1.9 Fourier Integral 107
Now we modify Eq. (1). Letλn=nπ/aand define two functionsAa(λ)=^1
π∫a−af(x)cos(λx)dx, Ba(λ)=^1
π∫a−af(x)sin(λx)dx. (3)Notice that
an=π
aAa(λn), bn=π
aBa(λn).Because of this, the Fourier series Eq. (1) becomes
f(x)=a 0 +∑∞
n= 1[
Aa(λn)cos(λnx)+Ba(λn)sin(λnx)]
· λ, −a<x<a, (4)where λ=π/a=λn+ 1 −λn.
The form in which Eq. (4) is written is chosen to suggest an integral with
respect toλover the interval 0<λ<∞. We may imagineaincreasing to
infinity, so λ→0and
Aa(λ)→A(λ)=^1
π∫∞
−∞f(x)cos(λx)dx, (5)Ba(λ)→B(λ)=^1
π∫∞
−∞f(x)sin(λx)dx, (6)anda 0 →0. Then Eq. (4) suggests
f(x)=∫∞
0[
A(λ)cos(λx)+B(λ)sin(λx)]
dλ, −∞<x<∞. (7)Example 1 (continued).
Forf(x)as in Example 1, we find
A(λ)=π^1∫∞
0e−xcos(λx)dx=π( 11 +λ (^2) ),
B(λ)=^1
π
∫∞
0e−xsin(λx)dx= λ
π( 1 +λ^2 ),
and therefore we expect that
∫∞
0[
1
π( 1 +λ^2 )cos(λx)+λ
π( 1 +λ^2 )sin(λx)]
dx={
e−x, 0 <x,
0 , x<0.
The foregoing derivation is not a proof, but it does suggest the following
theorem.