Section 6.6 Improper Integrals
2010 Kiryl Tsishchanka
Improper Integrals
Type 1: Infinite Intervals
Consider the infinite region S that lies under the curve y = 1/x2 , above the x-axis, and to the right of the
line x = 1. You might think that, since S is infinite in extend, its area must be infinite. However, this is
not true. In fact, the area of the part of S that lies to the left of the line x = t is
A(t) =
Zt
1
1
dx =
x2
Zt
1
x−2+1
x dx =
−2 + 1
−2
t
1
x−1
=
−1
t
1
1
=−
x
t
1
=1−
1
t
Notice that A(t) < 1 no matter how large t is chosen. Moreover, since
1
lim A(t) = lim 1 −
=1
t→∞
t→∞
t
we can say that the area of the infinite region S is equal to 1 and we write
Z∞
1
dx = lim
t→∞
x2
1
Zt
1
dx = 1
x2
1
DEFINITION OF AN IMPROPER INTEGRAL OF TYPE 1:
Zt
Z∞
Zt
(a) If f (x)dx exists for every number t ≥ a, then f (x)dx = lim f (x)dx provided this limit exists
t→∞
a
a
a
(as a finite number).
Zb
Zb
Zb
f (x)dx provided this limit exists
f (x)dx = lim
(b) If f (x)dx exists for every number t ≤ b, then
t→−∞
t
t
−∞
(as a finite number).
Z∞
Zb
The improper integrals
f (x)dx and
f (x)dx are called convergent if the corresponding limit exists
a
−∞
and divergent if the limit does not exist.
Z∞
Z∞
Za
Z∞
(c) The improper integral
f (x)dx is defined as
f (x)dx =
f (x)dx + f (x)dx, where a is any real
−∞
−∞
−∞
a
number. It is said to converge if both terms converge and diverge if either term diverges.
1
Section 6.6 Improper Integrals
2010 Kiryl Tsishchanka
EXAMPLES:
Z∞
1
1. Evaluate
dx if possible.
x
1
Solution: We have
Z∞
1
dx = lim
t→∞
x
Zt
1
dx = lim ln |x|]t1 = lim (ln t − ln 1) = lim ln t = ∞
t→∞
t→∞
t→∞
x
1
1
The limit does not exist as a finite number and so the improper integral
Z∞
1
dx is divergent.
x
1
2. Evaluate
Z∞
1
dx if possible.
x2
2
Solution: We have
Z∞
1
dx = lim
t→∞
x2
2
3. Evaluate
Z∞
Zt
1
1
dx
=
lim
−
t→∞
x2
x
2
t
2
= lim
t→∞
1 1
− +
t 2
=0+
1
1
=
2
2
(convergent)
1
√ dx if possible.
x
4
Solution: We have
Z∞
4
1
√ dx = lim
t→∞
x
Zt
1
√ dx = lim
t→∞
x
4
Zt
4
x
−1/2
x−1/2+1
dx = lim
t→∞ −1/2 + 1
t
4
√
√ √ t
= lim 2 x 4 = lim 2 t − 2 4 = ∞
t→∞
t→∞
The limit does not exist as a finite number and so the improper integral
Z∞
4
4. For what values of p is
Z∞
1
dx convergent?
xp
1
2
1
√ dx is divergent.
x
Section 6.6 Improper Integrals
4. For what values of p is
2010 Kiryl Tsishchanka
Z∞
1
dx convergent?
xp
1
Solution: We know that if p = 1, then the integral is divergent, so let’s assume that p 6= 1. Then
Z∞
1
dx = lim
t→∞
xp
Zt
1
dx = lim
t→∞
xp
1
1
Zt
1
x−p+1
x dx = lim
t→∞ −p + 1
−p
If p > 1, then p − 1 > 0, so as t → ∞, tp−1 → ∞ and
Z∞
1
t
1
1
= lim
t→∞ (1 − p)xp−1
1
tp−1
1
1
dx =
p
x
p−1
t
1
1
= lim
t→∞ 1 − p
1
tp−1
−1
→ 0. Therefore
if p > 1
and so the integral converges. On the other hand, if p < 1, then p − 1 < 0 and so
1
= t1−p → ∞ as t → ∞
p−1
t
and the integral diverges. So,
Z∞
1
dx is convergent if p > 1 and divergent if p ≤ 1.
xp
1
EXAMPLES: Determine whether each integral is convergent or divergent. Evaluate those that are convergent.
1.
Z0
ex dx
−∞
2.
Z∞
ex dx
0
3.
Z∞
xdx
−∞
Z∞
4. (1 − x)e−x dx
0
5.
Z∞
x2
Z∞
ex
dx
+4
0
6.
dx
+ e−x
−∞
3
Section 6.6 Improper Integrals
2010 Kiryl Tsishchanka
SOLUTIONS:
1. We have
Z0
ex dx = lim
t→−∞
Z0
ex dx = lim ex ]0t = lim (e0 − et ) = (1 − 0) = 1 (convergent)
t→−∞
t→−∞
t
−∞
2. We have
Z∞
x
e dx = lim
t→∞
Zt
ex dx = lim ex ]t0 = lim (et − e0 ) = ∞ (divergent)
t→∞
3. We have
Z∞
xdx =
The integral
Z1
xdx +
Z∞
xdx
1
−∞
−∞
Z∞
t→∞
0
0
xdx is divergent by the p-test, since p = −1 ≤ 1. Therefore
1
Z∞
xdx is divergent.
−∞
4. We first note that
Z

