On the zeros of the Riemann Zeta function

0.2 MB·English

by  Lazhar Fekih-Ahmed

ON THE ZEROS OF THE RIEMANN ZETA FUNCTION
1 LAZHARFEKIH-AHMED
1
0
2 Abstract. This paper is divided into two independent parts. The firstpart
presentsnewintegralandseriesrepresentationsoftheRiemaanzetafunction.
n AnequivalentformulationoftheRiemannhypothesisisgivenandfewresults
a on this formulation are briefly outlined. The second part exposes a totally
J
differentapproach. Usingthenewseriesrepresentationofthezetafunctionof

2 thefirstpart,exactinformationonitszerosisprovided.
2

]
M
G PART I
. 1. Introduction
h
t
a It is well known that the Riemaan zeta function defined by the Dirichlet series
m
1 1 1 ∞
[ (1.1) ζ(s)= + + + + = n−s
1s 2s ··· ks ···
nX=1
3
v converges for (s) > 1, and can be analytically continued to the whole complex
ℜ
3 plane with one singularity, a simple pole with residue 1 at s = 1. It is also well
4
known that ζ(s) satisfies the functional equation:
1
4
. s s
4 (1.2) χ(s)ζ(s)=ζ(1 s)χ(1 s) with χ(s)=π−2Γ( ),
− − 2
0
0 and that the zeros of ζ(s) come into two types. The trivial zeros which occur at
1 all negative even integers s = 2, 4, , and the nontrivial zeros which occur at
v: certain values of s C, 0< (−s)<−1. ···
∈ ℜ
i The Riemann hypothesis states that the nontrivial zeros of ζ(s) all have real
X
part (s) = 1. From the functional equation (1.2), the Riemann hypothesis is
r ℜ 2
a equivalent to ζ(s) not having any zeros in the strip 0< (s)< 1.
ℜ 2
2. An Analytic Continuation of ζ(s)
Let
n 1 n 1
(2.1) Sn(s)=1 − 2−s+ − 3−s +( 1)n−1(n)−s,n 2
−(cid:18) 1 (cid:19) (cid:18) 2 (cid:19) −··· − ≥
with S (s)=1.
1
Using the well-known identity, valid for (s)>0:
ℜ
Date:January21,2011.
2000 Mathematics Subject Classification. Primary11M26;Secondary11M06,11M45.
Key words and phrases. NumberTheory,RiemannZetafunction,RiemannHypothesis.
1
2 LAZHARFEKIH-AHMED
(2.2) n s = 1 ∞e ntts 1dt,
− − −
Γ(s)Z
0
we can rewrite S (s) in (2.1) as:
n
n 1
− n 1
Sn(s) = ( 1)k − (k+1)−s
− (cid:18) k (cid:19)
kX=0
n 1
(2.3) = 1 ∞ − ( 1)k n−1 e−(k+1)tts−1dt
Γ(s)Z − (cid:18) k (cid:19)
0 Xk=0
= 1 ∞(1 e−t)n−1e−tts−1dt,
Γ(s)Z −
0
n 1
− n 1
since ( 1)k − e−(k+1)t =e−t(1 e−t)n−1.
− (cid:18) k (cid:19) −
kX=0
We have
(2.4) ∞ Sn(s) = 1 ∞ ∞ (1−e−t)n−1e−tts−1dt
n+1 Γ(s) Z n+1
nX=1 nX=1 0
(2.5) = 1 ∞ ∞ (1−e−t)n−1e−tts−1dt
Γ(s)Z n+1
0 nX=1
(2.6) = 1 ∞ t 1 e−tts−1dt.
Γ(s)Z (cid:18)(1 e t)2 − 1 e t(cid:19)
0 − − − −
Before we proceed further, some remarks are in order:
Remark 2.1. The interchange of the summation and integration in equation (2.5)
is valid because the series ∞n=1 0∞ (1−en−+t1)n−1e−tts−1dt converges absolutely and
uniformly for 0 < t < P. ToRsee this, we show uniform convergence for the
dominating series ∞n=1∞0∞ (1−en−+t1)n−1e−ttσ−1dt, σ = ℜ(s). Indeed, let K =
max((1 e t)n 1ePt/2),R0 < t < . A straightforward calculation of the de-
− − −
− ∞
rivative shows that K = (1 1 )n 1 1 and is attained when e t = 1 .
