Proof that the real part of all non-trivial zeros of Riemann zeta function is 1/2

0.19 MB·

by  Kimichika Fukushima

Proof that the real part of all non-trivial zeros of Riemann zeta
function is 1/2
6
KimichikaFukushima
1 ∗
0 AdvancedReactor SystemEngineeringDepartment,
2 ToshibaNuclearEngineeringService Corporation,

t 8, Shinsugita-cho,Isogo-ku,Yokohama235-8523,Japan
c
O

1
ThisarticleprovestheRiemannhypothesis,whichstatesthatallnon-trivialzerosofthezetafunctionhaveareal
3
partequalto1/2.Weinspectindetailtheintegralformofthe(symmetrized)completedzetafunction,whichisaproduct

] between the zeta and gamma functions. It is known that two integrallines, expressing the completed zeta function,
M
rotatedfromtherealaxisintheoppositedirections,canbeshiftedwithoutaffectingthecompletedzetafunctionowing
G totheresiduetheorem.Thecompletedzetafunctionisregularintheregionofthecomplexplaneunderconsideration.
. Forconvenienceinthesubsequentsingularityanalysisoftheaboveintegral,wefirsttransform(convert)thisintegralto
h
thatalongthelineparalleltotherealaxis. Wetheninvestigatethesingularitiesofthecompositeelements(causedby
t
a polynomialintegralsinoppositedirections),whichappearonlyinthecaseforwhichthedistancebetweenthecontours
m
andtheoriginofthecoordinatesapproacheszero. Therealpartofthezerosofthezetafunctionisdeterminedtobe
[ 1/2alongasymmetrylinefromthesingularityremovalcondition.(Intheotherpoints,thesingularitiesareadequately

cancelledasawholetoleadtoafinitevalue.)
6
v
3
5
5 1 Introduction
3
0 Byconnectingcomplexanalysiswithnumbertheory,Riemannobserved[1]that(denotingasetofrealnumbersbyR
2. andlettingx R)thefunctionp (x), whichdenotesthenumberofprimenumbersbelowagivennumberx, contains
0 thesummatio∈novernon-trivialzeros(pointsatwhichthefunctionvanishes)ofthezetafunction. Riemannexpected
6 (denotingasetofcomplexnumbersbyCandlettingz C)therealpartofthenon-trivialzerosofthezetafunction
1 z (z) to be 1/2, which is known as the Riemann hypot∈hesis. Furthermore, von Koch showed [2] that p (x) is well
:
v approximatedbytheoffsetlogarithmicintegralfunctionLi(x)as
i
X
p (x)=Li(x)+O(x12logx), (1)
r
a
which is equivalentto the Riemann hypothesis. We denote a set of naturalnumbersby N and let n N and z C,
thenthe zeta functionz (z) is definedas a function,which is analyticallycontinuatedin the complex∈planefrom∈the
expressiondefinedbelow[3-5]
¥
z (z):= (cid:229) 1, (2)
nz
n=1
forzthatsatisfiesRe(z)>1(wedenotetherealandimaginarypartsofzasRe(z)andIm(z), respectively.) Thezeta
functionis also obtained with the help of the gammafunction G (z), and, lettingt R, then the gamma functionis
′
∈
definedasafunctionthatisalsoanalyticallycontinuatedintoallpointsinthecomplexplanefrom[3,6-9]
¥
G (z):= dt′(t′)z−1exp( t′), (3)
0 −
Z
E-mail:[email protected];[email protected]:[phone]Phone/Fax:[phone]
∗
1
forRe(z)>0.
Concerningthezerosofthezetafunction,whichstatesz (z)=0,thereexisttrivialzeros,suchasnegativeintegers
2, 4, [3]. Incontrast,Hardyshowedthatnumerousnon-trivialzerosofthezetafunctionexistalongthelinewith
− − ····
the realpartequalto 1/2[10]; however,notall the realpartsof the non-trivialzerosare known. Thecomputational
approach[11]stronglysuggeststhattherealpartofzerosofthezetafunctionis1/2.
