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20 Riemann Zeta Function
20.1 Equations and Evaluation
20.1.1 Equations
The Riemann zeta function ζ is defined by
ζ(s) =
∞
n=1
1
n s =
p prime
1
1 − p −s with (s) > 1 .
(20.1)
The ζ -function represented by Eq. (20.1) is absolute convergent in the complex
half-plane, and in the complex plane, except for the pole at s = 1, by analytic
continuation.
The ζ(s)-function can be evaluated in closed form for even integer values of s
and for negative integers:
ζ(2 n) =
(2π) 2n
2(2n)!
|B 2n | ,
(20.2a)
ζ(−n) = −
B n+1
n + 1
, thus
(20.2b)
ζ(−2n) = 0 ,
for n = 1, 2, 3, · · · .
(20.2c)
The last equation corresponds to the so-called trivial zeros of the Riemann ζ -
function. From Eq. (20.1) we get formally for s = −1, ζ(−1) = 1+2+3+· · · , thus
a divergent series; but from the analytic continuation, Eq. (20.2b), ζ(−1) = −
B 2
2 =
−
1
12 – a simple example for the ζ regularization.
For (s) > 0 the evaluation in SPECFUNPHYS class riemzeta is based on [2].
With s = σ + i t ζ(s) is given by
e k =
n
j =k
n
j
(20.3a)
ζ(s) =
1
1 − 2 1−s
n
k=1
(−1) k−1
k s
+
1
2 n
2n
k=n+1
(−1) k−1 e k−n
k s
+ γ n (s) (20.3b)
with |γ n (s)| ≤
1
8 n
1 + |
t
σ |
|1 − 2 1−s |
exp
|t|
π
2
for σ > 0
(20.3c)
and |γ n (s)| ≤
1
8 n
4 |σ |
|1−2 1−s ||Γ (s)|
for − (n − 1) < σ < 0 . (20.3d)
20 Riemann Zeta Function
20.1 Equations and Evaluation
20.1.1 Equations
The Riemann zeta function ζ is defined by
ζ(s) =
∞
n=1
1
n s =
p prime
1
1 − p −s with (s) > 1 .
(20.1)
The ζ -function represented by Eq. (20.1) is absolute convergent in the complex
half-plane, and in the complex plane, except for the pole at s = 1, by analytic
continuation.
The ζ(s)-function can be evaluated in closed form for even integer values of s
and for negative integers:
ζ(2 n) =
(2π) 2n
2(2n)!
|B 2n | ,
(20.2a)
ζ(−n) = −
B n+1
n + 1
, thus
(20.2b)
ζ(−2n) = 0 ,
for n = 1, 2, 3, · · · .
(20.2c)
The last equation corresponds to the so-called trivial zeros of the Riemann ζ -
function. From Eq. (20.1) we get formally for s = −1, ζ(−1) = 1+2+3+· · · , thus
a divergent series; but from the analytic continuation, Eq. (20.2b), ζ(−1) = −
B 2
2 =
−
1
12 – a simple example for the ζ regularization.
For (s) > 0 the evaluation in SPECFUNPHYS class riemzeta is based on [2].
With s = σ + i t ζ(s) is given by
e k =
n
j =k
n
j
(20.3a)
ζ(s) =
1
1 − 2 1−s
n
k=1
(−1) k−1
k s
+
1
2 n
2n
k=n+1
(−1) k−1 e k−n
k s
+ γ n (s) (20.3b)
with |γ n (s)| ≤
1
8 n
1 + |
t
σ |
|1 − 2 1−s |
exp
|t|
π
2
for σ > 0
(20.3c)
and |γ n (s)| ≤
1
8 n
4 |σ |
|1−2 1−s ||Γ (s)|
for − (n − 1) < σ < 0 . (20.3d)