− 2n 1 − √2n 1 − 2n 1
Now, for n 2 we have − − −
≥
1 ∞(1 e−t)n−1e−ttσ−1dt = 1 ∞(1 e−t)n−1e−t/2(e−t/2tσ−1)dt
n+1Z − n+1Z −
0 0
K ∞e t/2tσ 1dt
− −
≤ n+1Z
0
1 1 2σΓ(σ)
(2.7) = (1 )n 1
−
n+1 − 2n 1 √2n 1
− −
2σΓ(σ)
.
≤ (n+1)√2n 1
−
The lastinequality implies that the dominating series convergesby the compar-
ison test.
ON THE ZEROS OF THE RIEMANN ZETA FUNCTION 3
Remark 2.2. To get equation (2.6) we used the identity:
∞ (1 e−t)n−1 t 1
(2.8) − = ,
n+1 (1 e t)2 − 1 e t
nX=1 − − − −
whichcanbeobtainedbyputtingX =1 e tinto log(1−X) 1 = ∞ Xn−1.
− − − X2 − X n+1
nX=1
Now, since
d te t te t e t
− − −
(2.9) − = ,
dt1 e t (1 e t)2 − 1 e t
− − − − − −
an integration by parts in (2.6) yields
∞ Sn(s) s 1 ∞ e−tts−1
(2.10) = − dt=(s 1)ζ(s),
n+1 Γ(s) Z 1 e t −
nX=1 0 − −
which is valid when (s)>1. And since the integral (2.6) is valid for (s)>0,
ℜ ℜ
then we have proved
Theorem 2.3. Let φ(t)= t 1 , then for (s)>0,
(1 e−t)2 − 1 e−t ℜ
− −
(2.11) (s 1)ζ(s)= 1 ∞φ(t)e tts 1dt.
− −
− Γ(s)Z
0
Remark 2.4. Although we will not need it in the rest of the paper, we can also
obtain an analytic continuation of (s 1)ζ(s) when (s) 0. We simply rewrite
− ℜ ≤
(2.11) as a contour integral
Γ(1 s)
(2.12) − φ(t)e−tts−1dt,
2πi Z
C
where isthe Hankelcontourconsistingofthe threepartsC =C C C : a
ǫ +
C −∪ ∪
pathwhichextends from( , ǫ),aroundthe origincounterclockwiseonacircle
−∞ −
of center the originand of radius ǫ and back to ( ǫ, ), where ǫ is anarbitrarily
small positive number. The integral (2.12) now d−efin−es∞(s 1)ζ(s) for all s C.
− ∈
Remark 2.5. In particular, when s = k is a positive integer, we have yet another
formula for ζ(k):
(2.13) (k 1)ζ(k)= 1 ∞φ(t)e ttk 1dt.
− −
− (k 1)!Z
− 0
Remark 2.6. The above integral formula for (s 1)ζ(s), although obtained by
−
elementary means, does not seem to be found in the literature. As for the series
formula,ithasbeenobtainedbyadifferentmethodin[12]. Aseriesformulathatis
different but similar in form and often mentioned in the literature is that of Hasse
[8].
3. A Series Expansion of (s 1)ζ(s)Γ(s)
−
Theanalyticfunction(s 1)ζ(s)Γ(s)canberepresentedbyaTaylorseriesaround
−
anypoint s =1+iy on the verticalline σ =1. In particular,for s =1 we obtain
0 0
the well-known power series [2]:
4 LAZHARFEKIH-AHMED
(3.1) (s 1)ζ(s)Γ(s)=a +a (s 1)+a (s 1)2+a (s 1)3+
0 1 2 3
− − − − ···
where the coefficients a =1 and a are defined by
0 n
(3.2) an = n1!sli→m1ddsnn(cid:26)Z0∞φ(t)e−tts−1dt(cid:27)
(3.3) = 1 ∞φ(t)e t lim dn ts 1 dt
− −
n!Z0 s→1dsn(cid:8) (cid:9)
(3.4) = 1 ∞φ(t)e−t(logt)ndt,
n!Z
0
with φ(t) being the function defined in Theorem 2.3.
