Lecture 20Section 10.7 Improper Integrals
Jiwen He
1
Improper Integrals
What are Improper Integrals?
Z ∞
1
dx =?,
x2
1
Z
0
1
1
dx =?
x2
b
1
1
1 1
dx = −
Known:
f (x) dx =
= − , 0 < a < b,
2
x
x
a
b
a
a
a
• the interval of integration [a, b], 0 < a < b, is bounded,
Z
b
Z
b
• the function being integrated f (x) =
1
x2
is bounded over [a, b].
By a limit process, we can extend the integration process to
Z b
Z ∞
1
1 1
1
dx
=
lim
dx
=
lim
−
=
• unbounded intervals (e.g., [1, ∞)):
b→∞ 1 x2
b→∞ 1
x2
b
1
1
Z 1
1
1
+
dx =
• unbounded functions (e.g., as x → 0 , f (x) = x2 → ∞):
2
x
0
Z 1
1
1 1
lim
dx = lim+
−
=∞
2
+
a 1
a→0
a→0
a x
1.1
Integrals Over Unbounded Intervals
Integrals Over Unbounded Intervals
1
Z
Let f be continuous on [a, ∞). We define[0.5ex]
∞
Z
f (x) dx = lim
a
b→∞
b
f (x) dx
a
The improper integral converges if the limit exists. The improper integral diverges if the limit doesn’t exist.
Z ∞
If lim f (x) 6= 0, then
f (x) dx diverges.
b→∞
a
Z b
Z b
If f is continuous on (−∞, b],
f (x) dx = lim
f (x) dx. If f is cont.
a→−∞ a
−∞
Z ∞
Z 0
Z ∞
f (x) dx =
f (x) dx +
f (x) dx
on (−∞, ∞),
−∞
−∞
0
Fundamental Theorem of Integral Calculus
Let f (x) = F 0 (x) for ∀x ≥ a. Then
Z ∞
Z b
∞
f (x) dx = lim
f (x) dx = lim F (x)|ba = lim F (b)−F (a) = F (x)a = lim F (x)−F (a)
b→∞
a
b→∞
a
b→∞
x→∞
0
Let f (x) = F (x) for ∀x ≤ b. Then
Z b
b
f (x) dx = F (x)−∞ = F (b) − lim F (x)
x→−∞
−∞
Let f (x) = F 0 (x) for ∀x. Then
Z 0
Z ∞
Z ∞
∞
f (x) dx = F (x)−∞ = lim F (x) − lim F (x)
f (x) dx =
f (x) dx +
x→∞
x→−∞
−∞
−∞
0
= F (0) − lim F (x) + lim F (x) − F (0)
x→−∞
x→∞
2
Examples
Z
∞
1
∞
Z
1
−2x
e
0
∞
−∞
1
∞
dx = ln x|1 = lim (ln x) − ln 1 = ∞
x→∞
x
∞
Z
Z
∞
1
2 2
dx = − 2 = lim − 2 − (−2) = 2
x→∞
x3
x 1
x
∞
−2x e−2x 1
1
e
dx = −
=
lim
−
−
=
−
x→∞
2 0
2
2
2
∞
c
−1 x dx
=
tan
c2 + x2
c −∞
x x − lim tan−1
= lim tan−1
x→−∞
x→∞
c
c
π π
= − −
=π
2
2
What is Wrong with This?
What is wrong with the following argument?
Z b
Z b
Z ∞
2x
2x
2x
2 b
∞=
dx
=
lim
dx
=
lim
ln(1
+
x
)
=
6
lim
dx = 0
2
2
−b
b→∞
b→∞
b→∞
1
+
x
1
+
x
1
+
x2
−∞
−b
−b
Z ∞
∞
2x
dx
= lim ln(1 + b2 ) − ln(1 + (−b)2 ) = lim (0) = 0 = ln(1 + x2 )0 = lim ln(1 + x2 ) − ln 1
2
x→∞
b→∞
b→∞
1+x
0
Z ∞
Z b
f (x) dx as lim
f (x) dx.
• Note that we did not define
Z ∞
Z b −b
b→∞
−∞
f (x) dx = L ⇒
lim
f (x) dx = L
b→∞
−∞
Z
b
lim
b→∞
−b
∞
Z
f (x) dx = L
f (x) dx = L
;
−b
Z
• For every odd function f , we have lim
b→∞
Z ∞
f (x) dx.
certainly not the case for
−∞
R∞
1
1
xα
dx, α > 0
3
−∞
b
f (x) dx = 0;
−b
but this is
Z
1
If α 6= 1, then
Z ∞
1
∞