−x
1−x=u
e−x dx = dv

(1 − x)e dx =  d(1 − x) = du
−dx = du
= (x − 1)e
−x
We also note that
lim xe
x→∞
−x
−
Z
−e
−x


= v  = (1 − x)(−e−x ) −
Z
(−e−x )(−dx)
e−x dx = (x − 1)e−x + e−x + C = xe−x + C
x′
1
x
= lim x = lim x ′ = lim x = 0
x→∞ (e )
x→∞ e
x→∞ e
by L’Hospital’s Rule. Therefore
Zt
Z∞
−x
(1 − x)e dx = lim (1 − x)e−x dx = lim xe−x ]t0 = lim (te−t − 0 · e0 ) = 0 − 0 = 0 (convergent)
t→∞
t→∞
0
t→∞
0
4
Section 6.6 Improper Integrals
2010 Kiryl Tsishchanka
5. We have
Z∞
Zt
t
dx
1
−1 x
= lim tan
x2 + 4 t→∞ 2
2 0
0
π
1 π
t
1
−1
−1
−0 =
tan
− tan 0 =
= lim
t→∞ 2
2
2 2
4
dx
= lim
2
t→∞
x +4
0
6. We first note that
Z∞
dx
=
x
e + e−x
−∞
Z∞
ex dx
=
e2x + 1
Z0
−∞
−∞
ex dx
+
e2x + 1
Z∞
(convergent)
ex dx
e2x + 1
0
We also note that
Z


