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436 • Improper Integrals: Basic Concepts
did above. The only difference is that this time we have to approach b from
the left instead of the right. So
if f(x) is unbounded for x near b only, then set
Z
b
a
f(x) dx = lim
ε→0
+
Z
b−ε
a
f(x) dx,
if the limit exists; if it doesn’t exist, then as before, the integral diverges.
Ah, but what if f has a blow-up point at some number c in the interior
of the interval? In this case, if f is bounded everywhere on [a, b] except near
some point c in the interior (a, b), we have to split the integral into the two
pieces
Z
c
a
f(x) dx and
Z
b
c
f(x) dx.
We actually know how to define both of these by using limits—using the
formulas from the boxes above, we can see that the above integrals are
lim
ε→0
+
Z
c−ε
a
f(x) dx and lim
ε→0
+
Z
b
c+ε
f(x) dx,
respectively. Here’s the essential point: the whole integral
R
b
a
f(x) dx only
PSfrag
replacements
(
a, b)
[
a, b]
(
a, b]
[
a, b)
(
a, ∞)
[
a, ∞)
(
−∞, b)
(
−∞, b]
(
−∞, ∞)
{
x : a < x < b}
{
x : a ≤ x ≤ b}
{
x : a < x ≤ b}
{
x : a ≤ x < b}
{
x : x ≥ a}
{
x : x > a}
{
x : x ≤ b}
{
x : x < b}
R
a
b
shado
w
0
1
4
−
2
3
−
3
g(
x) = x
2
f(
x) = x
3
g(
x) = x
2
f(
x) = x
3
mirror
(y = x)
f
−
1
(x) =
3
√
x
y = h
(x)
y = h
−
1
(x)
y =
(x − 1)
2
−
1
x
Same
height
−
x
Same
length,
opp
osite signs
y = −
2x
−
2
1
y =
1
2
x − 1
2
−
1
y =
2
x
y =
10
x
y =
2
−x
y =
log
2
(x)
4
3
units
mirror
(x-axis)
y = |
x|
y = |
log
2
(x)|
θ radians
θ units
30
◦
=
π
6
45
◦
=
π
4
60
◦
=
π
3
120
◦
=
2
π
3
135
◦
=
3
π
4
150
◦
=
5
π
6
90
◦
=
π
2
180
◦
= π
210
◦
=
7
π
6
225
◦
=
5
π
4
240
◦
=
4
π
3
270
◦
=
3
π
2
300
◦
=
5
π
3
315
◦
=
7
π
4
330
◦
=
11
π
6
0
◦
=
0 radians
θ
hypotenuse
opp
osite
adjacen
t
0
(≡ 2π)
π
2
π
3
π
2
I
I
I
I
II
IV
θ
(
x, y)
x
y
r
7
π
6
reference
angle
reference
angle =
π
6
sin
+
sin −
cos
+
cos −
tan
+
tan −
A
S
T
C
7
π
4
9
π
13
5
π
6
(this
angle is
5π
6
clo
ckwise)
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y =
sin(x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y =
sin(x)
y =
cos(x)
−
π
2
π
2
y =
tan(x), −
π
2
<
x <
π
2
0
−
π
2
π
2
y =
tan(x)
−
2π
−
3π
−
5
π
2
−
3
π
2
−
π
−
π
2
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
y =
sec(x)
y =
csc(x)
y =
cot(x)
y = f (
x)
−
1
1
2
y = g(
x)
3
y = h
(x)
4
5
−
2
f(
x) =
1
x
g(
x) =
1
x
2
etc.
0
1
π
1
2
π
1
3
π
1
4
π
1
5
π
1
6
π
1
7
π
g(
x) = sin
1
x
1
0
−
1
L
10
100
200
y =
π
2
y = −
π
2
y =
tan
−1
(x)
π
2
π
y =
sin(
x)
x
,
x > 3
0
1
−
1
a
L
f(
x) = x sin (1/x)
(0 <
x < 0.3)
h
(x) = x
g(
x) = −x
a
L
lim
x
→a
+
f(x) = L
lim
x
→a
+
f(x) = ∞
lim
x
→a
+
f(x) = −∞
lim
x
→a
+
f(x) DNE
lim
x
→a
−
f(x) = L
lim
x
→a
−
f(x) = ∞
lim
x
→a
−
f(x) = −∞
lim
x
→a
−
f(x) DNE
M
}
lim
x
→a
−
f(x) = M
lim
x
→a
f(x) = L
lim
x
→a
f(x) DNE
lim
x
→∞
f(x) = L
lim
x
→∞
f(x) = ∞
lim
x
→∞
f(x) = −∞
lim
x
→∞
f(x) DNE
lim
x
→−∞
f(x) = L
lim
x
→−∞
f(x) = ∞
lim
x
→−∞
f(x) = −∞
lim
x
→−∞
f(x) DNE
lim
x →a
+
f(
x) = ∞
lim
x →a
+
f(
x) = −∞
lim
x →a
−
f(
x) = ∞
lim
x →a
−
f(
x) = −∞
lim
x →a
f(
x) = ∞
lim
x →a
f(
x) = −∞
lim
x →a
f(
x) DNE
y = f (
x)
a
y =
|
x|
x
1
−
1
y =
|
x + 2|
x +
2
1
−
1
−
2
1
2
3
4
a
a
b
y = x sin
1
x
y = x
y = −
x
a
b
c
d
C
a
b
c
d
−
1
0
1
2
3
time
y
t
u
(
t, f(t))
(
u, f(u))
time
y
t
u
y
x
(
x, f(x))
y = |
x|
(
z, f(z))
z
y = f (
x)
a
tangen
t at x = a
b
tangen
t at x = b
c
tangen
t at x = c
y = x
2
tangen
t
at x = −
1
u
v
uv
u +
∆u
v +
∆v
(
u + ∆u)(v + ∆v)
∆
u
∆
v
u
∆v
v∆
u
∆
u∆v
y = f (
x)
1
2
−
2
y = |
x
2
− 4|
y = x
2
− 4
y = −
2x + 5
y = g(
x)
1
2
3
4
5
6
7
8
9
0
−
1
−
2
−
3
−
4
−
5
−
6
y = f (
x)
3
−
3
3
−
3
0
−
1
2
easy
hard
flat
y = f
0
(
x)
3
−
3
0
−
1
2
1
−
1
y =
sin(x)
y = x
x
A
B
O
1
C
D
sin(
x)
tan(
x)
y =
sin
(x)
x
π
2
π
1
−
1
x =
0
a =
0
x
> 0
a
> 0
x
< 0
a
< 0
rest
position
+
−
y = x
2
sin
1
x
N
A
B
H
a
b
c
O
H
A
B
C
D
h
r
R
θ
1000
2000
α
β
p
h
y = g(
x) = log
b
(x)
y = f (
x) = b
x
y = e
x
5
10
1
2
3
4
0
−
1
−
2
−
3
−
4
y =
ln(x)
y =
cosh(x)
y =
sinh(x)
y =
tanh(x)
y =
sech(x)
y =
csch(x)
y =
coth(x)
1
−
1
y = f (
x)
original
function
in
verse function
slop
e = 0 at (x, y)
slop
e is infinite at (y, x)
−
108
2
5
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y =
sin(x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y =
sin(x)
y =
sin(x), −
π
2
≤ x ≤
π
2
−
2
−
1
0
2
π
2
−
π
2
y =
sin
−1
(x)
y =
cos(x)
π
π
2
y =
cos
−1
(x)
−
π
2
1
x
α
β
y =
tan(x)
y =
tan(x)
1
y =
tan
−1
(x)
y =
sec(x)
y =
sec
−1
(x)
y =
csc
−1
(x)
y =
cot
−1
(x)
1
y =
cosh
−1
(x)
y =
sinh
−1
(x)
y =
tanh
−1
(x)
y =
sech
−1
(x)
y =
csch
−1
(x)
y =
coth
−1
(x)
(0
, 3)
(2
, −1)
(5
, 2)
(7
, 0)
(
−1, 44)
(0
, 1)
(1
, −12)
(2
, 305)
y =
1
2
(2
, 3)
y = f (
x)
y = g(
x)
a
b
c
a
b
c
s
c
0
c
1
(
a, f(a))
(
b, f(b))
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y =
sin(x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
c
OR
Lo
cal maximum
Lo
cal minimum
Horizon
tal point of inflection
1
e
y = f
0
(
x)
y = f(
x) = x ln(x)
−
1
e
?
y = f(
x) = x
3
y = g(
x) = x
4
x
f(
x)
−
3
−
2
−
1
0
1
2
1
2
3
4
+
−
?
1
5
6
3
f
0
(
x)
2 −
1
2
√
6
2
+
1
2
√
6
f
00
(
x)
7
8
g
00
(
x)
f
00
(
x)
0
y =
(
x − 3)(x − 1)
2
x
3
(
x + 2)
y = x ln
(x)
1
e
−
1
e
5
−
108
2
α
β
2 −
1
2
√
6
2
+
1
2
√
6
y = x
2
(
x − 5)
3
−
e
−
1/2
√
3
e
−
1/2
√
3
−
e
−3/2
e
−
3/2
−
1
√
3
1
√
3
−
1
1
y = xe
−
3x
2
/2
y =
x
3
− 6
x
2
+ 13x − 8
x
28
2
600
500
400
300
200
100
0
−
100
−
200
−
300
−
400
−
500
−
600
0
10
−
10
5
−
5
20
−
20
15
−
15
0
4
5
6
x
P
0
(
x)
+
−
−
existing
fence
new
fence
enclosure
A
h
b
H
99
100
101
h
dA/dh
r
h
1
2
7
shallo
w
deep
LAND
SEA
N
y
z
s
t
3
11
9
L
(11)
√
11
y = L
(x)
y = f (
x)
11
y = L
(x)
y = f (
x)
F
P
a
a +
∆x
f(
a + ∆x)
L
(a + ∆x)
f(
a)
error
d
f
∆
x
a
b
y = f (
x)
true
zero
starting
approximation
b
etter approximation
v
t
3
5
50
40
60
4
20
30
25
t
1
t
2
t
3
t
4
t
n
−2
t
n
−1
t
0
= a
t
n
= b
v
1
v
2
v
3
v
4
v
n
−1
v
n
−
30
6
30
|
v|
a
b
p
q
c
v(
c)
v(
c
1
)
v(
c
2
)
v(
c
3
)
v(
c
4
)
v(
c
5
)
v(
c
6
)
t
1
t
2
t
3
t
4
t
5
c
1
c
2
c
3
c
4
c
5
c
6
t
0
=
a
t
6
=
b
t
16
=
b
t
10
=
b
a
b
x
y
y = f (
x)
1
2
y = x
5
0
−
2
y =
1
a
b
y =
sin(x)
π
−
π
0
−
1
−
2
0
2
4
y = x
2
0
1
2
3
4
2
n
4
n
6
n
2(
n−2)
n
2(
n−1)
n
2
n
n
=
2
width
of each interval =
2
n
−
2
1
3
0
I
I
I
I
II
IV
4
y
dx
y = −
x
2
− 2x + 3
3
−
5
y = |−
x
2
− 2x + 3|
I
I
I
I
Ia
5
3
0
1
2
a
b
y = f (
x)
y = g(
x)
y = x
2
a
b
5
3
0
1
2
y =
√
x
2
√
2
2
2
dy
x
2
a
b
y = f (
x)
y = g(
x)
M
m
1
2
−
1
−
2
0
y = e
−
x
2
1
2
e
−
1/4
f
a
v
y = f
a
v
c
A
M
0
1
2
a
b
x
t
y = f(t)
F (x )
y = f(t)
F (x + h )
x + h
F (x + h) − F (x)
f(x)
1
2
y = sin(x)
π
−π
−1
−2
y =
1
x
y = x
2
1
2
1
−1
y = ln|x|
θ
a
x
a
x
p
a
2
− x
2
3
x
p
9 − x
2
p
x
2
+ a
2
x
a
p
x
2
+ 15
x
√
15
x
p
x
2
− a
2
a
x
p
x
2
− 4
2
x
−
p
x
2
− a
2
a
x
−
p
x
2
− 4
2
y = f (x)
a
b
a + ε
ε
Z
b
a+ε
f(x) dx
small
even smaller
converges if both pieces above converge. If either piece diverges, so does the
whole thing. After all, how can you add something that doesn’t exist to
anything else, whether that other quantity exists or not? It can’t be done.
This example inspires our first main technique: to investigate an improper
integral, split it up into pieces, if necessary. Each piece has to have at most
one problem spot, which must be at one of the limits of integration. (For the
moment, the term “problem spot” means the same thing as “blow-up point,”
but in the next section we’ll see a different sort of problem spot that isn’t a
blow-up point.)
For example, to analyze the integral
PSfrag
replacements
(
a, b)
[
a, b]
(
a, b]
[
a, b)
(
a, ∞)
[
a, ∞)
(
−∞, b)
(
−∞, b]
(
−∞, ∞)
{
x : a < x < b}
{
x : a ≤ x ≤ b}
{
x : a < x ≤ b}
{
x : a ≤ x < b}
{
x : x ≥ a}
{
x : x > a}
{
x : x ≤ b}
{
x : x < b}
R
a
b
shado
w
0
1
4
−
2
3
−
3
g(
x) = x
2
f(
x) = x
3
g(
x) = x
2
f(
x) = x
3
mirror
(y = x)
f
−
1
(x) =
3
√
x
y = h
(x)
y = h
−
1
(x)
y =
(x − 1)
2
−
1
x
Same
height
−
x
Same
length,
opp
osite signs
y = −
2x
−
2
1
y =
1
2
x − 1
2
−
1
y =
2
x
y =
10
x
y =
2
−x
y =
log
2
(x)
4
3
units
mirror
(x-axis)
y = |
x|
y = |
log
2
(x)|
θ radians
θ units
30
◦
=
π
6
45
◦
=
π
4
60
◦
=
π
3
120
◦
=
2
π
3
135
◦
=
3
π
4
150
◦
=
5
π
6
90
◦
=
π
2
180
◦
= π
210
◦
=
7
π
6
225
◦
=
5
π
4
240
◦
=
4
π
3
270
◦
=
3
π
2
300
◦
=
5
π
3
315
◦
=
7
π
4
330
◦
=
11
π
6
0
◦
=
0 radians
θ
hypotenuse
opp
osite
adjacen
t
0
(≡ 2π)
π
2
π
3
π
2
I
I
I
I
II
IV
θ
(
x, y)
x
y
r
7
π
6
reference
angle
reference
angle =
π
6
sin
+
sin −
cos
+
cos −
tan
+
tan −
A
S
T
C
7
π
4
9
π
13
5
π
6
(this
angle is
5π
6
clo
ckwise)
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y =
sin(x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y =
sin(x)
y =
cos(x)
−
π
2
π
2
y =
tan(x), −
π
2
<
x <
π
2
0
−
π
2
π
2
y =
tan(x)
−
2π
−
3π
−
5
π
2
−
3
π
2
−
π
−
π
2
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
y =
sec(x)
y =
csc(x)
y =
cot(x)
y = f(
x)
−
1
1
2
y = g(
x)
3
y = h
(x)
4
5
−
2
f(
x) =
1
x
g(
x) =
1
x
2
etc.
0
1
π
1
2
π
1
3
π
1
4
π
1
5
π
1
6
π
1
7
π
g(
x) = sin
1
x
1
0
−
1
L
10
100
200
y =
π
2
y = −
π
2
y =
tan
−1
(x)
π
2
π
y =
sin(
x)
x
,
x > 3
0
1
−
1
a
L
f(
x) = x sin (1/x)
(0 <
x < 0.3)
h
(x) = x
g(
x) = −x
a
L
lim
x
→a
+
f(x) = L
lim
x
→a
+
f(x) = ∞
lim
x
→a
+
f(x) = −∞
lim
x
→a
+
f(x) DNE
lim
x
→a
−
f(x) = L
lim
x
→a
−
f(x) = ∞
lim
x
→a
−
f(x) = −∞
lim
x
→a
−
f(x) DNE
M
}
lim
x
→a
−
f(x) = M
lim
x
→a
f(x) = L
lim
x
→a
f(x) DNE
lim
x
→∞
f(x) = L
lim
x
→∞
f(x) = ∞
lim
x
→∞
f(x) = −∞
lim
x
→∞
f(x) DNE
lim
x
→−∞
f(x) = L
lim
x
→−∞
f(x) = ∞
lim
x
→−∞
f(x) = −∞
lim
x
→−∞
f(x) DNE
lim
x →a
+
f(
x) = ∞
lim
x →a
+
f(
x) = −∞
lim
x →a
−
f(
x) = ∞
lim
x →a
−
f(
x) = −∞
lim
x →a
f(
x) = ∞
lim
x →a
f(
x) = −∞
lim
x →a
f(
x) DNE
y = f (
x)
a
y =
|
x|
x
1
−
1
y =
|
x + 2|
x +
2
1
−
1
−
2
1
2
3
4
a
a
b
y = x sin
1
x
y = x
y = −
x
a
b
c
d
C
a
b
c
d
−
1
0
1
2
3
time
y
t
u
(
t, f(t))
(
u, f(u))
time
y
t
u
y
x
(
x, f(x))
y = |
x|
(
z, f(z))
z
y = f(
x)
a
tangen
t at x = a
b
tangen
t at x = b
c
tangen
t at x = c
y = x
2
tangen
t
at x = −
1
u
v
uv
u +
∆u
v +
∆v
(
u + ∆u)(v + ∆v)
∆
u
∆
v
u
∆v
v∆
u
∆
u∆v
y = f(
x)
1
2
−
2
y = |
x
2
− 4|
y = x
2
− 4
y = −
2x + 5
y = g(
x)
1
2
3
4
5
6
7
8
9
0
−
1
−
2
−
3
−
4
−
5
−
6
y = f (
x)
3
−
3
3
−
3
0
−
1
2
easy
hard
flat
y = f
0
(
x)
3
−
3
0
−
1
2
1
−
1
y =
sin(x)
y = x
x
A
B
O
1
C
D
sin(
x)
tan(
x)
y =
sin(
x)
x
π
2
π
1
−
1
x =
0
a =
0
x
> 0
a
> 0
x
< 0
a
< 0
rest
position
+
−
y = x
2
sin
1
x
N
A
B
H
a
b
c
O
H
A
B
C
D
h
r
R
θ
1000
2000
α
β
p
h
y = g(
x) = log
b
(x)
y = f(
x) = b
x
y = e
x
5
10
1
2
3
4
0
−
1
−
2
−
3
−
4
y =
ln(x)
y =
cosh(x)
y =
sinh(x)
y =
tanh(x)
y =
sech(x)
y =
csch(x)
y =
coth(x)
1
−
1
y = f(
x)
original
function
in
verse function
slop
e = 0 at (x, y)
slop
e is infinite at (y, x)
−
108
2
5
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y =
sin(x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y =
sin(x)
y =
sin(x), −
π
2
≤ x ≤
π
2
−
2
−
1
0
2
π
2
−
π
2
y =
sin
−1
(x)
y =
cos(x)
π
π
2
y =
cos
−1
(x)
−
π
2
1
x
α
β
y =
tan(x)
y =
tan(x)
1
y =
tan
−1
(x)
y =
sec(x)
y =
sec
−1
(x)
y =
csc
−1
(x)
y =
cot
−1
(x)
1
y =
cosh
−1
(x)
y =
sinh
−1
(x)
y =
tanh
−1
(x)
y =
sech
−1
(x)
y =
csch
−1
(x)
y =
coth
−1
(x)
(0
, 3)
(2
, −1)
(5
, 2)
(7
, 0)
(
−1, 44)
(0
, 1)
(1
, −12)
(2
, 305)
y =
1
2
(2
, 3)
y = f(
x)
y = g(
x)
a
b
c
a
b
c
s
c
0
c
1
(
a, f(a))
(
b, f(b))
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y =
sin(x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
c
OR
Lo
cal maximum
Lo
cal minimum
Horizon
tal point of inflection
1
e
y = f
0
(
x)
y = f (
x) = x ln(x)
−
1
e
?
y = f(
x) = x
3
y = g(
x) = x
4
x
f(
x)
−
3
−
2
−
1
0
1
2
1
2
3
4
+
−
?