Ontheotherhand,lettingz,w C,forthecompletedzetafunctiondefinedby
∈
zˆ(z):=p −2zG (z)z (z), (4)
2
integralformofthe(completed)zetafunctionisexpressedas[12]
p −1−2zG (1−z)z (1 z)
2 −
=p −1−2zG (1−2 z) 0 1dwexpw(pz−i1we)xp(e−xpp(iw2p)iw)
Z ց − −
+p −2zG (2z) 0 1dwexp(wp −iwz)expe(pxpiw(2)p iw). (5)
Z ւ − −
The above integralis performedalong the integral lines 0 1 and 0 1 with the slopes 1 and +1, respectively,
ց ւ −
whichpassthroughanarbitrarypointintheregionbetween0and1oftherealaxis. Sincetheresiduetheoremexhibits
theaboveequation,theintegralformisindependentoftheshiftofthisintersectionpointbetween0and1.Furthermore,
intheoriginalform[12]oftheaboveequation,thefunctionG (z/2)inthesecondtermontherightisproportionalto
theregularfunctionforRe(z)<1. ThefunctionG ((1 z)/2)ontheleftisregularintheregionRe(z)<1,whilethe
rightside is also regularbecause ofthe existenceof th−e derivative[13, 14], andthe functionz (1 z) is analytically
continuateduniquely[13]intotheregion0<Re(z)<1(therealpartofzerosofz (z)existsonlyin−thisregion). This
papertakesintoaccounttheformmentionedabove.
Sincethegammafunctionisregular,thenon-trivialzerosofthecompletedzetafunctionzˆ(z),whichistheproduct
betweenthegammafunctionG (z)andzetafunctionz (z), coincidewiththoseofthezetafunctionz (z) intheregion
being considered with the real part between 0 and 1. As is described in this paper, each of the two line integrals
expressingthe completedzeta functionhas a singularitywhen the integrallines approachthe axis origin. However,
thecompletedzetafunctionzˆ(z)doesnotdependonaspecificpointoftheintersectionpoint(shiftedbetween0and
1alongtherealaxis)betweentheaboveintegrallineandtherealaxisduetotheresiduetheorem,andzˆ(z)isregular
intheconsideringregion. Then,thesesingularitiesmustexactlycanceleachotherforzˆ(z)=0,whichisexpectedto
leadtothedeterminationoftherealpartofthezerosofthezetafunctionz (z).
Consideringthestatusmentionedabove,thispaperisaimedatprovingtheRiemannhypothesis.Wefirsttransform
(convert)theintegral(intheintegralformofthecompletedzetafunction)alongtheintegrallinerotatedfromthereal
axis to the integralalong the line parallelto the coordinate(real/imaginary)axis, for conveniencein the subsequent
analysisofthesingularityoftheintegralinacomplexplane.Bythisconversionoftheintegral,thesingularityanalysis
canbeconcentratedonthecomponentoftheintegralalongthelineparalleltotherealaxis(ortheequivalentintegral
component),droppingthecomponentoftheintegralinthedirectionoftheimaginaryaxis.
Thisresearchthenaddressesthesingularitiesthatappearinthetwointegrallinesoftheintegralformofthecom-
pletedzetafunction. Thesingularitiesoftheintegrandsforthecompositeelementsneartheoriginoftherealaxisare
causedby polynomials, only in the case whenthe contour-origindistance approacheszero. These singularitiesade-
quatelycanceleachotheryieldingafinitevalueindependentoftheintegralcontourasawhole. Incontrast,fromthe
equationzˆ(z)=0forcompletedzetafunctionzˆ(z),therealpartofthezerosofzˆ(z)isdeterminedbyrequiringthese
singularitiesto be an identical order power of the integral variable in the integrandsleading to the exact singularity
cancellation(givenbyTheorem2). Thisrequirementresultsinavalueof1/2fortherealpartofzerosofthecompleted
zetafunctionzˆ(z)(andtheoriginalzetafunctionz (z))duetothesymmetrywithrespecttothe1/2realpart,whichis
the originalityof the presentstudy andprovesthe Riemann hypothesis. TheRiemann hypothesisis oneof the most
2
important unproved problems in mathematics, and has its equivalent and advanced (extended) conjectures in other
relatedfields. ThepositiveproofoftheRiemannhypothesisadvancesmathematicsinotherrelatedfields[15,16].
