3 Proof of the Prime Number Theorem 375
of the function
k(t)= 1 −|t|for|t|≤ 1 ,=0for|t|≥ 1 ,
has the properties
kˆ(u)≥0for−∞<u<∞, C:=
∫∞
−∞
kˆ(u)du<∞.
Indeed
kˆ(u)=
∫ 1
− 1
eiut( 1 −|t|)dt
= 2
∫ 1
0
( 1 −t)cosut dt
= 2 ( 1 −cosu)/u^2.
Letε,λ,ybe arbitrary positive numbers. Ifs=ε+iλt,then
λ
∫ 1
− 1
eiλtyk(t)g(s)dt=λ
∫ 1
− 1
eiλtyk(t)
∫∞
0
e−εxe−iλtx{α(x)−A}dxdt
=λ
∫∞
0
e−εx{α(x)−A}
∫ 1
− 1
eiλt(y−x)k(t)dtdx
=λ
∫∞
0
e−εxα(x)kˆ(λ(y−x))dx
−λA
∫∞
0
e−εxkˆ(λ(y−x))dx.
Whenε→+0 the left side has the limit
χ(y):=λ
∫ 1
− 1
eiλtyk(t)γ(λt)dt
and the second term on the right has the limit
λA
∫∞
0
kˆ(λ(y−x))dx.
Consequently the first term on the right also has a finite limit. It follows that
λ
∫∞
0
α(x)kˆ(λ(y−x))dx
is finite and is the limit of the first term on the right. Thus
χ(y)=λ
∫∞
0
{α(x)−A}kˆ(λ(y−x))dx
=
∫λy
−∞
{α(y−v/λ)−A}kˆ(v)dv.