Riemann Zeta Function and Prime Number
Distribution
Mingrui Xu
June 2, 2014
Contents
1 Introduction
2
2 Definition of zeta function and Functional Equation
2.1 Definition and Euler Product . . . . . . . . . . . . . .
2.2 Analytic Continuation . . . . . . . . . . . . . . . . . .
2.3 The Functional Equation . . . . . . . . . . . . . . . .
2.4 Zeros of Riemann zeta-function . . . . . . . . . . . . .
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3 Prime Number Distribution
3.1 Related Functions and Gauss’ Conjecture . . . . . . . . . . . . .
3.2 Connection between Riemann zeta-function and the distribution
of prime numbers. . . . . . . . . . . . . . . . . . . . . . . . . . .
3.3 Formula for prime number function with zeros in zeta function .
3.4 Riemann Hypothesis and Prime Number distribution . . . . . . .
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11
1
Introduction
The Riemann Zeta Function is a function of complex variable which plays an
important role in analytic number theory and prime number theorem. The
function was first studied by Leonhard Euler as a function of real variable and
then extended by Bernhard Riemann to the entire complex plane.
2
Definition of zeta function and Functional Equation
2.1
Definition and Euler Product
The Riemann Zeta function could be defined by either as Dirichlet Series or as
an Euler Product.
Definition 1.
For all s such that s = σ + it and σ > 1,the Riemann zeta-function ζ(s) is
defined as
∞
X
1
ζ(s) =
(1)
s
n
n=1
For all s such that σ ≥ σ0 > 1, we have that the series always converges
uniformly because
|
∞
X
n=1
1
n
|≤
σ+it
∞
X
n=1
1
|nσ+it |
=
∞
∞
X
X
1
1
≤
σ
σ0
n
n
n=1
n=1
(2)
Using Weierstrass’s theorem, since all the terms in (1) is analytic, the converging function σ(s) is holomorphic for all s such that Res = σ ≥ 1
Historically, the zeta-function arose from the need for an analytic tool to deal
with prime numbers, as Euler has found an equation that could give an alternating definition for zeta-function.
Theorem 1. (Euler Product) [2]
For s = σ + it, σ > 1 ,the following holds:
ζ(s) = Π(
1
)
1 − p1s
(3)
where the product on the right is taken over all prime numbers p.
Proof.Let X ≥ 2, we define the function ζX (s) as
ζX (s) = Π( p ≤ X)(
1
)
1 − p1s
Then with each of the factors on the right,
≥ 1, we can always rewrite
P∞as Res
1
1
the term as a geometric series: 1−p
s =
k=0 pks where each geometric series
2
converges absolutely. Hence, we could multiply term by term and replace (3)
with the new equation:
∞
∞
∞
X
X
X
1
1
)
=
...
mj s
m1
pks
...p
(p
1
J )
k =0
k=0
k =0
ζ(s) = Π(
1
(4)
j
when 2 = p1 < p2 < ... < pj as pj represents all the prime numbers up to X.
Using fundamental theorem of arithmetic, we know that for any positive integer
n, n is uniquely determined by
mj
1
n = pm
1 ...pj
when m1 ...mj are nonnegative integers and p1 ...pj are prime numbers. Consequently we can take the right side of (4) as the form
X 1
X0 1
+
ns
ns
n>X
n≤X
when the 0 stands for the summation over those natural numbers n > X whose
prime divisors are all ≤ X. We can give an upper bound of this sum by
Z ∞
X 1
X 1
X 1
1
σ
du
|
|
≤
<
≤
+
≤
X1 − σ
(5)
σ
ns
nσ
nσ
Xσ
u
σ
−
1
X
n>X
n>X
n>X
The right formula gives an upper bound of the sum. Now combining the definition of ζX (s) with equation (4) and (5), we obtain the relation between ζX (s)
and the zeta-function as
X 1
1
σ
+ O(
X 1−σ ).
(6)
ζX (s) = Π(1 − s )( − 1) =
p
ns
σ−1
n≤X
σ
Here O( σ−1
X 1−σ ) represents the upper bound. As we take the limit as X → ∞,
1−σ
we have X
→ 0 since σ > 1, hence the upper bound of the difference vanished
and we proved
∞
X
1
= Π(
)
(7)
1 − p1s
n=1
2.2
Analytic Continuation
The original way of defining zeta-function as sum of geometric series only works
when Re(s) = σ > 1 as the series converges. But Riemann’s basic philosophy
is that analytic functions should be dealt with globally on the complex plane.
