IMPROPER INTEGRALS
ELECTRONIC VERSION OF LECTURE
HoChiMinh City University of Technology
Faculty of Applied Science, Department of Applied Mathematics
HCM — 2021.
(HCMUT-OISP)
IMPROPER INTEGRALS
HCM — 2021.
1 / 39
OUTLINE
1
T YPE 1: INFINITE INTERVALS
2
T YPE 2: INFINITY DISCONTINUOUS INTEGRANDS
3
MATL AB
(HCMUT-OISP)
IMPROPER INTEGRALS
HCM — 2021.
2 / 39
R
Definition of an improper integral of type 1 a+∞ f (x)d x
Type 1: Infinite intervals
T YPE 1: INFINITE INTERVALS
DEFINITION 1.1
Let f (x) be defined for every number x Ê a and be
integrable
on every interval [a, b]. Then
Rb
Φ(b) = a f (x)d x is defined on the interval [a, +∞). The
limit
Z
I = lim Φ(b) = lim
b→+∞
b→+∞ a
b
(1)
f (x)d x
is called an improper integral of type 1 of function
Zf (x) on the interval [a, +∞) and denoted by
+∞
f (x)d x.
a
(HCMUT-OISP)
IMPROPER INTEGRALS
HCM — 2021.
3 / 39
Type 1: Infinite intervals
R
Definition of an improper integral of type 1 a+∞ f (x)d x
DEFINITION 1.2
1
If the limit I = lim
Z
b
f (x)d x exists (as a finite
Z +∞
number) then the improper integrals
f (x)d x
b→+∞ a
a
are called convergent.
2
If the limit I = lim
Z
b→+∞ a
b
f (x)d x does not exist or is
equal
to ∞ then the improper integrals
Z
+∞
f (x)d x are called divergent.
a
(HCMUT-OISP)
IMPROPER INTEGRALS
HCM — 2021.
4 / 39
Type 1: Infinite intervals
Geometric meaning
GEOMETRIC MEANING
If f (x) Ê 0, ∀x ∈ [a, +∞) and the integral
Z
convergent then the improper integrals
+∞
f (x)d x is
aZ
+∞
f (x)d x
a
can be interpreted as an area of the region
S = {(x, y)|x Ê a, 0 É y É f (x)}.
(HCMUT-OISP)
IMPROPER INTEGRALS
HCM — 2021.
5 / 39
Type 1: Infinite intervals
Geometric meaning
According to geometric
meaning of an improper
Z
+∞
integral of type 1:
f (x)d x , if
a
lim f (x) = A 6= 0
x→+∞
and f (x) is integrable on every interval
[a,
Z b] ⊂ [a, +∞), then the improper integrals
+∞
f (x)d x are divergent.
a
(HCMUT-OISP)
IMPROPER INTEGRALS
HCM — 2021.
6 / 39
Type 1: Infinite intervals
Rb
Definition of an improper integral of type 1 −∞
f (x)d x
DEFINITION 1.3
Let f (x) be defined for every number x É b and be
integrable on every interval [a, b]. Then
Ψ(a) =
b
Z
f (x)d x is defined on the interval (−∞, b].
a
The limit
I = lim Ψ(a) = lim
a→−∞
Z
a→−∞ a
b
(2)
f (x)d x
is called an improper integral of type 1 of function
f (x) on the interval (−∞, b] and denoted by
Z
b
f (x)d x.
−∞
(HCMUT-OISP)
IMPROPER INTEGRALS
HCM — 2021.
7 / 39
Type 1: Infinite intervals
Rb
Definition of an improper integral of type 1 −∞
f (x)d x
DEFINITION 1.4
1
If the limit I = lim
Z
b
f (x)d x exists (as a finite
Z b
number) then the improper integrals
f (x)d x
a→−∞ a
are called convergent.
2
If the limit I = lim
Z
−∞
b
f (x)d x does not exist or is
Z b
equal to ∞ then the improper integrals
f (x)d x
a→−∞ a
are called divergent.
(HCMUT-OISP)
IMPROPER INTEGRALS
−∞
HCM — 2021.
8 / 39
Type 1: Infinite intervals
Geometric meaning
GEOMETRIC MEANING
If f (x) Ê 0, ∀x ∈ (−∞, b], and the integral
Z
convergent then the improper integrals
b
f (x)d x is
Z−∞b
f (x)d x
−∞
can be interpreted as an area of the region
S = {(x, y)|x É b, 0 É y É f (x)}.
(HCMUT-OISP)
IMPROPER INTEGRALS
HCM — 2021.
