Notes on the zeta function
The Riemann zeta function is defined to be
∞
X
1
ζ(s) =
ns
n=1
for s > 1. The “integral test” from calculus shows that this infinite series converges (absolutely)
when s > 1. One goal is to get a definition for the zeta function that is valid for all values of s
except 1 (it’s not hard to show - and we will - that the zeta function blows up as s approaches 1).
But first, we give some motivation for why a number theorist would care.
Euler’s Product Formula: For all s > 1 we have
ζ(s) =
Y
(1 − p−s )−1 ,
p
where the (infinite) product is over all prime numbers p.
Thus, the zeta function is intimately connected with the prime numbers. Before we prove this
formula for the zeta function, we need a little result.
Lemma 1: Suppose a1 , . . . , an are real numbers satisfying 1 ≤ ai ≤ 2 for all i = 1, . . . , n.
Then for any positive b < 1 we have
n
Y
(ai + b) <
i=1
n
Y
ai + 22n b.
i=1
Proof: Exercise 1.
Proof of Product Formula: Since every prime p ≥ 2, we see that the geometric series
P∞
n=0
p−sn converges whenever s > 0; in fact
∞
X
p−sn =
n=0
1
.
1 − p−s
Moreover, we even have an exact formula for the partial sums:
N
X
n=0
p−sn =
1 − p−s(N +1)
1
p−s(N +1)
=
−
.
1 − p−s
1 − p−s
1 − p−s
1
Thus, assuming that s > 1 and using p ≥ 2,
0 < (1 − p−s )−1 −
N
X
2−(N +1)
= 2−N .
1 − (1/2)
p−sn <
n=0
Note that we also have
N
X
1≤
p−sn < 2
n=0
as well.
Next, notice what happens when you multiply these finite sums together:
N
Y X
p−sn =
0
X
m−s ,
m
p≤M n=0
where the sum on the right is over all integers m which have a factorization of the form
m = pe11 · · · perr ,
pi ≤ M and ei ≤ N for all i.
In particular,
ζ(s) >
N
Y X
X
p−sn ≥ ζ(s) −
p≤M n=0
n−s .
m>M
Now fix an > 0 and suppose s > 1. Then there is an M0 such that
P
n>M0
n−s < (since
the series for ζ(s) converges). If M ≥ M0 , then get an N such that 22π(M )−N < . (That’s the
function π(M ) in the exponent, not the product of M and the number π.) By Lemma 1 and what
we have shown,
N
Y X
p−sn <
p≤M n=0
Y
Y
N
X
p≤M
n=0
(1 − p−s )−1 <
p≤M

ζ(s) −
X
n>M
n−s <
Y
(1 − p−s )−1 < 
N
Y X
!
p−sn + 2−N

p−sn  + 22π(M )−N
p≤M n=0
n≤M
ζ(s) − <
Y
(1 − p−s )−1 < ζ(s) + .
p≤M
The last two inequalities are valid whenever M ≥ M0 , and since > 0 was arbitrary, this proves
the product formula.
We really would like the zeta function to be defined for more values of s than just s > 1. The
simplest way to do something like this is the following. Set
f (s) =
∞
X
(−1)n+1
.
s
n
n=1
2
By the alternating series test from calculus, this series converges whenever s > 0.
Lemma 2: For s > 1
(1 − 21−s )ζ(s) = f (s).
Proof: Exercise 2.
This prompts us to define
ζ(s) =
f (s)
1 − 21−s
for all positive s 6= 1. Of course, we just showed that this is the same as our original definition when
s > 1. We can use this expanded definition of the zeta function to compute a limit which gives us
more precise information on how zeta “blows up” at s = 1 (note that the following two-sided limit
would not be possible using our original definition).
Lemma 3: We have
lim (s − 1)ζ(s) = 1.
s→1
Proof: Exercise 3.
In particular, lims→1+ ζ(s) = ∞. This together with the product formula shows that there
must be infinitely many primes (otherwise the product would be finite when s = 1). This may
seem like a rather silly way to prove there are infinitely many primes, but these sorts of ideas can
be used to prove more opaque results (e.g., there are infinitely many primes congruent to 3 modulo
7).
While we have managed to expand the domain of the zeta function somewhat, we can do
much better. One generally goes about this using complex analysis (meaning analysis involving the
complex numbers). That’s out of our league here in Math 480, so we’ll try to just use calculus as
much as possible and rely on a few “facts.”
We take a temporary diversion and look at the Gamma function. This may seem to be rather
tangential, but in fact the Gamma function and zeta function are closely related.
As with the zeta function, we start with a definition for only some values of s:
Z ∞
Γ(s) =
e−x xs−1 dx
for s > 0.
0
When we write an improper integral such as this, it is shorthand for the appropriate one-sided
limit(s). Note that the infinite limit of integration is really no problem at all, since e−x goes to
zero quite rapidly. The zero lower limit of integration is what forces us to only allow s > 0.
3
The Gamma function has a fairly simple “functional equation” which allows us to define it for
all s except non-positive integers.