The way of extending ζ(s) to the whole complex plane involves a new function
defined by taking a contour integral over a curve γ
Z
1
(−z)s−1
(8)
φ(s) =
2πi γ ez − 1
3
when the curve γ is defined by following a branch on the top with real part from
∞ to , detouring along a circle of radius around the origin, the returning from
to ∞ along the bottom edge of the branch. The integral function converges
and is entire on the whole complex plane.
To evaluate that,
we split the curve into 3 pieces, and first assume that Re(s) > 1, get the equation
Z
1
(
φ(s) =
2πi
Z
1
(
2πi
e(s−1)(logx−iπ)
dx +
ex − 1
(−z)s−1
dz +
e− 1
∞
e(s−1)(logx+iπ)
dx)
ex − 1
|z|=
∞
(9)
(as the on curve from ∞ to and from to ∞ are horizontal lines, dy = 0 so
we could replace dz with dx.) consider the integrand of the second integral, as
ez − 1 =P
0 on a simple point as z = 0, we can use Taylor expansion for ez − 1
∞
and get n=1 xn . By that, we can bound the integrand of the second integral
on the circle of |z| = by C ∗ s−2 , and hence the second integral is bounded by
C2 ∗ Re(s)−1 . Now as → 0 we have that the second integral also approaches
0. Passing the limit to the formula, we now obtain (as y = 0, we have that
dz = dx)
φ(s) =
0
∞
Z
Z
Z ∞ s−1
xs−1
x
s−1
dx ∗ (−1)
−
dx ∗ (−1)s−1 )
x
e −1
ex − 1
0
Z ∞ s−1
1 iπ(s−1)
x
−iπ(s−1)
=
(e
−e
)
dx
2πi
ex − 1
0
now rewrite the term in the parentheses, we obtain that
Z
sin(πs) ∞ xs−1
φ(s) = −
dx
π
ex − 1
0
(10)
(11)
then treat the integrand as a geometric series,which is justified as we are now
under the assumption that Re(s) > 1, we then get that
Z
0
∞
xs−1
dx =
ex − 1
∞
Z
(
0
4
∞
X
n=1
e−nx )xs−1 dx
(12)
as the geometric series converges uniformly for all the interval [, ∞), we can
justify interchanging the summation and integral sign, and hence get
Z
∞
(
0
∞
X
n=1
e−nx )xs−1 dx =
∞ Z
X
n=1
=
∞
e−nx xs−1 dx
0
∞ Z
∞
X
1 X ∞ −t s−1
Γ(s)
e
t
dt
=
= Γ(s)ζ(s)
ns n=1 0
ns
n=1
(13)
for Re(s) > 1.Now substitute this equation to (11), we get the following
φ(s) = −
sin(πs)
ζ(s)
Γ(s)ζ(s) = −
π
Γ(1 − s)
(14)
with the property of the Gamma Function. Now we can express ζ(s) with the
Gamma function and φ(s) which we just made. We obtain
Z
Γ(1 − s)
(−z)s−1
ζ(s) = −Γ(1 − s)φ(s) =
dz
(15)
z
2πi
γ e −1
here, both φ(s) and Γ(s) are entire functions except for Γ(s) has a simple pole
at s = 0 with residue 1. Hence, ζ(s) is analytic on the whole complex plane
except for a simple pole at s = 1.
2.3
The Functional Equation
The zeta-function has a very nice functional equation which would help people
compute it’s value and analyze it properties. For the zeta-function ζ(s), we have
the functional equation as
ζ(s) = 2s π s−1 sin(
πs
)Γ(1 − s)ζ(1 − s)
2
(16)
One way to get this functional equation is to slightly modify the curve of the
contour integral we use to define φ(s):we change the whole path from n going to
origin to n going to infinity, to a rectangle centered at original with horizontal
length 2n and vertical length sπn. We define such a path as γn ,and define
Z
1
(−z)s−1
φn (s) =
dz.
