1) The document discusses two types of improper integrals: Type I integrals over infinite intervals, and Type II integrals with a discontinuous integrand. 2) For a Type I integral over an infinite interval, the integral is defined as the limit of definite integrals over finite intervals as the interval extends to infinity. 3) For a Type II integral with a discontinuity, the integral is defined as the limit of definite integrals over intervals approaching the point of discontinuity.

1. Chapter 7: Techniques of Integration
Section 7.8: Improper Integrals
Alea Wittig
SUNY Albany

2. Outline
Type I: Infinite Intervals
Type II: Discontinuous Integrands
A Comparison Test for Integrals

3. Improper Integral
▶ When defining the definite integral
Z b
a
f (x)dx
we assume function f (x) defined on a a finite interval [a, b]
has no infinite discontinuity in the interval.
▶ In this section we extend the concept of a definite integral to
I. the case where the interval is infinite.
eg
Z ∞
a
f (x)dx,
Z b
−∞
f (x)dx,
Z ∞
−∞
f (x)dx
II. the case where f has an infinite discontinuity in [a, b].
eg
Z 1
−1
1
x
dx,
Z 1
0
ln xdx,
Z π/2
0
tan xdx
▶ In either case, the integral is called an improper integral.

4. Type I: Infinite Intervals

5. Example 1
▶ Consider the unbounded region S that lies under the curve
y =
1
x2
, above the x-axis, and to the right of x = 1.
▶ The area of the part of S to the left of t (where t ≥ 1) is
A(t) =
Z t
1
1
x2
dx = −
1
x

8. t
1
= 1 −
1
t

9. Example 1
▶ As t increases, A(t) appears to approach 1.
t 1 2 3 4 5 6 . . .
A(t) 0
1
2
2
3
3
4
4
5
5
6
. . .
▶ Taking the limit as t → ∞ we have
lim
t→∞
1 −
1
t
= lim
t→∞
1 − lim
t→∞
1
t
= 1 − 0
= 1
▶ So we say that the area of the infinite region S is 1
Z ∞
1
1
x2
dx = lim
t→∞
Z t
1
1
x2
dx = 1

10. Type I: Infinite Intervals Definition
(a) If
Z t
a
f (x)dx exists for every number t ≥ a, then
Z ∞
a
f (x)dx = lim
t→∞
Z t
a
f (x)dx
provided that this limit exists (as a finite number).
(b) If
Z b
t
f (x)dx exists for every number t ≤ b, then
Z b
−∞
f (x)dx = lim
t→−∞
Z b
t
f (t)dt
provided this limit exists (as a finite number).

11. Type I: Infinite Intervals Definition
The improper integrals
Z ∞
a
f (x)dx and
Z b
−∞
f (x)dx
are called convergent if the corresponding limit exists and
divergent if the corresponding limit does not exist.
(c) If both
Z ∞
a
f (x)dx and
Z a
−∞
f (x)dx are convergent then we
define
Z ∞
−∞
f (x)dx =
Z a
−∞
f (x)dx +
Z ∞
a
f (x)dx
for any real number a.

12. Example 2
1/5
Determine whether the integral
Z ∞
1
1
x
dx
is convergent or divergent. How does this compare to example 1?

13. Example 2
2/5
▶ Step 1 is to compute the definite integral
Z t
1
1
x
dx where t is
a real number greater than 1.
Z t
1
1
x
dx = ln |x|

16. t
1
= ln |t| − ln 1
= ln t (since t ≥ 1, |t| = t)
▶ Step 2 is to take the limit of the integral we computed:
lim
t→∞
Z t
1
1
x
dx = lim
t→∞
ln t

17. Example 2
3/5
▶ Intuitively, by looking at the graph of ln x, we probably know
lim
t→∞
ln t = ∞
▶ lim
x→∞
f (x) = ∞ means that for any M > 0, there is a real
value d such that f (x) > M for all x > d.
▶ M should be thought of as a really large real number . . .
maybe something like 1010101010
.
▶ No matter how large M is, we can always find a value d such
that for all x > d, f (x) is even greater than M.
▶ To prove that lim
t→∞
ln t = ∞, let M be any positive number.
ln t > M for all t > eM
, so just take d = eM
✓

18. Example 2
4/5
▶ So since Z ∞
1
1
x
dx = lim
t→∞
ln t = ∞
the integral is divergent.
▶ Compare this to the integral
Z ∞
1
1
x2
dx from example 1, which
we found to be convergent.
▶ What do you think accounts for the fact that one of these
integrals converges while the other does not?

