πiu θ πu πθ πiu θ πu e e e dθ e e e dθ θ= + iu π( ) πiu θ πu = e e e u = 0. 0 By Cauchy s Theorem, πθ πiθu πu 0 e e e dθ = 39

40 Gauge Insiue Journal, Volume 5, No 4, November 009 θ= + iu θ= πu πθ πiθu π(0) πθ πiθ(0) = e e e dθ + e e e dθ Therefore, Tha is, θ= + iu θ= θ= + iu θ= πu πθ πiθu πθ θ = θ =. e e e d e d θ= + iu θ= θ= + iu πu πθ πiθu e = e e dθ θ= + iu θ= πθ πiu θ = e e dθ. θ= For m =..., 3,,, 0,,, 3,..., subsiue and obain he equaions The change of variable u = m, θ= m π πθ πim θ e = e e dθ. θ= gives ω = θ 40

41 Gauge Insiue Journal, Volume 5, No 4, November 009 ω= m π ω π/ πimω e = e e dω. ω= Thus, for m =..., 3,,, 0,,, 3,..., e e e.... ω= e πi( 3) ω d = ω ( 3) π ω π/ ω= e ω= e πi( ) ω d = ω ( ) π ω π/ ω= e ω= e πi( ) ω d = ω ( ) π ω π/ ω= e e ω= e πi(0) ω = d ω (0) π ω π/ ω= e ω= e πi() ω = d ω e π ω π/ ω= e ω= e πi() ω = d ω e π ω π/ ω= e ω= e πi(3) ω = d ω. e 3 π ω π/ ω= e. 4

42 Gauge Insiue Journal, Volume 5, No 4, November 009 Summaion of boh sides gives + π = e π 3 π e e ω= ω= πi( 3) ω πi( ) ω ω ωπ/ ωπ/ ω= ω= =... + e d + e dω e e ω= ω= πi( ) ω πi(0) ω ω ωπ/ ωπ/ ω= e ω= e + e d ++ e dω ω= ω= πi() ω πi() ω. + e dω + e dω +... ωπ/ ωπ/ ω= e ω= e By Poisson Summaion Formula [Spieg, p.09], ω= ω= πi( 3) ω πi( ) ω... + F( ω) e dω + F( ω) e dω ω= ω= ω= ω= πi( ) ω πi(0) ω + F( ω) e dω + F( ω) e dω ω= ω= ω= ω= πi() ω πi() ω + F( ω) e dω + F( ω) e dω +... ω= ω= =... + F( 3) + F( ) + F( ) + F(0) + F() + F() + F(3) +... Applying o 4

43 Gauge Insiue Journal, Volume 5, No 4, November 009 we have π e π 3 π e e F( ω ) =, / e ωπ = e e e ( 3) π/ ( ) π/ ( ) π/ (0) π/ () π/ () π/ (3) π/ e e e e = π/. e π/ 3 π/ e e Tha is, + ψ( ) = ( + ψ(/ ) )

44 Gauge Insiue Journal, Volume 5, No 4, November Riemann's Formula for Zea 7. Riemann s Formula for Zea For any, = ( + )/ / η() = + ψ()( + ) d ( ) = where ψ () is Poisson Ellipic Funcion, and / η() = π Γ ( /)() ζ. Riemann wroe By using known properies of Γ () we obain he following relaion beween ζ (),and ζ( ) π / Γ ( /) ζ( ) is unchanged if replaces. This propery of Zea le me subsiue Γ ( /) insead of Γ (), ino he general erm of / n. This subsiuion gives a very convenien formula for ζ () We have 44

45 Gauge Insiue Journal, Volume 5, No 4, November 009 If we se we ge Since π = / n π / Γ ( /) = n = 0 e d n π e = ψ(), n= = / / π Γ ( /) ζ( ) = ψ( ) d. = 0 / ψ( ) + = [ ψ(/ ) + ] (Jacobi, Fund., p.84), i follows ha = / / π Γ ( /) ζ( ) = ψ( ) d = = + = 0 ψ( / ) = = 0 ( 3)/ d ( 3)/ / + ( ) d = / ( + )/ ψ()( ) d. = = + + ( ) Proof: Riemann generaed he parial sums of he Poisson Ellipic Funcion from Euler s Gamma funcion wrien for 45

46 Gauge Insiue Journal, Volume 5, No 4, November 009 u= Γ ( ) = u u du, for x > 0. e u= 0 The change of variable gives for x > 0, u n π =, Γ ( ) = π n d. Thus, for n,, 3,... N = n 0 e π = =, we wrie he N equaliies = Γ ( ) = d, π e π = 0 = 0 e π = Γ ( ) = d π, = 3 = 0 e Γ ( ) = d 3 π, π., = ( ) N N = 0 e π Γ = d. π 46

47 Gauge Insiue Journal, Volume 5, No 4, November 009 Summing he N equaliies, we have for x > 0, ( ) ( )... N π Γ = = e e e e π... d π 3 π N π. = 0 Leing N, he series converges only in he half plane x >. Thus, only for x >, Γ ( )... lim = + + d 3 N π e N π π. = 0 e ζ() η() = Now, he uniform convergence of he Poisson Ellipic for δ > 0, allows us o ake he limi under he inegral sign, = lim π e π N π = + e e N 0 = N = 0+ = lim π e π N π e e d d 47

