improper integrals

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 theline x = 1. You might think that, since S is infinite in extend, its area must be infinite. However, this isnot true. In fact, the area of the part of S that lies to the left of the line x = t is

A(t) =

t∫

1

1

x2dx = −1

x

]t

1

= 1 − 1

t

Notice that A(t) < 1 no matter how large t is chosen. Moreover, since

limt→∞

A(t) = limt→∞

(

1 − 1

t

)

= 1

we can say that the area of the infinite region S is equal to 1 and we write

∞∫

1

1

x2dx = lim

t→∞

t∫

1

1

x2dx = 1

DEFINITION OF AN IMPROPER INTEGRAL OF TYPE 1:

(a) If

t∫

a

f(x)dx exists for every number t ≥ a, then

∞∫

a

f(x)dx = limt→∞

t∫

a

f(x)dx provided this limit exists

(as a finite number).

(b) If

b∫

t

f(x)dx exists for every number t ≤ b, then

b∫

−∞

f(x)dx = limt→−∞

b∫

t

f(x)dx provided this limit exists

(as a finite number).

The improper integrals

∞∫

a

f(x)dx and

b∫

−∞

f(x)dx are called convergent if the corresponding limit exists

and divergent if the limit does not exist.

(c) The improper integral

∞∫

−∞

f(x)dx is defined as

∞∫

−∞

f(x)dx =

a∫

−∞

f(x)dx+

∞∫

a

f(x)dx, where a is any real

number. It is said to converge if both terms converge and diverge if either term diverges.

1


EXAMPLES:

1. Evaluate

∞∫

1

1

xdx if possible.

Solution: We have

∞∫

1

1

xdx = lim

t→∞

t∫

1

1

xdx = lim

t→∞

ln |x|]t1 = limt→∞

(ln t − ln 1) = limt→∞

ln t = ∞

The limit does not exist as a finite number and so the improper integral

∞∫

1

1

xdx is divergent.

2. Evaluate

∞∫

2

1

x2dx if possible.

Solution: We have

∞∫

2

1

x2dx = lim

t→∞

t∫

2

1

x2dx = lim

t→∞

−1

x

]t

2

= limt→∞

(

−1

t+

1

2

)

= 0 +1

2=

1

2(convergent)

3. Evaluate

∞∫

4

1√xdx if possible.

Solution: We have

∞∫

4

1√xdx = lim

t→∞

t∫

4

1√xdx = lim

t→∞

t∫

4

x−1/2dx = limt→∞

x−1/2+1

−1/2 + 1

]t

4

= limt→∞

2√

x]t

4= lim

t→∞

(

2√

t − 2√

4)

= ∞

The limit does not exist as a finite number and so the improper integral

∞∫

4

1√xdx is divergent.

4. For what values of p is

∞∫

1

1

xpdx convergent?

2


4. For what values of p is

∞∫

1

1

xpdx convergent?

Solution: We know that if p = 1, then the integral is divergent, so let’s assume that p 6= 1. Then

∞∫

1

1

xpdx = lim

t→∞

t∫

1

1

xpdx = lim

t→∞

t∫

1

x−pdx = limt→∞

x−p+1

−p + 1

]t

1

= limt→∞

1

(1 − p)xp−1

]t

1

= limt→∞

1

1 − p

(

1

tp−1− 1

)

If p > 1, then p − 1 > 0, so as t → ∞, tp−1 → ∞ and1

tp−1→ 0. Therefore

∞∫

1

1

xpdx =

1

p − 1if p > 1

and so the integral converges. On the other hand, if p < 1, then p − 1 < 0 and so

1

tp−1= t1−p → ∞ as t → ∞

and the integral diverges. So,

∞∫

1

1

xpdx is convergent if p > 1 and divergent if p ≤ 1.

EXAMPLES: Determine whether each integral is convergent or divergent. Evaluate those that are con-vergent.

1.

0∫

−∞

exdx

2.

∞∫

0

exdx

3.

∞∫

−∞

xdx

4.

∞∫

0

(1 − x)e−xdx

5.

∞∫

0

dx

x2 + 4

6.

∞∫

−∞

dx

ex + e−x

3


SOLUTIONS:

1. We have0

∫

−∞

exdx = limt→−∞

0∫

t

exdx = limt→−∞

ex]0t = limt→−∞

(e0 − et) = (1 − 0) = 1 (convergent)

2. We have∞

∫

0

exdx = limt→∞

t∫

0

exdx = limt→∞

ex]t0 = limt→∞

(et − e0) = ∞ (divergent)

3. Since∞

∫

−∞

xdx =

0∫

−∞

xdx +

∞∫

0

xdx

and∞

∫

0

xdx = limt→∞

t∫

0

xdx = limt→∞

x2

2

]t

0

= limt→∞

(

t2

2− 02

2

)

= ∞ (divergent)

it follows that

∞∫

−∞

xdx is divergent.