Thecontentsofthispaperareasfollows. Section2providesthetransformationoftheintegralintheintegralform
ofthecompletedzetafunctiontothelineparalleltotherealaxisforconvenienceinthesubsequentsingularityanalysis
oftheintegral.Section3presentstheproofthattherealpartofallnon-trivialzerosofthezetafunctionisequalto1/2,
aswasconjecturedbyRiemann,followedbytheconclusion.
2 Transformation of the integral to that along the line parallel to a coordi-
nate axis for the singularity analysis in a complex plane
Thissectionprovidesthetransformation(conversion)oftheintegral(intheintegralformofthecompletedzetafunc-
tion)alongtheline rotatedfromthe realaxisto theintegralalongthe line parallelto the realaxis(orthe equivalent
integral component)for conveniencein the subsequent analysis (in Section 3) of the singularity of the integral in a
complexplane. Bythistransformation,theanalysisoftheintegralinacomplexplanebecomesindependentfromthe
integralcomponentinthedirectionoftheimaginaryaxis. Notationsusedinthispaperareasfollows. Letz Cand
x,y Randletibetheimaginaryunit,then ∈
∈
(x,y):=z=x+iy. (6)
Wedenotetherealandimaginarycomponentsas
z :=Re(z)=x, (7)
R
z :=Im(z)=y. (8)
I
We first define the integrands, the main points in the complex plane, and integrals with contours for use in the
subsequentlemmainthissection.
Definition1. Letw,z C,thentheintegrandsI andI inw-planearedefinedby
N P
∈
IN:=p −1−2zG (1−2 z)expw(pz−i1we)xp(e−xpp(iw2p)iw), (9)
− −
IP:=p −2zG (2z)exp(wp −iwz)expe(pxpiw(2)p iw). (10)
− −
Definition2. Leta ,a ,a Rbepositivefinitenumbersbetween0and1,witha =a forI inEq.(9)anda =a
0 0l 0r 0 0l N 0 0r
forI inEq. (10),respective∈ly. Letb Rwithb >0beasmallnumber. Thenwedefinethepointsw ,w Cin
P 0 0 lU lL
∈ ∈
thecomplexplane,indicatedinFig. 1,as
w :=(a ,b ), (11)
lU 0 0
w :=(a , b ). (12)
lL 0 0
−
Lete¯ Cbedefinedby
3
∈
3
e¯ :=exp( p i), (13)
3
4
andletr Rwithr >0betheradiusofarcscenteredatw andw ,thenwedefinethefollowingw ,w ,w ,w
2 2 lU lL 11 12 21 22
CinFig.∈1 ∈
3 3
w :=r e¯ +w =(r cos( p )+a ,r sin( p )+b ), (14)
11 2 3 lU 2 0 2 0
4 4
3
✻
w C Im
11 1
❅❅
ց
❅
❅
I1f
I ❅
1
ց❅
I
w12↑ 1← ❅wlU R✲e
← O w w
lL❅ I ← 22
2
❅ ← ↑
❅❅I2 I2f
ց
❅
❅❅
ց
w
C 21
2
w : (a ,b )
lU 0 0
w : ( √2r +a ,√2r +b )
11 − 2 2 0 2 2 0
w : ( r +a ,b )
12 2 0 0
−
w : (a , b )
lL 0 0
−
w : (√2r +a , √2r b )
21 2 2 0 − 2 2− 0
w : (r +a , b )
22 2 0 0
−
r :radiusofarcscenteredatw andw
2 lU lL
a =a (between0and1)andr >0
0 0l 2
Note:b >0
0
Figure1: Locationofcomplexnumbers(indicatedatthebottomofthefigure)andcontoursinthecomplexw-plane.
