NOTES ON THE GAMMA FUNCTION AND THE RIEMANN
ZETA FUNCTION
1. Some results from analysis
Lemma 1. Suppose (fn ) is a sequence of functions analytic on an open subset D of
C. If (fn ) converges uniformly on every compact (closed and bounded) subset of D to
the limit function f then f is analytic on D. Moreover, the sequence of derivatives
(fn0 ) converges uniformly on compact subsets of D to f 0 .
Proof. Since fn is analytic on D we have by Cauchy’s integral formula
Z
1
fn (z)
fn (w) =
dz
2πi γ z − w
where γ is any closed and positively oriented contour in D and w is any interior point.
The region interior to and including γ is closed and bounded and hence compact.
So (fn ) converges uniformly on this region and hence we can pass to the limit under
the integral sign giving
Z
1
f (z)
f (w) =
dz.
2πi γ z − w
This implies that f is analytic on the region defined by γ and hence on the whole of
D.
For the derivatives we have
Z
fn (z)
1
0
dz
fn (w) =
2πi γ (z − w)2
and
1
f (w) =
2πi
0
Z
γ
f (z)
dz.
(z − w)2
Hence
|fn0 (w)
Z
1 fn (z) − f (z) − f (w)| =
dz 2π γ (z − w)2
fn (z) − f (z) ≤ (length of γ) × sup (z − w)2 z∈γ
0
and this tends to 0 as n → ∞ for any w on the interior of γ, and hence for any compact
subsets of D (just choose γ appropriately which is possible since D is open).
1
NOTES ON THE GAMMA FUNCTION AND THE RIEMANN ZETA FUNCTION
2
Lemma 2 (Differentiating under the integral sign). Let D be an open set and let γ
be a contour of finite length L(γ). Suppose ϕ : {γ} × D → C is a continuous function
and define g : D → C by
Z
ϕ(w, z)dw.
g(z) =
γ
Then g is continuous. Also, if
analytic with derivative
0
∂ϕ
∂z
exists and is continous on {γ} × D then g is
Z
g (z) =
γ
∂ϕ
(w, z)dw.
∂z
Proof. Let z1 , z2 ∈ D with z2 fixed. Since ϕ is continuous, given ε > 0 we can find a
δ > 0 such that |z1 − z2 | < δ ⇒ |ϕ(w, z1 ) − ϕ(w, z2 )| < ε/L(γ). Hence, given ε > 0
choose δ as above then by linearity of the integral and the estimation lemma
Z
|g(z1 ) − g(z2 )| = (ϕ(w, z1 ) − ϕ(w, z2 ))dw
γ
≤ L(γ) max |ϕ(w, z1 ) − ϕ(w, z2 )|
w∈γ
< ε.
Hence g is continuous. If ∂ϕ
exists and is continuous then
∂z
ϕ(w, z + h) − ϕ(w, z) ∂ϕ
→0
−
(w,
z)
h
∂z
with h. Then again by linearity of the integral and the estimation lemma
Z
g(z + h) − g(z)
∂ϕ
−
(w, z)dw
h
γ ∂z
Z ϕ(w,
z
+
h)
−
ϕ(w,
z)
∂ϕ
= −
(w, z) dw
h
∂z
γ
ϕ(w, z + h) − ϕ(w, z) ∂ϕ
≤ L(γ) max −
(w, z)
w∈γ
h
∂z
and this tends to 0 with h.
Corollary 3. Let D be an open set and ϕ : [a,
∞] × D → C be continuous with
R∞
∂ϕ
continuous partial derivative ∂z . If the integral a ϕ(t, z)dt converges uniformly on
subsets of D then it defines an analytic function there and has derivative
Rcompact
∞ ∂ϕ
(t,
z)dt.
a ∂z
Rn
Proof. Let fn (z) = a ϕ(t, z)dt (so γ is the straight line joining a and n) . By the
Rn
above lemma each fn is analytic (with fn0 (z) = a ∂ϕ
(t, z)dt) and by hypothesis
∂z
NOTES ON THE GAMMA FUNCTION AND THE RIEMANN ZETA FUNCTION
3
R∞
fn → f = a ϕ(t, z)dt uniformly on compact subsets of D. Applying Lemma 1 gives
us the result.
2. The Gamma function
Let s = σ + it with σ, t ∈ R. We define the Γ-function for σ > 0 by
Z ∞
(1)
Γ(s) =
e−t ts−1 dt.
0
Note that in the integral we have the two ‘bad’ points; 0 and ∞. Also, we cannot
immediately apply Corollary 3 since the integrand is not always continuous at 0. It
turns out none of this matters and we have the following.
Proposition 4. Γ(s) is analytic for σ > 0.
R∞
Proof. First, note for a > 0 the function defined by a e−t ts−1 dt is analytic. To see
this we only need show it is uniformly convergent (on compact blah) and then we can
apply Corollary 3 since all other hypotheses are met. As expected, the exponential
dominates the tail of the integral giving
Z n
Z ∞
Z ∞
−t
s−1
−t
s−1
−t
s−1
= e
t
dt
−
e
t
dt
e
t
dt
a
a
n
Z ∞
≤
e−t tσ−1 dt
n
Z ∞
1
≤ C
e− 2 t dt
n
− 12 n
= 2Ce
and this → 0 as n → ∞ giving uniform convergence. Now for σ > 0 define,
Z ∞
e−t ts−1 dt.
fn (s) =
1
n
By the above argument each fn is analytic. Suppose σ ≥ c > 0. For 0 < t ≤ 1 we
have e−t < 1 and tσ−1 ≤ tc−1 . Hence, for n > m,
Z 1
Z 1
m
m
1
−t s−1 e
t
dt
tc−1 d = (m−a − n−a ).
<
1
1
c
n
n
Given ε > 0 we can choose 0 < δ < 1 such that 1c (m−a − n−a ) < ε whenever
|m−1 − n−1 | < δ. Hence the fn satisfy the Cauchy condition for uniform convergence
in compact subsets of the halfplane σ > 0. Applying Lemma 1 we see that the
gamma function is analytic for σ > 0.
NOTES ON THE GAMMA FUNCTION AND THE RIEMANN ZETA FUNCTION
4
We can show the Γ function is an extension of factorial function to complex arguments, via the following functional equation
Proposition 5. For σ > 0 we have
(2)
Γ(s + 1) = sΓ(s).
