236 CHAPTER 10. IMPROPER INTEGRALS

+2π

∫∞

δ

sin(ru)u

f (u)du− 2π

∫∞

δ

sin(ru)u

f (0)du (10.7)

Now f (u)u ∈ L1 (δ ,∞) because

∣∣∣ f (u)u

∣∣∣≤ 1δ| f (u)| and so, by the Riemann Lebesgue lemma,

the first integral of 10.7 converges to 0 as r→ ∞. The second integral of 10.7 convergesto 0 as r→ ∞ because of the second part of Theorem 10.2.1. Now consider the integralin 10.6. s→ f (u)− f (0)

u is in L1 ([0,δ ]) because∣∣∣ f (u)− f (0)

u

∣∣∣ ≤ uγ−1 which has finite integral

since γ > 0. Therefore, limr→∞2π

∫δ

0sin(ru)

u ( f (u)− f (0))du = 0 by the Riemann Lebesguelemma.

Theorem 10.2.5 Suppose that g ∈ L1 (R) and that at some x, g is locally Holdercontinuous from the right and from the left. This means there exist constants K,δ > 0 andr ∈ (0,1] such that for |x− y|< δ ,

|g(x+)−g(y)|< K |x− y|r (10.8)

for y > x and|g(x−)−g(y)|< K |x− y|r (10.9)

for y < x. Then

limr→∞

2π

∫∞

0

sin(ur)u

(g(x−u)+g(x+u)

2

)du =

g(x+)+g(x−)2

.

Proof: The function u→ g(x−u)+g(x+u)2 ≡ f (u) is in L1 (0,∞) as noted earlier. Also for

f (0) defined as g(x+)+g(x−)2 , the conditions of Lemma 10.2.4 are obtained. Therefore, from

that lemma,

limr→∞

2π

∫∞

0

sin(ur)u

f (u)du = limr→∞

2π

∫∞

0

sin(ur)u

(g(x−u)+g(x+u)

2

)du

=g(x+)+g(x−)

2as claimed.

There is an other condition which will allow the same conclusion as the above condi-tion. It is that g is of bounded variation on [x−δ ,x+δ ] for some δ > 0. This is calledthe Jordan condition whereas the more common assumption used above is the Dini condi-tion. This Jordan condition implies that u→ (g(x−u)−g(x−)+g(x+u)−g(x+)) is ofbounded variation on [0,δ ].

First, here is a little review. If H is continuous, Theorem 9.9.1, the mean value theoremfor integrals, implies that if g is monotone, (either increasing or decreasing) then∫ b

aH (x)dg = H (c)(g(b)−g(a))

for some c ∈ [a,b]. Suppose then that H (x) ≡∫ x

a h(t)dt where h is continuous and g ismonotone on [a,b] .

Suppose H (x)≡∫ x

a h(t)dt where h is continuous and g is monotone on [a,b] . Then byintegration by parts, Theorem 9.4.1,

∫ ba gdH +

∫ ba Hdg = g(b)H (b). From the first mean

value theorem for integrals, there is c∈ [a,b] such that∫ b

a Hdg=H (c)(g(b)−g(a)) . Then∫ b

aghdt =

∫ b

agdH =−

∫ b

aHdg+g(b)H (b) =−H (c)(g(b)−g(a))+g(b)H (b)