(17)
2πi γn ez − 1
in this case, we could generally bound the term |ez − 1| on the edge of the
rectangle by the following:
On the horizontal edge, we have that |ez − 1| ≥ |Im(ez )| = |eimpi | = 1
On the vertical edge, we have that |ez − 1| ≥ min{|e1 − 1|, |e−1 − 1|} > 12
Now we have that |ez − 1| > 21 on edges of the rectangle, we could bound the
integrand by 2ns−1 . By ML-estimate, the integral over the edges are bounded
by Cns . In this case, as s < 0, we have that this integral over edges tends to
5
R
R
0 as n → ∞. Now, the difference between γn and γ is the integral around
the closed loop with the boundary of the rectangle Rn and a detour along an
elongated indentation that doesn’t cross over the branch cut. The poles in
this closed contour are at the points of z = 2πki when 0 < k ≤ n, such that
ez − 1 = 0. Every pole is a simple pole with residue of
(−z)s−1 = (2π)s−1 |k|s−1 e(s−1)Log(±i)
ez z=±2πki
Now we apply residue theorem with the close contour of γn − γ and obtain that
φn (s) − φ(s) =
∞
X
(2π)s−1 k s−1 ∗ (e(s−1)iπ/2 + e−(s−1)iπ/2 )
(18)
k=1
Now we combine terms and get
π
πs
e(s−1)iπ/2 + e−(s−1)iπ/2 = 2cos((s − 1) ) = 2sin( )
2
2
(19)
Now as know that φn (s) → 0 as n → ∞. With the substitution of formula (19)
we get the following equation:
−φ(s) = 2s π s−1 sin(
πs
)ζ(1 − s)
2
(20)
Combining this with (15), this eventually becomes the final form of functional
equation of
πs
ζ(s) = 2s π s−1 sin( Γ(1 − s)ζ(1 − s)
(21)
2
2.4
Zeros of Riemann zeta-function
The values s when ζ(s) attains zero are called zeros of Riemann zeta-function.
From the functional equation (16), one can easily deduce that ζ(s) = 0 when
s = −2, −4, −6.... Those zeros are called trivial zeros since they have much
smaller significance.
The rest of zeros, are all at the critical strip in complex plane, which is the
strip of complex numbers with real parts 0 < Re(s) = σ < 1. Those zeros are
called non-trivial zeros and their distributions is an important topic in analyzing
number theories and prime number distribution. In 1859, Riemann stated a
formula in his paper about the number of zeros in critical strip, which was later
proved by von Mangoldt in 1905 [1], namely
N (T ) =
T
T
T
log
−
+ O(logT )
2π
2π 2π
(22)
here N (t) represents the number of zeros in the region 0 < σ < 1, 0 < t ≤ T
as s = σ + it. The precise location of zeros of ζ(s) still remain to this day an
unsolved mystery in math, and the Riemann hypothesis could be recast in the
formula N (T ) = N0 (T ) for all T > 0, where N0 (T ) is the number of zeros of
6
ζ(s) of the form s = 12 + it, 0 < t ≤ T ,in other words, all non-trivial zeros are
on the line with Re(s) = 12 , which is called the critial line.
The first nontrivial result is proved by G.H.Hardy in 1914, which gives
N0 (T ) > CT
(C > 0, T ≥ T0 )
(23)
This indicates that there are infinitely many zeros on the critical line. The result
was improved in 1942 by A.Selberg to
(C > 0, T ≥ T0 )
N0 (T ) > CT logT
(24)
based on equation 8, since the number of total zeros on critical strip is bounded,
this is already the best lower bound we can give. In 1974, N.Levinson proved
in his paper that
N − 0(T + U ) − N0 (T ) > V (N − 0(T + U ) − N (T ))
(25)
for C = 31 , which states that at least 13 of all non-trivial zeros are on the critical
line. Although this was a quite impressive result, it’s still much weaker than
Riemann Hypothesis and this method it self seems not to be able to yield the
result N0 (T ) ∼ N (T ) for T → ∞.
On the other hand, huge effort is also devoted to compute explicitly the nontrivial zeros using computers and try to find counter examples for Riemann
hypothesis. [3] The initial computations were done by Cray supercomputer using Riemann-Siegel formula and computed that the first 1012 non-trivial zeros
are all on the critical line. Then an improved algorithm, jointly with Arnold
Schonhage, helped to compute the first 1020 non-trivial zeros. By far, people has
already worked through the first 1022 non-trivial zeros, and no counter example of Riemann hypothesis has been found. The computation is still continued
using spare cycles on machines at AT&T labs. There are some statements that
the counter examples may lie far away from the values we could compute with
current algorithm.
