The Chemistry Maths Book, Second Edition

15.6 Fourier transforms 435

so that, by the discussion in Section 5.4 of the Riemann integral as the limit of a sum,

Equation (15.65) is then

(15.69)

where, after taking the limit in (15.66) and (15.67),

(15.70)

(15.71)

The representation (15.69) of f(x)is called a Fourier integral(the name Fourier

transform is normally reserved for a slightly different form, as described below).*

EXAMPLE 15.8For the square wave

(15.72)

Find (i) the Fourier series expansion in interval−l 1 ≤ 1 x 1 ≤ 1 +lwithk 1 < 1 l, (ii) the Fourier

integral in the limitl 1 → 1 ∞

(i) The Fourier series

Figure 15.13 shows that the function is an

evenfunction of x, and it can therefore be

represented in the interval−l 1 ≤ 1 x 1 ≤ 1 +lby the

Fourier cosine series

where and, forn 1 > 10 ,

a

l

fx

nx

l

dx

A

l

nx

l

dx

Ak

l

n

lk

===

222

00

ZZ()cos cos

ππsiin( )

()

nkl

nkl

π

π

a

Ak

l

0

2

=

fx

a

a

nx

l

n

n

()=+ cos

=

∑

0

1

2

∞

π

fx

Akxk

()=

−< <+











for

0 otherwise

v()yfxxydx= ()sin

−

+

1

π

Z

∞

∞

uy()= fx xydx()cos

−

+

1

π

Z

∞

∞

fx()=+uy xy y xydy()cos ()sin













Z

0

∞

v

fx Fy y Fydy

l

n

nn

( ) lim=∆( ) =( )

→

=

∑

∞

∞

∞

0 0

Z

*The ‘derivation’ of (15.69) given here is only an outline, but is sufficient for most purposes; all problems

associated with discontinuities of the function and with the existence of the limit have been ignored.

−l −k 0 +k +l

x

f(x)

A

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Figure 15.13