4. We first note that

∫

(1 − x)e−xdx =

1 − x = u e−xdx = dv

d(1 − x) = du −e−x = v

−dx = du

= (1 − x)(−e−x) −

∫

(−e−x)(−dx)

= (x − 1)e−x −∫

e−xdx = (x − 1)e−x + e−x + C = xe−x + C

We also note that

limx→∞

xe−x = limx→∞

x

ex= lim

x→∞

x′

(ex)′= lim

x→∞

1

ex= 0

by L’Hospital’s Rule. Therefore

∞∫

0

(1 − x)e−xdx = limt→∞

t∫

0

(1 − x)e−xdx = limt→∞

xe−x]t0 = limt→∞

(te−t − 0 · e0) = 0 − 0 = 0 (convergent)

4


5. We have

∞∫

0

dx

x2 + 4= lim

t→∞

t∫

0

dx

x2 + 4= lim

t→∞

1

2tan−1

(x

2

)

]t

0

= limt→∞

1

2

(

tan−1

(

t

2

)

− tan−1 0

)

=1

2

(π

2− 0

)

=π

4(convergent)

6. We first note that∞

∫

−∞

dx

ex + e−x=

∞∫

−∞

exdx

e2x + 1=

0∫

−∞

exdx

e2x + 1+

∞∫

0

exdx

e2x + 1

We also note that

∫

exdx

e2x + 1=

ex = u

d(ex) = du

exdx = du

=

∫

1

u2 + 1du = tan−1 u + C = tan−1(ex) + C

Therefore

0∫

−∞

exdx

e2x + 1= lim

t→−∞

0∫

t

exdx

e2x + 1= lim

t→−∞

tan−1(ex)]0t = limt→−∞

(tan−1(e0) − tan−1(et)) =π

4− 0 =

π

4

and

∞∫

0

exdx

e2x + 1= lim

t→∞

t∫

0

exdx

e2x + 1= lim

t→∞

tan−1(ex)]t0 = limt→∞

(tan−1(et) − tan−1(e0)) =π

2− π

4=

π

4

hence∞

∫

−∞

dx

ex + e−x=

π

4+

π

4=

π

2(convergent)

5


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

b∫

a

f(x)dx = limt→b−

t∫

a

f(x)dx

if this limit exists (as a finite number).

(b) If f is continuous on (a, b] and is discontinuous at a, then

b∫

a

f(x)dx = limt→a+

b∫

t

f(x)dx

if this limit exists (as a finite number).

The improper integral

b∫

a

f(x)dx is called convergent if the corresponding limit exists and divergent if

the limit does not exist.

(c) If f has a discontinuity at c, where a < c < b, then the improper integral

b∫

a

f(x)dx is defined as

b∫

a

f(x)dx =

c∫

a

f(x)dx +

b∫

c

f(x)dx

It is said to converge if both terms converge and diverge if either term diverges.

EXAMPLES:

1. Evaluate

2∫

1

dx

1 − xif possible.

Solution: We first note that the given integral is improper because f(x) =1

1 − xhas the vertical asymptote

x = 1. We have

2∫

1

dx

1 − x= lim

t→1+

2∫

t

dx

1 − x= lim

t→1+− ln |1 − x|]2t = lim

t→1+(− ln 1 + ln |1 − t|) = −∞ (divergent)

2. Evaluate

3∫

0

dx√9 − x2

if possible.

6


2. Evaluate

3∫

0

dx√9 − x2

if possible.

Solution: We first note that the given integral is improper because f(x) =1√

9 − x2has the vertical

asymptotes x = ±3. We have

3∫

0

dx√9 − x2

= limt→3−

t∫

0

dx√9 − x2

= limt→3−

sin−1(x

3

)]t

0

= limt→3−

(

sin−1

(

t

3

)

− sin−1 0

)

= sin−1 1 − 0 =π

2(convergent)

3. Evaluate

1∫

−1

dx

xif possible.

Solution: We first note that the given integral is improper because f(x) =1

xhas the vertical asymptote

x = 0. We have1

∫

−1

dx

x=

0∫

−1

dx

x+

1∫

0

dx

x

Since1

∫

0

dx

x= lim

t→0+

1∫

t

dx

x= lim

t→0+ln |x|]1t = lim

t→0+(ln 1 − ln |t|) = ∞

it follows that

1∫

−1

dx

xis divergent.

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.

(a) If

∞∫

a

f(x)dx is convergent, then

∞∫

a

g(x)dx is convergent.

(b) If

∞∫

a

g(x)dx is divergent, then

∞∫

a

f(x)dx is divergent.

EXAMPLE: The integral

∞∫

1

dx

ex + x2is convergent, because

1

x2>

1

ex + x2> 0 and

∞∫

1

dx

x2is convergent by

the p-test, since p = 2 > 1.

7


EXAMPLE: Does the integral

∞∫

1

1

xexdx converge?

Solution: We have

0 <1

xex<

1

ex

Note that

∞∫

1

1

exdx is convergent, since

∞∫

1

1

exdx =

∞∫

1

e−xdx = limt→∞

t∫

1

e−xdx = limt→∞

[−e−x]t1 = limt→∞

(−e−t + e−1) = e−1

Therefore the integral

∞∫

1

1

xexdx converges.


∞∫

1

dx√x3 + 1

converge?

Solution: We have

0 <1√

x3 + 1<

1√x3

Note that

∞∫

1

1√x3

dx is convergent by the p-test, since p = 3/2 > 1. Therefore the integral

∞∫

1

dx√x3 + 1

converges.


∞∫

3

dx5√

x2 − x − 3converge?

Solution: We have

0 <1

5√

x2<

15√

x2 − x − 3

Note that

∞∫

3

15√

x2dx is divergent by the p-test, since p = 2/5 ≤ 1. Therefore the integral

∞∫

3

dx5√

x2 − x − 3

diverges.


∞∫

2

2 + sin x

x − 1dx converge?

Solution: We have

0 <1

x<

2 + sin x

x − 1

Note that

∞∫

2

1

xdx is divergent by the p-test, since p = 1 ≤ 1. Therefore the integral

∞∫

2

2 + sin x

x − 1dx diverges.

8

improper integrals

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