19. Example 2
5/5
▶ Notice that 1
x2 decreases more rapidly than 1
x .
x 1 2 3 4 5 6 . . .
1
x
1
1
2
1
3
1
4
1
5
1
6
. . .
1
x2
1
1
4
1
9
1
16
1
25
1
49
. . .
▶ So the area of the region under 1
x2 is smaller than the area of
the region under 1
x .
▶ In a minute we will find the values of p for which the integral
Z ∞
1
1
xp
dx converges and diverges.
▶ Do you have any guesses?

20. Example 3
1/6
Z 0
−∞
xex
dx
▶ Step 1: Compute the integral
Z 0
t
xex
dx:
▶ This looks like an integration by parts problem:
Z
udv = uv −
Z
vdu
u = x dv = ex
dx
du = dx v = ex
Z 0
t
xex
dx = xex

23. 0
t
−
Z 0
t
ex
dx
= (0e0
− tet
) − ex

25. 0
t
= −tet
− (e0
− et
)
= −tet
− 1 + et

26. Example 3
2/6
▶ Step 2: Take the limit of the integral we computed:
lim
t→−∞
Z 0
t
xex
dx = lim
t→−∞
(−tet
− 1 + et
)
= − lim
t→−∞
tet
| {z }
(a)
− lim
t→−∞
1
| {z }
(b)
+ lim
t→−∞
et
| {z }
(c)
▶ Let’s start with (b) since it’s the easiest:
(b) Since 1 is a constant, ie, not dependent on t,
lim
t→−∞
1 = 1

27. Example 3
3/6
(c) We have lim
t→−∞
et
= lim
t→−∞
1
e|t|
= lim
t→∞
1
et
▶ Intuitively or by looking at the graph of 1
et we probably know
lim
t→∞
1
et
= 0
▶ lim
x→∞
f (x) = L where L is a finite real number if for any value
ϵ > 0, there is a value d such that |f (x)−L| < ϵ for all x > d.
▶ ϵ should be though of as a very small positive number. . .
maybe something like .00000000000001
▶ No matter how small ϵ is, we can always find a value d such
that for all x > d, f (x) is within a distance ϵ of L.
▶ To prove lim
t→∞
1
et
= 0:

30. 1
et
− 0

33. =
1
et
< ϵ for all t > ln
1
ϵ
so take d = ln
1
ϵ
✓

34. Example 3
4/6
(a) lim
t→−∞
tet
is the limit of the product of t and et where
lim
t→−∞
t = −∞ and lim
t→−∞
et
= 0 from part (c).
▶ So we have an indeterminate form 0 · ∞.
▶ Use L’Hospitals Rule!
▶ L’Hospitals Rule: Suppose we have one of the following:
1. lim
x→a
f (x)
g(x)
=
0
0
2. lim
x→a
f (x)
g(x)
=
±∞
±∞
where a can be any real number, infinity, or negative infinity.
lim
x→a
f (x)
g(x)
= lim
x→a
f ′(x)
g′(x)

35. Example 3
5/6
▶ Now we have 0 · ∞ form, not 0
0 or ±∞
±∞ but we can easily
rewrite as follows:
lim
t→−∞
tet
= lim
t→−∞
t
1
et
=
−∞
∞
or alternatively, lim
t→−∞
tet
= lim
t→−∞
et
1
t
=
0
0
▶ Applying L.H to the first form we have
lim
t→−∞
t
1
et
= lim
t→−∞
1
− 1
e−t
= − lim
t→−∞
et
= − lim
t→∞
1
et
= 0
▶ Putting everything together we have
Z 0
−∞
xex
dx = −1
so the integral is convergent.