1
5
6
3
f
0
(
x)
2 −
1
2
√
6
2
+
1
2
√
6
f
00
(
x)
7
8
g
00
(
x)
f
00
(
x)
0
y =
(
x − 3)(x − 1)
2
x
3
(
x + 2)
y = x ln
(x)
1
e
−
1
e
5
−
108
2
α
β
2 −
1
2
√
6
2
+
1
2
√
6
y = x
2
(
x − 5)
3
−
e
−
1/2
√
3
e
−
1/2
√
3
−
e
−3/2
e
−
3/2
−
1
√
3
1
√
3
−
1
1
y = xe
−
3x
2
/2
y =
x
3
− 6
x
2
+ 13x − 8
x
28
2
600
500
400
300
200
100
0
−
100
−
200
−
300
−
400
−
500
−
600
0
10
−
10
5
−
5
20
−
20
15
−
15
0
4
5
6
x
P
0
(
x)
+
−
−
existing
fence
new
fence
enclosure
A
h
b
H
99
100
101
h
dA/dh
r
h
1
2
7
shallo
w
deep
LAND
SEA
N
y
z
s
t
3
11
9
L
(11)
√
11
y = L
(x)
y = f (
x)
11
y = L
(x)
y = f(
x)
F
P
a
a +
∆x
f(
a + ∆x)
L
(a + ∆x)
f(
a)
error
d
f
∆
x
a
b
y = f(
x)
true
zero
starting
approximation
b
etter approximation
v
t
3
5
50
40
60
4
20
30
25
t
1
t
2
t
3
t
4
t
n
−2
t
n
−1
t
0
= a
t
n
= b
v
1
v
2
v
3
v
4
v
n
−1
v
n
−
30
6
30
|
v|
a
b
p
q
c
v(
c)
v(
c
1
)
v(
c
2
)
v(
c
3
)
v(
c
4
)
v(
c
5
)
v(
c
6
)
t
1
t
2
t
3
t
4
t
5
c
1
c
2
c
3
c
4
c
5
c
6
t
0
=
a
t
6
=
b
t
16
=
b
t
10
=
b
a
b
x
y
y = f(
x)
1
2
y = x
5
0
−
2
y =
1
a
b
y =
sin(x)
π
−
π
0
−
1
−
2
0
2
4
y = x
2
0
1
2
3
4
2
n
4
n
6
n
2(
n−2)
n
2(
n−1)
n
2
n
n
=
2
width
of each interval =
2
n
−
2
1
3
0
I
I
I
I
II
IV
4
y
dx
y = −
x
2
− 2x + 3
3
−
5
y = |−
x
2
− 2x + 3|
I
I
I
I
Ia
5
3
0
1
2
a
b
y = f (
x)
y = g(
x)
y = x
2
a
b
5
3
0
1
2
y =
√
x
2
√
2
2
2
dy
x
2
a
b
y = f(
x)
y = g(
x)
M
m
1
2
−
1
−
2
0
y = e
−x
2
1
2
e
−1/4
f
av
y = f
av
c
A
M
0
1
2
a
b
x
t
y = f (t)
F (x )
y = f (t)
F (x + h)
x + h
F (x + h) − F (x)
f(x)
1
2
y = sin(x)
π
−π
−1
−2
y =
1
x
y = x
2
1
2
1
−1
y = ln|x|
θ
a
x
a
x
p
a
2
− x
2
3
x
p
9 − x
2
p
x
2
+ a
2
x
a
p
x
2
+ 15
x
√
15
x
p
x
2
− a
2
a
x
p
x
2
− 4
2
x
−
p
x
2
− a
2
a
x
−
p
x
2
− 4
2
y = f(x)
a
b
a + ε
ε
Z
b
a+ε
f(x) dx
small
even smaller
I =
Z
3
0
1
x(x − 1)(x + 1)(x − 2)
dx,
we see that the integrand has problem spots at x = 0, 1, 2, and −1. The last
of these doesn’t matter since we’re only integrating from 0 to 3. The other
three do matter. We need to pick numbers between the problem spots—we
may as well pick 1/2 and 3/2, since it doesn’t matter—and now we need to
split the original integral into the following five integrals:
I
1
=
Z
1/2
0
1
x(x − 1)(x + 1)(x − 2)
dx, I
2
=
Z
1
1/2
1
x(x − 1)(x + 1)(x − 2)
dx,
I
3
=
Z
3/2
1
1
x(x − 1)(x + 1)(x − 2)
dx, I
4
=
Z
2
3/2
1
x(x − 1)(x + 1)(x − 2)
dx,
and I
5
=
Z
3
2
1
x(x − 1)(x + 1)(x − 2)
dx.
Section 20.2: Integrals over Unbounded Regions • 437
Notice that none of these integrals has more than one problem spot, and that
all the problem spots are at one of the limits of integration. The integrals I
1
,
I
3
, and I
5
have their only problem spot at their left-hand limits of integration,
while I
2
and I
4
have their problem spot on the right. The only way that
original integral I can converge is if all five pieces I
1
through I
5
converge. If
they do all converge, then the value of I is the sum of the values of I
1
through
I
5
. (In fact, none of the five pieces converge! We’ll see why in Section 21.5 of
the next chapter.)
20.2 Integrals over Unbounded Regions
Now, we still have to look at what happens when one or both of the limits of
integration are infinite; this means that the region of integration is unbounded.
To handle
Z
∞
a
f(x) dx,
where a is any finite number and f has no blow-up points in [a, ∞), let’s use
another limiting technique. This time, we integrate over the region [a, N],
where N is a massively large number. This will give us a nice finite value.
Then repeat but with an even larger N to get a new value. Continue onward
and see what happens to the values of the integrals. If they have a limit, then
the integral converges. Otherwise, it diverges. In symbols, we are defining
Z
∞
a
f(x) dx = lim
N→∞
Z
N
a
f(x) dx,
provided that the limit exists; in this case, the integral converges. Otherwise,
it diverges. For reasons similar to those described at the end of Section 20.1.1
above, the value of a is irrelevant. So long as you don’t pick up any new blow-
up points of f , the value of a doesn’t affect whether the improper integral
converges or diverges. The only thing that really matters is how f(x) behaves
when x is very large indeed.
In a similar manner to the above definition, if f has no blow-up points in
(−∞, b], then
Z
b
−∞
f(x) dx = lim
N→∞
Z
b
−N
f(x) dx.
What if f has no blow-up points anywhere and we want to find
Z
∞
−∞
f(x) dx?
Although there are no blow-up points, there are still two problem spots: ∞
and −∞. That’s right: we are regarding ∞ and −∞ as problem spots when-
ever they show up, since we have to treat them separately. So we have to
split the above integral into two pieces so that each one has only one problem
spot. Pick your favorite number (mine is 0 for the moment), and consider the
integrals
Z
0
−∞
f(x) dx and
Z
∞
0
f(x) dx.
438 • Improper Integrals: Basic Concepts
We know what both of these mean, and of course the whole integral converges
if and only if both pieces do. It doesn’t matter if you have a different favorite
number from 0, since the convergence or divergence of the above integrals
doesn’t depend on the endpoint.
Here are some examples involving an unbounded region of integration.
Consider the integrals
PSfrag
replacements
(
a, b)
[
a, b]
(
a, b]
[
a, b)
(
a, ∞)
[
a, ∞)
(
−∞, b)
(
−∞, b]
(
−∞, ∞)
{
x : a < x < b}
{
x : a ≤ x ≤ b}
{
x : a < x ≤ b}
{
x : a ≤ x < b}
{
x : x ≥ a}
{
x : x > a}
{
x : x ≤ b}
{
x : x < b}
R
a
b
shado
w
0
1
4
−
2
3
−
3
g(
x) = x
2
f(
x) = x
3
g(
x) = x
2
f(
x) = x
3
mirror
(y = x)
f
−
1
(x) =
3
√
x
y = h
(x)
y = h
−
1
(x)
y =
(x − 1)
2
−
1
x
Same
height
−
x
Same
length,
opp
osite signs
y = −
2x
−
2
1
y =
1
2
x − 1
2
−
1
y =
2
x
y =
10
x
y =
2
−x
y =
log
2
(x)
4
3
units
mirror
(x-axis)
y = |
x|
y = |
log
2
(x)|
θ radians
θ units
30
◦
=
π
6
45
◦
=
π
4
60
◦
=
π
3
120
◦
=
2
π
3
135
◦
=
3
π
4
150
◦
=
5
π
6
90
◦
=
π
2
180
◦
= π
210
◦
=
7
π
6
225
◦
=
5
π
4
240
◦
=
4
π
3
270
◦
=
3
π
2
300
◦
=
5
π
3
315
◦
=
7
π
4
330
◦
=
11
π
6
0
◦
=
0 radians
θ
hypotenuse
opp
osite
adjacen
t
0
(≡ 2π)
π
2
π
3
π
2
I
I
I
I
II
IV
θ
(
x, y)
x
y
r
7
π
6
reference
angle
reference
angle =
π
6
sin
+
sin −
cos
+
cos −
tan
+
tan −
A
S
T
C
7
π
4
9
π
13
5
π
6
(this
angle is
5π
6
clo
ckwise)
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y =
sin(x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y =
sin(x)
y =
cos(x)
−
π
2
π
2
y =
tan(x), −
π
2
<
x <
π
2
0
−
π
2
π
2
y =
tan(x)
−
2π
−
3π
−
5
π
2
−
3
π
2
−
π
−
π
2
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
y =
sec(x)
y =
csc(x)
y =
cot(x)
y = f(
x)
−
1
1
2
y = g(
x)
3
y = h
(x)
4
5
−
2
f(
x) =
1
x
g(
x) =
1
x
2
etc.
0
1
π
1
2
π
1
3
π
1
4
π
1
5
π
1
6
π
1
7
π
g(
x) = sin
1
x
1
0
−
1
L
10
100
200
y =
π
2
y = −
π
2
y =
tan
−1
(x)
π
2
π
y =
sin(
x)
x
,
x > 3
0
1
−
1
a
L
f(
x) = x sin (1/x)
(0 <
x < 0.3)
h
(x) = x
g(
x) = −x
a
L
lim
x
→a
+
f(x) = L
lim
x
→a
+
f(x) = ∞
lim
x
→a
+
f(x) = −∞
lim
x
→a
+
f(x) DNE
lim
x
→a
−
f(x) = L
lim
x
→a
−
f(x) = ∞
lim
x
→a
−
f(x) = −∞
lim
x
→a
−
f(x) DNE
M
}
lim
x
→a
−
f(x) = M
lim
x
→a
f(x) = L
lim
x
→a
f(x) DNE
lim
x
→∞
f(x) = L
lim
x
→∞
f(x) = ∞
lim
x
→∞
f(x) = −∞
lim
x
→∞
f(x) DNE
lim
x
→−∞
f(x) = L
lim
x
→−∞
f(x) = ∞
lim
x
→−∞
f(x) = −∞
lim
x
→−∞
f(x) DNE
lim
x →a
+
f(
x) = ∞
lim
x →a
+
f(
x) = −∞
lim
x →a
−
f(
x) = ∞
lim
x →a
−
f(
x) = −∞
lim
x →a
f(
x) = ∞
lim
x →a
f(
x) = −∞
lim
x →a
f(
x) DNE
y = f (
x)
a
y =
|
x|
x
1
−
1
y =
|
x + 2|
x +
2
1
−
1
−
2
1
2
3
4
a
a
b
y = x sin
1
x
y = x
y = −
x
a
b
c
d
C
a
b
c
d
−
1
0
1
2
3
time
y
t
u
(
t, f(t))
(
u, f(u))
time
y
t
u
y
x
(
x, f(x))
y = |
x|
(
z, f(z))
z
y = f(
x)
a
tangen
t at x = a
b
tangen
t at x = b
c
tangen
t at x = c
y = x
2
tangen
t
at x = −
1
u
v
uv
u +
∆u
v +
∆v
(
u + ∆u)(v + ∆v)
∆
u
∆
v
u
∆v
v∆
u
∆
u∆v
y = f(
x)
1
2
−
2
y = |
x
2
− 4|
y = x
2
− 4
y = −
2x + 5
y = g(
x)
1
2
3
4
5
6
7
8
9
0
−
1
−
2
−
3
−
4
−
5
−
6
y = f (
x)
3
−
3
3
−
3
0
−
1
2
easy
hard
flat
y = f
0
(
x)
3
−
3
0
−
1
2
1
−
1
y =
sin(x)
y = x
x
A
B
O
1
C
D
sin(
x)
tan(
x)
y =
sin(
x)
x
π
2
π
1
−
1
x =
0
a =
0
x
> 0
a
> 0
x
< 0
a
< 0
rest
position
+
−
y = x
2
sin
1
x
N
A
B
H
a
b
c
O
H
A
B
C
D
h
r
R
θ
1000
2000
α
β
p
h
y = g(
x) = log
b
(x)
y = f(
x) = b
x
y = e
x
5
10
1
2
3
4
0
−
1
−
2
−
3
−
4
y =
ln(x)
y =
cosh(x)
y =
sinh(x)
y =
tanh(x)
y =
sech(x)
y =
csch(x)
y =
coth(x)
1
−
1
y = f(
x)
original
function
in
verse function
slop
e = 0 at (x, y)
slop
e is infinite at (y, x)
−
108
2
5
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y =
sin(x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y =
sin(x)
y =
sin(x), −
π
2
≤ x ≤
π
2
−
2
−
1
0
2
π
2
−
π
2
y =
sin
−1
(x)
y =
cos(x)
π
π
2
y =
cos
−1
(x)
−
π
2
1
x
α
β
y =
tan(x)
y =
tan(x)
1
y =
tan
−1
(x)
y =
sec(x)
y =
sec
−1
(x)
y =
csc
−1
(x)
y =
cot
−1
(x)
1
y =
cosh
−1
(x)
y =
sinh
−1
(x)
y =
tanh
−1
(x)
y =
sech
−1
(x)
y =
csch
−1
(x)
y =
coth
−1
(x)
(0
, 3)
(2
, −1)
(5
, 2)
(7
, 0)
(
−1, 44)
(0
, 1)
(1
, −12)
(2
, 305)
y =
1
2
(2
, 3)
y = f(
x)
y = g(
x)
a
b
c
a
b
c
s
c
0
c
1
(
a, f(a))
(
b, f(b))
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y =
sin(x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
c
OR
Lo
cal maximum
Lo
cal minimum
Horizon
tal point of inflection
1
e
y = f
0
(
x)
y = f (
x) = x ln(x)
−
1
e
?
y = f(
x) = x
3
y = g(
x) = x
4
x
f(
x)
−
3
−
2
−
1
0
1
2
1
2
3
4
+
−
?
1
5
6
3
f
0
(
x)
2 −
1
2
√
6
2
+
1
2
√
6
f
00
(
x)
7
8
g
00
(
x)
f
00
(
x)
0
y =
(
x − 3)(x − 1)
2
x
3
(
x + 2)
y = x ln
(x)
1
e
−
1
e
5
−
108
2
α
β
2 −
1
2
√
6
2
+
1
2
√
6
y = x
2
(
x − 5)
3
−
e
−
1/2
√
3
e
−
1/2
√
3
−
e
−3/2
e
−
3/2
−
1
√
3
1
√
3
−
1
1
y = xe
−
3x
2
/2
y =
x
3
− 6
x
2
+ 13x − 8
x
28
2
600
500
400
300
200
100
0
−
100
−
200
−
300
−
400
−
500
−
600
0
10
−
10
5
−
5
20
−
20
15
−
15
0
4
5
6
x
P
0
(
x)
+
−
−
existing
fence
new
fence
enclosure
A
h
b
H
99
100
101
h
dA/dh
r
h
1
2
7
shallo
w
deep
LAND
SEA
N
y
z
s
t
3
11
9
L
(11)
√
11
y = L
(x)
y = f (
x)
11
y = L
(x)
y = f(
x)
F
P
a
a +
∆x
f(
a + ∆x)
L
(a + ∆x)
f(
a)
error
d
f
∆
x
a
b
y = f(
x)
true
zero
starting
approximation
b
etter approximation
v
t
3
5
50
40
60
4
20
30
25
t
1
t
2
t
3
t
4
t
n
−2
t
n
−1
t
0
= a
t
n
= b
v
1
v
2
v
3
v
4
v
n
−1
v
n
−
30
6
30
|
v|
a
b
p
q
c
v(
c)
v(
c
1
)
v(
c
2
)
v(
c
3
)
v(
c
4
)
v(
c
5
)
v(
c
6
)
t
1
t
2
t
3
t
4
t
5
c
1
c
2
c
3
c
4
c
5
c
6
t
0
=
a
t
6
=
b
t
16
=
b
t
10
=
b
a
b
x
y
y = f(
x)
1
2
y = x
5
0
−
2
y =
1
a
b
y =
sin(x)
π
−
π
0
−
1
−
2
0
2
4
y = x
2
0
1
2
3
4
2
n
4
n
6
n
2(
n−2)
n
2(
n−1)
n
2
n
n
=
2
width
of each interval =
2
n
−
2
1
3
0
I
I
I
I
II
IV
4
y
dx
y = −
x
2
− 2x + 3
3
−
5
y = |−
x
2
− 2x + 3|
I
I
I
I
Ia
5
3
0
1
2
a
b
y = f (
x)
y = g(
x)
y = x
2
a
b
5
3
0
1
2
y =
√
x
2
√
2
2
2
dy
x
2
a
b
y = f(
x)
y = g(
x)
M
m
1
2
−
1
−
2
0
y = e
−
x
2
1
2
e
−
1/4
f
a
v
y = f
a
v
c
A
M
0
1
2
a
b
x
t
y = f (
t)
F (
x )
y = f (
t)
F (
x + h)
x + h
F (
x + h) − F (x)
f(
x)
1
2
y =
sin(x)
π
−π
−1
−2
y =
1
x
y = x
2
1
2
1
−1
y = ln|x|
θ
a
x
a
x
p
a
2
− x
2
3
x
p
9 − x
2
p
x
2
+ a
2
x
a
p
x
2
+ 15
x
√
15
x
p
x
2
− a
2
a
x
p
x
2
− 4
2
x
−
p
x
2
− a
2
a
x
−
p
x
2
− 4
2
y = f(x)
a
b
a + ε
ε
Z
b
a+ε
f(x) dx
small
even smaller
Z
∞
1
1
x
dx and
Z
∞
1
1
x
2
dx.
The first one is
lim
N→∞
Z
N
1
1
x
dx = lim
N→∞
ln|x|
N
1
= lim
N→∞
(ln(N) − ln(1)) = ∞,
while the second one is
lim
N→∞
Z
N
1
1
x
2
dx = lim
N→∞
−
1
x
N
1
= lim
N→∞
−
1
N
+ 1
= 1.
So the first integral diverges while the second converges.
