Advanced Calculus (II)

(1)

W^EN-C^HINGL^IEN

Department of Mathematics National Cheng Kung University

2009

(2)

Ch11: Differentiability on R

ⁿ

11.1: Partial Derivatives and Partial Integrals

Notation:

(1) E1× E₂× . . . × E_n

:= {(x1,x2, . . . ,xn) :xj ∈ E_j, for j = 1, . . . , n}

(2) The partial derivative fxj exists at a pointa if and only if the limit

∂f

∂xj

(a) = lim

h→0

f (a + hej) −f (a) h

exists.

(3)

fxjxk := ∂²f

∂xk∂xj

:= ∂

∂xk

∂f

∂xj

.

WEN-CHINGLIEN Advanced Calculus (II)

(3)

Let V be a nonempty, open subset ofRⁿ, let f : V →R^m, and let p ∈N.

(i) f is said to be C^p on V if and only if each partial derivative of f of order k ≤ p exists and is continuous on V .

(ii) f is said to be C^∞on V if and only if f is C^p on V for all p ∈N.

(4)

Theorem (11.2)

Suppose that V is open inR², that (a, b) ∈ V , and that f : V →R. If f is C¹on V , and if one of the mixed second partial derivatives of f exists on V and is continuous at the point (a, b), then the other mixed second partial derivative exists at (a, b) and

∂²f

∂y ∂x(a, b) = ∂²f

∂x ∂y(a, b).

Note: These hypotheses are met if f ∈ C²(H).

(5)

Suppose that fyx exists on V and is continuous at the point (a, b). Consider ∆(h, k ) := f (a + h, b + k ) − f (a + h, b)

−f (a, b + k ) + f (a, b), define for |h|, |k | < r /√

2, where r > 0 is so small that Br(a, b) ⊂ V .Apply the Mean Value Theorem twice to choose scalars s, t ∈ (0, 1) such that

∆(h, k ) = k∂f

∂y(a + h, b + tk ) − k ∂f

∂y(a, b + tk )

=hk ∂f²

∂x ∂y(a + sh, b + tk ).

Since this last mixed partial derivative is continuous at the point (a, b), we have

(1) lim lim ∆(h, k )

= ∂²f

(a, b).

(6)

Proof.

−f (a, b + k ) + f (a, b),define for |h|, |k | < r /√

2, where r > 0 is so small that Br(a, b) ⊂ V . Apply the Mean Value Theorem twice to choose scalars s, t ∈ (0, 1) such that

∆(h, k ) = k∂f

∂y(a + h, b + tk ) − k ∂f

∂y(a, b + tk )

=hk ∂f²

∂x ∂y(a + sh, b + tk ).

(1) lim

k →0lim

h→0

∆(h, k )

hk = ∂²f

∂x ∂y(a, b).

(7)

2, where r > 0 is so small that Br(a, b) ⊂ V .Apply the Mean Value Theorem twice to choose scalars s, t ∈ (0, 1) such that

∆(h, k ) = k∂f

∂y(a + h, b + tk ) − k ∂f

∂y(a, b + tk )

=hk ∂f²

∂x ∂y(a + sh, b + tk ).

= ∂²f

(a, b).

(8)

Proof.

∆(h, k ) = k∂f

∂y(a + h, b + tk ) − k ∂f

∂y(a, b + tk )

=hk ∂f²

∂x ∂y(a + sh, b + tk ).

(1) lim

k →0lim

h→0

∆(h, k )

hk = ∂²f

∂x ∂y(a, b).

(9)

∆(h, k ) = k∂f

∂y(a + h, b + tk ) − k ∂f

∂y(a, b + tk )

=hk ∂f²

∂x ∂y(a + sh, b + tk ).

Since this last mixed partial derivative is continuous at the point (a, b),we have

= ∂²f

(a, b).

(10)

Proof.

∆(h, k ) = k∂f

∂y(a + h, b + tk ) − k ∂f

∂y(a, b + tk )

=hk ∂f²

∂x ∂y(a + sh, b + tk ).

(1) lim

k →0lim

h→0

∆(h, k )

hk = ∂²f

∂x ∂y(a, b).

(11)

∆(h, k ) = k∂f

∂y(a + h, b + tk ) − k ∂f

∂y(a, b + tk )

=hk ∂f²

∂x ∂y(a + sh, b + tk ).

Since this last mixed partial derivative is continuous at the point (a, b),we have

= ∂²f

(a, b).

