Number Theory: An Introduction to Mathematics

138 III More on Divisibility

by Dirichlet’s convergence criterion in the theory of Fourier series,

{F(+ 0 )+F(− 0 )}/ 2 = lim

N→∞

∑N

h=−N

∫ 1

0

e−^2 πihtF(t)dt.

But
∫ 1

0

e−^2 πihtF(t)dt=

∑∞

k=−∞

∫ 1

0

e−^2 πihtf(t+k)dt

=

∑∞

k=−∞

∫k+ 1

k

e−^2 πihtf(t)dt=

∫n

0

e−^2 πihtf(t)dt.

Thus we obtain

f( 0 )/ 2 +f( 1 )+···+f(n− 1 )+f(n)/ 2 = lim

N→∞

∑N

h=−N

∫n

0

e−^2 πihtf(t)dt. (∗)

This is a simple form ofPoisson’s summation formula(which makes an appearance
also in Chapters IX and X).

In particular, if we takef(t)=e^2 πit

(^2) m/n
( 0 ≤t≤n),wheremis also a positive
integer, then the left side of(∗)is just the Gauss sumG(m,n). We will now evaluate
the right side of(∗)for this case. Puth= 2 mq+μ,whereqandμare integers and
0 ≤μ< 2 m.Then
e−^2 πihtf(t)=e^2 πim(t−nq)
(^2) /n
e−^2 πiμt.
Ashruns through all the integers,qdoes also andμruns independently through the
integers 0,..., 2 m−1. Hence
lim
N→∞

∑N

h=−N

∫n

0

e−^2 πihtf(t)dt=

(^2) ∑m− 1
μ= 0
lim
Q→∞

∑Q

q=−Q

∫n

0

e^2 πim(t−nq)

(^2) /n
e−^2 πiμtdt

=

(^2) ∑m− 1
μ= 0
lim
Q→∞

∑Q

q=−Q

∫−(q− 1 )n

−qn

e^2 πit

(^2) m/n
e−^2 πiμtdt

=

(^2) ∑m− 1
μ= 0

∫∞

−∞

e^2 πit

(^2) m/n
e−^2 πiμtdt

=

(^2) ∑m− 1
μ= 0

∫∞

−∞

e^2 πim(t−μn/^2 m)

(^2) /n
e−πiμ
(^2) n/ 2 m
dt

=

(^2) ∑m− 1
μ= 0
e−πiμ
(^2) n/ 2 m

∫∞

−∞

e^2 πit

(^2) m/n
dt

=

√

n

m

C

(^2) ∑m− 1
μ= 0
e−πiμ
(^2) n/ 2 m
,