Here’s a question: do the following integrals converge or diverge?
Z
∞
0
1
x
dx and
Z
∞
0
1
x
2
dx?
Since both integrals have problem spots at 0 and ∞, we’ll have to split them
both up. For the first one, we can look at
Z
1
0
1
x
dx and
Z
∞
1
1
x
dx.
Note that the choice of 1 as the split point is up to you. It doesn’t matter at
all what you pick (so long as it’s a positive number)! Anyway, we’ve already
seen that both of these integrals diverge, so certainly the integral
R
∞
0
1/x dx
diverges.
As for
R
∞
0
1/x
2
dx, we split it up into the two pieces
Z
1
0
1
x
2
dx and
Z
∞
1
1
x
2
dx.
Now we already saw that the second of these pieces converges. To examine the
first piece, we could use our formula involving limits, but there’s a sneakier
way. The idea is that we already saw that
R
1
0
1/x dx diverges to infinity. But
if you think about it, 1/x
2
is greater than 1/x when x is between 0 and 1.
(Right? Hmm . . . x
2
is less than x in the region (0, 1), so the reciprocals are
the other way around.) So if the area under 1/x above [0, 1] is infinite, the area
under 1/x
2
above [0, 1] is even bigger—so still infinite! Without doing any
more work, we can already say that
R
1
0
1/x
2
dx diverges. It follows that the
whole integral
R
∞
0
1/x
2
dx diverges, and that the real problem is due to the
left-hand endpoint 0, not the right-hand endpoint ∞ in this case. Notice how
we compared 1/x
2
with 1/x; this is a special case of the so-called comparison
test, which we’ll look at now.
Section 20.3: The Comparison Test (Theory) • 439
20.3 The Comparison Test (Theory)
Suppose we have two functions which are never negative, at least in some
region of interest. If the first function is bigger than the second function, and
the integral of the second function (over our region) diverges, then the integral
of the first function (over the same region) also diverges. Mathematically, it
looks like this. Let’s say we want to know something about
R
b
a
f(x) dx, but
we only know something about
R
b
a
g(x) dx. If f(x) ≥ g(x) ≥ 0 for x in the
interval (a, b), and we know that
R
b
a
g(x) dx diverges, then so does
R
b
a
f(x) dx.
In fact, since f(x) ≥ g(x), we can write
Z
b
a
f(x) dx ≥
Z
b
a
g(x) dx = ∞.
So the first integral also diverges. In our example above, we’d just write
Z
1
0
1
x
2
dx ≥
Z
1
0
1
x
dx = ∞,
and conclude that the left-hand integral diverges. Of course, we had to know
that the right-hand integral diverges, but we already saw that earlier.
The situation is even clearer when one looks at a picture:
PSfrag replacements
(
a, b)
[
a, b]
(
a, b]
[
a, b)
(
a, ∞)
[
a, ∞)
(
−∞, b)
(
−∞, b]
(
−∞, ∞)
{
x : a < x < b}
{
x : a ≤ x ≤ b}
{
x : a < x ≤ b}
{
x : a ≤ x < b}
{
x : x ≥ a}
{
x : x > a}
{
x : x ≤ b}
{
x : x < b}
R
a
b
shadow
0
1
4
−
2
3
−
3
g(
x) = x
2
f(
x) = x
3
g(
x) = x
2
f(
x) = x
3
mirror (
y = x)
f
−
1
(x) =
3
√
x
y = h
(x)
y = h
−
1
(x)
y = (
x − 1)
2
−
1
x
Same height
−
x
Same length,
opposite signs
y = −
2x
−
2
1
y =
1
2
x − 1
2
−
1
y = 2
x
y = 10
x
y = 2
−
x
y = log
2
(
x)
4
3 units
mirror (
x-axis)
y = |
x|
y = |
log
2
(x)|
θ radians
θ units
30
◦
=
π
6
45
◦
=
π
4
60
◦
=
π
3
120
◦
=
2
π
3
135
◦
=
3
π
4
150
◦
=
5
π
6
90
◦
=
π
2
180
◦
= π
210
◦
=
7
π
6
225
◦
=
5
π
4
240
◦
=
4
π
3
270
◦
=
3
π
2
300
◦
=
5
π
3
315
◦
=
7
π
4
330
◦
=
11
π
6
0
◦
= 0 radians
θ
hypotenuse
opposite
adjacent
0 (
≡ 2π)
π
2
π
3
π
2
I
II
III
IV
θ
(
x, y)
x
y
r
7
π
6
reference angle
reference angle =
π
6
sin +
sin −
cos +
cos −
tan +
tan −
A
S
T
C
7
π
4
9
π
13
5
π
6
(this angle is
5
π
6
clockwise)
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y = sin(
x)
y = cos(
x)
−
π
2
π
2
y = tan(
x), −
π
2
< x <
π
2
0
−
π
2
π
2
y = tan(
x)
−
2π
−
3π
−
5
π
2
−
3
π
2
−
π
−
π
2
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
y = sec(
x)
y = csc(
x)
y = cot(
x)
y = f(
x)
−
1
1
2
y = g(
x)
3
y = h
(x)
4
5
−
2
f(
x) =
1
x
g(
x) =
1
x
2
etc.
0
1
π
1
2
π
1
3
π
1
4
π
1
5
π
1
6
π
1
7
π
g(
x) = sin
1
x
1
0
−
1
L
10
100
200
y =
π
2
y = −
π
2
y = tan
−
1
(x)
π
2
π
y =
sin(
x)
x
, x > 3
0
1
−
1
a
L
f(
x) = x sin (1/x)
(0 < x < 0
.3)
h
(x) = x
g(
x) = −x
a
L
lim
x
→a
+
f(x) = L
lim
x
→a
+
f(x) = ∞
lim
x
→a
+
f(x) = −∞
lim
x
→a
+
f(x) DNE
lim
x
→a
−
f(x) = L
lim
x
→a
−
f(x) = ∞
lim
x
→a
−
f(x) = −∞
lim
x
→a
−
f(x) DNE
M
}
lim
x
→a
−
f(x) = M
lim
x
→a
f(x) = L
lim
x
→a
f(x) DNE
lim
x
→∞
f(x) = L
lim
x
→∞
f(x) = ∞
lim
x
→∞
f(x) = −∞
lim
x
→∞
f(x) DNE
lim
x
→−∞
f(x) = L
lim
x
→−∞
f(x) = ∞
lim
x
→−∞
f(x) = −∞
lim
x
→−∞
f(x) DNE
lim
x →a
+
f(
x) = ∞
lim
x →a
+
f(
x) = −∞
lim
x →a
−
f(
x) = ∞
lim
x →a
−
f(
x) = −∞
lim
x →a
f(
x) = ∞
lim
x →a
f(
x) = −∞
lim
x →a
f(
x) DNE
y = f (
x)
a
y =
|
x|
x
1
−
1
y =
|
x + 2|
x + 2
1
−
1
−
2
1
2
3
4
a
a
b
y = x sin
1
x
y = x
y = −
x
a
b
c
d
C
a
b
c
d
−
1
0
1
2
3
time
y
t
u
(
t, f(t))
(
u, f(u))
time
y
t
u
y
x
(
x, f(x))
y = |
x|
(
z, f(z))
z
y = f(
x)
a
tangent at x = a
b
tangent at x = b
c
tangent at x = c
y = x
2
tangent
at x = −
1
u
v
uv
u + ∆
u
v + ∆
v
(
u + ∆u)(v + ∆v)
∆
u
∆
v
u
∆v
v∆
u
∆
u∆v
y = f(
x)
1
2
−
2
y = |
x
2
− 4|
y = x
2
− 4
y = −
2x + 5
y = g(
x)
1
2
3
4
5
6
7
8
9
0
−
1
−
2
−
3
−
4
−
5
−
6
y = f(
x)
3
−
3
3
−
3
0
−
1
2
easy
hard
flat
y = f
0
(
x)
3
−
3
0
−
1
2
1
−
1
y = sin(
x)
y = x
x
A
B
O
1
C
D
sin(
x)
tan(
x)
y =
sin(
x)
x
π
2
π
1
−
1
x = 0
a = 0
x > 0
a > 0
x < 0
a < 0
rest position
+
−
y = x
2
sin
1
x
N
A
B
H
a
b
c
O
H
A
B
C
D
h
r
R
θ
1000
2000
α
β
p
h
y = g(
x) = log
b
(x)
y = f (
x) = b
x
y = e
x
5
10
1
2
3
4
0
−
1
−
2
−
3
−
4
y = ln(
x)
y = cosh(
x)
y = sinh(
x)
y = tanh(
x)
y = sech(
x)
y = csch(
x)
y = coth(
x)
1
−
1
y = f(
x)
original function
inverse function
slope = 0 at (
x, y)
slope is infinite at (
y, x)
−
108
2
5
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y = sin(
x)
y = sin(
x), −
π
2
≤ x ≤
π
2
−
2
−
1
0
2
π
2
−
π
2
y = sin
−
1
(x)
y = cos(
x)
π
π
2
y = cos
−
1
(x)
−
π
2
1
x
α
β
y = tan(
x)
y = tan(
x)
1
y = tan
−
1
(x)
y = sec(
x)
y = sec
−
1
(x)
y = csc
−
1
(x)
y = cot
−
1
(x)
1
y = cosh
−
1
(x)
y = sinh
−
1
(x)
y = tanh
−
1
(x)
y = sech
−
1
(x)
y = csch
−
1
(x)
y = coth
−
1
(x)
(0
, 3)
(2
, −1)
(5
, 2)
(7
, 0)
(
−1, 44)
(0
, 1)
(1
, −12)
(2
, 305)
y = 1
2
(2
, 3)
y = f(
x)
y = g(
x)
a
b
c
a
b
c
s
c
0
c
1
(
a, f(a))
(
b, f(b))
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
c
OR
Local maximum
Local minimum
Horizontal point of inflection
1
e
y = f
0
(
x)
y = f(
x) = x ln(x)
−
1
e
?
y = f (
x) = x
3
y = g(
x) = x
4
x
f(
x)
−
3
−
2
−
1
0
1
2
1
2
3
4
+
−
?
1
5
6
3
f
0
(
x)
2 −
1
2
√
6
2 +
1
2
√
6
f
00
(
x)
7
8
g
00
(
x)
f
00
(
x)
0
y =
(
x − 3)(x − 1)
2
x
3
(
x + 2)
y = x ln(
x)
1
e
−
1
e
5
−
108
2
α
β
2 −
1
2
√
6
2 +
1
2
√
6
y = x
2
(
x − 5)
3
−
e
−
1/2
√
3
e
−
1/2
√
3
−
e
−3/2
e
−
3/2
−
1
√
3
1
√
3
−
1
1
y = xe
−
3x
2
/2
y =
x
3
− 6
x
2
+ 13x − 8
x
28
2
600
500
400
300
200
100
0
−
100
−
200
−
300
−
400
−
500
−
600
0
10
−
10
5
−
5
20
−
20
15
−
15
0
4
5
6
x
P
0
(
x)
+
−
−
existing fence
new fence
enclosure
A
h
b
H
99
100
101
h
dA/dh
r
h
1
2
7
shallow
deep
LAND
SEA
N
y
z
s
t
3
11
9
L
(11)
√
11
y = L
(x)
y = f(
x)
11
y = L
(x)
y = f(
x)
F
P
a
a + ∆
x
f(
a + ∆x)
L
(a + ∆x)
f(
a)
error
df
∆
x
a
b
y = f(
x)
true zero
starting approximation
better approximation
v
t
3
5
50
40
60
4
20
30
25
t
1
t
2
t
3
t
4
t
n
−2
t
n
−1
t
0
= a
t
n
= b
v
1
v
2
v
3
v
4
v
n
−1
v
n
−
30
6
30
|
v|
a
b
p
q
c
v(
c)
v(
c
1
)
v(
c
2
)
v(
c
3
)
v(
c
4
)
v(
c
5
)
v(
c
6
)
t
1
t
2
t
3
t
4
t
5
c
1
c
2
c
3
c
4
c
5
c
6
t
0
=
a
t
6
=
b
t
16
=
b
t
10
=
b
a
b
x
y
y = f(
x)
1
2
y = x
5
0
−
2
y = 1
a
b
y = sin(
x)
π
−
π
0
−
1
−
2
0
2
4
y = x
2
0
1
2
3
4
2
n
4
n
6
n
2(
n−2)
n
2(
n−1)
n
2
n
n
= 2
width of each interval =
2
n
−
2
1
3
0
I
II
III
IV
4
y
dx
y = −
x
2
− 2x + 3
3
−
5
y = |−
x
2
− 2x + 3|
I
II
IIa
5
3
0
1
2
a
b
y = f(
x)
y = g(
x)
y = x
2
a
b
5
3
0
1
2
y =
√
x
2
√
2
2
2
dy
x
2
a
b
y = f(
x)
y = g(
x)
M
m
1
2
−
1
−
2
0
y = e
−x
2
1
2
e
−1/4
f
av
y = f
av
c
A
M
0
1
2
a
b
x
t
y = f(t)
F (x )
y = f(t)
F (x + h )
x + h
F (x + h) − F (x)
f(x)
1
2
y = sin(x)
π
−π
−1
−2
y =
1
x
y = x
2
1
2
1
−1
y = ln|x|
θ
a
x
a
x
p
a
2
− x
2
3
x
p
9 − x
2
p
x
2
+ a
2
x
a
p
x
2
+ 15
x
√
15
x
p
x
2
− a
2
a
x
p
x
2
− 4
2
x
−
p
x
2
− a
2
a
x
−
p
x
2
− 4
2
y = f(x)
a b
a + ε
ε
Z
b
a+ε
f(x) dx
small
even smaller
y = g(x)
infinite area
In this picture, the area under y = g(x) between x = a and x = b is supposed
to be infinite. The curve y = f(x) sits above y = g(x), so the area under it
(between x = a and x = b) should be even greater. More than infinite is still
infinite, so
R
b
a
f(x) dx also diverges.
What if
R
b
a
g(x) dx diverges but f(x) ≤ g(x) instead? What can you say
about
R
b
a
f(x) dx? The answer is: diddly-squat. Bubkes. Nothing at all. Let’s
see how the math would go:
Z
b
a
f(x) dx ≤
Z
b
a
g(x) dx = ∞.
So the integral we are interested in,
R
b
a
f(x) dx, is less than or equal to infinity.
That is, either it is less than infinity, so it converges, or it is equal to infinity,
440 • Improper Integrals: Basic Concepts
so it diverges. Great—we now know that it either converges or diverges.
Whoop-di-doo. Yup, we haven’t accomplished anything. So don’t do this.
On the other hand, for convergence, it is the other way around. Here, if
we want to know about
R
b
a
f(x) dx and we know that
R
b
a
g(x) dx converges,
we’d better hope that f(x) ≤ g(x). You might say that we want f to be
“controlled” by g. Well, then we’d get convergence (still assuming that both
functions are positive). So, if 0 ≤ f (x) ≤ g(x) on (a, b) and
R
b
a
g(x) dx con-
verges, then so does
R
b
a
f(x) dx. Mathematically,
Z
b
a
f(x) dx ≤
Z
b
a
g(x) dx < ∞,
so both integrals converge (noting that the left-hand integral is positive, so it
can’t diverge down to −∞). The picture looks like this:
PSfrag
replacements
(
a, b)
[
a, b]
(
a, b]
[
a, b)
(
a, ∞)
[
a, ∞)
(
−∞, b)
(
−∞, b]
(
−∞, ∞)
{
x : a < x < b}
{
x : a ≤ x ≤ b}
{
x : a < x ≤ b}
{
x : a ≤ x < b}
{
x : x ≥ a}
{
x : x > a}
{
x : x ≤ b}
{
x : x < b}
R
a
b
shado
w
0
1
4
−
2
3
−
3
g(
x) = x
2
f(
x) = x
3
g(
x) = x
2
f(
x) = x
3
mirror
(y = x)
f
−
1
(x) =
3
√
x
y = h
(x)
y = h
−
1
(x)
y =
(x − 1)
2
−
1
x
Same
height
−
x
Same
length,
opp
osite signs
y = −
2x
−
2
1
y =
1
2
x − 1
2
−
1
y =
2
x
y =
10
x
y =
2
−x
y =
log
2
(x)
4
3
units
mirror
(x-axis)
y = |
x|
y = |
log
2
(x)|
θ radians
θ units
30
◦
=
π
6
45
◦
=
π
4
60
◦
=
π
3
120
◦
=
2
π
3
135
◦
=
3
π
4
150
◦
=
5
π
6
90
◦
=
π
2
180
◦
= π
210
◦
=
7
π
6
225
◦
=
5
π
4
240
◦
=
4
π
3
270
◦
=
3
π
2
300
◦
=
5
π
3
315
◦
=
7
π
4
330
◦
=
11
π
6
0
◦
=
0 radians
θ
hypotenuse
opp
osite
adjacen
t
0
(≡ 2π)
π
2
π
3
π
2
I
I
I
I
II
IV
θ
(
x, y)
x
y
r
7
π
6
reference
angle
reference
angle =
π
6
sin
+
sin −
cos
+
cos −
tan
+
tan −
A
S
T
C
7
π
4
9
π
13
5
π
6
(this
angle is
5π
6
clo
ckwise)
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y =
sin(x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y =
sin(x)
y =
cos(x)
−
π
2
π
2
y =
tan(x), −
π
2
<
x <
π
2
0
−
π
2
π
2
y =
tan(x)
−
2π
−
3π
−
5
π
2
−
3
π
2
−
π
−
π
2
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
y =
sec(x)
y =
csc(x)
y =
cot(x)
y = f(
x)
−
1
1
2
y = g(
x)
3
y = h
(x)
4
5
−
2
f(
x) =
1
x
g(
x) =
1
x
2
etc.
