Lecture 7: Improper Integrals
Today: Improper Integrals of Type I and Type II
We recall the statement of the Evaluation Theorem.
Theorem (Evaluation Theorem). If f is continuous on the interval [a, b], then
Z
a
b
f (x) dx = F (b) − F (a),
where F is an antiderivative of f .
In today’s class, we relax the condition on the Evaluation Theorem in two different ways,
and introduce the improper integrals.
Improper Integrals of Type I: Infinite Intervals
First, we relax the condition on the finite interval by looking at the following example
Example. Find the area of the region that lies under the curve y = x−2 , above the x-axis,
and to the right of the line x = 1.
y
y
0
1
x2
1
x
It may seem that the region would have infinite area because the region itself is infinite. But
let’s take a closer look. If we want to know the area between x = 1 and x = 2, we can set
up the definite integral
Z 2
h
i2
1
1
x−2 dx = − x−1 = 1 − = .
2
2
1
1
Again, if we want to know the area between x = 1 and x = 3,
Z 3
h
i3
1
x−2 dx = − x−1 = 1 − =
3
1
1
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we set up the definite integral
2
.
3
Similarly, if we just draw an arbitrary vertical line x = t with t > 1, and want to know the
area between x = 1 and x = t, the definite integral would tell us that the area is
Z t
h
it
t−1
1
−2
−1
x dx = − x
.
=1− =
t
t
1
1
The following figures shows our computation results.
y
y
y
Area = 1/2
0
1
Area =
Area = 2/3
2
0
x
1
3
x
0
1
t-1
t
t
x
Notice that if we take t → ∞, the area of the shaded region would approach
t−1
= 1.
t→∞
t
lim
So we say that the area of the infinite region is equal to 1, and write that
Z ∞
Z t
t−1
−2
= 1.
x dx = lim
x−2 dx = lim
t→∞
t→∞
t
1
1
With this example in mind, we can define the integral of a function over an infinite interval
in the following way.
Rt
Definition (Improper Integral, Type I). (a) If a f (x) dx exists for every number t ≥ a,
then
Z ∞
Z t
f (x) dx = lim
t→∞
a
f (x) dx,
a
provided this limit exists (as a finite number).
Rb
(b) If t f (x) dx exists for every number t ≤ b, then
Z
b
f (x) dx = lim
t→−∞
−∞
Z
b
f (x) dx,
t
provided this limit exists (as a finite number).
We say an improper integral is convergent if the corresponding limit exists, and divergent
if the limit does not exist.
R∞
Ra
(c) If both a f (x) dx and −∞ f (x) dx are convergent for a number a, then we define
Z ∞
Z ∞
Z a
f (x) dx =
f (x) dx +
f (x) dx.
−∞
−∞
a
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Z
∞
1
dx is convergent or divergent.
Example. (a) Determine whether the improper integral
x
1
Z ∞
1
(b) For what values of p is the improper integral
dx convergent?
xp
1
(a) Use the definition, we have
Z t
Z ∞
h
it
1
1
dx = lim
dx = lim ln |x| = lim ln(t) = ∞.
t→∞ 1 x
t→∞
t→∞
x
1
1
So the improper integral is divergent.
(b) We know from part (a) that when p = 1, the integral is divergent. Now let’s assume that
p 6= 1, then
Z t
Z ∞
h x1−p it
1
t1−p − 1
1
dx
=
lim
dx
=
lim
=
lim
.
t→∞ 1 xp
t→∞ 1 − p 1
t→∞ 1 − p
xp
1
Here we have two cases:
Case 1. If p < 1, then 1 − p > 0, and
t1−p − 1
= ∞.
t→∞ 1 − p
lim
Case 2. If p > 1, then 1 − p < 0, and
t1−p − 1
−1
1
=
=
.
t→∞ 1 − p
1−p
p−1
lim
Thus the improper integral
R∞
1
1
xp
dx is convergent if p > 1, and is divergent if p ≤ 1.
Example. Evaluate the improper integral
Z ∞
−∞
1
dx
1 + x2
Let’s choose a = 0 to evaluate this improper integral.
Z ∞
Z 0
Z ∞
1
1
1
dx =
dx +
dx
2
2
1 + x2
−∞ 1 + x
−∞ 1 + x
0
Z 0
Z t
1
1
= lim
dx + lim
dx
s→−∞ s 1 + x2
t→∞ 0 1 + x2
h
i0
h
it
= lim arctan(x) + lim arctan(x)
s→−∞
s
t→∞
0
= arctan(0) − lim arctan(s) + lim arctan(t) − arctan(0, )
s→−∞
t→∞
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We look at the graph of y = arctan(x), and notice that there are two horizontal asymptotes
at y = ±π/2.
π/2
y  tan-1 (x)
-π/2
Thus
Z
1
π π
− −
= π.
dx
=
lim
arctan(t)
−
lim
arctan(s)
=
2
t→∞
s→−∞
2
2
−∞ 1 + x
Z ∞
e−x dx convergent or divergent?
Exercise. Is the improper integral
∞
0
By definition of improper integral,
Z ∞
Z t
h
it
−x
e dx = lim
e−x dx = lim − e−x = lim (−e−t ) − (−e0 ) = 1.
0
t→∞
t→∞
0
t→∞
0
Thus this improper integral is convergent.
Improper Integrals of Type II: Discontinuous Integrand
The first type of improper integrals concerns the area of a region that extends infinitely on
the horizontal direction. We now introduce the second type of improper integral on functions
that have vertical asymptotes.
Definition (Improper Integral, Type II). (a) If f is continuous on [a, b) and is discontinuous
at b, then
Z
Z
b
a
t
f (x) dx = lim−
t→b
f (x) dx,
a
provided that this limit exists (as a finite number).
(b) If f is continuous on (a, b] and is discontinuous at a, then
Z b
Z b
f (x) dx = lim+
f (x) dx,
t→a
a
t
provided that this limit exists (as a finite number).
Rc
Rb
(c) If f is continuous on [a, b] except at c ∈ (a, b), and both a f (x) dx and c f (x) dx are
convergent, then we define
Z b
Z c
Z b
f (x) dx =
f (x) dx +
f (x) dx.
a
a
c
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Example. (a) For what values of p is the integral
(b) For what values of p is the integral
Z
Z
0
1
0
1
1
dx improper?
xp
1
dx divergent?
xp
(a) The only type of discontinuity on [0, 1] that could arise in the integrand is when xp = 0.
However, we notice that when p ≤ 0, −p ≥ 0, and
1
= x−p
p
x
is continuous throughout [0, 1]. When p > 0, the integrand is not continuous at 0. Thus the
integral is improper if and only if p > 0.
(b) We assume that p > 0 first. Then
Z
Z
x1−p
1
−p
dx
=
x
dx
=
+ C.
xp
1−p
Now we take the limit
lim+
t→0
Z
t
1
h 1
1
t1−p i
dx
=
lim
−
.
t→0+ 1 − p
xp
1−p
We shall notice that
lim t1−p =
t→0+




0,
1,



∞,
if p < 1,
if p = 1,
if p > 1.
However, when p = 1, the denominator 1 − p would be zero. Thus the improper integral is
divergent if and only if p ≥ 1.
Exercise. Find the mistake in the following evaluation:
Z 2
h
i2
1
dx = ln |x − 1| = ln(1) − ln(1) = 0.
0
0 x−1
The integrand is discontinuous at x = 1, which is inside the range of integration. Thus this
should be an improper integral
Z 2
Z 1
Z 2
1
1
1
dx =
dx +
dx.
0 x−1
1 x−1
0 x−1
We know the second integral is divergent from the previous example, thus this integral should
also be divergent.
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