Number Theory: An Introduction to Mathematics

406 X A Character Study

Proof Since fis holomorphic atβ, we can chooseδ>0sothatfis holomorphic
in the disc|s−(β+δ)|< 2 δ. Thus its Taylor series converges in this disc. But for
Rs>βthen-th derivative offis given by

f(n)(s)=(− 1 )n

∫∞

0

e−sxxndφ(x).

Hence, for anyσsuch thatβ−δ<σ<β+δ,

f(σ)=

∑∞

n= 0

(σ−β−δ)nf(n)(β+δ)/n!

=

∑∞

n= 0

(σ−β−δ)n(− 1 )n

∫∞

0

e−(β+δ)xxndφ(x)/n!

=

∑∞

n= 0

∫∞

0

e−(β+δ)x(β+δ−σ)nxn/n!dφ(x).

Since the integrands are non-negative, we can interchange the orders of summation
and integration, obtaining

f(σ)=

∫∞

0

e−(β+δ)x

∑∞

n= 0

(β+δ−σ)nxn/n!dφ(x)

=

∫∞

0

e−(β+δ)xe(β+δ−σ)xdφ(x)

=

∫∞

0

e−σxdφ(x).

Thus the integral in (†) converges for reals>β−δ.
Letγ be the greatest lower bound of all reals∈(α,β)for which the integral
in (†) converges. Then the integral in (†) is also convergent forRs>γand defines
there a holomorphic function. Since this holomorphic function coincides withf(s)for
Rs>β, it follows that (†) holds forRs>γ.Moreoverγ=α, since ifγ>αwe
could replaceβbyγin the preceding argument and thus obtain a contradiction to the
definition ofγ.

The punch-line is the following proposition:

Proposition 7L( 1 +it,χ)= 0 for every real t and everyχ=χ 1.

Proof Assume on the contrary thatL( 1 +iα,χ)=0 for some realαand some
χ=χ 1 .ThenalsoL( 1 −iα,χ) ̄ =0. If we put

f(s)=ζ^2 (s)L(s+iα,χ)L(s−iα,χ), ̄

thenfis holomorphic and nonzero forσ>1. Furthermorefis holomorphic on the
real segment [1/ 2 ,1], since the double pole ofζ^2 (s)ats=1 is cancelled by the zeros
of the other two factors. By logarithmic differentiation we obtain, forσ>1,