Improper Integrals
Rb
Our definition of the definite integral a f (x)dx requires that [a, b] is a finite interval and that f has
no infinite discontinuities on [a, b]. When the interval is infinite or f has an infinite discontinuity in
[a, b], we are dealing with an improper integral.
Infinite Intervals
1. We define
∞
Z t
f (x)dx = lim
f (x)dx
t→∞ a
a
Rt
if the limit exists. (Implicit Assumption: a f (x)dx exists for every number t ≥ a.)
Z
2. We define
b
Z
b
Z
f (x)dx = lim
t→−∞ t
−∞
if the limit exists. (Implicit Assumption:
3. We define
Z ∞
Z
f (x)dx =
−∞
c
Z
f (x)dx +
−∞
Rb
t
f (x)dx
f (x)dx exists for every number t ≤ b.)
∞
Z
f (x)dx = lim
c
if both limits exist for some number c. (Implicit
all real numbers t.)
c
Z
f (x)dx + lim
t→−∞ −t
Rt
Assumption: c f (x)dx
=
t
f (x)dx
t→∞ c
Rc
− t f (x)dx
exists for
Remember: We say a limit exists if it is a finite number.
An improper integral is called convergent if the corresponding limit exists; it is called divergent if
the limit does not exist.
Rc
Remark. You can use any convenient number c in Case 3. This is because if −∞ f (x)dx and
R∞
R c0
R∞
0
c f (x)dx both exist for some number c, then, for any other number c , −∞ f (x)dx and c0 f (x)dx
both exist and
Z c
Z ∞
Z c0
Z ∞
f (x)dx +
f (x)dx =
f (x)dx +
f (x)dx.
−∞
c
−∞
∞
Z
c0
Remark. We did NOT define
Z
f (x) dx = lim
−∞
t
t→∞ −t
f (x) dx
Rt
R∞
When it exists, limt→∞ −t f (x) dx is called the principal value of −∞ f (x)dx. It is possible for
Rt
R∞
limt→∞ −t f (x)dx to exist when −∞ f (x) dx does not (for example, take f (x) = x). However,
R∞
Rc
Rt
if −∞ f (x)dx exists (that is, if limt→−∞ −t f (x) dx and limt→∞ c f (x) dx both exist for some
R∞
Rt
number c), then −∞ f (x) dx = limt→∞ −t f (x) dx. To reiterate, you CANNOT conclude that
R∞
Rt
−∞ f (x) dx exists by simply checking that limt→∞ −t f (x)dx exists.
1
2
Z
∞
1
dx
x
Example 1.
1
Z
t
t
1
dx = ln |x| = ln |t| − ln |1| = ln t
x
1
1
Since we will be letting t → ∞, we assume t > 0, so that ln |t| = ln t.
Then
Z
lim
t→∞ 1
∞
1
dx diverges.
x
Z
∞
Z
Therefore
1
t
1
dx = lim ln t = ∞.
t→∞
x
1
dx
xp
Example 2.
1
The case p = 1 is the previous example, so we assume p 6= 1.
Z t
t
1
1
1
1
−p+1 lim
dx
=
lim
x
(t−p+1 − 1) =
( lim t−p+1 − 1)
= lim
p
t→∞ 1 x
t→∞ −p + 1
t→∞ −p + 1
−p + 1 t→∞
1
If p > 1, then −p + 1 < 0, and so limt→∞ t−p+1 = 0.
If p < 1, then −p + 1 < 0, and so limt→∞ t−p+1 = ∞.
Therefore
Z

