<<

. 11
( 14)



>>


π/4 π/6
3
cos4 3x sin 3x dx
49. sin 2x cos 2x dx 50.
0 0

1 1/2
earctan x earcsin x

51. dx 52. dx
1 + x2 2
1’x
0 0
276 CHAPTER 6. TECHNIQUES OF INTEGRATION


3 1
earcsec x dx
√ √
53. dx 54.
2’1 1 + x2
xx
2 0



6.3 Integration by Parts
Theorem 6.3.1 Let f (x), g(x), f (x) and g (x) be continuous on an interval
[a, b]. Then

f (x)g (x)dx = f (x)g(x) ’
(i) g(x)f (x)dx

b b
f (x)g (x)dx = (f (b)g(b) ’ f (a)g(a)) ’
(ii) g(x)f (x)dx
a a


udv = uv ’
(iii) vdu

where u = f (x) and dv = g (x)dx are the parts of the integrand.
Proof. See the proof of Theorem 5.4.1.


Exercises 6.3 Evaluate each of the following integrals.

1. x sin x dx 2. x cos x dx


x ex dx
3. x ln x dx 4.


x 4x dx x2 ln x dx
5. 6.


x2 sin x dx x2 cos x dx
7. 8.


x2 ex dx x2 10x dx
9. 10.
6.3. INTEGRATION BY PARTS 277


ex sin x dx (Let u = ex twice and solve.)
11.


ex cos x dx (Let u = ex twice and solve.)
12.


e2x sin 3x dx (Let u = e2x twice and solve.)
13.


x2 cos(2x)dx
14. x sin(3x)dx 15.


x2 e4x dx x3 ln(2x)dx
16. 17.


x sec2 x dx x csc2 x dx
18. 19.


x2 cosh x dx
20. x sinh(4x)dx 21.


22. x cos(5x)dx 23. sin(ln x)dx


24. cos(ln x)dx 25. x arcsin x dx


26. x arccos x dx 27. x arctan x dx


28. x arcsec x dx 29. arcsin x dx


30. arccos x dx 31. arctan x dx


32. arcsec x dx


Verify the following integration formulas:
278 CHAPTER 6. TECHNIQUES OF INTEGRATION

sinn’1 (ax) cos(ax) n ’ 1
n
(sinn’2 ax)dx
sin (ax)dx = ’
33. +
na n
n’1
1
cosn (ax)dx = cosn’1 (ax) sin(ax) + (cosn’2 ax)dx
34.
na n

xn ex dx = xn ex ’ n xn’1 ex dx
35.


xn sin x dx = ’xn cos x + n xn’1 cos x dx
36.


xn cos x dx = xn sin x ’ n xn’1 sin x dx
37.

1
eax [a sin(bx) ’ b cos(bx)] + C
eax sin(bx)dx =
38. 2 + b2
a
1
eax cos(bx) dx = eax [a cos(bx) + b sin(bx)] + C
39.
a2 + b2
1 1
xn ln x dx = xn+1 ln x ’ xn+1 + C, n = ’1, x > 0
40.
(n + 1)2
n+1
n’2
1
secn x dx = secn’2 x tan x + secn’2 x dx, n = 1, n > 0
41.
n’1 n’1

’1 n’2
cscn x dx = cscn’2 x cot x + cscn’2 x dx, n = 1, n > 0
42.
n’1 n’1
Use the formulas 33“42 to evaluate the following integrals:

sin4 x dx cos5 x dx
43. 44.


x3 ex dx x4 sin x dx
45. 46.


x3 cos x dx e2x sin 3x dx
47. 48.


e3x cos 2x dx x5 ln x dx
49. 50.
6.3. INTEGRATION BY PARTS 279



sec3 x dx csc3 x dx
51. 52.