3
3.1
Prime Number Distribution
Related Functions and Gauss’ Conjecture
To begin the discussion about the distribution of prime number, we have to
start with the definition of several functions that’s related to prime numbers on
the real line.First, the prime counting function
π(x) =
X
1
(26)
p≤x
shows the number of primes not exceeding x.Then, the von Mangoldt function
(
log p,
n = pk (p as prime, k ≥ 1)
Λ(n) =
(27)
0,
otherwise
7
and the more important Chebyshev function which is defined as
X
ψ(x) =
Λ(n).
(28)
n≤x
There is a closed connection between π(x) and φ(s), which is given by the
formula
φ(x)
1
π(x) =
(1 + O(
))
(29)
logx
logx
this equation shows that the prime number counting function π(s) roughly differ the Chebyshev-φ function by a factor of logx. Gauss made an important
conjecture with an estimation about prime counting function as
Z x
dt
(30)
π(x) ∼
logt
2
where as the formula on the right of (11) is sometimes also denote as function
Li(x). The prime number theorem, which states that
φ(x) ∼ x
(31)
is roughly equivalent to the estimate that π(x) ∼
3.2
x
logx .
Connection between Riemann zeta-function and the
distribution of prime numbers.
One convenient way to connect Riemann zeta-function to prime number distribution is to cast the Euler identity (2) into another form by taking the logarithm
on both sides and then differentiating, which leads to the formula
∞
−
XX
ζ 0 (s) X log p
=
=
(log p)p−ns
s
ζ(s)
p −1
n=1
(32)
The procedure is justified as Res = σ > 1 ,we can always write ps − 1 in the
form of geometric series. Then with the right formula, change the order of
summation, we obtain that for Res = σ > 1,
Z ∞
∞
ζ 0 (s) X
−s
−
=
Λ(n)n =
x−s dψ(s)
(33)
ζ(s)
1
n=1
equation (31) is important in the sense of connecting the zeta-function and the
distribution of prime numbers. Evaluate this integral by parts, we get that
Z ∞
1
ζ 0 (s)
= x−s ψ(x)∞ −
ψ(s)d(x−s )
(34)
ζ(s)
1
by definition, we have that ψ(1) = 0 and ψ(x) = O(xlogx) as x → ∞, we can
simplify this equation down to
Z ∞
ζ 0 (s)
−
=s
x−s−1 ψ(s)dx.
(35)
ζ(s)
1
8
which points a deep connection between the zeta function and the distribution
of prime numbers.
3.3
Formula for prime number function with zeros in zeta
function
There is one formula which link zeros of ζ(s) to Chebyshev ψ function:
ψ(x) = x −
X xρ
+ O(xT −1 (logxT )2 ) + O(logx)
ρ
(36)
|γ|≤T
when ρ denotes zeros for zeta function. To prove this formula, we need to prove
first a lemma. [1]
Lemma 1.
Let δ = 0 if 0 < y < 1, δ(1) = 12 , δ(y) = 1 if y > 1, and
Z
1
I(y, T ) =
2πi
c+iT
y s s−1 ds
(37)
c−iT
Then for y, c, T > 0 we have
(
|I(y, T ) − δ(y)| <
y c min(1, T −1 |logy|−1 )
cT −1 ,
if y = 1
if y 6= 1,
(38)
After the Lemma is proved, we could start prove this theorem by set c = 1 +
(logx)−1 and consider the following integral
1
J(x, T ) =
2πi
Z
c+iT
−
c−iT
ζ 0 (s)xs
ds
ζ(s)s
(39)
For Re(s) = σ > 1, we can apply equation (32) and get that the series converges
absolutely. Then apply Lemma 1, we get the following equation:
X0
n≤x
Λ(n) = J(x, T ) + O(
∞
X
Λ(n)(x/n)c min(1, T −1 |logx/n|−1 ))
n−1,n6=x
+ O(T −1 logx)
(40)
To estimate this series we consider first the terms with n ≤ 43 x or n ≥ 45 x. Then
|logx/n| 1, and since xc = ex the contribution of these term is
xT −1
∞
X
Λ(n)n−c =
1
xT −1 ζ 0 (c)
