Scholrs Journl of Phsics, Mhemics nd Sisics Sch. J. Phs. Mh. S. 5; (B:5-6 Scholrs Acdemic nd Scienific Publishers (SAS Publishers (An Inernionl Publisher for Acdemic nd Scienific Resources *Corresonding Auhor: Fez Fok Al Adeh Emil: hf@scs-ne.org How o rove he Riemnn Hohesis Fez Fok Al Adeh Presiden of he Srin Cosmologicl Socie, P.O.Bo:387,Dmscus, Sri ISSN 393-856 (Prin ISSN 393-864 (Online Absrc: I hve lred discovered simle roof of he Riemnn Hohesis. The hohesis ses h he nonrivil zeros of he Riemnn ze funcion hve rel r equl o.5. I ssume h n such zero is s + bi.i use inegrl clculus in he firs r of he roof. In he second r I emlo vriionl clculus. Through equions (5 o (59 I consider ( s fied eonen, nd verif h.5.from equion (6 onwrd I view ( s rmeer ( <.5 nd rrive conrdicion. A he end of he roof (from equion (73 nd hrough he ssumion h ( is rmeer, I verif gin h.5. Kewords: Definie Inegrel, Indefinie Inegrl, Vriionl Clculus. INTRODUCTION The Riemnn ze funcion is he funcion of he comle vrible s + bi (i, defined in he hlf lne > b he bsolue convergen series ( ( s s n nd in he whole comle lne b nlic coninuion. The funcion (s hs zeros he negive even inegers -, -4, nd one refers o hem s he rivil zeros. The Riemnn hohesis ses h he nonrivil zeros of (s hve rel r equl o.5. PROOF OF THE HYPOTHESIS We begin wih he equion ( (s And wih (3 s +bi (4 ( bi I is known h he nonrivil zeros of (s re ll comle. Their rel rs lie beween zero nd one. If < < hen (5 [ ] (s s s ( < < [] is he ineger funcion Hence (6 [ ] s Therefore Avilble Online: h://ssjournls.com/sjms 5
(7 ([ ] bi (8 ([ ] bi (9 ([ ] (cos( b log i sin( b log Sering he rel nd imginr rs we ge ( ([ ] cos( b log ( ([ ] sin( b log According o he funcionl equion, if (s hen ( s ( ([ ] sin( In equion ( relce he dumm vrible b he dumm vrible (3 ([ ] sin( b log d. Hence we ge besides equion ( We form he roduc of he inegrls (nd (3.This is jusified b he fc h boh inegrls ( nd (3 re bsoluel convergen.as o inegrl ( we noice h ([ ] sin( (( ( where ((z is he frcionl r of z, ((z< lim( + lim ( (( ( is ver smll osiive number (since (( whenever < + lim ( < + lim ( And s o inegrl (3 (( + ([ ] sin( d ([ ] sin( d ([]- sin ( Avilble Online: h://ssjournls.com/sjms 6
(( d lim ( d + lim( (( d ( is ver smll osiive number (since (( whenever < + lim( (( d < + d + Since he limis of inegrion do no involve or, he roduc cn be eressed s he double inegrl (4 ([ ] ([ ] sin( sin( d Thus (5 ([ ] ([ ] (cos( b log b log cos( b log b log d (6 ([ ] ([ ] (cos( cos( d Th is (7 ([ ] ([ ] cos( d ([ ] ([ ] cos( d Consider he inegrl on he righ-hnd side of equion (7 (8 ([ ] ([ ] cos( d dz In his inegrl mke he subsiuion z z The inegrl becomes dz (9 z ([ ] ([ ] cos( z d z z z Th is ( - z ([ ] ([ ] cos( z dzd z z This is equivlen o Avilble Online: h://ssjournls.com/sjms 7