Lemma 4: For s > 0
Z
∞
e
Γ(s + 1) =
Z
−x s
x dx = s
∞
e−x xs−1 dx = sΓ(s).
0
0
Proof: Exercise 4.
Using this gives us a way to define Γ(s) for all s 6= 0, −1, −2, . . . For example,
R ∞ −x −.4
e x dx
Γ(−1.4)
Γ(−.4)
Γ(.6)
0
Γ(−2.4) =
=
=
=
.
−2.4
(−1.4)(−2.4)
(−.4)(−1.4)(−2.4)
(−.4)(−1.4)(−2.4)
It isn’t difficult to determine that Γ(1) = 1 (see below). Thus,
(1)
lim sΓ(s) = 1,
s→0
so that the Gamma function “blows up” like 1/s for s near 0.
Exercise 5: Use (1) and the functional equation for the Gamma function to evaluate
lim (s + 1)Γ(s)
s→−1
lim (s + 2)Γ(s)
s→−2
lim (s + 3)Γ(s).
s→−3
Find a general formula for
lim (s + n)Γ(s)
s→−n
for all positive integers n.
We explicitly evaluate the Gamma function for a couple values of s. First,
Z
Γ(1) =
0
∞
b
e−x dx = lim −e−x x=0 = lim 1 − e−b = 1.
b→∞
b→∞
Using this, if n is a positive integer we get
(2)
Γ(n + 1) = nΓ(n) = n(n − 1)Γ(n − 1) = · · · = n!Γ(1) = n!.
This also explains why 0! should be 1, since 0! should be Γ(1) if we set n = 0 in our equation above.
4
We next evaluate Γ(1/2). To do this we’ll use more advanced calculus (double integrals, to be
precise, and changing to polar coordinates). First, if u2 = x, then 2u du = dx and we have
∞
Z
e
Γ(1/2) =
−x −1/2
x
Z
∞
dx =
2
e−u 2 du
0
0
since u = 0 when x = 0 and u → ∞ when x → ∞. Squaring and passing to polar coordinates gives
us
∞
Z
2
Γ(1/2) = 4
−u2
e
0
Z ∞Z
0
e
−u2 −v 2
du dv
0
∞
Z
2
e−v dv
du
∞
=4
0
∞
Z
π/2
Z
0
2
e−r r dθ dr
=4
0
∞
Z
2
e−r r dr
= 2π
0
2 b
= π lim −e−r b→∞
r=0
= π.
In other words,
(3)
Γ(1/2) =
√
π.
Fact 1: For all s with neither s = −n nor s + 1/2 = −n for a non-negative integer n,
Γ(s)Γ(s + 1/2) = π 1/2 21−2s Γ(2s).
This is called Legendre’s formula.
Euler’s constant is the number γ defined by
γ = lim
n→∞
!
n
X
1
− log n .
m
m=1
Fact 2: For all s
∞ Y
1
s −s/n
= seγs
1+
e
.
Γ(s)
n
n=1
This is called Euler’s formula.
This infinite product converges for all s, and clearly is equal to zero precisely when s =
0, −1, −2, . . .
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Fact 3: For all s not an integer,
Γ(s)Γ(1 − s) = −sΓ(s)Γ(−s) =
π
.
sin(πs)
One more thing before we get back to the zeta function. We will use a certain type of function
called a theta function in what follows. Theta functions are used frequently in number theory, and
are a subject of study all by themselves. We will use a particular theta function,
θ(z) =
X
2
e−πn
z
=1+2
∞
X
2
e−πn z .
n=1
n∈Z
Fact 4: For all positive z
√
zθ(z) = θ(1/z).
We now return to the zeta function. So far we’ve seen how to define ζ(s) for all positive s 6= 1,
and we’ve seen how ζ(s) behaves as s approaches 1. We’ll prove a functional equation which will
allow us to define ζ(s) for all s 6= 1.
Consider the function
F (s) = π −s/2 Γ(s/2)ζ(s).
For now, we just say this defines F for all positive s 6= 1. We know, however, what F (s) should
behave like as s approaches 1; via Lemma 3 and (3)
lim (s − 1)F (s) = π −1/2 Γ(1/2) lim (s − 1)ζ(s) = 1.
s→1
s→1
For the moment, assume s > 1. Then using our original representation of the zeta function
F (s) = π −s/2
Z ∞
∞
X
1
dx
e−x xs/2
.
s
n 0
x
n=1
Take the integral here and make the substitution y = ax, where a is some positive number. Then
dy = a dx and
dy
y
=
a dx
ax
=
dx
x .
Also, y = 0 when x = 0 and y → ∞ when x → ∞. We get
Z
0
∞
e−x xs/2
dx
=
x
Z
6
0
∞
e−y/a
y s/2 dy
.
as/2 y
In particular, if we set a = 1/πn2 , we get
F (s) = π −s/2
=
=
=
∞ Z
X
Z ∞
∞
X
dx
1
e−x xs/2
s
n
x
0
n=1
∞
e−x
n=1 0
∞ Z ∞
X
n=1 0
Z ∞X
∞
0
x s/2 dx
πn2
x
2
e−πn y y s/2
2
e−πn y y s/2
n=1
dy
y
dy
.
y
The last step where we interchange the infinite sum with the integral is justified since everything
P∞
2
is absolutely convergent. If we set g(z) = n=1 e−πn z , then
Z ∞
dy
.