0
1
π
1
2
π
1
3
π
1
4
π
1
5
π
1
6
π
1
7
π
g(
x) = sin
1
x
1
0
−
1
L
10
100
200
y =
π
2
y = −
π
2
y =
tan
−1
(x)
π
2
π
y =
sin
(x)
x
,
x > 3
0
1
−
1
a
L
f(
x) = x sin (1/x)
(0 <
x < 0.3)
h
(x) = x
g(
x) = −x
a
L
lim
x
→a
+
f(x) = L
lim
x
→a
+
f(x) = ∞
lim
x
→a
+
f(x) = −∞
lim
x
→a
+
f(x) DNE
lim
x
→a
−
f(x) = L
lim
x
→a
−
f(x) = ∞
lim
x
→a
−
f(x) = −∞
lim
x
→a
−
f(x) DNE
M
}
lim
x
→a
−
f(x) = M
lim
x
→a
f(x) = L
lim
x
→a
f(x) DNE
lim
x
→∞
f(x) = L
lim
x
→∞
f(x) = ∞
lim
x
→∞
f(x) = −∞
lim
x
→∞
f(x) DNE
lim
x
→−∞
f(x) = L
lim
x
→−∞
f(x) = ∞
lim
x
→−∞
f(x) = −∞
lim
x
→−∞
f(x) DNE
lim
x →a
+
f(
x) = ∞
lim
x →a
+
f(
x) = −∞
lim
x →a
−
f(
x) = ∞
lim
x →a
−
f(
x) = −∞
lim
x →a
f(
x) = ∞
lim
x →a
f(
x) = −∞
lim
x →a
f(
x) DNE
y = f (
x)
a
y =
|
x|
x
1
−
1
y =
|
x + 2|
x +
2
1
−
1
−
2
1
2
3
4
a
a
b
y = x sin
1
x
y = x
y = −
x
a
b
c
d
C
a
b
c
d
−
1
0
1
2
3
time
y
t
u
(
t, f(t))
(
u, f(u))
time
y
t
u
y
x
(
x, f(x))
y = |
x|
(
z, f(z))
z
y = f(
x)
a
tangen
t at x = a
b
tangen
t at x = b
c
tangen
t at x = c
y = x
2
tangen
t
at x = −
1
u
v
uv
u +
∆u
v +
∆v
(
u + ∆u)(v + ∆v)
∆
u
∆
v
u
∆v
v∆
u
∆
u∆v
y = f(
x)
1
2
−
2
y = |
x
2
− 4|
y = x
2
− 4
y = −
2x + 5
y = g(
x)
1
2
3
4
5
6
7
8
9
0
−
1
−
2
−
3
−
4
−
5
−
6
y = f(
x)
3
−
3
3
−
3
0
−
1
2
easy
hard
flat
y = f
0
(
x)
3
−
3
0
−
1
2
1
−
1
y =
sin(x)
y = x
x
A
B
O
1
C
D
sin(
x)
tan(
x)
y =
sin
(x)
x
π
2
π
1
−
1
x =
0
a =
0
x
> 0
a
> 0
x
< 0
a
< 0
rest
position
+
−
y = x
2
sin
1
x
N
A
B
H
a
b
c
O
H
A
B
C
D
h
r
R
θ
1000
2000
α
β
p
h
y = g(
x) = log
b
(x)
y = f (
x) = b
x
y = e
x
5
10
1
2
3
4
0
−
1
−
2
−
3
−
4
y =
ln(x)
y =
cosh(x)
y =
sinh(x)
y =
tanh(x)
y =
sech(x)
y =
csch(x)
y =
coth(x)
1
−
1
y = f(
x)
original
function
in
verse function
slop
e = 0 at (x, y)
slop
e is infinite at (y, x)
−
108
2
5
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y =
sin(x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y =
sin(x)
y =
sin(x), −
π
2
≤ x ≤
π
2
−
2
−
1
0
2
π
2
−
π
2
y =
sin
−1
(x)
y =
cos(x)
π
π
2
y =
cos
−1
(x)
−
π
2
1
x
α
β
y =
tan(x)
y =
tan(x)
1
y =
tan
−1
(x)
y =
sec(x)
y =
sec
−1
(x)
y =
csc
−1
(x)
y =
cot
−1
(x)
1
y =
cosh
−1
(x)
y =
sinh
−1
(x)
y =
tanh
−1
(x)
y =
sech
−1
(x)
y =
csch
−1
(x)
y =
coth
−1
(x)
(0
, 3)
(2
, −1)
(5
, 2)
(7
, 0)
(
−1, 44)
(0
, 1)
(1
, −12)
(2
, 305)
y =
1
2
(2
, 3)
y = f(
x)
y = g(
x)
a
b
c
a
b
c
s
c
0
c
1
(
a, f(a))
(
b, f(b))
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y =
sin(x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
c
OR
Lo
cal maximum
Lo
cal minimum
Horizon
tal point of inflection
1
e
y = f
0
(
x)
y = f(
x) = x ln(x)
−
1
e
?
y = f (
x) = x
3
y = g(
x) = x
4
x
f(
x)
−
3
−
2
−
1
0
1
2
1
2
3
4
+
−
?
1
5
6
3
f
0
(
x)
2 −
1
2
√
6
2
+
1
2
√
6
f
00
(
x)
7
8
g
00
(
x)
f
00
(
x)
0
y =
(
x − 3)(x − 1)
2
x
3
(
x + 2)
y = x ln(
x)
1
e
−
1
e
5
−
108
2
α
β
2 −
1
2
√
6
2
+
1
2
√
6
y = x
2
(
x − 5)
3
−
e
−
1/2
√
3
e
−
1/2
√
3
−
e
−3/2
e
−
3/2
−
1
√
3
1
√
3
−
1
1
y = xe
−
3x
2
/2
y =
x
3
− 6
x
2
+ 13x − 8
x
28
2
600
500
400
300
200
100
0
−
100
−
200
−
300
−
400
−
500
−
600
0
10
−
10
5
−
5
20
−
20
15
−
15
0
4
5
6
x
P
0
(
x)
+
−
−
existing
fence
new
fence
enclosure
A
h
b
H
99
100
101
h
dA/dh
r
h
1
2
7
shallo
w
deep
LAND
SEA
N
y
z
s
t
3
11
9
L
(11)
√
11
y = L
(x)
y = f(
x)
11
y = L
(x)
y = f(
x)
F
P
a
a +
∆x
f(
a + ∆x)
L
(a + ∆x)
f(
a)
error
d
f
∆
x
a
b
y = f(
x)
true
zero
starting
approximation
b
etter approximation
v
t
3
5
50
40
60
4
20
30
25
t
1
t
2
t
3
t
4
t
n
−2
t
n
−1
t
0
= a
t
n
= b
v
1
v
2
v
3
v
4
v
n
−1
v
n
−
30
6
30
|
v|
a
b
p
q
c
v(
c)
v(
c
1
)
v(
c
2
)
v(
c
3
)
v(
c
4
)
v(
c
5
)
v(
c
6
)
t
1
t
2
t
3
t
4
t
5
c
1
c
2
c
3
c
4
c
5
c
6
t
0
=
a
t
6
=
b
t
16
=
b
t
10
=
b
a
b
x
y
y = f(
x)
1
2
y = x
5
0
−
2
y =
1
a
b
y =
sin(x)
π
−
π
0
−
1
−
2
0
2
4
y = x
2
0
1
2
3
4
2
n
4
n
6
n
2(
n−2)
n
2(
n−1)
n
2
n
n
=
2
width
of each interval =
2
n
−
2
1
3
0
I
I
I
I
II
IV
4
y
dx
y = −
x
2
− 2x + 3
3
−
5
y = |−
x
2
− 2x + 3|
I
I
I
I
Ia
5
3
0
1
2
a
b
y = f(
x)
y = g(
x)
y = x
2
a
b
5
3
0
1
2
y =
√
x
2
√
2
2
2
dy
x
2
a
b
y = f(
x)
y = g(
x)
M
m
1
2
−
1
−
2
0
y = e
−
x
2
1
2
e
−
1/4
f
a
v
y = f
a
v
c
A
M
0
1
2
a
b
x
t
y = f(t)
F (x )
y = f(t)
F (x + h )
x + h
F (x + h) − F (x)
f(x)
1
2
y = sin(x)
π
−π
−1
−2
y =
1
x
y = x
2
1
2
1
−1
y = ln|x|
θ
a
x
a
x
p
a
2
− x
2
3
x
p
9 − x
2
p
x
2
+ a
2
x
a
p
x
2
+ 15
x
√
15
x
p
x
2
− a
2
a
x
p
x
2
− 4
2
x
−
p
x
2
− a
2
a
x
−
p
x
2
− 4
2
y = f(x)
a b
a + ε
ε
Z
b
a+ε
f(x) dx
small
even smaller
y = g(x)
infinite area
finite area
The shaded area under y = g(x) between x = a and x = b is assumed to be
finite. You can clearly see from the picture that the area we want, which is
under y = f(x) between x = a and x = b, is less than the finite shaded area.
Since the area we want is positive and less than a finite number, it must also
be finite.
Beware: suppose you know that
R
b
a
g(x) dx converges, but you have the
inequality f (x) ≥ g(x) instead. Now the curve you want (y = f(x)) sits above
the other curve (y = g(x)). This is no good: you’d only be able to say that
Z
b
a
f(x) dx ≥
Z
b
a
g(x) dx.
So the integral we are interested in on the left-hand side is greater than
or equal to some finite number. Our integral is therefore finite or infinite.
Great—no info whatsoever. Hey, we’re in the whoop-di-doo case again! So
don’t go there.
It’s true that I haven’t really justified the comparison test, mathematically
PSfrag
replacements
(
a, b)
[
a, b]
(
a, b]
[
a, b)
(
a, ∞)
[
a, ∞)
(
−∞, b)
(
−∞, b]
(
−∞, ∞)
{
x : a < x < b}
{
x : a ≤ x ≤ b}
{
x : a < x ≤ b}
{
x : a ≤ x < b}
{
x : x ≥ a}
{
x : x > a}
{
x : x ≤ b}
{
x : x < b}
R
a
b
shado
w
0
1
4
−
2
3
−
3
g(
x) = x
2
f(
x) = x
3
g(
x) = x
2
f(
x) = x
3
mirror
(y = x)
f
−
1
(x) =
3
√
x
y = h
(x)
y = h
−
1
(x)
y =
(x − 1)
2
−
1
x
Same
height
−
x
Same
length,
opp
osite signs
y = −
2x
−
2
1
y =
1
2
x − 1
2
−
1
y =
2
x
y =
10
x
y =
2
−x
y =
log
2
(x)
4
3
units
mirror
(x-axis)
y = |
x|
y = |
log
2
(x)|
θ radians
θ units
30
◦
=
π
6
45
◦
=
π
4
60
◦
=
π
3
120
◦
=
2
π
3
135
◦
=
3
π
4
150
◦
=
5
π
6
90
◦
=
π
2
180
◦
= π
210
◦
=
7
π
6
225
◦
=
5
π
4
240
◦
=
4
π
3
270
◦
=
3
π
2
300
◦
=
5
π
3
315
◦
=
7
π
4
330
◦
=
11
π
6
0
◦
=
0 radians
θ
hypotenuse
opp
osite
adjacen
t
0
(≡ 2π)
π
2
π
3
π
2
I
I
I
I
II
IV
θ
(
x, y)
x
y
r
7
π
6
reference
angle
reference
angle =
π
6
sin
+
sin −
cos
+
cos −
tan
+
tan −
A
S
T
C
7
π
4
9
π
13
5
π
6
(this
angle is
5π
6
clo
ckwise)
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y =
sin(x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y =
sin(x)
y =
cos(x)
−
π
2
π
2
y =
tan(x), −
π
2
<
x <
π
2
0
−
π
2
π
2
y =
tan(x)
−
2π
−
3π
−
5
π
2
−
3
π
2
−
π
−
π
2
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
y =
sec(x)
y =
csc(x)
y =
cot(x)
y = f(
x)
−
1
1
2
y = g(
x)
3
y = h
(x)
4
5
−
2
f(
x) =
1
x
g(
x) =
1
x
2
etc.
0
1
π
1
2
π
1
3
π
1
4
π
1
5
π
1
6
π
1
7
π
g(
x) = sin
1
x
1
0
−
1
L
10
100
200
y =
π
2
y = −
π
2
y =
tan
−1
(x)
π
2
π
y =
sin(
x)
x
,
x > 3
0
1
−
1
a
L
f(
x) = x sin (1/x)
(0 <
x < 0.3)
h
(x) = x
g(
x) = −x
a
L
lim
x
→a
+
f(x) = L
lim
x
→a
+
f(x) = ∞
lim
x
→a
+
f(x) = −∞
lim
x
→a
+
f(x) DNE
lim
x
→a
−
f(x) = L
lim
x
→a
−
f(x) = ∞
lim
x
→a
−
f(x) = −∞
lim
x
→a
−
f(x) DNE
M
}
lim
x
→a
−
f(x) = M
lim
x
→a
f(x) = L
lim
x
→a
f(x) DNE
lim
x
→∞
f(x) = L
lim
x
→∞
f(x) = ∞
lim
x
→∞
f(x) = −∞
lim
x
→∞
f(x) DNE
lim
x
→−∞
f(x) = L
lim
x
→−∞
f(x) = ∞
lim
x
→−∞
f(x) = −∞
lim
x
→−∞
f(x) DNE
lim
x →a
+
f(
x) = ∞
lim
x →a
+
f(
x) = −∞
lim
x →a
−
f(
x) = ∞
lim
x →a
−
f(
x) = −∞
lim
x →a
f(
x) = ∞
lim
x →a
f(
x) = −∞
lim
x →a
f(
x) DNE
y = f (
x)
a
y =
|
x|
x
1
−
1
y =
|
x + 2|
x +
2
1
−
1
−
2
1
2
3
4
a
a
b
y = x sin
1
x
y = x
y = −
x
a
b
c
d
C
a
b
c
d
−
1
0
1
2
3
time
y
t
u
(
t, f(t))
(
u, f(u))
time
y
t
u
y
x
(
x, f(x))
y = |
x|
(
z, f(z))
z
y = f(
x)
a
tangen
t at x = a
b
tangen
t at x = b
c
tangen
t at x = c
y = x
2
tangen
t
at x = −
1
u
v
uv
u +
∆u
v +
∆v
(
u + ∆u)(v + ∆v)
∆
u
∆
v
u
∆v
v∆
u
∆
u∆v
y = f(
x)
1
2
−
2
y = |
x
2
− 4|
y = x
2
− 4
y = −
2x + 5
y = g(
x)
1
2
3
4
5
6
7
8
9
0
−
1
−
2
−
3
−
4
−
5
−
6
y = f (
x)
3
−
3
3
−
3
0
−
1
2
easy
hard
flat
y = f
0
(
x)
3
−
3
0
−
1
2
1
−
1
y =
sin(x)
y = x
x
A
B
O
1
C
D
sin(
x)
tan(
x)
y =
sin(
x)
x
π
2
π
1
−
1
x =
0
a =
0
x
> 0
a
> 0
x
< 0
a
< 0
rest
position
+
−
y = x
2
sin
1
x
N
A
B
H
a
b
c
O
H
A
B
C
D
h
r
R
θ
1000
2000
α
β
p
h
y = g(
x) = log
b
(x)
y = f(
x) = b
x
y = e
x
5
10
1
2
3
4
0
−
1
−
2
−
3
−
4
y =
ln(x)
y =
cosh(x)
y =
sinh(x)
y =
tanh(x)
y =
sech(x)
y =
csch(x)
y =
coth(x)
1
−
1
y = f(
x)
original
function
in
verse function
slop
e = 0 at (x, y)
slop
e is infinite at (y, x)
−
108
2
5
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y =
sin(x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y =
sin(x)
y =
sin(x), −
π
2
≤ x ≤
π
2
−
2
−
1
0
2
π
2
−
π
2
y =
sin
−1
(x)
y =
cos(x)
π
π
2
y =
cos
−1
(x)
−
π
2
1
x
α
β
y =
tan(x)
y =
tan(x)
1
y =
tan
−1
(x)
y =
sec(x)
y =
sec
−1
(x)
y =
csc
−1
(x)
y =
cot
−1
(x)
1
y =
cosh
−1
(x)
y =
sinh
−1
(x)
y =
tanh
−1
(x)
y =
sech
−1
(x)
y =
csch
−1
(x)
y =
coth
−1
(x)
(0
, 3)
(2
, −1)
(5
, 2)
(7
, 0)
(
−1, 44)
(0
, 1)
(1
, −12)
(2
, 305)
y =
1
2
(2
, 3)
y = f(
x)
y = g(
x)
a
b
c
a
b
c
s
c
0
c
1
(
a, f(a))
(
b, f(b))
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y =
sin(x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
c
OR
Lo
cal maximum
Lo
cal minimum
Horizon
tal point of inflection
1
e
y = f
0
(
x)
y = f (
x) = x ln(x)
−
1
e
?
y = f(
x) = x
3
y = g(
x) = x
4
x
f(
x)
−
3
−
2
−
1
0
1
2
1
2
3
4
+
−
?
1
5
6
3
f
0
(
x)
2 −
1
2
√
6
2
+
1
2
√
6
f
00
(
x)
7
8
g
00
(
x)
f
00
(
x)
0
y =
(
x − 3)(x − 1)
2
x
3
(
x + 2)
y = x ln
(x)
1
e
−
1
e
5
−
108
2
α
β
2 −
1
2
√
6
2
+
1
2
√
6
y = x
2
(
x − 5)
3
−
e
−
1/2
√
3
e
−
1/2
√
3
−
e
−3/2
e
−
3/2
−
1
√
3
1
√
3
−
1
1
y = xe
−
3x
2
/2
y =
x
3
− 6
x
2
+ 13x − 8
x
28
2
600
500
400
300
200
100
0
−
100
−
200
−
300
−
400
−
500
−
600
0
10
−
10
5
−
5
20
−
20
15
−
15
0
4
5
6
x
P
0
(
x)
+
−
−
existing
fence
new
fence
enclosure
A
h
b
H
99
100
101
h
dA/dh
r
h
1
2
7
shallo
w
deep
LAND
SEA
N
y
z
s
t
3
11
9
L
(11)
√
11
y = L
(x)
y = f (
x)
11
y = L
(x)
y = f(
x)
F
P
a
a +
∆x
f(
a + ∆x)
L
(a + ∆x)
f(
a)
error
d
f
∆
x
a
b
y = f(
x)
true
zero
starting
approximation
b
etter approximation
v
t
3
5
50
40
60
4
20
30
25
t
1
t
2
t
3
t
4
t
n
−2
t
n
−1
t
0
= a
t
n
= b
v
1
v
2
v
3
v
4
v
n
−1
v
n
−
30
6
30
|
v|
a
b
p
q
c
v(
c)
v(
c
1
)
v(
c
2
)
v(
c
3
)
v(
c
4
)
v(
c
5
)
v(
c
6
)
t
1
t
2
t
3
t
4
t
5
c
1
c
2
c
3
c
4
c
5
c
6
t
0
=
a
t
6
=
b
t
16
=
b
t
10
=
b
a
b
x
y
y = f(
x)
1
2
y = x
5
0
−
2
y =
1
a
b
y =
sin(x)
π
−
π
0
−
1
−
2
0
2
4
y = x
2
0
1
2
3
4
2
n
4
n
6
n
2(
n−2)
n
2(
n−1)
n
2
n
n
=
2
width
of each interval =
2
n
−
2
1
3
0
I
I
I
I
II
IV
4
y
dx
y = −
x
2
− 2x + 3
3
−
5
y = |−
x
2
− 2x + 3|
I
I
I
I
Ia
5
3
0
1
2
a
b
y = f (
x)
y = g(
x)
y = x
2
a
b
5
3
0
1
2
y =
√
x
2
√
2
2
2
dy
x
2
a
b
y = f(x)
y = g(x)
M
m
1
2
−1
−2
0
y = e
−x
2
1
2
e
−1/4
f
av
y = f
av
c
A
M
0
1
2
a
b
x
t
y = f (t)
F (x )
y = f (t)
F (x + h)
x + h
F (x + h) − F (x)
f(x)
1
2
y = sin(x)
π
−π
−1
−2
y =
1
x
y = x
2
1
2
1
−1
y = ln|x|
θ
a
x
a
x
p
a
2
− x
2
3
x
p
9 − x
2
p
x
2
+ a
2
x
a
p
x
2
+ 15
x
√
15
x
p
x
2
− a
2
a
x
p
x
2
− 4
2
x
−
p
x
2
− a
2
a
x
−
p
x
2
− 4
2
y = f(x)
a
b
a + ε
ε
Z
b
a+ε
f(x) dx
small
even smaller
y = g(x)
infinite area
finite area
speaking. Actually there’s not that much to it. Some splitting of integrals
(and hairs) is required, but we’ve already seen the basic idea. For example, if
f and g both have a vertical asymptote at x = a and have no blow-up points
anywhere else, and 0 ≤ f (x) ≤ g(x) for all x in the interval [a, b], then we can
Section 20.4: The Limit Comparison Test (Theory) • 441
say that
0 ≤
Z
b
a+ε
f(x) dx ≤
Z
b
a+ε
g(x) dx
for any ε > 0. Now take limits. If the improper integral
R
b
a
g(x) dx converges,
then the right-hand side becomes finite. Everything now depends on the
middle integral. Since f(x) is always positive, the middle integral gets bigger
as ε tends toward 0 from above. It’s getting bigger and bigger, but it can’t
get past the barrier at
R
b
a
g(x) dx, which is a nice finite number. The only
possibility is that the middle integral converges to some finite number as
ε → 0
+
.