(12)

Proof.

∆(h, k ) = k∂f

∂y(a + h, b + tk ) − k ∂f

∂y(a, b + tk )

=hk ∂f²

∂x ∂y(a + sh, b + tk ).

(1) lim

k →0lim

h→0

∆(h, k )

hk = ∂²f

∂x ∂y(a, b).

(13)

On the other hand, the Mean Value Theorem also implies that there is a scalar u ∈ (0, 1) such that

∆(h, k ) = f (a + h, b + k ) − f (a, b + k ) − f (a + h, b) + f (a, b)

=h∂f

∂x(a + uh, b + k ) − h∂f

∂x(a + uh, b).

Hence, it follows from (1) that

k →0lim lim

h→0

1 k

∂f

∂x(a + uh, b + k ) − ∂f

∂x(a + uh, b)

= lim

k →0lim

h→0

∆(h, k )

hk = ∂²f

∂x ∂y(a, b).

(14)

Proof.

=h∂f

∂x(a + uh, b + k ) − h∂f

∂x(a + uh, b).

k →0lim lim

h→0

1 k

∂f

∂x(a + uh, b + k ) − ∂f

∂x(a + uh, b)

= lim

k →0lim

h→0

∆(h, k )

hk = ∂²f

∂x ∂y(a, b).

(15)

=h∂f

∂x(a + uh, b + k ) − h∂f

∂x(a + uh, b).

k →0lim lim

h→0

1 k

∂f

∂x(a + uh, b + k ) − ∂f

∂x(a + uh, b)

= lim

k →0lim

h→0

∆(h, k )

hk = ∂²f

∂x ∂y(a, b).

(16)

Proof.

=h∂f

∂x(a + uh, b + k ) − h∂f

∂x(a + uh, b).

k →0lim lim

h→0

1 k

∂f

∂x(a + uh, b + k ) − ∂f

∂x(a + uh, b)

= lim

k →0lim

h→0

∆(h, k )

hk = ∂²f

∂x ∂y(a, b).

(17)

=h∂f

∂x(a + uh, b + k ) − h∂f

∂x(a + uh, b).

k →0lim lim

h→0

1 k

∂f

∂x(a + uh, b + k ) − ∂f

∂x(a + uh, b)

= lim

k →0lim

h→0

∆(h, k )

hk = ∂²f

∂x ∂y(a, b).

(18)

Proof.

Since fx is continuous on Br(a, b), we can let h = 0 in the first expression. We conclude by definition that

∂²f

∂y ∂x(a, b)= lim

k →0

1 k

∂f

∂x(a, b+k )−∂f

∂x(a, b)

= ∂²f

∂x ∂y(a, b).

(19)

Since fx is continuous on Br(a, b), we can let h = 0 in the first expression.We conclude by definition that

∂²f

∂y ∂x(a, b) = lim

k →0

1 k

∂f

∂x(a, b+k )−∂f

∂x(a, b)

= ∂²f

∂x ∂y(a, b).

(20)

Proof.

∂²f

∂y ∂x(a, b)= lim

k →0

1 k

∂f

∂x(a, b+k )−∂f

∂x(a, b)

= ∂²f

∂x ∂y(a, b).

(21)

∂²f

k →0

1 k

∂f

∂x(a, b+k )−∂f

∂x(a, b)

= ∂²f

∂x ∂y(a, b).

(22)

Proof.

∂²f

k →0

1 k

∂f

∂x(a, b+k )−∂f

∂x(a, b)

= ∂²f

∂x ∂y(a, b).

(23)

Prove that

f (x , y ) = (

xy ^x_x²2^−y+y²²

(x , y ) 6= 0

0 (x , y ) = 0

(24)

Theorem (11.4)

Let H = [a, b] × [c, d ] be a rectangle and suppose that f : H →R is continuous. If

F (y ) = Z b

a

f (x , y )dx , then F is continuous on [c, d ]; i.e.,

y →y0lim

y ∈[c,d ]

Z b a

f (x , y )dx = Z b

a

y →y0lim

y ∈[c,d ]

f (x , y )dx

for all y0∈ [c, d].

(25)

For each y ∈ [c, d ], f (·, y ) is continuous on [a, b].Hence, by Theorem 5.10, F (y ) exists for y ∈ [c, d ].