Proof. Integration by parts gives
Z
Z ∞
∞
−t s
−t s e t dt = −e t + s
0
0
∞
e−t ts−1 dt = sΓ(s)
0
By direct computation we see Γ(1) = 1 and hence by induction Γ(n+1) = n! for all
positive integers n. The functional equation also gives us the analytic continuation
of Γ.
Theorem 1. The Γ function can be extended over the whole complex plane to a meromorphic function with simple poles at the negative integers and zero. The residues
of these poles are given by
(−1)n
(3)
Ress=−n Γ(s) =
n!
Proof. By (2) we have
(4)
Γ(s) =
Γ(s+n)
s(s+1)(s+2) · · · (s+n−1)
for any positive integer n. Now Γ(s+n) is analytic for σ > −n so the function on
the right is meromorphic for σ > −n with simple poles at 0, −1, −2, . . . , −(n − 1).
Since n is arbitrary we’re done. By construction this extension of Γ satisfies (2). To
calculate the residues we rewrite (4) as
Γ(s) =
Γ(s+n + 1)
s(s+1)(s+2) · · · (s+n)
and proceed directly viz:
(s + n)Γ(s+n + 1)
s→−n s(s+1)(s+2) · · · (s+n)
Γ(s+n + 1)
= lim
s→−n s(s+1)(s+2) · · · (s+n− 1)
(−1)n
=
n!
where we have used Γ(1) = 1 in the numerator.
Res(Γ; −n) =
lim
NOTES ON THE GAMMA FUNCTION AND THE RIEMANN ZETA FUNCTION
5
From now on when we refer to the Γ-function we mean the meromorphic continuation. Heuristically we can think of this as the limit in n of the right hand side of
(4), and this is in fact not too far from the truth.
n
it is not
2.1. Product Representations of Γ(s). Since ex = limn→∞ 1 + nx
unreasonable to expect
n
Z n
t
(5)
Γ(s) = lim
1−
ts−1 dt.
n→∞ 0
n
We first prove this and then use it to give an alternative representation of Γ(s), which
can be thought of as the limit in n of (4).
Lemma 6. Formula (5) holds for σ > 0.
Proof. Denote
Z
fn (s) =
0
Then
n
t
1−
n
n
ts−1 dt.
n Z ∞
t
s−1
Γ(s) − fn (s) =
t dt +
e−t ts−1 dt.
e − 1−
n
0
n
The second integral is just the tail of the Γ-function which → 0 as n → ∞. We
want to show the first integral also → 0. This seems feasible since the first factor of
the integrand gets arbitrarily small for increasing n. Now, for 0 ≤ y ≤ 1 we have
1 + y ≤ ey ≤ (1 − y)−1 . For n large set y = t/n then
n
−n
t
t
−t
≤e ≤ 1+
.
1−
n
n
Z
n
−t
Hence
0≤e
−t
t
− 1−
n
n
n t
t
= e
1−e 1−
n
n n t
t
−t
≤ e
1− 1+
1−
n
n
2 n
t
= e−t 1 − 1 − 2
.
n
−t
Now, if 0 ≤ a ≤ 1 then (1 − a)n ≥ 1 − na when na < 1. Letting a = t2 /n2 then for
large n
n
t2
t2
1− 1− 2
≤ .
n
n
NOTES ON THE GAMMA FUNCTION AND THE RIEMANN ZETA FUNCTION
6
Therefore
0≤e
−t
t
− 1−
n
n
≤ n−1 t2 e−t .
This gives
1
|Γ(s) − fn (s)| ≤
n
Z
n
e−t tσ+1 dt <
0
1
Γ(σ + 2) → 0
n
as n → ∞ since Γ(σ + 2) is finite.
We deduce the alternative representation of Γ(s) as follows. Substituting u = t/n
we have
n
Z n
Z 1
t
s−1
s
1−
t dt = n
(1 − u)n us−1 du
n
0
0
1 n Z 1
1 s
s
n−1 s
n
= n
(1 − u) u du
u (1 − u) +
s
s 0
0
Z 1
n
s
n−1 s
= n
(1 − u) u du
s 0
= ······
Z 1
n(n − 1) . . . 1
s
= n
u(s−1)+n du
s(s + 1) . . . ((s − 1) + n) 0
n!
=
ns .
s(s + 1) · · · (s + n)
Taking the limit as n → ∞ gives
Proposition 7. For s 6= 0, −1, . . .
(6)
n!
ns .
n→∞ s(s + 1) · · · (s + n)
Γ(s) = lim
This converges for all other s so gives us another meromorphic continuation of Γ.
This formula is quite useful and has a few consequences. The first of which is
Corollary 8 (Weierstrass Product). For s 6= 0, −1, . . . we have
(7)
∞
e−γs Y s −1 z/k
Γ(s) =
1+
e
s k=1
k
where γ is the Euler-Mascheroni constant.
NOTES ON THE GAMMA FUNCTION AND THE RIEMANN ZETA FUNCTION
7
Proof. For s 6= 0, −1, . . . we have
n!
ns
n→∞ s(s + 1) · · · (s + n)
1
= lim
es log n
n→∞ s(1 + s)(1 + s/2) · · · (1 + s/n)
Γ(s) =
lim
1
1
1
1
es(log n−1− 2 −···− n )
es(1+ 2 +···+ n )
= lim
n→∞
s
(1 + s)(1 + s/2) · · · (1 + s/n)
n
−γs
Y
s −1 z/k
e
lim
1+
=
e .
s n→∞ k=1
k
This formula clearly demonstrates the poles as well as giving us the fact that Γ(s)
has no zeros. Also, taking the logarithm of the product, differentiating and then
evaluating at s = 1 gives Γ0 (1) = −γ.
2.2. The Reflection and Duplication Formulae. We can use (6) to prove the
famous reflection and duplication formulae. For the first of these we need a lemma.
Lemma 9. We have the following expansions
∞
X
1
1
(8)
π cot(πs) = + 2s
2
s
s − n2
n=1
and
(9)
Proof. Let
∞ s2
sin(πs) Y
1− 2 .
=
πs
n
n=1
∞
X
1
1
F (s) = + 2s
.