ContourC passes throughw , w and w as in the integraldirections , and , in order. ContourC passes
1 11 lU 12 2
ց ← ↑
throughwlL,w21 andw22 asinthedirections , and ,inorder. SymbolsI1 ,I1 andI1f (givenbyEq. (26))are
integralsofIN (ofEq. (9))alongC1,whileI2ց,I↑2f and←I2 (givenbyEq. (31)ց)are←integralsofIP alongC2. (I1f and
I2f areangularintegralsaroundwlU andwlL,rցespectively.)←
w :=r ( 1,0)+w =( r +a ,b ), (15)
12 2 lU 2 0 0
− −
1 1
w := r e¯ +w =(r cos( p )+a ,r sin( p ) b ), (16)
21 2 3 lL 2 0 2 0
− −4 −4 −
w :=r (1,0)+w =(r +a , b ). (17)
22 2 lL 2 0 0
−
Inasimilarway,usinga (a =a forI inEq. (10))andb (withb >0)definedinthesentenceaboveEq. (11)
0 0 0r P 0 0
andlettingw ,w C,thenwedefinethepointsinFig.2as
rU rL
∈
w :=(a ,b ), (18)
rU 0 0
w :=(a , b ). (19)
rL 0 0
−
Additionally,lete¯ Cbedefinedby
∈
1
e¯ :=exp( p i), (20)
4
4
Im ✻ C3 w31
(cid:0)(cid:0)
(cid:0)ւ
(cid:0)
(cid:0) I3ւ I3f
(cid:0)
I
OwrU(cid:0) 3→ ↑w32✲
w42 → I (cid:0)wrL → Re
4
↑ → (cid:0)
I4f I4 (cid:0)
ւ(cid:0)
(cid:0)
(cid:0)(cid:0)ւ
w
41
C
4
w : (a ,b )
rU 0 0
w : (√2r +a ,√2r +b )
31 2 2 0 2 2 0
w : (r +a ,b )
32 2 0 0
w : (a , b )
rL 0 0
−
w : ( √2r +a , √2r b )
41 − 2 2 0 − 2 2− 0
w : ( r +a , b )
42 2 0 0
− −
r :radiusofarcscenteredatw andw
2 rU rL
a =a (between0and1)andr >0
0 0r 2
Note: b >0
0
Figure2: Locationofcomplexnumbers(indicatedatthebottomofthefigure)andcontoursinthecomplexw-plane.
ContourC passes throughw , w andw as in the integraldirections , and , in order. ContourC passes
3 31 rU 32 4
ւ → ↑
throughwrL,w41 andw42 asinthedirections , and ,inorder. SymbolsI3 ,I3 andI3f (givenbyEq. (33))are
integralsofIP(ofEq. (10))alongC3,whileI4ւ,I↑4f and→I4 (givenbyEq. (35)ւ)are→integralsofIPalongC4. (I3f and
I4f areangularintegralsaroundwrU andwrL,rւespectively.)→
and let r R with r > 0 be the radius of arcs centered at w and w , then we further define the following
2 2 rU rL
w ,w ,w∈,w CinFig. 2
31 32 41 42
∈
1 1
w :=r e¯+w =(r cos( p )+a ,r sin( p )+b ), (21)
31 2 rU 2 0 2 0
4 4
w :=r (1,0)+w =(r +a ,b ), (22)
32 2 rU 2 0 0
3 3
w := r e¯+w =(r cos( p )+a ,r sin( p ) b ), (23)
41 2 rL 2 0 2 0
− −4 −4 −
w :=r ( 1,0)+w =( r +a , b ). (24)
42 2 lL 2 0 0
− − −
Definition3. UsingthedefinitionsofI ,w ,w andw inEqs. (9),(11),(14)and(15),respectively,andwiththe
N lU 11 12
helpofFig.1,wedefinetheintegralsandcontoursas
dwIN:=I1 +I1 +I1f , (25)
IC1 ց ←
5
wherecontourC passesthroughw ,w andw asintheintegraldirections , and ,inorderinFig. 1,and
1 11 lU 12
ց ← ↑
wlU w12 w11
I1 := dwIN, I1 := dwIN, I1f := dwIN. (26)
ց Zw11 ← ZwlU Zw12
In the last integral above, let f R be an angle of circular coordinates(centered at w ) of the integral variable,
1 lU
∈
wheretheangleismeasuredcounterclockwisefromtheaxisparalleltotherealaxis,denotedas
w=r exp(if )+w =(r cos(f ),r sin(f ))+(a ,b ), (27)
2 1 lU 2 1 2 1 0 0
withitsderivative
dw
=ir exp(if ). (28)
df 2 1
1
ThenthelastintegralinEq.(26)isdefinedby
I1f =Zww1211dwIN:=Zp 34p df 1ddfw1IN. (29)
In a similar way, let f ,f ,f R be anglesof circular coordinatescentered at w , w and w , respectively.
