About the Riemann Hypothesis

Transcription

1 Journal of Applied Matematics and Pysics, 6, 4, Publised Online Marc 6 in SciRes. ttp:// ttp://dx.doi.org/.46/jamp.6.46 About te Riemann Hypotesis Jinua Fei Cangling Company of Electronic Tecnology, Baoji, Cina Received February 6; accepted 7 Marc 6; publised Marc 6 Copyrigt 6 by autor and Scientific Researc Publising Inc. Tis work is licensed under te Creative Commons Attribution International License (CC BY). ttp://creativecommons.org/licenses/by/4./ Abstract Te Riemann ypotesis is part of Hilbert s eigt problem in David Hilbert s list of unsolved problems. It is also one of te Clay Matematics Institute s Millennium Prize Problems. Some matematicians consider it te most important unresolved problem in pure matematics. Many matematicians made a lot of efforts; tey don t ave to prove te Riemann ypotesis. In tis paper, I use te analytic metods to deny te Riemann Hypotesis; if tere s someting wrong, please criticize and correct me. Keywords Riemann Hypotesis, Disavowal. Introduction Riemann Hypotesis was posed by Riemann in early 5 s of te 9t century in is tesis titled Te Number of Primes Less tan a Given Number. It is one of te unsolved super problems of matematics. Te Riemann Hypotesis is closely related to te well-known Prime Number Teorem. Te Riemann Hypotesis states tat all te nontrivial zeros of te zeta-function lie on te critical line s: Re s =. In tis paper, we use te analytical metods, and refute te Riemann Hypotesis. For convenience, we will abbreviate te Riemann Hypotesis as RH.. Some Teorems in te Classic Teory In tis paper, Γ ( s) is te Euler gamma function, ( s) Lemma.. If Re w >, ten πi ζ is te Riemann zeta function. s Γ s w ds = exp w How to cite tis paper: Fei, J.H. (6) About te Riemann Hypotesis. Journal of Applied Matematics and Pysics, 4, ttp://dx.doi.org/.46/jamp.6.46

2 were Re w is te real part of complex number w. Let η > be given, wen s η and arg s π η, ten If 4 σ 4, t, ten Γ ( s) = log s+ O. Γ s π π π ( it) π t σ σ Γ σ + = exp t it ( log t ) iλ σ O t exp t were λ = if t, λ = if t. See [] page 5, page 55. Lemma.. If Re s >, ten were ( n) Λ is te Mangol function. Let s is any complex number, we ave ζ ζ ( s) = n= Λ ( n) ζ ( s ) c Γ ζ s s s = ρ ρ ρ Γ were ρ be te nontrivial zeros of ζ ( s), c be te positive constant. We write s = σ + it. If σ, π < Im{ log ( s ) } π, { ( s ρ )} were Im s is te imaginary part of complex number s. n s π < Im log π, ten ( s) = ( s ) + ( s ρ) + O( ( t + )) logζ log log log γ t See [] page 4, page, page 8. Lemma.. Let N( T ) is te number of zeros of ( s) = arg ζ + it. π See [] page 98. Lemma.4. Assume tat RH, If x, ten were S ( T ) were R x x log x. See [] page.. Some Preparation Work Lemma.. Assume tat RH, and were Proof. By Lemma. and RH, we ave ζ in te rectangle < σ <, < t < T. ten N ( T 7 ) T log T T π π π 8 S T O = T ψ ( x) = Λ ( n) = x+ R( x) n x < δ, ten 5 δ ( + i ) ( + i ) logζ σ γ dσ and logζ σ γ dσ + δ γ is te ordinate of nontrivial first zero of ( s) ζ, γ

3 because and logζ σ γ log σ γ γ log γ ( + i ) + i i + O γγ log σ log σ ( γ γ) iarg σ + = log σ + log σ + ( γ γ) + 9 log σ log σ + ( γ γ) log + 4 terefore And because terefore Similarly, we ave log σ + ( γ γ) log σ + log σ + log σ +. log σ dσ = log σ dσ + log σ dσ + δ + δ Tis completes te proof of Lemma.. Trougout te paper, we write + δ = log σ dσ + log σ dσ = logσdσ + logσdσ δ O = δ logδ + dσ + O = logζ ( σ + ) dσ log σ + dσ + γγ + δ + δ δ + δ δ log σ dσ +. ( + i ) logζ σ γ dσ. z = a+ i b, a =, T 5, b = π. T 56