36. Example 4
1/2
Z ∞
−∞
1
1 + x2
dx
▶ Using the definition we start by determining if
Z a
−∞
1
1 + x2
dx
and
Z ∞
a
1
1 + x2
dx are both convergent for any real number a.
▶ Choose a = 0 for simplicity.
▶ The curve f (x) = 1
1+x2 is symmetric about the y-axis so
Z 0
−∞
1
1 + x2
dx =
Z ∞
0
1
1 + x2
dx

37. Example 4
2/2
▶ Step 1: For t ≤ 0 we compute
Z 0
t
1
1 + x2
dx = tan−1
x

40. 0
t
= tan−1
0 − tan−1
t
= − tan−1
t (where t ≤ 0)
▶ Step 2: lim
t→−∞
Z 0
t
1
1 + x2
dx = − lim
t→−∞
tan−1
t =
π
2
▶ This limit can be computed by using the definition of the
inverse tangent function, or by looking at the graph.
▶ Now we have
Z 0
−∞
1
1 + x2
dx =
Z ∞
0
1
1 + x2
dx =
π
2
=⇒
Z ∞
−∞
1
1 + x2
dx =
Z 0
−∞
1
1 + x2
dx +
Z ∞
0
1
1 + x2
dx = 2 ·
π
2
= π

41. Example 5
1/4
For what values of p is the integral
Z ∞
1
1
xp
convergent?
▶ We already determined that the integral diverges for p = 1.
▶ We will break the problem up into the remaining cases:
1. Case where p 1
2. Case where p 1

42. Example 5
2/4
Let p ̸= 1.
▶ Step 1: Compute the integral
Z t
1
1
xp
dx:
Z t
1
1
xp
dx =
Z t
1
x−p
dx
=
x1−p
1 − p

45. t
1
=
t1−p
1 − p
−
1
1 − p
=
1
1 − p
t1−p
− 1
▶ Step 2: Take the limit
lim
t→∞
1
1 − p
t1−p
− 1
=
1
1 − p
lim
t→∞
t1−p
− 1

46. Example 5
3/4
1. p 1
p 1 =⇒ −p −1 =⇒ 1 − p 0 =⇒
1
1 − p
lim
t→∞
(t1−p
− 1) =
1
1 − p
lim
t→∞
1
t|1−p|
− 1
=
1
1 − p
0 − 1
=
1
p − 1
So
Z ∞
1
1
xp
dx converges to 1
p−1 for all p 1.

47. Example 5
4/4
2. p 1
p 1 =⇒ −p −1 =⇒ 1 − p 0 =⇒
1
1 − p
lim
t→∞
t1−p
− 1
=
1
1 − p
lim
t→∞
t|1−p|
− 1
= ∞
So
Z ∞
1
1
xp
dx is divergent for all p 1.

48. General Rule for
Z ∞
1
1
xp
dx
Z ∞
1
1
xp
dx is convergent if p 1 and divergent if p ≤ 1
Remark:
▶ After examples 1 and 2 we may have had an intuition that
▶
Z ∞
1
1
xp
dx is divergent for any p 1 since
1
xp
1
x
for p 1
and x ≥ 1.
▶
Z ∞
1
1
xp
dx is convergent for any p 2 since 0
1
xp
1
x2
for
p 2 and x ≥ 1.
▶ We will see that that intuition in general is correct through the
comparison test for integrals in just a bit.

49. Type II: Discontinuous Integrands

50. Type II: Discontinuous Integrands
▶ Suppose f is a positive continuous function defined on a finite
interval [a, b), but f (x) has a vertical asymptote at x = b.
▶ Let S be the unbounded region under the graph of f and
above the x-axis between a and b.
▶ The area of the part of S between a and t where a t b is
A(t) =
Z t
a
f (x)dx

51. Type II: Discontinuous Integrands Definition
(a) If f is continuous on [a, b) and is discontinuous at b, then
Z b
a
f (x)dx = lim
t→b−
Z t
a
f (x)dx
provided that this limit exists (as a finite number).
(b) If f is continuous on (a, b] and is discontinuous at a, then
Z b
a
f (x)dx = lim
t→a+
Z b
t
f (x)dx
provided this limit exists (as a finite number).