Prove each of the following formulas:

1
tann x dx = tann’1 x ’ tann’2 x dx, n = 1
53.
n’1
1
cotn x dx = cotn’1 x ’ cotn’2 x dx, n = 1
54.
n’1

sin2n+1 x dx = ’ (1 ’ u2 )n du, u = cos x
55.


cos2n+1 x dx = ’ (1 ’ u2 )n du, u = sin x
56.


sin2n+1 x cosm x dx = ’ (1 ’ u2 )n um du, u = cos x
57.


cos2n+1 x sinm x dx = (1 ’ u2 )n um du, u = sin x
58.


sin2n x cos2m x dx = (sin x)2n (1 ’ sin2 x)m dx
59.


tann x sec2m x dx = un (1 + u2 )m’1 du, u = tan x
60.


cotn x csc2m x dx = ’ un (1 + u2 )m’1 du, u = cot x
61.


tan2n+1 x secm x dx = (u2 ’ 1)n um’1 du, u = sec x
62.


cot2n+1 x cscm x dx = ’ (u2 ’ 1)n um’1 du, u = csc x
63.

cos(m + n)x cos(m ’ n)x
1
+ C; m2 = n2
sin mx cos nx dx = ’
64. +
m’n
2 m+n
280 CHAPTER 6. TECHNIQUES OF INTEGRATION

1 sin(m ’ n)x sin(m + n)x
+ C; m2 = n2

65. sin mx sin nx dx =
m’n
2 m+n
1 sin(m ’ n)x sin(m + n)x
+ C; m2 = n2
66. cos mx cos nx dx = +
m’n
2 m+n

6.4 Trigonometric Integrals
The trigonometric integrals are of two types. The integrand of the ¬rst
type consists of a product of powers of trigonometric functions of x. The
integrand of the second type consists of sin(nx) cos(mx), sin(nx) sin(mx) or
cos(nx) cos(mx). By expressing all trigonometric functions in terms of sine
and cosine, many trigonometric integrals can be computed by using the fol-
lowing theorem.

Theorem 6.4.1 Suppose that m and n are integers, positive, negative, or
zero. Then the following reduction formulas are valid:

(n ’ 1)
’1
sinn’1 x cos x + sinn’2 x dx, n > 0
sinn x dx =
1.
n n
1 n
sinn’2 x dx = sinn’1 x cos x + sinn x dx, n ¤ 0
2.
n’1 n’1

(sin x)’1 dx = csc x dx = ln | csc x’cot x|+c or ’ ln | csc x+cot x|+c
3.

n’1
1
cosn x dx = cosn’1 x sin x + cosn’2 x dx, n > 0
4.
n n
’1 n
cosn’2 x dx = cosn’1 x sin x + cosn x dx, n ¤ 0
5.
n’1 n’1

(cos x)’1 dx = sec x dx = ln | sec x + tan x| + c
6.


sinn x cos2m+1 x dx = sinn x(1 ’ sin2 x)m cos x dx
7.
un (1 ’ u2 )m du, u = sin x, du = cos x dx
=
6.4. TRIGONOMETRIC INTEGRALS 281


sin2n+1 x cosm x dx = cosm x(1 ’ cos2 x)n sin x dx
8.
= ’ um (1 ’ u2 )n du, u = cos x, du = ’ sin x dx

sin2n x cos2m x dx = (1 ’ cos2 x)n cos2m x dx
9.
= (1 ’ sin2 x)m sin2n x dx

’1 cos(m + n)x cos(m ’ n)x
+ c, m2 = n2
10. sin(nx) cos(mx)dx = +
m’n m’n
2

1 sin(m ’ n)x sin(m + n)x
+ c, m2 = n2

11. sin(mx) sin(mx) dx =
m’n
2 m+n

1 sin(m ’ n)x sin(m + n)x
+ c, m2 = n2
12. cos(mx) cos(mx) dx = +
m’n
2 m+n

Corollary. The following integration formulas are valid:

tann’1 u
n
tann’2 u d

13. tan u du =
n’1

n’2
1
secn u du = secn’2 x tan x + secn’2 x dx
14.
n’1 n’1

’1 n’2
cscn u du = cscn’2 x cot x + cscn’2 x dx
15.
n’1 n’1



Exercises 6.4 Evaluate each of the following integrals.

sin5 x dx cos4 x dx
1. 2.


tan5 x dx cot4 x dx
3. 4.


sec5 x dx csc4 x dx
5. 6.
282 CHAPTER 6. TECHNIQUES OF INTEGRATION



sin5 x cos4 x dx sin3 x cos5 x dx
7. 8.


sin4 x cos3 x dx sin2 x cos4 x dx
9. 10.


tan5 x sec4 x dx cot5 x csc4 x dx
11. 12.


tan4 x sec5 x dx cot4 x csc5 x dx
13. 14.


tan4 x sec4 x dx cot4 x csc4 x dx
15. 16.


tan3 x sec3 x dx cot3 x csc3 x dx
17. 18.