48 Gauge Insiue Journal, Volume 5, No 4, November 009 = = π e π 3 π e e = 0+ d = = 0 = ψ() d, where ψ () is he Poisson Ellipic Funcion. We aim o remove he singulariy a = 0, from under he inegral sign. To ha end, we wrie he las inegral as he sum of wo inegrals = = = ψ() d= ψ() d+ ψ() d. = 0 = 0 = In = = 0, we replace ψ () wih / / ψ(/ ) +. = = / ( / ) ψ( / ) [ ] ψ( ) = + d+ d = 0 = ( ) = = = 3 3 = ψ( / ) d+ d+ ψ( ) d = 0 = 0 = = d wih 3 Replacing ( ) = 0 = ( ), 48

49 Gauge Insiue Journal, Volume 5, No 4, November 009 = = 3 = + ψ( / ) d + ψ( ) d ( ). = 0 = The change of variable τ = /, in = = 0 ψ(/ ) 3 d, gives τ= τ= + 3 ( ) + ψττ ( ) ( dτ/ τ ) ψττ ( ) = dτ. τ= τ= Therefore, If x > 0, = = ( ) ( ) ( + ) ψ() d ψ() = + + d. = 0 = ( + ) π ( ) π ( ) ψ() + = ( e + e +...) +. If x < 0, he change of noaion x = u, y = υ, s = u + iυ, gives he same inegrand ( + ) π ( ) π s s ( ) ψ( ) + = ( e + e +...) +. Therefore, he inegral 49

50 Gauge Insiue Journal, Volume 5, No 4, November 009 = = ( ( + ) + ) ψ() d is an enire funcion, and he funcion = ( ) ( + ) μ() + ψ() + d ( ) = is analyic in less a pole a =, saisfies he funcional equaion = ( ) ( + ) μ( ) = + ψ( ) + d ( )( ) equals η () in x >. = = μ(). Thus, μ () is he analyic coninuaion of η () in wihou =. 50

51 Gauge Insiue Journal, Volume 5, No 4, November The funcion ξ () 8. Definiion of ξ () ξ() ( )() η ( ) π ( ) ζ( ) = Γ 8. Formulas for ξ () = = ( ( + ) ) ξ( ) = + ( ) ψ( ) + d ξ (0) = 8.3 Riemann s Firs Formula for ξ (). = 3 4 ψ = ( ) ξ() = ( ) () cos [( i )]log d Proof: By 7., we have 5

52 Gauge Insiue Journal, Volume 5, No 4, November 009 = ( + )/ / ξ() = ( ) ψ()( ) d + + ( ) = = = + ( ) ( ) ξ() = ψ() + d = ( ) 3 i ( ) ( ) ( i[ ] i i[ ] = ψ() 4 + ) d = = ( ) 3 i ( ) ( ) ( i[( ] log i i[ ] log = ψ() 4 e + e ) d = = 3 4 ψ = ( ) = ( ) ( ) cos i[ ]log d. 8.4 Riemann s Second Formula for ξ (). Proof: = = 3 ψ 4 ( ) ξ() = 4 ( ()) cos [( i )]log d = = + ( ) ( ) ξ() = ψ() + d 5

53 Gauge Insiue Journal, Volume 5, No 4, November 009 = ( ) = ψ() d + = = = ( ) ( ) = ψ() dψ() = = = ( ) ψ ( ) = + + ( ) ( ) + + ( ) ψ ( ) d = = = = ( ) 3 = + ψ() + + ( ) ψ ( ) d = = 3 ( ) = + ψ() + ψ ( ) d = = ( ) ( ) ( ) 3 3 = + ψ() + ψ ( )( ) ( ) d + + ψ ( ) = = = = i i[ ] i i[ ] 3 ( ) ( ) = + ψ() + 4 ψ () d ψ ( ) = ( ) ( ) ( 3 i i[ ] log i i[ ] log = + ψ() + 4 ψ () + 4 e e + ) d( ψ ( ) ) = = 4 = 3 ( ) ( ) = + ψ() + 4 ψ () + 4 cos i[ ]log d ψ ( ) To evaluae ψ (), differeniae Poisson Funcional Equaion 53

54 Gauge Insiue Journal, Volume 5, No 4, November 009 / / ψ() = ψ(/ ) +. ψ ( ) = ψ ( / )( / ) + ψ( / )( ) 4 ψ () = ψ ()( ) + ψ()( ) 4 / 3/ 3/ Thus, = = 4 ψ () + ψ() + = 0. 3 ( ψ ) ξ() = 4 4 cos i[ ]log d () = = ( ) 3 4 = 4 cos i[ ]log ( ψ ( ))' d. 54

55 Gauge Insiue Journal, Volume 5, No 4, November The Funcion ξ ( + iw) We use ξ( + iw) ξ( ) o ge a correc facoriaion for ξ. 9. Formula for ξ ( + iw) = ( 3 ) + = ψ 4 = ξ( iw) 4 ( ) cos( w log ) d Riemann wroe Se and Then, or also = + iw ( ) π Γ ( ) ζ( ) = ξ( ) w = 3 ( ) ψ ( ) 4 ξ( w) = + w ( ) cos wlog d 4 = = = 3 d ( ψ ( )) ξ( w) = 4 4 cos( wlog ) d. d 55

56 Gauge Insiue Journal, Volume 5, No 4, November 009 Proof: Subsiuing w = i( ) = y + i( x) ino Riemann s Firs Formula of 8.3 = ( ) 3 ψ 4, ξ() = ( ) () cos [( i )]log d = we obain Riemann s Firs Formula here = 3 ( ) ψ ( ) 4 ξ( + iw) = + w ( ) cos w log d. 4 = Subsiuing w ino Riemann s Second Formula of 8.4 we obain = = 3 ( ψ 4 ) ( ) ξ() = 4 () cos [( i )]log d, 3 ( ψ ) = 4 = ξ( + iw) = 4 ( ) cos( w log ) d. 9. Taylor Series for ξ ( + iw) ξ ( + iw) = A A w A w where A n > 0. = A + A( ) + A ( )