52. Type II: Discontinuous Integrands Definition
The improper integral
Z b
a
f (x)dx
is called convergent if the corresponding limit exists and
divergent if the corresponding limit does not exist.
(c) If f has a discontinuity at c, where a c b, and both
R c
a f (x)dx and
R b
c f (x)dx are convergent, then we define
Z b
a
f (x)dx =
Z c
a
f (x)dx +
Z b
c
f (x)dx

53. Example 6
1/2
Z 5
2
1
√
x − 2
dx
▶ f (x) =
1
√
x − 2
has an infinite discontinuity at x = 2 which is
an endpoint of the interval [2, 5].
▶ Step 1: Compute
Z 5
t
1
√
x − 2
dx where 2 t 5
Z 5
t
1
√
x − 2
dx =
Z t−2
3
1
√
u
du u = x − 2, du = dx
=
Z t−2
3
u−1
2 du
=
u
1
2
1
2

56. 3
t−2
= 2
√
3 − 2
√
t − 2

57. Example 6
2/2
▶ Step 2: Take the limit as t goes to 2 from the right.
lim
t→2+
Z 5
t
1
√
x − 2
dx = 2
√
3 − 2 lim
t→2+
√
t − 2
= 2
√
3 − 0
= 2
√
3 =⇒
Z 5
2
1
√
x − 2
dx = 2
√
3
▶ Note that here since we are taking the limit from the right side
of 2, we are only considering values of t which are greater
than 2. This means t − 2 is positive and
√
t − 2 is well defined
for these values of t.

58. Example 7
1/2
Z π/2
0
sec xdx
▶ f (x) = sec x has infinite discontinuities whenever cos x = 0,
which occurs when x = nπ + π
2 for integers n = 0, ±1, ±2, . . ..
▶ So sec x has an infinite discontinuity at π/2 which is an
endpoint of the interval [0, π
2 ].
▶ Step 1: Compute
Z t
0
sec xdx 0 t π/2
Z t
0
sec xdx = ln | sec x + tan x|

61. t
0
= ln | sec t + tan t| − ln | sec 0 + tan 0|
= ln | sec t + tan t|

62. Example 7
2/2
▶ Step 2: Take the limit as t goes to π/2 from the left.
▶ Both sec t → ∞ and tan t → ∞ as t → π/2 from the left.
lim
t→π/2−
Z t
0
sec xdx = lim
t→π/2−
ln | sec t + tan t|
= ln
lim
t→π/2−
| sec t + tan t|
= ∞

63. Example 8
1/
Z 3
0
dx
x − 1
▶ f (x) = 1
x−1 has an infinite discontinuity at x = 1, which is
between x = 0 and x = 3.
▶ Step 1: Compute the two integrals
I1 =
Z t
0
dx
x − 1
| {z }
0t1
and I2 =
Z 3
t
dx
x − 1
| {z }
1t3

64. Example 8
2/
I1 =
Z t
0
dx
x − 1
| {z }
u=x−1, du=dx
u1=−1, u2=t−1
=
Z t−1
−1
du
u
= ln |u|

67. t−1
−1
= ln |t − 1| − ln | − 1|
= ln |t − 1| − ln 1
= ln |t − 1|
I2 =
Z 3
t
dx
x − 1
| {z }
u=x−1 du
u1=t−1, u2=2
= ln |u|

70. 2
t−1
= ln |2| − ln |t − 1|
= ln

73. 2
t − 1

77. Example 8
3/
▶ Step 2: Take the limits
▶ of I1 as t goes to 1 from the right
▶ of I2 as t goes to 1 from the left.
lim
t→1+
I1 = lim
t→1+
ln |t − 1|
= lim
v→0
ln |v|
since lim
t→1+
(t − 1) = 0
!
= −∞
lim
t→1−
I2 = lim
t→1−
ln

80. 2
t − 1

83. = lim
v→−∞
ln |v|
since lim
t→1−
2
t − 1
= −∞
!
= ∞

84. Example 8
3/
▶ It is important to note that we did not need to calculate both
I1 and I2 in this case to conclude that the integral
Z 3
0
dx
x − 1
is
divergent.
▶ Once we determined that I1 was divergent, we knew the whole
thing was divergent.
▶ If we found one was convergent and the other was divergent,
the whole thing would still be divergent.

85. Example 9
1/
Z 1
0
ln xdx
▶ f (x) = ln x has a vertical asymptote at the y-axis, x = 0
which is an endpoint of the interval [0, 1].
▶ Step 1: Compute
Z 1
t
ln xdx 0 t ≤ 1
Z 1
t
ln xdx = x ln x − x

88. 1
t
= (1 ln 1 − 1) − (t ln t − t)
= −1 − t ln t + t

89. Example 9
2/
▶ Step 2: Take the limit as t → 0 from the right.
lim
t→0+
Z 1
t
ln xdx = lim
t→0+
(−1 − t ln t + t)
= −1 − lim
t→0+
t ln t
= −1 − lim
t→0+
ln t
1
t
→
∞
∞
Use L’Hospitals
= −1 − lim
t→0+
1
t
− 1
t2
= −1 + lim
t→0+
t
= −1
This says that the area of the region between ln x and the x and y
axes is 1.