1

1
1
−
if p > 1
−p+1
dx
=
(
lim
t
−
1)
=
−p
+1
p

x
−p + 1 t→∞
∞ if p < 1
t
lim
t→∞ 1
p-Integral.
Z
∞
1
Z
∞
1
dx is convergent if p > 1 and divergent if p ≤ 1.
xp
e−x dx
Example 3.
−∞
Z
lim
t
t→∞ 0
Z 0
lim
t→−∞ t
Z
∞
Therefore
−∞
t
e−x dx = lim −e−x = lim (−e−t + 1) = −0 + 1
t→∞
t→∞
0
0
e−x dx = lim −e−x = lim (−1 + e−t ) = ∞
e−x dx diverges.
t→−∞
t
t→−∞
3
Z
∞
Example 4.
2
xe−x dx We first compute
−∞
Z
t
2
xe−x dx = −
0
1
2
Z
−t2
eu du
0
u = −x2
du = −2x dx
x=0⇒u=0
x = t ⇒ u = −t2
1 −t2
= − eu 2 0
1
2
= − (e−t − 1)
2
Therefore
Z
lim
t→∞ 0
t
2
xe−x dx = −
1
1
1
2
lim (e−t − 1) = − (0 − 1) = .
2 t→∞
2
2
With the same substitution, we find
Z
Z 0
1 u 0
1
1 0 u
2
−x2
e du = − e 2 = − (1 − e−t )
xe
dx = −
2
2
2
−t
2
−t
t
and so
Z 0
1
1
1
2
2
lim
xe−x dx = − lim (1 − e−t ) = − (1 − 0) = − .
t→−∞ t
2 t→−∞
2
2
Z ∞
2
Therefore
xe−x dx converges and
−∞
Z ∞
2
xe−x dx = lim
t→∞ 0
−∞
Z
Z
t
2
xe−x dx + lim
Z
t→−∞ t
0
2
xe−x dx =
1 1
− =0
2 2
∞
sin θ dθ
Example 5.
0
Z
t
t
sin θ dθ = lim − cos θ = lim (− cos t + cos(0)) = −( lim cos t) + 1
t→∞ 0
t→∞
t→∞
t→∞
0
Z ∞
The limit does not exist. Therefore
sin θ dθ diverges.
lim
0
4
Infinite Discontinuities
Remember: f has an infinite discontinuity at c if limx→c+ f (x) = ±∞ or limx→c− f (x) = ±∞.
Assume −∞ < a < b < ∞.
1. If limx→b− f (x) = ±∞, we define
b
Z
Z
t
f (x)dx
= lim
t→b−
a
if this limit exists. (Implicit Assumption:
Rt
a
a
f (x)dx exists for every number t in [a, b).)
2. If limx→a+ f (x) = ±∞, then we define
Z b
Z b
f (x)dx = lim
f (x)dx
t→a+
a
if this limit exists. (Implicit Assumption:
Rb
t
t
f (x)dx exists for every number t in (a, b].)
3. If f has a infinite discontinuity at c, where a < c < b, then we define
Z b
Z c
Z b
Z t
Z b
f (x)dx =
f (x)dx +
f (x)dx = lim
f (x)dx + lim
f (x)dx
a
a
t→c−
c
if both limits exist. (Implicit Assumption:
Rb
t f (x)dx exists for every number t in (c, b].)
Rt
a
a
t→c+
f (x)dx exists for every number t in [a, c) and
Remark. The formulas above are actually valid for any type of discontinuity.
Z 5
1
√
Example 6.
dx
x−2
2
1
= ∞.
x−2
Z 5
1
1
√
√
dx = lim
dx
+
t→2
x−2
x−2
t
Z 5
= lim
(x − 2)−1/2 dx
t→2+ t
5
1/2 = lim 2(x − 2) t
t→2+
√
√
= lim 2( 3 − t − 2)
t→2+
√
=2 3
The integral is improper because lim √
x→2+
Z
2
So the integral is convergent.
5
t
5
Z
Example 7.
0
3
1
dx
x−1
1
The integral is improper because f (x) = x−1
has an infinite discontinuity in the interval [0, 3] at 1.
We write
Z 3
Z 1
Z 3
Z t
Z 3
1
1
1
1
1
dx =
dx +
dx = lim
dx + lim
dx.
−
+
t→1
t→1
0 x−1
0 x−1
1 x−1
0 x−1
t x−1
We have
Z t
t
1
lim
dx = lim ln |x − 1|
0
t→1− 0 x − 1
t→1−
= lim (ln |t − 1| − ln | − 1|)
t→1−
= lim ln(1 − t)
t→1−
= −∞
Z 3
1
+
−
dx is divergent. (We don’t need to evaluate
because 1 − t → 0 as t → 1 . Therefore
0 x−1
Z 3
1
dx.)
lim
+
t→1
t x−1
Z 3
1
dx is an improper integral, we might have incorWRONG WAY: If we had not noticed
0 x−1
rectly calculated that
Z 3
3
1
dx = ln |x − 1| = ln 2 − ln 1 = ln 2.
x
−
1
0
0
Remark. Whenever you work with an integral
definite integral or an improper integral.
Z π/2
Example 8.
sec x dx
Rb
a
f (x)dx, you must check if it is an ordinary
0
The integral is improper because
Z
lim
t→(π/2)−
lim
1
= ∞. We have
cos x
t
lim ln | sec x + tan x|
−
x→(π/2)−
t
sec x dx =
sec x =
lim
x→(π/2)−
0
t→(π/2)
0
=
=
=
lim
(ln | sec t + tan t| − ln | sec 0 + tan 0|)
lim
(ln | sec t + tan t| − ln |1 + 0|)
lim
ln | sec t + tan t|
t→(π/2)−
t→(π/2)−
t→(π/2)−
=∞
because sec t → ∞ and tan t → ∞ as t → (π/2)− and ln x → ∞ as x → ∞.
Z π/2
Therefore
sec x dx diverges.
0