( z ([ ] ([ ] cos( z dzd z z If we relce he dumm vrible z b he dumm vrible, he inegrl kes he form ( ([ ] ([ ] cos( b log d Rewrie his inegrl in he equivlen form (3 ( [ ] ([ ] cos( d Thus equion 7 becomes (4 ([ ] ([ ] cos( d ( [ ] ([ ] cos( d Wrie he ls equion in he form (5 ([ ] cos( b log { ( [ ] Le < be n rbirr smll osiive number. We consider he following regions in he lne. (6 The region of inegrion I [, [, (7 The lrge region I [, [, (8 The nrrow sri I [, [, ] (9 The nrrow sri I 3 [,] [, Noe h (3 I I I I 3 -([]-} d Denoe he inegrnd in he lef hnd side of equion (5 b (3 F (, ([ ] cos( { Le us find he limi of F (, s nd. This limi is given b (3 Lim [ - (( ] cos ( [ - (( + (( ] ((z is he frcionl r of he number z, ((z < ( [ ] -([]-} The bove limi vnishes,since ll he funcions [ - (( ],cos (, - ((, nd (( remin bounded s nd Noe h he funcion F (, is defined nd bounded in he region I. We cn rove h he inegrl (33 F(, d is bounded s follows I (34 F(, d [ - (( ] cos ( [ - (( + (( I I Avilble Online: h://ssjournls.com/sjms 8
] d I ( ( < ( [ - (( ] cos ( [ - (( + (( cos ( [ - (( + (( ] cos ( [ - (( + (( ] ] d cos ( [ - (( + (( [ (( + (( ] { lim( [ (( + (( [ (( + (( ] [ - (( ]d ] [ (( + (( ] + lim( ] } where is ver smll rbirr osiive. number. Since he inegrl lim( show h lim( [ (( + (( [ (( + (( ] is bounded, i remins o ] is bounded. Since >,hen (( nd we hve lim( lim( lim( < lim( ( [ [ [ + + (( ] + (( ] ] [ - (( ] d [ - (( ] d Avilble Online: h://ssjournls.com/sjms 9 [ (( + (( ]
Hence he boundedness of he inegrl F(, d is roved. Consider he region (35 I4I I3 We know h (36 F(, d F(, d + F(, d nd h I I I I4 (37 F(, d is bounded I From which we deduce h he inegrl (38 F(, d is bounded I4 Remember h (39 F(, d F(, d + F(, d I4 I I3 Consider he inegrl (4 F(, d F(, d I ( I { (( - (( } cos( ( { (( - (( } cos( d d ( { (( - (( } cos( d { (( - (( } d (This is becuse in his region ((. I is eviden h he inegrl bounded, his ws roved in he course of roving h he inegrl he inegrl I d Avilble Online: h://ssjournls.com/sjms { (( - (( } is F(, d is bounded.also i is eviden h
is bounded. Thus we deduce h he inegrl (4 F(, d is bounded I Hence,ccording o equion(39,he inegrl F(, d is bounded. Now consider he inegrl (4 F(, d I3 We wrie i in he form (4 F(, d ( I3 ( This is becuse in his region (( ( ( ( (( cos (b log d (( d (( cos (b log d {(( I3 (( cos (b log d {(( } {(( } Now we consider he inegrl wih resec o (43 (lim (( d {(( } } d+ (lim (( d ( where is ver smll rbirr osiive number.( Noe h (( whenever <. Thus we hve (lim nd (lim Hence he inegrl (43 d (( d < (lim (( d is bounded. d Avilble Online: h://ssjournls.com/sjms
Since cos (b logd (( cos (b logd (( d, we conclude h he inegrl is bounded funcion of. Le his funcion be H(. Thus we hve (44 (( cos (b logd H ( K ( K is osiive number Now equion (44 gives us (45 K (( cos (b logd K According o equion (4 we hve (46 F(, d ( I3 {(( } ( K K Since F(, d is bounded, hen I3 he inegrl {(( } (47 G is bounded We denoe he inegrnd of (47 b (48 F { (( - } (( cos (b log d {(( } {(( } {(( } is lso bounded. Therefore Le G [F] be he vriion of he inegrl G due o he vriion of he inegrnd F. Since (49 G [F] F (he inegrl (49 is indefinie ( here we do no consider s rmeer, rher we consider i s given eonen G[ F] We deduce h F( h is (5 G [F] F ( Bu we hve (5 G[F] G[ F] F( ( he inegrl (5 is indefinie F( Using equion (5 we deduce h (5 G[F] F( ( he inegrl (5 is indefinie (( Avilble Online: h://ssjournls.com/sjms