ζ(c)
xT −1(c − 1)−1 = T −1 xlogx
9
Then the work left is to estimate the term with 43 < n < 54 .We get that the
terms with in this range have contributions T −1 xlog 2 x, while there are at
most five terms, the sum of them is logx. Therefore we obtain
X
Λ(n) = J(x, T ) + O(T −1 xlog 2 x) + O(logx)
(41)
n≤x
Then the next step of the proof is to replace the vertical segment from c − iT
to c + iT in J(x, T )by the other three sides of the rectangle with vertices c −
iT, c + iT, −U + iT, −U − iT , where U is a large odd integer. If T 6= γ for any
zero rho, the the residue theorem gives
ψ(x) = x −
X xρ
ζ 0 (0)
−
−
p
ζ(0)
|γ|<T
X
0<2m<U
x−2m
−2m
+ O(T −1 xlog 2 x) + O(logx) +
Z
1
ζ 0 (s)xs ds
∗( −
)
2πi
ζ(s)s
α
(42)
when α is the modified curve. Based on the fact that the number of zeros rho
for which —γ − T | < 1 is O(logT ), the difference of those zeros could not all
be o(1/logT ). Hence we can choose T so that |γ − T | (logT )−1 for all zeros
ρ. The integral in (41) would all contribute to error terms in (36), and we need
0
(s)
bounds for ζζ(s)
to estimate the error term, which could be stated as
ζ 0 (s)
=
ζ(s)
X
ρ,|γ−T |<1
1
+ O(logT ) log 2 T
s−ρ
(−1 ≤ σ ≤ 2)
(43)
then use the functional equation, take the derivative to obtain the bound for
ρ ≤ −1 we have that
ζ 0 (s)
log(2|s|)
ζ(s)
(σ ≤ −1)
(44)
Now using these two bounds we get have
Z −U ±
Z c
T −1 log 2 T
xσ dσ xlog 2 T /(T logx)
(45)
−∞
c±iT
while (43) gives
Z
−U −iT
U −1 logU
−U +iT
Z
T
x−U dt T logU/U xU = o(1)
(U → ∞)
(46)
−T
insert this result to Riemann-von Mangoldt formula (18), we eventually proved
that
X xρ
X xρ
=
+ O(xT −1 logT ).
(47)
ρ
ρ
|γ|<T
|γ|T +A
10
3.4
Riemann Hypothesis and Prime Number distribution
Riemann Hypothesis suggests that all the zeros on the critical strip was on the
critical line with σ = 12 , although it’s not been proved yet, it could yield a lot of
elegant results and could be wrote in the equivalent form of some nice bounds
for prime numbers distribution.
For example, there is a theorem states that if 21 ≤ θ < 1 is fixed, then
ψ(x) = x + O(xθ log 2 x)
(48)
ζ(s) 6= 0
(49)
if and only if
f or
σ>θ
To prove this equation, we set σ > 1 and then obtain
Z ∞
Z ∞
∞
ζ 0 (s) X
−s
−s
Λ(n)n =
=
y dφ(y) = s
φ(y)y −s−1 dy.
ζ(s)
1−0
1
n=1
if we set φ(x) = x + R(x), we then obtain
Z ∞
ζ 0 (s)
s
−
=
+s
R(y)y −s−1 dy
ζ(s)
s−1
1
(σ > 1)
(50)
(51)
Now if R(x) xθ log 2 x, then the integral (50) is just a regular function of s
for σ > θ, so ζ(s) would not vanish. Then we deduce (48) from (49) using the
theorem (36) we just proved. Then for ρ = β + iγ, |γ| ≤ T we have β ≤ θ and
choose T = x1−θ we obtain the following
φ(x) = x + O(xθ
X 1
) + O(xT −1 log 2 x)
|γ|
|γ|≤T
= x + O(xθ log 2 x) + O(xT −1 log 2 x) = x + O(xθ log 2 x). (52)
Now as we already proved this theorem, with Riemann Hypothesis, we can easily
deduce this formula to
1
ψ(x) = x + O(x 2 log 2 x)
(53)
which could be a nice bound for the prime number function.
11
References
[1] Ivic, A. The Riemann zeta-function: theory and applications. Courier
Dover Publications, 2003.
[2] Karatsuba, A. A., and Voronin, S. M. The Riemann zeta-function,
vol. 5. Walter de Gruyter, 1992.
[3] Odlyzko, A. M. The 10ˆ 2ˆ 2-nd zero of the riemann zeta function.
Contemporary Mathematics 290 (2001), 139–144.
12