57 Gauge Insiue Journal, Volume 5, No 4, November 009 Riemann wroe ξ ( w) is finie for all finie values of w, and can be expanded ino a very rapidly convergen series in powers of w. Proof: By he second formula for ξ ( w), 3 ( ψ ) = + = 4 = ξ( iw) 4 ( ) cos( w log ) d = = 3 ( ψ ) = 4 ( ) ( log ) w ( log ) w... + d! 4! where, = A Aw + Aw = = 3 ( ψ ) n A 4 ( ) 4 n = log d. ( n)! To obain ξ (), Subsiue w = i( ). Then, ξ ( ) = A A ([ i ]) + A ([ i ]) A ([ i ]) = A + A( ) + A ( ) + A ( ) +... The A n > 0 because 57

58 Gauge Insiue Journal, Volume 5, No 4, November ( 3 π π π ( ψ ( )) = e + e + e +...) 3 ( ( )) π π 3 ( ) 3 π π e + e e... = ( ) 3 ( ) π π e π e = ( π) π 3 π 3 ( e +...) = π [ π ] + e [ π ] > 0. Hadamard [Hada] proved ha he rapid convergence is equivalen o he facoriaion ξ() = ξ(0). ρ ρ= ero of ξ 58

59 Gauge Insiue Journal, Volume 5, No 4, November The Zeros of ξ () 0. The eros of ξ () Hadamard denoed he eros of by and ξ () ρ = x + iy, ρ ρ = x iy. ρ ρ ρ Since = + iw, he corresponding eros of are We have ξ ( + iw) α, and α. ρ = + iα and ρ = iα We shall use 4 4 ( ) = = ( ) = ( w + ) 4 ρ( ρ) = α + 59

60 Gauge Insiue Journal, Volume 5, No 4, November All he eros of ξ () are in 0 < x < Riemann wroe Proof: Since for = x + iy, wih x > is finie, log ζ ( ) = log( / p ) and since he same is rue for he logarihms of he oher facors of he funcion will vanish only if For x >, he Euler produc ξ ( w), ξ ( w) < Im( w) <. p= prime p = ζ() has no vanishing facor, and is non-ero. Namely, for x >, and ζ() 0. / ξ() = ( ) π Γ( /)() ζ 0 60

61 Gauge Insiue Journal, Volume 5, No 4, November 009 Thus, if ρ is a ero of ξ () Re( ρ ) <. Since ξ() = ξ( ), we have Tha is, ξρ ( ) = 0 ξ( ρ) = 0. if ρ is a ero of ξ () so is ρ Hence, Therefore, Thus, > Re( ρ) = Re( ρ) 0 < Re( ρ) Namely, in or, since x + iy = + iw, All he eros of ξ () are in 0 < x <. < x < < Im( w) <. 0.3 For R large enough, log ξ( ) RlogR Proof: By 9., in w = R 6

62 Gauge Insiue Journal, Volume 5, No 4, November ξ ( ) = A + A ( ) + A ( ) + A ( ) +... where A n > 0. Therefore, if R, and R is large, Then, and ξ() ξ( n) R + = n n = (n ) π Γ ( n + ) ζ( n) n nπ n! ζ() n n + Hence, for large enough R and ξ() R R log ξ( ) RlogR. This enabled Hadamard o obain his esimae for he number of eros of ξ (). 0.4 Hadamard Esimae nr= ( ) number of eros of ξ () in If R is large enough, hen nr ( ) RlogR R 6

63 Gauge Insiue Journal, Volume 5, No 4, November 009 Proof: By Jensen's Theorem [Ed, p. 40], applied o w = R n( R) θ= π ( R) iθ log ξ( / ) = log f ( Re ) dθ ( ρ /)...( ρ /) π Subsiue n( R) θ= 0 iθ log ξ( Re ) RlogR ρ k / = α k n( R) n( R) n( R) n( R) ( R) R ( R) log = nr ( )log + log + log α... α α... α α... α n( R) nr ( ) nr ( ) + n( R) Since for k =... n( R), and α R k α + nr ( ) k R we have Therefore, n( R) ( R) nr ( )log log α... α. n( R) Hence, for R large enough, log ξ ( / ) + nr ( )log RlogR. nr ( ) RlogR. 63

64 Gauge Insiue Journal, Volume 5, No 4, November 009 Hadamard esimae is sufficien for he derivaion of he facoriaion for ξ (). 0.5 Riemann Esimae: ny ( ) = number of eros of ξ () in 0 < y < Y, 0 < x <. Y Y Y ny ( ) log. π π π Riemann wroe Consider he couner-clockwise closed-pah inegral around (0,) (0, Y ) dlog ξ( + iw) Proof: around he domain wih 0 < x <, and 0 < y < Y. Wih relaive error of he order of /Y, he inegral is equal o Y iy log π On he oher hand, he inegral equals π i ny ( ). In 94, Backlund [Back] gave a proof. 64