∗
In other words,
R
b
a
f(x) dx converges. That proves the comparison
test in its convergence version (the second of the two versions we looked at
above), in the special case where f and g only have problems at x = a. It is
now up to you to prove the divergence version and also to work out how to
PSfrag replacements
(
a, b)
[
a, b]
(
a, b]
[
a, b)
(
a, ∞)
[
a, ∞)
(
−∞, b)
(
−∞, b]
(
−∞, ∞)
{
x : a < x < b}
{
x : a ≤ x ≤ b}
{
x : a < x ≤ b}
{
x : a ≤ x < b}
{
x : x ≥ a}
{
x : x > a}
{
x : x ≤ b}
{
x : x < b}
R
a
b
shadow
0
1
4
−
2
3
−
3
g(
x) = x
2
f(
x) = x
3
g(
x) = x
2
f(
x) = x
3
mirror (
y = x)
f
−
1
(x) =
3
√
x
y = h
(x)
y = h
−
1
(x)
y = (
x − 1)
2
−
1
x
Same height
−
x
Same length,
opposite signs
y = −
2x
−
2
1
y =
1
2
x − 1
2
−
1
y = 2
x
y = 10
x
y = 2
−
x
y = log
2
(
x)
4
3 units
mirror (
x-axis)
y = |
x|
y = |
log
2
(x)|
θ radians
θ units
30
◦
=
π
6
45
◦
=
π
4
60
◦
=
π
3
120
◦
=
2
π
3
135
◦
=
3
π
4
150
◦
=
5
π
6
90
◦
=
π
2
180
◦
= π
210
◦
=
7
π
6
225
◦
=
5
π
4
240
◦
=
4
π
3
270
◦
=
3
π
2
300
◦
=
5
π
3
315
◦
=
7
π
4
330
◦
=
11
π
6
0
◦
= 0 radians
θ
hypotenuse
opposite
adjacent
0 (
≡ 2π)
π
2
π
3
π
2
I
II
III
IV
θ
(
x, y)
x
y
r
7
π
6
reference angle
reference angle =
π
6
sin +
sin −
cos +
cos −
tan +
tan −
A
S
T
C
7
π
4
9
π
13
5
π
6
(this angle is
5
π
6
clockwise)
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y = sin(
x)
y = cos(
x)
−
π
2
π
2
y = tan(
x), −
π
2
< x <
π
2
0
−
π
2
π
2
y = tan(
x)
−
2π
−
3π
−
5
π
2
−
3
π
2
−
π
−
π
2
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
y = sec(
x)
y = csc(
x)
y = cot(
x)
y = f(
x)
−
1
1
2
y = g(
x)
3
y = h
(x)
4
5
−
2
f(
x) =
1
x
g(
x) =
1
x
2
etc.
0
1
π
1
2
π
1
3
π
1
4
π
1
5
π
1
6
π
1
7
π
g(
x) = sin
1
x
1
0
−
1
L
10
100
200
y =
π
2
y = −
π
2
y = tan
−
1
(x)
π
2
π
y =
sin(
x)
x
, x > 3
0
1
−
1
a
L
f(
x) = x sin (1/x)
(0 < x < 0
.3)
h
(x) = x
g(
x) = −x
a
L
lim
x
→a
+
f(x) = L
lim
x
→a
+
f(x) = ∞
lim
x
→a
+
f(x) = −∞
lim
x
→a
+
f(x) DNE
lim
x
→a
−
f(x) = L
lim
x
→a
−
f(x) = ∞
lim
x
→a
−
f(x) = −∞
lim
x
→a
−
f(x) DNE
M
}
lim
x
→a
−
f(x) = M
lim
x
→a
f(x) = L
lim
x
→a
f(x) DNE
lim
x
→∞
f(x) = L
lim
x
→∞
f(x) = ∞
lim
x
→∞
f(x) = −∞
lim
x
→∞
f(x) DNE
lim
x
→−∞
f(x) = L
lim
x
→−∞
f(x) = ∞
lim
x
→−∞
f(x) = −∞
lim
x
→−∞
f(x) DNE
lim
x →a
+
f(
x) = ∞
lim
x →a
+
f(
x) = −∞
lim
x →a
−
f(
x) = ∞
lim
x →a
−
f(
x) = −∞
lim
x →a
f(
x) = ∞
lim
x →a
f(
x) = −∞
lim
x →a
f(
x) DNE
y = f (
x)
a
y =
|
x|
x
1
−
1
y =
|
x + 2|
x + 2
1
−
1
−
2
1
2
3
4
a
a
b
y = x sin
1
x
y = x
y = −
x
a
b
c
d
C
a
b
c
d
−
1
0
1
2
3
time
y
t
u
(
t, f(t))
(
u, f(u))
time
y
t
u
y
x
(
x, f(x))
y = |
x|
(
z, f(z))
z
y = f(
x)
a
tangent at x = a
b
tangent at x = b
c
tangent at x = c
y = x
2
tangent
at x = −
1
u
v
uv
u + ∆
u
v + ∆
v
(
u + ∆u)(v + ∆v)
∆
u
∆
v
u
∆v
v∆
u
∆
u∆v
y = f(
x)
1
2
−
2
y = |
x
2
− 4|
y = x
2
− 4
y = −
2x + 5
y = g(
x)
1
2
3
4
5
6
7
8
9
0
−
1
−
2
−
3
−
4
−
5
−
6
y = f (
x)
3
−
3
3
−
3
0
−
1
2
easy
hard
flat
y = f
0
(
x)
3
−
3
0
−
1
2
1
−
1
y = sin(
x)
y = x
x
A
B
O
1
C
D
sin(
x)
tan(
x)
y =
sin(
x)
x
π
2
π
1
−
1
x = 0
a = 0
x > 0
a > 0
x < 0
a < 0
rest position
+
−
y = x
2
sin
1
x
N
A
B
H
a
b
c
O
H
A
B
C
D
h
r
R
θ
1000
2000
α
β
p
h
y = g(
x) = log
b
(x)
y = f(
x) = b
x
y = e
x
5
10
1
2
3
4
0
−
1
−
2
−
3
−
4
y = ln(
x)
y = cosh(
x)
y = sinh(
x)
y = tanh(
x)
y = sech(
x)
y = csch(
x)
y = coth(
x)
1
−
1
y = f(
x)
original function
inverse function
slope = 0 at (
x, y)
slope is infinite at (
y, x)
−
108
2
5
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y = sin(
x)
y = sin(
x), −
π
2
≤ x ≤
π
2
−
2
−
1
0
2
π
2
−
π
2
y = sin
−
1
(x)
y = cos(
x)
π
π
2
y = cos
−
1
(x)
−
π
2
1
x
α
β
y = tan(
x)
y = tan(
x)
1
y = tan
−
1
(x)
y = sec(
x)
y = sec
−
1
(x)
y = csc
−
1
(x)
y = cot
−
1
(x)
1
y = cosh
−
1
(x)
y = sinh
−
1
(x)
y = tanh
−
1
(x)
y = sech
−
1
(x)
y = csch
−
1
(x)
y = coth
−
1
(x)
(0
, 3)
(2
, −1)
(5
, 2)
(7
, 0)
(
−1, 44)
(0
, 1)
(1
, −12)
(2
, 305)
y = 1
2
(2
, 3)
y = f(
x)
y = g(
x)
a
b
c
a
b
c
s
c
0
c
1
(
a, f(a))
(
b, f(b))
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
c
OR
Local maximum
Local minimum
Horizontal point of inflection
1
e
y = f
0
(
x)
y = f (
x) = x ln(x)
−
1
e
?
y = f(
x) = x
3
y = g(
x) = x
4
x
f(
x)
−
3
−
2
−
1
0
1
2
1
2
3
4
+
−
?
1
5
6
3
f
0
(
x)
2 −
1
2
√
6
2 +
1
2
√
6
f
00
(
x)
7
8
g
00
(
x)
f
00
(
x)
0
y =
(
x − 3)(x − 1)
2
x
3
(
x + 2)
y = x ln(
x)
1
e
−
1
e
5
−
108
2
α
β
2 −
1
2
√
6
2 +
1
2
√
6
y = x
2
(
x − 5)
3
−
e
−
1/2
√
3
e
−
1/2
√
3
−
e
−3/2
e
−
3/2
−
1
√
3
1
√
3
−
1
1
y = xe
−
3x
2
/2
y =
x
3
− 6
x
2
+ 13x − 8
x
28
2
600
500
400
300
200
100
0
−
100
−
200
−
300
−
400
−
500
−
600
0
10
−
10
5
−
5
20
−
20
15
−
15
0
4
5
6
x
P
0
(
x)
+
−
−
existing fence
new fence
enclosure
A
h
b
H
99
100
101
h
dA/dh
r
h
1
2
7
shallow
deep
LAND
SEA
N
y
z
s
t
3
11
9
L
(11)
√
11
y = L
(x)
y = f (
x)
11
y = L
(x)
y = f(
x)
F
P
a
a + ∆
x
f(
a + ∆x)
L
(a + ∆x)
f(
a)
error
df
∆
x
a
b
y = f(
x)
true zero
starting approximation
better approximation
v
t
3
5
50
40
60
4
20
30
25
t
1
t
2
t
3
t
4
t
n
−2
t
n
−1
t
0
= a
t
n
= b
v
1
v
2
v
3
v
4
v
n
−1
v
n
−
30
6
30
|
v|
a
b
p
q
c
v(
c)
v(
c
1
)
v(
c
2
)
v(
c
3
)
v(
c
4
)
v(
c
5
)
v(
c
6
)
t
1
t
2
t
3
t
4
t
5
c
1
c
2
c
3
c
4
c
5
c
6
t
0
=
a
t
6
=
b
t
16
=
b
t
10
=
b
a
b
x
y
y = f(
x)
1
2
y = x
5
0
−
2
y = 1
a
b
y = sin(
x)
π
−
π
0
−
1
−
2
0
2
4
y = x
2
0
1
2
3
4
2
n
4
n
6
n
2(
n−2)
n
2(
n−1)
n
2
n
n
= 2
width of each interval =
2
n
−
2
1
3
0
I
II
III
IV
4
y
dx
y = −
x
2
− 2x + 3
3
−
5
y = |−
x
2
− 2x + 3|
I
II
IIa
5
3
0
1
2
a
b
y = f (
x)
y = g(
x)
y = x
2
a
b
5
3
0
1
2
y =
√
x
2
√
2
2
2
dy
x
2
a
b
y = f(
x)
y = g(
x)
M
m
1
2
−
1
−
2
0
y = e
−
x
2
1
2
e
−
1/4
f
av
y = f
av
c
A
M
0
1
2
a
b
x
t
y = f (
t)
F (
x )
y = f (
t)
F (
x + h)
x + h
F (x + h) − F (x)
f(x)
1
2
y = sin(x)
π
−π
−1
−2
y =
1
x
y = x
2
1
2
1
−1
y = ln|x|
θ
a
x
a
x
p
a
2
− x
2
3
x
p
9 − x
2
p
x
2
+ a
2
x
a
p
x
2
+ 15
x
√
15
x
p
x
2
− a
2
a
x
p
x
2
− 4
2
x
−
p
x
2
− a
2
a
x
−
p
x
2
− 4
2
y = f(x)
a
b
a + ε
ε
Z
b
a+ε
f(x) dx
small
even smaller
y = g(x)
infinite area
finite area
deal with problems at x = b. There’s really not much difference. Of course,
if the problems are in the middle somewhere, or there are multiple problems,
you have to split the integral into pieces before using the comparison test
anyway.
We’ll look at many examples involving the comparison test in the next
chapter. Now it’s time to look at another test.
20.4 The Limit Comparison Test (Theory)
The comparison test uses the improper integral of one function to get infor-
mation about an improper integral of another function. The limit comparison
test does the same thing, except that we don’t actually need one function to
be bigger than the other. Instead, we need the two functions to be just about
the same. Here’s the basic idea: suppose that two functions f and g are very
close to each other at the blow-up point x = a (and have no other blow-up
points). Then
R
b
a
f(x) dx and
R
b
a
g(x) dx either both diverge or both converge.
Their behavior is identical. Intuitively, it makes sense; let’s get down to de-
tails by specifying what we really mean when we say that two functions are
“very close” to each other.
20.4.1 Functions asymptotic to each other
Suppose we have two functions f and g such that
lim
x→a
f(x)
g(x)
= 1.
This means that when x is near a, the ratio f (x)/g(x) is close to 1. If the
ratio were equal to 1, then f(x) would equal g(x). Since the ratio is only close
to 1, then f(x) is “very close” to g(x). This doesn’t mean that the difference
between f (x) and g(x) is small! For example, f(x) could be a trillion and g(x)
∗
Actually, this statement, which seems obvious, is quite profound. The statement is
pretty much what distinguishes R from any of its proper subsets which contain every rational
number.
442 • Improper Integrals: Basic Concepts
could be a trillion plus a million (for the same value of x); in that case, the
ratio f(x)/g(x) would be a little under 1, while the difference between f(x)
and g(x) is still a million! On the other hand, the two numbers are relatively
close to each other, since a million is a small difference relative to the size of
the numbers.
So, we’ll say that f(x) ∼ g(x) as x → a if the limit of the ratio is 1. That
is,
f(x) ∼ g(x) as x → a means the same thing as lim
x→a
f(x)
g(x)
= 1.
This doesn’t mean that f(x) is approximately equal to g(x) when x is near
a: it means that the ratio of f(x) to g(x) is near 1 when x is near a. We say
that f and g are asymptotic to each other as x → a. Of course, you could
replace x → a by x → ∞, or even x → a
+
; all you have to do is make the
same replacement in the limit too.
All this is useless unless we have limits of the form
lim
x→a
f(x)
g(x)
= 1.
Actually, we’ve seen many of these types of limits! Here are some examples:
∗
PSfrag
replacements
(
a, b)
[
a, b]
(
a, b]
[
a, b)
(
a, ∞)
[
a, ∞)
(
−∞, b)
(
−∞, b]
(
−∞, ∞)
{
x : a < x < b}
{
x : a ≤ x ≤ b}
{
x : a < x ≤ b}
{
x : a ≤ x < b}
{
x : x ≥ a}
{
x : x > a}
{
x : x ≤ b}
{
x : x < b}
R
a
b
shado
w
0
1
4
−
2
3
−
3
g(
x) = x
2
f(
x) = x
3
g(
x) = x
2
f(
x) = x
3
mirror
(y = x)
f
−
1
(x) =
3
√
x
y = h
(x)
y = h
−
1
(x)
y =
(x − 1)
2
−
1
x
Same
height
−
x
Same
length,
opp
osite signs
y = −
2x
−
2
1
y =
1
2
x − 1
2
−
1
y =
2
x
y =
10
x
y =
2
−x
y =
log
2
(x)
4
3
units
mirror
(x-axis)
y = |
x|
y = |
log
2
(x)|
θ radians
θ units
30
◦
=
π
6
45
◦
=
π
4
60
◦
=
π
3
120
◦
=
2
π
3
135
◦
=
3
π
4
150
◦
=
5
π
6
90
◦
=
π
2
180
◦
= π
210
◦
=
7
π
6
225
◦
=
5
π
4
240
◦
=
4
π
3
270
◦
=
3
π
2
300
◦
=
5
π
3
315
◦
=
7
π
4
330
◦
=
11
π
6
0
◦
=
0 radians
θ
hypotenuse
opp
osite
adjacen
t
0
(≡ 2π)
π
2
π
3
π
2
I
I
I
I
II
IV
θ
(
x, y)
x
y
r
7
π
6
reference
angle
reference
angle =
π
6
sin
+
sin −
cos
+
cos −
tan
+
tan −
A
S
T
C
7
π
4
9
π
13
5
π
6
(this
angle is
5π
6
clo
ckwise)
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y =
sin(x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y =
sin(x)
y =
cos(x)
−
π
2
π
2
y =
tan(x), −
π
2
<
x <
π
2
0
−
π
2
π
2
y =
tan(x)
−
2π
−
3π
−
5
π
2
−
3
π
2
−
π
−
π
2
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
y =
sec(x)
y =
csc(x)
y =
cot(x)
y = f(
x)
−
1
1
2
y = g(
x)
3
y = h
(x)
4
5
−
2
f(
x) =
1
x
g(
x) =
1
x
2
etc.
0
1
π
1
2
π
1
3
π
1
4
π
1
5
π
1
6
π
1
7
π
g(
x) = sin
1
x
1
0
−
1
L
10
100
200
y =
π
2
y = −
π
2
y =
tan
−1
(x)
π
2
π
y =
sin(
x)
x
,
x > 3
0
1
−
1
a
L
f(
x) = x sin (1/x)
(0 <
x < 0.3)
h
(x) = x
g(
x) = −x
a
L
lim
x
→a
+
f(x) = L
lim
x
→a
+
f(x) = ∞
lim
x
→a
+
f(x) = −∞
lim
x
→a
+
f(x) DNE
lim
x
→a
−
f(x) = L
lim
x
→a
−
f(x) = ∞
lim
x
→a
−
f(x) = −∞
lim
x
→a
−
f(x) DNE
M
}
lim
x
→a
−
f(x) = M
lim
x
→a
f(x) = L
lim
x
→a
f(x) DNE
lim
x
→∞
f(x) = L
lim
x
→∞
f(x) = ∞
lim
x
→∞
f(x) = −∞
lim
x
→∞
f(x) DNE
lim
x
→−∞
f(x) = L
lim
x
→−∞
f(x) = ∞
lim
x
→−∞
f(x) = −∞
lim
x
→−∞
f(x) DNE
lim
x →a
+
f(
x) = ∞
lim
x →a
+
f(
x) = −∞
lim
x →a
−
f(
x) = ∞
lim
x →a
−
f(
x) = −∞
lim
x →a
f(
x) = ∞
lim
x →a
f(
x) = −∞
lim
x →a
f(
x) DNE
y = f (
x)
a
y =
|
x|
x
1
−
1
y =
|
x + 2|
x +
2
1
−
1
−
2
1
2
3
4
a
a
b
y = x sin
1
x
y = x
y = −
x
a
b
c
d
C
a
b
c
d
−
1
0
1
2
3
time
y
t
u
(
t, f(t))
(
u, f(u))
time
y
t
u
y
x
(
x, f(x))
y = |
x|
(
z, f(z))
z
y = f(
x)
a
tangen
t at x = a
b
tangen
t at x = b
c
tangen
t at x = c
y = x
2
tangen
t
at x = −
1
u
v
uv
u +
∆u
v +
∆v
(
u + ∆u)(v + ∆v)
∆
u
∆
v
u
∆v
v∆
u
∆
u∆v
y = f(
x)
1
2
−
2
y = |
x
2
− 4|
y = x
2
− 4
y = −
2x + 5
y = g(
x)
1
2
3
4
5
6
7
8
9
0
−
1
−
2
−
3
−
4
−
5
−
6
y = f (
x)
3
−
3
3
−
3
0
−
1
2
easy
hard
flat
y = f
0
(
x)
3
−
3
0
−
1
2
1
−
1
y =
sin(x)
y = x
x
A
B
O
1
C
D
sin(
x)
tan(
x)
y =
sin(
x)
x
π
2
π
1
−
1
x =
0
a =
0
x
> 0
a
> 0
x
< 0
a
< 0
rest
position
+
−
y = x
2
sin
1
x
N
A
B
H
a
b
c
O
H
A
B
C
D
h
r
R
θ
1000
2000
α
β
p
h
y = g(
x) = log
b
(x)
y = f(
x) = b
x
y = e
x
5
10
1
2
3
4
0
−
1
−
2
−
3
−
4
y =
ln(x)
y =
cosh(x)
y =
sinh(x)
y =
tanh(x)
y =
sech(x)
y =
csch(x)
y =
coth(x)
1
−
1
y = f(
x)
original
function
in
verse function
slop
e = 0 at (x, y)
slop
e is infinite at (y, x)
−
108
2
5
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y =
sin(x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y =
sin(x)
y =
sin(x), −
π
2
≤ x ≤
π
2
−
2
−
1
0
2
π
2
−
π
2
y =
sin
−1
(x)
y =
cos(x)
π
π
2
y =
cos
−1
(x)
−
π
2
1
x
α
β
y =
tan(x)
y =
tan(x)
1
y =
tan
−1
(x)
y =
sec(x)
y =
sec
−1
(x)
y =
csc
−1
(x)
y =
cot
−1
(x)
1
y =
cosh
−1
(x)
y =
sinh
−1
(x)
y =
tanh
−1
(x)
y =
sech
−1
(x)
y =
csch
−1
(x)
y =
coth
−1
(x)
(0
, 3)
(2
, −1)
(5
, 2)
(7
, 0)
(
−1, 44)
(0
, 1)
(1
, −12)
(2
, 305)
y =
1
2
(2
, 3)
y = f(
x)
y = g(
x)
a
b
c
a
b
c
s
c
0
c
1
(
a, f(a))
(
b, f(b))
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y =
sin(x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
c
OR
Lo
cal maximum
Lo
cal minimum
Horizon
tal point of inflection
1
e
y = f
0
(
x)
y = f (
x) = x ln(x)
−
1
e
?
y = f(
x) = x
3
y = g(
x) = x
4
x
f(
x)
−
3
−
2
−
1
0
1
2
1
2
3
4
+
−
?
