272 CHAPTER 11. FUNDAMENTAL TRANSFORMS
=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 11.3.5,
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.■
Observation 11.3.7 If t →∫
∞
−∞e−ityg(y)dy is itself in L1 (R) , then you don’t need to
do the inversion in terms of a principal value integral as above in which
limR→∞
∫ R
−Reixt∫
∞
−∞
e−ityg(y)dydt
was considered. Instead, you simply get
12π
∫∞
−∞
eixt∫
∞
−∞
e−ityg(y)dydt =g(x+)+g(x−)
2
Does this situation ever occur? Yes, it does. This is discussed a little later.
11.4 Inversion of Laplace TransformsHow does the Fourier transform relate to the Laplace transform? This is considered next.Recall that from Theorem 11.2.2 if g has exponential growth |g(t)| ≤Ceηt , then if Re(s)>η , one can define L g(s) as L g(s)≡
∫∞
0 e−sug(u)du and also s→L g(s) is differentiableon Re(s)> η in the sense that if h ∈ C and G(s)≡L g(s) , then
limh→0
G(s+h)−G(s)h
= G′ (s) =−∫
∞
0ue−sug(u)du
This is an example of an analytic function of the complex variable s. The next theoremshows how to invert the Laplace transform. One can prove similar theorems about Fourierseries. See my single variable analysis book on the web site for this.