The coefficients a are very important in the evaluation of ζ(k)(0) as given by
n
Apostol in [2]. Up to now the a are regarded as unknowns and as difficult to
n
approximate as ζ(k)(0) itself as pointed out by Lehmer [10]. The formula above
solvestheexactevaluationproblemoftheζ(k)(0)andmanyothervariantformulae.
The following proposition provides more information on the sequence a .
n
{ }
Proposition 3.1. For n large enough, the coefficients a are given by
n
1 1 1 1
a =( 1)n +O( ).
n − (cid:18)2 − 62n+1(cid:19) 4n
Proof. The expression of a can be split into the sum
n
a = 1 1φ(t)e t(logt)ndt+ 1 ∞φ(t)e t(logt)ndt
n − −
n!Z n!Z
0 1
( 1)n ( 1)n 1 1 1 n
(3.5) = − + − φ(t)e−t log dt
2 n! Z − 2 (cid:18)t(cid:19)
0 (cid:2) (cid:3)
+ 1 ∞φ(t)e t(logt)ndt.
−
n!Z
1
To obtain an estimate the first integralin (3.5), we use equation (2.9). A differ-
entiation with respect to t of the following expansion which defines the Bernoulli
numbers1
(3.6) te−t = ∞ Bntn, t <2π
1 e t n! | |
− − nX=0
gives
(3.7) φ(t)e−t 1 = ∞ −Bn tn−1 = 1t+ 1 t3 1 t5+
− 2 (n 1)! −6 180 − 5040 ···
nX=2 −
Now, since
1B0 = 1 B1 = −1/2, B2 = 1/6, B3 = 0, B4 = −1/30, B5 = 0, B6 = 1/42, B7 = 0,
B8=−1/30etc.
ON THE ZEROS OF THE RIEMANN ZETA FUNCTION 5
1 1 1 n 1
(3.8) tmlog dt=
n!Z (cid:18)t(cid:19) (m+1)n+1
0
for all n,m positive, the first integral in (3.5) without the factor ( 1)n has an
−
expansion
1 1 φ(t)e−t 1 log 1 ndt = ∞ −Bn
n!Z − 2 (cid:18)t(cid:19) (n 1)!(n+1)n+1
0 (cid:2) (cid:3) nX=2 −
1 1 1 1
(3.9) = +
−62n+1 1804n+1 −···
To estimate the second integral in (3.5), we use the bound
(3.10) (logt)n <eǫt
whichisvalidforall t n1+ǫ andforn largeenoughandwhereǫis anypositive
≥
small number. We split the integral into two parts
1 ∞φ(t)e t(logt)ndt = 1 n1+ǫφ(t)e t(logt)ndt+
− −
n!Z n!Z
1 1
(3.11) 1 ∞ φ(t)e t(logt)ndt
−
n!Zn1+ǫ
C 1 e (1 ǫ)n1+ǫ
(3.12) 0(1+ǫ)nlog(n)n+ − − ,
≤ n! 2n! 1 ǫ
−
where C0 = 1∞φ(t)e−tdt=0.58.
Clearly, theRterm 1 e−(1−ǫ)n1+ǫ is extremely small for n large enough. In par-
2n! 1 ǫ
ticular, for n 2, it is less t−han C where C is a positive constant and a is any
≥ an
positive constant greater than 2.
Using Stirling formula n! √2πn n n, we can verify that the term C0(1 +
∼ e n!
ǫ)nlog(n)n can also be made less than(cid:0) C(cid:1) for n large enough. Taking a = 4 for
an
example, we obtain
(3.13) 1 ∞φ(t)e t(logt)ndt C .
−
n!Z ≤ 4n
1
By combining the above estimates, we obtain
1 1 1 1
(3.14) a =( 1)n +O( ).
n − (cid:18)2 − 62n+1(cid:19) 4n
(cid:3)
For any s of the form s =1+iy the corresponding Taylor series is
0 0
(3.15) (s 1)ζ(s)Γ(s)=b +b (s s )+b (s s )2+b (s s )3+
0 1 0 2 0 3 0
− − − − ···
where the b is expressed as
n
6 LAZHARFEKIH-AHMED
(3.16) b = 1 ∞φ(t)e t(logt)ntiydt.
n −
n!Z
0
Anasymptoticestimateofthecoefficientsb isgivenbythefollowingproposition
n
whose proof is the same as the proof of Proposition 3.1.