F (s) =
g(y)y s/2
y
0
As we remarked about the Gamma function, the problem here isn’t the infinite limit of integration so much as the zero lower limit of integration. We’ll get around that by splitting the
integral up and doing another change of variables. First
Z ∞
Z 1
Z ∞
dy
dy
dy
=
g(y)y s/2
+
g(y)y s/2
.
F (s) =
g(y)y s/2
y
y
y
1
0
0
Now set z = 1/y. Then dz = −y −2 dy and dz/z = −dy/y. Also, z = 1 when y = 1 and z → ∞ as
y → 0+ . Thus,
Z ∞
Z 1
Z ∞
Z ∞
dz
s/2 dy
−s/2 dz
s/2 dy
F (s) =
−
=
+
g(y)y
g(1/z)z
g(y)y
g(1/z)z −s/2 .
y
z
y
z
1
∞
1
1
But notice that 2g(w) + 1 = θ(w) for any w, so that Fact 4 gives
2g(1/z) + 1 = θ(1/z) =
√
zθ(z) =
√
z (2g(z) + 1) .
Hence
g(1/z) = z 1/2 g(z) +
z 1/2 − 1
.
2
Using this, we have
Z
∞
F (s) =
g(y)y
s/2
1
Z
∞
g(y)y s/2
=
1
Z
=
1
∞
dy
+
y
Z
dy
+
y
Z
∞
g(1/z)z −s/2
dz
z
g(z)z (1−s)/2
dz
+
z
1
∞
1
Z
1
∞
z 1/2 − 1 −s/2 dz
z
2
z
dy Z ∞ z 1/2 − 1
dz
g(y) y s/2 + y (1−s)/2
+
z −s/2 .
y
2
z
1
7
This last integral is easily evaluated (remember s > 1):
Z ∞ 1/2
Z
z
− 1 −s/2 dz
1 ∞ −(s+1)/2
z
z
=
− z −1−s/2 dz
2
z
2 1
1
b
1
z −(s−1)/2
z −s/2 = lim
−
2 b→∞ −(s − 1)/2
−s/2 z=1
b−(s−1)/2
b−s/2
1
1
1
= lim
−
−
−
2 b→∞ −(s − 1)/2 −s/2
−(s − 1)/2 −s/2
1
1
1
=
−
2 (s − 1)/2 s/2
1
1
=
− .
s−1 s
This gives us what we sought:
1
1
F (s) =
− +
s−1 s
Z
∞
1
dy
.
g(y) y s/2 + y (1−s)/2
y
The integral here converges no matter what s is (since it starts at 1 instead of 0). Thus, we have
a definition for F (s) whenever s 6= 0, 1, and we readily see that
lim −sF (s) = lim (s − 1)F (s) = 1.
s→0
s→1
Since we already had a definition for Γ(s/2) for all s 6= 0, −2, −4, . . . (where it blows up), knowing
that
π −s/2 Γ(s/2)ζ(s) =
1
1
− +
s−1 s
Z
∞
1
dy
g(y) y s/2 + y (1−s)/2
y
gives us a definition of ζ(s) for all s 6= 1. Notice that to counteract where Γ(s/2) blows up, we see
that
ζ(s) = 0
for s = −2, −4, −6, . . .
Exercise 6: Determine
lim
s→0
F (s)
Γ(s/2)
and use this to compute ζ(0).
Crucially, we notice that
Z ∞
dy
1
1
F (1 − s) =
−
+
g(y) y (1−s)/2 + y (1−(1−s))/2
(1 − s) − 1 1 − s
y
1
Z ∞
1
1
dy
=
− +
g(y) y (1−s)/2 + y s/2
s−1 s
y
1
= F (s).
8
In other words,
π (s−1)/2 Γ (1 − s)/2 ζ(1 − s) = π −s/2 Γ(s/2)ζ(s).
This equation can be used to define ζ(s) for all s 6= 1. We can clean this up by using Legendre’s
formula with s/2 in plase of s and Euler’s formula with (s + 1)/2 in place of s. We get
Γ(s/2)
ζ(s)
Γ(1 − (s + 1)/2)
−s+1/2 Γ(s/2)Γ (1 + s)/2 sin π(1 + s)/2
ζ(s)
=π
π
ζ(1 − s) = π −s+1/2
= π −s−1/2 cos(πs/2)π 1/2 21−s Γ(s)ζ(s)
= π −s 21−s cos(πs/2)Γ(s)ζ(s).
(Here we used the identity from trigonometry sin(θ + π/2) = cos(θ).) We now have Riemann’s
functional equation for the zeta function:
ζ(1 − s) = π −s 21−s cos(πs/2)Γ(s)ζ(s).
Exericse 7: Use Lemma 3, (1) and the functional equation to compute ζ(0).
9