12.8. THE FOURIER TRANSFORM 293
One merely takes a constant outside the integral and then moves a constant inside an inte-gral. Consider the following manipulations.
12π
∫ R
−Reixt∫
∞
−∞
e−ityg(y)dydt =
12π
∫∞
−∞
∫ R
−Reixte−ityg(y)dtdy =
12π
∫∞
−∞
∫ R
−Rei(x−y)tg(y)dtdy
=1
2π
∫∞
−∞
g(y)(∫ R
0ei(x−y)tdt +
∫ R
0e−i(x−y)tdt
)dy
=1
2π
∫∞
−∞
g(y)(∫ R
02cos((x− y) t)dt
)dy
=1π
∫∞
−∞
g(y)sinR(x− y)
x− ydy =
1π
∫∞
−∞
g(x− y)sinRy
ydy
=1π
∫∞
0(g(x− y)+g(x+ y))
sinRyy
dy
=2π
∫∞
0
(g(x− y)+g(x+ y)
2
)sinRy
ydy
From Theorem 10.2.5 or Corollary 10.2.8,
limR→∞
12π
∫ R
−Reixt∫
∞
−∞
e−ityg(y)dydt
= limR→∞
2π
∫∞
0
(g(x− y)+g(x+ y)
2
)sinRy
ydy =
g(x+)+g(x−)2
.
Also we have the Fourier cosine formula. This is interesting because you might have afunction which is not periodic so there would be no hope of representing the function as aFourier series but this next theorem says you can represent it in terms of a Fourier integral.It is the Fourier integral theorem.
Theorem 12.8.3 Let f be piecewise continuous on every finite interval and∫∞
−∞
| f (y)|dy < ∞
and let x ∈ (−∞,∞) satisfy the conditions 10.8 and 10.9 or the Jordan condition that f isof finite total variation on [x−δ ,x+δ ] for some δ > 0. Then
limR→∞
1π
∫ R
0
∫∞
−∞
cos(t (x− y)) f (y)dydt =f (x+)+ f (x−)
2(12.14)
Proof: Consider the following:
1π
∫ R
0
∫∞
−∞
cos(t (x− y)) f (y)dydt =1π
∫∞
−∞
∫ R
0cos(t (x− y)) f (y)dtdy