Proposition 3.2. The radius of convergence of the series (3.15) is 1+y2 and
for n large enough, the coefficients b are given by p
n
1 ( 1)n 1 ( 1)n 1
b = − − +O( ).
n 2(1+iy)n+1 − 6(2+iy)n+1 ( 16+y2)n
p
4. The zeros of ζ(s)
The Taylorseriesexpansion(s 1)ζ(s)Γ(s) providesus with a toolto study the
−
zeros of ζ(s) is a neighborhood of s =1+iy. We have
0
(4.1) (s 1)ζ(s)Γ(s)=b +b (s s )+b (s s )2+b (s s )3+
0 1 0 2 0 3 0
− − − − ···
with
(4.2) b =iyζ(1+iy)Γ(1+iy),
0
(4.3) bn = 1 ∞φ(t)e−t(logt)ntiydt.
n!Z
0
It is a well-know fact [7] that ζ(1+iy)= 0 for all y. Therefore b = 0 for all y
0
6 6
andthe inverseof(s 1)ζ(s)Γ(s)is well-definedandcanbe expandedintoa power
−
series of the form
1
(4.4) =c +c (s s )+c (s s )2+c (s s )3+ ,
0 1 0 2 0 3 0
(s 1)ζ(s)Γ(s) − − − ···
−
where the coefficients c are given by
n
∆
(4.5) c =( 1)n n ,
n − bn+1
0
with
b b 0
1 0
··· ···
(cid:12) b2 b1 b0 0 (cid:12)
(cid:12) ··· (cid:12)
(4.6) ∆ =(cid:12) .. .. .. .. .. (cid:12).
n (cid:12) . . . . . (cid:12)
(cid:12) (cid:12)
(cid:12) b b b b (cid:12)
(cid:12) n−1 n−2 n−3 ··· 0 (cid:12)
(cid:12) b b b b (cid:12)
(cid:12) n n−1 n−2 ··· 1 (cid:12)
(cid:12) (cid:12)
Let D(s ,1) be the open(cid:12)disk of center s and radius 1(cid:12). The zeros of ζ(s) are
0 2 0 2
the same as those of (s 1)ζ(s)Γ(s) in the right half plane; therefore, ζ(s) = 0
− 6
in D(s ,1) for any y is equivalent to the radius of convergence of the series (4.4)
0 2
being at least 1 for any y.
2
ON THE ZEROS OF THE RIEMANN ZETA FUNCTION 7
Now, the union of the strips S = s C : 1 < σ < 1 and S = s C : 0 <
1 { ∈ 2 } 2 { ∈
σ < 1 form the critical strip minus the critical line σ = 1. Moreover, the strip
2} 2
S = s C: 1 <σ < 3 can be written as
{ ∈ 2 2}
1
(4.7) S = D(1+iy, )
2
[y
We concludefromthe above,thatifζ(s)doestnothaveazeroinsidethe stripS
and a fortiori does not have a zero in the strip S , then by the functional equation
1
ζ(s) cannot have a zero inside S neither. We thus have proved
2
Theorem 4.1. The Riemann hypothesis is equivalent to either
(1) the series (4.1) does not have any zero in the disk D(s ,1) for any y.
0 2
(2) the radius of convergence of the series (4.4) is at least 1 for any y.
2
Remark 4.2. We have been able to prove that the series (4.1) does not have any
zero in the disk D(1,1) (i.e y = 0). The proof is trivial and uses the criterion of
2
Petrovitch [9] for power series. We also have been able to prove the well-known
result that D(1,1) is a zero-free region. We have been unable to generalize the
prooftoanyy becauseasy getslargethevalueof b becomeverysmallcompared
0
| |
to that of b . The typical power series and polynomial non-zero regions criteria
1
| |
are inapplicable. More knowledge on the ratios b /b is needed.
n 0
| | | |
Remark 4.3. Anothercriterionofthe Riemannhypothesis canbe formulatedusing
the conformalmapping s= 1 which maps the plane (s)> 1 onto the unit disk
1 z ℜ 2
z <1. We can write −
| |
( z )ζ( 1 )Γ( 1 ),f(z) = ∞φ(t)e−tt1−zz dt
1 z 1 z 1 z Z
− − − 0
∞
(4.8) = a˜ zn,
n
nX=0
where the coefficients a˜ are given by
n
(4.9) a˜ = ∞φ(t)e tL ( lnt)dt,
n − n
Z −
0
L (x)=L0(x) being the Laguerre polynomial of order 0.
n n
The Riemann hypothesis is equivalent to the function f(z) having no zeros in
the unit disk.
