Section 8.3, Improper Integrals: Infinite Limits of Integration
Homework: 8.3 #1–19 odds, 25, 27, 31
In this section and the next, we will look at improper integrals, which are definite integrals where
either one or both of the limits of integration are infinite, or the integrand is infinite. In this section,
we will focus on definite integrals with infinite limits.
Definition of Improper Integrals with Infinite Limits
b
Z
b
Z
f (x) dx
f (x) dx = lim
a→−∞
−∞
Z ∞
Z
a
b
f (x) dx = lim
b→∞
a
f (x) dx
a
If the limit exists and have finite values, we say that the improper integral converges. Otherwise,
we say that the integral diverges.
Examples
Evaluate each of the following integrals or show that it diverges:
Z 2
1.
ex dx
−∞
Z
2
Z
x
e dx = lim
−∞
a→−∞
2
ex dx
a
2
= lim ex a→−∞
a
= lim
a→−∞
Z
2.
1
∞
e2 − ea ) = e2
dx
√
πx
Z
1
∞
dx
√
= lim
πx b→∞
Z
b
(πx)−1/2 dx
1
b
2
(πx)1/2 b→∞ π
1
2
1/2
(bπ) − π 1/2
= lim
b→∞ π
= ∞,
= lim
so this integral diverges.
Z
∞
3.
1
x
dx
(1 + x2 )2
Z
∞
1
x
dx = lim
b→∞
(1 + x2 )2
b
Z
x(1 + x2 )−2 dx
1
b
1
2 −1 = lim − (1 + x )
b→∞
2
1
1
1 1
2 −1
= lim − (1 + b ) + ·
b→∞
2
2 2
1
=
4
Z
4.
∞
1
dx, where p is a real number
xp
We will consider every possible value of p. First, for p = 1:
1
Z
∞
1
1
dx = lim
b→∞
x
Z
b
1
dx
x
1
b
= lim ln x
b→∞
1
= lim ln b = ∞,
b→∞
so the integral diverges when p = 1. Now, for p 6= 1, the power rule applies:
Z
∞
1
1
dx = lim
b→∞
xp
Z
b
x−p dx
1
b
1
1−p = lim
x b→∞ 1 − p
1
1 1−p
1
b
−
= lim
b→∞ 1 − p
1−p
(
∞
if p < 1
=
1
if
p>1
p−1
This means that for p ≤ 1, the integral diverges, and for p > 1, it equals
Definition
of Improper
Integrals with Two
R∞
R0
R ∞Infinite Limits
If 0 f (x) dx and −∞ f (x) dx converge, then −∞ f (x) dx converges and
Z
∞
Z
0
f (x) dx =
−∞
Otherwise,
Z
f (x) dx +
−∞
R∞
−∞
∞
f (x) dx
0
f (x) dx diverges.
Example
Z ∞
dx
2 + 16
x
−∞
Z
∞
−∞
Z b
dx
dx
+ lim
2
2
b→∞
a x + 16
0 x + 16
0
b
1
1
−1 x −1 x + lim tan
= lim
tan
a→−∞ 4
4 a b→∞ 4
4 0
1
1
−1 a
−1 b
= lim
0 − tan
+ lim
tan
−0
a→−∞
b→∞ 4
4
4
4
1 π 1 π
π
= · + · =
4 2
4 2
4
dx
= lim
2
x + 16 a→−∞
Z
0
1
p−1
1
Probability Density Functions
A probabilityRdensity function (PDF) is a non-negative function f (x), defined on the real num∞
bers such that −∞ f (x) dx = 1.
Rb
The probability that a random variable X is between a and b is a f (x) dx.
The mean of the random variable with the PDF of f (x) is
Z ∞
xf (x) dx.
µ=
−∞
This is a weighted average of all the value that the random variable can take.
The variance is
Z ∞
2
σ =
(x − µ2 )f (x) dx.
−∞
This describes how spread out the distribution is.
Example
A continuous random variable X has an exponential distribution with parameter θ for θ > 0 if
its PDF is:
(
1 −x/θ
e
if x > 0
f (x) = θ
0
if x < 0
1. Show that this is a valid PDF.
Z ∞
Z 0
Z
f (x) dx =
0 dx +
−∞
∞
1 −x/θ
e
dx
θ
−∞
0
b
−x/θ = 0 + lim (−e
)
b→∞
= lim
b→∞
−e
−b/θ
+e
0
0
=1
2. Find the mean µ and variance σ 2 of the distribution.
Z ∞
x −x/θ
µ=
e
dx
θ
0
∞ Z ∞
= −xe−x/θ +
e−x/θ dx
0
0
∞
−x/θ = 0 + −θe
0
= θ,
where we used integration by parts.
Z ∞
1
σ2 =
(x − θ)2 e−x/θ dx
θ
0
∞ Z ∞
= −(x − θ)2 e−x/θ +
2(x − θ)e−x/θ dx
0
0
∞ Z ∞
2
= θ + (−2)θ(x − θ)e−x/θ +
2θe−x/θ dx
0
∞ 0
= θ2 − 2θ2 − 2θ2 e−x/θ 0
= θ2 − 2θ2 + 2θ2 = θ2
3. If θ = 2, find the probability that X is less than ln 4.
Z
ln 4
1 −x/2
e
dx
2
0
ln 4
−x/2 = −e
= −e−(ln 4)/2 + 1
P (X < ln 4) =
0
1
=
2