1
,
1
α−1
dx
=

xα

∞,
if α > 1,
if α ≤ 1.
∞
1
1
1
1
1−α dx
=
x
lim x1−α −
= 1 − α x→∞
α
x
1−α
1
−
α
1
1
if α > 1,
α−1 ,
=
∞, if α < 1.
If α = 1, then
Z
1
∞
1
∞
dx = ln x|1 = lim (ln x) − ln 1 = ∞
x→∞
x
Quiz
Quiz
x2
x→∞ ex
(a) 0, (b) 1, (c) ∞.
1. Find lim
x
ln x
(b) 1, (c) ∞.
2. Find lim
x→∞
(a) 0,
4
1.2
Integrals of Unbounded Functions
Integrals of Unbounded Functions
−
Let f be continuous on [a,
but f (x) → ±∞
Z b),
Z c as x → b . We define
b
f (x) dx = lim−
f (x) dx
The improper integral converges
if thec→b
limit aexists. The improper integral dia
verges if the limit doesn’t exist.
If f is continuous on (a, Zb], but f (x) → ±∞Zas x → a+ .,
b
b
f (x) dx = lim+
f (x) dx.
If f is cont. on [a, b], except
at somec→a
point cc in (a, b) where f (x) → ±∞ as
a
x → c− or x → c+ , Z b
Z c
Z b
f (x) dx =
f (x) dx +
f (x) dx
a
a
c
Fundamental Theorem of Integral Calculus
Let f (x) = F 0 (x) for ∀x ∈ [a, b) and f (x) → ±∞ as x → b− . Then
Z b
Z c
b
f (x) dx = lim− f (x) dx = lim− (F (x)|ca ) = lim F (c)−F (a) = F (x)a = lim F (x)−F (a)
−
−
a
c→b
c→b
a
Z
If lim− F (x) = ±∞, then
x→b
c→b
b
f (x) dx diverges.
a
Let f (x) = F 0 (x) for ∀x ∈ (a, b] and f (x) → ±∞ as x → a+ . Then
Z b
b
f (x) dx = F (x)a = F (b) − lim+ F (x)
x→a
a
5
x→b
Examples
1
Z
1
x1/3
0
1
Z
0
1
3
3 3
3 2/3 lim x2/3 =
dx = x = −
+
2
2 2 x→0
2
0
1
1
dx = ln x|0 = ln 1 − lim (ln x) = ∞
x
x→0+
π
2
Z
π
(ln sec x) − ln 1 = ∞
tan x dx = ln sec x|02 = lim
π−
x→ 2
0
Examples
∞
Z
√
0
e−x
dx =
1 − e−2x
0
−du
=
1
− u2
1
1
π
= sin−1 u0 =
2
Z
√
Z
1
√
0
1
du
1 − u2
Set u = e−x , then du = −e−x dx and u2 = e−2x .
Z
0
π
3
sin x
√
dx =
2 cos x − 1
Z
0
1
− 12 du
1
√
=
2
u
Z
0
1
1
√ du
u
1 √ 1
1
=
u0 =
4
4
Set u = 2 cos x − 1, then du = −2 sin x dx.
What is Wrong with This?
What is wrong with the following argument?
4
Z 4
3
1
1 ∞=
dx
=
=
6
−
=−
2
x − 2 1
2
1 (x − 2)
2
Z 2
1
1
1 = lim −
dx = −
−1=∞
2
x − 2 1
x−2
x→2−
1 (x − 2)
6
4
1
or x → 2 . To evaluate
dx we
(x − 2)2
Z 2
Z1 4
1
1
need to calculate the improper integrals
dx and
dx.
2
(x
−
2)
(x
−
2)2
1
2
Note that
R1
1
0 xα
1
(x−2)2
−
+
→ ∞ as x → 2
dx, α > 0
7
Z
Z
0
1



1
,
1
1−α
dx
=

xα

∞,
if α < 1,
if α ≥ 1.
If α 6= 1, then
Z
0
1
1
1
1
1
1
1−α dx
=
x
lim+ x1−α
= 1 − α − 1 − α x→0
xα
1−α
0
1
if α < 1,
1−α ,
=
∞, if α > 1.
If α = 1, then
Z
0
1
1
1
dx = ln x|0 = ln 1 − lim (ln x) = ∞
x
x→0+
Quiz
Quiz
1
x→∞
x
(a) 0, (b) 1, (c) ∞.
3. Find lim x sin
x
ex
(b) 1, (c) ∞.
4. Find lim
x→0+
(a) 0,
8
2
Comparison Test for Convergence
2.1
Comparison Test for Convergence
Comparison Test for Convergence
Z
Let
be improper integration over an interval I, and suppose that f and g are
I
continuous on I such that
Z
• If
0 ≤ f (x) ≤ g(x), ∀x ∈ I
Z
g(x) dx converges, then so does
f (x) dx.
I
I
Z
• If
Z
f (x) dx diverges, then so does
I
g(x) dx.
I
Z
∞
Z
∞
1
dx converges since
1
+ x3
1
Z ∞
1
1
1
√
√
√ dx converges.
0≤
<
, ∀x ≥ 1 and
3
3
1+x
x
x3
1
√
The improper integral
1
dx diverges since
1 + x2
1
Z ∞
1
1
1
0≤
<√
, ∀x ≥ 1 and
dx diverges.
2
1+x
1+x
1+x
1
The improper integral
√
Outline
9
Contents
1 Improper Integrals
1.1 Unbounded Intervals . . . . . . . . . . . . . . . . . . . . . . . . .
1.2 Unbounded Functions . . . . . . . . . . . . . . . . . . . . . . . .
1
1
5
2 Comparison Test
2.1 Comparison Test . . . . . . . . . . . . . . . . . . . . . . . . . . .
9
9
10