Improper Integrals

Improper Integrals b

In defining the definite integral

∫

f ( x) dx , we dealt with a function f defined on a

a

finite interval [a, b] and we assumed that f does not have an infinite discontinuity. However, we can extend the concept of a definite integral to the case where the interval is infinite and also to the case where f has an infinite discontinuity in [a, b]. In either case the integral is called an improper integral. I. Improper Integral on an Infinite Interval A. Definition ∞

M

1. If

∫

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

a

∫

f ( x) dx =

a

M

lim

M →∞

∫

f ( x) dx provided this limit exists (as a finite number).

a

b

2. If

b

∫

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

∫

f ( x) dx =

−∞

M

b

lim

M → −∞

∫

f ( x) dx provided this limit exists (as a finite number).

M

∞

[Note: The integrals

b

∫

f ( x) dx and

∫

f ( x) dx are said to be convergent

−∞

a

if the corresponding limit exists and divergent if the limit does not exist.] ∞

3. If both

∫

a

f ( x) dx and

∞

−∞

used.]

f ( x) dx are convergent, then we define

−∞

a

∫

∫

∞

a

f ( x) dx =

∫

−∞

f ( x) dx +

∫ a

f ( x ) dx . [Note: Any real number a can be

1 B. Examples ∞

1. Evaluate

∫ 1

∞

∫ 1

1 dx . x2 M

∫

1 dx = Mlim →∞ x2

M

x

−2

1

−1   − 1 + 1 = 1 ⇒ dx = Mlim = Mlim  →∞  →∞  M    x 1

∞

∫ 1

1 dx x2

converges to 1. ∞

2. Evaluate

∫ 1

∞

∫ 1

∞

∫ 1

1 x

dx

M

1 x 1 x

dx = Mlim →∞

∫

1

1

x

{

dx = Mlim 2 x →∞

}

M 1

{

}

= Mlim 2 M −2 =∞⇒ →∞

dx diverges.

0

3. Evaluate

∫

e x dx .

−∞ 0

∫

0

e dx = Mlim → −∞ x

−∞

∫e

x

dx = Mlim → −∞

{e } x

0 M

{

e0 − e M = Mlim → −∞

}=

1–0=1 ⇒

M

0

∫

e x dx converges to 1.

−∞ ∞

4. Evaluate

∫

−∞ ∞

∫

−∞

1 dx 1+ x2

1 dx = 1+ x2

0

∫

−∞

1 dx + 1+ x2

∞

∫ 0

∞

∫

1 1 dx = 2 dx (by symmetry) = 2 1+ x 1+ x2 0

2 M

2 Mlim →∞

∫ 0

1 dx = 2 Mlim { arctan x } →∞ 1+ x2

π  arctan M = 2  =π ⇒ 2 Mlim →∞ 2

∞

∫

−∞

M 0

{ arctan M − arctan 0} = = 2 Mlim →∞

1 dx converges to π . 1+ x2

∞

5. Evaluate

∫ ln x dx . 1

∞

∫

M

ln x dx = Mlim →∞

1

∫

ln x dx = Mlim { x ln x − x} 1M = Mlim [ { M ln M − M } – →∞ →∞

1

{1ln 1 − 1} ] = Mlim { M (ln M − 1)} + 1 =  lim M  ⋅ lim (ln M − 1) + 1 = →∞  M →∞

 M →∞



∞

∞ ⋅ ∞ +1 = ∞ ⇒

∫ ln x dx diverges. 1

∞

6. Evaluate

∫

xe − x dx .

0

∞

∫

M

xe

−x

dx = Mlim →∞

0

∫

{

xe − x dx = Mlim − xe − x − e − x →∞

0

}

M 0

−x 1 M = Mlim  x − x0 = →∞ e   e

1  − 0 1  −M − M  lim lim − − − . By L’Hospital’s Rule ,      M = M M 0 0 M →∞ M →∞  e

e 

e

e 

e



1  − 0 1   −1 −M lim lim = 0 − −  0 − 0  = (0 − 0) − (0 − 1) = 1 . . Thus,    M M M  M →∞ M →∞ e 

 e

∞

Thus,

∫ 0

xe − x dx converges to 1.

e 

e

e 

3 II. Improper Integral with Discontinuous Integrand A. Definition b

1. If f is continuous on [ a, b ) and is discontinuous at b, then

∫

f ( x) dx =

a

M

lim

M →b −

∫

f ( x) dx if this limit exists (as a finite number).

a

b

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

∫

f ( x) dx =

a

b

lim

M →a +

∫

f ( x) dx if this limit exists (as a finite number).