1
5
6
3
f
0
(
x)
2 −
1
2
√
6
2
+
1
2
√
6
f
00
(
x)
7
8
g
00
(
x)
f
00
(
x)
0
y =
(
x − 3)(x − 1)
2
x
3
(
x + 2)
y = x ln
(x)
1
e
−
1
e
5
−
108
2
α
β
2 −
1
2
√
6
2
+
1
2
√
6
y = x
2
(
x − 5)
3
−
e
−
1/2
√
3
e
−
1/2
√
3
−
e
−3/2
e
−
3/2
−
1
√
3
1
√
3
−
1
1
y = xe
−
3x
2
/2
y =
x
3
− 6
x
2
+ 13x − 8
x
28
2
600
500
400
300
200
100
0
−
100
−
200
−
300
−
400
−
500
−
600
0
10
−
10
5
−
5
20
−
20
15
−
15
0
4
5
6
x
P
0
(
x)
+
−
−
existing
fence
new
fence
enclosure
A
h
b
H
99
100
101
h
dA/dh
r
h
1
2
7
shallo
w
deep
LAND
SEA
N
y
z
s
t
3
11
9
L
(11)
√
11
y = L
(x)
y = f (
x)
11
y = L
(x)
y = f(
x)
F
P
a
a +
∆x
f(
a + ∆x)
L
(a + ∆x)
f(
a)
error
d
f
∆
x
a
b
y = f(
x)
true
zero
starting
approximation
b
etter approximation
v
t
3
5
50
40
60
4
20
30
25
t
1
t
2
t
3
t
4
t
n
−2
t
n
−1
t
0
= a
t
n
= b
v
1
v
2
v
3
v
4
v
n
−1
v
n
−
30
6
30
|
v|
a
b
p
q
c
v(
c)
v(
c
1
)
v(
c
2
)
v(
c
3
)
v(
c
4
)
v(
c
5
)
v(
c
6
)
t
1
t
2
t
3
t
4
t
5
c
1
c
2
c
3
c
4
c
5
c
6
t
0
=
a
t
6
=
b
t
16
=
b
t
10
=
b
a
b
x
y
y = f(
x)
1
2
y = x
5
0
−
2
y =
1
a
b
y =
sin(x)
π
−
π
0
−
1
−
2
0
2
4
y = x
2
0
1
2
3
4
2
n
4
n
6
n
2(
n−2)
n
2(
n−1)
n
2
n
n
=
2
width
of each interval =
2
n
−
2
1
3
0
I
I
I
I
II
IV
4
y
dx
y = −
x
2
− 2x + 3
3
−
5
y = |−
x
2
− 2x + 3|
I
I
I
I
Ia
5
3
0
1
2
a
b
y = f (
x)
y = g(
x)
y = x
2
a
b
5
3
0
1
2
y =
√
x
2
√
2
2
2
dy
x
2
a
b
y = f(
x)
y = g(
x)
M
m
1
2
−
1
−
2
0
y = e
−
x
2
1
2
e
−
1/4
f
a
v
y = f
a
v
c
A
M
0
1
2
a
b
x
t
y = f (t)
F (x )
y = f (t)
F (x + h)
x + h
F (x + h) − F (x)
f(x)
1
2
y = sin(x)
π
−π
−1
−2
y =
1
x
y = x
2
1
2
1
−1
y = ln|x|
θ
a
x
a
x
p
a
2
− x
2
3
x
p
9 − x
2
p
x
2
+ a
2
x
a
p
x
2
+ 15
x
√
15
x
p
x
2
− a
2
a
x
p
x
2
− 4
2
x
−
p
x
2
− a
2
a
x
−
p
x
2
− 4
2
y = f(x)
a
b
a + ε
ε
Z
b
a+ε
f(x) dx
small
even smaller
y = g(x)
infinite area
finite area
lim
x→∞
3x
3
− 1000x
2
+ 5x − 7
3x
3
= 1, lim
x→0
sin(x)
x
= 1,
lim
x→0
e
x
− 1
x
= 1, and lim
x→0
ln(1 + x)
x
= 1.
The first limit above can be written as 3x
3
−1000x
2
+5x−7 ∼ 3x
3
as x → ∞.
That is, 3x
3
−1000x
2
+5x−7 and 3x
3
are asymptotic to each other as x → ∞.
Similarly, the second limit says that sin(x) ∼ x as x → 0. The third and fourth
limits show that e
x
−1 and ln(1 + x) are also both asymptotic to x as x → 0;
that is, e
x
− 1 ∼ x and ln(1 + x) ∼ x as x → 0.
All we’ve done is to rewrite each limit in a different form, but it is a
very convenient form. Indeed, you can take powers of asymptotic relations
and get new ones. For example, knowing that sin(x) ∼ x as x → 0, we can
immediately write that sin
3
(x) ∼ x
3
as x → 0, or even that 1/ sin(x) ∼ 1/x as
x → 0. You can also replace x by any other quantity that goes to 0 as x does,
such as a power of x. For example, starting with sin(x) ∼ x as x → 0 once
again, we can replace x by 4x
7
to see that sin(4x
7
) ∼ 4x
7
as x → 0. You can
even multiply or divide two relations by each other, provided that the limit is
at the same value of x for both asymptotic relations. For example, we know
that tan(x) ∼ x as x → 0 since
lim
x→0
tan(x)
x
= 1.
So we can multiply tan(x) ∼ x and sin(x) ∼ x (both as x → 0) together to
get the asymptotic relation tan(x) sin(x) ∼ x
2
as x → 0.
∗
The examples can be found in Section 4.3 of Chapter 4, Section 7.1.1 of Chapter 7,
and Sections 9.4.2 and Sections 9.4.3 of Chapter 9, respectively.
Section 20.4.2: The statement of the test • 443
What you cannot do is add or subtract these relations. For example, if
PSfrag replacements
(
a, b)
[
a, b]
(
a, b]
[
a, b)
(
a, ∞)
[
a, ∞)
(
−∞, b)
(
−∞, b]
(
−∞, ∞)
{
x : a < x < b}
{
x : a ≤ x ≤ b}
{
x : a < x ≤ b}
{
x : a ≤ x < b}
{
x : x ≥ a}
{
x : x > a}
{
x : x ≤ b}
{
x : x < b}
R
a
b
shadow
0
1
4
−
2
3
−
3
g(
x) = x
2
f(
x) = x
3
g(
x) = x
2
f(
x) = x
3
mirror (
y = x)
f
−
1
(x) =
3
√
x
y = h
(x)
y = h
−
1
(x)
y = (
x − 1)
2
−
1
x
Same height
−
x
Same length,
opposite signs
y = −
2x
−
2
1
y =
1
2
x − 1
2
−
1
y = 2
x
y = 10
x
y = 2
−
x
y = log
2
(
x)
4
3 units
mirror (
x-axis)
y = |
x|
y = |
log
2
(x)|
θ radians
θ units
30
◦
=
π
6
45
◦
=
π
4
60
◦
=
π
3
120
◦
=
2
π
3
135
◦
=
3
π
4
150
◦
=
5
π
6
90
◦
=
π
2
180
◦
= π
210
◦
=
7
π
6
225
◦
=
5
π
4
240
◦
=
4
π
3
270
◦
=
3
π
2
300
◦
=
5
π
3
315
◦
=
7
π
4
330
◦
=
11
π
6
0
◦
= 0 radians
θ
hypotenuse
opposite
adjacent
0 (
≡ 2π)
π
2
π
3
π
2
I
II
III
IV
θ
(
x, y)
x
y
r
7
π
6
reference angle
reference angle =
π
6
sin +
sin −
cos +
cos −
tan +
tan −
A
S
T
C
7
π
4
9
π
13
5
π
6
(this angle is
5
π
6
clockwise)
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y = sin(
x)
y = cos(
x)
−
π
2
π
2
y = tan(
x), −
π
2
< x <
π
2
0
−
π
2
π
2
y = tan(
x)
−
2π
−
3π
−
5
π
2
−
3
π
2
−
π
−
π
2
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
y = sec(
x)
y = csc(
x)
y = cot(
x)
y = f (
x)
−
1
1
2
y = g(
x)
3
y = h
(x)
4
5
−
2
f(
x) =
1
x
g(
x) =
1
x
2
etc.
0
1
π
1
2
π
1
3
π
1
4
π
1
5
π
1
6
π
1
7
π
g(
x) = sin
1
x
1
0
−
1
L
10
100
200
y =
π
2
y = −
π
2
y = tan
−
1
(x)
π
2
π
y =
sin(
x)
x
, x > 3
0
1
−
1
a
L
f(
x) = x sin (1/x)
(0 < x < 0
.3)
h
(x) = x
g(
x) = −x
a
L
lim
x
→a
+
f(x) = L
lim
x
→a
+
f(x) = ∞
lim
x
→a
+
f(x) = −∞
lim
x
→a
+
f(x) DNE
lim
x
→a
−
f(x) = L
lim
x
→a
−
f(x) = ∞
lim
x
→a
−
f(x) = −∞
lim
x
→a
−
f(x) DNE
M
}
lim
x
→a
−
f(x) = M
lim
x
→a
f(x) = L
lim
x
→a
f(x) DNE
lim
x
→∞
f(x) = L
lim
x
→∞
f(x) = ∞
lim
x
→∞
f(x) = −∞
lim
x
→∞
f(x) DNE
lim
x
→−∞
f(x) = L
lim
x
→−∞
f(x) = ∞
lim
x
→−∞
f(x) = −∞
lim
x
→−∞
f(x) DNE
lim
x →a
+
f(
x) = ∞
lim
x →a
+
f(
x) = −∞
lim
x →a
−
f(
x) = ∞
lim
x →a
−
f(
x) = −∞
lim
x →a
f(
x) = ∞
lim
x →a
f(
x) = −∞
lim
x →a
f(
x) DNE
y = f (
x)
a
y =
|
x|
x
1
−
1
y =
|
x + 2|
x + 2
1
−
1
−
2
1
2
3
4
a
a
b
y = x sin
1
x
y = x
y = −
x
a
b
c
d
C
a
b
c
d
−
1
0
1
2
3
time
y
t
u
(
t, f(t))
(
u, f(u))
time
y
t
u
y
x
(
x, f(x))
y = |
x|
(
z, f(z))
z
y = f (
x)
a
tangent at x = a
b
tangent at x = b
c
tangent at x = c
y = x
2
tangent
at x = −
1
u
v
uv
u + ∆
u
v + ∆
v
(
u + ∆u)(v + ∆v)
∆
u
∆
v
u
∆v
v∆
u
∆
u∆v
y = f (
x)
1
2
−
2
y = |
x
2
− 4|
y = x
2
− 4
y = −
2x + 5
y = g(
x)
1
2
3
4
5
6
7
8
9
0
−
1
−
2
−
3
−
4
−
5
−
6
y = f (
x)
3
−
3
3
−
3
0
−
1
2
easy
hard
flat
y = f
0
(
x)
3
−
3
0
−
1
2
1
−
1
y = sin(
x)
y = x
x
A
B
O
1
C
D
sin(
x)
tan(
x)
y =
sin(
x)
x
π
2
π
1
−
1
x = 0
a = 0
x > 0
a > 0
x < 0
a < 0
rest position
+
−
y = x
2
sin
1
x
N
A
B
H
a
b
c
O
H
A
B
C
D
h
r
R
θ
1000
2000
α
β
p
h
y = g(
x) = log
b
(x)
y = f (
x) = b
x
y = e
x
5
10
1
2
3
4
0
−
1
−
2
−
3
−
4
y = ln(
x)
y = cosh(
x)
y = sinh(
x)
y = tanh(
x)
y = sech(
x)
y = csch(
x)
y = coth(
x)
1
−
1
y = f (
x)
original function
inverse function
slope = 0 at (
x, y)
slope is infinite at (
y, x)
−
108
2
5
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y = sin(
x)
y = sin(
x), −
π
2
≤ x ≤
π
2
−
2
−
1
0
2
π
2
−
π
2
y = sin
−
1
(x)
y = cos(
x)
π
π
2
y = cos
−
1
(x)
−
π
2
1
x
α
β
y = tan(
x)
y = tan(
x)
1
y = tan
−
1
(x)
y = sec(
x)
y = sec
−
1
(x)
y = csc
−
1
(x)
y = cot
−
1
(x)
1
y = cosh
−
1
(x)
y = sinh
−
1
(x)
y = tanh
−
1
(x)
y = sech
−
1
(x)
y = csch
−
1
(x)
y = coth
−
1
(x)
(0
, 3)
(2
, −1)
(5
, 2)
(7
, 0)
(
−1, 44)
(0
, 1)
(1
, −12)
(2
, 305)
y = 1
2
(2
, 3)
y = f (
x)
y = g(
x)
a
b
c
a
b
c
s
c
0
c
1
(
a, f(a))
(
b, f(b))
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
c
OR
Local maximum
Local minimum
Horizontal point of inflection
1
e
y = f
0
(
x)
y = f(
x) = x ln(x)
−
1
e
?
y = f(
x) = x
3
y = g(
x) = x
4
x
f(
x)
−
3
−
2
−
1
0
1
2
1
2
3
4
+
−
?
1
5
6
3
f
0
(
x)
2 −
1
2
√
6
2 +
1
2
√
6
f
00
(
x)
7
8
g
00
(
x)
f
00
(
x)
0
y =
(
x − 3)(x − 1)
2
x
3
(
x + 2)
y = x ln(
x)
1
e
−
1
e
5
−
108
2
α
β
2 −
1
2
√
6
2 +
1
2
√
6
y = x
2
(
x − 5)
3
−
e
−
1/2
√
3
e
−
1/2
√
3
−
e
−3/2
e
−
3/2
−
1
√
3
1
√
3
−
1
1
y = xe
−
3x
2
/2
y =
x
3
− 6
x
2
+ 13x − 8
x
28
2
600
500
400
300
200
100
0
−
100
−
200
−
300
−
400
−
500
−
600
0
10
−
10
5
−
5
20
−
20
15
−
15
0
4
5
6
x
P
0
(
x)
+
−
−
existing fence
new fence
enclosure
A
h
b
H
99
100
101
h
dA/dh
r
h
1
2
7
shallow
deep
LAND
SEA
N
y
z
s
t
3
11
9
L
(11)
√
11
y = L
(x)
y = f (
x)
11
y = L
(x)
y = f (
x)
F
P
a
a + ∆
x
f(
a + ∆x)
L
(a + ∆x)
f(
a)
error
df
∆
x
a
b
y = f (
x)
true zero
starting approximation
better approximation
v
t
3
5
50
40
60
4
20
30
25
t
1
t
2
t
3
t
4
t
n
−2
t
n
−1
t
0
= a
t
n
= b
v
1
v
2
v
3
v
4
v
n
−1
v
n
−
30
6
30
|
v|
a
b
p
q
c
v(
c)
v(
c
1
)
v(
c
2
)
v(
c
3
)
v(
c
4
)
v(
c
5
)
v(
c
6
)
t
1
t
2
t
3
t
4
t
5
c
1
c
2
c
3
c
4
c
5
c
6
t
0
=
a
t
6
=
b
t
16
=
b
t
10
=
b
a
b
x
y
y = f (
x)
1
2
y = x
5
0
−
2
y = 1
a
b
y = sin(
x)
π
−
π
0
−
1
−
2
0
2
4
y = x
2
0
1
2
3
4
2
n
4
n
6
n
2(
n−2)
n
2(
n−1)
n
2
n
n
= 2
width of each interval =
2
n
−
2
1
3
0
I
II
III
IV
4
y
dx
y = −
x
2
− 2x + 3
3
−
5
y = |−
x
2
− 2x + 3|
I
II
IIa
5
3
0
1
2
a
b
y = f (
x)
y = g(
x)
y = x
2
a
b
5
3
0
1
2
y =
√
x
2
√
2
2
2
dy
x
2
a
b
y = f (
x)
y = g(
x)
M
m
1
2
−
1
−
2
0
y = e
−
x
2
1
2
e
−
1/4
f
av
y = f
av
c
A
M
0
1
2
a
b
x
t
y = f(
t)
F (
x )
y = f(
t)
F (
x + h )
x + h
F (
x + h) − F (x)
f(
x)
1
2
y = sin(
x)
π
−
π
−
1
−
2
y =
1
x
y = x
2
1
2
1
−1
y = ln|x|
θ
a
x
a
x
p
a
2
− x
2
3
x
p
9 − x
2
p
x
2
+ a
2
x
a
p
x
2
+ 15
x
√
15
x
p
x
2
− a
2
a
x
p
x
2
− 4
2
x
−
p
x
2
− a
2
a
x
−
p
x
2
− 4
2
y = f (x)
a
b
a + ε
ε
Z
b
a+ε
f(x) dx
small
even smaller
y = g(x)
infinite area
finite area
you start with tan(x) ∼ x and sin(x) ∼ x as x → 0, you can’t just subtract
the second relation from the first to get tan(x) −sin(x) ∼ x −x. Indeed, x −x
is just 0, and nothing can be asymptotic to 0. Why not? Well, if f(x) ∼ 0 as
x → a, then we’d need
lim
x→a
f(x)
0
= 1.
That’s clearly garbage, since the left-hand side doesn’t make any sense. So,
by all means, multiply, divide, and take powers of asymptotic relations, but
don’t add or subtract them.
