10.2. CONVERGENCE 237
=−(∫ c
ah(t)dt
)(g(b)−g(a))+g(b)
∫ b
ah(t)dt = g(b)
∫ b
ch(t)dt +g(a)
∫ c
ah(t)dt
This is sometimes called the second mean value theorem for integrals. Sufficient conditionsare that h is continuous and g monotone. This is stated as the following lemma.
Lemma 10.2.6 Let h be continuous and g monotone. Then there is c ∈ [a,b] such that∫ b
ag(x)h(x)dx = g(b)
∫ b
ch(x)dx+g(a)
∫ c
ah(x)dx
The conclusion is exactly the same if g(a) is replaced with g(a+) with maybe a differentc ∈ [a,b]. c only needs to satisfy H (c)(g(b)−g(a)) =
∫ ba Hdg where H (x) =
∫ xa h(t)dt.
Proof: The last claim follows from a repeat of the above argument using g̃(x) definedas g(x) for x > a and g(a+) when x = a. Such a change does nothing to the Riemannintegral on the left in the above formula and g̃ is still monotone. Hence, for some c ∈ [a,b] ,∫ b
ag(x)h(x)dx =
∫ b
ag̃(x)h(x)dx = g(b)
∫ b
ch(x)dx+g(a+)
∫ c
ah(x)dx
Lemma 10.2.7 Suppose g is of bounded variation on [0,δ ] ,δ > 0 and suppose g ∈L1 (0,a) where δ < a≤ ∞. Then
limr→∞
2π
∫ a
0g(t)
sin(rt)t
dt = g(0+)
Proof: Since every bounded variation function is the difference of two increasing func-tions, it suffices to assume that g is increasing on [0,δ ] and so this will be assumed. Notethat this also shows that g(0+) exists. Recall the second mean value theorem of Problem45 on Page 223 or the above Lemma 10.2.6 applied to g(0) defined as g(0+).
From the material on the Dirichlet integral, there exists C with∣∣∫ r
0sin t
t dt∣∣<C indepen-
dent of r. Indeed,∣∣∫ r
0sin t
t dt− π
2
∣∣< 1 if r is sufficiently large. For r not this large, one hasthe integral of a continuous function.
Let h ∈ (0,δ ) be such that if t ≤ h, |g(t)−g(0+)|< ε
2C+1 . Then split up the integral asfollows.
2π
∫ a
0g(t)
sin(rt)t
dt =
I1︷ ︸︸ ︷2π
∫ h
0(g(t)−g(0+))
sin(rt)t
dt +g(0+)
I2︷ ︸︸ ︷2π
∫ h
0
sin(rt)t
dt
+
I3︷ ︸︸ ︷∫ a
hg(t)
sin(rt)t
dt
Use the second mean value theorem on I1. It equals
(g(h)−g(0+))2π
∫ h
cr
sin(rt)t
dt = (g(h)−g(0+))2π
∫ rh
rcr
sin(u)u
du
the integral is(∫ hr
0sinu
u du−∫ rcr
0sinu
u du)
and both terms are bounded by some constant Cso the integral is bounded independent of large r by 2C. Then∣∣∣∣(g(h)−g(0+))
2π
∫ rh
rcr
sin(u)u
du∣∣∣∣≤ ε
2C+12C < ε