19. sin 2x cos 3x dx 20. sin 4x cos 4x dx


21. sin 3x cos 3x dx 22. sin 2x sin 3x dx


23. sin 4x sin 6x dx 24. sin 3x sin 5x dx


25. cos 3x cos 5x dx 26. cos 2x cos 4x dx


27. cos 3x cos 4x dx 28. sin 4x cos 4x dx




6.5 Trigonometric Substitutions

Theorem 6.5.1 (a2 ’ u2 Forms). Suppose that u = a sin t, a > 0. Then
6.5. TRIGONOMETRIC SUBSTITUTIONS 283




2 2 2 2
du = a cos tdt, a ’ u = a cos t, a2 ’ u2 = a cos t, t = arcsin(u/a),

a2 ’ u2
u u
, tan t = √
sin t = , cos t = ,
a a a2 ’ u2

a2 ’ u2 a a
, sec t = √
cot t = , csc t .
u u
a2 ’ u2




graph



The following integration formulas are valid:

udu 1
= ’ ln |a2 ’ u2 | + c
1.
a2 ’ u2 2

a’u 1
du 1 u
+ c = arctanh
2. = ln +c
a2 ’ u2 2a a+u a a


udu
√ = ’ a2 ’ u2 + c
3.
2 ’ u2
a

du u

4. = arcsin +c
a
a2 ’ u2

a2 ’ u2
a
du 1
√ = ln ’ +c
5.
a u u
u a2 ’ u2

√ 1 √2
a2 u
2 ’ u2 du = + u a ’ u2 + c
6. a arcsin
2 a 2

Proof. The proof of this theorem is left as an exercise.
284 CHAPTER 6. TECHNIQUES OF INTEGRATION

Theorem 6.5.2 (a2 + u2 Forms). Suppose that u = a tan t, a > 0. Then
√ u
2 2 2 2 2
du = a sec tdt, a + u = a sec t, a2 + u2 = a sec t, t = arctan ,
a
u a u
sin t = √ , cos t = √ , tan t =
a
a2 + u2 a2 + u2
√ √
a2 + u2 a2 + u2 a
csc t = , sec t = , cot t = .
u a u

graph



Proof. The proof of this theorem is left as an exercise.
The following integration formulas are valid:
1
udu
= ln a2 + u2 + c
1.
a2 + u2 2
du 1 u
2. = arctan +c
a2 + u2 a a

udu
√ = a2 + u2 + c
3.
2 + u2
a

du
√ = ln u + a2 + u2 + c
4.
2 + u2
a

a2 + u2 a
du 1
√ ’ +c
5. = ln
a u u
u a2 + u2
√ √
1 √2 a2
2 + u2 du = 2+ ln u + a2 + u2 + c
6. a u a +u
2 2

Theorem 6.5.3 (u2 ’ a2 Forms) Suppose that u = a sec t, a > 0. Then
√ u
2 2 2 2
du = a sec t tan t dt, u ’ a = a tan t, u2 ’ a2 = a tan t, t = arcsec ,
a
√ √
2 ’ a2 u2 ’ a2
u a
sin t = , cos t = , tan t = ,
u u a
u u a
csc t = √ , sec t = , cot t = √ .
a
2 ’ a2 2 ’ a2
u u
6.5. TRIGONOMETRIC SUBSTITUTIONS 285

graph



Proof. The proof of this theorem is left as an exercise.

The following integration formulas are valid:
udu 1
= ln u2 ’ a2 + c
1.
u2 ’ a2 2
u’a
du 1
2. +c
= ln
u2 ’ a2 2a u+a

udu
√ = u2 ’ a2 + c
3.
2 ’ a2
u

du
√ = ln u + u2 ’ a2 + c
4.
u2 ’ a2
du 1 u

5. = arcsec +c
a a
u u2 ’ a2

√ 1 √2 a2
2 ’ a2 du = 2’ ln u + u2 ’ a2 + c
u u ’a
6. u
2 2

Exercises 6.5 Prove each of the following formulas:
u du 1
= ’ ln |a2 ’ u2 | + C
1.
a2 ’ u2 2
a’u
du 1
+C
2. = ln
a2 ’ u2 2a a+u

u du
√ = ’ a2 ’ u2 + C
3.
2 ’ u2
a
du u

4. = arcsin + C, a > 0
a
a2 ’ u2

a2 ’ u2
a
du 1
√ = ln ’ +C
5.
a u u
2 ’ u2
ua
286 CHAPTER 6. TECHNIQUES OF INTEGRATION

