Joe Foster
Improper Integrals
Definition 1: Integrals with infinite limits of integration are called improper integrals of Type I.
ˆ ∞ ˆ b
1. If f (x) is continuous on [a, ∞), then f (x) dx = lim f (x) dx.
a b→∞ a
ˆ b ˆ b
2. If f (x) is continuous on (−∞, b], then f (x) dx = lim f (x) dx.
−∞ a→∞ −a
ˆ ∞ ˆ c ˆ ∞
3. If f (x) is continuous on (−∞, ∞), then f (x) dx = f (x) dx + f (x) dx, where c is any real number.
−∞ −∞ c
Definition 2: Integrals of functions that become infinite at a point within the interval of integration are called improper
integrals of Type II.
ˆ b ˆ a
1. If f (x) is continuous on (a, b] and discontinuous at a, then f (x) dx = lim+ f (x) dx.
a c→a c
ˆ b ˆ c
2. If f (x) is continuous on [a, b) and discontinuous at b, then f (x) dx = lim f (x) dx.
a c→b− a
ˆ b ˆ c ˆ b
3. If f (x) is discontinuous at c ∈ (a, b) and continuous elsewhere, then f (x) dx = f (x) dx + f (x) dx.
a a c
In each case, if the limit is finite we sat that the improper integral converges and that the limit is the value of the improper
integral. If the limit fails to exist, the improper integral diverges.
Example 1:Evaluate
∞
ln(x)
ˆ
dx.
1 x2
b
b b
ln(x) ln(x) 1
ˆ ˆ
dx = − − − dx
1 x2 x 1 x2
1
b
1 ln(x) 1
u = ln(x) dv = dx =− −
x2 x x
1 1 1
du = dx v=− ln(b) 1
ln(1) 1
x x =− − − − −
b b 1 1
ln(b) 1
=− − +1
b b
Now we take a limit,
∞ b
ln(x) ln(x) ln(b) 1 ln(b) 1/b
ˆ ˆ
L’H
dx = lim dx = lim − − + 1 = lim − − 0 + 1 = lim − +1=0+1= 1
1 x2 b→∞ 1 x2 b→∞ b b b→∞ b b→∞ 1
L’Hôpital’s Rule Suppose that f (a) = g(a) = 0, that f (x) and g(x) are differentiable on an open interval I containing
a and that g ′ (x) 6= 0 on I if x 6= a. Then
f (x) f ′ (x)
lim = lim ′ ,
x→a g(x) x→a g (x)
assuming that the limit on the left and right both exist.
Page 1 of 4
Improper Integrals
Definition 1: Integrals with infinite limits of integration are called improper integrals of Type I.
ˆ ∞ ˆ b
1. If f (x) is continuous on [a, ∞), then f (x) dx = lim f (x) dx.
a b→∞ a
ˆ b ˆ b
2. If f (x) is continuous on (−∞, b], then f (x) dx = lim f (x) dx.
−∞ a→∞ −a
ˆ ∞ ˆ c ˆ ∞
3. If f (x) is continuous on (−∞, ∞), then f (x) dx = f (x) dx + f (x) dx, where c is any real number.
−∞ −∞ c
Definition 2: Integrals of functions that become infinite at a point within the interval of integration are called improper
integrals of Type II.
ˆ b ˆ a
1. If f (x) is continuous on (a, b] and discontinuous at a, then f (x) dx = lim+ f (x) dx.
a c→a c
ˆ b ˆ c
2. If f (x) is continuous on [a, b) and discontinuous at b, then f (x) dx = lim f (x) dx.
a c→b− a
ˆ b ˆ c ˆ b
3. If f (x) is discontinuous at c ∈ (a, b) and continuous elsewhere, then f (x) dx = f (x) dx + f (x) dx.
a a c
In each case, if the limit is finite we sat that the improper integral converges and that the limit is the value of the improper
integral. If the limit fails to exist, the improper integral diverges.
Example 1:Evaluate
∞
ln(x)
ˆ
dx.
1 x2
b
b b
ln(x) ln(x) 1
ˆ ˆ
dx = − − − dx
1 x2 x 1 x2
1
b
1 ln(x) 1
u = ln(x) dv = dx =− −
x2 x x
1 1 1
du = dx v=− ln(b) 1
ln(1) 1
x x =− − − − −
b b 1 1
ln(b) 1
=− − +1
b b
Now we take a limit,
∞ b
ln(x) ln(x) ln(b) 1 ln(b) 1/b
ˆ ˆ
L’H
dx = lim dx = lim − − + 1 = lim − − 0 + 1 = lim − +1=0+1= 1
1 x2 b→∞ 1 x2 b→∞ b b b→∞ b b→∞ 1
L’Hôpital’s Rule Suppose that f (a) = g(a) = 0, that f (x) and g(x) are differentiable on an open interval I containing
a and that g ′ (x) 6= 0 on I if x 6= a. Then
f (x) f ′ (x)
lim = lim ′ ,
x→a g(x) x→a g (x)
assuming that the limit on the left and right both exist.
Page 1 of 4