2 3 4 lL rU rL
∈
Using I , I (Eqs. (9), (10)), w , w (Eqs. (11), (12)), w , w (Eqs. (16), (17)), w , w (Eqs. (18), (19)) and
N P lU lL 21 22 rU rL
w -w (Eqs. (21)-(24)),andwiththehelpofFigs. 1-2,theotherintegralsandcontoursaredefinedby
31 42
dwIN:=I2 +I2f +I2 , (30)
IC2 ց ←
contourC passesthroughw ,w andw asinthedirections , and ,inorderinFig. 1,and
2 lL 21 22
ց ↑ ←
,
I2ց:= wwlL21dwIN, I2f := ww2212dwIN:= −041p df 2ddfw2INwithw=r2exp(if 2)+wlL, I2←:= ww2l2LdwIN (31)
 R R R R 
dwIP:=I3 +I3 +I3f , (32)
IC3 ւ →
contourC passesthroughw ,w andw asinthedirections , and ,inorderinFig. 2,and
3 31 rU 32
ւ → ↑
1p ,
I3ւ:= ww3r1UdwIP, I3→:= wwr3U2dwIP, I3f := ww3321dwIP:= 04 df 3ddfw3IPwithw=r2exp(if 3)+wrU! (33)
R R R R
dwIP:=I4 +I4f +I4 , (34)
IC4 ւ →
contourC passesthroughw ,w andw asinthedirections , and ,inorderinFig. 2,and
4 rL 41 42
ւ ↑ →
.
I4ւ:= wwr4L1dwIP, I4f := ww4412dwIP:= −−34pp df 4ddfw4IPwithw=r2exp(if 4)+wrL, I4→:= ww4r2LdwIP (35)
 R R R R 
Wethenpresentthelemmaconcerningthetransformationmentionedatthebeginningofthissection.
Lemma 1. Leta andb be numbersthatspecify thex andy coordinatesofw , w , w andw asexpressed in
0 0 lL lU rU rL
Eqs. (11)-(12)and(18)-(19)with thehelpofFigs. 1-2. Letr betheradiusofthecontourforthecircularintegrals
2
indicatedinEqs. (27), (31), (33)and(35). UsingtheintegralsI , I ,I , I , I , I , I andI definedin
1 2 3 4 1 2 3 4
Eqs. (26),(31),(33)and(35)withthehelpofFigs. 1-2,wederiveց ց ւ ւ ← ← → →
(I +I +I +I )= (I +I +I +I )
1 2 3 4 1 2 3 4
← ← → → − ց ց ւ ւ
forapositivefinitevalueofa inthelimitb 0whichfollowsr ¥ (withtherelationsuchasb >2/r ). (36)
0 0 2 0 2
→ →
6
Theaboveequationimpliesinthee d notationthat
−
( e >0)( d >0, d >0)( b , r R)
0 0 2
∀ ∃ ∃ ∀ ∀ ∈
2
( <d <b <d (I +I +I +I )+(I +I +I +I ) <e ). (37)
0 0 1 2 3 4 1 2 3 4
r2 ⇒| ← ← → → ց ց ւ ւ |
The sumson theleft andrightofthe aboveequationtake a definitefinitevalue, statingthatthe sum ofthe integrals
inthe e¯¥ direction(indicatedby )andinthe e¯¥ direction(indicatedby )aretransformed(converted)tothe
3
− ց − ւ
sumoftheintegralsalongthelineparalleltotherealaxis(indicatedby and ). Theaboverelationholdsforthe
← →
exchangez 1 z.