2
s
s − n2
n=1
with s ∈ C\Z. If s is near an integer we expect to see some fairly large terms in
the series but these will die out as n increases. This is enough to guarantee absolute
convergence: for n > √12 |s| we have |s2 − n2 | ≥ n2 − |s|2 > n2 /2 and hence
X
X 1
1
<
|s2 − n2 |
n2
1
1
n> √ |s|
2
n> √ |s|
2
which converges. Adding in the finite number of other terms gives that the series
in (8) converges absolutely. Note this also implies the series converges uniformly
NOTES ON THE GAMMA FUNCTION AND THE RIEMANN ZETA FUNCTION
8
on compact subsets and hence defines an analytic function on C\Z by Lemma 1.
Splitting the summand into partial fractions we see F(s) is periodic in σ with period
1. The pole at 0 is simple and has residue 1. By periodicity every poles is simple
with residue 1. Therefore, the function defined by f (s) = π cot(πs) − F (s) is entire
and periodic in σ with period 1. We show f (s) is bounded then apply Liouville’s
theorem. By periodicity it suffices to show f is bounded when 0 ≤ σ < 1 and since
f is entire we need to show it’s bounded as t = =(s) → ±∞. Now,
π cot(πs) = πi
eπis + e−πis
2πi
.
= πi + 2πis
πis
−πis
e −e
e
−1
Since |e2πis | = e−2πt we have limt→±∞ π cot(πs) = ∓πi. For F (s) note that in
the region 0 ≤ σ < 1 we have |t| ≤ |s| < |t| + 1. We also have |s2 − n2 | =
|σ 2 − t2 − n2 + 2iσt| ≥ |σ 2 − t2 − n2 | = |σ 2 − (t2 + n2 )| ≥ |t2 + n2 | − σ 2 > t2 + n2 − 1.
Hence
∞
X
1
1
|F (s)| ≤
+ 2(|t| + 1)
|t|
n2 + t2 − 1
n=1
Z
∞
1
dx
≤
+ 2(|t| + 1)
|t|
x2 + t 2 − 1
0
√
∞
tan−1 x/ t2 − 1 1
√
+ 2(|t| + 1)
=
|t|
t2 − 1
0
|t| + 1
1
+ π√
.
=
|t|
t2 − 1
So F (s) is bounded. Therefore f (s) is bounded and hence constant by Liouville. At
s = 1/2 we have π cot(π/2) = 0 and
∞ X
1
1
−
F (1/2) = 2 −
=0
n − 1/2 n + 1/2
n=1
hence f ≡ 0.
Q
To see (9) consider g(s) = sin(πs)/(πs n≥1 (1−s2 /n2 )). The product is absolutely
convergent so g exists for s ∈ C\Z. g(s) tends to 1 as s tends to 0 and g has period
1 implying g(s) tends to 1 as s tends to any integer. The logarithmic derivative is
given by
!
∞
0
X
g (s)
1
1
= π cot(πs) −
+ 2s
=0
2
g(s)
s
s − n2
n=1
hence g is constant and since g(0) = 1 we have g(s) = 1 for all s.
NOTES ON THE GAMMA FUNCTION AND THE RIEMANN ZETA FUNCTION
9
Proposition 10 (Reflection Formula).
Γ(s)Γ(1 − s) =
(10)
π
.
sin πs
Proof. By (6) and (9) we have
n!
n!
ns
n−s
n→∞ s(s + 1) · · · (s + n)
−s(−s + 1) · · · (−s + n)
1
Qn
= lim
s
2
n→∞ −z
1 − ns
k=1 1 + n
π
= −
.
s sin πs
Γ(s)Γ(−s) =
lim
So by (2)
Γ(s)Γ(1 − s) = Γ(s)(−s)Γ(−s) =
π
.
sin πs
Setting s = 1/2 in the reflection formula gives Γ(1/2) =
√
π.
Proposition 11 (Duplication Formula). We have the following formula
1
(11)
Γ(s)Γ s +
= 21−2s π 1/2 Γ(2s)
2
Proof. The trick here is to use a convenient expression for Γ(2s). By (6) we have
(
n!ns
n!ns+1/2
Γ(s)Γ(s + 1/2)
= lim
n→∞
Γ(2s)
s(s + 1) · · · (s + n) (s + 12 )(s + 23 ) · · · (s + 12 + n)
)
2s(2s + 1) · · · (2s + 2n)
×
(2n)!(2n)2s
(
)
1
(n!)2 n1/2 22n+1
= 2s lim
2 n→∞ (2n)!(z + n + 12 )
(
)
1
(n!)2 22n+1
= 2s lim
1
2 n→∞ (2n)!n1/2 (1 + z/n + 2n
)
(
)
1
(n!)2 22n+1
= 2s lim
2 n→∞ (2n)!n1/2
1
C
22s
√
Setting s = 1/2 gives C = 2Γ(1/2) = 2 π and we’re done.
=
NOTES ON THE GAMMA FUNCTION AND THE RIEMANN ZETA FUNCTION
10
We finish this section on the Γ function with a formula that is very useful for
estimating Γ(s).
2.3. Stirling’s Formula. The following theorem characterises the Γ function uniquely
and will prove useful.
Theorem 2 (Uniqueness Theorem). Let F be analytic in the right half-plane A =
{s ∈ C : σ > 0}. Suppose F (s + 1) = sF (s) and that F is bounded in the strip
1 ≤ σ < 2. Then F (s) = aΓ in A with a = F (1).
Proof. Consider f (s) = F (s) − aΓ(s). The equation f (s + 1) = sf (s) holds in A
and so we can extend f meromorphically to the whole plane as we did for Γ. If
any poles occur these must be at the negative integers. Since f (1) = 0 we have
lims→0 sf (s) = 0, hence f doesn’t have a pole, or anything worse, at 0 and we can
thus continue f analytically to 0. This gives the analytic continuation of f to the
negative integers via f (s + 1) = sf (s).
Now, |Γ(s)| ≤ Γ(σ) and this is bounded for 1 ≤ σ < 2. Since F is bounded here
by hypothesis, f is also. Now consider the region with 0 ≤ σ ≤ 1. If t = =(s) ≤ 1
then f is bounded since it’s analytic here. If t > 1 then f is bounded here since
f (s) = f (s + 1)/s and f is bounded for 1 ≤ σ < 2. Since f (s) and f (1 − s) assume
the same values for 0 ≤ σ ≤ 1 we have that g(s) = f (s)f (1 − s) is bounded and
analytic. By Liouville g(s) ≡ g(1) = 0 and hence f ≡ 0.
