Función Zeta De Riemann

NOTES ON RIEMANN’S ZETA FUNCTION
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.
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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

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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...