Chapter 8
Euler’s Gamma function
The Gamma function plays an important role in the functional equation for ζ(s)
that we will derive in the next chapter. In the present chapter we have collected
some properties of the Gamma function.
For t ∈ R>0 , z ∈ C, define tz := ez log t , where log t is he ordinary real logarithm.
Euler’s Gamma function is defined by the integral
Z ∞
Γ(z) :=
e−t tz−1 dt (z ∈ C, Re z > 0).
0
Lemma 8.1. Γ(z) defines an analytic function on {z ∈ C : Re z > 0}.
Proof. We prove that Γ(z) is analytic on Uδ,R := {z ∈ C, δ < Re z < R} for
every δ, R with 0 < δ < R. This is standard using Theorem 2.29. First, the function
e−t tz−1 is continuous, hence measurable on R>0 ×Uδ,R . Second, for any fixed t ∈ R>0 ,
the function z 7→ e−t z t−1 is analytic on Uδ,R . Lastly, for z ∈ Uδ,R we have
δ−1
t
for 0 6 t 6 1,
−t z−1
|e t | 6 M (t) :=
−t R−1
−t/2
6 Ce
for t > 1,
e t
where C is some constant. Now we have
Z ∞
Z 1
Z
δ−1
M (t)dt =
t dt + C
0
0
∞
e−t/2 dt = δ −1 + 2C < ∞.
1
Hence all conditions of Theorem 2.29 are satisfied, and thus, Γ(z) is analytic on
Uδ,R .
115
We now show that Γ has a meromorphic continuation to C.
Theorem 8.2. There exists a unique meromorphic function Γ on C with the following properties:
R∞
(i) Γ(z) = 0 e−t tz−1 dt for z ∈ C, Re z > 0;
(ii) the function Γ is analytic on C \ {0, −1, −2, . . .};
(iii) Γ has a simple pole with residue (−1)n /n! at z = −n for n = 0, 1, 2, . . .;
(iv) Γ(z + 1) = zΓ(z) for z ∈ C \ {0, −1, −2, . . .};
(v) Γ(n) = (n − 1)! for n ∈ Z>0 .
R∞
Proof. The function Γ has already been defined for Re z > 0 by 0 e−t tz−1 dt. By
Corollary 2.22, Γ has at most one analytic continuation to any larger connected
open set, hence there is at most one function Γ with properties (i)–(v). We proceed
to construct such a function.
Let z ∈ C with Re z > 0. Then using integration by parts,
Z ∞
Z ∞
−t z−1
(8.1)
e−t z −1 dtz
e t dt =
Γ(z) =
0
0
Z ∞
Z ∞
h
i∞
−1 −t z
−1
−t z
−1
= z e t
+z
e t dt = z
e−t tz dt
0
0
0
−1
= z Γ(z + 1).
Now by induction on n it follows that
(8.2)
Γ(z) =
1
· Γ(z + n) for Re z > 0, n = 1, 2, . . . .
z(z + 1) · · · (z + n − 1)
We continue Γ to B := C \ {0, −1, −2, . . .} as follows. For z ∈ B, choose n ∈ Z>0
such that Re z + n > 0 and define Γ(z) by the right-hand side of (8.2). This does not
depend on the choice of n. For if m, n are any two integers with m > n > −Re z,
then by (8.2) with z + n, m − n instead of z, n we have
Γ(z + n) =
1
· Γ(z + m),
z + n) · · · (z + m − 1)
and so
1
1
· Γ(z + n) =
· Γ(z + m).
z(z + 1) · · · (z + n − 1)
z(z + 1) · · · (z + m − 1)
116
So Γ is well-defined on B, and it is analytic on B since the right-hand side of (8.2)
is analytic if Re z + n > 0. This proves (ii).
We prove (iii). By (8.2) we have
1
Γ(z + n + 1)
z→−n
z(z + 1) · · · (z + n)
1
(−1)n
=
Γ(1) =
.
(−n)(−n + 1) · · · (−1)
n!
lim (z + n)Γ(z) =
z→−n
lim (z + n)
Hence Γ has a simple pole at z = −n of residue (−1)n /n!.
We prove (iv). Both functions Γ(z + 1) and zΓ(z) are analytic on B, and by
(8.1), they are equal on the set {z ∈ C : Re z > 0} which has limit points in B. So
by Corollary 2.21, Γ(z + 1) = zΓ(z) for z ∈ B.
R∞
Identity (v) follows easily by observing that Γ(1) = 0 e−t dt = 1, and by repeatedly applying (iv).
Theorem 8.3. We have Γ(z)Γ(1 − z) =
π
for z ∈ C \ Z.
sin πz
Proof. We prove that zΓ(z)Γ(1 − z) = πz/ sin πz, or equivalently,
(8.3)
Γ(1 + z)Γ(1 − z) =
πz
for z ∈ A := (C \ Z) ∪ {0},
sin πz
which implies Theorem 8.3. Notice that by Theorem 8.2 the left-hand side is analytic
on A, while by limz→0 πz/ sin πz = 1 the right-hand side is also analytic on A. By
Corollary 2.19, it suffices to prove that (8.3) holds for z ∈ S, where S is any subset
1
of A having a limit point in A. For the set S we take { 2n
: n ∈ Z>0 }; this set has
limit point 0 in A. Thus, (8.3), and hence Theorem 8.3, follows once we have proved
that
(8.4)
Γ(1 +
1
)
2n
· Γ(1 −
1
)
2n
=
π/2n
sin π/2n
Z
∞
(n = 1, 2, . . .).
Notice that
Γ(1 +
1
)
2n
· Γ(1 −
1
)
2n
=
−s 1/2n
e s
0
Z ∞Z ∞
=
117
0
ds ·
∞
e−t t−1/2n dt
0
−s−t
e
0
Z
(s/t)1/2n dsdt.
Define new variables u = s + t, v = s/t. Then s = uv/(v + 1), t = u/(v + 1). The
Jacobian of the substitution (s, t) 7→ (u, v) is
v
u
∂s ∂s
(v + 1)2
∂(s, t)
∂u ∂v = v + 1
=
1
u
∂t ∂t
∂(u, v)
−
v+1
(v + 1)2
∂u ∂v
−uv − u
−u
=
=
.
3
(v + 1)
(v + 1)2
It follows that
Γ(1 +
1
)
2n
Z
· Γ(1 −
∞
Z
Z
e−u v 1/2n
0
∞
Z
∞
0
∂(s, t)
· dudv
∂(u, v)
Z ∞
Z
−u
· dudv =
e udu ·
e−u v 1/2n
=
∞
=
0
1
)
2n
0
u
(v + 1)2
0
0
∞
v 1/2n
dv.
(v + 1)2
In the last product, the first integral is equal to 1, while for the second integral we
have, by homework exercise 4,
Z ∞ 1/2n
Z ∞
1 v
1/2n
v
d
dv
=
−
(v + 1)2
v+1
0
0
1/2n ∞ Z ∞
Z ∞
v
dw
π/2n
1
1/2n
· dv
=
=
.
=−
+
2n
v+1 0
v+1
w +1
sin π/2n
0
0
This implies (8.4), hence Theorem 8.3.
√
Corollary 8.4. Γ( 21 ) = π.
Proof. Substitute z =
1
2
in Theorem 8.3, and use Γ( 12 ) > 0.
Corollary 8.5. (i) Γ(z) 6= 0 for z ∈ C \ {0, −1, −2, . . .}.
(ii) 1/Γ is analytic on C, and 1/Γ has simple zeros at z = 0, −1, −2, . . ..
Proof. (i) Recall that Γ(n) = (n − 1)! 6= 0 for n = 1, 2, . . .. Further, Γ(z) 6= 0 for
z ∈ C \ Z by Theorem 8.3.
(ii) By (i), the function 1/Γ is analytic on C \ {0, −1, −2, . . .}. Further, at
z = 0, −1, −2, . . ., Γ has a simple pole, hence 1/Γ is analytic and has a simple
zero.
118
We give two other expressions for the Gamma function. Recall that the EulerMascheroni constant γ is defined by
γ := lim
N →∞
N
X
1
− log N.
n
n=1
Theorem 8.6. For z ∈ Z \ {0, −1, −2, . . .} we have
∞
Y
n! · nz
ez/n
= e−γz z −1
.
n→∞ z(z + 1) · · · (z + n)
1
+
z/n
n=1
Γ(z) = lim
Proof. We first show that the second and third expression are equal, assuming that
either the limit exists, or the product converges. Indeed,
e−γz z −1
∞
Y
N
Y
1
1
ez/n
ez/n
= lim e(log N −1− 2 −···− N )z z −1
1 + z/n N →∞
1 + z/n
n=1
n=1
z log N −1
= lim e
N →∞
z
N
Y
Nz · N!
1
= lim
.
1 + z/n N →∞ z(z + 1) · · · (z + N )
n=1
The remainder of the proof of Theorem 8.6 is a combination of a few lemmas.
Define for the moment
∞
Y
n! · nz
ez/n
−γz −1
g(z) := lim
=e z
.
n→∞ z(z + 1) · · · (z + n)
1
+
z/n
n=1
Lemma 8.7. g(z) defines an analytic function on C \ {0, −1, −2, . . .}.
Q
Proof. It suffices to prove that h(z) := ∞
1 + nz e−z/n is analytic on C and
n=1
that h(z) 6= 0 on B := C \ {0, −1, −2, . . .}. For this, it is sufficient to prove that
h(z) is analytic on D(0, R) = {z ∈ C : |z| < R} and h(z) 6= 0 for z 6= 0, −1, −2, . . ..
We proceed to show that there is a sequence {Mn }∞
n=1 such that
(8.5)
z
(1 + )e−z/n − 1 6 Mn for z ∈ D(0, R), n > 1,
n
∞
X
Mn < ∞.
n=1
Then the infinite product defining h is pointwise absolutely convergent, which implies that h(z) 6= 0 whenever any of the factors in the product is 6= 0; that is,
119
whenever z 6= 0, −1, −2, . . .. Further, by Corollary 2.27, the infinite product defines
an analytic function on D(0, R).
We now show that there is a sequence {Mn }∞
n=1 with (8.5). Notice that
1+
z −z/n
z
z z 2 /n2 z 3 /n3
e
= 1+
· 1− +
−
+ ···
n
n
n
2!
3!
1 z2
1 z3
1 z4
1
1
1− · 2 +
−
−
−
+ ···
2 n
2! 3! n3
3! 4! n4
Hence for z ∈ C with |z| < R we have
∞
X 1
1 |z| k
z
−
1 + e−z/n − 1 6
n
(k − 1)! k!
n
k=2
∞
∞
|z| 2 X
|z| 2 X
k−1
1
k−2
· |z|
6
· |z|k−2
6
n
(k
−
1)!
n
(k
−
2)!
k=2
k=2
6
Clearly,
P∞
n=1
R2 R
·e .
n2
R2 eR /n2 converges. This proves Lemma 8.7.
To prove that Γ(z) = g(z) for z ∈ B := C \ {0, −1, −2, . . .}, it suffices to prove
that Γ(z) = g(z) for s aubset of B with a limit point in B. For this subset, we take
R>0 . So it suffices to prove that
(8.6)
n! · nx
n→∞ x(x + 1) · · · (x + n)
for x ∈ R>0 .
Γ(x) = lim
Lemma 8.8. We have
n! · nx
=
x(x + 1) · · · (x + n)
Z
n
1−
0
t n x−1
t dx.
n
Proof. By substituting s = t/n, the integral becomes
n
x
Z
1
(1 − s)n sx−1 ds.
0
The rest is left as an exercise.
120
Lemma 8.9. For every integer n > 2 and every real t with 0 6 t 6 n we have
0 6 e−t − 1 −
t n
t2
6 e−t · 2 .
n
n
Proof. This is equivalent to
t n
t2
6 et 1 −
6 1 (0 6 t 6 n, n > 2).
n
n
Recall that if f, g are continuously differentiable, real functions with f (0) = g(0)
and f 0 (x) 6 g 0 (x) for 0 6 x 6 A, say, then f (x) 6 g(x) for 0 6 x 6 A. From this
observation, one easily deduces that
1−
1 + x 6 ex , 1 − x 6 e−x , (1 − x)r > 1 − rx for 0 6 x 6 1, r > 0.
This implies on the one hand, for n > 2, 0 6 t 6 n,
t n
et 1 −
6 et (e−t/n )n 6 1,
n
on the other hand
t n
t n
t2 n
t2
t n
et 1 −
> 1+
· 1−
= 1− 2 >1− .
n
n
n
n
n
We prove (8.6) and complete the proof of Theorem 8.6. We have for x > 0, by
the integral expression for Γ(x) for x > 0 and by Lemma 8.8,
n! · nx
n→∞ x(x + 1) · · · (x + n)
Z n
Z n
t n x−1
−t x−1
t dt
= lim
e t dt −
1−
n→∞
n
0
0
Z n
t n x−1
= lim
e−t − 1 −
t dt.
n→∞ 0
n
Γ(x) − lim
Now Lemma 8.9 implies
Z n
n! · nx
t2
0 6 Γ(x) − lim
6 lim
e−t · tx−1 dt
n→∞ x(x + 1) · · · (x + n)
n→∞ 0
n
Z ∞
1
Γ(x + 2)
6 lim
e−t tx+1 dt = lim
= 0.
n→∞ n 0
n→∞
n
121
We deduce some consequences.
Corollary 8.10. We have
∞ Y
z2
sin πz = πz ·
1− 2
for z ∈ C.
n
n=1
Proof. For z ∈ C we have by Theorem 8.3, Corollary 8.5 and Theorem 8.6,
sin πz =
π
π
=
Γ(z)Γ(1 − z)
Γ(z)(−z)Γ(−z)
∞ ∞ Y
Y
z z −z/n
−γz
−z/n
e
1−
· (−z)e
= π(−z) e z
e
1+
n
n
n=1
n=1
−1 γz
= πz
∞ Y
n=1
∞ Y
z z
z2
1−
1+
= πz ·
1− 2 .
n
n
n
n=1
Recall that the Bernoulli numbers Bn (n > 0) are given by
∞
X Bn
z
=
· z n (|z| < 2π).
ez − 1 n=0 n!
Corollary 8.11. We have B0 = 1, B1 = − 12 , B3 = B5 = · · · = 0 and
ζ(2n) = (−1)n−1 22n−1
B2n 2n
π
for n = 1, 2, . . . .
(2n)!
Proof. Let z ∈ C with 0 < |z| < 1. Then sin πz 6= 0 and so, by taking the logarithmic
derivative of sin πz,
π cos πz
π(eπiz + e−πiz )/2
sin0 πz
=
= πiz
sin πz
sin πz
(e − e−πiz )/2i
1
2πiz
· 2πiz
z e
−1
∞
X
1
Bn
= πi +
· (2πi)n z n .
z n=0 n!
= πi +
(8.7)
122
We obtain another expression for the logarithmic derivative of sin πz by applying
Corollary 2.28 to the product identity from Corollary 8.10. Note that for z ∈ C with
P
−2
|z| < 1 we have |z 2 /n2 | < n−2 and that ∞
converges. Hence the logarithmic
n=1 n
derivative of the infinite product is the infinite sum of the logarithmic derivatives of
the factors, i.e.,
∞
sin0 πz
(πz)0 X (1 − z 2 /n2 )0
=
+
sin πz
πz
1 − z 2 /n2
n=1
∞
∞
X
z
1
1 X −2z/n2
+
−
2
=
=
z n=1 1 − z 2 /n2
z
n2
n=1
∞ 2
X
z k
k=0
!
n2
∞ X
∞
X
1
z 2k+1
=
−2
(by absolute convergence)
z
n2k+2
k=0 n=1
∞
X
1
ζ(2k + 2)z 2k+1 .
−2
=
z
k=0
(8.8)
Now Corollary 8.11 easily follows by comparing the coefficients of the Laurent series
in (8.7) and (8.8).
We finish with another important consequence of Theorem 8.6, the so-called
duplication formula.
Corollary 8.12. We have
√
π · Γ(2z) = 22z−1 Γ(z)Γ(z + 21 ) for z ∈ C, z 6= 0, − 21 , −1, − 32 , −2, . . . .
Proof. Let A be the set of z indicated in the lemma. We show that the function
F (z) := 22z Γ(z)Γ(z + 21 )/Γ(2z) is constant on A. Substituting z = 12 gives that the
√
constant is 2 π, and then Corollary 8.12 follows.
Let z ∈ A. To get nice cancellations in the numerator and denominator, we use
123
the expressions
Γ(z) =
Γ(z +
1
)
2
2n+1 · n! · nz
n! · nz
= lim
,
n→∞ 2z(2z + 2) · · · (2z + 2n)
n→∞ z(z + 1) · · · (z + n)
lim
n! · nz+1/2
= lim
n→∞ (z + 1/2)(z + 3/2) · · · (z + n + 1/2)
2n+1 · n! · nz+1/2
,
n→∞ (2z + 1)(2z + 3) · · · (2z + 2n + 1)
= lim
Γ(2z) =
(2n + 1)! · (2n + 1)2z
n→∞ 2z(2z + 1) · · · (2z + 2n + 1)
lim
(take in Theorem 8.6 the limit over the odd integers). Thus,
22z Γ(z)Γ(z + 12 )
Γ(2z)
2z(2z + 1) · · · (2z + 2n + 1)
22n+2 (n!)2 n2z+1/2
2z
·
= 2 lim
n→∞
2z(2z + 1) · · · (2z + 2n + 1)
(2n + 1)! · (2n + 1)2z
2n+2
√ 2
(n!)2 n
= lim
n→∞
(2n + 1)!
F (z) =
since
22z · n2z
= lim e2z log(2n/(2n+1)) = 1.
n→∞
n→∞ (2n + 1)2z
lim
This shows that indeed F (z) is constant. Substituting z = 21 , we get
F (z) = F ( 12 ) =
√
2Γ( 21 )Γ(1)
= 2 π.
Γ(1)
Remark. More generally, one can derive the multiplication formula of LegendreGauss,
(2π)(n−1)/2 Γ(nz) = nnz−1/2 Γ(z)Γ(z + n1 ) · · · Γ(z + n−1
)
n
for every integer n > 2. The idea of the proof is similar to that of Theorem 8.12
(exercise).
124