Section 8.4, Improper Integrals: Infinite Integrands
Homework: 8.4 #1–33 odds
Z
2
Consider the definite integral
−1
Z
2
−1
dx
. Can we calculate it as
x3
1
dx
1 −2 1
3
1
= − x = − · 2−2 + · 1 = ?
x3
2
2
2
8
−2
No, we can’t calculate the definite integral this way! Part of the “fine print” in the Second Fundamental Theorem of Calculus is that the integrand must be a continuous function on the interval of
integration (here, [−1, 2]). However, x13 is discontinuous at x = 0.
Definition of Definite Integrals with Infinite Integrands
Let f be continuous on the interval [a, b) and let lim− |f (x)| = ∞. Then,
x→b
Z
b
t→b
a
t
Z
f (x) dx = lim−
f (x) dx
a
provided that this limit exists and is finite. In this case, we say that the integral converges. Otherwise, we say that the integral diverges.
An analogous statement holds if |f (x)| → ∞ at the lower end of the interval.
Examples
Z
1.
3
dx
(x
−
1)4/3
1
The integrand has a vertical asymptote at x = 1, so:
Z 3
Z 3
dx
= lim+
(x − 1)−4/3 dx
4/3
t→1
1 (x − 1)
t
3
= lim − 3(x − 1)−1/3 t→1+
t
= lim+ − 3 · 2−1/3 + 3(t − 1)−1/3 ,
t→1
which is infinite, so this integral diverges.
Z 4
dx
√
2.
4−x
0
The integrand has a vertical asymptote at x = 4, so
Z 4
Z t
dx
√
= lim−
(4 − x)−1/2 dx
4 − x t→4 0
0
t
= lim − 2(4 − x)1/2 t→4−
0
= lim− − 2(t − x)1/2 + 2 · 41/2
t→4
=4
Z
3.
0
1
x−p dx if p > 0.
If p = 1,
Z 1
x−1 dx = lim
t→0+
0
1
Z
x−1 dx
t
1
= lim+ ln |x| = lim+ (ln 1 − ln) = ∞
t→0
t→0
t
so the integral diverges.
If p 6= 1,
Z 1
0
x−p dx = lim+
Z
t→0
1
x−p dx
t
1
1
1−p = lim+
x t→0 1 − p
t
1
t1−p
= lim
−
1−p 1−p
t→0+
(
1
if 0 < p < 1
= 1−p
∞
if p > 1
This means that
(
Z 1
1
−p
x dx = 1−p
∞
0
if 0 < p < 1
if p ≥ 1
If f is continuous on [a, b] except at a number c such that a < c < b, and suppose that lim |f (x)| = ∞.
x→c
Then define
Z b
Z c
Z b
f (x) dx =
f (x) dx +
f (x) dx
a
a
c
provided that both integrals on the right converge. Otherwise, we say that
Examples
Z 2
dx
1.
3
−1 x
The integrand has a vertical asymptote at x = 0.
Z 2
Z 0
Z 2
dx
dx
dx
=
+
3
3
3
x
x
−1
−1
0 x
By the last example, this diverges.
Z 1
5
2.
dx
(x
+
2)3/5
−3
This integrand has a vertical asymptote at x = −2
Z 1
Z 1
Z −2
5
−3/5
5(x
+
2)
dx
+
5(x + 2)−3/5 dx
dx
=
3/5
(x
+
2)
−3
−3
−2
−2
1
25
25
2/5 2/5 =
(x + 2) + (x + 2) 2
2
−3
25 2/5
25
=
0 − (−1)2/5 +
3 −0
2
2
25(32/5 − 1)
=
2
−2
Rb
a
f (x) dx diverges.