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

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Chapter 6 Integration Techniques and Improper Integrals

(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

6.1 Integration by Parts 6.2 Trigonometric Integrals 6.3 Trigonometric Substitution 6.4 Partial Fractions

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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Section 6.1

Integration by Parts

(分部、分步或部分積分)

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Observation

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

(uv)^′ = u^′v + uv^′ = uv ^′+ vu ^′. Thus, we immediately obtain

∫

(uv)^′dx =

∫

uv ^′dx +

∫

vu^′dx

=⇒uv =

∫

u dv +

∫ v du

=⇒

∫

u dv = uv−

∫ v du.

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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

∫

xⁿe^axdx,

∫

xⁿsin(ax) dx,

∫

xⁿcos(ax) dx with n∈ R and a ̸= 0.

How to select u and dv for this case?

u = xⁿ.

dv = e^axdx, sin(ax)dx, cos(ax)dx.

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Sol: (第二種更簡潔的寫法)

∫

xe^xdx=

∫

xd(e^x)≡

∫ udv

= uv−

∫

vdu (使用 I.B.P. 公式)

= xe^x−

∫ e^xdx

= xe^x− e^x+ C, where C is a constant of integration.

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Type II of I.B.P.

∫

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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Type III of I.B.P.

∫

e^axsin(bx) dx,

∫

e^axcos(bx) dx with a,̸= 0 and b, ̸= 0.

How to select u and dv for this case?

u = e^ax anddv = sin(bx)dx, cos(bx)dx.

u = sin(bx), cos(bx)anddv = e^axdx.

通常使用兩次 I.B.P. 公式!

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Another Solution

Applying the tabular method, we immediately obtain + cos x e^xdx

− − sin x e^x + − cos x e^x Thus the original integral satisfies

∫

e^xcos x dx = e^xcos x + e^xsin x +

∫

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

∫ e^x

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Section 6.2

Trigonometric Integrals (三角積分)

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Main Goals

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

∫

sin^mx cosⁿx dx, where m, n∈ Q.

(INT-2) :

∫

sec^mx 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

∫

sin^2k+1x cosⁿx dx =

∫

(sin²x)^kcosⁿx·sin x dx

=−

∫

(1− cos²x)^kcosⁿxd(cos x)=−

∫

(1− u²)^kuⁿdu, where we letu = cos x.

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Type II of (INT-1)

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

∫

sin^mx cos^2k+1x dx =

∫

sin^mx(cos²x)^k· cos x dx

=

∫

sin^mx(1− sin²x)^kd(sin x) =

∫

u^m(1− u²)^kdu, where we letu = sin x.

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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!)

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Type I of (INT-2)

Ifm = 2kis even for some k∈ N, then

∫

sec^2kx tanⁿx dx =

∫

(sec²x)^k⁻¹tanⁿx· sec²x dx

=

∫

(1 + tan²x)^k⁻¹tanⁿx d(tan x) =

∫

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

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Type II of (INT-2)

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

∫

sec^mx tan^2k+1x dx =

∫

sec^m⁻¹x(tan²)^k·sec x tan x dx

=

∫

sec^m⁻¹x(sec²−1)^kd(sec x)=

∫

u^m⁻¹(u²− 1)^kdu, where we letu = sec x.

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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!)

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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)?

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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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Section 6.3

Trigonometric Substitution

(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?

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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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Type II of Trigonometric Substitution

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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Type III of Trigonometric Substitution

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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Section 6.4 Partial Fractions

(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.

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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 +· · · + A_m (px + q)^m + B₁x + C₁

ax²+ bx + c+ B₂x + C₂

(ax²+ bx + c)² +· · · + B_nx + C_n (ax²+ bx + c)ⁿ, where A_i (i = 1, . . . , m) and B_j, Cj (j = 1, . . . , n) are unknowns.

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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 + q_m) repeated linear factors (重複的㇐次因式)

(px + q)^m,

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

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

The denominator D(x) contains

distinct quadratic factors

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

(ax²+ bx + c)ⁿ,

whereb²_j − 4ajc_j< 0 (j = 1, 2, . . . , n) andb²− 4ac < 0for some n∈ N.

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Section 6.7 Improper Integrals

(瑕積分)

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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→c⁻f(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]

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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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Type II: Infinite Discontinuities (1/2)

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

x→b⁻f(x) =±∞, the integral

∫ b a

f(x) dx = lim

c→b⁻

[ ∫ ^c

a

f(x) dx]

(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]

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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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Thm 6.7 (A Special Type of Improper Integral)

∫ _∞

1

1 x^pdx =

{ 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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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, ∞).

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∫ _∞

a

g(x) dx converges =⇒

∫ _∞

a

f(x) dx converges.

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∫ _∞

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.

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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 I_b≡

∫ _b

1

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

b→∞I_b≈ 0.1268694335685518.

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Numerical Results (2/2)

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

Chapter 6

Integration Techniques and Improper Integrals

(積分技巧與瑕積分)

Hung-Yuan Fan (范洪源)

本章預定授課範圍

6.1 Integration by Parts 6.2 Trigonometric Integrals 6.3 Trigonometric Substitution 6.4 Partial Fractions

6.7 Improper Integrals

Section 6.1

Integration by Parts

(分部、分步或部分積分)

Observation

Thm 6.1 (I.B.P. 公式)

How to choose u and dv?

Type I of I.B.P.

How to select u and dv for this case?

Sol: (第二種更簡潔的寫法)

Type II of I.B.P.

How to select u and dv for this case?

Type III of I.B.P.

How to select u and dv for this case?

Another Solution

Section 6.2

Trigonometric Integrals (三角積分)

Main Goals

Type I of (INT-1)

Type II of (INT-1)

Type III of (INT-1)

Type I of (INT-2)

Type II of (INT-2)

Type III of (INT-2)

Integrals of Sine-Cosine Products

Product-to-Sum Identities

Section 6.3

Trigonometric Substitution

(三角代換)

Main Goal

Type I of Trigonometric Substitution

Type II of Trigonometric Substitution

Type III of Trigonometric Substitution

Section 6.4 Partial Fractions

(部分分式)

Long Division (⻑除法)

Note: integrating w.r.t. x =

Remarks

Partial Fraction Decomposition

Type I: Distinct or Repeated Linear Factors

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

Section 6.7 Improper Integrals

(瑕積分)

Two Types of Improper Integrals

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

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

Note: we say that the improper integral

Type II: Infinite Discontinuities (1/2)

Type II: Infinite Discontinuities (2/2)

Note: we say that the improper integral

Thm 6.7 (A Special Type of Improper Integral)

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

Comparison Theorem for Improper Integrals

Thm (瑕積分的比較定理)

Example (補充題)

Sol: Since x

Numerical Results (1/2)

Numerical Results (2/2)

Thank you for your attention!