↔ −
Proof. UsingEqs. (25),(26)and(30)-(35),eachclosedcontourofthefollowingintegralsdoesnotenclosethesingu-
larityat the coordinatesoriginin the complexplane whena and b (indicatedin Eqs. (11)-(12) and(18)-(19) with
0 0
thehelpofFigs. 1-2)arepositiveandfinite. Then,theassociatedintegralsalongtheclosedcontourvanish,duetothe
residuetheorem,asexpressedby
dwI = dwI = dwI = dwI =0 forb >0 (seealsonext6lines). (38)
N N P P 0
IC1 IC2 IC3 IC4
(Wenoteinadvancethatbecauseofb >0theaboveequationholdsforapositivefinitevalueofa inthelimitb 0
0 0 0
thatfollowsr ¥ withtherelationsuchasb >2/r ,whichimpliesthat →
2 0 2
→
( e >0)( d >0, d >0)( b , r R)
0 0 2
∀ ∃ ∃ ∀ ∀ ∈
2
( <d <b <d dwI <e , dwI <e , dwI <e , dwI <e ). (39)
0 0 N N P P
r2 ⇒ |IC1 | |IC2 | |IC3 | |IC4 |
Namely,becauseofb >0,nopolesexistwithintheregionsenclosedbytheintegralcontoursC ,C ,C ,C including
0 1 2 3 4
contours,andtheresiduetheoremworks.)
Accordingly,withtheuseofEqs. (25),(30),(32)and(34),wehave
I = dwI + dwI + dwI + dwI
S N N P P
IC1 IC2 IC3 IC4
=(I1 +I1 +I1f )+(I2 +I2f +I2 )+(I3 +I3 +I3f )+(I4 +I4f +I4 )=0. (40)
ց ← ց ← ւ → ւ →
Eachoftheaboveintegralterms(suchasI1 ,I1 andI1f inEqs. (26),(31),(33)and(35))hasadefinitefinitevalue
withinafiniteregionthatdoesnotcontainanցysin←gularitiesoftheintegrandwhena andb (suchasthoseinEq.(11))
0 0
arepositiveandfinite,andtheorderinthesumischangedtoproduce
(I1 +I2 +I3 +I4 )+(I1 +I2 +I3 +I4 )+(I1f +I2f +I3f +I4f )=0. (41)
ց ց ւ ւ ← ← → →
For each of the last four terms I1f , I2f , I3f and I4f above, the real part of term p iw2 in the leading function
exp( p iw2) (in the integrand IN given by Eq. (9) for I1f and I2f , such as that in∓(28)-(29), and the integrand IP
given∓byEq. (10)forI3f andI4f )yields 2p Re(w) Im(w) inthelimitofr2 ¥ (thatis,forthelarge w)andthe
functionexp( p iw2)vanishesforthelarg−e w|. Then|,|weder|ive → | |
∓ | |
I1f =I2f =I3f =I4f =0 inthelimitofr2 ¥ (thatis,forthelarge w)forb0>0. (42)
→ | |
Thisimpliesthatlettingb >0
0
(∀e >0)(∃d >0)(∀r2∈R)(r22 <d ⇒ |I1f |<e , |I2f |<e , |I3f |<e , |I4f |<e ).! (43)
In contrast, with the use of Eqs. (4) and (5), the sum of the first four terms I , I , I and I in Eq. (41)
1 2 3 4
amountsto ց ց ւ ւ
(I +I +I +I )=zˆ(1 z)
1 2 3 4
ց ց ւ ւ −
forapositivefinitevalueofa inthelimitb 0whichfollowsr ¥ (suchasb >2/r ). (44)
0 0 2 0 2
→ →
7
Thisimpliesthat
(∀e >0)(∃d >0,∃d 0>0)(∀b0,∀r2∈R)(r22 <d <b0<d 0 ⇒ |(I1ց+I2ց+I3ւ+I4ւ)−zˆ(1−z)|<e ).!(45)
Thus,Eqs. (41)-(44)leadto
(I +I +I +I )= (I +I +I +I )= zˆ(1 z)
1 2 3 4 1 2 3 4
← ← → → − ց ց ւ ւ − −
forapositivefinitevalueofa inthelimitb 0whichfollowsr ¥ (suchasb >2/r ). (46)
0 0 2 0 2
→ →
Thismeansthat
( e >0)( d >0, d >0)( b , r R)
(2 <d <b <d∀ (I ∃+I +∃I0 +I∀)0+∀(I2∈+I +I +I )
 r2 0 0⇒|=1(←I1 +2←I2 +3→I3 +4→I4 )+1zցˆ(1 2ցz) <3eւ). 4ւ | (47)
 | ← ← → → − | 
 
The above equation implies that the sum of the integrals in the e¯¥ direction (indicated by ) and in the e¯¥
3
− ց −
direction(indicatedby )istransformed(converted)tothesumoftheintegralsalongthelineparalleltotherealaxis
ւ
(indicatedby and ),owingtothatthecompletedzetafunctionisregular. Namely,theintegralofthecompleted
← →
zetafunctionalongthelinerotatedfromtherealaxiscanbetransformedtotheintegralalongthelineparalleltothe
realaxisasawholeforconvenience.Wenotethattheaboverelationshipsholdfortheexchangez 1 z,keepingthe
↔ −
region0<Re(z)<1underthisexchange.