Our goal is to use the uniqueness theorem to prove
√
1
Γ(s) = 2πss− 2 e−s eµ(s)
for s ∈ C− = C\R≤0 where
Z
µ(s) := −
0
∞
P1 (x)
dx
s+x
1
.
2
and P1 (x) = x − bxc −
We need to show µ is analytic and that it posseses an
appropriate functional equation (so that the above representation of Γ satisfies the
functional equation (2)). For this we need a lemma.
Lemma 12 (‘Twisted’ 4 -inequality). For s = reiθ and x ≥ 0 we have
(12)
|s + x| ≥ (|s| + x) cos(θ/2).
This gives
(13)
|s + x| ≥ (|s| + x) sin(δ/2)
when |θ| ≤ π − δ, 0 < δ ≤ π (since cos(θ/2) ≥ sin(δ/2)).
NOTES ON THE GAMMA FUNCTION AND THE RIEMANN ZETA FUNCTION
11
Proof. Using cos θ = 1 − 2 sin2 (θ/2) and (r + x)2 ≥ 4rx we have
|s + x|2 = r2 + 2rx cos θ + x2 = (r + x)2 − 4rx sin2 (θ/2) ≥ (r + x)2 cos2 (θ/2)
Proposition 13. µ(s) is analytic in C− .
Proof. Define Q(x) = 21 (x − bxc − (x − bxc)2 ). Then Q(t) is an antiderivative of
−P1 (x) (so continuous) and 0 ≤ Q(x) ≤ 1/8. We have
β Z β
Z β
P1 (x)
Q(x)
Q(x) (14)
−
dx =
+
dx
2
s+x α
α s+x
α (s + x)
for 0 < α < β < ∞. Now let 0 < δ ≤ π and ε > 0. Then for x ≥ 0, s = reiθ with
r > ε and |θ| ≤ π − δ we have (by the above lemma)
Q(x) 1
|Q(t)|
(s + x)2 ≤ sin2 (δ/2)(ε + x)2 ≤ 8 sin2 (δ/2)(ε + x)2 .
Hence the integral
Z
∞
0
Q(x)
dx
(s + x)2
is uniformly convergent in compact subsets of C− and therefore defines an analytic
function by Corollary 3. But by (14) we have
Z ∞
Q(x)
dx.
(15)
µ(s) =
(s + x)2
0
Note by (15) and (12), (13) we have
(16)
(17)
1
8 cos2 (θ/2)|s|
1
|µ(s)| ≤
2
8 sin (δ/2)|s|
|µ(s)| ≤
for s = reiθ , |θ| ≤ π − δ, 0 < δ ≤ π.
Proposition 14. For s ∈ C− we have
1
s+1
(18)
µ(s) − µ(s + 1) = s +
log
− 1.
2
s
NOTES ON THE GAMMA FUNCTION AND THE RIEMANN ZETA FUNCTION
12
Proof. Since P1 (x + 1) = P1 (x) we have
Z ∞
Z ∞
P1 (x + 1)
P1 (x)
dx = −
dx
µ(s + 1) = −
(s + 1) + x
s+x+1
0
0
Z 1
Z ∞
P1 (x)
x − bxc − 1/2
dx = µ(s) − −
dx
= −
s+x
s+x
1
0
Z 1
x − 1/2
= µ(s) +
dx
s+x
0
Z 1
s + 1/2
= µ(s) +
1−
dx.
s+x
0
Theorem 3 (Complex Stirling’s Formula). For s ∈ C− we have
√
1
(19)
Γ(s) = 2πss− 2 e−s eµ(s) .
Proof. By Proposition 13 the function
1
F (s) = ss− 2 e−s eµ(s)
1
1
is analytic in C− (we define ss− 2 = e(s− 2 ) log s where log is the principal branch of
the logarithm). By Proposition 14 we have
1
F (s + 1) = (s + 1)s+1/2 e−s−1 eµ(s)−(s+ 2 ) log(
s+1
s
)+1 = ss+1/2 e−s eµ(s) = sF (s).
Also F (s) is bounded in the region A = {s ∈ C : σ > 0}: Clearly, eµ(s) is bounded by
1
(16). Writing s = σ + it = |s|eiθ ∈ C we have |ss− 2 e−s | = |s|σ−1/2 e−θt e−σ . Then for
s ∈ A with |t| ≥ 2 we have σ − 1/2 ≤ 2, |s| ≤ 2t and −tθ ≤ −π|t|/2. Hence, in this
1
region we have |ss− 2 e−s | ≤ 4t2 e−π|t|/2 e−1 → 0 as |t| → ∞. Hence F (s) is bounded in
A. By the Uniqueness Theorem we must have
1
Γ(s) = ass− 2 e−s eµ(s)
for some a. Substituting this into the duplication formula (11) gives
√
a2 ss−1/2 e−s eµ(s) (s + 1/2)s e−s−1/2 eµ(s+1/2) = a21−2s π(2s)2s−1/2 e−2s eµ(2s)
√
= a 2πs2s−1/2 e−2s eµ(2s) .
Hence
√
a(1 + 1/2s)s eµ(s)+µ(s+1/2) = 2πeeµ(2s) .
Now let s be real and approach infinity.√By (16) the exponentials tend to 1 and since
√
limx→∞ (1 + 1/2x)x = e we have a = 2π.
This result immediately gives the original form of stirling’s formula
NOTES ON THE GAMMA FUNCTION AND THE RIEMANN ZETA FUNCTION
Corollary 15. As x → ∞ in R we have Γ(x) ∼
√
13
2πxx−1/2 e−x .
Another consequence is the following.
Corollary 16 (Rapid decay in vertical strips). Let σ ∈ R be fixed. Then as |t| → ∞
we have
√
(20)
|Γ(σ + it)| ∼ 2π|t|σ−1/2 e−π|t|/2 .
Proof. Assume t ≥ 0. By (19) and (16) we have
log |Γ(σ + it)| = <(log(Γ(σ + it)))
1
log 2π + O(1/t)
2
= <((σ + it − 1/2)(log(σ 2 + t2 )/2 + i arg(σ + it)))
1
−σ + log 2π + O(1/t)
2 p
= (σ − 1/2) log |t| 1 + σ 2 /t2 − t arg(σ + it)
= <((σ + it − 1/2) log(σ + it) − σ − it) +
1
log 2π + O(1/t)
2
= (σ − 1/2)(log |t| + o(1)) − t(π/2 − tan−1 (σ/t))
1
−σ + log 2π + o(1)
2
1
= (σ − 1/2) log |t| − πt/2 + log 2π + o(1).
