SOME REMARKS ON LIMITS AND IMPROPER INTEGRALS
LANCE D. DRAGER
1. Introduction
This note contains some remarks that will help dealing with improper integrals,
including Problem 7.12.
Suppose that f : [a, ∞) → R is a function such that f is Riemann integrable on
each interval [a, b] for b > a. Then, we define
Z ∞
Z b
f (x) dx = lim
f (x) dx,
b→∞
a
a
if the limit exists. If the limit exists (as a real number), we say
Z ∞
f (x) dx
a
converges, otherwise it diverges.
In dealing with this, it’s useful to have some material on limits and some material
on integrals.
2. Limits
For limits, we have the following propositions.
Proposition 2.1. Let F : [a, ∞) → R be a function. Then, the following conditions
are equivalent:
(1)
lim F (x) = L,
x→∞
where L is a real number.
(2) For every sequence {xn } ⊆ [a, ∞) such that xn → ∞,
lim F (xn ) = L.
n→∞
Proof. First suppose that (1) holds, and let {xn } ⊆ [a, ∞) be a sequence such that
xn → ∞. Let ε > 0 be given. Then we can find some number A so that
(2.1)
x ≥ A =⇒ |F (x) − L| < ε.
Since xn → ∞, there is some integer N so that n ≥ N =⇒ xn > A. Then, by
(2.1),
n ≥ N =⇒ |F (xn ) − L| < ε.
This shows that
lim F (xn ) = L.
n→∞
1
2
LANCE D. DRAGER
For the second part, we want to prove that (2) =⇒ (1) To do this, we prove the
contrapositive. Thus, assume that (1) does not hold. Then, there is some ε > 0
such that for every number A, there is an x ∈ [a, ∞) with x > A so that
|F (x) − L| ≥ ε.
Letting A = n for n = 1, 2, 3, . . . , we get numbers xn > n so that
|F (xn ) − L)| ≥ ε.
Since xn → ∞, (2) fails.
Proposition 2.2. Let F : [a, ∞) → R be a function, and suppose that
lim F (xn )
n→∞
exists for every sequence xn → ∞. Then
lim F (x)
x→∞
exists.
Proof. Suppose that F (xn ) converges for every sequence xn → ∞.
What we have to prove is that F (xn ) converges to the same thing for every
sequence with xn → ∞. So, let {xn } and {yn } be sequences in [a, ∞) that converge
to infinity. Then
lim F (xn ) = L1 ,
lim F (yn ) = L2 ,
n→∞
n→∞
for some numbers L1 and L2 .
Let {zn } be the sequence
x1 , y2 , x3 , y4 , . . . ,
It’s left as an exercise for the reader to show that zn → ∞.
We have
lim F (zn ) = L
n→∞
for some number L. Any subsequence of {F (zn )} must also converge to L. Thus
F (z2k ) → L as k → ∞. But F (z2k ) = F (x2k ) → L1 , since {F (x2k )} is a subseqence
of {F (xn )}. Thus L1 = L.
Similarly, F (z2k−1 ) → L, but F (z2k−k ) = F (y2k−1 ) → L2 , so L2 = L.
We conclude that L1 = L2 . Thus,
lim F (xn )
n→∞
is the same number, say M , for every sequence xn → ∞. By the previous proposition,
lim F (x) = M.
x→∞
Similar remarks apply to taking a limit at a finite point. In fact, we have the
following Proposition, whose proof is left to the reader.
Proposition 2.3. Let X be a metric space and let f : X ⊇ E → R be a function.
Suppose that p ∈ X is a limit point of E. Then, the following statements are
equivalent.
SOME REMARKS ON LIMITS AND IMPROPER INTEGRALS
3
(1)
lim f (x) = L.
x→p
(2) For every sequence {xn } ⊆ E \ {p} that converges to p, f (xn ) → L.
If
lim f (xn )
n→∞
exists for every sequence {xn } ⊆ E \ {p} that converges to p, then
lim f (x)
x→p
exists.
Let’s return to the case of taking the limit of F (x) as x → ∞. It’s easier to
analyze what happens to F (x) as x → ∞ in the case where F is increasing.
You probably remember the following Theorem about sequences, which is often
used when studying series.
Theorem 2.4 (Bounded Montone Convergence Theorem for Sequences). Let {tn } ⊆
R be an increasing sequence, i.e., tn ≤ tn+1 for all n. Of course, {tn } is either
bounded above or not it’s not.
(1) If {tn } is not bounded above, then tn → ∞.