Fix y0 ∈ [c, d] and let ε > 0.Since H is compact, f is uniformly continuous on H. Hence, choose δ > 0 such that k(x , y ) − (z, w )k < δ and (x , y ), (z, w ) ∈ H imply

|f (x, y ) − f (z, w )| < ε b − a. Since |y − y0| = k(x, y ) − (x, y₀)k, it follows that

|F (y ) − F (y0)| ≤ Z b

a

|f (x, y ) − f (x, y0)|dx < ε for all y ∈ [c, d ] that satisfy |y − y0| < δ. We conclude that F is continuous on [c, d ].

(26)

Proof.

For each y ∈ [c, d ], f (·, y ) is continuous on [a, b]. Hence, by Theorem 5.10, F (y ) exists for y ∈ [c, d ].

Fix y0 ∈ [c, d] and let ε > 0. Since H is compact, f is uniformly continuous on H.Hence, choose δ > 0 such that k(x , y ) − (z, w )k < δ and (x , y ), (z, w ) ∈ H imply

|F (y ) − F (y0)| ≤ Z b

a

(27)

Fix y0 ∈ [c, d] and let ε > 0.Since H is compact, f is uniformly continuous on H. Hence, choose δ > 0 such thatk(x, y ) − (z, w )k < δ and (x, y ), (z, w ) ∈ H imply

|F (y ) − F (y0)| ≤ Z b

a

(28)

Proof.

Fix y0 ∈ [c, d] and let ε > 0. Since H is compact, f is uniformly continuous on H.Hence, choose δ > 0 such that k(x , y ) − (z, w )k < δ and (x , y ), (z, w ) ∈ H imply

|F (y ) − F (y0)| ≤ Z b

a

(29)

Fix y0 ∈ [c, d] and let ε > 0. Since H is compact, f is uniformly continuous on H. Hence, choose δ > 0 such thatk(x, y ) − (z, w )k < δ and (x, y ), (z, w ) ∈ H imply

|f (x, y ) − f (z, w )| < ε b − a. Since |y − y0| = k(x, y ) − (x, y₀)k,it follows that

|F (y ) − F (y0)| ≤ Z b

a

(30)

Proof.

Fix y0 ∈ [c, d] and let ε > 0. Since H is compact, f is uniformly continuous on H. Hence, choose δ > 0 such that k(x , y ) − (z, w )k < δ and (x , y ), (z, w ) ∈ H imply

|F (y ) − F (y0)| ≤ Z b

a

|f (x, y ) − f (x, y0)|dx < ε for all y ∈ [c, d ] that satisfy |y − y0| < δ.We conclude that F is continuous on [c, d ].

(31)

|f (x, y ) − f (z, w )| < ε b − a. Since |y − y0| = k(x, y ) − (x, y₀)k,it follows that

|F (y ) − F (y0)| ≤ Z b

a

(32)

Proof.

|F (y ) − F (y0)| ≤ Z b

a

|f (x, y ) − f (x, y0)|dx < ε for all y ∈ [c, d ] that satisfy |y − y0| < δ.We conclude that F is continuous on [c, d ].

(33)

|F (y ) − F (y0)| ≤ Z b

a

(34)

Theorem (11.5)

Let H = [a, b] × [c, d ] be a rectangle inR²and let f : H →R. Suppose that f (·, y ) is integrable on [a, b] for each y ∈ [c, d ] and that the partial derivative fy(x , ·) exists on [c, d ] for each x ∈ [a, b]. If the two-variable function fy(x , y ) is continuous on H, then

d dy

Z b a

f (x , y )dx = Z b

a

∂f

∂y(x , y )dx for all y ∈ [c, d ].

Note: These hypotheses are met if f ∈ C¹(H).

(35)

Recall that ”fy(x , ·) exists on [c, d ]” means that fy(x , ·) exists on (c, d ), and

fy(x , c) := lim

h→0+

f (x , c + h) − f (x , c)

h ,

fy(x , d ) := lim

h→0−

f (x , d + h) − f (x , d ) h

both exist (see Definition 4.6). Hence, it suffices to show that

h→0+lim Z b

a

f (x , y + h) − f (x , y )

h dx =

Z b a

∂f

∂y(x , y )dx for y ∈ [c, d ), and

(36)

Proof.