65 Gauge Insiue Journal, Volume 5, No 4, November The Hypohesis All he eros of ξ () are on x = Equivalenly, All he eros of ξ ( + iw) have Im( w ) = 0. Riemann wroe I is very likely ha all of he eros of ξ ( w) are real. One would like o have a rigorous proof of his, bu afer several fleeing aemps o no avail, I have emporarily se aside he search for his proof because i appeared o be unnecessary for he immediae purpose of my invesigaion. The Hypohesis is required in he applicaion of he facoriaion o he coun of he primes, and Riemann's erroneous facoriaion did no indicae he need for he Hypohesis. While he correc facoriaion is needed for a derivaion, he formula for he coun of he primes may be arrived a empirically, and Riemann realied laer ha he needed he Hypohesis for a derivaion, and ried o prove i. Afer he publicaion of he 859 paper, Riemann wroe 65

66 Gauge Insiue Journal, Volume 5, No 4, November The Theorem which I merely cied, ha beween 0, and Y, here are around Y Y log π π real eros of he funcion ξ, follows from a new developmen of ξ, which I had no simplified enough o repor i... Apparenly, Riemann had no proof for he Hypohesis. 66

67 Gauge Insiue Journal, Volume 5, No 4, November 009 Facoriaion of ξ. Weiersrass Facoriaion of ξ () where h ( ) ξ() = e e ρ ρ Q ( )( ) n n= n n u he eros of ξ (), he ρ ' s are sequenced by heir sie and increase o n ( ) u he polynomials Q () guaranee he uniform convergence of he produc in he open plane u h () is an enire funcion n Proof: Since ξ () is an enire funcion so ha ξ(0) 0, by Weiersrass [Sak, Chaper 7,.3] h ( ) ξ() = e e ρ n = Q ( ) n n () Since he ρ n ' s are sequenced by sie, and since ρ n is a ero oo, he produc represenaion is 67

68 Gauge Insiue Journal, Volume 5, No 4, November 009 h ( ) ξ() = e e ρ ρ Q ( )( ) n n = n n ( ). ( )( ). n= n n h ( ). ξ() = e ρ ρ Proof: To obain Q ( ) e n =, we show ha ( ) = ρ ρ ρ ( ρ ) n= n n n n n converges. Since he convergence of ( ) ρ ( ρ ) m m m is equivalen o he convergence of and since ρ ( ρ ), m m m m m m m ρ ( ρ ) = ( ρ / ) / 4 > ρ /, i is sufficien o show ha = <. m ρ / m α m m 68

69 Gauge Insiue Journal, Volume 5, No 4, November 009 Hence, we need o show ha <. m> N αm The α m are arranged so ha is decreasing. For m large enough, α m m = N, N +, N +,... define posiive numbers R m > so ha Then, and logr m > m = 4 R log R. m m logm > logr m. By Hadamard Esimae of 0.3, he number of eros of ξ () in w R m is bounded by R log R. Hence, and we have m m m> N α m> N R m m m α > Rm = 4 (log ) R m m> N m 69

70 Gauge Insiue Journal, Volume 5, No 4, November 009 ( logm Since ) / m 4 m> N = 4 m> N ( logm ) m m ( logm ) 3/ / m 0, as m, we have and Therefore, and ( logm ) m / <, for m > N <. m> N α m ρ ρ converges, n= n n Q ( ) e n =, h ( ) ξ() = e ρ ρ ( )( ) n= n n..3 Facoriaion of ξ ( + iw), and ξ () + ξ( + = w 4 iw) ξ(0) n= αn

71 Gauge Insiue Journal, Volume 5, No 4, November 009 ξ() = ξ(0) n α + n 4 Proof: Firs, he facoriaion facors are ( ) w + 4 ( ) n n n n n 4 = = ρ ρ ρ ρ α +. ( )( ) h ( ) Second, we show ha e = ξ(0). The Hadamard Facoriaion Theorem [Holl, p. 68] applies o a funcion f () for which lim sup log log max ( ) logr f R = R < is called he order of f (). Hadamard replaced Weiersrass wih a polynomial so ha h () Q () deg Q ( ) By 0., if R is large enough, log ξ( ) RlogR in w R. Hence, log log max ξ( ) log R log R logr w = R logr ( ) 7

72 Gauge Insiue Journal, Volume 5, No 4, November 009 = + log logr logr log logr logrr Since by L'Hospial lim = lim = 0, R logr R R lim sup log log max ξ( ) =. logr w = R R Tha is, Hence, and Therefore, ξ() is of order = deg Q ( ). Qw ( ) = A+ Bw A+ Bw w + /4 ξ( / + iw) = e, α + /4 n n where he produc is an even funcion of w. Indeed, by 9., is an even funcion of w. 4 0 ξ ( + iw ) = A A w + A w +... Consequenly, and B = 0, /4 ξ( ) + + /4. A w + iw = e n αn 7

73 Gauge Insiue Journal, Volume 5, No 4, November 009 Seing w = i ξ (0) = e A, and + /4 ξ( ) (0) + /4 w + iw = ξ n αn ξ() = ξ(0) ρ ρ n n n = ξ(0) + n α n Riemann s Facoriaion Error Riemann wroe Since he densiy of he eros of sie w increases like he series log w π, α= ero of ξ( w) log( w ) α converges, and grows like w log w. log ξ( w) log( w ) α= ero of ξ( w) α 73

