756 CHAPTER 37. SOME FUNDAMENTAL FUNCTIONS AND TRANSFORMS
Proof: As in the proof of Theorem 37.3.2, changing variables shows that 2π
∫∞
0sin(ru)
u du=1.Therefore,
2π
∫∞
0
sin(ur)u
(g(x−u)+g(x+u)
2
)du− g(x+)+g(x−)
2
=2π
∫∞
0
sin(ur)u
(g(x−u)−g(x−)+g(x+u)−g(x+)
2
)du
=2π
∫δ
0sin(ur)
(g(x−u)−g(x−)
2u+
g(x+u)−g(x+)
2u
)du
+2π
∫∞
δ
sin(ur)u
(g(x−u)−g(x−)
2+
g(x+u)−g(x+)
2
)du (37.4)
Second Integral: It equals
2π
∫∞
δ
sin(ur)u
(g(x−u)+g(x+u)
2− g(x−)+g(x+)
2
)du
=2π
∫∞
δ
sin(ur)u
(g(x−u)+g(x+u)
2
)− 2
π
∫∞
δ
sin(ur)u
(g(x−)+g(x+)
2
)(37.5)
From part 2 of Theorem 37.3.2,
limr→∞
2π
∫∞
δ
sin(ur)u
g(x−)+g(x+)
2du = 0
Thus consider the first integral in 37.4.
2π
∫∞
δ
sin(ur)u
(g(x−u)+g(x+u)
2
)du
=1π
∫∞
δ
sin(ur)u
g(x−u)du+1π
∫∞
δ
sin(ur)u
g(x+u)du
=1π
(∫ −δ
−∞
sin(ur)u
g(x+u)du+∫
∞
δ
sin(ur)u
g(x+u)du)
Now ∫ −δ
−∞
sin(ur)u
g(x+u)du =∫ −δ
−∞
sin(ur)g(x+u)
udu
and∣∣∣ g(x+u)
u
∣∣∣≤ 1δ|g(x+u)| for u <−δ . Thus u→ g(x+u)
u is in L1 ((−∞,−δ )) . Indeed,
∫ −δ
−∞
∣∣∣∣g(x+u)u
∣∣∣∣du≤ 1δ
∫R|g(x+u)|du =
1δ
∫R|g(y)|dy < ∞
It follows from the Riemann Lebesgue lemma
limr→∞
∫ −δ
−∞
sin(ur)g(x+u)
udu = lim
r→∞
∫∞
δ
sin(ur)g(x+u)
udu = 0