(2) If {tn } is bounded, say tn ≤ M for all n, then
lim tn = sup {tn | n ∈ N} ≤ M.
n→∞
There is a similar theorem for increasing functions. The proof is left to the
reader.
Theorem 2.5 (Bounded Montone Convergence Theorem for Functions). Let
F : [a, ∞) → R be an increasing function. Of course, either F is bounded above, or
it’s not.
(1) If F is not bounded above, then
lim F (x) = ∞.
x→∞
(2) If F is bounded above, say F (x) ≤ M for all x, then
lim F (x) = sup {F (x) | x ∈ [a, ∞)} ≤ M.
x→∞
In the case of an increasing function, it suffices to look at F (tn ) for only one
sequence tn → ∞.
Proposition 2.6. Let F : [a, ∞) → R be an increasing function and let {tn } ⊆
[a, ∞) be a sequence that goes to infinity. Let
E = lim F (tn ),
n→∞
−∞ < E ≤ ∞.
Then we have
lim F (x) = E.
x→∞
Proof. If E = ∞, the values F (tn ) of F can be made arbitrarily large. Thus, F is
not bounded above, so F (x) → ∞ as x → ∞.
If E < ∞, then
E = sup {F (tn ) | n ∈ N} .
4
LANCE D. DRAGER
For any x, we can find an n so that x < tn , since tn → ∞. But then
F (x) ≤ F (tn ) ≤ E,
since F is increasing. Thus, F is bounded above by E. Let S = {F (x) | x ∈ [a, ∞)},
so S ≤ E. One the other hand, F (tn ) is a value of F , so F (tn ) ≤ S. Taking the
sup over n, we get E ≤ S. Thus, S = E.
Of course one can apply similar reasoning to the case where F is decreasing, and
to the case of limits at −∞. Similar ideas can also be applied to one-sided limits
at a point p ∈ R, we leave it to the reader to formulate the theorems.
3. Integrals
Suppose that f : [α, β] → R is Riemann integrable. If α ≤ a < b ≤ β, the integral
Z b
f (t) dt
a
exits. If a, b, c ∈ [α, β] and a < b < c then we know from the book that
Z c
Z b
Z c
(3.1)
f (t) dt =
f (t) dt +
f (t) dt.
a
a
b
It’s convenient to extend the notation for the integral as follows. Define
Z a
f (t) dt = 0,
for any a ∈ [α, β].
a
If a < b, we define
Z
a
f (t) dt
(“endpoints out of order”)
b
b
Z
=−
f (t) dt.
a
We want to show that Equation (3.1) holds for any three points a, b, c ∈ [α, β],
no matter what order they’re in and even if some of them coincide.
Indeed, we can show more. We can define a function F on [α, β] by
Z x
F (x) =
f (t) dt.
α
If α < a < b, we have (by the book),
Z b
Z
f (t) dt =
α
a
Z
f (t) dt +
α
b
f (t) dt.
a
But this can be rewritten as
Z
b
F (b) = F (a) +
f (t) dt.
a
Thus, we have
Z
F (b) − F (a) =
b
f (t) dt.
a
when a < b. But, if we multiply this equation by −1, we get
Z b
Z a
F (a) − F (b) = −
f (t) dt =
f (t) dt.
a
b
SOME REMARKS ON LIMITS AND IMPROPER INTEGRALS
5
We certainly also have
a
Z
F (a) − F (a) = 0 =
f (t) dt.
a
So, in summary, we have
Z y
(3.2)
f (t) dt = F (y) − F (x),
for all x, y ∈ [α, β],
x
regardless of the ordering of the points x and y.
Suppose that x, y, z ∈ [α, β] and we don’t know anything about the order of the
points. We want to show that the equation
Z z
Z y
Z z
(3.3)
f (t) dt =
f (t) dt +
f (t) dt
x
x
y
holds. But, we have
z
Z
f (t) dt = F (z) − F (x),
Z xy
f (t) dt = F (y) − F (x),
x
z
Z
f (t) dt = F (z) − F (y).
y
But then (3.3) becomes
F (z) − F (x) = [F (y) − F (x)] + [F (z) − F (y)],
and you can’t argue with that!
From the book, we know that if a < b, we have
Z b
Z b
f (t) dt ≤
|f (t)| dt.
a
a
If we don’t know the order of the points, the corresponding inequality is
Z y
Z y
(3.4)
f (t) dt ≤ |f (t) dt|,
for any x, y ∈ [α, β],
x
x
where the outside absolute value bars on the right are needed to take care of the
case y < x (exercise!).
