Improper integrals
D. DeTurck
University of Pennsylvania
October 10, 2016
D. DeTurck
Math 104 002 2015C: Improper integrals
1 / 11
Improper integrals
These are a special kind of limit. An improper integral is one
where either
1
the interval of integration is infinite, or
2
the interval of integration includes a singularity of the
function being integrated
or both.
D. DeTurck
Math 104 002 2015C: Improper integrals
2 / 11
Improper integrals
Examples of the first kind are
ˆ ∞
e −x dx
and
ˆ
∞
−∞
0
Examples of the second kind are
ˆ 1
1
√ dx
and
x
0
ˆ
1
dx
1 + x2
2π/3
tan x dx
π/4
The second of these is subtle because the singularity of the
integrand occurs in the interior of the interval of integration rather
than at one of the endpoints.
D. DeTurck
Math 104 002 2015C: Improper integrals
3 / 11
Dealing with improper integrals
No matter which kind of improper integral (or combination of
improper integrals) we are faced with, the method of dealing with
them is the same:
ˆ ∞
ˆ b
−x
e dx means the same thing as lim
e −x dx
b→∞ 0
0
ˆ
∞
−∞
1
dx means the same thing as
1 + x2
ˆ
lim
a→−∞ a
0
1
dx + lim
b→∞
1 + x2
D. DeTurck
ˆ
0
b
1
dx
1 + x2
Math 104 002 2015C: Improper integrals
4 / 11
Dealing with improper integrals
No matter which kind of improper integral (or combination of
improper integrals) we are faced with, the method of dealing with
them is the same:
ˆ 1
ˆ 1
1
1
√ dx means the same thing as lim
√ dx
a→0+
x
x
0
a
ˆ
2π/3
tan x dx means the same thing as
π/4
ˆ
lim
π
ˆ
b
b→ 2 − π/4
tan x dx + lim
π
2π/3
tan x dx
a→ 2 + a
The limit is basically a way of “sneaking up on the infinity” in the
problem.
D. DeTurck
Math 104 002 2015C: Improper integrals
5 / 11
OK, let’s calculate some improper integrals
ˆ
What is the value of the limit lim
b→∞
ˆ ∞
improper integral
e −x dx ?
b
e −x dx and hence of the
0
0
A. 0
B. 1
D. e −b
C. e
ˆ
What is the value of the limit lim
a→0+
ˆ 1
1
√ dx?
improper integral
x
0
A. 0
B. 1
C. 2
D. DeTurck
a
1
E. ∞
1
√ dx and hence of the
x
D.
√
2
E. ∞
Math 104 002 2015C: Improper integrals
6 / 11
Convergent and divergent improper integrals
It is possible that the limit used to define an improper integral fails
to exist or is infinite. In this case, the improper integral is said to
diverge (or be divergent).
If the limit does exist and is finite, then the improper integral
converges.
For example, the two integrals you just did both converge.
And
ˆ
0
1
1
dx = lim ln 1 − ln a = ∞
a→0+
x
is an example of a divergent improper integral.
D. DeTurck
Math 104 002 2015C: Improper integrals
7 / 11
The other examples
ˆ
1
dx = arctan x + C .
1 + x2
ˆ
If it is convergent, what is the value of
Recall that
∞
−∞
A. 0
π
B.
4
π
C.
2
1
dx ?
1 + x2
D. π
ˆ
E. divergent
2π/3
If it is convergent, what is the value of
tan x dx ?
π/4
A. 0
B.
1
2
ln 32
C.
D. DeTurck
1
2
ln 6
D. 1
E. divergent
Math 104 002 2015C: Improper integrals
8 / 11
Using estimation to show convergence or divergence
Sometimes it is possible to show that an improper integral
converges or diverges without actually evaluating it:
Since
x4
1
1
< 4 for all positive values of x, we have that
+x +7
x
ˆ b
ˆ b
1
1
1
1
dx <
dx = − 3
4
4
3 3b
1 x +x +7
1 x
for all b > 1. So the limit of the first integral as b → ∞ must be
finite because it increases as b does but it is bounded above (by 31 ).
ˆ
This shows that the improper integral
1
convergent.
D. DeTurck
∞
x4
1
dx is
+x +7
Math 104 002 2015C: Improper integrals
9 / 11
A puzzling example
1
Consider the surface obtained by rotating the graph of y = for
x
x > 1 around the x-axis.
We can calculate the volume of
the surface via disks as
ˆ ∞
π
V =
dx
x2
1
and
ˆ
V =
1
∞
π
dx = lim
b→∞
x2
ˆ
1
b
π π
π
dx
=
lim
− +
= π cubic units
b→∞
x2
b
1
D. DeTurck
Math 104 002 2015C: Improper integrals
10 / 11
But what about the surface area?
The surface area is equal to
ˆ
σ=
∞
ˆ
q
2πf (x) 1 + (f 0 (x))2 dx =
1
1
∞
2π
x
r
1+
1
dx.
x4
This integral is difficult (impossible) to evaluate directly, but it’s
easy to see that the integrand can be bounded from below by
something simpler:
r
2π
1
2π
1+ 4 ≥
x
x
x
for all x ≥ 1.
ˆ ∞
2π
But the integral
dx is divergent, so the surface has infinite
x
1
surface area.
This surfaces is sometimes called “Gabriel’s horn” — it is a surface
that can be “filled with water” but not “painted”.
D. DeTurck
Math 104 002 2015C: Improper integrals
11 / 11