2
Doing similar stuff for t ≤ 0 gives the result.
−σ +
3. The Riemann zeta function
3.1. Analytic continuation and functional equation. Let s = σ + it with σ, t ∈
R and let
−1
∞
X
Y
1
1
=
1− s
.
(21)
ζ(s) =
s
n
p
n=1
p
The series is absolutely and locally uniformly convergent for σ > 1 and hence ζ(s)
is analytic in this region. We plan to extend the domain on which ζ can be defined
(hopefully this will allow us to exploit its connection with the primes). More precisely,
we wish to find a function which agrees with the zeta function on the domain σ > 1,
NOTES ON THE GAMMA FUNCTION AND THE RIEMANN ZETA FUNCTION
14
but that also makes sense outside of this region. This process is known as analytic
continuation.
As an example of analytic continuation, consider the two functions
f (z) =
∞
X
zn,
n=0
and
1
.
1−z
Now, f is analytic in the domain D = {z ∈ C : |z| < 1} and g is analyic in the larger
domain z ∈ C\{1}. However, the restriction of g to D, g|D , is equal to f and so g is
an analytic continuation of f . Also, suppose we have another analytic continuation
of f i.e. an analytic function h : C\{1} → C for which h|D = f . Then on D we
must have g(z) − h(z) ≡ 0. Hence g(z) = h(z) for all z ∈ C\{1} since holomorphic
functions are essentially determined by their local behaviour. Generalising these
ideas we see that an analytic continuation of a given function is unique, and so we
may talk of THE analytic continuation.
g(z) =
Theorem 4 (Analytic continuation and functional equation). The Riemann zeta
function can be continued to a function analytic on all of C with the exception of a
simple pole at s = 1 with residue 1. The continuation satisfies the functional equation
s
1−s
− 2s
− 1−s
(22)
π Γ
ζ(s) = π 2 Γ
ζ(1 − s).
2
2
Remark. Here we have denoted the analytic continuation of the zeta function also
by ζ(s), really, this is the zeta function proper. When referring to the zeta function
from now on we will mean its analytic continuation.
A key tool in our proof of Theorem 4 is the Poisson summation formula. In order to
prove this, amongst other things, we need a suitable condition under which integrals
and sums can be interchanged. This is given by a special case of Fubini’s Theorem.
Proposition 17. Suppose fn ∈ L1 (R) for all n ∈ Z and that fn (t) ∈ `1 for all t ∈ R.
If either
Z X
XZ
|fn (t)|dt < ∞ or
|fn (t)|dt < ∞
R n∈Z
n∈Z
R
then
Z X
R n∈Z
|fn (t)|dt =
XZ
n∈Z
R
|fn (t)|dt.
NOTES ON THE GAMMA FUNCTION AND THE RIEMANN ZETA FUNCTION
15
Recall that the Schwartz space consists of smooth functions (infinitely differentiable) of rapid decay (derivatives of all order decay faster than any negative power
of x: for all k ≥ 0 and m ≥ 1, xm f (k) (x) → 0 as |x| → ∞ ).
Theorem 5 (Poisson summation). Suppose f is in Schwartz space. Define the
fourier transform of f by
Z
f˜(t) =
f (x)e−2πitx dx.
R
Then
X
X
f (n) =
n∈Z
f˜(n).
n∈Z
Proof. We first periodise f . Let
S(t) =
X
f (n + t).
n∈Z
Since f is in Schwarz space, S(t) is absolutely convergent. Also notice S(t) is periodic
of period ≤ 1. Hence we can represent S as a fourier series
X
ck e2πikt .
S(t) =
k∈Z
The fourier coefficients ck are given by
Z 1
ck =
S(x)e−2πikx dx
0
Z 1X
=
f (n + x) e−2πikx dx
0
=
XZ
f (n + x)e−2πikx dx
0
n∈Z
=
n∈Z
1
XZ
n+1
f (t)e−2πik(t−n) dt
n∈Z n
∞
Z
=
f (t)e−2πikt dt
−∞
= f˜(k).
Note the interchange of integration and summation is valid since S(x) is absolutely
convergent. Hence
X
X
f (n + t) =
f˜(n)e2πint
n∈Z
and setting t = 0 gives the result.
n∈Z
NOTES ON THE GAMMA FUNCTION AND THE RIEMANN ZETA FUNCTION
16
Proof of Theorem 4. Our starting point is an identity involving the gamma function.
Recall that
Z ∞
Γ(s) =
e−t ts−1 dt.
0
Letting s 7→ s/2 and substituting t = πn2 x gives
Z ∞
s
2
x 2 −1 e−πn x dx = n−s π −s/2 Γ(s/2).
0
Hence for σ > 1,
π
−s/2
ζ(s)Γ(s/2) =
∞ Z
X
n=1
∞
x
s
−1
2
−πn2 x
e
∞
Z
dx =
x
s
−1
2
0
0
∞
X
2
e−πn x dx
n=1
if the interchange of summation and integration can be justified. This is clear since
for σ > 1
∞ Z ∞
X
s
2
|x 2 −1 e−πn x |dx = π −σ/2 ζ(σ)Γ(σ/2) < ∞.
n=1
0
We define
ψ(x) =
∞
X
2x
e−πn
n=1
so that
π
−s/2
Z
∞
s
x 2 −1 ψ(x)dx
ζ(s)Γ(s/2) =
0
Z
1
x
=
s
−1
2
Z
∞
ψ(x)dx +
s
x 2 −1 ψ(x)dx.