Since G[F] is bounded cross he elemenr inervl [,], we mus hve h (53 G[F] is bounded cross his inervl From (5 we conclude h (54 G Since he vlue of [ F (55 lim ( Thus we mus hve h (56 (lim [ df F( [ F ] ( - [ F ] ( ] ( is bounded, we deduce from equion (54 h F mus remin bounded. { (( - }] is bounded. Firs we comue (57 (lim Aling L 'Hosil ' rule we ge (58 (lim (lim d( We conclude from (56 h he roduc (59 (lim { (( - }mus remin bounded. Assume h.5.( remember h we considered s given eonen This vlue.5 will gurnee h he quni { (( - }.Therefore, in his cse (.5 (56 will roch zero s ( will remin bounded in he limi s ( hence remin bounded. Now suose h <.5.In his cse we consider s rmeer. Hence we hve (6 G [] F (, (he inegrl (6 is indefinie Thus G (6 [ ] F (, Bu we hve h (6 G [] G [ ] ( he inegrl (6 is indefinie Subsiuing from (6 we ge (63 G [] F(, ( he inegrl (63 is indefinie We reurn o equion (49 nd wrie (64 G lim ( F ( is ver smll osiive number << { F ( lim ( F ( } - lim ( Le us comue df nd Avilble Online: h://ssjournls.com/sjms 3
(65 lim ( F ( lim ( Thus equion (64 reduces o (66 G F ( - lim ( df (( Noe h he lef hnd side of equion (66 is bounded. Equion (63 gives us (67 G lim ( F ( is he sme smll osiive number << We cn esil rove h he wo inegrls df nd F re bsoluel convergen.since he limis of inegrion do no involve n vrible, we form he roduc of (66 nd (67 (68 K lim( df F lim( FdF ( K is bounded quni Th is F F (69 K lim( [ ( - ( ] [ ( - ( ] We conclude from his equion h F F (7 { [ ( - lim( ( ] [ ( ] } is bounded. ( since lim(, which is he sme hing s lim( F F Since ( is bounded, we deduce once h mus remin bounded in he limi s (, which is he sme hing s sing h F mus remin bounded in he limi s (. Therefore. (7 lim ( (( mus remin bounded Bu (( (( (7 lim ( lim( (( lim( lim( I is eviden h his ls limi is unbounded. This conrdics our conclusion (7 h lim ( (( mus remin bounded (for <.5 Avilble Online: h://ssjournls.com/sjms 4
Therefore he cse <.5 is rejeced.we verif here h,for.5 (7remins bounded s (. We hve h (73 (( < - Therefore (( (74 lim(.5 ( We consider he limi (75 lim(.5 ( We wrie (76 (lim (.5 + Hence we ge (77 lim(.5 ( lim ( < lim(.5 ( (.5 Avilble Online: h://ssjournls.com/sjms 5 lim ( (Since lim( Therefore we mus l L 'Hosil ' rule wih resec o in he limiing rocess (75 ( (78 lim(.5 ( lim(.5 ( ( lim(.5 ( Now we wrie gin (79 (lim (.5 + Thus he limi (78 becomes ( (8 lim(.5 ( (.5 lim (.5 lim ( (.5 ( Since lim ( lim ( (.5.5.5 We mus l L 'Hosil ' rule.5 (.5 (.5 (8 lim ( lim ( lim (.5. 5.5 (.5 Thus we hve verified here h,for.5 (7 roches zero s ( nd hence remins bounded. We consider he cse >.5. This cse is lso rejeced, since ccording o he funcionl equion, if ( (s (s + bi hs roo wih >.5,hen i mus hve noher roo wih noher vlue of <.5. Bu we hve lred rejeced his ls cse wih <.5 Thus we re lef wih he onl ossible vlue of which is.5 Therefore.5 This roves he Riemnn Hohesis. REFERENCES. Tich mrsh EC; The Theor of he Riemnn ze funcion.london : Oford Universi Press,999.
. Aosol,Tom M; Mhemicl Anlsis. Reding,Msschuses ;Addison wesle Publishing Comn,974. 3. Edwrds HM; Riemnn s ze funcion. New York : Acdemic Press,Inc,974. 4. Aosol TM; Inroducion o Anlic Number Theor, New York: Sringer Verlg,976. 5. Koblis N, P- dic Numbers, P dic Anlsis, nd Ze Funcions. New York : Sringer Verlg,984. 6. Geiner W, Reinhrd J; Field Qunizion. Berlin :Sringer Verlg,996. Avilble Online: h://ssjournls.com/sjms 6