1116 CHAPTER 32. FOURIER TRANSFORMS
7. ↑ Let g ∈ L1 (R) and suppose g is locally Holder continuous from the right and fromthe left at x. Show that then
limR→∞
12π
∫ R
−Reixt∫
∞
−∞
e−ityg(y)dydt =g(x+)+g(x−)
2.
This is very interesting. If g ∈ L2 (R), this shows F−1 (Fg)(x) = g(x+)+g(x−)2 , the
midpoint of the jump in g at the point, x. In particular, if g ∈ G , F−1 (Fg) = g. Hint:Show the left side of the above equation reduces to
2π
∫∞
0
sin(ur)u
(g(x−u)+g(x+u)
2
)du
and then use Problem 6 to obtain the result.
8. ↑ A measurable function g defined on (0,∞) has exponential growth if |g(t)| ≤Ceηt
for some η . For Re(s)> η , define the Laplace Transform by
Lg(s)≡∫
∞
0e−sug(u)du.
Assume that g has exponential growth as above and is Holder continuous from theright and from the left at t. Pick γ > η . Show that
limR→∞
12π
∫ R
−ReγteiytLg(γ + iy)dy =
g(t+)+g(t−)2
.
This formula is sometimes written in the form
12πi
∫γ+i∞
γ−i∞estLg(s)ds
and is called the complex inversion integral for Laplace transforms. It can be used tofind inverse Laplace transforms. Hint:
12π
∫ R
−ReγteiytLg(γ + iy)dy =
12π
∫ R
−Reγteiyt
∫∞
0e−(γ+iy)ug(u)dudy.
Now use Fubini’s theorem and do the integral from −R to R to get this equal to
eγt
π
∫∞
−∞
e−γug(u)sin(R(t−u))
t−udu
where g is the zero extension of g off [0,∞). Then this equals
eγt
π
∫∞
−∞
e−γ(t−u)g(t−u)sin(Ru)
udu