We’ll now apply this material to improper integrals. Let f : [a, ∞) → R be a
function which is Riemann integrable on every interval of the form [a, b] for b > a
(and hence on every compact subinterval of [a, ∞)). As we said in the introduction,
we define
Z x
Z ∞
f (t) dt,
f (t) dt = lim
x→∞
a
a
if this limit exists. The integral on the left is called an improper integral since it’s
not (properly speaking) a Riemann integral.
We can describe this in a somewhat different way. We can define a function
F : [a, ∞) → R by
Z
x
F (x) =
f (t) dt.
a
Then
Z
∞
f (t) dt = lim F (x),
a
x→∞
6
LANCE D. DRAGER
if the limit exists.
The function |f | is also Riemann integrable on every compact subinterval of
[a, ∞) and we can define a function G by
Z x
G(x) =
|f (t)| dt.
a
But, since |f | ≥ 0, G is an increasing function (exercise), so limx→∞ G(x) is either
finite or ∞. Thus, we can say that
Z ∞
|f (t)| dt = lim G(x)
x→∞
a
is either finite and converges, or is ∞ and diverges.
Proposition 3.1. Let f : [a, ∞) → R be Riemann integrable on every interval of
the form [a, b] for b > a. If
Z
∞
|f (t)| dt < ∞,
a
(i.e., it converges), then
b
Z
f (t) dt
a
converges and we have
Z
(3.5)
∞
Z
f (t) dt ≤
a
∞
|f (t)| dt.
a
In this case, we say that
Z
∞
f (t) dt
a
is absolutely convergent.
Proof. Let
Z
∞
|f (t)| dt = lim G(x) < ∞.
I=
x→∞
a
To prove that
lim F (x)
x→∞
exits, it will suffice to show that
lim F (xn )
n→∞
exists for every sequence [a, ∞) 3 xn → ∞.
So, let {xn } be such a sequence. The sequence G(xn ) converges to I, and so
{G(xn )} is a Cauchy sequence. For any indices n and m, we have
Z xn
Z xn
|f (t)| dt = |G(xn ) − G(xm )|.
|F (xn ) − F (xm )| = f (t) dt ≤ xm
xm
Thus,
|F (xn ) − F (xm )| ≤ |G(xn ) − G(xm ))|,
but the right hand side of this inequality goes to zero as n, m → ∞, since {G(xn )}
is Cauchy. We conclude that {F (xn )} is a Cauchy sequence, so
lim F (xn )
n→∞
SOME REMARKS ON LIMITS AND IMPROPER INTEGRALS
exists. This completes the proof that
Z
7
∞
f (t) dt
a
converges. To get (3.5), note that for every x > a,
Z x
Z x
f (t) dt ≤
|f (t)| dt
a
a
and take the limit on both sides as x → ∞.
Note that if b > a, our function f is also Riemann integrable on every interval
of the form [b, c], for c > b, so it makes sense to consider
Z ∞
f (t) dt.
b
Proposition 3.2. Let f : [a, ∞) be Riemann integrable on every interval of the
form [a, c] for c > a. Suppose that
Z ∞
f (t) dt
a
converges.
Then, for every b > a, the improper integral
Z ∞
f (t) dt
b
converges and we have
Z
(3.6)
∞
b
Z
f (t) dt =
∞
Z
f (t) dt +
a
a
f (t) dt.
b
We also have
∞
Z
(3.7)
lim
b→∞
f (t) dt = 0.
b
Proof. For any x ∈ [a, ∞), we have
Z x
Z
f (t) dt =
a
b
x
Z
f (t) dt +
a
f (t) dt.
b
Thus,
Z
(3.8)
x
Z
b
Z
f (t) dt −
f (t) dt =
b
x
a
f (t) dt.
a
The second term on the right is a constant, independent of x, so the limit of the
right hand side as x → ∞ is
Z ∞
Z b
f (t) dt −
f (t) dt
a
a
We conclude that the limit of the left side of (3.8) exists, and
Z ∞
Z b
Z ∞
f (t) dt,
(3.9)
f (t) dt =
f (t) dt −
b
which is equivalent to (3.6).
a
a
8
LANCE D. DRAGER
To get (3.7), notice that
Z
lim
b→∞
b
Z
f (t) dt =
a
∞
f (t) dt
a
and take the limit of both sides of (3.9) as b → ∞.
I hope these remarks will be helpful with Problem 7.12, and in general.
Department of Mathematics and Statistics, Texas Tech University, Lubbock, TX
79409-1042
E-mail address: lance.drager@ttu.edu