a limit theorem for the hurwitz zeta-function in the space of analytic ...

82 A limit theorem for the Hurwitz zeta-function...
Proof. By choice of σ 1 and σ we have that σ+σ 1 > 1. Consequently, the function
ζ(s + z, α) for Rez = σ 1 , is represented by an absolutely convergent Dirichlet series
Let
Consider the series
Since
ζ(s + z, α) =
a n (m, α) = 1
2πi
a n (m, α) = B(m + α) −σ 1
∞∑
m=1
∫ ∞
−∞
∞∑
m=1
1
(m + α) s+z .
σ 1+i∞ ∫
σ 1 −i∞
l n (s, α)ds
s(m + α) s .
a n (m, α)
(m + α) s . (7)
|l n (σ 1 + it, α)|dt = B(m + α) −σ 1
,
the series (7) converges absolutely for σ > 1 2
. Thus, interchanging sum and integral
in the definition of ζ 2,n , we find
By Lemma 6
ζ 2,n (s, α) =
∞∑
m=1
{
a n (m, α) = exp −
Therefore the equality (8) can be writen as
ζ 2,n (s, α) =
∞∑
m=1
{
1
(m + α) s exp −
a n (m, α)
(m + α) s . (8)
( ) σ1
}
(m + α)
.
(n + α)
( ) σ1
}
(m + α)
,
(n + α)
the series being absolutely convergent for σ > 1 2 .
Now we change the contour in the integral for ζ 2,n (s, α). The integrand has
simple poles at z = 0 and z = 1 − s. Let σ belong to σ ∈ [ 1 2
+ ε, 1 − ε] when s ∈ K.
We put
Then by the residue theorem
ζ 2,n (s, α) = 1
2πi
∫
σ 2−σ+i∞
σ 2−σ−i∞
σ 2 = 1 2 + ε 2 .
ζ(s + z, α)l n (z, α) dz
z + ζ(s, α) + l n(1 − s, α)
. (9)
1 − s