Zeta Functions in Number Theory
Prof. Don Bernard Zagier
University of Maryland, College Park, MD
and
Max-Planck-Institut für Mathematik, Bonn, F.R.G.
Hedrick Lectures given at the MAA/AMS meeting, Phoenix AZ, Jan 11-13, 1989
1. The Riemann Zeta Function and the Prime Number Theorem
The prototype of all zeta functions is the Riemann zeta function defined by the Dirichlet
series
X
n−s
(1)
ζ(s) =
n≥1
for complex numbers s = σ + it with σ > 1 and by analytic continuation for other values of
s. In these lectures we pursue two main themes: the arithmetical properties fo special values
of the zeta function; and the role of zeta-functions in analytic number theory, especially their
application to questions about the distribution of primes.
The first of these topics, which forms the material of the secnd and third lectures, orginated
with Euler. In 1735 he evaluated ζ(2k) for k = 1, 2, . . .. Here are a few of these remarkable
identities:
ζ(2) = 1 + 2−2 + 3−2 + . . . = π 2 /6,
ζ(4) = 1 + 2−4 + 3−4 + . . . = π 4 /90,
ζ(12) = 691π 12 /638512875.
In 1749 he published a paper entitled ‘Remarques sur un beau rapport entre les séries de
puissances tant directes que réciproques’ (or ‘Remarks on a beautiful relation between series of
direct and inverse powers’) in which he was able to attach a meaning to the zeta function at
negative integral arguments, deducing these additional formulae:
ζ(−1) = −1/12, ζ(−3) = 1/120, ζ(−1/11) = −691/32760.
These identies were established via a relationship between ζ(s) and ζ(1−s) which is a harbinger
of the functional equation proven by Riemann in 1859.
The topic of this lecture originated with Riemann. In his paper of 1859 ‘Über die Anzahl
der Primzahlen unter einer gegebenen Grosse’ he worked out the analytic properties of the
zeta function and established its connection with prime number theory. In particular, Riemann
proved that the zeta function has an analytic continuation to the entire complex plan and that
this extended function is meromorphic with a single pole at s = 1. Furthermore, he established
the functional equation relating the value of ζ(s) to ζ(1 − s) as adumbrated in Euler’s work.
Lastly, but most importantly for number theory, the holomorphic behavior of the zeta function
was linked to the distribution of prime numbers.
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To get a glimpse of how the zeta function is continued, let us see how the continuation to
the domain σ = <(s) > 0 is effected. This is all we need for now; the continuation to all s will
be taken up in lecture 2. On the one hand, for <(s) > 1 we have
Z ∞
∞
X
X Z n+1
1
1
dx
=
−
=
(n−s − x−s ) dx,
ζ(s) −
s
s − 1 n=1 ns
x
1
n
n≥1
and on the other hand, we have
Z
Z n+1
−s
−s
≤
(n
−
x
)
dx
n
1
|n−s − (x + n)−s | dx
0
|s|
n1+σ
by the Mean Value Theorem. So we see that the series in (2) converges absolutely and locally
uniformly for σ > 0, thus furnishing an analytic continuation of the zeta function to the domain
<(s) > 0.
The relationship of the zeta function to prime numbers stems from the Euler factorization
of ζ(s): For σ = <(s) > 1
X
n≥1
n−s =
Y
(1 + p−s + p−2s + . . .)
p
Y
(1 − p−s )−1
p
where the product extends over all primes p. (From now on the letter
P p always denotes a
prime.) The proof is easy: expand (1 − p−s )−1 as a geometric series p p−ks and use the fact
that every natural number can be written uniquely as a product of primes. Thus the Euler
factorization of the zeta function may be regarded as an analytic equivalent of the fundamental
theorem of arithmetic.
Rather then attempting a survey of the developments in analytic number theory in the
140 years since Riemann’s paper, we have decided to give a complete proof of a single result in
this field, namely, the Prime Number Theorem. Our reason for this choice is two-fold: first of
all, the Prime Number Theorem is one of the most famous achievements of mathematics, and
secondly its proof has recently become much more accessible.