9 / 39
Type 1: Infinite intervals
R +∞
Definition of an improper integral of type 1 −∞
f (x)d x
DEFINITION 1.5
If f (x) is defined on R and integrable on every interval
[a, b] then for every number c ∈ R, the improper
integral of type 1 of function f (x) on (−∞, +∞) is
defined by
Z
+∞
Z
c
f (x)d x =
f (x)d x +
−∞
The improper integral of type 1
convergent if both
convergent.
(HCMUT-OISP)
f (x)d x.
Z
+∞
f (x)d x is
−∞
Z
c
f (x)d x and
−∞
IMPROPER INTEGRALS
(3)
c
−∞
Z
+∞
Z
+∞
f (x)d x are
c
HCM — 2021.
10 / 39
Type 1: Infinite intervals
Newton-Leibniz’s Formula
NEWTON-LEIBNIZ’S FORMULA
Suppose that f (x) has antiderivative F (x) on the
interval [a, +∞) and is intergrable on every
interval
Z
+∞
f (x)d x is
[a, b]. The improper integral of type 1
a
convergent if and only if lim F (b) = F (+∞) exists as a
b→+∞
finite number. Then
+∞
Z
a
(HCMUT-OISP)
¯+∞
¯
f (x)d x = F (+∞) − F (a) = F (x)¯
a
IMPROPER INTEGRALS
HCM — 2021.
(4)
11 / 39
Type 1: Infinite intervals
Newton-Leibniz’s Formula
NEWTON-LEIBNIZ’S FORMULA
Suppose that f (x) has antiderivative F (x) on the
interval (−∞, b] and is intergrable on every interval
[a, b]. The improper integral of type 1
Z
b
f (x)d x is
−∞
convergent if and only if lim F (a) = F (−∞) exists as
a→−∞
a finite number. Then
Z
b
−∞
(HCMUT-OISP)
¯b
¯
f (x)d x = F (b) − F (−∞) = F (x)¯
−∞
IMPROPER INTEGRALS
HCM — 2021.
(5)
12 / 39
Type 1: Infinite intervals
Z
+∞
Z
Newton-Leibniz’s Formula
c
f (x)d x =
Z
f (x)d x
¶
³
´ µ
= F (c) − lim F (a) + lim F (b) − F (c)
−∞
f (x)d x +
+∞
−∞
c
a→−∞
The improper integral of type 1
b→+∞
Z
+∞
f (x)d x is
−∞
convergent if and only if lim F (a) and lim F (b)
a→−∞
b→+∞
exist as finite numbers
Z
+∞
−∞
¯+∞
¯
f (x)d x = F (+∞) − F (−∞) = F (x)¯
(HCMUT-OISP)
−∞
IMPROPER INTEGRALS
HCM — 2021.
(6)
13 / 39
Type 1: Infinite intervals
Newton-Leibniz’s Formula
EXAMPLE 1.1
Evaluate I =
Z
+∞
cos xd x.
0
SOLUTION
¯+∞
¯
I = sin x ¯ = lim sin b − sin 0 = lim sin b.
0
b→+∞
b→+∞
The limit lim sin b does not exist. Therefore the
b→+∞
improper integral I is divergent.
(HCMUT-OISP)
IMPROPER INTEGRALS
HCM — 2021.
14 / 39
Type 1: Infinite intervals
Newton-Leibniz’s Formula
EXAMPLE 1.2
Evaluate I =
−1
dx
2
−∞ x
Z
SOLUTION
¯
1
1 ¯¯−1
I = − ¯ = 1 + lim
= 1.
a→−∞ a
x −∞
Therefore, the improper integral I is convergent.
(HCMUT-OISP)
IMPROPER INTEGRALS
HCM — 2021.
15 / 39
Type 1: Infinite intervals
Newton-Leibniz’s Formula
EXAMPLE 1.3
Evaluate I =
Z
+∞
−∞
dx
1 + x2
SOLUTION
¯+∞
¯
I = arctan x ¯ = lim arctan b − lim arctan a
a→−∞
−∞
b→+∞
π ³ π´
= − − = π.
2
2
So the given integral I is convergent.
(HCMUT-OISP)
IMPROPER INTEGRALS
HCM — 2021.
16 / 39
Type 1: Infinite intervals
Newton-Leibniz’s Formula
EXAMPLE 1.4
Evaluate I =
Z
+∞
2
xe −x d x
0
SOLUTION
1
I =−
2
+∞
Z
e
0
−x 2
¯+∞
¯
1
1
1 1
2
2
d (−x 2 ) = − e −x ¯¯ = lim − e −b + =
b→+∞ 2
2
2 2
0
So the given integral I is convergent.