90. A Comparison Test for Integrals

91. Comparison Theorem
▶ Theorem: Suppose that f and g are continuous functions
with f (x) ≥ g(x) ≥ 0 for x ≥ a.
(a) If
Z ∞
a
f (x)dx is convergent, then
Z ∞
a
g(x)dx is convergent.
(b) If
Z ∞
a
g(x)dx is divergent, then
Z ∞
a
f (x)dx is divergent.
▶ In the case that we need to know whether an improper integral
is convergent or divergent, but it is impossible to find its exact
value, comparison test can be useful.
▶ Compare to the Squeeze Theorem regarding limits of functions
in Calc I.

92. Example 10
1/3
Show that Z ∞
0
e−x2
dx
is convergent.
▶ The integral of f (x) = e−x2
is not possible to compute in
terms of elementary functions.
▶ So we want to compare f (x) = e−x2
to a function we can
integrate, which also has a convergent integral from x = 1 to
∞.

93. Example 10
2/3
▶ Since x2 x for x 1, ex2
ex for x 1, and thus
e−x2
e−x for x 1.
▶ Let’s determine if
Z ∞
1
e−x
dx converges.
▶ Step 1: Compute
Z t
1
e−x
dx
Z t
1
e−x
dx
| {z }
u=−t, du=−dt
u1=−1 u2=−t
= −
Z −t
−1
eu
du
= −eu

96. −t
−1
= −e−t
+ e−1
=
1
e
−
1
et

97. Example 10
3/3
▶ Step 2: Take the limit:
lim
t→∞
Z t
1
e−x
dx = lim
t→∞
1
e
−
1
et
=
1
e
▶ Thus by the Comparison theorem,
Z ∞
1
1
ex2 dx converges to
some number below 1
e although we do not know the exact
value.

98. Example 11
1/3
Does the following integral converge or diverge?
Z 1
0
sec2 x
x
√
x
dx

99. Example 11
2/3
▶ On the interval [0, 1], sec x ≥ 1, so sec2 x ≥ 1.
sec2 x
x
√
x
≥
1
x
√
x
=
1
x3/2
▶ Now,
Z 1
0
1
x3/2
dx is an improper integral of type II, with
infinite discontinuity at x = 0.
▶ Let’s determine if it diverges.

100. Example 11
3/3
▶ Step 1:
Z 1
t
1
x3/2
dx =
x1−3/2
1 − 3/2

103. 1
t
(t 0)
= −
2
√
x

106. 1
t
=
2
√
t
− 2
▶ Step 2: Take the limit
lim
t→0+
Z 1
t
1
x3/2
dx = lim
t→0+
2
√
t
− 2
= ∞
▶ So
Z 1
0
1
x3/2
dx is divergent
▶ Therefore by comparison,
Z 1
0
sec2 x
x
√
x
dx is divergent as well.

107. Example 12
1/2
Does the following integral converge or diverge?
Z ∞
0
7x
x5 + 1
dx

108. Example 12
1/2
▶ We can use the fact that 7x
x5+1
7x
x5 = 7
x4 for all x 0.
▶ We know
R ∞
1
7
x4 dx converges since it is an integral of the form
R ∞
1
1
xp dx with p = 4 1.
▶ The constant multiple 7 does not affect the convergence of the
integral. 7 times a constant is a constant. . .and 7 times ±∞ is
not going to be finite.
▶ So by (a) of comparison test,
R ∞
1
7x
x5+1
dx converges.
▶ Our original integral starts at 0 not 1. But we know
R 1
0
7x
x5+1
dx
is finite because the integrand is continuous on [0, 1]. So since
Z ∞
0
7x
x5 + 1
dx =
Z 1
0
7x
x5 + 1
dx +
Z ∞
1
7x
x5 + 1
dx
the integral Z ∞
0
7x
x5 + 1
dx converges