1
0
Since ψ(x) = O(e−πx ) we see by the usual argument that the second integral is
absolutely convergent for all values of s, and uniformly convergent on compact subsets
of C. Since the integrand is differentiable the integral defines an analytic function
by Corollary 3. We seek similar behaviour in the first integral so that we can give
meaning to ζ(s) for σ < 1. To this end we introduce
θ(x) =
X
n∈Z
2x
e−πn
= 2ψ(x) + 1
NOTES ON THE GAMMA FUNCTION AND THE RIEMANN ZETA FUNCTION
2x
Writing f (u) = e−πu
17
we see
f˜(n) =
Z
∞
2
e−πu x e−2πinu du
Z−∞
∞
2
2 /x
e−πx(u+in/x) e−πn
−∞
Z ∞
2
−πn2 /x
e−πxu du
=e
=
du
−∞
1
2
= √ e−πn /x .
x
Since f is a Schwartz function we can apply Poisson summation to give
1
1
θ(x) = √ θ
x
x
and hence
1
2ψ(x) + 1 = √ (2ψ(1/x) + 1).
x
Therefore for σ > 1 we obtain
Z 1
Z ∞
s
s
1
1
1
−1
−s/2
2
√ ψ(1/x) + √ −
π
ζ(s)Γ(s/2) =
x
dx +
x 2 −1 ψ(x)dx
x
2 x 2
0
1
Z 1
Z ∞
s
s
3
1
1
x 2 − 2 ψ(1/x)dx +
x 2 −1 ψ(x)dx
− +
=
s−1 s
(23)
Z ∞0
Z ∞1
1−s
s
1
=
+
y 2 −1 ψ(y)dy +
x 2 −1 ψ(x)dx
s(s − 1)
1
Z1 ∞ 1−s
s
1
=
+
x 2 −1 + x 2 −1 ψ(x)dx.
s(s − 1)
1
Once again we see the integral represents an entire function and so the right hand
side is analytic on C\{0, 1}. Since the factor π s/2 Γ(s/2) never vanishes we may use
the above expression to define ζ(s) for C\{0, 1}. Since Γ(s/2)−1 has a simple zero
at s = 0 we conclude the resulting expression for ζ(s) is analytic at s = 0. Also
note that the expression on the right is invariant under s 7→ 1 − s and hence the
functional equation follows. Finally, to find the
√ residue at s = 1 we use (23) to
calculate lims→1 (s − 1)ζ(s) and note Γ(1/2) = π.
The nature of the functional equation
s
1−s
− 2s
− 1−s
π Γ
ζ(s) = π 2 Γ
ζ(1 − s)
2
2
NOTES ON THE GAMMA FUNCTION AND THE RIEMANN ZETA FUNCTION
18
allows us to infer the behaviour of ζ(s) for σ < 0 from its behaviour when σ > 1. In
particular, the left hand side is non-zero and holomorphic for σ > 1. Since Γ(s) has
simple poles at the negative integers, ζ(s) must have simple zeros at the negative
even integers, and these are the only zeros for σ < 0. We refer to these as the ‘trivial’
zeros of ζ(s). If ζ(s) were to have any other zeros then these must lie in the region
0 ≤ σ ≤ 1, known as the critical strip.
3.2. The Hadamard Product. Similarly to the product expression for sin:
∞ z2
sin πz Y
=
1− 2 ,
πz
n
n=1
we wish to express the zeta function as a product over its zeros. Thus, we must first
investigate infinite products in general.
3.2.1. Infinite products.
Definition 1. Let (zn ) be a sequence of complex numbers and let
P (N ) =
N
Y
zn .
n=1
Then if theQlimit limN →∞ P (N ) exists it is called the infinite product of (zn ) and it
is denoted ∞
j=1 zn .
If no term in the sequence (zn ) is zero then P (N ) is non-zero for all N . If
lim P (N ) 6= 0 then
P (N )
lim zN = lim
= 1.
P (N − 1)
Therefore, a necessary condition for the convergence of a non-zero infinite product is
that the nth term goes to 1. Consequently, we may assume throughout the following
that <(zn ) > 0 for all n. This allows us to take logarithms of the zn (using the
principal branch) and hence convert products into sums.
Q∞
Proposition 18. Suppose <(zn ) > 0 for all n.
P∞Then the infinite product n=1 zn
converges to a non-zero number if and only if n=1 log zn converges.
P
Proof. If N
n=1 log zn converges to some limit, z say, then by continuity
X
∞
N
N
Y
X
zn = lim exp
log zn = exp lim
log zn = ez 6= 0.
n=1
N →∞
n=1
N →∞
n=1
NOTES ON THE GAMMA FUNCTION AND THE RIEMANN ZETA FUNCTION
Suppose
Q∞
n=1 zn
19
= z = reiθ with −π < θ ≤ π and r > 0. Let
N
Y
P (N ) =
zn ,
S(N ) =
n=1
N
X
log zn
n=1
and let Log P (N ) = log |P (N )| + iθN with −π + θ < θN ≤ π + θ. Since P (N ) =
exp(S(N )) we must have S(N ) = Log P (N ) + 2πikN for some integer kN . Now,
since P (N ) → z, we have zN → 1 and hence S(N ) − S(N − 1) = log zN → 0. Also,
Log P (N ) − Log P (N − 1) → 0 which implies θN − θN −1 → 0 and so kN − kN −1 → 0.
Therefore, since kN is an integer there exists some fixed integer k for which kn = k
for all n after some point n0 . Consequently, S(N ) → log r + iθ + 2πik.
Q∞
DefinitionP
2. If <(zn ) > 0 for all n we say the product n=1 zn converges absolutely
iff the sum ∞
n=1 log zn converges absolutely. We may always refer to finite products
as absolutely convergent (no restriction on <(zn ) needed).
The following inequality will prove useful. By the power series expansion for the
logarithm we have for |z| < 1
log(1
+
z)
=|z/2 − z 2 /3 + · · · |
1 −
z
1
≤ (|z| + |z|2 + · · · )
2
1 |z|
=
.
2 1 − |z|
If |z| < 1/2 then this last quantity is < 1/2. Hence, for |z| < 1/2 we have
1
|z|
2
(24)
≤ | log(1 + z)| ≤ 23 |z|.
This immediately gives the following.
Proposition 19. Suppose
P <zn > −1. Then the series
absolutely if and only if ∞
n=1 zn converges absolutely.
P∞
n=1
log(1 + zn ) converges
Examples.
(i) The Euler product
−1
Y
1
1− s
p
p
converges absolutely for σ > 1: This follows from
P −σ
< ∞.
nn
P
p
|p−s | =
P
p
p−σ <
NOTES ON THE GAMMA FUNCTION AND THE RIEMANN ZETA FUNCTION
Q∞
20
2
z
n2
converges absolutely for all z ∈ C: Fix z ∈
Q
z2
C and let |z| = R. Then the product 2R
1
−
is finite and hence
n=1
n2
Q
z2
absolutely convergent. In the remaining product ∞
1
−
we have
2
n=2R+1
n
(ii) The product
n=1
1−
2
2
|z
1/2 and hence <(1 − z 2 /n2 ) > 0. The result now follows from
P /n | <
2
2
2
n>2R z /n R .