74 Gauge Insiue Journal, Volume 5, No 4, November 009 is a funcion ha is coninuous, and finie for finie w. For w, log ξ ( w ) log( w ) 0 w Therefore, α. α= ero of ξ( w) w cons. log ξ( w) log( ) = Seing w = 0, gives α= ero of ξ( w) α log ξ (0) = cons. Evidenly, w = 0 =, and Riemann should have log ξ ( ) = cons, as observed by Genocchi in 860. Riemann s derivaion does no obain he correc coefficien of he facoriaion which is ξ (0). Now, exponeniaing boh sides, Riemann s facoriaion is w ξ( w) = ξ(0) α α= ero of ξ( w) Riemann s facor, w α, is irreconcilable wih he correc facor w α + /4. + /4 74

75 Gauge Insiue Journal, Volume 5, No 4, November 009. The Number of Primes <. π (), and F (). Riemann wroe...we can now deermine π (), he number of primes less han. Le We define Thus, be equal o and o so ha if F () jumps a, F () π () if is no a prime, π () +, if is a prime. F () = [ F ( + 0) + F ( 0]. π() number of prime numbers p so ha p <. π () = 0 π ( ) = 0 75

76 Gauge Insiue Journal, Volume 5, No 4, November 009 π () = 0 π () e = π (3) = ππ ( ) = To apply Fourier Inegral Theorem, we need he auxiliary funcion π() if prime F () π() + / if = prime ha saisfies he Dirichle condiion Thus, [ F ( 0) F ( 0] F ( ) + + = π() F() < 0 0 = 0 / < < 3 = 3 + / 3 < < 5 = 5 + / 76

77 Gauge Insiue Journal, Volume 5, No 4, November Zea in erms of f () 3. Definiion of f () / /3 f () F () + F ( ) + F ( ) ζ () in erms of f () = log ζ( ) = = + fd () Riemann wroe If x >, log ζ( ) = log p Subsiue p= prime = p= prime p p= prime p p= prime p 77

78 Gauge Insiue Journal, Volume 5, No 4, November 009 p = = = p d + p = = d + = p. Proof: Then, where = log ζ( ) = = + f () d / /3 f () = F () + F ( ) + F ( ) For x >, ζ ( ) = log ζ( ) = log log log =

79 Gauge Insiue Journal, Volume 5, No 4, November 009 = Since The Series 0, if prime df() = F( + ) F( ), if = prime is he Riemann-Sieljes inegral = = df() and we have Similarly, = = df( ). 3 5 = u= = 5 3 u u= df( u) and = = = ( ) df 79

80 Gauge Insiue Journal, Volume 5, No 4, November 009 = = = 3 5 df( 3).. Therefore, = = = / /3 3 = = = log ζ( ) = df ( ) + df ( ) + df ( ) +... Inegraing by pars, = = = / n / n / n ( ) = ( ) ( )( ) = = = df F F d Since F () = 0, and = = 0, Consequenly, Tha is, = / n = F( ) d. + = = = / + + = = log ζ( ) = F( ) d + F( ) d +... = log ζ( ) / /3 d = F ( ) F ( ) F ( ) = +. 80

81 Gauge Insiue Journal, Volume 5, No 4, November f () in erms of Zea 4. () f in erms of ζ () f () y= log ζ( ) = d πi, y= Riemann wroe, If for x >, u= g () = huu () d(log), u= 0 hen by he Fourier Theorem h can be wrien in erms of If hu ( ) is real gx ( + iy) = g( y) + g( y), g (). hen he equaion splis ino 8

82 Gauge Insiue Journal, Volume 5, No 4, November 009 and g () y = h() u u cos( ylog u)(log) d u, u= u= 0 u= u= 0 x x g () y = i h() u u sin( ylog u)(log) d u. Muliply boh equaions by [cos( ylog ) + isin( ylog )] dy, and inegrae from y = o y =. Then, he righ hand side of eiher equaion is x πh (). Adding he equaions, and muliplying by x+ i πih( ) = g( ) d, x i where x is fixed hrough he inegraion. Thus, if h () has a jump a, hen, h () = [ h ( + 0) + h ( 0)]. x i, Since f () has he same propery, we ge wih complee generaliy f () x+ i log ζ( ) d πi =. x i 8

83 Gauge Insiue Journal, Volume 5, No 4, November 009 Proof: By 3., For 0 u we have u= log ζ( ) = f ( udu ). + u u= 0 Fu ( ) = 0, and Hence, f ( u ) = 0. u= log ζ( ) = u u= + f ( udu ) u= = f ( uu ) d(log u). u= 0 u= = u= 0 x iy logu fu ( ) u e d(log u) We apply Fourier Inversion o wrie f in erms of log ζ ( ). To ha end, fix x, muliply boh sides of he equaion above by x iy x iy log d= dx ( + iy) = ie dy and inegrae from y = o y =. Then, 83

84 Gauge Insiue Journal, Volume 5, No 4, November 009 d = i e fuu ( ) e d (log u ) dy y= y= u= log ζ( ) x iy log x iy logu. y= y= u= 0 Since we have F () = [ F ( + 0) + F ( 0)] f () = [ f ( + 0) + f ( 0)] and by Fourier inegral Theorem, we can change he order of inegraion. Therefore, for fixed x. u= y= x x iy(log log u) = i f( u) u e dy d(log u) u= 0 y= u= x x = i f( u) u πδ(log log u) d(log u) = u= 0 πif( ) y= log ζ( ) f () = d πi y= y= log ζ( ) dy = π, y= 84

85 Gauge Insiue Journal, Volume 5, No 4, November f () in erms of ξ () 5. () f in erms of ξ () f () y= ( ) log( ) log Γ / + = d πi + log y= log ξ(0) log + α α + /4 Riemann wroe For we may subsiue x+ i log ζ( ) f () = d πi. x i log ζ ( ) log π log( ) log Γ + ( /) + log + + log ξ(0) α= ero of ξ α 85