4 It is easy to see tat k + b π a arctg,,. k a = = + b 4πT πt k k = ( k+ ) Lemma.. We calculate te tree complex numbers. Because terefore wen t is te real number, we ave b π a + ib = a + b exp i arctg = a + b exp i i a 8 i t π π π it 4 z = a + b exp i i + t t exp t t i t π π π it z = a + b exp i + i + t t exp t t 4 t + it + it + i 4 π π π z = ( a + ib) = ( a + b ) exp i + i t + t exp t + t 4 te tree complex numbers required below. Lemma.. ( ) Proof. By Lemma. and Lemma., we ave ( ) Tis completes te proof of Lemma.. Lemma ζ s Γ ( s) ( s)( a + ib) ds ζ ζ s ζ it Γ ( s) ( s)( a + ib) ds = i it it ( a ib) 4 ζ Γ ζ 4 Proof. By Lemma. and Lemma., we ave it t Γ it ( a ib) + + log π ζ it Γ it it ( a ib) ζ 4 5 ( + ) 4 ( + ) ( ) t log t exp t 5 ( + ) 4 ( + ) t log t. t Γ + ( + ) log d log π it it a ib t T t 4 π i t = π ( a + b ) exp i + i exp( t + it ( log t ) )( a + b ) log 4 π ( ) 4 + O a + b t exp( t) log t π = I ( a + b ) i + i + I 4 4 π exp 564

5 we write ( t) ( t) r = a + b, π r π + t I = exp( t + it ( log t log r ) ) log π exp t = log d exp log log ilog t ilog r π ( it ( t r )) exp t = i log t log π d expit log log t log r re ( t) exp t = i exp( t) + log t log π d expit log γ log t log r re exp t( log t log r) exp t t r t = O + i exp( t) + log exp it log γ log t log r π re exp t( log t log r) exp t t r t = O + i exp( t) + log exp it log γ log t log r π re exp t + γ t ( log t log r) I t exp( t) log t + t exp( t) log t γ Tis completes te proof of Lemma.4. Lemma.5. t log t + log exp t log log T Proof. Wen t γ, by Lemma., we ave By Lemma. and Lemma., we ave it it Γ + it z Γ + it z log z t log T Γ + it Γ + it log + it + Γ + it + it π π π + exp t logt t exp t exp t log t. it it Γ + it z Γ + it z log z t ( ) ( ) + ( ) γ γ t t t + t t T t exp t logt t exp t logt t exp t logt log d log exp d log. 565

6 Tis completes te proof of Lemma.5. Lemma.6. Assume tat RH, ten were S ( t) it Γ + it z S ( t) and Γ + it z S ( t) it γ γ = arg ζ + it. π Proof. By Lemma., it is easy to see tat We write t it 4 π π i Γ + it z =Γ + it a + b exp i + i + t t 4 It is easy to see tat Assume tat RH and t 4 π π π i =Γ + it a + b exp t tcos + + isin =Γ ( + ) H s s a b s π π G ( s) = H ( s) exp i + is + exp i + i( s ) π π G ( s) = H ( s) exp i + is exp i + i( s ) π π G ( s) = H ( s) exp i + is + exp i + i( s ) π π G4 ( s) = H ( s) exp i + is exp i + i( s ). t i 4 π π G + it = Γ + it a + b exp t t cos + 4 t i 4 π π G + it = iγ + it a + b exp t t sin + 4 t + i 4 π π G + it = Γ it a + b exp t t cos + 4 π π G4 + it = iγ it a + b t t i t 4 exp sin. < δ, by te contour integration metod, we ave 5 δ+ i + δ+ ζ ζ ζ G s log s ds+ G s log s ds+ G s log s ds = δ+ i + δ+ i + δ+ ζ ζ ζ G s log s ds = G s log s ds G s log s d s = J + J. δ+ i + 566