M

b

[Note: The improper integral

∫

f ( x) dx is called convergent if the

a

corresponding limit exists and divergent if the limit does not exist.] c

3. If f has a discontinuity at c, where a < c < b , and both

∫

f ( x) dx and

a

b

∫

b

f ( x) dx are convergent, then we define

c

∫ a

c

f ( x) dx =

∫

f ( x) dx +

a

b

∫

f ( x) dx .

c

B. Examples 1

1. Evaluate

∫ 0

1

∫ 0

1 dx . x2

1 dx = lim+ M →0 x2

1

∫ M

1  − 1 1  − 1 +  = +∞ ⇒ x −2 dx = Mlim   M = Mlim +  →0 + → 0 M  x 

4 1

∫ 0

1 dx diverges. x2 4

∫

2. Evaluate

1

0

4

∫

1

0

x

dx .

x

4

∫

dx = lim+ M →0

x

4−2 0 = 4⇒

∫

1 x

0

1

3. Evaluate

∫ 0

∫ 0

2

{ }

dx = Mlim 2 x →0 +

4 M

{

}

= Mlim 4−2 M = →0 +

M

4

1

−1

1− x2

dx .

M

1 1− x

1

dx converges to 4.

dx = lim− M →1

2

∫ 0

1 1− x

2

dx = lim− { arcsin x} M →1

lim { arcsin M − arcsin 0} = arcsin 1 − 0 = π ⇒ M →1− 2

1

∫ 0

M 0

1 1− x

=

dx converges to

2

1

4. Evaluate

∫ ln x dx . 0

1

1

∫ ln x dx = lim ∫ ln x dx = lim { x ln x − x} M →0 +

0

M →0 +

1 M

= (1 ln 1 − 1) −

M

lim ( M ln M − M ) = −1 − lim ( M ln M ) + 0 = −1 + lim M ln M . By M →0 M →0

M →0 +

+

+

1

ln M

L’Hospital’s Rule, Mlim M ln M = Mlim 1 →0 →0 +

+

M = = Mlim →0 − 1 M M2 +

π . 2

5

lim (−M ) = 0 . Thus, lim M →0

M →0 +

+

2

∫

5. Evaluate

1

2

∫

1

M

0

∫ ln x dx = −1 ⇒ ∫ ln x dx converges to – 1.

1 dx . x ln x 2

∫

1 dx = Mlim →1+ x ln x

1

1

M

1 dx = Mlim { ln(ln x)} →1+ x ln x

2 M

ln(ln 2) − = Mlim →1+

lim ln(ln M ) = ln(ln 2) − ln(ln(1)) = ln(ln 2) − ln(0) = ln(ln 2) − (−∞) ⇒

M →1+ 2

∫

1 dx diverges. x ln x

1

4

∫

6. Evaluate

0

4

∫

e

x

x

dx .

4

x

x

0

e

dx = lim+ M →0

∫

e

M

x

x 4

2e − 2e = 2e − 2 ⇒ 2

0

2

∫ 0

{

dx = Mlim 2e →0 +

e

x

}

4 M

{

= Mlim 2e →0 +

4

− 2e

x

x

dx converges to 2e 2 − 2 .

Practice Sheet for Improper Integrals 1

(1)

∫ 0

1 1− x2

dx = 6

M

}=

∞ 1

(2)

∫

e x

dx =

x2

1

4

(3)

∫

1

dx =

4− x

0

4

(4)

∫

e x

dx =

x

0

∞

(5)

∫(

1

dx =

x ln x ) 2

e

3

(6)

∫

x

0

∞

(7)

∫

0

dx =

9 − x2

x2 1 + x6

dx =

2

(8)

∫

1

dx =

x x2 − 1

1

∞

(9)

∫

xe − x

dx =

0

∞

(10)

∫ 1

1 1 + x2

dx = 7

∞

(11)

∫

x

dx =

1 + x2

0

∞

(12)

∫

1 xln x

dx =

e

∞

(13)

∫ 0

arctan x

dx =

1+ x2

e

(14)

∫

1 dx = x ln x

1

e

(15)

∫ 1

1 x(ln x) 2

dx =

Solution Key for Improper Integrals 1

(1)

∫

M

1 1− x

0

2

dx =

∫

lim

M → 1−

0

arcsin 1 − arcsin 0 =

π 2

1 1− x

2

dx =

1

. Thus,

∫ 0

∞

(2)

M

1

∫

e x dx = x2

1

∞

∫ 1

lim M→ ∞

∫ 1

1

e x dx = x2

1 1− x

2

{

converges to

e−1

. 8

1

= lim {arcsin M −arcsin 0} M→ 1−

converges to

dx

lim −e M →∞

1

e x dx x2

{arcsin x} 0M

lim

M → 1−

x

}

M

1

{

= lim − e M →∞

1

M

π 2

.