86 Gauge Insiue Journal, Volume 5, No 4, November 009 Proof: Bu he inegrals of hese erms are divergen a infiniy. So we have o inegrae he equaion for f by pars f () x+ i d log ζ( ) d πi log d =. x i Since / ξ() = ( ) π Γ ( /)() ζ, we have log ζ( ) = log π log( ) log Γ + + log ξ( ). Replacing ξ () by is facoriaion of.3, we have ( ) log ζ( ) log log( ) Γ + = log π + log ξ(0) + log + + α + /4 If we subsiue his expression for log ζ ( ) f (), he firs erm, gives ha diverges. α x+ i log π, log π log π x iy d πi = 4πilog x i ino 4. o obain y= y= 86

87 Gauge Insiue Journal, Volume 5, No 4, November 009 On he oher hand, inegraing by pars gives Now, y= Y x+ i log ζ( ) log ζ( ) f () = lim d πi logy πi log y= Y x i log ζ( ) = log p= prime p = log p= prime p log p= prime p = p= prime p p p p= prime p p p = x x 3 3x p= prime p p p = log x p= prime p = log x = log p= prime p p= prime x p 87

88 Gauge Insiue Journal, Volume 5, No 4, November 009 = log ζ( x) log ζ( x) Thus, log ζ ( ) is bounded independen of y, and he boundary erm vanishes as Y. We have Therefore, y = Y x log ζ( ) log ζ( x) 0, as Y. x + Y y= Y x+ i log ζ( ) f () = d πi log y= x i = πi log ( ) y= log( ) log Γ + log ξ(0) d log + + α α + /4. 88

89 Gauge Insiue Journal, Volume 5, No 4, November The 6. log ξ(0) erm y= y= d d πi = = log πi y= y= Proof: Inegraion by pars gives y= y= Y y= d lim d πi = + log πi logy πi y= The boundary erm vanishes because y = Y x y= Y y= 0, as Y. x + Y y= Y Subsiuing in he inegral d = idy, x iylog = e, u= = u du, u= 89

90 Gauge Insiue Journal, Volume 5, No 4, November 009 The inegral becomes y= u= x iylog = u du e dy π y= u= By he change of variable u = e ω we have du = e ω dω y= ω= x ω iylog = e dω e dy π y= ω= 0 y= ω= x ωx iωy iylog = e e dω e dy π y= ω= 0 By Fourier Inegral Theorem for x e ω, ω > 0, we change 0, ω < 0 inegraion order, and obain ω= y= x ωx iy(log ω) = e e dy dω π ω= 0 y= ω= x = e δω ( log ) dω ω= 0 x x log = e =. ωx 90

91 Gauge Insiue Journal, Volume 5, No 4, November y= log ξ(0) d log ξ(0) log πi = = log y= Proof: y= y= log ξ(0) d log ξ(0) d πi = log πi log y= y= =, by 6. By 8., ξ (0) =, = = log ξ(0) log = log. 9

92 Gauge Insiue Journal, Volume 5, No 4, November Terms wih log β 7. d( log Γ ( + ) ) = d( log( + ) ) n n= Proof: For he Gamma funcion Hence, ( ) Γ + = lim N ( + ) n= N log ( ) lim log( ) Γ + = logn + N n. n = N n Dividing boh sides by, and differeniaing, N d d log Γ ( + ) ( ) = lim log + n log N d d N n = By he uniform convergence of he series, 9

93 Gauge Insiue Journal, Volume 5, No 4, November 009 N d lim log ( ) = logn N d + n n = N = d lim log d + N n = n= ( ( )) d = ( log( + )). d n n Therefore, d log Γ + = d log + n. n= 7. Le β = σ + iτ be a consan. Then u= β u du + cons, if σ < 0 y= logu u = d log πi = log β y= u= β u + > u= 0 du$ cons, if σ 0 logu Riemann wroe consider 93

94 Gauge Insiue Journal, Volume 5, No 4, November 009 x+ i d log ± d πi log d β x i Now, d log = dβ β ( β ) β If x > σ, u= β u du, σ < 0 x+ i πi ( β ) β = β = Therefore, x i β u= d u= β σ > x+ i u= 0 d log d πi log = d β x i x+ i log d = πi β x i u du, 0 u= β u du + cons, σ < 0 logu u= = u= β u du + cons, σ > 0 logu u= 0. Proof: Inegraion by pars gives 94

95 Gauge Insiue Journal, Volume 5, No 4, November 009 y= d log πi = log β y= y= Y y= = lim log log d πilogy β πi β y= Y y= We firs show ha he boundary erm vanishes. Since 3 log = β β β 3 β we have β β 3 β = log β, x Y log log x Y y= Y x + β β y= Y + and by L Hospial, 0 as Y. Therefore, he boundary erm vanishes, and y= y= d log log d πi = log β πi β y= y= Subsiuing d = idy x iylog = e 95

96 Gauge Insiue Journal, Volume 5, No 4, November 009 log = dβ β, β( β ) we have y= x iylog = dβ e dy π β( β ). y= β y= x iylog e dy dβ π β β β y= = y= s= x ( β) iylog = s ds e dy dβ π β β y= s= By he change of variable s ds = e ω ω = e dω, y= ω= x ω( β) iylog = e dω e dy dβ π β β y= ω= 0 y= ω= x ω( x σ) iω( y τ) iylog = e e dω e dy dβ π β β y= ω= 0 If x > σ, by Fourier Inegral Theorem for ( x e ω σ), ω > 0, 0, ω < 0 we change inegraion order 96

