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8.8 Improper Integrals
lim
b
b
a a
f x dx f x dx
∞
→∞
2
1
x
e dx
∞
−
2
1
lim
b
x
b
e dx
−
→∞
2
1
lim
b
x
b
e
−
→∞
2
2 e
2 2
lim
b
b
e e
→∞
2
1
lim
b
x
b
e
→∞
2
1
since lim 0
2
b
b e
→∞
−
=
Infinite Upper LimitInfinite Upper LimitInfinite Upper LimitInfinite Upper Limit
Math 104 – Rimmer
8.8 Improper Integrals
1
x
x
e
dx
e
∞
1
lim
b x
x
b
e
dx
e
→∞
1
lim ln 1
b
x
b
e
→∞
DIVERGENT
Infinite Upper LimitInfinite Upper LimitInfinite Upper LimitInfinite Upper Limit
1
x
x
u e
du e dx
= +
=
1
ln
u
du = u +C
∫
lim ln 1 ln 1
b
b
e e
→∞
8.8 Improper Integrals
lim
b b
a
a
f x dx f x dx
→−∞
−∞
1
x
xe dx
−∞
1
lim
x
a
a
xe dx
→−∞
1
lim
x x
a a
xe e
→−∞
lim
a a
a
e e ae e
→−∞
Infinite Lower LimitInfinite Lower LimitInfinite Lower LimitInfinite Lower Limit
x
D I
x e
x
e
x
e
+
-
lim 1
a
a
e a
→−∞
= − = 0 ⋅ ∞(indeterminate)
lim
a
a
a
e
−
→−∞
L'Hospital
'
lim
L H
a
a
e
−
→−∞
Math 104 – Rimmer
8.8 Improper Integrals
; any real number
c
c
f x dx f x dx f x dx c
∞ ∞
−∞ −∞
3
π
=
Infinite Upper and Lower LimitInfinite Upper and Lower LimitInfinite Upper and Lower LimitInfinite Upper and Lower Limit
lim lim
c b
a b
a c
f x dx f x dx f x dx
∞
→−∞ →∞
−∞
2
6
x
dx
x
∞
−∞
0 2 2
6 6
0
lim lim
b
a b
a
x x
dx dx
x x
→−∞ →∞
3
2 2 1
3
3
u x
du x dx du x dx
=
= =
2
1 1 1
3 3 1
arctan
u
du u C
= +
∫
0
3 3
1 1
3 3
0
lim arctan lim arctan
b
a b a
x x
→−∞ →∞
3 3 1 1
3 3
lim arctan lim arctan
a b
a b
→−∞ →∞
1 1
3 2 3 2
π π
8.8 Improper Integrals
Infinite Discontinuity inside the intervalInfinite Discontinuity inside the intervalInfinite Discontinuity inside the intervalInfinite Discontinuity inside the interval
( )
( )
( ) ( ) ( ) ( )
infinite
discontinuity
lim lim
b c b t b
t c t c
a a c a t
f c
a c b
f x dx f x dx f x dx f x dx f x dx
− +
→ →
→
< <
∫ ∫ ∫ ∫ ∫
( )
3
4
2
0 infinite
discontinuity
-2 0 3
f
dx
x
−
→
< <
∫
0 3
4 4
2 0
dx dx
x x
−
∫ ∫
3
4 4
0 0
2
lim lim
t
t t
t
dx dx
x x
− +
→ →
−
∫ ∫
3
3 3
0 0
2
lim lim
t
t t
t
x x
− +
→ →
−
Both are
actually only need one of them to be divergent
for the entire integral to be divergent
DIVERGENT
Math 104 – Rimmer
8.8 Improper Integrals
Doubly ImproperDoubly ImproperDoubly ImproperDoubly Improper
( )
( )
( ) ( ) ( ) ( )
0
0 0
0 infinite
discontinuity
lim lim
c c b
b a
c a c
f
f x dx f x dx f x dx f x dx f x dx
∞ ∞
→∞ →
→
∫ ∫ ∫ ∫ ∫
( )
1/
2
0
0 infinite
discontinuity
x
f
e
dx
x
∞ −
→
∫
1 1/ 1/
2 2
0 1
x x
e e
dx dx
x x
∞ − −
∫ ∫
1 1/ 1/
2 2
0
1
lim lim
b x x
a b
a
e e
dx dx
x x
− −
→ →∞
∫ ∫
1
1/ 1/
1
0
lim lim
b
x x
a b a
e e
− −
→∞ →
1 / 1 /
1
1 1
1
0
lim lim
x x
b
e e a b a
→∞ →
1/ 1/
1 1 1 1
0
lim lim
a b
e e
e e
b a
→∞ →
= 1
0
1 1
lim lim
1 1
b
a
b a
e
e
→∞ +
→
= −
= 1 − 0
2
1
1
x
u u
x
u
du dx e du e C
−
=
= = +
∫
8.8 Improper Integrals
Comparison TheoremComparison TheoremComparison TheoremComparison Theorem
Suppose that f x and g x are continuous functions with f x ≥ g x ≥ 0 for x ≥a.
) If ( ) is , then ( ) is.
a a
a f x dx g x dx
∞ ∞
∫ ∫
convergent convergent
) If ( ) is , then ( ) is.
a a
b g x dx f x dx
∞ ∞
∫ ∫
divergent divergent
See problems 49-54 in section 8.
3
0
1
x
dx
x
∞
∫
1
2
x
e
dx
x
∞ −
∫
4
1
1
x
dx
x x
∞
−
∫
0
arctan
2
x
x
dx
e
∞
∫
1 2
0
sec
x
dx
x x
∫
2
0
sin
x
dx
x
π
∫
Math 104 – Rimmer
8.8 Improper Integrals
3
0
1
x
dx
x
∞
∫
3
1
x
x +
3
for 1
x
x
x
≤ ≥
3 2
1
1
g f
x
x x
≤
1
3 3 3
0 0 1
1 1 1
x x x
dx dx dx
x x x
∞ ∞
= +
∫ ∫ ∫
for x ≥ 1
2
1
1
dx
x
∞
∫
2
1
lim
b
b
x dx
−
→∞
=
∫
1
1
lim
b
b
x
→∞
−
=
1
lim 1
b
b
→∞
−
= +
= 1 Convergent
3
1
Thus is by the Comparison Theorem.
1
x
dx
x
∞
∫
convergent
convergent constant
3
0
is.
1
x
dx
x
∞
∫
convergent