}

+ e = −1 + e

. Thus,

=

4

(3)

∫

M

1

dx =

4− x

0

∫

Thus,

1 4− x

0

(4)

∫ 0

4

∫

e

0

∞

(5)

dx =

x

∫

lim

M → 0+

x

e

dx =

x

M

M

M →4

−

M →4 −

{

x

lim 2e

M →0

+

}

4 M

{

= lim 2e + M →0

4

− 2e

M

} = 2e

2

−2

. Thus,

x

converges to

dx

x

∫ x( ln x)

2e 2 −2.

M

1

2

∫ x( ln x)

lim M→ ∞

=

dx

e

∞

4− x

lim {−2 4 − x }0 = lim {−2 4 − M + 4} = 4

dx =

converges to 4. 4

x

e

1

0

4

4

∫

lim

M → 4−

1

M

 −1   −1  +1 = 1  = lim  lim  M →∞ ln x e M →∞ ln M 

=

2

e

∫ x( ln x) 1

dx

2

. Thus,

converges to 1.

e

3

(6)

∫

M

x 9− x

0

2

∫ 0

∞

(7)

∫ 0

x2 dx 1 + x6

(8)

∫x 1

1 x −1 2

x 9 − x2

x 9− x

=

∫

lim M→ ∞

0

. Thus,

∫ 0

2

dx

=

M → 1+

dx =

x2 dx = 1+ x6

∞

lim

2

{

2 lim − 9 − x

M →3

−

}

M

0

{

= lim − 9 − M M →3−

2

}

+3 =

converges to 3.

dx

M

1 π  π 0=  = 3 2  6

2

M → 3−

0

3

3. Thus,

∫

lim

=

dx

x2 dx 1 + x6

∫x M

1 x −1 2

M

1 1 3  3   arctan(M ) − lim  arctan( x ) = Mlim M →∞ 3 →∞ 3 0 

converges to

dx =

9

π 6

.

2 lim {arc sec x} M = arc sec 2 −

M→ 1+

.

π

2

π

lim {arc sec M } = 3 − 0 = 3

∫x

. Thus,

M →1+

1

∞

(9)

∞

∫

xe − x dx

∫

lim M→ ∞

=

0

∞

(10)

∫ 1

π π π − = 2 4 4

π 3

.

lim  M →∞

{0 – 1}=

∞

∫ xe

. Thus,

−x

converges to 1.

dx

0

∫ 1+ x

lim

=

x −1

converges to

1  1  − x − M − x  = lim  M − M  − x M → ∞ e e e e  0  

=

M

1 dx 1+ x2

dx

2

M

xe − x dx

0

 −1  lim  M − 0 + 1 = 1 M →∞ e 

1

M→ ∞

1

2

M {arctan M } −arctan 1 = lim {arctan x}1 = M lim →∞

dx =

M →∞

1

∞

∫ 1+ x

. Thus,

1

2

π

converges to

dx

.

4

1

∞

(11)

∫ 0

M

x dx 1+ x2

∫

lim M→ ∞

=

0

x dx = 1+ x2

∞

0 = ∞ − 0 = ∞

∫ 1+ x

. Thus,

x

2

dx

M

1 1 2  2   ln 1 + x  = lim  ln 1 + M  − lim M →∞ 2 M →∞ 2 

0





diverges.

0

∞

(12)

∫ e

M

1 dx x ln x

∫ x ln x dx =

lim M→ ∞

=

lim {ln ln x

M →∞

}

M

e

{

= lim ln ln M M →∞

e

∞

∞ −0 =∞

1

∫ x ln x dx

. Thus,

1

diverges.

e

∞

(13)

∫ 0

arctan x dx 1 + x2

M

=

lim M→ ∞

∫ 0

arctan x dx = 1+ x2

M

 2  (arctan x )  = lim M →∞ 2 1



10

0

} −ln ln e

=

∞

∫

2

1 π  π2 1 2  (arctan M )  − 0 =   = lim M →∞ 2 22  8 

e

(14)

∫

e

1 dx = x ln x

∫

lim

M → 1+

M

1

0 −( − ∞ ) =∞

M →1+

∫ x ln x dx

. Thus,

0

lim {ln ln x

1 dx = x ln x

e

. Thus,

1

}

e M

arctan x dx 1 + x2

converges to

{

= ln ln e − lim ln ln M + M →1

}=

diverges.

1

e

(15)

∫ 1

e

1 x(ln x)

2

dx =

∫

lim

M → 1+

M

e

− 1+ ∞ = ∞

. Thus,

−1

1 dx = x(ln x) 2

∫ x(ln x) 1

2

dx

e

−1

−1

    − lim   = = lim  1 M →1 ln x M M →1 ln M  +

diverges.

1

11

+

π2 8

.