Number Theory: An Introduction to Mathematics

378 IX The Number of Prime Numbers

Proof Putf(v)=e−v

(^2) πy
and let
g(u)=

∫∞

−∞

f(v)e−^2 πiuvdv

be the Fourier transform off(v). We are going to show that

∑∞

n=−∞

f(v+n)=

∑∞

n=−∞

g(n)e^2 πinv.

Let

F(v)=

∑∞

n=−∞

f(v+n).

This infinite series is uniformly convergent for 0≤v≤1, and so also is the series
obtained by term by term differentiation. HenceF(v)is a continuously differentiable
function. Consequently, since it is periodic with period 1, it is the sum of its own
Fourier series:

F(v)=

∑∞

m=−∞

cme^2 πimv,

where

cm=

∫ 1

0

F(v)e−^2 πimvdv.

We can evaluatecmby term by term integration:

cm=

∑∞

n=−∞

∫ 1

0

f(v+n)e−^2 πimvdv=

∑∞

n=−∞

∫n+ 1

n

f(v)e−^2 πimvdv

=

∫∞

−∞

f(v)e−^2 πimvdv=g(m).

The argument up to this point is an instance ofPoisson’s summation formula.To

evaluateg(u)in the casef(v)=e−v

(^2) πy
we differentiate with respect touand integrate
by parts, obtaining
g′(u)=− 2 πi

∫∞

−∞

e−v

(^2) πy
ve−^2 πiuvdv
=(i/y)

∫∞

−∞

e−^2 πiuvde−v

(^2) πy
=−(i/y)

∫∞

−∞

e−v

(^2) πy
de−^2 πiuv
=−( 2 πu/y)g(u).