Class Activity #5: Improper Integrals
David Amos
MATH 152 – Engineering Math II, Spring 2015
1
Type I Improper Integrals
As the name of this section suggests, there is more than one kind of improper integral.
The first type involves integrating over and infinite interval. The main idea here is to
think of the integral as a function of some parameter, t say, and then determine the
behavior as t becomes infinite. For example, suppose we want to determine the area A
under the curve y = x12 to the right of the line x = 1. Since x12 > 0 for every x ≥ 1, we
might expect this area to be infinite. In this case, our intuition is misguiding us, as the
area is in fact finite. To see why, consider first the area A(t) under the curve to the right
of x = 1 and to the left of x = t for some t > 1. This area is given by the integral
t
Z t
1 1
1
dx = − = 1 − .
(1)
A(t) =
2
x
t
1 x
1
To find the area of the entire region under the curve and to the right of x = 1, we just
evaluate the limit as t → ∞ in (1):
1
A = lim A(t) = lim 1 −
= 1.
t→∞
t→∞
t
Our intuition was way off!
Something surprising happens when we consider the curve y = x1 . In light of our
discovery in the previous paragraph, we may now expect that the area under y to the
right of x = 1 is finite. Or maybe we just don’t know what to expect. Repeating the
same process as above, the area A(t) under y and between x = 1 and x = t is given by
Z t
t
1
A(t) =
dx = ln |x| = ln |t|.
1
1 x
Thus the total area to the right of x = 1 is:
A = lim A(t) = lim ln |x| = ∞.
t→∞
t→∞
R∞ 1
Rt
R∞
Rt
As a shorthand, we write 1 x2 dx for limt→∞ 1 x12 dx and 1 x1 for limt→∞ 1 x1 dx.
Both of these are examples of Type I improper integrals. For the first case, we say that the
improper integral converges, while the latter diverges. In general, Type I imporoper
integrals are defined as follows
1
D. Amos
Class Activity #5
2
Rt
Definition 1 (Type I improper integral). (a) If a f (x)dx exists for every number t ≥ a,
then
Z t
Z ∞
f (x)dx = lim
f (x)dx
t→∞
a
a
provided this limit is finite.
Rb
(b) If
t
f (x)dx exists for every number t ≤ b, then
Z b
Z b
f (x)dx
f (x)dx = lim
t→−∞
−∞
t
provided this limit is finite.
In either case, if the limit exists and is finite, we say the the improper integral converges.
Otherwise, if the limit does not exists, the integral diverges.
(c) If both
R∞
a
Ra
f (x)dx and −∞ f (x)dx are convergent, then we define
Z ∞
Z a
Z ∞
f (x)dx.
f (x)dx +
f (x)dx =
∞
−∞
a
Problem 1. Evaluate each integral. Is it convergent or divergent?
Z ∞
(a)
sin(θ)dθ
2π
Z
∞
(b)
e
Z
∞
(c)
1
dx
x(ln x)3
2
xe−x dx
−∞
2
Type II improper integrals
A Type II improper integral involvesRintegrating over an interval for which the integrand
5
1
1
. Here, √x−2
is undefined at x = 2. We
has a discontinuity. For example, 2 √x−2
take
R 5 1a similar approach as that for Type 1I improper integrals and consider the integral
√
dx for 2 < t ≤ 5 (the integrand √x−2 is continuous for all such t):
t
x−2
Z 5
5
√
√
√
1
√
dx = 2 x − 2 = 2 3 − 2 t − 2.
t
x−2
t
Now we take the limit as t approaches 2 from the right:
Z 5
√
√
√
1
√
dx = lim+ 2 3 − 2 t − 2 = 2 3.
t→2
x−2
2
In general, we have the following definition (compare this to the definition for Type I
improper integrals).
D. Amos
Class Activity #5
3
Definition 2 (Type II improper integrals). (a) If f is continuous on [a, b) and is discontinuous at b, then
Z b
Z t
f (x)dx = lim−
f (x)dx
t→b
a
a
provided this limit is a finite number.
(b) If f is continuous on (a, b] and is discontinuous at a, then
b
Z
a
Z
f (x)dx = lim+
t→a
b
f (x)dx
t
provides this limit is a finite number.
In both cases, the improper integral is convergent if the corresponding limit is finite.
Otherwise, it is divergent.
Rc
Rb
(c) If f has a discontinuity at c, where a < c < 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
Problem 2. Evaluate each integral. Is it convergent or divergent?
Z 3
dx
(a)
2
0 x − 6x + 5
Z 3
1
(b)
dx
4
−2 x
Z 0 1/x
e
(c)
dx
3
−1 x
3
Recognizing improper integrals
In the previous two sections, we were introduced to two types of improper integrals.
Then are the only two types. It is extremely important, when confronted with
any integral, to be able to recognize whether or not the integral is improper.
In fact, this is the first thing to check for any integral from now on. ToRsee what
1
can happen if an integral is not recognized as improper, consider the integral −2 x12 dx.
Since x12 is discontinuous as x = 0, this is an improper integral. To evaluate it, we write
Z
1
−2
1
dx =
x2
Z
0
−2
1
dx +
x2
Z
0
1
1
dx.
x2
D. Amos
Class Activity #5
4
We now evaluate the first term on the right hand side of the above equation:
Z 0
Z t
1
1
dx = lim−
dx
2
2
t→0
−2 x
−2 x
t
1 = lim− − t→0
x
−2
1 1
= lim−
−
t→0
2 t
= −∞.
Since this term diverges, the original integral also diverges (see part (c) in Definition 2).
However, if we had not recognized the discontinuity at x = 0, we would have been misled
into performing the following incorrect computation:
1
Z 1
1
1
1 3
dx = − = −1 − = − .
2
x
2
2
−2 x
−2
Recognizing Type I improper integrals is easy, as there will be an ∞ or −∞ in the
limits of integration. Recognizing a Type II improper integral is a little more subtle. One
must look for asymptotes of the integrand, or any jump or removable discontinuities.
Problem 3. Which of the following integrals are improper. Why?
Z 2
1
dx
(a)
1 2x − 1
Z 1
1
(b)
dx
0 2x − 1
Z ∞
sin x
(c)
dx
2
−∞ 1 + x
Z 2
(d)
ln(x − 1)dx
1