Q
z
(iii) The product ∞
n=1 1 − n does not converge for any non-zero z ∈ C: This
follows from the divergence of the harmonic series.
z/n
Q
z
(iv) The product ∞
converges absolutely for all z ∈ C: Fix z and
n=1 1 − n e
choose n0 such that |z/n| < 1/2 and <((1 − z/n)ez/n ) > 0 for all n ≥ n0 .
Then
∞
X
n=n0
z/n
log((1 − z/n)e
)=
∞
X
log(1 − z/n) + z/n
n=n0
∞
X
=−
n=n0
∞
X
2
z2
z3
+
+ ···
2n2 3n3
|z|
=|z|2
n=n0
∞
X
n=n0
|z| |z|2
1
1+
+ 2 + ···
n2
n
n
1
1
< 2ζ(2)|z|2 .
2
n 1 − |z|/n
P
−2
< ∞ the product
Similarly,
for any sequence (zn ) which satisfies ∞
n=1 |zn |
Q∞
z/zn
is absolutely convergent for all z ∈ C. This is important
n=1 (1 − z/zn )e
because it allows us to create a function whose only zeros are at the points
zn . The main goal of this section is to show that this function is also analytic.
Lemma 20. Let K be a compact (closed and bounded) subset of C and let f, fn :
K → C be a set of functions such that fn (z) → f (z) uniformly for z ∈ K. If there
is a constant A such that <f (z) ≤ A for all z ∈ K then exp(fn (z)) → exp(f (z))
uniformly for z ∈ K.
Proof. Given > 0 choose δ > 0 such that |ez − 1| < e−A for |z| < δ. Also, choose
N such that |fn (z) − f (z)| < δ for all z ∈ K whenever n > N . With these choices
n (z))
− 1|.
e−A > | exp(fn (z) − f (z)) − 1| = | exp(f
exp(f (z))
NOTES ON THE GAMMA FUNCTION AND THE RIEMANN ZETA FUNCTION
21
Therefore,
| exp(fn (z)) − exp(f (z))| < e−A | exp(f (z))| < .
Proposition 21. Let (zn ) be a sequence of non-zero complex numbers and suppose
P
∞
−2
converges. Then the product
n=1 |zn |
∞ Y
z
(25)
P (z) =
1−
ez/zn
z
n
n=1
is absolutely and uniformly convergent on compact subsets of C and is therefore
analytic (by Lemma 1).
Proof. We have already shown the absolute convergence. It remains to show that
Sm (z) =
m
X
log((1 − z/zn )ez/zn )
n=1
converges uniformly to
S(z) =
∞
X
log((1 − z/zn )ez/zn )
n=1
on compact subsets K and that also in such regions there exists an A such that
<S(z) ≤ A. So fix a compact K and let maxz∈K |z| = R. Since we’ve already shown
absolute convergence then it must follow that <S(z) is bounded for z ∈ K. Also,
following a similar analysis to before
∞
∞
X
X
1
z/zn 2
|S(z) − Sm (z)| = log((1 − z/zn )e
) R
→0
2
|z
|
n
n=m+1
n=m+1
as m → ∞ and clearly this estimate holds uniformly for z ∈ K.
P
So given a sequence (zn ) satisfying
|zn |−2 < ∞ we can construct an analytic
function whose only zeros are the points zn . Conversely, given an analytic function f (z) with zeros (zn ), we would like to know what conditions on f guarantee
P
|zn |−2 < ∞. Assuming that f satisfies these conditions we could then write
f (z) = F (z)P (z)
where P (z) is given by (25) and where F (z) is manifestly some analytic function
with no zeros. It turns out that the behaviour of a functions zeros is closely related
to its order of growth.
NOTES ON THE GAMMA FUNCTION AND THE RIEMANN ZETA FUNCTION
22
3.2.2. Entire functions of order 1.
Definition 3. A function is called entire if it is analytic on all of C. An entire
function f (z) is said to be of finite order if there exists an α such that
α
f (z) = O(e|z| ),
|z| → ∞
The order of f is defined as the infimum of all such α.
The behaviour of an analytic functions zeroes and its order of growth are related
through the following result of Jensen, which we quote without proof.
Theorem 6 (Jensen’s formula). Suppose f is analytic on the disc |z| ≤ R and
suppose z1 , . . . , zn are its zeros on the interior |z| < R. Also, suppose f does not
have any zeros on the boundary |z| = R. Then
Z 2π
Rn
1
log |f (Reiθ )|dθ − log |f (0)| = log
(26)
2π 0
|z1 | · · · |zn |
Corollary 22. Let f be an entire function of order α and let n(R) denote the number
of zeros of f in |z| < R. Then
n(R) Rβ
(27)
for any β > α.
Proof. Let z1 , . . . , zn denote the number of zeros of f in |z| < r and let |zi | = ri .
Then the right hand side of (26) can be written as
Z R
n(r)
dr
r
0
since this is equal to
log r2 /r1 + 2 log r3 /r2 + · · · + n log R/rn = log(Rn /r1 · · · rn ).
Suppose β > α. Then
log |f (Reiθ | Rβ
and hence
Z
0
R
n(r)
dr Rβ − log |f (0)| Rβ .
r
On the other hand
Z
2R
R
n(r)
dr ≥ n(R)
r
Z
2R
r−1 dr = n(R) log 2.
R
Corollary 23. Suppose
f is an entire function of order 1 with non-zero zeros (zn ).
P∞
Then the series n=1 |zn |−1− converges for any > 0.
NOTES ON THE GAMMA FUNCTION AND THE RIEMANN ZETA FUNCTION
Proof. By partial summation we have
Z
Z ∞
∞
X
−2−
−1−
r
n(r)dr |zn |
= (1 + )
∞
r−1−/2 dr < ∞.
0
0
n=1
23
A consequence of this is that for an entire function f of order 1 with zeros (zn ),
the product
∞ Y
z
(28)
P (z) =
1−
ez/zn
z
n
n=1
is an entire function. Writing
f (z) = F (z)P (z)
we see that F (z) is an entire function without zeros.