Although the aboveformulationsof the Riemann hypothesis seemto be promis-
ing since exact information on the coefficients is known, we will not pursue this
approach. The new approach that we will adopt is presented next.
PART II
Inthispartwewillpursueacompletelydifferentapproachfromtheonepresented
inPARTI.Usingthenewseriesrepresentationofthezetafunctionofthefirstpart,
exact information on its zeros is provided based on Tauberian-like results.
8 LAZHARFEKIH-AHMED
5. The Series representation of (s 1)ζ(s)
−
In PART I, we showed that (s 1)ζ(s) process both an integral and a series
−
representation valid for (s) > 0. In the remaining of the paper we will only
ℜ
consider the series representation. We recall the series representation valid when
(s)>1:
ℜ
∞ Sn(s) s 1 ∞ e−tts−1
(5.1) = − dt=(s 1)ζ(s),
n+1 Γ(s) Z 1 e t −
nX=1 0 − −
where S (s) is given by
n
n 1
− n 1
(5.2) Sn(s)= ( 1)k − (k+1)−s.
− (cid:18) k (cid:19)
Xk=0
First, we provide another proof of the validity of the series representation for
(s) > 0. To prove the analytic continuation when (s) > 0, we need to evaluate
ℜ ℜ
the sum when (s) > 0. The next lemma, which will also be needed in the rest
ℜ
of the paper, provides such an estimation. It provides an estimate of the exact
asymptotic order of growth of Sn(s) when n is large.
n+1
S (s) 1
Lemma 5.1. n ∼ for n large enough and for all s =
n+1 n(n+1)(logn)1 sΓ(s)
−
σ+it, (s)>0, s / 1,2, .
ℜ ∈{ ···}
Proof. By putting k =m 1 in (5.2), we have by definition
−
n n
n 1 m n
Sn(s) = − ( 1)m−1m−s = ( 1)m−1m−s
(cid:18)m 1(cid:19) − n(cid:18)m(cid:19) −
mX=1 − mX=1
n
1 n 1
(5.3) = − ( 1)mm1−s = − ∆n(s 1),
n (cid:18)m(cid:19) − n −
mX=1
n
n
where ∆n(λ), ( 1)mm−λ.
(cid:18)m(cid:19) −
mX=1
The asymptotic expansion of sums of the form ∆ (λ), with λ C being nonin-
n
∈
tegralhasbeengiveninTheorem3ofFlajoletetal.[5]. Withaslightmodification
ofnotation,the authorsin[5]haveshownthat∆ (λ)hasanasymptoticexpansion
n
in descending powers of logn of the form
(5.4) ∆ (λ)∼(logn)λ ∞ ( 1)j Γ(j)(1) 1
− n − j!Γ(1+λ j)(logn)j
Xj=0 −
We apply the theorem to ∆ (λ) with λ=s 1 to get
n
−
(logn)s 1
(5.5) ∆ (s 1)∼ − − ,
n
− Γ(s)
which leads to the result
1
(5.6) S (s)∼ .
n n(logn)1 sΓ(s)
−
ON THE ZEROS OF THE RIEMANN ZETA FUNCTION 9
The Lemma follows from dividing equation (5.6) by n+1. (cid:3)
Now to obtain an analytic continuation when (s)>0, we simply observe that
ℜ
the logarithmic test of series in combination with the asymptotic value of S (s)
n
providedby Lemma 5.1imply thatthe absolutevalue ofthe seriesonthe left hand
sideof(5.1)isdominatedbyauniformlyconvergentseriesforallfiniteswhosereal
part is greater than 0.
Remark 5.2. ByWeierstrasstheorem,wecanseethatthefunction(s 1)ζ(s)canbe
−
extended outside of the domain (s)>1 and that it does not have any singularity
when (s)>0. Moreover, by reℜpeating the same process for (s)> k, k N, it
is cleaℜr that the series defines an analytic continuation of ζ(s)ℜvalid fo−r all s∈ C.