(HCMUT-OISP)
IMPROPER INTEGRALS
HCM — 2021.
17 / 39
Type 1: Infinite intervals
Newton-Leibniz’s Formula
EXAMPLE 1.5
For what values of α is the integral
+∞
Z
I=
1
dx
xα
convergent?
If α 6= 1 then
1
I =−
α−1
If α > 1, then lim
x→+∞
1
x α−1
µ
lim
x→+∞
1
x α−1
¶
1α−1
= 0. Therefore I =
the integral I converges.
(HCMUT-OISP)
−
1
IMPROPER INTEGRALS
1
and so
α−1
HCM — 2021.
18 / 39
Type 1: Infinite intervals
If α < 1, then lim
x→+∞
1
x α−1
Newton-Leibniz’s Formula
= +∞ and so the integral I
diverges.
If α = 1, then I = lim ln |x| − ln 1 = +∞ and so the
x→+∞
integral I diverges.
SUMMARY
1
2
+∞
dx
converges.
xα
1
Z +∞
dx
If α É 1 then I =
diverges.
xα
1
If α > 1 then I =
(HCMUT-OISP)
Z
IMPROPER INTEGRALS
HCM — 2021.
19 / 39
Type 1: Infinite intervals
A comparison test for improper integrals of type 1
A COMPARISON TEST FOR IMPROPER INTEGRALS OF
TYPE 1
THEOREM 1.1
Suppose that f and g are continuous functions on
every interval [a, b] ⊂ [a, +∞) with 0 É g (x) É f (x),
∀x Ê a.
Z
1
+∞
If
f (x)d x is convergent, then
a
If
+∞
Z
g (x)d x is divergent then
a
+∞
Z
f (x)d x is
a
divergent.
(HCMUT-OISP)
g (x)d x is
a
convergent.
2
+∞
Z
IMPROPER INTEGRALS
HCM — 2021.
20 / 39
Type 1: Infinite intervals
1
A comparison test for improper integrals of type 1
If the area under the top curve y = f (x) is finite,
then so is the area under the bottom curve
y = g (x).
2
If the area under y = g (x) is infinite, then so is the
area under y = f (x).
(HCMUT-OISP)
IMPROPER INTEGRALS
HCM — 2021.
21 / 39
Type 1: Infinite intervals
A comparison test for improper integrals of type 1
EXAMPLE 1.6
DetermineZ whether the integral is convergent or
+∞
divergent
1
SOLUTION
Since
+∞
Z
1
Z
1
+∞
1 + e −x
dx
x
1 + e −x 1
>
x
x
1
d x is divergent, so the integral
x
1 + e −x
d x is divergent.
x
(HCMUT-OISP)
IMPROPER INTEGRALS
HCM — 2021.
22 / 39
Type 2: Infinity discontinuous integrands
R
Definition of an improper integral of type 2 ab f (x)d x on [a, b)
T YPE 2: INFINITY DISCONTINUOUS INTEGRANDS
Suppose that f is defined on a finite interval [a, b)
but has a vertical asymptote as x → b − and f is
integrable
on every interval [a, η] ⊂ [a, b). Then
Z
Φ(η) =
η
f (x)d x is defined on the interval [a, b).
a
DEFINITION 2.1
The limit of function Φ(η) as η → b − is called an
improper integral of type 2 on the interval [a, b)
b
Z
a
(HCMUT-OISP)
f (x)d x = lim− Φ(η) = lim−
η→b
η→b
IMPROPER INTEGRALS
η
Z
f (x)d x
(7)
a
HCM — 2021.
23 / 39
Type 2: Infinity discontinuous integrands
R
Definition of an improper integral of type 2 ab f (x)d x on [a, b)
DEFINITION 2.2
1
If lim− Φ(η) = lim−
η→b
η→b
η
Z
f (x)d x exists (as a finite
a
number) then the improper integral of type 2
b
Z
f (x)d x converges.
Z η
If lim− Φ(η) = lim− f (x)d x does not exist or is
a
2
η→b
η→b
a
equal to ∞ then the improper integral of type 2
b
Z
f (x)d x diverges.
a
(HCMUT-OISP)
IMPROPER INTEGRALS
HCM — 2021.
24 / 39
Type 2: Infinity discontinuous integrands
Geometric meaning
GEOMETRIC MEANING
If f (x) Ê 0, ∀x ∈ [a, b) and the integral
b
Z
f (x)d x is
Z b
convergent then the improper integrals
f (x)d x
a
a
can be interpreted as an area of the region
S = {(x, y)|a É x < b, 0 É y É f (x)}, where x = b is the
vertical asymptote of the graph of function f (x)
(HCMUT-OISP)
IMPROPER INTEGRALS
HCM — 2021.
25 / 39
Type 2: Infinity discontinuous integrands
R
Definition of an improper integral of type 2 ab f (x)d x on (a, b]
Suppose that f is defined on a finite interval (a, b]
but has a vertical asymptote as x → a + and f is
integrable on every interval [ξ, b] ⊂ (a, b]. Then
Ψ(ξ) =
b
Z
ξ
f (x)d x is defined on the interval (a, b].
DEFINITION 2.3
The limit of function Ψ(ξ) as ξ → a + is called an
improper integral of type 2 on the interval (a, b]
b
Z
a
(HCMUT-OISP)
f (x)d x = lim+ Ψ(ξ) = lim+
ξ→a
ξ→a
IMPROPER INTEGRALS
Z
ξ
b
f (x)d x
HCM — 2021.
(8)
26 / 39
Type 2: Infinity discontinuous integrands
R
Definition of an improper integral of type 2 ab f (x)d x on (a, b]
DEFINITION 2.4
1
If lim+ Ψ(ξ) = lim+
ξ→a
ξ→a
b
Z
ξ
f (x)d x exists (as a finite
number) then the improper integral of type 2
b
Z
f (x)d x converges.
Z b
If lim+ Ψ(ξ) = lim+
f (x)d x does not exist or is
a
2
ξ→a
ξ→a
ξ
equal to ∞ then the improper integral of type 2
b
Z
f (x)d x diverges.
a
(HCMUT-OISP)
IMPROPER INTEGRALS
HCM — 2021.
27 / 39
Type 2: Infinity discontinuous integrands
Geometric meaning
GEOMETRIC MEANING
If f (x) Ê 0, ∀x ∈ (a, b] and the integral
b
Z
f (x)d x is
Z b
convergent then the improper integrals
f (x)d x
a
a
can be interpreted as an area of the region
S = {(x, y)|a < x É b, 0 É y É f (x)}, where x = a is the
vertical asymptote of the graph of function f (x)
(HCMUT-OISP)
IMPROPER INTEGRALS
HCM — 2021.
28 / 39
Definition of an improper integral of type 2 a f (x)d x, c ∈ [a, b] is
Type 2: Infinity discontinuous integrands point of discontinuity
If f has a infinity discontinuity as x → c, where
c ∈ (a, b) then
b
Z
f (x)d x is defined by
a
b
Z
a
c
Z
f (x)d x =
a
Z
f (x)d x +
b
f (x)d x
(9)
c
DEFINITION 2.5
The improper integral of type 2
both integrals
convergent.
(HCMUT-OISP)
c
Z
f (x)d x and
a
Z
b
Z
f (x)d x converges if
a
b
f (x)d x are
c
IMPROPER INTEGRALS
HCM — 2021.
29 / 39
Definition of an improper integral of type 2 a f (x)d x, c ∈ [a, b] is
Type 2: Infinity discontinuous integrands point of discontinuity
(HCMUT-OISP)
IMPROPER INTEGRALS
HCM — 2021.
30 / 39
Type 2: Infinity discontinuous integrands
Newton-Leibniz’s formula
NEWTON-LEIBNIZ’S FORMULA
Suppose that f (x) has antiderivative F (x) on every
intervals [a, η] ⊂ [a, b) and lim− f (x) = ∞. The
x→b
improper integral of type 2
b
Z
f (x)d x is convergent if
a
and only if lim− F (η) = F (b − 0) exists as a finite
η→b
number. Then
b
Z
a
¯b −
¯
f (x)d x = F (b − 0) − F (a) = F (x)¯ .
(HCMUT-OISP)
(10)
a
IMPROPER INTEGRALS
HCM — 2021.
31 / 39
Type 2: Infinity discontinuous integrands
Newton-Leibniz’s formula
NEWTON-LEIBNIZ’S FORMULA
Suppose that f (x) has antiderivative F (x) on every
intervals [ξ, b] ⊂ (a, b] and lim+ f (x) = ∞. The improper
x→a
integral of type 2
b
Z
f (x)d x is convergent if and only
a
if lim+ F (ξ) = F (a + 0) exists as a finite number. Then
ξ→a
b
Z
a
¯b
¯
f (x)d x = F (b) − F (a + 0) = F (x)¯ + .
(HCMUT-OISP)
(11)
a
IMPROPER INTEGRALS
HCM — 2021.
32 / 39
Type 2: Infinity discontinuous integrands
Newton-Leibniz’s formula
NEWTON-LEIBNIZ’S FORMULA
Suppose that f (x) has antiderivative F (x) on every
subintervals [a, η] ⊂ [a, c), antiderivative G(x) on
every subintervals [ξ, b] ⊂ (c, b] and lim f (x) = ∞. The
x→c
improper integral of type 2
b
Z
f (x)d x is convergent if
a
and only if lim− F (η) = F (c − 0) and lim+ G(ξ) = G(c + 0)
η→c
ξ→c
exist as a finite number. Then
b
Z
a
f (x)d x = F (c − 0) − F (a) +G(b) −G(c + 0).
(HCMUT-OISP)
IMPROPER INTEGRALS
HCM — 2021.
(12)
33 / 39
Type 2: Infinity discontinuous integrands
Newton-Leibniz’s formula
EXAMPLE 2.1
Evaluate I =
Z
1
0
dx
x
SOLUTION
1
We have lim = +∞. Therefore x = 0 is vertical
x→0+
x
asymptote. Since
¯1
¯
I = ln |x|¯ = ln 1 − lim ln |a| = +∞.
0
a→0+
so improper integral I is divergent.
(HCMUT-OISP)
IMPROPER INTEGRALS
HCM — 2021.
34 / 39
Type 2: Infinity discontinuous integrands
Newton-Leibniz’s formula
EXAMPLE 2.2
Evaluate I =
1
arccos x
dx
p
2
−1
1−x
Z
arccos x
SOLUTION We have lim p
x→−1+
1 − x2
= +∞. Therefore
x = −1 is vertical asymptote.In other hand, x = 1 is
arccos x
not vertical asymptote because lim p
= 1.
x→1−
1 − x2
Since
1
¯1
1
¯
2
I =−
arccos xd (arccos x) = − · (arccos x)¯
−1
2
−1
2
1
π
= − (arccos2 1 − lim arccos2 x) =
x→−1+
2
2
so improper integral I is convergent.
Z
(HCMUT-OISP)
IMPROPER INTEGRALS
HCM — 2021.
35 / 39
Type 2: Infinity discontinuous integrands
Newton-Leibniz’s formula
EXAMPLE 2.3
Evaluate I =
b
Z
a
dx
, (a < b)
(b − x)α
SOLUTION
b−ε
¯b−ε
dx
1
−α+1 ¯
I = lim+
=−
lim (b − x)
¯
a
ε→0
(b − x)α
−α + 1 ε→0+
a
1
1
=
lim+ ε−α+1 +
(b − a)−α+1
α − 1 ε→0
−α + 1
Z
If α < 1 then lim+ ε−α+1 = 0.
ε→0
If α > 1 then lim+ ε−α+1 = ∞.
ε→0
(HCMUT-OISP)
IMPROPER INTEGRALS
HCM — 2021.
36 / 39
Type 2: Infinity discontinuous integrands
Newton-Leibniz’s formula
If α = 1 then
Zb−ε
I = lim+
ε→0
a
¯b−ε
dx
¯
= − lim+ ln |b − x|¯
a
ε→0
b−x
= − lim+ ln |ε| + ln(b − a) = ∞.
ε→0
SUMMARY
1
2
If α < 1, then improper integral
converges.
Z
If α Ê 1, then improper integral
Z
diverges.
(HCMUT-OISP)
IMPROPER INTEGRALS
b
dx
(b − x)α
b
dx
(b − x)α
a
a
HCM — 2021.
37 / 39
MatLab
EVALUATING INTEGRALS
1
>> s yms x; >> i nt (1/(1 + x 2 ), 0, i n f ) ⇒ Ans = pi /2
2
>> s yms x; >> i nt (1/(1 + x), 0, i n f ) ⇒ Ans = I n f
3
>> s yms x; >> i nt (1/sqr t (x), 0, 1) ⇒ Ans = 2
4
>> s yms x; >> i nt (1/x, 0, 1) ⇒ Ans = I n f
(HCMUT-OISP)
IMPROPER INTEGRALS
HCM — 2021.
38 / 39
MatLab
THANK YOU FOR YOUR ATTENTION
(HCMUT-OISP)
IMPROPER INTEGRALS
HCM — 2021.
39 / 39