Improper integrals (Sect. 8.7) - Michigan State University
Improper integrals (Sect. 8.7)
Review: Improper integrals type I and II.
dx
1 dx
Examples: I =
1
xp , and I = 0 xp .
Convergence test: Direct comparison test.
Convergence test: Limit comparison test.
Improper integrals (Sect. 8.7)
Review: Improper integrals type I and II.
dx
1 dx
Examples: I =
1
xp , and I = 0 xp .
Convergence test: Direct comparison test.
Convergence test: Limit comparison test.
Review: Improper integrals type I
Definition (Type I)
Improper integrals of Type I are integrals of continuous functions
on infinite domains; these include:
The improper integral of a continuous function f on [a, ),
b
f (x) dx = lim f (x) dx.
a
b a
The improper integral of a continuous function f on (-, b],
b
b
f (x) dx = lim f (x) dx.
-
a- a
The improper integral of a continuous function f on (-, ),
c
f (x) dx = f (x) dx + f (x) dx.
-
-
c
Review: Improper integrals type II
Definition (Type II)
Improper integrals of Type II are integrals of functions with vertical asymptotes within the integration interval; these include:
If f is continuous on (a, b] and discontinuous at a, then
b
b
f (x) dx = lim f (x) dx.
a
ca+ c
If f is continuous on [a, b) and discontinuous at b, then
b
c
f (x) dx = lim f (x) dx.
a
cb- a
If f is continuous on [a, c) (c, b] and discontinuous at c, then
b
c
b
f (x) dx = f (x) dx + f (x) dx.
a
a
c
Improper integrals (Sect. 8.7)
Review: Improper integrals type I and II.
dx
1 dx
Examples: I =
1
xp , and I = 0 xp .
Convergence test: Direct comparison test.
Convergence test: Limit comparison test.
1 dx
dx
The cases 0 xp and 1 xp
Summary: In the case p = 1 both integrals diverge,
1 dx = diverges,
0x In the case p = 1 we have:
dx = diverges.
1x
1 0
dx xp
=
1 =
1-p diverges
p < 1, p > 1.
y
y = 1/x1/2
y = 1/x
y = 1/x 2
dx
diverges p < 1,
1
1
xp
=
=
p
1 -1
p > 1.
0
1
x
Improper integrals (Sect. 8.7)
Review: Improper integrals type I and II.
dx
1 dx
Examples: I =
1
xp , and I = 0 xp .
Convergence test: Direct comparison test.
Convergence test: Limit comparison test.
Convergence test: Direct comparison test
Remark: Convergence tests determine whether an improper
integral converges or diverges.
Theorem (Direct comparison test)
If functions f , g : [a, ) R are continuous and 0 f (x) g (x) for every x [a, ), then holds
0
f (x) dx
g (x) dx.
a
a
The inequalities above imply the following statements:
(a) g (x) dx converges
f (x) dx converges;
a
a
(b) f (x) dx diverges
g (x) dx diverges.
a
a
Convergence test: Direct comparison test
Example
Determine whether I = e-x2 dx converges or diverges.
1
Solution: Notice that e-x2 dx does not have an expression in terms of elementary functions. However,
1 x x x 2 -x 2 -x e-x2 e-x .
The last inequality follows because exp is an increasing function.
0
e-x2 dx
e-x dx
= -e-x
=
1 .
1
1
1e
Since 0
e-x2 dx
1 , the integral converges.
1
e
Convergence test: Direct comparison test
Example
dx
Determine whether I =
converges or diverges.
1 x6 + 1
Solution: We need to find an appropriate function to compare with the integrand above. We need to find either
a bigger function with convergent integral; or a smaller function with divergent integral.
Notice: x6 < x6 +1
x3 <
x6 + 1
1
1
x6
+
1
<
x3.
dx
dx
x -2 1
Therefore, 0 <
1
<
x6 + 1 1
x3 = - 2
=. 12
Since 0
dx
1
, the integral converges.
1 x6 + 1 2
................
................
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