3 ProofoftheRiemann’sconjecturethattherealpartofallnon-trivialzeros
of the zeta function is 1/2
ThissectionfirstconvertstheintegralformofthecompletedzetafunctionexpressedbyEq. (5)totheusualformto
giveTheorem1. Then,inTheorem2, weprovethattherealpartofallnon-trivialzerosofthezetafunctionmustbe
1/2.Asitisknownthatallnon-trivialzerosofthezetafunctionexistintheregion0<Re(z)<1inliterature[17,18],
weconcentrateonthisregion.Furthermore,itisalsoknownthatthenumberofzeros(ofthezetafunction)withareal
partof1/2isinfinite[10]. Theusualintegralformofthezetafunctionisasfollows.
Theorem1. (the(third)integralform ofthe(completed)zetafunction)Letz,w C. Letzˆ(z)bethecompletedzeta
∈
functiondefinedbyEq. (4). Let
zˆl(z):=p −2zG (2z) 0 1dwexpw(p−izwe)xp(e−xpp(iw2p)iw), (48)
Z ց − −
andlet
zˆr(z):=p −1−2zG (1−2 z) 0 1dwexp(wpzi−w1)expe(xppi(w2p)iw), (49)
Z ւ − −
intermsofthegammafunctionG (z). Thenzˆ(z)isexpressedby
zˆ(z)=zˆ(z)+zˆ(z), (50)
l r
whichiscalledthe(third)integralformofthe(completed)zetafunction.
Proof. FromEq. (5),theabovecompletedzetafunctionzˆ(z)isobtainedbytheexchangez 1 z,wheretheregion
↔ −
0<Re(z)<1iskeptunderthisexchange.Theaboveregion0<Re(z)<1isconsistentwiththeregion0<Re(z)<1
weareconsideringinthispaper.
8
✻
Im
C1c w14
←
C C
w16❤❤❤1n❤w❤15 ✭w1✭3 ✭1✭p✭wlU
← ↓ ↑ ← R✲e
O
w : (a ,b )
lU 0 0
w : (a /2,b /2) w : (0,r )
13 0 0 14 1
w : ( a /2,b /2) w : ( a ,b )
15 0 0 16 0 0
− −
C : contourfromw tow
1p lU 13
C :arcfromw tow withradiusr centeredat(0,0)
1c 13 15 1
C : contourfromw tow
1n 15 16
a =a (between0and1)andr =[(a /2)2+(b /2)2]1/2
0 0l 1 0 0
Note:0<b <<a
0 0
Figure 3: A part of the componentparallel to the real axis in the contour C in Fig. 1, deformedinto the contour
1
composedofC ,C andC (withoutdisconnectionsfromotherparts)tobypasstheoriginofthecomplexw-plane.
1p 1c 1n
Locationsofcomplexnumbersinthecomplexplaneareindicatedatthebottomofthefigure. Theanglef ofcircular
c
coordinatescenteredattheorigin(0,0)ismeasuredcounterclockwisefromtherealaxiswiththepositivesign.