√ 1 √2
a2 u
2 ’ u2 du = + u a ’ u2 + C, a > 0
6. a arcsin
2 a 2
u du 1
= ln a2 + u2 + C
7.
a2 + u2 2
du 1 u
8. = arctan +C
a2 + u2 a a

u du
√ = a2 + u2 + C
9.
a2 + u2

du
√ = ln u + a2 + u2 + C
10.
a2 + u2

a2 + u2 a
du 1
√ ’ +C
11. = ln
a u u
u a2 + u2

√ √
1 √2 a2
2 + u2 du = 2+ ln u + a2 + u2 + C
12. a u a +u
2 2
u du 1
= ln u2 ’ a2 + C
13.
u2 ’ a2 2

u’a
du 1
14. +C
= ln
u2 ’ a2 2a u+a

u du
√ = u2 ’ a2 + C
15.
u2 ’ a2

du
√ = ln u + u2 ’ a2 + C
16.
u2 ’ a2
du 1 u

17. = arcsec +C
a a
u u2 ’ a2

√ 1 √2 a2
2 ’ a2 du = 2’ ln u + u2 ’ a2 + C
u u ’a
18. u
2 2
Evaluate each of the following integrals:
6.5. TRIGONOMETRIC SUBSTITUTIONS 287

x dx dx x dx
√ √
19. 20. 21.
4 ’ x2
4 ’ x2 4 ’ x2

x dx dx
dx
23. 24.
22.
4 ’ x2 9 + x2 9 + x2

x dx dx x dx
√ √
25. 26. 27.
x2 ’ 16
9 + x2 9 + x2

dx
dx x dx
√ √
30.
28. 29.
x2 ’ 16 x2 ’ 16 x2 ’ 16

dx
dx dx

√ √
32.
31. 33.
x x2 ’ 4 x 9 ’ x2 x x2 + 16
√ √ x2

9 ’ x2 dx 4 ’ 9x2
34. 35. 36. dx
1 ’ x2

x2 x2 dx
√ √ dx 39.
37. dx 38.
(9 + x2 )2
x2 ’ 16
4 + x2

dx dx dx
40. 41. 42.
(9 ’ x2 )2 (x2 ’ 16)2 (4 + x2 )3/2
√ √
4 + x2 x2 ’ 4 dx

43. 44. dx 45.
x x x2 x2 + 4

dx
dx dx
√ √ 48.
46. 47.
x2 ’ 2x + 5
x2 4 ’ x2 x2 x2 ’ 4

dx dx dx
√ √
49. 50. 51.
x2 ’ 4x + 12 4x ’ x2 x2 ’ 4x + 12

x dx
dx dx
√ 54.
52. 53.
4x ’ x2 x2 ’ 4x ’ 12
x2 ’ 2x + 5

x
x dx
(5 ’ 4x ’ x2 )1/2 dx
√ 56.
55. dx 57.
x2 + 4x + 13
x2 ’ 2x + 5
288 CHAPTER 6. TECHNIQUES OF INTEGRATION

2x + 7 x+3 dx
√ √
58. dx 59. dx 60.
x2 + 4 + 13 4x2 ’ 1
x2 + 2x + 5

e2x dx
x+4 x+2
√ √
61. dx 62. dx 63.
(5 ’ e2x + e4x )1/2
16 ’ 9x2
9x2 + 16

e3x dx
64.
(e6x + 4e3x + 3)1/2


6.6 Integration by Partial Fractions
A polynomial with real coe¬cients can be factored into a product of powers
of linear and quadratic factors. This fact can be used to integrate rational
functions of the form P (x)/Q(x) where P (x) and Q(x) are polynomials that
have no factors in common. If the degree of P (x) is greater than or equal to
the degree of Q(x), then by long division we can express the rational function
by
r(x)
P (x)
= q(x) +
Q(x) Q(x)
where q(x) is the quotient and r(x) is the remainder whose degree is less than
the degree of Q(x). Then Q(x) is factored as a product of powers of linear
and quadratic factors. Finally r(x)/Q(x) is split into a sum of fractions of
the form
A2 An
A1
+ ··· +
+
ax + b (ax + b)2 (ax + b)n
and
B2 x + c2 Bm x + cm
B1 x + c1
+ ··· +
+ .
ax2 + bx + c (ax2 + bxc)2 (ax2 + bx + c)m
Many calculators and computer algebra systems, such as Maple or Mathe-
matica, are able to factor polynomials and split rational functions into partial
fractions. Once the partial fraction split up is made, the problem of inte-
grating a rational function is reduced to integration by substitution using
linear or trigonometric substitutions. It is best to study some examples and
do some simple problems by hand.


Exercises 6.6 Evaluate each of the following integrals:
6.7. FRACTIONAL POWER SUBSTITUTIONS 289

dx dx
1. 2.
(x ’ 1)(x ’ 2)(x + 4) (x ’ 4)(10 + x)

dx dx
3. 4.
(x ’ a)(x ’ b) (x ’ a)(b ’ x)

dx dx
5. 6.
(x2 + 1)(x2 + 4) (x ’ 1)(x2 + 1)

2x dx x dx
7. 8.
x2 ’ 5x + 6 (x + 3)(x + 4)

x+1 (x + 2)dx
9. dx 10.
(x + 2)(x2 + 4) (x + 3)(x2 + 1)

2 dx dx
11. 12.
(x2 + 4)(x2 + 9) (x2 ’ 4)(x2 ’ 9)

x2 dx x dx
13. 14.
(x2 + 4)(x2 + 9) (x2 ’ 4)(x2 ’ 9)

dx x dx
15. 16.
x4 ’ 16 x4 ’ 81

6.7 Fractional Power Substitutions
If the integrand contains one or more fractional powers of the form xs/r ,
then the substitution, x = un , where n is the least common multiple of the
denominators of the fractional exponents, may be helpful in computing the
integral. It is best to look at some examples and work some problems by
hand.


Exercises 6.7 Evaluate each of the following integrals using the given sub-
stitution.

4x3/2 dx
dx; x = u6 ; x = u3
2.
1.
1 + x1/3 1 + x1/3
290 CHAPTER 6. TECHNIQUES OF INTEGRATION

dx dx
; u2 = x3 ’ 8
; u2 = 1 + e2x
√ √
3. 4.
x x3 ’ 8
1 + e2x
Evaluate each of the following by using an appropriate substitution:
x2 dx
x dx
√ √
5. 6.
x+2 x+4

1 x dx
√ dx √
7. 8.
4+ x 1+ x

x2/3
x

9. 10.
8 + x1/2
1+ 3x

dx
1

dx 12.
11.
x2/3 + 1 1+ x

x dx 1+ x
√ dx
13. 14.
1 + x2/3 2+ x
√ √
1’ x 1+ x
15. dx 16. dx
1 + x3/2 1 ’ x3/2

6.8 Tangent x/2 Substitution
If the integrand contains an expression of the form (a+b sin x) or (a+b cos x),
then the following theorem may be helpful in evaluating the integral.

Theorem 6.8.1 Suppose that u = tan(x/2). Then

1 ’ u2
2u 2
sin x = , cos x = and dx = du.
1 + u2 1 + u2 1 + u2
Furthermore,
(2/(1 + u2 ))du
dx 2du
= =
2u
a(1 + u2 ) + 2bu
a + b sin x a + b 1+u2
(2/(1 + u2 ))du
dx 2du
= .
= 2
a(1 + u2 ) + b(1 ’ u2 )
a + b 1’u2
a + b cos x 1+u
6.9. NUMERICAL INTEGRATION 291

Proof. The proof of this theorem is left as an exercise.


Exercises 6.8

1. Prove Theorem 6.8.1

Evaluate the following integrals:

dx dx
2. 3.
2 + sin x sin x + cos x

dx dx
4. 5.
sin x ’ cos x 2 sin x + 3 cos x

dx dx
6. 7.
2 ’ sin x 3 + cos x

dx sin x dx
8. 9.
3 ’ cos x sin x + cos x

cos x dx (1 + sin x)dx
10. 11.
sin x ’ cos x (1 ’ sin x)

1 ’ cos x 2 ’ cos x
12. dx 13. dx
1 + cos x 2 + cos x

2 ’ sin x
2 + cos x
14. dx 15. dx
2 ’ sin x 3 + cos x

dx
16.
1 + sin x + cos x

6.9 Numerical Integration
Not all integrals can be computed in the closed form in terms of the elemen-
tary functions. It becomes necessary to use approximation methods. Some
of the simplest numerical methods of integration are stated in the next few
theorems.
292 CHAPTER 6. TECHNIQUES OF INTEGRATION

Theorem 6.9.1 (Midpoint Rule) If f, f and f are continuous on [a, b],
then there exists some c such that a < c < b and
b
a+b f (c)
(b ’ a)3 .
f (x)dx = (b ’ a)f +
2 24
a

Proof. The proof of this theorem is omitted.

Theorem 6.9.2 (Trapezoidal Rule) If f, f and f are continuous on [a, b],
then there exists some c such that a < c < b and
b
1 f (c)
(b ’ a)3 .
f (x)dx = (b ’ a) (f (a) + f (b)) ’
2 12
a

Proof. The proof of this theorem is omitted.

Theorem 6.9.3 (Simpson™s Rule) If f, f , f , f (3) and f (4) are continuous
on [a, b], then there exists some c such that a < c < b and
b
f (4) (c)
b’a a+b
(b ’ a)5 .
+ f (b) ’
f (x)dx = f (a) + 4f
6 2 2880
a

These basic numerical formulas can be applied on each subinterval [xi , xi+1 ]
of a partition P = {a = x0 < x1 < · · · < xn = b} of the interval [a, b]
to get composite numerical methods. We assume that h = (b ’ a)/n, xi =
a + ih, i = 0, 1, 2, · · · , n.
Proof. The proof of this theorem is omitted.

Theorem 6.9.4 (Composite Trapezoidal Rule) If f, f and f are continu-
ous on [a, b], then there exists some c such that a < c < b and
n’1
b
b’a 2
h
f (xi ) + f (b) ’
f (a) + 2 h f (c).
f (x)dx =
2 12
a i=1

Proof. The proof of this theorem is omitted.

Theorem 6.9.5 (Composite Simpson™s Rule) If f, f , f , f ( 3) and f (4) are
continuous on [a, b], then there exists some c such that a < c < b and
® 
n/2’1 n/2
b
b ’ a 4 (4)
h
f (x2i’1 ) + f (b)»’
f (x)dx = °f (a) + 2 f (x2i ) + 4 h f (c).
3 180
a i=1 i=1

where n is an even natural number.
6.9. NUMERICAL INTEGRATION 293

Proof. The proof of this theorem is omitted.

Remark 22 In practice, the composite Trapezoidal and Simpson™s rules can
be applied when the value of the function is known at the subdivision points
xi , i = 0, 1, 2, · · · , n.


Exercises 6.9 Approximate the value of each of the following integrals for
a given value of n and using
n
f (xi’1 )(xi ’ xi’1 )
(a) Left-hand end point approximation:
i=1
n
f (xi )(xi ’ xi’1 )
(b) Right-hand end point approximation:
i=1
n
xi’1 + xi
(xi ’ xi’1 )
(c) Mid point approximation: f
2
i=1
(d) Composite Trapezoidal Rule
(e) Composite Simpson™s Rule

4
3
1
1
√ dx, n = 10
dx, n = 10 2.
1.
x x
2
1

1 2
1 1
√ dx, n = 10
3. 4. dx, n = 10
1 + x2
1+ x
0 1


1 2
1+ x
x3 dx, n = 10
5. dx, n = 10 6.
1+x
0 0

2 1
2
(1 + x2 )1/2 dx, n = 10
(x ’ 2x) dx, n = 10
7. 8.
0 0

1 1
3 1/2
(1 + x4 )1/2 dx, n = 10
9. (1 + x ) dx, n = 10 10.
0 0
Chapter 7

Improper Integrals and
Indeterminate Forms

7.1 Integrals over Unbounded Intervals
De¬nition 7.1.1 Suppose that a function f is continuous on (’∞, ∞).
Then we de¬ne the following improper integrals when the limits exist
∞ b
f (x)dx = lim f (x)dx (1)
b’∞
a a
b
b
f (x)dx (2)
f (x)dx = lim
a’’∞
’∞ a
∞ ∞
c
f (x)dx = f (x)dx + f (x)dx (3)
’∞ ’∞ c

provided the integrals on the right hand side exist for some c. If these im-
proper integrals exist, we say that they are convergent; otherwise they are
said to be divergent.


De¬nition 7.1.2 Suppose that a function f is continuous on [0, ∞). Then
the Laplace transform of f , written L(f ) or F (s), is de¬ned by

e’st f (t)dt .
L(f ) = F (s) =
0




294
7.1. INTEGRALS OVER UNBOUNDED INTERVALS 295

Theorem 7.1.1 The Laplace transform has the following properties:
c
L(c) = (4)
s
1
L(eat ) = (5)
s’a
s
L(cosh at) = (6)
s2 ’ a2
a
L(sinh at) = 2 (7)
s ’ a2
s
L(cos ωt) = 2 (8)
s + ω2
ω
L(sin ωt) = 2 (9)
s + ω2
1
L(t) = 2 (10)
s
Proof.

ce’st dt
(i) L(c) =
0


ce’st
=
’s 0

c
=.
s




eat e’st dt
at
(ii) L(e ) =
0


e’(s’a)t dt
=
0


e’(s’a)t
=
’(s ’ a) 0

1
=
s’a
296CHAPTER 7. IMPROPER INTEGRALS AND INDETERMINATE FORMS

provided s > a.

eat + e’at
e’st dt
(iii) L(cosh at) =
2
0

1
= [L(eat ) + L(e’at )]
2
1 1 1
= +
2 s’a s+a
s
, s > |a|.
=
s2 ’ a2


1 at
(e ’ e’at )e’st dt
(iv) L(sinh at) =
2
0

1 1 1
’ , s > |a|
=
2 s’a s+a
a
, s > |a|.
=
s2 ’ a2


cos ωte’st dt
(v) L(cos ωt) =
0

1 ∞
e’st (’s cos ωt + ω sin ωt)
= 0
ω 2 + s2
s
.
=
ω 2 + s2


sin ωte’st dt
(vi) L(sin ωt) =
0

1 ∞
e’st (’s sin ωt ’ ω cos ωt)
= 0
2 + s2
ω
ω
.
=
ω 2 + s2
7.1. INTEGRALS OVER UNBOUNDED INTERVALS 297

te’st dt; (u = t, dv = e’st dt)
(vii) L(t) =
0

∞ ∞
e’st
te’st
+
= dt
’s s
0
0


e’st
=
’s2 0

1
.
=
s2

This completes the proof of Theorem 7.1.1.


Theorem 7.1.2 Suppose that f and g are continuous on [a, ∞) and 0 ¤
f (x) ¤ g(x) on [a, ∞).
∞ ∞
(i) If g(x)dx converges, then f (x)dx converges.
a a

∞ ∞
(ii) If f (x)dx diverges, then g(x)dx diverges.
a a

Proof. The proof of this follows from the order properties of the integral
and is omitted.

De¬nition 7.1.3 For each x > 0, the Gamma function, denoted “(x), is
de¬ned by

tx’1 e’t dt.
“(x) =
0



Theorem 7.1.3 The Gamma function has the following properties:

“(1) = 1 (11)
“(x + 1) = x“(x) (12)
“(n + 1) = n!, n = natural number (13)
298CHAPTER 7. IMPROPER INTEGRALS AND INDETERMINATE FORMS

Proof.

e’t dt
“(1) =
0

= ’e’t 0
=1

tx e’t dt; (u = tx , dv = e’t dt)
“(x + 1) =
0


tx’1 e’t dt
’tx e’t 0 +x
=
0
= x“(x), x > 0
“(2) = 1“(1) = 1
“(3) = 2“(2) = 1 · 2 = 2!
If “(k) = (k ’ 1)!, then
“(k + 1) = k“(k)
= k((k ’ 1)!)
= k!.

By the principle of mathematical induction,

“(n + 1) = n!

for all natural numbers n. This completes the proof of this theorem.

Theorem 7.1.4 Let f be the normal probability distribution function de¬ned
by
2
1 ’ x’µ

f (x) = √ e 2σ
σ 2π
where µ is the constant mean of the distribution and σ is the constant stan-
dard deviation of the distribution. Then the improper integral

f (x)dx = 1.
’∞

Let F be the normal distribution function de¬ned by
x
F (x) = f (x)dx.
’∞
7.2. DISCONTINUITIES AT END POINTS 299

<<

. 11
( 14)



>>