Section 13.7, Improper Integrals
When we have looked at definite integrals so far, there have been distinct upper and lower bounds
for the interval. Here, we will look at what happens when one of the bounds is infinite. If a and b
are real numbers, we can find definite integrals with infinite bounds as follows:
Z ∞
Z b
f (x) dx = lim
f (x) dx
Z
b→∞
a
b
a
Z
b
f (x) dx
f (x) dx = lim
a→−∞
−∞
a
If the limit is a unique finite number, we say that the integral converges. Otherwise, we say that
the integral diverges.
R∞
Ra
R∞
Note: −∞ f (x) dx = −∞ f (x) dx + a f (x) dx for any real number a, and converges exactly when
both integrals in the sum converge.
Examples
R −1
1. −∞ x12 dx
Z −1
−∞
2.
R∞
1
−1
Z
1
dx = lim
a→−∞
x2
x
−2
−x
dx = lim
a→−∞
a
−1 = lim (1 + a−1 ) = 1
a→−∞
−1 a
2
xe−x dx
Z ∞
Z
−x2
xe
dx = lim
b→∞
1
b
xe
−x2
dx = lim
b→∞
1
b 1 −b2 1 −1
1 −x2 lim − e
+ e
− e
= b→∞
2
2
2
1
1
=
2e
3.
R∞
2
1
e3x
dx
Z
∞
2
4.
R∞
1
√1
x
dx
Z
∞
1
b
1
dx = lim
b→∞
e3x
Z
1
√ dx = lim
b→∞
x
Z
e
2
1
b
−3x
b
1 −3x 1
dx = lim
e
= 3e6
b→∞ −3
2
b
x−1/2 dx = lim 2x1/2 = lim (2b1/2 − 2)
b→∞
b→∞
1
Since the first term goes to infinity, the limit is infinite, so this integral diverges.
R∞
5. −∞ ex dx
0
b
Z ∞
Z 0
Z ∞
x
x
x
x
x
e dx = lim e + lim e = 1 + a term that diverges
e dx =
e dx +
−∞
−∞
0
a→−∞
a
b→∞
0
Since the second integral in the sum diverges, the original integral diverges as well.
6. Find the area under the curve f (x) = x21√x and to the right of x = 1.
∞
Z ∞
2 −3/2 2
−5/2
x
dx = − x
=3
3
1
1