∈
6. Preparation Lemmas
Throughout this section, we suppose that 0< (s)<1. For a fixed s =σ+it,
ℜ
we associate with ζ(s) the following power series:
S (s) S (s) S (s) S (s)
(6.1) (s 1)ζ(s,x), 1 x+ 2 x2+ + n−1 xn−1+ n xn+
− 2 3 ··· n n+1 ···
x R.
∈
Let’s also further define the “comparison” power series by
s
log(1 x)
(6.2) Φ(x),(1 x) − =φ +φ x+φ x2+ +φ xn+
0 1 2 n
− (cid:18) x (cid:19) ··· ···
−
It is easy to verify that for σ >0
s
log(1 x)
(6.3) lim(1 x) − =0.
x 1 − (cid:18) x (cid:19)
→ −
Furthermore, direct calculation of Φ(x) yields the expression
′
s
log(1 x) s s
(6.4) Φ′(x)= − 1+s .
(cid:18) x (cid:19) − − log(1 x) − x
− (cid:0) − (cid:1)
Clearly, Φ(x) is well-defined for all x [0,1) and satisfies
′
∈
(6.5) lim Φ(x) = .
′
x 1| | ∞
→
In other words, the function Φ(x) is a continuous well-defined function of x,
′
converges for all values of x [0,1) and diverges when x 1. Moreover, because
∈ →
Φ(x) is analytic at x=0, Φ(x) must possess the following power series expansion
′
around x=0:
(6.6) Φ′(x)=φ1+2φ2x+ +nφnxn−1+
··· ···
Finally, we associate to the series (6.6) the following positive coefficients power
series:
(6.7) Φ˜(x), φ1 +2φ2 x+ +nφn xn−1+
| | | | ··· | | ···
10 LAZHARFEKIH-AHMED
The proofs in the remaining of this section will be based on two theorems. The
firsttheorem,whichisduetoNo¨rlund[11]andmorerecentlygeneralizedbyFlajolet
etal.,estimatestheasymptoticbehaviorofthecoefficientsofcertainpowersseries:
Theorem 6.1 ([6]). Let α be a positive integer and β be a real or complex number,
β / 0,1,2, . Define the function f(z) by
∈{ ···}
(6.8) f(z)=(1 z)α(1 log 1 β.
− z 1 z
(cid:0) − (cid:1)
Then, the Taylor coefficients f of f(z) satisfy
n
e (β) e β(β 1)
(6.9) fn ∼n−α−1(logn)β 11! (−logn) + 22! (log−n)2 +··· ,
(cid:0) (cid:1)
with
dk 1
(6.10) e = .
k dsk(cid:18)Γ( s)(cid:19)(cid:12)
− (cid:12)s=α
(cid:12)
The derivatives in (6.10) when s=α is a posi(cid:12)tive integer can be evaluated with
the help of the identity:
1 sin(πs)
(6.11) = Γ(1+s).
Γ( s) − π
−
For example, the value of e when α is a positive integer is given by
1
d sin(πs)
(6.12) e = Γ(1+s) = cos(πα)Γ(1+α),
1 −ds(cid:18) π (cid:19)(cid:12) −
(cid:12)s=α
(cid:12)
and in this case (cid:12)
βcos(πα)Γ(1+α)
(6.13) f ∼ .
n n1+α(logn)1 β
−
The secondtheorem, due to Appell [1], is the counterpartofl’Hospital’s rule for
divergent positive coefficients power series:
Theorem 6.2 ([3] p. 66). Let f(x),g(x) be two real power series of the form
∞ ∞
(6.14) f(x)= a xn,g(x)= b xn,a ,b >0 for all n>N,0<x<1.
n n n n
nX=1 nX=1
We further suppose that
• the series ∞n=1an, ∞n=1bn are both divergent so that x=1 is a singular
point of boPth f(x) anPd g(x).
a
n
lim =l,
• n→∞ bn
then,
f(x)
(6.15) lim =l.
x 1 g(x)
→
Our first result establishes an important property on the behavior of the deriv-
ative of the function Φ(x) when x is close to 1: