Función Zeta De Riemann
ˇ ´ DRAGAN MILICIC
1. Gamma function 1.1. Definition of the Gamma function. The integral
∞
Γ(z) =
0
tz−1 e−t dt
is well-defined and defines a holomorphic function in the right half-plane {z ∈ C | Re z > 0}. This function is Euler’s Gamma function. First, by integration by parts
∞ ∞
Γ(z + 1) =
0
tz e−t dt = −tz e−t
0
∞
+z
0
tz−1 e−tdt = zΓ(z)
for any z in the right half-plane. In particular, for any positive integer n, we have Γ(n) = (n − 1)Γ(n − 1) = (n − 1)!Γ(1). On the other hand,
∞ ∞
Γ(1) =
0
e−t dt = −e−t
0
= 1;
and we have the following result. 1.1.1. Lemma. Γ(n) = (n − 1)! for any n ∈ Z. Therefore, we can view the Gamma function as a extension of the factorial. 1.2. Meromorphic continuation. Now wewant to show that Γ extends to a meromorphic function in C. We start with a technical lemma. 1.2.1. Lemma. Let cn , n ∈ Z+ , be complex numbers such such that converges. Let S = {−n | n ∈ Z+ and cn = 0}. Then f (z) = cn z+n n=0
∞ ∞ n=0
|cn |
converges absolutely for z ∈ C − S and uniformly on bounded subsets of C − S. The function f is a meromorphic function on C with simple poles at thepoints in S and Res(f, −n) = cn for any −n ∈ S.
1
2
ˇ ´ D. MILICIC
Proof. Clearly, if |z| < R, we have |z + n| ≥ |n − R| for all n ≥ R. Therefore, we 1 1 have | z+n | ≤ n−R for |z| < R and n ≥ R. It follows that for n0 > R, we have
∞ n=n0
|cn | |cn | 1 cn ≤ ≤ ≤ z+n |z + n| n=n n − R n0 − R n=n
0 0
∞
∞
∞
|cn |.
n=n0
cn Hence, the series n>R z+n converges absolutely anduniformly on the disk {z | ∞ cn |z| < R} and defines there a holomorphic function. It follows that n=0 z+n is a meromorphic function on that disk with simple poles at the points of S in cn {z | |z| < R}. Therefore, ∞ z+n is a meromorphic function with simple poles n=0 at the points in S. Therefore, for any −n ∈ S we have cm cn cn + = + g(z) f (z) = z+n z+m z+n −m∈S−{n}
where g is holomorphic at−n. This implies that Res(f, −n) = cn . Going back to Γ, we have
∞ 1 0 ∞ 1
Γ(z) =
0
tz−1 e−t dt =
tz−1 e−t dt +
tz−1 e−t dt.
Clearly, the second integral converges for any complex z and represents an entire function. On the other hand, since the exponential function is entire, its Taylor series converges uniformly on compact sets in C, and we have
1 0
tz−1 e−t dt =
0
1
∞tz−1
p=0
(−1)p p t p!
∞
dt (−1)p p!
1 ∞
=
p=0
tp+z−1 dt =
0 p=0
(−1)p 1 p! z + p
for any z ∈ C. Therefore,
∞ ∞
Γ(z) =
1
t
z−1 −t
e
dt +
p=0
(−1)p 1 p! z + p
for any z in the right half-plane. By 1.2.1, the right side of this equation defines a meromorphic function on the complex plane with simple poles at 0, −1, −2, −3, − · · · . Hence, we have thefollowing result. 1.2.2. Theorem. The function Γ extends to a meromorphic function on the complex plane. It has simple poles at 0, −1, −2, −3, · · · . The residues of Γ are −p are given by (−1)p Res(Γ, −p) = p! for any p ∈ Z+ . This result combined with the above calculation immediately implies the following functional equation. 1.2.3. Proposition. For any z ∈ C we have Γ(z + 1) = zΓ(z).
NOTESON RIEMANN’S ZETA FUNCTION
3
1.3. Another functional equation. Let Re z > 0. Then
∞
Γ(z) =
0
tz−1 e−t dt.
Let Re p > 0 and Re q > 0. Then, by change of variable t = u2 , we get
∞
Γ(p) =
0
tp−1 e−t dt = 2
0 ∞
2
∞
e−u u2p−1 du.
2
Analogously we have Γ(q) = 2
0
e−v v 2q−1 dv.
Hence, it follows that
∞ ∞ 0
Γ(p)Γ(q) = 4
0
e−(u
2
+v 2 ) 2p−12q−1
u
v
du dv,
and by passing to the polar coordinates by u = r cos ϕ, v = r sin ϕ, we have
∞
Γ(p)Γ(q) = 4
0 ∞ 0
π 2
e−r r2(p+q)−1 cos2p−1 ϕ sin2q−1 ϕ dr dϕ r dr · 2
0
π 2
2
=
2
0
e
−r 2 2(p+q)−1
cos2p−1 ϕ sin2q−1 ϕ dϕ
π 2
= 2Γ(p + q)
0
cos2p−1 ϕ sin2q−1 ϕ dϕ.
We put s = sin2 ϕ in the integral. Then we have
π 2
2
0
cos2p−1 ϕ sin2q−1...
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