fy(x , c) := lim

h→0+

f (x , c + h) − f (x , c)

h ,

fy(x , d ) := lim

h→0−

f (x , d + h) − f (x , d ) h

both exist (see Definition 4.6).Hence, it suffices to show that

h→0+lim Z b

a

f (x , y + h) − f (x , y )

h dx =

Z b a

∂f

(37)

fy(x , c) := lim

h→0+

f (x , c + h) − f (x , c)

h ,

fy(x , d ) := lim

h→0−

f (x , d + h) − f (x , d ) h

both exist (see Definition 4.6). Hence, it suffices to show that

h→0+lim Z b

a

f (x , y + h) − f (x , y )

h dx =

Z b a

∂f

(38)

Proof.

h→0−lim Z b

a

f (x , y + h) − f (x , y )

h dx =

Z b a

∂f

∂y(x , y )dx for y ∈ (c, d ].The arguments are similar; we provide the details only for the first identity.

Fix x ∈ [a, b] and y ∈ [c, d ), and let h > 0 be so small that y + h ∈ [c, d ).Let ε > 0. By uniform continuity, choose a δ >0 so small that |y − c| < δ and x ∈ [a, b] imply

|fy(x , y ) − fy(x , c)| < ε/(b − a). By the Mean Value Theorem, choose a point c(x ; h) between y and y + h such that

F (x , y , h) := f (x , y + h) − f (x , y )

h = ∂f

∂y(x , c(x ; h)).

(39)

h→0−lim Z b

a

f (x , y + h) − f (x , y )

h dx =

Z b a

∂f

∂y(x , y )dx for y ∈ (c, d ]. The arguments are similar; we provide the details only for the first identity.

Fix x ∈ [a, b] and y ∈ [c, d ), and let h > 0 be so small that y + h ∈ [c, d ). Let ε > 0. By uniform continuity, choose a δ >0 so small that |y − c| < δ and x ∈ [a, b] imply

F (x , y , h) := f (x , y + h) − f (x , y )

h = ∂f

∂y(x , c(x ; h)).

(40)

Proof.

h→0−lim Z b

a

f (x , y + h) − f (x , y )

h dx =

Z b a

∂f

Fix x ∈ [a, b] and y ∈ [c, d ), and let h > 0 be so small that y + h ∈ [c, d ).Let ε > 0. By uniform continuity, choose a δ >0 so small that |y − c| < δ and x ∈ [a, b] imply

|fy(x , y ) − fy(x , c)| < ε/(b − a).By the Mean Value Theorem, choose a point c(x ; h) between y and y + h such that

F (x , y , h) := f (x , y + h) − f (x , y )

h = ∂f

∂y(x , c(x ; h)).

(41)

h→0−lim Z b

a

f (x , y + h) − f (x , y )

h dx =

Z b a

∂f

F (x , y , h) := f (x , y + h) − f (x , y )

h = ∂f

∂y(x , c(x ; h)).

(42)

Proof.

h→0−lim Z b

a

f (x , y + h) − f (x , y )

h dx =

Z b a

∂f

|fy(x , y ) − fy(x , c)| < ε/(b − a).By the Mean Value Theorem, choose a point c(x ; h) between y and y + h such that

F (x , y , h) := f (x , y + h) − f (x , y )

h = ∂f

∂y(x , c(x ; h)).

(43)

h→0−lim Z b

a

f (x , y + h) − f (x , y )

h dx =

Z b a

∂f

F (x , y , h) := f (x , y + h) − f (x , y )

h = ∂f

∂y(x , c(x ; h)).

(44)

Proof.

h→0−lim Z b

a

f (x , y + h) − f (x , y )

h dx =

Z b a

∂f

F (x , y , h) := f (x , y + h) − f (x , y )

h = ∂f

∂y(x , c(x ; h)).

(45)

Since |c(x ; h) − y | = c(x ; h) − y ≤ h,it follows that if 0 < h < δ, then

F (x , y , h) − Z b

a

∂f

∂y(x , y )dx

≤ Z b

a

∂f

∂y(x , c(x ; h)) − ∂f

∂y(x , y )

dx < ε.

Therefore, d dy

Z b a

f (x , y )dx = Z b

a

∂f

∂y(x , y )dx .

(46)

Proof.

Since |c(x ; h) − y | = c(x ; h) − y ≤ h, it follows that if 0 < h < δ, then

F (x , y , h) − Z b

a

∂f

∂y(x , y )dx

≤ Z b

a

∂f

∂y(x , c(x ; h)) − ∂f

∂y(x , y )

dx < ε.

Therefore, d dy

Z b a

f (x , y )dx = Z b

a

∂f

∂y(x , y )dx .

