Improper Integrals
John E. Gilbert, Heather Van Ligten, and Benni Goetz
If something is improper about integration, we’d better find out what’s proper! You’re already
familiar with using definite integrals and the Fundamental Theorem of Calculus to find the area under
the graphs as in
1
2
3
4
5
1
x
y
f
(
x
) =
1
x
2
A
1
area
(
A
1
) =
4
1
1
x
2
dx
=
3
4
.
2
1
x
y
1
4
1
2
3
4
1
f
(
x
) =
1
√
x
A
2
area
(
A
2
) =
1
1
4
1
√
x
dx
= 1
.
In both cases the interval of integration is finite and does not contain a vertical asymptote of
f
. The
area is always finite  things are behaving well: call these
Proper Integrals
. If they are not, they are
Improper
!
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Definition:
an
Improper Integral
is one EITHER of the form
•
I
1
=
I
f
(
x
)
dx
where
I
=
[
a,
∞
)
,
(
∞
, b
]
,
or
(
∞
,
∞
)
,
OR
•
I
2
=
b
a
g
(
x
)
dx
where
g
(
x
)
has a vertical asymptote
in
[
a, b
]
.
As we shall soon see, ones of type
I
1
like
Int
1
=
∞
1
1
x
2
dx
occur when dealing with infinite series, and learning to deal with integrals over an infinite interval is a
good ‘warmup’ for studying infinite series. They are important too in probability and in many
applications as you’ll see in later courses. Integrals of type
I
2
like
Int
2
=
1
0
1
√
x
dx
often occur also  some functions blow up!
It’s easy to make Int
1
‘proper’: cut o
ff
integration at a large
t
. Since
t
1
1
x
2
dx
=

1
x
t
1
= 1

1
t
is always finite, we can think of the area under the graph of
1
/x
2
as being finite on all of
[1
,
∞
)
in the
sense that
Int
1
=
∞
1
1
x
2
dx
=
lim
t
→ ∞
t
1
1
x
2
dx
=
lim
t
→ ∞
1

1
t
= 1
.
But the area is
not
finite for all
f
(
x
)
on
[1
,
∞
)
; this is the crucial point of improper integrals! Indeed,
lim
t
→ ∞
t
1
1
x
dx
=
ln
t
t
1
=
lim
t
→ ∞
ln
t
=
∞
,
so the area under the graph is
infinite
on
[1
,
∞
)
when
f
(
x
) = 1
/x
.
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 Spring '07
 Sadler
 Fundamental Theorem Of Calculus, Improper Integrals, Integrals, Multivariable Calculus, lim, Mathematical analysis, dx, Riemann integral, Henstock–Kurzweil integral

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