Proposition 24. Let f be an entire function of order 1 with zeros (zn ) and let
f (z) = F (z)P (z) where P (z) is given by (28). Then F (z) is a non-zero entire
function of order at most 1.
Proof. We need to show that F has order at most 1. We do this by lower bounding
P (z) on a sequence of circles |z| = R. These must be kept away from the values
rn = |zn |, since otherwise we may obtain
lower bounds e.g. from an occurrence
P trivial
−2
r
converges,
these points cannot occur
of 1 − |z|/|zn | = 0. However, since ∞
n=1 n
too densely. Indeed, the combined length of the intervals (rn − rn−2 , rn + rn−2 ) is finite
and hence we may always choose an arbitrarily large R such that
|R − rn | > rn−2
(29)
for all n.
Write P (z) = P1 (z)P2 (z)P3 (z) where the subproducts are extended over the following sets of n:
P1 :
rn < R/2
P2 :
P3 :
(30)
R/2 ≤ rn ≤ 2R
rn > 2R.
Throughout the following, ’s may vary from line to line. For P1 , on |z| = R we have
|(1 − z/zn )ez/zn | ≥ (|z/zn | − 1)e−|z|/|zn | > e−R/rn
and since
X
r≤R/2
rn−1 =
X
r≤R/2
rn rn−1− ≤ (R/2)
∞
X
n=1
rn−1− R
NOTES ON THE GAMMA FUNCTION AND THE RIEMANN ZETA FUNCTION
24
it follows that
P1 (z) exp(−R1+ ).
For P2 we have
|(1 − z/zn )ez/zn | ≥ |z/zn − 1|e−|z|/|zn | ≥ e−2 |z − zn |/2R R−3
by (29). By (27), the number of factors in P2 is R1+ and hence
|P2 (z)| (R−3 )R
1+
exp(−R1+2 ).
Finally, In P3 we have |z/zn | < 1/2 and hence
|(1 − z/zn )ez/zn | = |1 −
z2
2
2zn
2
+ O((z/zn )3 )| > e−c(R/rn )
for some constant c > 0. Also,
∞
X
X
X
rn−2 =
r−1+ r−1− < (2R)−1+
rn−1− R−1+ .
n>2R
n=1
n>2R
Therefore,
P3 (z) exp(−cR1+ ) exp(−R1+2 ).
Consequently, for |z| = R we have
P (z) exp(−R1+ )
and hence
F (z) exp(R1+ )
since f is of order 1.
Proposition 25. Let F (z) be an entire function of order α with no zeros. Then
F (z) = eG(z) for some polynomial G. The order of F is the degree of G and hence is
an integer.
Proof. Since F is entire and non-zero the function G(z) := log F (z) is itself entire
and can be defined to be single valued. It satisfies
<(G(z)) = log |F (z)| Rα
for |z| = R. Writing
G(z) =
∞
X
(an + ibn )z n
n=0
then
<(G(z)) =
∞
X
n=0
an Rn cos nθ −
∞
X
bn Rn sin nθ
n=1
for z = Reiθ . Multiplying the above by 1 ± cos mθ and integrating over θ ∈ [0, 2π]
gives π(2a0 ± am rm ) on the right side and O(Rα ) on the left side. Consequently,
NOTES ON THE GAMMA FUNCTION AND THE RIEMANN ZETA FUNCTION
25
am = O(Rα−m ) and hence it follows, by taking R arbitrarily large, that am = 0 if
m > α. A similar argument gives the same for bm and hence G(z) is a polynomial
of degree at most α. Since F is of order α the degree is in fact equal to α, which
consequently is an integer.
Combining the above two Propositions with our previous results gives the following.
Theorem 7. Let f (z) be an entire function of order 1 with zeros (zn ). Then
∞ Y
z
a+bz
1−
(31)
f (z) = e
ez/zn
z
n
n=1
P∞
for some constants a, b. Moreover, the series n=1 |zn |−1− converges for all > 0.
P
−1
converges then f (z) ≤ ec|z| for some constant c. This
Note that if ∞
n=1 |zn |
follows on applying the inequality
|(1 − z)ez | = |1 −
z2
2
−
z3
3
−
z4
8
− · · · | ≤ e2|z|
to the product representation
Consequently, if the inequality f (z) ≤ ec|z| does
P∞ (31). −1
not hold, then the series n=1 |zn | diverges and so f must have infinitely many
zeros.
3.3. The Hadamard product for the Riemann zeta function. Let
(32)
ξ(s) = 12 s(s − 1)π −s/2 Γ(s/2)ζ(s).
Then this is an entire function whose only zeros are the non-trivial zeros of the zeta
function. We will show that ξ(s) is of order 1. Since ξ(s) = ξ(1 − s) we only need to
look at the case σ ≥ 1/2. Clearly,
1
s(s
2
− 1)π −s/2 exp(c|s|)
and by Stirling’s formula we have
Γ(s/2) exp(c|s| log |s|).
On applying partial summation to the series representation of ζ(s) when σ > 1 we
find
Z ∞
s
ζ(s) =
−s
(x − bxc)x−s−1 dx.
s−1
1
Note that the integral converges for σ > 0 and hence this representation acts as an
analytic continuation of ζ(s) to σ > 0. In particular, the integral is bounded for
σ ≥ 1/2 and therefore
ζ(s) |s|
NOTES ON THE GAMMA FUNCTION AND THE RIEMANN ZETA FUNCTION
26
for |s| large. Putting these together gives
(33)
ξ(s) exp(c|s| log |s|)
as |s| → ∞. Note that since log Γ(s) ∼ s log s and ζ(s) → 1 the above inequality
α
for ξ(s) is essentially best possible. Therefore ξ(s) = O(e|s| ) for all α > 1 and the
infimum of all such α is 1 i.e. ξ(s) is of order 1. In particular, ξ(s) is not ec|z| for
any c. On applying the results of the previous section we have the following.
Theorem 8. The zeta function has an infinity of non-trivial zeros, ρ1 , ρ2 , . . ., such
that the series
∞
X
|ρn |−1−
n=1
converges for any > 0 and
∞
X
|ρn |−1
n=1
diverges. It satisfies the product representation
Y
s s/ρ
eA+Bs
1−
e
(34)
ζ(s) =
(s − 1)Γ( 2s + 1) ρ
ρ
for some constants A, B.