(47)

F (x , y , h) − Z b

a

∂f

∂y(x , y )dx

≤ Z b

a

∂f

∂y(x , c(x ; h)) − ∂f

∂y(x , y )

dx < ε.

Therefore, d dy

Z b a

f (x , y )dx = Z b

a

∂f

∂y(x , y )dx .

(48)

Proof.

F (x , y , h) − Z b

a

∂f

∂y(x , y )dx

≤ Z b

a

∂f

∂y(x , c(x ; h)) − ∂f

∂y(x , y )

dx < ε.

Therefore, d dy

Z b a

f (x , y )dx = Z b

a

∂f

∂y(x , y )dx .

(49)

F (x , y , h) − Z b

a

∂f

∂y(x , y )dx

≤ Z b

a

∂f

∂y(x , c(x ; h)) − ∂f

∂y(x , y )

dx < ε.

Therefore, d dy

Z b a

f (x , y )dx = Z b

a

∂f

∂y(x , y )dx .

(50)

Definition (11.6)

Let a < b be extended real numbers, let I be an interval in R, and suppose that f : (a, b) × I → R. The improper integral

Z b a

f (x , y )dx

is said to converge uniformly on I if and only if f (·, y ) is improperly integrable on (a, b) for each y ∈ I and given ε >0 there exist real numbers A, B ∈ (a, b) such that

Z b a

f (x , y )dx − Z β

α

f (x , y )dx

< ε for all a < α < A, B < β < b and all y ∈ I.

(51)

Suppose that a < b are extended real numbers, that I is an interval inR, that f : (a, b) × I → R, and that f (·, y ) is locally integrable on the interval (a, b) for each y ∈ I. If there is a function M : (a, b) →R, absolutely integrable on (a, b), such that

|f (x, y )| ≤ M(x) for all x ∈ (a, b) and y ∈ I, then

Z b a

f (x , y )dx converges uniformly on I.

(52)

Proof.

Let ε > 0.By hypothesis and the Comparison Test for improper integrals,Rb

a f (x , y )dx exists and is finite for each y ∈ I.Moreover, since M(x ) is improperly integrable on (a, b), there exist real numbers A, B such that

a < A < B < b and

Z A α

M(x )dx + Z b

B

M(x )dx < ε.

Thus for each a < α < A < B < β < b and each y ∈ I, we have

Z b a

f (x , y )dx − Z β

α

f (x , y )dx

≤ Z α

a

|f (x, y )|dx+

Z b β

|f (x, y )|dx

≤ Z A

a

M(x )dx + Z b

B

M(x )dx < ε

(53)

Let ε > 0. By hypothesis and the Comparison Test for improper integrals,Rb

a f (x , y )dx exists and is finite for each y ∈ I. Moreover, since M(x ) is improperly integrable on (a, b), there exist real numbers A, B such that

a < A < B < b and

Z A α

M(x )dx + Z b

B

M(x )dx < ε.

Z b a

f (x , y )dx − Z β

α

f (x , y )dx

≤ Z α

a

|f (x, y )|dx+

Z b β

|f (x, y )|dx

≤ Z A

M(x )dx + Z b

M(x )dx < ε

(54)

Proof.

a f (x , y )dx exists and is finite for each y ∈ I.Moreover, since M(x ) is improperly integrable on (a, b), there exist real numbers A, B such that

a < A < B < b and

Z A α

M(x )dx + Z b

B

M(x )dx < ε.

Z b a

f (x , y )dx − Z β

α

f (x , y )dx

≤ Z α

a

|f (x, y )|dx+

Z b β

|f (x, y )|dx

≤ Z A

a

M(x )dx + Z b

B

M(x )dx < ε

(55)

a < A < B < b and

Z A α

M(x )dx + Z b

B

M(x )dx < ε.

Thus for each a < α < A < B < β < b and each y ∈ I,we have

Z b a

f (x , y )dx − Z β

α

f (x , y )dx

≤ Z α

a

|f (x, y )|dx+

Z b β

|f (x, y )|dx

≤ Z A

M(x )dx + Z b

M(x )dx < ε

(56)

Proof.

a < A < B < b and

Z A α

M(x )dx + Z b

B

M(x )dx < ε.

Z b a

f (x , y )dx − Z β

α

f (x , y )dx

≤ Z α

a

|f (x, y )|dx+

Z b β

|f (x, y )|dx

≤ Z A

a

M(x )dx + Z b

B

M(x )dx < ε

(57)

a < A < B < b and

Z A α

M(x )dx + Z b

B

M(x )dx < ε.

