Improper Integrals
We use limits to evaluate definite integrals over an “infinite” interval. Let’s consider a motivating example.
Motivating Example: Evaluate and interpret A(t) =
Rt
1
0 1+x2
dx.
Intuition: The number A(t) represents the area enclosed between the curve y =
x = 0 to x = t. See the image below.
Z
1
1+x2
and the x-axis from
t
1
dx
2
0 1+x
t
= arctan(x)
A(t) =
0
= arctan(t)
Observation: Since lim A(t) = lim arctan(t) =
t→∞
the curve y =
1
1+x2
t→∞
π
, it follows that
2
π
2
units2 of area are enclosed between
and the x-axis when x ≥ 0.
Improper Integral of Type I:
Rt
• If a f (x) dx exists for every t ≥ a, then
∞
Z
t
Z
f (x) dx = lim
• If
Rb
t
f (x) dx.
t→∞
a
a
f (x) dx exists for every t ≤ b, then
Z
b
b
Z
f (x) dx = lim
f (x) dx.
t→−∞
−∞
t
We call the improper integral convergent if the limit exists and divergent if the limit does not exist.
Observation: If both
R∞
a
f (x) dx and
Z
Ra
−∞
f (x) dx converge, then
∞
Z
f (x) dx =
−∞
Z
Example 1. Evaluate
∞
∞
Z
a
f (x) dx +
f (x) dx.
−∞
a
2
xe−x dx.
0
Intuition: This is an improper integral of Type I. We MUST rewrite as a limit.
Z
0
∞
2
xe−x dx = lim
t→∞
Z
t
2
xe−x dx
0
Let u = −x2 so that du = −2xdx and − 12 du = xdx. As this is a definite integral, we also need to change
the limits of integration. Using u = −x2 , we have x = 0 → u = 0 and x = t → u = −t2 . Therefore,
1
Calculus II Resources
Integration Techniques
∞
Z
xe
−x2
t
Z
t→∞
0
=−
2
xe−x dx
dx = lim
0
1
lim
2 t→∞
Z
−t2
eu du
0
−t2
1
u
= − lim e 2 t→∞ 0
h 2
i
1
= − lim e−t − 1
2 t→∞
1
= − [0 − 1]
2
1
= .
2
Observation: When evaluating an improper integral, the first step is to rewrite as a limit.
See solution video
Z
Example 2. Is
1
∞
1
dx convergent or divergent?
x
∞
Z
1
Z
t
1
dx
x
1
t
= lim ln(x)
t→∞
1
dx = lim
t→∞
x
1
= lim ln(t)
t→∞
=∞
Therefore,
R∞
1
1
x
dx is divergent.
Z
Observation: One can show that
1
∞
1
dx converges for p > 1 and diverges for p ≤ 1.
xp
See solution video
Z
∞
x
dx is convergent or divergent.
3+1
x
1
R∞
R∞
Intuition: If we can find a function f (x) with f (x) ≥ x3x+1 and 1 f (x) dx convergent, then 1
must be convergent as well. See the below image.
Example 3. Without evaluating, determine if
Since
R∞
1
1
x2
dx is convergent (i.e. f (x) =
1
x2
R∞
1
x
x3 +1
dx
encloses a finite amount of area) and
0<
it follows that
x
x3 +1
x
x
1
≤ 3 = 2
x3 + 1
x
x
dx is convergent.
Observations:
• Even though we know the improper integral is convergent, we do not know its value. To determine the
value you need to evaluate the integral.
2
Calculus II Resources
Integration Techniques
• We have applied a property known as the Comparison Test.
See solution video
Comparison Test: Suppose f and g are continuous and 0 ≤ g(x) ≤ f (x) for x ≥ a.
R∞
R∞
• If a f (x) dx converges, then a g(x) dx converges.
R∞
R∞
• If a g(x) dx diverges, then a f (x) dx diverges.
Z
1
Example 4. Evaluate
1/4
1
dx
4x − 1
1
is not continuous at x = 1/4, we cannot apply the FTOC. This is an
Observation: Since f (x) = 4x−1
Improper Integral of Type II. We need to rewrite as a limit.
1
Z
1/4
1
dx = lim
4x − 1
t→1/4+
Z
1
1
dx
t 4x − 1
1
1
= lim
ln |4x − 1|
+
t→1/4 4
t
1
1
= lim +
ln |3| − ln |4t − 1|
4
4
t→1/4
1
1
= ln |3| −
lim ln |4t − 1|
4
4 t→1/4+
=∞
The improper integral
Observations:
•
R1
1
1/4 4x−1
dx is divergent.
lim ln |4t − 1| = −∞ because 4t − 1 → 0 and lim ln(t) = −∞.
t→0
t→1/4+
Rb
• If f (x) is NOT continuous at x = a, then we must view
b
Z
a
a
b
Z
f (x) dx = lim+
t→a
f (x) dx as the limit
f (x) dx.
t
See solution video
Summary:
• We view
R∞
a
f (x) dx as the limit
Z
∞
Z
f (x) dx = lim
t→∞
a
• If both
R∞
a
f (x) dx and
Ra
−∞
∞
Z
∞
f (x) dx =
−∞
R∞
1
1
xp
f (x) dx.
a
f (x) dx converge, then
Z
• The improper integral
t
Z
a
f (x) dx +
f (x) dx.
−∞
a
dx converges for p > 1 and diverges for p ≤ 1.
3
Calculus II Resources
Integration Techniques
Rb
• If f (x) is NOT continuous at x = a, then a f (x) dx is an improper integral of type II and must be
viewed as the limit
Z b
Z b
f (x) dx = lim
f (x) dx.
t→a+
a
t
See Improper Integrals overview video
Practice Problems:
Z
∞
(1) Evaluate
e
1
dx.
x(ln(x))3
Z
(2) Determine if
1
Z
0
(3) Evaluate
−1
∞
ex
q
x2 −
dx converges or diverges.
1
2
e1/x
dx.
x2
See Solutions
4