x
e
=
u
Z
1
ex dx


x
=  d(e ) = du  =
du = tan−1 u + C = tan−1 (ex ) + C
2x
2
e +1
u
+
1
ex dx = du
Therefore
Z0
ex dx
= lim
e2x + 1 t→−∞
Z0
ex dx
π
π
= lim tan−1 (ex )]0t = lim (tan−1 (e0 ) − tan−1 (et )) = − 0 =
2x
t→−∞
t→−∞
e +1
4
4
Zt
ex dx
π π
π
= lim tan−1 (ex )]t0 = lim (tan−1 (et ) − tan−1 (e0 )) = − =
2x
t→∞
e + 1 t→∞
2
4
4
t
−∞
and
Z∞
0
hence
ex dx
= lim
e2x + 1 t→∞
0
Z∞
ex
π π
π
dx
= + =
−x
+e
4
4
2
−∞
5
(convergent)
Section 6.6 Improper Integrals
2010 Kiryl Tsishchanka
Type 2: Discontinuous Integrands
DEFINITION OF AN IMPROPER INTEGRAL OF TYPE 2:
(a) If f is continuous on [a, b) and is discontinuous at b, then
Zb
f (x)dx = lim−
t→b
Zt
f (x)dx
a
a
if this limit exists (as a finite number).
(b) If f is continuous on (a, b] and is discontinuous at a, then
Zb
f (x)dx = lim+
t→a
Zb
f (x)dx
t
a
if this limit exists (as a finite number).
Zb
The improper integral
f (x)dx is called convergent if the corresponding limit exists and divergent if
a
the limit does not exist.
(c) If f has a discontinuity at c, where a < c < b, then the improper integral
Zb
f (x)dx is defined as
a
Zb
a
f (x)dx =
Zc
f (x)dx +
Zb
f (x)dx
c
a
It is said to converge if both terms converge and diverge if either term diverges.
EXAMPLES:
1. Evaluate
Z2
1
dx
if possible.
1−x
Solution: We first note that the given integral is improper because f (x) =
x = 1. We have
Z2
1
dx
= lim
1 − x t→1+
2. Evaluate
Z3
0
√
Z2
t
1
has the vertical asymptote
1−x
dx
= lim − ln |1 − x|]2t = lim+ (− ln 1 + ln |1 − t|) = −∞ (divergent)
t→1
1 − x t→1+
dx
if possible.
9 − x2
6
Section 6.6 Improper Integrals
2. Evaluate
Z3
√
0
2010 Kiryl Tsishchanka
dx
if possible.
9 − x2
Solution: We first note that the given integral is improper because f (x) = √
asymptotes x = ±3. We have
Z3
0
3. Evaluate
Z1
Zt
x it
dx
= lim− sin−1
3 0
9 − x2 t→3
0
t
π
−1
−1
= lim− sin
− sin 0 = sin−1 1 − 0 =
t→3
3
2
dx
√
= lim−
t→3
9 − x2
1
has the vertical
9 − x2
√
(convergent)
dx
if possible.
x
−1
1
Solution: We first note that the given integral is improper because f (x) = has the vertical asymptote
x
x = 0. We have
Z0
Z1
Z1
dx
dx
dx
=
+
x
x
x
−1
Since
Z1
dx
= lim+
t→0
x
it follows that
dx
= lim+ ln |x|]1t = lim+ (ln 1 − ln |t|) = ∞
t→0
t→0
x
t
0
Z1
Z1
0
−1
dx
is divergent.
x
−1
A Comparison Test for Improper Integrals
COMPARISON TEST: Suppose that f and g are continuous functions with f (x) ≥ g(x) ≥ 0 for x ≥ a.
Z∞
Z∞
(a) If f (x)dx is convergent, then g(x)dx is convergent.
a
(b) If
Z∞
a
g(x)dx is divergent, then
a
EXAMPLE: The integral
Z∞
f (x)dx is divergent.
a
Z∞
dx
1
1
is convergent, because 2 > x
> 0 and
x
2
e +x
x
e + x2
1
Z∞
1
the p-test, since p = 2 > 1.
7
dx
is convergent by
x2
Section 6.6 Improper Integrals
2010 Kiryl Tsishchanka
EXAMPLE: Does the integral
Z∞
1
dx converge?
xex
1
Solution: We have
0<
Note that
Z∞
1
1
< x
x
xe
e
1
dx is convergent, since
ex
1
Z∞
1
dx =
ex
1
Therefore the integral
Z∞
−x
e dx = lim
t→∞
e−x dx = lim [−e−x ]t1 = lim (−e−t + e−1 ) = e−1
t→∞
t→∞
1
1
Z∞
Zt
1
dx converges.
xex
1
EXAMPLE: Does the integral
Z∞
1
√
dx
converge?
x3 + 1
Solution: We have
0< √
Note that
converges.
Z∞
1
Z∞
3
Solution: We have
Note that
diverges.
3
√
5
1
1
0< √
< √
5
5
2
2
x −x−3
x
Z∞
2
1
√
dx
x3 + 1
Z∞
3
dx
√
5
2
x −x−3
2 + sin x
dx converge?
x−1
Solution: We have
0<
Note that
Z∞
dx
converge?
x2 − x − 3
1
√
dx is divergent by the p-test, since p = 2/5 ≤ 1. Therefore the integral
5
x2
EXAMPLE: Does the integral
Z∞
1
<√
+1
x3
1
√ dx is convergent by the p-test, since p = 3/2 > 1. Therefore the integral
x3
EXAMPLE: Does the integral
Z∞
1
x3
1
2 + sin x
<
x
x−1
1
dx is divergent by the p-test, since p = 1 ≤ 1. Therefore the integral
x
Z∞
2
2
8
2 + sin x
dx diverges.
x−1