Thus for each a < α < A < B < β < b and each y ∈ I,we have

Z b a

f (x , y )dx − Z β

α

f (x , y )dx

≤ Z α

a

|f (x, y )|dx+

Z b β

|f (x, y )|dx

≤ Z A

M(x )dx + Z b

M(x )dx < ε

(58)

Proof.

a < A < B < b and

Z A α

M(x )dx + Z b

B

M(x )dx < ε.

Z b a

f (x , y )dx − Z β

α

f (x , y )dx

≤ Z α

a

|f (x, y )|dx+

Z b β

|f (x, y )|dx

≤ Z A

a

M(x )dx + Z b

B

M(x )dx < ε

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a < A < B < b and

Z A α

M(x )dx + Z b

B

M(x )dx < ε.

Z b a

f (x , y )dx − Z β

α

f (x , y )dx

≤ Z α

a

|f (x, y )|dx+

Z b β

|f (x, y )|dx

≤ Z A

M(x )dx + Z b

M(x )dx < ε

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Proof.

a < A < B < b and

Z A α

M(x )dx + Z b

B

M(x )dx < ε.

Z b a

f (x , y )dx − Z β

α

f (x , y )dx

≤ Z α

a

|f (x, y )|dx+

Z b β

|f (x, y )|dx

≤ Z A

a

M(x )dx + Z b

B

M(x )dx < ε

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Suppose that a < b are extended real numbers, that c < d are finite real numbers, and that

f : (a, b) × [c, d ] →R is continuous. If F (y ) =

Z b a

f (x , y )dx converges uniformly on [c, d ]; i.e.,

y →y0lim

y ∈[c,d ]

Z b a

f (x , y )dx = Z b

a

y →y0lim

y ∈[c,d ]

f (x , y )dx

for all y0∈ [c, d].

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Proof.

Let ε > 0 and y0∈ [c, d].Choose real numbers A, B such that a < A < B < b and

F (y ) − Z B

A

f (x , y )dx

< ε 3

for all y ∈ [c, d ]. By Theorem 11.4, choose δ > 0 such that

Z B A

(f (x , y ) − f (x , y0))dx

< ε 3 for all y ∈ [c, d ] that satisfy |y − y0| < δ. Then

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Let ε > 0 and y0∈ [c, d]. Choose real numbers A, B such that a < A < B < b and

F (y ) − Z B

A

f (x , y )dx

< ε 3

for all y ∈ [c, d ].By Theorem 11.4, choose δ > 0 such that

Z B A

(f (x , y ) − f (x , y0))dx

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Proof.

Let ε > 0 and y0∈ [c, d]. Choose real numbers A, B such that a < A < B < b and

F (y ) − Z B

A

f (x , y )dx

< ε 3

for all y ∈ [c, d ]. By Theorem 11.4, choose δ > 0 such that

Z B A

(f (x , y ) − f (x , y0))dx

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|F (y ) − F (y₀)|≤

F (y ) − Z B

A

f (x , y )dx +

Z B A

(f (x , y ) − f (x , y0))dx +

F (y0) − Z B

A

f (x , y0)dx

< ε 3 + ε

3+ ε 3 = ε for all y ∈ [c, d ] that satisfy |y − y0| < δ.

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Proof.

|F (y ) − F (y₀)| ≤

F (y ) − Z B

A

f (x , y )dx +

Z B A

(f (x , y ) − f (x , y0))dx +

F (y0) − Z B

A

f (x , y0)dx

< ε 3 + ε

3+ ε 3 = ε for all y ∈ [c, d ] that satisfy |y − y0| < δ.

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Suppose that a < b are extended real numbers, that c < d are finite real numbers, that f : (a, b) × [c, d ] →R is continuous, and that the improper integral

F (y ) = Z b

a

f (x , y )dx

exists for all y ∈ [c, d ]. If fy(x , y ) exists and is continuous on (a, b) × [c, d ] and if

φ(y ) = Z b

a

∂f

∂y(x , y )dx

converges uniformly on [c, d ], then F is differentiable on [c, d ], and F⁰(y ) = φ(y ); i.e.,

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Theorem (11.9) d dy

Z b a

f (x , y )dx = Z b

a

∂f

∂y(x , y )dx for all y ∈ [c, d ].

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