In addition, the completed zeta function satisfies the following known symmetry relation [1, 16, 15] for the ex-
changez 1 z
↔ −
zˆ(z)=zˆ(1 z). (51)
−
To derive the real part of non-trivial zeros of the zeta function, the present approach uses (in addition to the above
symmetrygivenbyEq.(51))theproperty(withmerits)thataquantityinoneterminahighly(attainable)symmetrized
integralformgeneratesacorresponding(paired)quantityinanotherterm.
Here, for the Riemann-Siegelintegralform, each partof the contour(for the integralvariablew C) parallelto
∈
therealaxisin thecontoursC andC in Figs. 1-2 isfurtherdeformedto bypasstheorigininthe complexw-plane
1 4
as in Figs. 3-4 according to Definition 4 below. Each contour part deformedas in Figs. 3-4 is connected (without
disconnections)withtheotherremainingcontoursinFigs1-2. Theintegrandshavepolynomialsofwwiththepower
of z C, and the absolute value of the integrandis expressed using w and the argument(angle) arg(w) in circular
∈ | |
coordinates. Then, bymakingarg(w) independentof w near the originof the w-plane, singularityanalysisbecomes
simple. In Figs. 3-4, each linear contour employed has a constant arg(w), while arg(w) along each arc (of circle)
contourisintegratedtothecircularintegral. Therefore,singularityanalysisfocusesonthepowerforthepolynomials
of w. (Wenotethatnopoleexistsaroundthepolelocatedattheoriginofthecomplexw-plane,whentheconcerned
| |
regionissmall.)
9
✻
Im
O R✲e
w43✭✭C→✭4n✭w✭44 ↓ ↑ ❤w4❤6 C❤→4❤p❤wrL
C→4c w45
w :(a , b )
rL 0 0
−
w : ( a , b ) w : ( a /2, b /2)
43 0 0 44 0 0
− − − −
w : (0, r ) w : (a /2, b /2)
45 1 46 0 0
− −
C : contourfromw tow
4n 43 44
C :arcfromw tow withradiusr centeredat(0,0)
4c 44 46 1
C : contourfromw tow
4p 46 rL
a =a (between0and1)andr =[(a /2)2+(b /2)2]1/2
0 0r 1 0 0
Note:0<b <<a
0 0
Figure 4: A part of the componentparallel to the real axis in the contour C in Fig. 2, deformedinto the contour
4
composedofC ,C andC (withoutdisconnectionsfromotherparts)tobypasstheoriginofthecomplexw-plane.
4n 4c 4p
Locationsofcomplexnumbersinthecomplexplaneareindicatedatthebottomofthefigure. Theanglef ofcircular
c
coordinatescenteredattheorigin(0,0)ismeasuredcounterclockwisefromtherealaxiswiththepositivesign.
Definition4. Letb ,a Rwithb anda beingsmall(b >0asindicatedsucasinEq.(45);b <<a asdescribed
0 0 0 0 0 0 0
inNote belowEq. (65))∈. Letw C asdefinedin Definition2 with Figs. 1-2. Letting f R, we definethe angle
lU c
f ofcircularcoordinatescentered∈attheorigin(0,0)ofthecomplexw-plane,thatismeasure∈dcounterclockwisefrom
c
therealaxiswiththepositivesign. Letr R(radiusofthearc(ofcircle)contoursdefinedbelowandinFigs. 3-4)
1
suchthatr =a /2. Letw ,w ,w ,w ∈ C, then,asindicatedinFig. 3, wedefinethesenumbersinthecomplex
1 0 13 14 15 16
∈
w-planeby
a b
0 0
w :=( , ), w :=(0,r ),
13 14 1
2 2
a b
0 0
w :=( , ), w :=( a ,b ), (52)
15 16 0 0
− 2 2 −
where
a 2 b 2 1/2
0 0
r := + . (53)
1
2 2
"(cid:16) (cid:17) (cid:18) (cid:19) #
UsingthequantitiesgiveninFig. 1,apartofthecomponent,paralleltotherealaxis,ofthecontourC inFig. 1is
1
deformedintothecontourcomposedofC ,C andC tobypassthecoordinateoriginasinFig. 3. Thecontoursin
1p 1c 1n
Fig.3aredefinedasfollows.
10