Math 104: Improper Integrals (With Solutions)
Ryan Blair
University of Pennsylvania
Tuesday March 12, 2013
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Math 104: Improper Integrals
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Outline
1
Improper Integrals
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Math 104: Improper Integrals
Tuesday March 12, 2013
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Improper Integrals
Improper integrals
Definite integrals
Z
b
f (x)dx were required to have
a
finite domain of integration [a, b]
finite integrand f (x) < ±∞
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Math 104: Improper Integrals
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Improper Integrals
Improper integrals
Definite integrals
Z
b
f (x)dx were required to have
a
finite domain of integration [a, b]
finite integrand f (x) < ±∞
Improper integrals
1
Infinite limits of integration
2
Integrals with vertical asymptotes i.e. with infinite discontinuity
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Math 104: Improper Integrals
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Improper Integrals
Infinite limits of integration
Definition
Improper integrals are said to be
convergent if the limit is finite and that limit is the value of the
improper integral.
divergent if the limit does not exist.
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Math 104: Improper Integrals
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Improper Integrals
Infinite limits of integration
Definition
Improper integrals are said to be
convergent if the limit is finite and that limit is the value of the
improper integral.
divergent if the limit does not exist.
Each integral on the previous page is defined as a limit.
If the limit is finite we say the integral converges, while if the limit is
infinite or does not exist, we say the integral diverges.
Convergence is good (means we can do the integral); divergence is
bad (means we can’t do the integral).
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Math 104: Improper Integrals
Tuesday March 12, 2013
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Improper Integrals
Example 1
Find
Z
∞
e −x dx.
0
(if it even converges)
Solution:
Z
∞
e
0
−x
dx = lim
b→∞
Z
b
b→∞
0
= lim −e
b→∞
e −x dx = lim
−b
h
− e −x
ib
0
0
+ e = 0 + 1= 1.
So the integral converges and equals 1.
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Math 104: Improper Integrals
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Improper Integrals
Example 2
Find
(if it even converges)
Z
∞
−∞
1
dx.
1 + x2
Solution: By definition,
Z c
Z ∞
Z ∞
1
1
1
dx =
dx +
dx,
2
2
1 + x2
−∞ 1 + x
c
−∞ 1 + x
where we get to pick whatever c we want. Let’s pick c = 0.
Z 0
h
i0
1
dx = lim arctan(x) = lim [arctan(0) − arctan(b)]
2
b→−∞
b→−∞
b
−∞ 1 + x
π
= 0 − lim arctan(b) =
b→−∞
2
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Math 104: Improper Integrals
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Improper Integrals
Example 2, continued
Similarly,
Z
0
Therefore,
Z
∞
−∞
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∞
π
1
dx = .
2
1+x
2
π π
1
dx
+ = π.
=
1 + x2
2
2
Math 104: Improper Integrals
Tuesday March 12, 2013
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Improper Integrals
Example 3, the p-test
The integral
Z
1
∞
1
dx
xp
Converges if p > 1;
Diverges if p ≤ 1.
For example:
Z ∞
h 2 ib
1
dx
=
lim
−
= 2,
b→∞
x 3/2
x 1/2 1
1
while
Z ∞
h √ ib
√
1
x
dx
2
=
lim
2
=
lim
b − 2 = ∞,
b→∞
b→∞
x 1/2
1
1
and
Z ∞
h
ib
1
dx = lim ln(x) = lim ln(b) − 0 = ∞.
b→∞
b→∞
x
1
1
1
2
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Math 104: Improper Integrals
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Improper Integrals
Convergence vs. Divergence
In each case, if the limit exists (or if both limits exist, in case 3!), we
say the improper integral converges.
If the limit fails to exist or is infinite, the integral diverges. In case 3,
if either limit fails to exist or is infinite, the integral diverges.
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Math 104: Improper Integrals
Tuesday March 12, 2013
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Improper Integrals
Example 4
Find
Z
(if it converges)
2
0
2x
dx.
−4
x2
Solution: The denominator of x 22x−4 is 0 when x = 2, so the function
is not even defined when x = 2. So
Z 2
Z c
h
ic
2x
2x
2
dx = lim
dx = lim ln |x − 4|
2
c→2− 0 x 2 − 4
c→2−
0
0 x −4
2
= lim ln |x − 4| − ln(4) = −∞,
c→2−
so the integral diverges.
