Riemann Zeta Function
Ben Dulaney
Jonathan Gerhard
March 19, 2015
Ben Dulaney Jonathan Gerhard
Riemann Zeta Function
What is the ζ function?
Consider the series
Z (s) =
∞
X
1
ns
n=1
Notice that for <(s) ≤ 1, Z (s) will diverge.
Ben Dulaney Jonathan Gerhard
Riemann Zeta Function
What is the ζ function?
Consider the series
Z (s) =
∞
X
1
ns
n=1
Notice that for <(s) ≤ 1, Z (s) will diverge.
Example
1 1
+ + ...
2 3
This is the harmonic series, which we know diverges.
Z (1) = 1 +
Ben Dulaney Jonathan Gerhard
Riemann Zeta Function
What is the ζ function?
Euler was able to show that
Z (2) =
π2
, and
6
π4
.
90
Euler also noticed a connection between this series and prime
numbers.
Z (4) =
Ben Dulaney Jonathan Gerhard
Riemann Zeta Function
What is the ζ function?
Euler was able to show that
Z (2) =
π2
, and
6
π4
.
90
Euler also noticed a connection between this series and prime
numbers.
Z (4) =
Z (s) =
Y
prime p
Ben Dulaney Jonathan Gerhard
Riemann Zeta Function
1
1 − p −s
What is the ζ function?
Euler was able to show that
Z (2) =
π2
, and
6
π4
.
90
Euler also noticed a connection between this series and prime
numbers.
Z (4) =
Z (s) =
Y
prime p
1
1 − p −s
This product formula can be used to calculate the probability that
a random set of integers are relatively prime.
Ben Dulaney Jonathan Gerhard
Riemann Zeta Function
What is the ζ function?
Euler was able to show that
Z (2) =
π2
, and
6
π4
.
90
Euler also noticed a connection between this series and prime
numbers.
Z (4) =
Z (s) =
Y
prime p
1
1 − p −s
This product formula can be used to calculate the probability that
a random set of integers are relatively prime.
1
This probability actually ends up being Z (s)
.
Ben Dulaney Jonathan Gerhard
Riemann Zeta Function
The Riemann ζ Function
Riemann worked to extend this function to the entire complex
plane (except where s = 1, where there is a pole).
Ben Dulaney Jonathan Gerhard
Riemann Zeta Function
The Riemann ζ Function
Riemann worked to extend this function to the entire complex
plane (except where s = 1, where there is a pole).
First, the Gamma function Γ(s) is the analytic continuation of the
factorial, and was found by Euler. That is, for integer s > 1, we
have Γ(s) = (s − 1)!.
Ben Dulaney Jonathan Gerhard
Riemann Zeta Function
The Riemann ζ Function
Riemann worked to extend this function to the entire complex
plane (except where s = 1, where there is a pole).
First, the Gamma function Γ(s) is the analytic continuation of the
factorial, and was found by Euler. That is, for integer s > 1, we
have Γ(s) = (s − 1)!.
Z
Γ(s) =
0
∞
dt s −1
t e , <(s) > 0.
t
Ben Dulaney Jonathan Gerhard
Riemann Zeta Function
The Riemann ζ Function
Consider the integral
Z ∞ s−1 −x
Z ∞ s−1
x
x e
dx
=
dx
L=
x −1
e
1 − e −x
0
0
∞ Z ∞
X
=
x s−1 e −(n+1)x dx
=
n=0 0
∞
X
n=0
Γ(s)
(n + 1)s
= Γ(s)ζ(s).
Ben Dulaney Jonathan Gerhard
Riemann Zeta Function
The Riemann ζ Function
Consider the integral
Z ∞ s−1 −x
Z ∞ s−1
x
x e
dx
=
dx
L=
x −1
e
1 − e −x
0
0
∞ Z ∞
X
=
x s−1 e −(n+1)x dx
=
n=0 0
∞
X
n=0
Γ(s)
(n + 1)s
= Γ(s)ζ(s).
Since we know L = Γ(s)ζ(s), we have
ζ(s) =
Ben Dulaney Jonathan Gerhard
Riemann Zeta Function
L
.
Γ(s)
The Riemann ζ Function
Consider the contour integral
Z
(−z)s−1
dz.
K=
z
C e −1
Ben Dulaney Jonathan Gerhard
Riemann Zeta Function
The Riemann ζ Function
Consider the contour integral
Z
(−z)s−1
K=
dz.
z
C e −1
Then
K = −2i sin(π(s − 1))L.
Ben Dulaney Jonathan Gerhard
Riemann Zeta Function
The Riemann ζ Function
Consider the contour integral
Z
(−z)s−1
dz.
K=
z
C e −1
Then
K = −2i sin(π(s − 1))L.
From this, we obtain the fantastically amazing result that
Z
1
1
(−z)s−1
ζ(s) =
dz.
2i sin(πs) Γ(s) C e z − 1
Ben Dulaney Jonathan Gerhard
Riemann Zeta Function
The Riemann ζ Function
Consider the contour integral
Z
(−z)s−1
dz.
K=
z
C e −1
Then
K = −2i sin(π(s − 1))L.
From this, we obtain the fantastically amazing result that
Z
1
1
(−z)s−1
ζ(s) =
dz.
2i sin(πs) Γ(s) C e z − 1
Since Γ(s)Γ(1 − s) =
function as
π
sin(πs) ,
we get the final form of the ζ
Z
Γ(1 − s)
(−z)s−1
ζ(s) =
dz.
z
2πi
C e −1
Ben Dulaney Jonathan Gerhard
Riemann Zeta Function
Some Surprising Results
Using this analytic continuation, we can obtain some fascinating
results.
1
ζ(0) = −
2
1
ζ(−1) = −
12
ζ(−2m) = 0
1
ζ(−3) =
120
(2π)2m
ζ(2m) =
|B2m |
2(2m)!
B2m
ζ(1 − 2m) = −
2m
Ben Dulaney Jonathan Gerhard
Riemann Zeta Function
References
http://www.nhn.ou.edu/~milton/p5013/zeta.pdf
http://www.math.jhu.edu/~wright/RH2.pdf
http:
//mathworld.wolfram.com/RiemannZetaFunction.html
http:
//en.wikipedia.org/wiki/Riemann_zeta_function
Ben Dulaney Jonathan Gerhard
Riemann Zeta Function
Thank You!
Ben Dulaney Jonathan Gerhard
Riemann Zeta Function