7 By Lemma. and Lemma., + ζ ζ J = G s log s ds G + it log + it i γ γ π Γ ( + it ) exp t t( logt) t ( logt) exp( t) t log t. By Lemma., Lemma. and Lemma., we ave δ+ δ J = G s logζ s ds G σ + logζ σ + dσ + Wen δ, we ave Similarly, Assume tat RH and δ ( + i ) logζ σ γ dσ. G + it logζ + it. G + it logζ + it. < δ, by te contour integration metod, we ave 5 i + i δ + + i ζ ζ ζ G s log s ds+ G s log s ds+ G s log s ds = + i + δ+ i + δ+ same as above Wen δ, we ave Similarly, + δ+ + δ i i logζ G s s ds + ζ ζ = G s log s ds G s log s ds + i + δ+ + δ i + δ+ ζ G s log s ds. G + it logζ + it. G4 + it logζ + it. 567

8 Syntesize te above conclusion, we ave terefore Similarly, terefore Similarly, Terefore G + it + G + it logζ + it π π = H + it + H it exp t t cos + logζ + it 4 π π = 4 Re H + it exp t t cos + log ζ + it + i argζ + it γ 4 π π Re H + it exp t t cos + S ( t). 4 G + it G + it logζ + it π π = 4i Im H + it exp t t cos + log ζ + it + i argζ + it 4 π π Im H + it exp t t cos + S ( t). 4 π π Re H + it exp t t sin + S ( t) 4 π π Im H + it exp t t sin + S ( t). 4 We use te same process, we can get Tis completes te proof of Lemma.6. Lemma.7. Assume tat RH, we ave < γ < it Γ + it z S ( t). it Γ + it z S ( t). were γ be te ordinates of te nontrivial zeros of ( s) Proof. < γ < Γ + a + ib Γ + a + ib log T ζ. + = ( a ib) ( a ib) Γ Γ + = A + A γ< γ< γ< γ< 568

9 by Lemma., te above formula By Lemma.4, te above formula A = Γ + a + ib = Γ + it z N t γ γ< γ γ it d it t t t 7 = Γ + it z d log + + S ( t) + O π π π 8 t it t it it z log it z = d S ( t) O. π Γ + + Γ + + π t it it = i it z i it z Γ + Γ + log z S ( t) it it + O Γ + it z +Γ + it z log z + O log T γ t by Lemma.5 and Lemma.6, above formulas By Lemma. and Lemma., we ave log T. γ γ< γ γ< + i A = Γ a + ib γ exp( πγ + γ). Tis completes te proof of Lemma.7. Lemma.8. Assume tat RH, if T, ten Proof. By Lemma.4, we ave n= n Λ( n) = T + OT T T exp log. n x x T T T Λ( n) exp = exp dψ ( x) = exp d ( x+ R( x) ) n= Tis completes te proof of Lemma Conclusions Wen x x = exp dx+ exp R x dx+ O T T T exp x = T + O x log x exp dx+ O T T T = T + O T x x+ T x x+ O = T + OT ( log log ) exp d log T. a =, T 5, b = π, n is te positive integer; by Lemma., we ave T By Lemma., we ave πi Γ ( s)( a + ib) n ds = exp( an ibn) = exp. T s s n 569

10 By Lemma. and RH, te above formula is n ζ s Λ( n) exp = Γ ( s) ( s)( a + ib) d s. T πi ζ n= ζ ζ s = a + ib + Γ + i a + ib + + Γ s s a + ib d. s γ < γ < ζ πi ζ ( ) By Lemma. and Lemma.7, te above formula is log T. By Lemma.8, we get a contradiction; terefore te RH is incorrect. References [] Montgomery, H.L. and Vaugan, R.C. (6) Multiplicative Number Teory I. Classical Teory. Cambridge University Press, Cambridge. ttp://dx.doi.org/.7/cbo [] Titcmars, E.C. (988) Te Teory of te Riemann Zeta Function. Oxford University Press, Oxford. [] Davenport, H. (967) Multiplicative Number Teory. Springer Verlag, Berlin. 4 57