20.4.2 The statement of the test
OK, so we have this notion of two functions being asymptotic to each other,
and we have some examples too (like sin(x) ∼ x as x → 0). So what? Well,
suppose you have some function f with a problem spot only at a, and you’re
trying to see if the improper integral
R
b
a
f(x) dx converges or diverges. If you
can find a function g which behaves like f when the argument x is near a,
then you can just replace f by g and see if
R
b
a
g(x) dx converges or diverges.
Whatever you find for g also holds for f.
More formally, if f(x) ∼ g(x) as x → a, and neither function has any
problem spots anywhere else on the interval [a, b], then the integrals
R
b
a
f(x) dx
and
R
b
a
g(x) dx both diverge or both converge. (If they both converge, then
the values they converge to may be different.) This is one case of the limit
comparison test. Here’s a sneak preview of its power; we’ll see many more
examples in the next chapter. Suppose we want to know whether
PSfrag replacements
(
a, b)
[
a, b]
(
a, b]
[
a, b)
(
a, ∞)
[
a, ∞)
(
−∞, b)
(
−∞, b]
(
−∞, ∞)
{
x : a < x < b}
{
x : a ≤ x ≤ b}
{
x : a < x ≤ b}
{
x : a ≤ x < b}
{
x : x ≥ a}
{
x : x > a}
{
x : x ≤ b}
{
x : x < b}
R
a
b
shadow
0
1
4
−
2
3
−
3
g(
x) = x
2
f(
x) = x
3
g(
x) = x
2
f(
x) = x
3
mirror (
y = x)
f
−
1
(x) =
3
√
x
y = h
(x)
y = h
−
1
(x)
y = (
x − 1)
2
−
1
x
Same height
−
x
Same length,
opposite signs
y = −
2x
−
2
1
y =
1
2
x − 1
2
−
1
y = 2
x
y = 10
x
y = 2
−
x
y = log
2
(
x)
4
3 units
mirror (
x-axis)
y = |
x|
y = |
log
2
(x)|
θ radians
θ units
30
◦
=
π
6
45
◦
=
π
4
60
◦
=
π
3
120
◦
=
2
π
3
135
◦
=
3
π
4
150
◦
=
5
π
6
90
◦
=
π
2
180
◦
= π
210
◦
=
7
π
6
225
◦
=
5
π
4
240
◦
=
4
π
3
270
◦
=
3
π
2
300
◦
=
5
π
3
315
◦
=
7
π
4
330
◦
=
11
π
6
0
◦
= 0 radians
θ
hypotenuse
opposite
adjacent
0 (
≡ 2π)
π
2
π
3
π
2
I
II
III
IV
θ
(
x, y)
x
y
r
7
π
6
reference angle
reference angle =
π
6
sin +
sin −
cos +
cos −
tan +
tan −
A
S
T
C
7
π
4
9
π
13
5
π
6
(this angle is
5
π
6
clockwise)
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y = sin(
x)
y = cos(
x)
−
π
2
π
2
y = tan(
x), −
π
2
< x <
π
2
0
−
π
2
π
2
y = tan(
x)
−
2π
−
3π
−
5
π
2
−
3
π
2
−
π
−
π
2
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
y = sec(
x)
y = csc(
x)
y = cot(
x)
y = f(
x)
−
1
1
2
y = g(
x)
3
y = h
(x)
4
5
−
2
f(
x) =
1
x
g(
x) =
1
x
2
etc.
0
1
π
1
2
π
1
3
π
1
4
π
1
5
π
1
6
π
1
7
π
g(
x) = sin
1
x
1
0
−
1
L
10
100
200
y =
π
2
y = −
π
2
y = tan
−
1
(x)
π
2
π
y =
sin(
x)
x
, x > 3
0
1
−
1
a
L
f(
x) = x sin (1/x)
(0 < x < 0
.3)
h
(x) = x
g(
x) = −x
a
L
lim
x
→a
+
f(x) = L
lim
x
→a
+
f(x) = ∞
lim
x
→a
+
f(x) = −∞
lim
x
→a
+
f(x) DNE
lim
x
→a
−
f(x) = L
lim
x
→a
−
f(x) = ∞
lim
x
→a
−
f(x) = −∞
lim
x
→a
−
f(x) DNE
M
}
lim
x
→a
−
f(x) = M
lim
x
→a
f(x) = L
lim
x
→a
f(x) DNE
lim
x
→∞
f(x) = L
lim
x
→∞
f(x) = ∞
lim
x
→∞
f(x) = −∞
lim
x
→∞
f(x) DNE
lim
x
→−∞
f(x) = L
lim
x
→−∞
f(x) = ∞
lim
x
→−∞
f(x) = −∞
lim
x
→−∞
f(x) DNE
lim
x →a
+
f(
x) = ∞
lim
x →a
+
f(
x) = −∞
lim
x →a
−
f(
x) = ∞
lim
x →a
−
f(
x) = −∞
lim
x →a
f(
x) = ∞
lim
x →a
f(
x) = −∞
lim
x →a
f(
x) DNE
y = f (
x)
a
y =
|
x|
x
1
−
1
y =
|
x + 2|
x + 2
1
−
1
−
2
1
2
3
4
a
a
b
y = x sin
1
x
y = x
y = −
x
a
b
c
d
C
a
b
c
d
−
1
0
1
2
3
time
y
t
u
(
t, f(t))
(
u, f(u))
time
y
t
u
y
x
(
x, f(x))
y = |
x|
(
z, f(z))
z
y = f(
x)
a
tangent at x = a
b
tangent at x = b
c
tangent at x = c
y = x
2
tangent
at x = −
1
u
v
uv
u + ∆
u
v + ∆
v
(
u + ∆u)(v + ∆v)
∆
u
∆
v
u
∆v
v∆
u
∆
u∆v
y = f(
x)
1
2
−
2
y = |
x
2
− 4|
y = x
2
− 4
y = −
2x + 5
y = g(
x)
1
2
3
4
5
6
7
8
9
0
−
1
−
2
−
3
−
4
−
5
−
6
y = f (
x)
3
−
3
3
−
3
0
−
1
2
easy
hard
flat
y = f
0
(
x)
3
−
3
0
−
1
2
1
−
1
y = sin(
x)
y = x
x
A
B
O
1
C
D
sin(
x)
tan(
x)
y =
sin(
x)
x
π
2
π
1
−
1
x = 0
a = 0
x > 0
a > 0
x < 0
a < 0
rest position
+
−
y = x
2
sin
1
x
N
A
B
H
a
b
c
O
H
A
B
C
D
h
r
R
θ
1000
2000
α
β
p
h
y = g(
x) = log
b
(x)
y = f(
x) = b
x
y = e
x
5
10
1
2
3
4
0
−
1
−
2
−
3
−
4
y = ln(
x)
y = cosh(
x)
y = sinh(
x)
y = tanh(
x)
y = sech(
x)
y = csch(
x)
y = coth(
x)
1
−
1
y = f(
x)
original function
inverse function
slope = 0 at (
x, y)
slope is infinite at (
y, x)
−
108
2
5
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y = sin(
x)
y = sin(
x), −
π
2
≤ x ≤
π
2
−
2
−
1
0
2
π
2
−
π
2
y = sin
−
1
(x)
y = cos(
x)
π
π
2
y = cos
−
1
(x)
−
π
2
1
x
α
β
y = tan(
x)
y = tan(
x)
1
y = tan
−
1
(x)
y = sec(
x)
y = sec
−
1
(x)
y = csc
−
1
(x)
y = cot
−
1
(x)
1
y = cosh
−
1
(x)
y = sinh
−
1
(x)
y = tanh
−
1
(x)
y = sech
−
1
(x)
y = csch
−
1
(x)
y = coth
−
1
(x)
(0
, 3)
(2
, −1)
(5
, 2)
(7
, 0)
(
−1, 44)
(0
, 1)
(1
, −12)
(2
, 305)
y = 1
2
(2
, 3)
y = f(
x)
y = g(
x)
a
b
c
a
b
c
s
c
0
c
1
(
a, f(a))
(
b, f(b))
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
c
OR
Local maximum
Local minimum
Horizontal point of inflection
1
e
y = f
0
(
x)
y = f (
x) = x ln(x)
−
1
e
?
y = f(
x) = x
3
y = g(
x) = x
4
x
f(
x)
−
3
−
2
−
1
0
1
2
1
2
3
4
+
−
?
1
5
6
3
f
0
(
x)
2 −
1
2
√
6
2 +
1
2
√
6
f
00
(
x)
7
8
g
00
(
x)
f
00
(
x)
0
y =
(
x − 3)(x − 1)
2
x
3
(
x + 2)
y = x ln(
x)
1
e
−
1
e
5
−
108
2
α
β
2 −
1
2
√
6
2 +
1
2
√
6
y = x
2
(
x − 5)
3
−
e
−
1/2
√
3
e
−
1/2
√
3
−
e
−3/2
e
−
3/2
−
1
√
3
1
√
3
−
1
1
y = xe
−
3x
2
/2
y =
x
3
− 6
x
2
+ 13x − 8
x
28
2
600
500
400
300
200
100
0
−
100
−
200
−
300
−
400
−
500
−
600
0
10
−
10
5
−
5
20
−
20
15
−
15
0
4
5
6
x
P
0
(
x)
+
−
−
existing fence
new fence
enclosure
A
h
b
H
99
100
101
h
dA/dh
r
h
1
2
7
shallow
deep
LAND
SEA
N
y
z
s
t
3
11
9
L
(11)
√
11
y = L
(x)
y = f (
x)
11
y = L
(x)
y = f(
x)
F
P
a
a + ∆
x
f(
a + ∆x)
L
(a + ∆x)
f(
a)
error
df
∆
x
a
b
y = f(
x)
true zero
starting approximation
better approximation
v
t
3
5
50
40
60
4
20
30
25
t
1
t
2
t
3
t
4
t
n
−2
t
n
−1
t
0
= a
t
n
= b
v
1
v
2
v
3
v
4
v
n
−1
v
n
−
30
6
30
|
v|
a
b
p
q
c
v(
c)
v(
c
1
)
v(
c
2
)
v(
c
3
)
v(
c
4
)
v(
c
5
)
v(
c
6
)
t
1
t
2
t
3
t
4
t
5
c
1
c
2
c
3
c
4
c
5
c
6
t
0
=
a
t
6
=
b
t
16
=
b
t
10
=
b
a
b
x
y
y = f(
x)
1
2
y = x
5
0
−
2
y = 1
a
b
y = sin(
x)
π
−
π
0
−
1
−
2
0
2
4
y = x
2
0
1
2
3
4
2
n
4
n
6
n
2(
n−2)
n
2(
n−1)
n
2
n
n
= 2
width of each interval =
2
n
−
2
1
3
0
I
II
III
IV
4
y
dx
y = −
x
2
− 2x + 3
3
−
5
y = |−
x
2
− 2x + 3|
I
II
IIa
5
3
0
1
2
a
b
y = f (
x)
y = g(
x)
y = x
2
a
b
5
3
0
1
2
y =
√
x
2
√
2
2
2
dy
x
2
a
b
y = f(
x)
y = g(
x)
M
m
1
2
−
1
−
2
0
y = e
−
x
2
1
2
e
−
1/4
f
av
y = f
av
c
A
M
0
1
2
a
b
x
t
y = f (t)
F (x )
y = f (t)
F (x + h)
x + h
F (x + h) − F (x)
f(x)
1
2
y = sin(x)
π
−π
−1
−2
y =
1
x
y = x
2
1
2
1
−1
y = ln|x|
θ
a
x
a
x
p
a
2
− x
2
3
x
p
9 − x
2
p
x
2
+ a
2
x
a
p
x
2
+ 15
x
√
15
x
p
x
2
− a
2
a
x
p
x
2
− 4
2
x
−
p
x
2
− a
2
a
x
−
p
x
2
− 4
2
y = f(x)
a
b
a + ε
ε
Z
b
a+ε
f(x) dx
small
even smaller
y = g(x)
infinite area
finite area
Z
1
0
1
sin(
√
x)
dx
converges or diverges. It seems difficult to find an antiderivative of 1/ sin(
√
x).
Luckily, we don’t have to. Since sin(x) ∼ x as x → 0, we can replace the small
quantity x by another small quantity
√
x to see that sin(
√
x) ∼
√
x as x → 0
+
.
(We need to use x → 0
+
because
√
x only makes sense when x ≥ 0.) Taking
reciprocals, we have
1
sin(
√
x)
∼
1
√
x
as x → 0
+
.
Also note that 1/ sin(
√
x) and 1/
√
x have no blow-up points in (0, 1]. So, the
limit comparison test says that the two integrals
Z
1
0
1
sin(
√
x)
dx and
Z
1
0
1
√
x
dx
either both converge or both diverge. We have replaced a difficult integral
with a much easier one,
R
1
0
1/
√
x dx. We already know from Section 20.1.1
above that this easier integral converges, so we can immediately conclude that
the integral we want (on the left) also converges.
444 • Improper Integrals: Basic Concepts
Of course, there are cases of the test which apply when the blow-up point
is at b, or when the region of integration is unbounded. We’ll list all the
versions in Section 21.2 of the next chapter. In the meantime, let’s see why
PSfrag replacements
(
a, b)
[
a, b]
(
a, b]
[
a, b)
(
a, ∞)
[
a, ∞)
(
−∞, b)
(
−∞, b]
(
−∞, ∞)
{
x : a < x < b}
{
x : a ≤ x ≤ b}
{
x : a < x ≤ b}
{
x : a ≤ x < b}
{
x : x ≥ a}
{
x : x > a}
{
x : x ≤ b}
{
x : x < b}
R
a
b
shadow
0
1
4
−
2
3
−
3
g(
x) = x
2
f(
x) = x
3
g(
x) = x
2
f(
x) = x
3
mirror (
y = x)
f
−
1
(x) =
3
√
x
y = h
(x)
y = h
−
1
(x)
y = (
x − 1)
2
−
1
x
Same height
−
x
Same length,
opposite signs
y = −
2x
−
2
1
y =
1
2
x − 1
2
−
1
y = 2
x
y = 10
x
y = 2
−
x
y = log
2
(
x)
4
3 units
mirror (
x-axis)
y = |
x|
y = |
log
2
(x)|
θ radians
θ units
30
◦
=
π
6
45
◦
=
π
4
60
◦
=
π
3
120
◦
=
2
π
3
135
◦
=
3
π
4
150
◦
=
5
π
6
90
◦
=
π
2
180
◦
= π
210
◦
=
7
π
6
225
◦
=
5
π
4
240
◦
=
4
π
3
270
◦
=
3
π
2
300
◦
=
5
π
3
315
◦
=
7
π
4
330
◦
=
11
π
6
0
◦
= 0 radians
θ
hypotenuse
opposite
adjacent
0 (
≡ 2π)
π
2
π
3
π
2
I
II
III
IV
θ
(
x, y)
x
y
r
7
π
6
reference angle
reference angle =
π
6
sin +
sin −
cos +
cos −
tan +
tan −
A
S
T
C
7
π
4
9
π
13
5
π
6
(this angle is
5
π
6
clockwise)
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y = sin(
x)
y = cos(
x)
−
π
2
π
2
y = tan(
x), −
π
2
< x <
π
2
0
−
π
2
π
2
y = tan(
x)
−
2π
−
3π
−
5
π
2
−
3
π
2
−
π
−
π
2
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
y = sec(
x)
y = csc(
x)
y = cot(
x)
y = f(
x)
−
1
1
2
y = g(
x)
3
y = h
(x)
4
5
−
2
f(
x) =
1
x
g(
x) =
1
x
2
etc.
0
1
π
1
2
π
1
3
π
1
4
π
1
5
π
1
6
π
1
7
π
g(
x) = sin
1
x
1
0
−
1
L
10
100
200
y =
π
2
y = −
π
2
y = tan
−
1
(x)
π
2
π
y =
sin(
x)
x
, x > 3
0
1
−
1
a
L
f(
x) = x sin (1/x)
(0 < x < 0
.3)
h
(x) = x
g(
x) = −x
a
L
lim
x
→a
+
f(x) = L
lim
x
→a
+
f(x) = ∞
lim
x
→a
+
f(x) = −∞
lim
x
→a
+
f(x) DNE
lim
x
→a
−
f(x) = L
lim
x
→a
−
f(x) = ∞
lim
x
→a
−
f(x) = −∞
lim
x
→a
−
f(x) DNE
M
}
lim
x
→a
−
f(x) = M
lim
x
→a
f(x) = L
lim
x
→a
f(x) DNE
lim
x
→∞
f(x) = L
lim
x
→∞
f(x) = ∞
lim
x
→∞
f(x) = −∞
lim
x
→∞
f(x) DNE
lim
x
→−∞
f(x) = L
lim
x
→−∞
f(x) = ∞
lim
x
→−∞
f(x) = −∞
lim
x
→−∞
f(x) DNE
lim
x →a
+
f(
x) = ∞
lim
x →a
+
f(
x) = −∞
lim
x →a
−
f(
x) = ∞
lim
x →a
−
f(
x) = −∞
lim
x →a
f(
x) = ∞
lim
x →a
f(
x) = −∞
lim
x →a
f(
x) DNE
y = f (
x)
a
y =
|
x|
x
1
−
1
y =
|
x + 2|
x + 2
1
−
1
−
2
1
2
3
4
a
a
b
y = x sin
1
x
y = x
y = −
x
a
b
c
d
C
a
b
c
d
−
1
0
1
2
3
time
y
t
u
(
t, f(t))
(
u, f(u))
time
y
t
u
y
x
(
x, f(x))
y = |
x|
(
z, f(z))
z
y = f(
x)
a
tangent at x = a
b
tangent at x = b
c
tangent at x = c
y = x
2
tangent
at x = −
1
u
v
uv
u + ∆
u
v + ∆
v
(
u + ∆u)(v + ∆v)
∆
u
∆
v
u
∆v
v∆
u
∆
u∆v
y = f(
x)
1
2
−
2
y = |
x
2
− 4|
y = x
2
− 4
y = −
2x + 5
y = g(
x)
1
2
3
4
5
6
7
8
9
0
−
1
−
2
−
3
−
4
−
5
−
6
y = f (
x)
3
−
3
3
−
3
0
−
1
2
easy
hard
flat
y = f
0
(
x)
3
−
3
0
−
1
2
1
−
1
y = sin(
x)
y = x
x
A
B
O
1
C
D
sin(
x)
tan(
x)
y =
sin(
x)
x
π
2
π
1
−
1
x = 0
a = 0
x > 0
a > 0
x < 0
a < 0
rest position
+
−
y = x
2
sin
1
x
N
A
B
H
a
b
c
O
H
A
B
C
D
h
r
R
θ
1000
2000
α
β
p
h
y = g(
x) = log
b
(x)
y = f(
x) = b
x
y = e
x
5
10
1
2
3
4
0
−
1
−
2
−
3
−
4
y = ln(
x)
y = cosh(
x)
y = sinh(
x)
y = tanh(
x)
y = sech(
x)
y = csch(
x)
y = coth(
x)
1
−
1
y = f(
x)
original function
inverse function
slope = 0 at (
x, y)
slope is infinite at (
y, x)
−
108
2
5
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y = sin(
x)
y = sin(
x), −
π
2
≤ x ≤
π
2
−
2
−
1
0
2
π
2
−
π
2
y = sin
−
1
(x)
y = cos(
x)
π
π
2
y = cos
−
1
(x)
−
π
2
1
x
α
β
y = tan(
x)
y = tan(
x)
1
y = tan
−
1
(x)
y = sec(
x)
y = sec
−
1
(x)
y = csc
−
1
(x)
y = cot
−
1
(x)
1
y = cosh
−
1
(x)
y = sinh
−
1
(x)
y = tanh
−
1
(x)
y = sech
−
1
(x)
y = csch
−
1
(x)
y = coth
−
1
(x)
(0
, 3)
(2
, −1)
(5
, 2)
(7
, 0)
(
−1, 44)
(0
, 1)
(1
, −12)
(2
, 305)
y = 1
2
(2
, 3)
y = f(
x)
y = g(
x)
a
b
c
a
b
c
s
c
0
c
1
(
a, f(a))
(
b, f(b))
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
c
OR
Local maximum
Local minimum
Horizontal point of inflection
1
e
y = f
0
(
x)
y = f (
x) = x ln(x)
−
1
e
?
y = f(
x) = x
3
y = g(
x) = x
4
x
f(
x)
−
3
−
2
−
1
0
1
2
1
2
3
4
+
−
?