97 Gauge Insiue Journal, Volume 5, No 4, November 009 ω= y= x ω( x σ) iωτ iy(log ω) = e e e dy dω dβ π β β ω= 0 y= δω ( log ) ω= x ω( x σ) iωτ = e e δω ( log ) dω dβ β β ω= 0 x ( x σ)log iτlog β = e e d β β x x log ( σ+ iτ)log = e e dβ β x β β = β d β β β u= β u du dβ σ < β u = = u= β u du dβ σ > β u= u= β u dβ du, σ < 0 u= β = u= β u dβ du, σ > 0 u= 0 β u= β u du + cons, σ < 0 logu u= = u= β u du + cons, σ > 0 log u u= 0 97

98 Gauge Insiue Journal, Volume 5, No 4, November If σ < 0, y= u= β u d log du πi = log β logu y= u= Riemann wroe If σ < 0, x+ i u= β d log u d du cons πi log = + d β logu x i u= The consan of inegraion drops ou by leing β. Proof: By 7., if σ < 0, x+ i u= β d log u d du cons πi log = + d β logu x i u= To esablish ha he consan vanishes, we ll show ha boh inegrals vanish as σ. As in 7., inegraion by pars of he lef-side-inegral gives y= y= d log log d πi = log β πi β y= y= 98

99 Gauge Insiue Journal, Volume 5, No 4, November 009 The inegral is bounded by y= y= x log d log dy πi β π β. y= If we le σ, y= log log = 0. β Therefore, by Lebesgue Dominan Convergence, y= d log 0 πi log β, as σ. y= For he righ-side-inegral, logu 0, in < u <, for >. And If we le σ, u= β u= β u u du du logu. logu u= u= σ β u u u = = 0, logu logu logu Therefore, by Lebesgue Dominan Convergence, u= u= β u du logu 0, as σ. Consequenly, and cons = 0. 99

100 Gauge Insiue Journal, Volume 5, No 4, November 009 y= u= β u d log = du πi log β logu. y= u= 7.4 If σ > 0, y= u= β u d log du πi = log logu β y= u= 0 Riemann wroe If σ > 0, x+ i u= d β log u d= du+ cons πi log d β logu. x i u= 0 The inegral from u = 0 o u = will be infiniesimal, if he pah of inegraion is in he upper half-plane, and we le τ, or if he pah of inegraion is in he lower half-plane, and we le Then, we evaluae τ. log β 00

101 Gauge Insiue Journal, Volume 5, No 4, November 009 on he lef side so ha he inegraion consan drops ou. Proof: By 7., if σ > 0, x+ i u= d β log u d= du+ cons πi log d β logu. x i u= 0 To esablish ha he consan vanishes, we ll show ha boh inegrals vanish as τ, or τ. The lef-side-inegral As in 7.3, inegraion by pars of he lef-side-inegral leads o y= y= x d log log dy πi log β π β. y= y= If we le τ, log log = 0. β Therefore, by Lebesgue Dominan Convergence, 0

102 Gauge Insiue Journal, Volume 5, No 4, November 009 y= d log 0 πi log, as τ. β y= The Righ-Side-Inegral To show ha he righ-side-inegral vanishes oo, we make he change of variable u = e ω Then, du = e ω dω u= β ω= log u log du = u 0 u= ω= ωβ e ω dω And we ake a pah in he upper half-plane ha runs from ω = o ω = + iδ, along ω = + iε runs from ω = + iδ o ω = log + iδ, along ω = υ + iδ runs from ω = log + iδ o ω = log, along ω = log + iε Then, ω = log ωβ ω = + i δ ωβ ω = log+ i δ ωβ ω = log ωβ e e e e dω = dω + dω + dω ω ω ω ω ω= ω= ω= + iδ ω= log+ iδ 0

103 Gauge Insiue Journal, Volume 5, No 4, November 009 The Firs Inegral ω= + iδ ωβ ε= δ e ω ω= ε= 0 ( + iε)( σ+ iτ) dω e dε + iε = ε= δ ε= 0 e ( σ+ ετ) d ε Since σ > 0, and ε > 0, hen for τ, σ ετ ( σ+ ε) e e = Hence, by Lebesgue Dominan Convergence 0. ω= + iδ ω= ωβ e ω dω 0, as τ. The Second Inegral ω= log+ iδ ωβ υ= log e ( υ+ iδ)( σ+ iτ) dω e dυ ω υ + iδ ω= + iδ υ= = υ= log υ= υ + iδ e υσ δτ dυ Since σ > 0, and δ > 0, hen for τ, 03

104 Gauge Insiue Journal, Volume 5, No 4, November 009 υσ δτ υσ δ e e = e = υ + iδ υ + iδ υ + iδ 0. Hence, by Lebesgue Dominan convergence, ω= log+ iδ ω= + iδ ωβ e ω dω 0, as τ. The Third Inegral ω= log ωβ ε= δ e (log + iε)( σ+ iτ) dω log e dε ω + iε log i 0 ω= + δ ε= = ε= δ ε= 0 log + iε e σlog ετ dε Since σ > 0, and ε > 0, hen for τ, exp( σlog ετ) exp( ) = log + iε log + iε 0. Hence, by Lebesgue Dominan convergence, ω= log e dω 0, as τ ω ω= log+ iδ ωβ Thus, he righ-side-inegral vanishes. Similarly, if we ake a pah in he lower half plane and le 04