The Prime Number Theorem gives an asymptomic for π(x), the number of primes less
than or equal to x. More precisely, it says
π(x) ∼
x
as x → ∞.
log x
(The ‘∼ ’ means that the ratio of the functions to the left and right of the ∼ tends to 1
as x tends to infinity.) This theorem was conjectured on the basis of numerical evidence by
Legendre and Gauss around 1800 and on the basis of theoretical evidence by Riemann in his
1859 paper, but was first proved only in 1896 by the French mathematician Hadamard and
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the Belgian mathematician de la Vallée Poussin (independently). Their proof was analytic
and sufficiently difficult that, I daresay, a majority of mathematicians, even number theorists,
could not reconstruct it unaided. A proof that can be considered ’elementary’ in that no use is
made of complex function theory or the holomorphic properties of ζ was found by Selberg and
Erdös in 1949, but this proof is even more complicated and recondite than the analytic proofs.
Finally, in 1980 D. J. Newman found a new proof which, although not ‘elementary’ in the sense
of the last sentence, is, nonetheless, far less complicated than the previous analytic proofs. In
fact, Newman’s proof is sufficiently simple that it can be presented in an hour to an audience
familiar with a basis course in complex variables. It is his proof that we describe now.
We begin with some well known preliminaties. First of all it is preferable to deal with the
function φ(x) defined by
X
log(p)
φ(x) =
p≤x
and to prove
φ(x) ∼ x as x → ∞.
(3)
This relation implies the Prime Number Theorem because a one-line calculation shows that
φ(x)
φ(x)
≤ π(x) ≤
+ x1−
log x
(1 − ) log x
for > 0. To establish equation (3) we need to show that, for any positive δ, we have |φ(x)−x| <
δx for sufficiently large x. As a first step we prove the special case δ = 1, i.e.
φ(x) < 2x,
by a method introduced by Chebyshev in 1849. (Actually, Chebyshev proved the case δ = 0.11
by a refinement requiring much more work.) Consider the middle binomial coefficient 2n
n =
(2n)!
2n
2n
= 2 by the
n!n! where n is a positive integer. On the one hand, it is less than (1 + 1)
binomial theorem, and on Q
the other hand it is an integer divisible by every prime between n
and 2n. This implies that n<p≤2n p ≤ 22n and, by taking logarithms, that
φ(2n) − φ(n) ≤ 2n log 2 ≤ 1.4n.
Assuming inductively that (4) is true for x = n, we find
φ(2n) ≤ φ(n) + 1.4n ≤ 3.4n < 2 × (2n).
and
φ(2n + 1) ≤ φ(2n) + log(2n + 1) ≤ 3.4n + log(2n + 1) < 2 × (2n + 1).
This proves (4) for all integral, and hence for all real, x > 0. The next step is to show that
ζ(s) has no zeros in the closed half-plane σ ≥ 1. This key analytic fact is common to all nonelementary proofs of the Prime Number Theorem. For its proof it is convenient to introduce
the function
X log p
,
(5)
F (s) =
ps
p
3
(the sum being over all primes in accordance with our convention about the letter p). Clearly,
the sum is absolutely convergent for σ > 1; moreover, in this range we have
−
X log p
X log p
ζ 0 (s)
d
= − log ζ(s) =
=
F
(s)
+
ζ(s)
ds
ps − 1
ps (ps − 1)
p
p
(6)
by virtue of the Euler factorization of ζ(s). The final sum is convergent for σ > 12 , so in view
of the properties of the zeta function which have been proven already, F(s) has a meromorphic
continuation to this half-plane with poles exactly at s = 1 and at the zeros of ζ(s). In particular,
this shows that ζ(s) has no zeros with σ > 1.
Let µ ≥ 0 be ther order of the zero of ζ(s) at s = 1 + iα, and let ν ≥ 0 be the order of the
zero of ζ(s) at s = 1 + 2iα, α 6= 0. Since ζ(s) = ζ(s), the orders of ζ(s) at 1 − iα and 1 + 2iα
are also µ and ν. We want to show µ = 0. Equation (6) implies the following estimates as ↓ 0
F (1 + ) ∼ 1/,
F (1 ± iα + ) ∼ −µ/,
F (1 ± 2iα) ∼ −ν/.
Consider the following linear combination of values of F
F (1 + 2iα + ) + 4F (1 + iα + ) + 6F (1 + ) + 4F (1 − iα + ) + F (1 − 2iα + ).
Noting that the coefficients are the binomial coefficients 4j , we see that this sum equals
i4
X log(p) h
iα/2
−iα/2
+
np
,
p
1+s
p
p
and so is greater than 0. Taking the limit as ↓ 0 we have
−ν − 4µ + 6 − 4µ − ν ≥ 0,
but since ν ≥ 0, this implies µ ≤ 3/4, and so in fact µ = 0, as required. This ingenious
argument was found by Mertens in 1898.