1
5
6
3
f
0
(
x)
2 −
1
2
√
6
2 +
1
2
√
6
f
00
(
x)
7
8
g
00
(
x)
f
00
(
x)
0
y =
(
x − 3)(x − 1)
2
x
3
(
x + 2)
y = x ln(
x)
1
e
−
1
e
5
−
108
2
α
β
2 −
1
2
√
6
2 +
1
2
√
6
y = x
2
(
x − 5)
3
−
e
−
1/2
√
3
e
−
1/2
√
3
−
e
−3/2
e
−
3/2
−
1
√
3
1
√
3
−
1
1
y = xe
−
3x
2
/2
y =
x
3
− 6
x
2
+ 13x − 8
x
28
2
600
500
400
300
200
100
0
−
100
−
200
−
300
−
400
−
500
−
600
0
10
−
10
5
−
5
20
−
20
15
−
15
0
4
5
6
x
P
0
(
x)
+
−
−
existing fence
new fence
enclosure
A
h
b
H
99
100
101
h
dA/dh
r
h
1
2
7
shallow
deep
LAND
SEA
N
y
z
s
t
3
11
9
L
(11)
√
11
y = L
(x)
y = f (
x)
11
y = L
(x)
y = f(
x)
F
P
a
a + ∆
x
f(
a + ∆x)
L
(a + ∆x)
f(
a)
error
df
∆
x
a
b
y = f(
x)
true zero
starting approximation
better approximation
v
t
3
5
50
40
60
4
20
30
25
t
1
t
2
t
3
t
4
t
n
−2
t
n
−1
t
0
= a
t
n
= b
v
1
v
2
v
3
v
4
v
n
−1
v
n
−
30
6
30
|
v|
a
b
p
q
c
v(
c)
v(
c
1
)
v(
c
2
)
v(
c
3
)
v(
c
4
)
v(
c
5
)
v(
c
6
)
t
1
t
2
t
3
t
4
t
5
c
1
c
2
c
3
c
4
c
5
c
6
t
0
=
a
t
6
=
b
t
16
=
b
t
10
=
b
a
b
x
y
y = f(
x)
1
2
y = x
5
0
−
2
y = 1
a
b
y = sin(
x)
π
−
π
0
−
1
−
2
0
2
4
y = x
2
0
1
2
3
4
2
n
4
n
6
n
2(
n−2)
n
2(
n−1)
n
2
n
n
= 2
width of each interval =
2
n
−
2
1
3
0
I
II
III
IV
4
y
dx
y = −
x
2
− 2x + 3
3
−
5
y = |−
x
2
− 2x + 3|
I
II
IIa
5
3
0
1
2
a
b
y = f (
x)
y = g(
x)
y = x
2
a
b
5
3
0
1
2
y =
√
x
2
√
2
2
2
dy
x
2
a
b
y = f(
x)
y = g(
x)
M
m
1
2
−
1
−
2
0
y = e
−
x
2
1
2
e
−
1/4
f
av
y = f
av
c
A
M
0
1
2
a
b
x
t
y = f (
t)
F (
x )
y = f (
t)
F (
x + h)
x + h
F (
x + h) − F (x)
f(
x)
1
2
y = sin(
x)
π
−
π
−
1
−
2
y =
1
x
y = x
2
1
2
1
−1
y = ln|x|
θ
a
x
a
x
p
a
2
− x
2
3
x
p
9 − x
2
p
x
2
+ a
2
x
a
p
x
2
+ 15
x
√
15
x
p
x
2
− a
2
a
x
p
x
2
− 4
2
x
−
p
x
2
− a
2
a
x
−
p
x
2
− 4
2
y = f(x)
a
b
a + ε
ε
Z
b
a+ε
f(x) dx
small
even smaller
y = g(x)
infinite area
finite area
the test works in the above case. Since f(x) ∼ g(x) as x → a, we know that
lim
x→a
f(x)
g(x)
= 1.
In particular, provided we get close enough to a, the ratio f(x)/g(x) must be
at least
1
2
and no more than 2. That is, we can pick some c between a and b
such that
1
2
≤
f(x)
g(x)
≤ 2 for all x in (a, c].
This inequality can be rewritten as
1
2
g(x) ≤ f(x) ≤ 2g(x) for all x in (a, c].
Now we can use the comparison test. For example, if
R
b
a
g(x) dx diverges, then
so does
R
c
a
g(x) dx (as we’ve seen above). In fact, so does
1
2
R
c
a
g(x) dx, infor-
mally since one-half of infinity is still infinity! So, the fact that f(x) is greater
than
1
2
g(x) means that the integral
R
c
a
f(x) dx diverges, and it follows that
R
b
a
f(x) dx diverges too. On the other hand, if
R
b
a
g(x) dx actually converges,
then so does 2
R
c
a
g(x) dx and we can again use the comparison test (you can
fill in the details) to show that
R
b
a
f(x) dx converges as well.
PSfrag replacements
(
a, b)
[
a, b]
(
a, b]
[
a, b)
(
a, ∞)
[
a, ∞)
(
−∞, b)
(
−∞, b]
(
−∞, ∞)
{
x : a < x < b}
{
x : a ≤ x ≤ b}
{
x : a < x ≤ b}
{
x : a ≤ x < b}
{
x : x ≥ a}
{
x : x > a}
{
x : x ≤ b}
{
x : x < b}
R
a
b
shadow
0
1
4
−
2
3
−
3
g(
x) = x
2
f(
x) = x
3
g(
x) = x
2
f(
x) = x
3
mirror (
y = x)
f
−
1
(x) =
3
√
x
y = h
(x)
y = h
−
1
(x)
y = (
x − 1)
2
−
1
x
Same height
−
x
Same length,
opposite signs
y = −
2x
−
2
1
y =
1
2
x − 1
2
−
1
y = 2
x
y = 10
x
y = 2
−
x
y = log
2
(
x)
4
3 units
mirror (
x-axis)
y = |
x|
y = |
log
2
(x)|
θ radians
θ units
30
◦
=
π
6
45
◦
=
π
4
60
◦
=
π
3
120
◦
=
2
π
3
135
◦
=
3
π
4
150
◦
=
5
π
6
90
◦
=
π
2
180
◦
= π
210
◦
=
7
π
6
225
◦
=
5
π
4
240
◦
=
4
π
3
270
◦
=
3
π
2
300
◦
=
5
π
3
315
◦
=
7
π
4
330
◦
=
11
π
6
0
◦
= 0 radians
θ
hypotenuse
opposite
adjacent
0 (
≡ 2π)
π
2
π
3
π
2
I
II
III
IV
θ
(
x, y)
x
y
r
7
π
6
reference angle
reference angle =
π
6
sin +
sin −
cos +
cos −
tan +
tan −
A
S
T
C
7
π
4
9
π
13
5
π
6
(this angle is
5
π
6
clockwise)
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y = sin(
x)
y = cos(
x)
−
π
2
π
2
y = tan(
x), −
π
2
< x <
π
2
0
−
π
2
π
2
y = tan(
x)
−
2π
−
3π
−
5
π
2
−
3
π
2
−
π
−
π
2
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
y = sec(
x)
y = csc(
x)
y = cot(
x)
y = f(
x)
−
1
1
2
y = g(
x)
3
y = h
(x)
4
5
−
2
f(
x) =
1
x
g(
x) =
1
x
2
etc.
0
1
π
1
2
π
1
3
π
1
4
π
1
5
π
1
6
π
1
7
π
g(
x) = sin
1
x
1
0
−
1
L
10
100
200
y =
π
2
y = −
π
2
y = tan
−
1
(x)
π
2
π
y =
sin(
x)
x
, x > 3
0
1
−
1
a
L
f(
x) = x sin (1/x)
(0 < x < 0
.3)
h
(x) = x
g(
x) = −x
a
L
lim
x
→a
+
f(x) = L
lim
x
→a
+
f(x) = ∞
lim
x
→a
+
f(x) = −∞
lim
x
→a
+
f(x) DNE
lim
x
→a
−
f(x) = L
lim
x
→a
−
f(x) = ∞
lim
x
→a
−
f(x) = −∞
lim
x
→a
−
f(x) DNE
M
}
lim
x
→a
−
f(x) = M
lim
x
→a
f(x) = L
lim
x
→a
f(x) DNE
lim
x
→∞
f(x) = L
lim
x
→∞
f(x) = ∞
lim
x
→∞
f(x) = −∞
lim
x
→∞
f(x) DNE
lim
x
→−∞
f(x) = L
lim
x
→−∞
f(x) = ∞
lim
x
→−∞
f(x) = −∞
lim
x
→−∞
f(x) DNE
lim
x →a
+
f(
x) = ∞
lim
x →a
+
f(
x) = −∞
lim
x →a
−
f(
x) = ∞
lim
x →a
−
f(
x) = −∞
lim
x →a
f(
x) = ∞
lim
x →a
f(
x) = −∞
lim
x →a
f(
x) DNE
y = f (
x)
a
y =
|
x|
x
1
−
1
y =
|
x + 2|
x + 2
1
−
1
−
2
1
2
3
4
a
a
b
y = x sin
1
x
y = x
y = −
x
a
b
c
d
C
a
b
c
d
−
1
0
1
2
3
time
y
t
u
(
t, f(t))
(
u, f(u))
time
y
t
u
y
x
(
x, f(x))
y = |
x|
(
z, f(z))
z
y = f(
x)
a
tangent at x = a
b
tangent at x = b
c
tangent at x = c
y = x
2
tangent
at x = −
1
u
v
uv
u + ∆
u
v + ∆
v
(
u + ∆u)(v + ∆v)
∆
u
∆
v
u
∆v
v∆
u
∆
u∆v
y = f(
x)
1
2
−
2
y = |
x
2
− 4|
y = x
2
− 4
y = −
2x + 5
y = g(
x)
1
2
3
4
5
6
7
8
9
0
−
1
−
2
−
3
−
4
−
5
−
6
y = f (
x)
3
−
3
3
−
3
0
−
1
2
easy
hard
flat
y = f
0
(
x)
3
−
3
0
−
1
2
1
−
1
y = sin(
x)
y = x
x
A
B
O
1
C
D
sin(
x)
tan(
x)
y =
sin(
x)
x
π
2
π
1
−
1
x = 0
a = 0
x > 0
a > 0
x < 0
a < 0
rest position
+
−
y = x
2
sin
1
x
N
A
B
H
a
b
c
O
H
A
B
C
D
h
r
R
θ
1000
2000
α
β
p
h
y = g(
x) = log
b
(x)
y = f(
x) = b
x
y = e
x
5
10
1
2
3
4
0
−
1
−
2
−
3
−
4
y = ln(
x)
y = cosh(
x)
y = sinh(
x)
y = tanh(
x)
y = sech(
x)
y = csch(
x)
y = coth(
x)
1
−
1
y = f(
x)
original function
inverse function
slope = 0 at (
x, y)
slope is infinite at (
y, x)
−
108
2
5
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y = sin(
x)
y = sin(
x), −
π
2
≤ x ≤
π
2
−
2
−
1
0
2
π
2
−
π
2
y = sin
−
1
(x)
y = cos(
x)
π
π
2
y = cos
−
1
(x)
−
π
2
1
x
α
β
y = tan(
x)
y = tan(
x)
1
y = tan
−
1
(x)
y = sec(
x)
y = sec
−
1
(x)
y = csc
−
1
(x)
y = cot
−
1
(x)
1
y = cosh
−
1
(x)
y = sinh
−
1
(x)
y = tanh
−
1
(x)
y = sech
−
1
(x)
y = csch
−
1
(x)
y = coth
−
1
(x)
(0
, 3)
(2
, −1)
(5
, 2)
(7
, 0)
(
−1, 44)
(0
, 1)
(1
, −12)
(2
, 305)
y = 1
2
(2
, 3)
y = f(
x)
y = g(
x)
a
b
c
a
b
c
s
c
0
c
1
(
a, f(a))
(
b, f(b))
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
c
OR
Local maximum
Local minimum
Horizontal point of inflection
1
e
y = f
0
(
x)
y = f (
x) = x ln(x)
−
1
e
?
y = f(
x) = x
3
y = g(
x) = x
4
x
f(
x)
−
3
−
2
−
1
0
1
2
1
2
3
4
+
−
?
1
5
6
3
f
0
(
x)
2 −
1
2
√
6
2 +
1
2
√
6
f
00
(
x)
7
8
g
00
(
x)
f
00
(
x)
0
y =
(
x − 3)(x − 1)
2
x
3
(
x + 2)
y = x ln(
x)
1
e
−
1
e
5
−
108
2
α
β
2 −
1
2
√
6
2 +
1
2
√
6
y = x
2
(
x − 5)
3
−
e
−
1/2
√
3
e
−
1/2
√
3
−
e
−3/2
e
−
3/2
−
1
√
3
1
√
3
−
1
1
y = xe
−
3x
2
/2
y =
x
3
− 6
x
2
+ 13x − 8
x
28
2
600
500
400
300
200
100
0
−
100
−
200
−
300
−
400
−
500
−
600
0
10
−
10
5
−
5
20
−
20
15
−
15
0
4
5
6
x
P
0
(
x)
+
−
−
existing fence
new fence
enclosure
A
h
b
H
99
100
101
h
dA/dh
r
h
1
2
7
shallow
deep
LAND
SEA
N
y
z
s
t
3
11
9
L
(11)
√
11
y = L
(x)
y = f (
x)
11
y = L
(x)
y = f(
x)
F
P
a
a + ∆
x
f(
a + ∆x)
L
(a + ∆x)
f(
a)
error
df
∆
x
a
b
y = f(
x)
true zero
starting approximation
better approximation
v
t
3
5
50
40
60
4
20
30
25
t
1
t
2
t
3
t
4
t
n
−2
t
n
−1
t
0
= a
t
n
= b
v
1
v
2
v
3
v
4
v
n
−1
v
n
−
30
6
30
|
v|
a
b
p
q
c
v(
c)
v(
c
1
)
v(
c
2
)
v(
c
3
)
v(
c
4
)
v(
c
5
)
v(
c
6
)
t
1
t
2
t
3
t
4
t
5
c
1
c
2
c
3
c
4
c
5
c
6
t
0
=
a
t
6
=
b
t
16
=
b
t
10
=
b
a
b
x
y
y = f(
x)
1
2
y = x
5
0
−
2
y = 1
a
b
y = sin(
x)
π
−
π
0
−
1
−
2
0
2
4
y = x
2
0
1
2
3
4
2
n
4
n
6
n
2(
n−2)
n
2(
n−1)
n
2
n
n
= 2
width of each interval =
2
n
−
2
1
3
0
I
II
III
IV
4
y
dx
y = −
x
2
− 2x + 3
3
−
5
y = |−
x
2
− 2x + 3|
I
II
IIa
5
3
0
1
2
a
b
y = f (
x)
y = g(
x)
y = x
2
a
b
5
3
0
1
2
y =
√
x
2
√
2
2
2
dy
x
2
a
b
y = f(
x)
y = g(
x)
M
m
1
2
−
1
−
2
0
y = e
−
x
2
1
2
e
−
1/4
f
av
y = f
av
c
A
M
0
1
2
a
b
x
t
y = f (t)
F (x )
y = f (t)
F (x + h)
x + h
F (x + h) − F (x)
f(x)
1
2
y = sin(x)
π
−π
−1
−2
y =
1
x
y = x
2
1
2
1
−1
y = ln|x|
θ
a
x
a
x
p
a
2
− x
2
3
x
p
9 − x
2
p
x
2
+ a
2
x
a
p
x
2
+ 15
x
√
15
x
p
x
2
− a
2
a
x
p
x
2
− 4
2
x
−
p
x
2
− a
2
a
x
−
p
x
2
− 4
2
y = f(x)
a
b
a + ε
ε
Z
b
a+ε
f(x) dx
small
even smaller
y = g(x)
infinite area
finite area
A quick comment: most textbooks have a different statement of the limit
comparison test. In particular, the limit of f (x)/g(x) doesn’t actually have
to be 1—it could be any positive number and the above argument would still
work (after a slight modification). On the other hand, allowing a limit other
than 1 doesn’t really gain anything, and it loses the ability to use the intuitive
∼ notation. As we’ll see in the next chapter, we’ll get by very nicely with our
version of the test.
20.5 The p-test (Theory)
Now that we have the comparison test and limit comparison test, we need to
know how to use them. Our basic strategy, which will be greatly elaborated
upon in the next chapter, will be to pick a function g which we can compare our
function f with. Hopefully g is simple enough that we can at least say whether
its integral (over the region under consideration) converges or diverges.
The question is, what are some functions we could choose as g? Well,
the most useful are the functions 1/x
p
for some p > 0. For example, we
have already looked at some integrals involving 1/x, 1/
√
x, and 1/x
2
, which
correspond to p = 1,
1
2
, and 2, respectively. Since these functions are so easy
to integrate, we can use the limit formulas to get the p-test:
Section 20.5: The p-test (Theory) • 445
• (p-test,
R
∞
version) For any finite a > 0, the integral
Z
∞
a
1
x
p
dx
converges if p > 1 and diverges if p ≤ 1.
• (p-test,
R
0
version) For any finite a > 0, the integral
Z
a
0
1
x
p
dx
converges if p < 1 and diverges if p ≥ 1.
Notice that the two versions of the test are basically opposites: except for
when p = 1, one of the integrals
Z
a
0
1
x
p
dx and
Z
∞
a
1
x
p
dx
converges and the other one diverges. The case p = 1 corresponds to 1/x, and
as we already know, both of the integrals diverge in this case.
Now, this p-test is really useful and comes up often in practice, so it’s
really important that you don’t get the two versions of the test mixed up!
One way to remember the correct version of the test is to remember what
happens with 1/x
2
and 1/
√
x. I just remember the two little facts:
Z
∞
a
1
x
2
dx converges, and so does
Z
a
0
1
√
x
dx.
From these two facts, I can remember the whole of the p-test! How does it
work? Well, from the first fact, and the knowledge that what goes on near ∞
is opposite from what goes on near 0, I know that
Z
a
0
1
x
2
dx
diverges. Similarly, from the second fact, I know that
Z
∞
a
1
√
x
dx
also diverges. What about other exponents? Well, any exponent higher than
1 (for example,
3
2
, 2, or 70) behaves in the same manner as 1/x
2
, and any
exponent lower than 1 (for example,
1
2
,
2
3
, or 0.999) behaves exactly like 1/
√
x
(remember, this is the same as 1/x
1/2
).
It might also help to examine the following diagram: