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# Chapter 6 Integration Techniques and Improper Integrals (積分技巧與瑕積分)

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## (積分技巧與瑕積分)

### Hung-Yuan Fan (范洪源)

Department of Mathematics, National Taiwan Normal University, Taiwan

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## 本章預定授課範圍

### 6.7 Improper Integrals

Hung-Yuan Fan (范洪源), Dep. of Math., NTNU, Taiwan Chapter 6, Calculus B 2/77

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## Observation

If u and v are diff. functions of x, then

(uv) = uv + uv = uv + vu . Thus, we immediately obtain

(uv)dx =

uv dx +

vudx

=⇒uv =

u dv +

v du

=

u dv = uv−

v du.

Hung-Yuan Fan (范洪源), Dep. of Math., NTNU, Taiwan Chapter 6, Calculus B 4/77

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### Thm 6.1 (I.B.P. 公式)

If u = u(x) and v = v(x) are functions of x havingg conti.

derivatives, then we have

u dv = uv−

v du.

### How to choose u and dv?

u = u(x): easily differentiable function of x. (u: 好微分函數) v = v(x): easily integrable function of x. (v: 好積分函數)

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### Type I of I.B.P.

The integrals of the form

xneaxdx,

xnsin(ax) dx,

xncos(ax) dx with n∈ R and a ̸= 0.

### How to select u and dv for this case?

u = xn.

dv = eaxdx, sin(ax)dx, cos(ax)dx.

Hung-Yuan Fan (范洪源), Dep. of Math., NTNU, Taiwan Chapter 6, Calculus B 6/77

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### Sol: (第二種更簡潔的寫法)

xexdx=

xd(ex)

udv

= uv−

vdu (使用 I.B.P. 公式)

= xex

exdx

= xex− ex+ C, where C is a constant of integration.

Hung-Yuan Fan (范洪源), Dep. of Math., NTNU, Taiwan Chapter 6, Calculus B 8/77

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Hung-Yuan Fan (范洪源), Dep. of Math., NTNU, Taiwan Chapter 6, Calculus B 10/77

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### Type II of I.B.P.

The integrals of the form

xnln x dx,

xnsin−1x dx,

xntan−1x dx with n̸= −1.

### How to select u and dv for this case?

u = ln x, sin−1x, tan−1x.

dv = xndx.

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Hung-Yuan Fan (范洪源), Dep. of Math., NTNU, Taiwan Chapter 6, Calculus B 12/77

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Hung-Yuan Fan (范洪源), Dep. of Math., NTNU, Taiwan Chapter 6, Calculus B 14/77

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### Type III of I.B.P.

The integrals of the form

eaxsin(bx) dx,

eaxcos(bx) dx with a,̸= 0 and b, ̸= 0.

### How to select u and dv for this case?

u = eax anddv = sin(bx)dx, cos(bx)dx.

u = sin(bx), cos(bx)anddv = eaxdx.

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Hung-Yuan Fan (范洪源), Dep. of Math., NTNU, Taiwan Chapter 6, Calculus B 16/77

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## Another Solution

Applying the tabular method, we immediately obtain + cos x exdx

− sin x ex + − cos x ex Thus the original integral satisfies

excos x dx = excos x + exsin x +

ex(− cos x) dx, and hence the indefinite integral is given by

ex

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## Trigonometric Integrals(三角積分)

Hung-Yuan Fan (范洪源), Dep. of Math., NTNU, Taiwan Chapter 6, Calculus B 18/77

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### Main Goals

Try to find the integrals of the form (INT-1) :

sinmx cosnx dx, where m, n∈ Q.

(INT-2) :

secmx tannx dx, where m, n∈ Q.

(INT-3) :

sin α cos β dx,

sin α sin β dx,

cos α cos β dx, where α = α(x) and β = β(x) are functions of x.

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### Type I of (INT-1)

Ifm = 2k + 1is odd for some k∈ N, then

sin2k+1x cosnx dx =

(sin2x)kcosnx·sin x dx

=

(1− cos2x)kcosnxd(cos x)=

(1− u2)kundu, where we letu = cos x.

Hung-Yuan Fan (范洪源), Dep. of Math., NTNU, Taiwan Chapter 6, Calculus B 20/77

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### Type II of (INT-1)

Ifn = 2k + 1is odd for some k∈ N, then

sinmx cos2k+1x dx =

sinmx(cos2x)k· cos x dx

=

sinmx(1− sin2x)kd(sin x) =

um(1− u2)kdu, where we letu = sin x.

Hung-Yuan Fan (范洪源), Dep. of Math., NTNU, Taiwan Chapter 6, Calculus B 22/77

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### Type III of (INT-1)

Ifm and n are even and nonnegative, try to use the identities sin2x = 1− cos(2x)

2 , cos2x = 1 + cos(2x)

2 .

(當 m, n 為偶數或零，試用倍角公式!)

Hung-Yuan Fan (范洪源), Dep. of Math., NTNU, Taiwan Chapter 6, Calculus B 24/77

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### Type I of (INT-2)

Ifm = 2kis even for some k∈ N, then

sec2kx tannx dx =

(sec2x)k−1tannx· sec2x dx

=

(1 + tan2x)k−1tannx d(tan x) =

(1 + u2)k−1undu, where we letu = tan x.

Hung-Yuan Fan (范洪源), Dep. of Math., NTNU, Taiwan Chapter 6, Calculus B 26/77

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### Type II of (INT-2)

Ifn = 2k + 1is odd for some k∈ N, then

secmx tan2k+1x dx =

secm−1x(tan2)k·sec x tan x dx

=

secm−1x(sec2−1)kd(sec x)=

um−1(u2− 1)kdu, where we letu = sec x.

Hung-Yuan Fan (范洪源), Dep. of Math., NTNU, Taiwan Chapter 6, Calculus B 28/77

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### Type III of (INT-2)

Ifm is odd or n is even, try to use the identity tan2x = sec2x− 1.

(當 m 是奇數或 n 是偶數時，試用上述等式!)

Hung-Yuan Fan (范洪源), Dep. of Math., NTNU, Taiwan Chapter 6, Calculus B 30/77

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### Integrals of Sine-Cosine Products

If α = α(x) and β = β(x) are conti. functions of x, how to evaluate

(INT-3) :

sin α cos β dx,

sin α sin β dx,

cos α cos β dx using the product-to-sum identities (積化和差等式)?

Hung-Yuan Fan (范洪源), Dep. of Math., NTNU, Taiwan Chapter 6, Calculus B 32/77

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### Product-to-Sum Identities

(1) sin α sin β = 12[

cos(α− β) − cos(α + β)] . (2) cos α cos β = 12[

cos(α− β) + cos(α + β)] . (3) sin α cos β = 12[

sin(α− β) + sin(α + β)] .

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Hung-Yuan Fan (范洪源), Dep. of Math., NTNU, Taiwan Chapter 6, Calculus B 34/77

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## (三角代換)

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### Main Goal

To deal with the integrals involving

a2− x2,

a2+ x2 or √ x2− a2 with a > 0.

More precisely, how to evaluate the three types of

aforementioned integrals using the technique of trigonometric substitution?

Hung-Yuan Fan (范洪源), Dep. of Math., NTNU, Taiwan Chapter 6, Calculus B 36/77

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### Type I of Trigonometric Substitution

For integrals involving

a2− x2 with a > 0, let x = a sin θ.

Thendx = a cos θ dθ and we see that

a2− x2 =

a2(1− sin2θ) =√

a2cos2θ =a cos θ for−π2 ≤θ≤ π2.

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Hung-Yuan Fan (范洪源), Dep. of Math., NTNU, Taiwan Chapter 6, Calculus B 38/77

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### Type II of Trigonometric Substitution

For integrals involving

a2+ x2 with a > 0, let x = a tan θ.

Thendx = a sec2θ dθand we see that

a2+ x2 =

a2(1 + tan2θ) =√

a2sec2θ =a sec θ for−π2 < θ < π2.

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Hung-Yuan Fan (范洪源), Dep. of Math., NTNU, Taiwan Chapter 6, Calculus B 40/77

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Hung-Yuan Fan (范洪源), Dep. of Math., NTNU, Taiwan Chapter 6, Calculus B 42/77

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### Type III of Trigonometric Substitution

For integrals involving

x2− a2 with a > 0, let x = a sec θ.

Thendx = a sec θ tan θ dθ and we see that

x2− a2 =√

a2(sec2θ− 1) =√

a2tan2θ =a| tan θ|

for0≤ θ < π2 or π

2 < θ≤ π.

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Hung-Yuan Fan (范洪源), Dep. of Math., NTNU, Taiwan Chapter 6, Calculus B 44/77

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## (部分分式)

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### Long Division (⻑除法)

If N(x) and D(x) are polynomials with deg(N)≥ deg(D), then N(x)

D(x) = Q(x) +N1(x) D(x),

where Q(x) and N1(x) are polynomials with deg(N1) < deg(D). In this case, N1(x)

D(x) is called aproperrational function of x (真分式).

### Note: integrating w.r.t. x =

N(x) D(x)dx =

Q(x) dx +

N1(x) D(x) dx.

Hung-Yuan Fan (范洪源), Dep. of Math., NTNU, Taiwan Chapter 6, Calculus B 46/77

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### Remarks

Q(x) dx can evaluated easily by the power rule.

But, how to evaluate the integral

N1(x)

D(x) dx using the technique of partial fraction decomposition?

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### Partial Fraction Decomposition

IfD(x) = (px + q)m(ax2+ bx + c)n with p̸= 0 andb2− 4ac < 0, may write

N1(x)

D(x) = A1

px + q + A2

(px + q)2 +· · · + Am (px + q)m + B1x + C1

ax2+ bx + c+ B2x + C2

(ax2+ bx + c)2 +· · · + Bnx + Cn (ax2+ bx + c)n, where Ai (i = 1, . . . , m) and Bj, Cj (j = 1, . . . , n) are unknowns.

Hung-Yuan Fan (范洪源), Dep. of Math., NTNU, Taiwan Chapter 6, Calculus B 48/77

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### Type I: Distinct or Repeated Linear Factors

The denominator (分⺟) D(x) contains

distinct linear factors (相異的㇐次因式)

(p1x + q1)(p2x + q2)· · · (pmx + qm) repeated linear factors (重複的㇐次因式)

(px + q)m,

where pj ̸= 0 (j = 1, 2, . . . , m) and p ̸= 0 for some m ∈ N.

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Hung-Yuan Fan (范洪源), Dep. of Math., NTNU, Taiwan Chapter 6, Calculus B 50/77

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Hung-Yuan Fan (范洪源), Dep. of Math., NTNU, Taiwan Chapter 6, Calculus B 52/77

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### Type II: Distinct or Repeated Quadratic Factors (二次因式)

The denominator D(x) contains

(a1x2+ b1x + c1)(a2x2+ b2x + c2)· · · (anx2+ bnx + cn) repeated quadratic factors

(ax2+ bx + c)n,

whereb2j − 4ajcj< 0 (j = 1, 2, . . . , n) andb2− 4ac < 0for some n∈ N.

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Hung-Yuan Fan (范洪源), Dep. of Math., NTNU, Taiwan Chapter 6, Calculus B 54/77

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Hung-Yuan Fan (范洪源), Dep. of Math., NTNU, Taiwan Chapter 6, Calculus B 56/77

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## (瑕積分)

Hung-Yuan Fan (范洪源), Dep. of Math., NTNU, Taiwan Chapter 6, Calculus B 58/77

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### Two Types of Improper Integrals

Type I : Infinite Limits of Integration (無窮積分上下限), e.g.,

a

f(x) dx or

b

−∞f(x) dx.

Type II : Infinite Discontinuities (無窮不連續點), e.g., we say that

b

a

f(x) dx is an improper integral of Type II if

xlim→c+f(x) =±∞ or lim

x→cf(x) =±∞

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### Type I: Infinite Limits of Integration (1/2)

(1) If f is conti. on [a,∞), the improper integral

a

f(x) dx = lim

b→∞

[ ∫ b

a

f(x) dx]

converges (收斂) whenever the limit exists. Otherwise, we say that the improper integral diverges (發散).

(2) If f is conti. on (−∞, b], the improper integral

b

−∞

f(x) dx = lim

a→−∞

[ ∫ b

a

f(x) dx]

converges (收斂) whenever the limit exists. Otherwise, we say that the improper integral diverges (發散).

Hung-Yuan Fan (范洪源), Dep. of Math., NTNU, Taiwan Chapter 6, Calculus B 60/77

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### Type I: Infinite Limits of Integration (2/2)

(3) If f is conti. on (−∞, ∞), the improper integral

−∞f(x) dx =

c

−∞f(x) dx +

c

f(x) dx

converges (收斂) whenever the improper integrals on the RHS both converge for all c∈ R.

### Note: we say that the improper integral

−∞

f(x) dx diverges if either of the improper integrals on the RHS diverges.

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Hung-Yuan Fan (范洪源), Dep. of Math., NTNU, Taiwan Chapter 6, Calculus B 62/77

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### Type II: Infinite Discontinuities (1/2)

(1) If f is conti. on [a, b)and lim

x→bf(x) =±∞, the integral

b a

f(x) dx = lim

c→b

[ ∫ c

a

f(x) dx]

converges (收斂) whenever the limit exists. Otherwise, we say that the improper integral diverges (發散).

(2) If f is conti. on (a, b]and lim

x→a+f(x) =±∞, the integral

b a

f(x) dx = lim

c→a+

[ ∫ b

c

f(x) dx]

converges (收斂) whenever the limit exists. Otherwise, we say that the improper integral diverges (發散).

Hung-Yuan Fan (范洪源), Dep. of Math., NTNU, Taiwan Chapter 6, Calculus B 64/77

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### Type II: Infinite Discontinuities (2/2)

(3) If f is conti. on [a, b]and has an infinite discontinuity at c∈ (a, b), the improper integral

b

a

f(x) dx =

c

a

f(x) dx +

b

c

f(x) dx

converges (收斂) whenever the improper integrals on the RHS both converge.

### Note: we say that the improper integral

b

a

f(x) dx diverges if

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Hung-Yuan Fan (范洪源), Dep. of Math., NTNU, Taiwan Chapter 6, Calculus B 66/77

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Hung-Yuan Fan (范洪源), Dep. of Math., NTNU, Taiwan Chapter 6, Calculus B 68/77

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Hung-Yuan Fan (范洪源), Dep. of Math., NTNU, Taiwan Chapter 6, Calculus B 70/77

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### Thm 6.7 (A Special Type of Improper Integral)

1

1 xpdx =

{ 1

p−1, p > 1 diverges, p≤ 1.

### Note: Thm 6.7 will be used in the Integral Test (積分測試法) for

determining the convergence of an infinite series (無窮級數的收斂 性), see Section 7.3 for details.

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Hung-Yuan Fan (范洪源), Dep. of Math., NTNU, Taiwan Chapter 6, Calculus B 72/77

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## Comparison Theorem for Improper Integrals

### Thm (瑕積分的比較定理)

Suppose that f and g areconti. on [a,∞)with f(x)≤ g(x) ∀ x ∈ [a, ∞).

(1)

a

g(x) dx converges =⇒

a

f(x) dx converges.

(2)

a

f(x) dx diverges =⇒

a

g(x) dx diverges.

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### Example (補充題)

Does the integral

1

e−x2tan−1x dx converge or diverge?

### Sol: Since x

2≥ x for all x ≥ 1, it follows that −x2≤ −x for all x≥ 1 and hence we know that

f(x)≡ e−x2tan−1x≤ π

2e−x≡ g(x) ∀ x ∈ [1, ∞).

In addition, since the improper integral

1

g(x) dx = π 2

1

e−xdx = π 2 lim

b→∞(−e−x) b

1= π 2e, it follows from the Comparison Thm that the improper integral

1

f(x) dx =

1

e−x2tan−1x dx converges.

Hung-Yuan Fan (范洪源), Dep. of Math., NTNU, Taiwan Chapter 6, Calculus B 74/77

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## Numerical Results (1/2)

What is the true value of I =

1

e−x2tan−1x dx?

Although it is usually difficult to get the true value of I, we may apply the technique of numerical integration (數值積分) to compute its approximate value.

If Ib

b

1

e−x2tan−1x dx for b≥ 1, we see that I = lim

b→∞Ib≈ 0.1268694335685518.

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## Numerical Results (2/2)

Hung-Yuan Fan (范洪源), Dep. of Math., NTNU, Taiwan Chapter 6, Calculus B 76/77

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## Thank you for your attention!

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[r]

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Thus the given improper integral is convergent and, since the integrand is positive, we can interpret the value of the integral as the area of the shaded region in Figure

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