Now we come to the heart of Newman’s proof: we are going to show that the integral
Z ∞
φ(x) − x
dx
(7)
x2
1
converges. This apparently modest assertion easily implies (3), which as we have seen, implies
the Prime Number Theorem. Indeed, suppose that for some δ > 0 there are arbitrarily large
numbers x0 for which φ(x0 ) > (1 + δ)x0 . Then, since φ(x) is monotone increasing, we can
bound the integrand in (7) by
1
δx0
φ(x) − x
2
≥
2
x
(1 + 12 δ)2 x20
2
δ/2
δ
for x between x0 and (1 + 2 )x0 , and so the integral over this range is ≥ (1+ δ , which
2
contradicts the convergence of the intergral in (7) if x0 is sufficiently large. The reader may
supply a similar argument, using the interval ((1 − 12 δ)x0 , x0 ), for the case φ(x0 ) < (1 − δ)x0 .
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Now let us turn to the proof of Newman’s theorem. For σ = <(s) > 1 define a function
g(s) by
Z ∞
g(s) =
[φ(x) − x]/xs+1 dx.
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The integral exists, since by Chebyshev’s lemma φ(x) < 2x. Let us also introduce the truncated
integral:
Z A
gA (s) =
[φ(x) − x]/xs+1 dx, A > 1
1
which defines a function holomorphic for all s. Then
g(s) =
=
1X
1
log(p)/ps −
s p
s−1
1
1
F (s) −
.
s
s−1
The function F (s), which was introduced above in equation (5), is holomorphic for <(s) > 12
except for a simple pole with residue 1 at s = 1. Hence g(s) is holomorphic for <(s) ≥ 1 and
therefore, actually holomorphic in a slightly larger open set.
Our goal is to show that gA (1) → g(1) as A → ∞. By the Cauchy integral formula we
have
Z
s−1
1
1
s−1
+
gA (s)A
ds.
gA (1) =
2πi |s−1|=R
s−1
R2
The contour of intergration is made up of two semi-circles: C− the semi-circle to the left of
the line <(s) = 1, and C+ the semi-circle to the right of the line. The presence of the factor
As−1 and of the additive term (s − 1)/R2 in the intergrand may seem artificial now, but the
significance of their role emerges in the balance of the proof. We also have
1
g(1) =
2πi
Z
s−1
g(s)A
C0 ∪C+
s−1
1
+
ds,
s−1
R2
where the contour C0 extends from 1 + iR to 1 − iR and lies to the left of the line <(s) = 1
but within the region of regularity of g.
Let us estimate the difference
s−1
1
+
[gA (s) − g(s)]A
ds
s−1
R2
C+
Z
1
s−1
1
s−1
+
+
gA (s)A
ds
2πi C−
s−1
R2
Z
1
s−1
1
s−1
ds
−
+
g(s)A
2πi C0
s−1
R2
1
gA (1) − g(1) =
2πi
Z
s−1
5
Note that |As−1 | = Aσ−1 . For σ > 1, we have by Chebyshev’s estimate |φ(x) − x| < x
Z ∞
[φ(x) − x]/xs+1 dx
|gA (s) − g(s)| = A
Z ∞
A1−σ
≤
.
x−σ dx =
σ−1
A
(by the way, this explains why As−1 was inserted in the integrand.) For σ < 1 we have similarly
Z
|gA (s)| = A
1
[φ(x) − x]/xs+1 dx ≤ A1−σ /(σ − 1).
Finally, for |s − 1| = R we have
1
s − 1 1
1 +
+
=
s−1
R2
s − 1 s̄ − 1
s + s̄ − 2 =
(s − 1)(s̄ − 1)
= 2|σ − 1|/R2 .
Thus the integral over C+ and C1 have integrands which are less than 2/R2 in absolute
value, and consequently the two corresponding terms in the formula for gA (1) − g(1) are each
less than 1/R in absolute value. The integral over C0 depends on A only via s − 1. However, on
this contour <(s) < 1, and so the integral can be made arbitrarily small by taking A sufficiently
large.
Just to be sure that the quantifiers are correctly chosen, let us take > 0 and then set
R = /3. For this fixed value of R choose A so large that the integral over C0 is less than /3.
Then the three integrals are all together not more than .
Note. This was retyped from the original handout fromthe 1989 MAA/AMS meeting.
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