108 Chapter 1 Fourier Series and Integrals
Fourier Integral Representation Theorem. Let f(x)be sectionally smooth on
every finite interval, and let
∫∞
−∞|f(x)|dx be finite. Then at every point x,
∫∞0(
A(λ)cos(λx)+B(λ)sin(λx))
dλ=1
2
(
f(x+)+f(x−))
,
−∞<x<∞, (8)where
A(λ)=π^1∫∞
−∞f(x)cos(λx)dx, B(λ)=π^1∫∞
−∞f(x)sin(λx)dx. (9)
Equation (8) is called theFourier integral representation of f(x);A(λ)andB(λ)
in Eq. (9) are theFourier integral coefficient functionsoff(x). The right-hand
sideofEq.(8)wasalsoseenintheFourierseriesconvergencetheorem.Since
f(x)is sectionally smooth, the expression in Eq. (8) is the same asf(x)almost
everywhere, so we often write Eq. (7) instead of Eq. (8).
Example 2.
The function
f(x)={
1 , |x|<1,
0 , |x|> 1has the Fourier integral coefficient functions
A(λ)=^1
π∫∞
−∞f(x)cos(λx)dx=^1
π∫ 1
− 1cos(λx)dx=2sin(λ)
πλ,
B(λ)= 0.Sincef(x)is sectionally smooth, the Fourier integral representation is legit-
imate, and we write
f(x)=∫∞
02sin(λ)
πλ cos(λx)dλ.(Actually the integral equals^12 atx=±1, so equality is not strictly correct at
these two points.)
Example 3.
Find the Fourier integral representation off(x)=exp(−|x|).
Solution:Direct integration gives
A(λ)=π^1∫∞
−∞exp(
−|x|)
cos(λx)dx, (10)