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Math 104: Improper Integrals
Tuesday March 12, 2013
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Improper Integrals
Example 5
Find
Solution:
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Z
3
0
1
dx,
(x − 1)2/3
if it converges.
Math 104: Improper Integrals
Tuesday March 12, 2013
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Improper Integrals
Example 5
Find
Z
3
0
1
dx,
(x − 1)2/3
if it converges.
Solution: We might think just to do
Z 3
h
i3
1
1/3
,
dx
=
3(x
−
1)
2/3
0
0 (x − 1)
Ryan Blair (U Penn)
Math 104: Improper Integrals
Tuesday March 12, 2013
11 / 15
Improper Integrals
Example 5
Find
Z
3
0
1
dx,
(x − 1)2/3
if it converges.
Solution: We might think just to do
Z 3
h
i3
1
1/3
,
dx
=
3(x
−
1)
2/3
0
0 (x − 1)
1
but this is not okay: The function f (x) = (x−1)
2/3 is undefined when
x = 1, so we need to split the problem into two integrals.
Z 1
Z 3
Z 3
1
1
1
dx
=
dx
+
dx.
2/3
2/3
2/3
0 (x − 1)
1 (x − 1)
0 (x − 1)
Ryan Blair (U Penn)
Math 104: Improper Integrals
Tuesday March 12, 2013
11 / 15
Improper Integrals
Example 5
Find
Z
3
0
1
dx,
(x − 1)2/3
if it converges.
Solution: We might think just to do
Z 3
h
i3
1
1/3
,
dx
=
3(x
−
1)
2/3
0
0 (x − 1)
1
but this is not okay: The function f (x) = (x−1)
2/3 is undefined when
x = 1, so we need to split the problem into two integrals.
Z 1
Z 3
Z 3
1
1
1
dx
=
dx
+
dx.
2/3
2/3
2/3
0 (x − 1)
1 (x − 1)
0 (x − 1)
The two integrals
R 3 on 1the right hand side1/3both converge and add up to
1/3
3[1 + 2 ], so 0 (x−1)2/3 dx = 3[1 + 2 ].
Ryan Blair (U Penn)
Math 104: Improper Integrals
Tuesday March 12, 2013
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Improper Integrals
Tests for convergence and divergence
The gist:
1
If you’re smaller than something that converges, then you
converge.
2
If you’re bigger than something that diverges, then you diverge.
Theorem
Let f and g be continuous on [a, ∞) with 0 ≤ f (x) ≤ g (x) for all
x ≥ a. Then
R∞
R∞
1
f
(x)
dx
converges
if
g (x) dx converges.
Ra∞
R ∞a
2
g (x) dx diverges if a f (x) dx diverges.
a
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Math 104: Improper Integrals
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Improper Integrals
Example 6
Which of the following integrals converge?
Z ∞
Z ∞ 2
sin (x)
−x 2
dx.
(a)
e
dx,
(b)
x2
1
1
Solution: Both integrals converge.
−x 2
(a) Note
that
0
<
e
≤ e −x for all x ≥ 1, and from example 1 we
R ∞ −x
R
2
∞
see 1 e dx = e1 , so 1 e −x dx converges.
(b) 0 ≤ sin2 (x) ≤ 1 for all x, so
0≤
for all x ≥ 1. Since
R ∞ sin2 (x)
dx.
1
x2
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R∞
1
1
x2
1
sin2 (x)
≤ 2
2
x
x
dx converges (by p-test), so does
Math 104: Improper Integrals
Tuesday March 12, 2013
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Improper Integrals
Limit Comparison Test
Theorem
If positive functions f and g are continuous on [a, ∞) and
f (x)
= L,
x→∞ g (x)
lim
then
Z
∞
f (x) dx
and
a
0 < L < ∞,
Z
∞
g (x) dx
a
BOTH converge or BOTH diverge.
Example 7: Let f (x) =
√ 1 ; consider
x+1
Z ∞
1
dx.
x +1
1
Does the integral converge or diverge?
√
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Math 104: Improper Integrals
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Improper Integrals
Example 7, continued
Solution: We note that f looks a lot like g (x) =
R∞
g (x) dx diverges by the p-test. Furthermore,
1
√
f (x)
x
lim
=√
= 1,
x→∞ g (x)
x +1
R∞ 1
dx diverges.
so the LCT says 1 √x+1
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Math 104: Improper Integrals
√1 ,
x
and
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