105 Gauge Insiue Journal, Volume 5, No 4, November 009 τ, he righ side inegral vanishes. Consequenly, if σ > 0, hen and cons = 0, y= u= β u d log = du πi log logu. β y= u= If σ > 0, hen u= u= 0 β u β Li( ) πi, for upper-half-plane pah du = logu β Li( ) + πi, for lower-half-plane pah Riemann wroe If σ > 0, he inegral u= u= β u log du u 0 akes on wo values which differ by π i depending on wheher he pah of inegraion is in he upper-half-plane or in he lower-half-plane. Proof: 05

106 Gauge Insiue Journal, Volume 5, No 4, November 009 The inegrand β u logu is singular a u =, and he pah of inegraion has o bypass he singulariy. Thus, an upper-half-plane pah will run from u = 0 o u = ε encircle he singulariy clockwise from u = ε o u = + ε run from u = + ε o u =. Then, u= β u= ε β ω= log( + ε) ωβ u= β u u e u du = du + dω + du logu logu ω logu u= 0 u= 0 ω= log( ε) u= + ε By he Residue Theorem for he clockwise semi-circle ω= log( + ε) ω= log( ε) ωβ e ωβ π e dω = πi Res ω π ω ω= 0 = πi lim ω e ω 0 ω ωβ Hence, = πi 06

107 Gauge Insiue Journal, Volume 5, No 4, November 009 u= β u= ε β u= β u u u du = du + du + i logu logu logu u= 0 u= 0 u= + ε By he change of variable ( π ) υ = u β β dυ = βu du log υ = βlogu dυ dυ dυ = + πi log υ log υ log υ β β β β υ= υ= ( ε) υ= υ= 0 υ= 0 υ= ( + ε) β Li( ), as ε 0 Leing ε 0, Hence, β = Li( ) πi u= u= 0 β u β du = Li( ) πi. logu Similarly, wih a lower half-plane pah ha encircles he singulariy couner-clockwise, u= u= 0 β u β du = Li( ) + πi. logu 07

108 Gauge Insiue Journal, Volume 5, No 4, November If σ > 0, y= β Li( ) πi, d log β πi = log β Li( ) + πi, y= for upper-half-plane pah for lower-half-plane pah Proof: By 7.4, and

109 Gauge Insiue Journal, Volume 5, No 4, November The ( ) log erm y= d log( ) = Li( ) πi log 8. ( ) Proof: y= log( ) is defined wih a cu along he posiive real numbers. Therefore, o obain in he main branch of log( ), we roae clockwise, by muliplying i by i e π Tha is, iπ ( e ) = and y= y= π y= y= iπ d log( ) d log( e )( ) πi log = i log y= y= π y= y= = d log( ) d ( iπ) πi log + i log 09

110 Gauge Insiue Journal, Volume 5, No 4, November 009 The Firs Inegral y= d πi log y= log( ) has a erm of he form where log β σ = > 0. Therefore, if we ake an upper half-plane pah wih clockwise oriened semicircle around he singulariy of he logarihmic inegral a u =, by 7.6 we have y= d log( ) Li( ) πi πi = log y= The Second Inegral y= y= d ( πi) ( πi) d = π y= y= πi log i log Consequenly, y= = πi =, by 6. d log( ) ( Li( ) πi) πi πi = + log y= = Li( ). 0

111 Gauge Insiue Journal, Volume 5, No 4, November The log + α α erm + 4 Riemann wroe, πi y= y= ( ) d log + α α d = d + log Li( i α iα = ) Li( + ) α The summaion is over all posiive eros of ξ (or all eros wih posiive real par), ordered by heir sie. Wih a more precise discussion of he funcion ξ i is easy o show ha he sum equals α /+ iα / iα Li( ) + Li( ) x+ iy d lim log Y π x iy α ( ) + d i d α If he eros are no sequenced by heir sie, he sum may have any arbirary real value.

112 Gauge Insiue Journal, Volume 5, No 4, November The assumpion of he Hypohesis The claim ha he series equals he inegral α + iα iα Li( ) Li( + ) y= d log πi log + y α α + = 4 includes he assumpion of he Hypohesis ha 0 = Im( α) = x 0 Tha is, he eros are assumed o be on he line x =. 9. Implici Claim The following claim is implici in Riemann s derivaion of his formula for he coun of he prime numbers d log + d log + log = log α α 4 α α + + y= 4 y= y= y=

113 Gauge Insiue Journal, Volume 5, No 4, November 009 In 908, Landau [Land] proved ha he summaion over he eros, and he inegraion can be inerchanged. 9.3 If he eros of ξ are on x = and are sequenced by sie y= + iα iα d log Li( ) Li( ) πi log + = + y α α = + 4 α Proof: Inegraing erm by erm by 9. y= d log + πi log = y α α + = 4 y= = d log + i log α π α + y= 4 y= = d log log i log i α π + + α iα y= The inegral y= d log πi log + iα y= 3

114 Gauge Insiue Journal, Volume 5, No 4, November 009 has a erm of he form where and log = log β σ + iτ, σ = > 0, τ = α > 0. By 7.6, on an upper half-plane pah, he inegral equals + iα Li( ) iπ. Similarly, using a pah in he lower half plane, he inegral y= i d log α Li( ) iπ πi log + + iα y= Therefore, y= πi log + = + iα iα y= + i i d + α α log log Li( ) Li( ) Consequenly, y= + iα iα d log + Li( ) Li( ) πi log = + y α α = + 4 α 9.4 The Hypohesis Origin u α, and α are eros of ξ u α n is increasing 4