Functions Defined by Improper Integrals

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FUNCTIONS DEFINED BY IMPROPER INTEGRALS

William F. Trench Trinity University

Trinity University Book: Functions Defined by Improper Integrals (Trench)

William F. Trench

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TABLE OF CONTENTS Licensing

1: Chapters 1.1: Introduction to Improper Functions 1.2: Preparation 1.3: Uniform convergence of improper integrals 1.4: Absolutely Uniformly Convergent Improper Integrals 1.5: Dirichlet’s Tests 1.6: Consequences of Uniform Convergence 1.7: Applications to Laplace transforms 1.8: Improper Functions (Exercises)

Index Glossary Detailed Licensing

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Licensing A detailed breakdown of this resource's licensing can be found in Back Matter/Detailed Licensing.

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CHAPTER OVERVIEW 1: Chapters 1.1: Introduction to Improper Functions 1.2: Preparation 1.3: Uniform convergence of improper integrals 1.4: Absolutely Uniformly Convergent Improper Integrals 1.5: Dirichlet’s Tests 1.6: Consequences of Uniform Convergence 1.7: Applications to Laplace transforms 1.8: Improper Functions (Exercises)

This page titled 1: Chapters is shared under a not declared license and was authored, remixed, and/or curated by William F. Trench via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.

1

1.1: Introduction to Improper Functions In Section 7.2 (pp. 462–484) we considered functions of the form b

F (y) = ∫

f (x, y) dx,

c ≤ y ≤ d.

(1.1.1)

a

We saw that if f is continuous on [a, b] × [c, d] , then F is continuous on [c, d] (Exercise 7.2.3, p. 481) and that we can reverse the order of integration in d

∫

d

F (y) dy = ∫

c

b

(∫

c

f (x, y) dx) dy

(1.1.2)

f (x, y) dy) dx

(1.1.3)

a

to evaluate it as d

∫

b

F (y) dy = ∫

c

d

(∫

a

c

(Corollary 7.2.3, p. 466). Here is another important property of F . [theorem:1] If f and f are continuous on [a, b] × [c, d], then y

b

F (y) = ∫

f (x, y) dx,

c ≤ y ≤ d,

(1.1.4)

a

is continuously differentiable on [c, d] and F that is,

′

(y)

can be obtained by differentiating [eq:1] under the integral sign with respect to y; b

′

F (y) = ∫

fy (x, y) dx,

c ≤ y ≤ d.

(1.1.5)

a

Here F

′

(a)

and f

y (x,

a)

are derivatives from the right and F

′

(b)

and f

y (x,

b)

are derivatives from the left.

If y and y + Δy are in [c, d] and Δy ≠ 0 , then b

F (y + Δy) − F (y)

f (x, y + Δy) − f (x, y)

=∫ Δy

dx.

(1.1.6)

Δy

a

From the mean value theorem (Theorem 2.3.11, p. 83), if x ∈ [a, b] and y , y + Δy ∈ [c, d] , there is a y(x) between y and y + Δy such that f (x, y + Δy) − f (x, y) = fy (x, y)Δy = fy (x, y(x))Δy + (fy (x, y(x) − fy (x, y))Δy.

(1.1.7)

From this and [eq:3], b

∣ F (y + Δy) − F (y) ∣

−∫ Δy

∣

b

∣ fy (x, y) dx ∣ ≤ ∫ ∣

a

| fy (x, y(x)) − fy (x, y)| dx.

Now suppose ϵ > 0 . Since f is uniformly continuous on the compact set between y and y + Δy , there is a δ > 0 such that if |Δ| < δ then y

| fy (x, y) − fy (x, y(x))| < ϵ,

(1.1.8)

a

[a, b] × [c, d]

(Corollary 5.2.14, p. 314) and

(x, y) ∈ [a, b] × [c, d].

y(x)

is

(1.1.9)

This and [eq:4] imply that b

∣ F (y + Δy − F (y)) ∣ ∣

−∫ Δy

∣ fy (x, y) dx ∣ < ϵ(b − a)

a

(1.1.10)

∣

if y and y + Δy are in [c, d] and 0 < |Δy| < δ . This implies [eq:2]. Since the integral in [eq:2] is continuous on 7.2.3, p. 481, with f replaced by f ), F is continuous on [c, d].

[c, d]

(Exercise

′

y

[example:1] Since

1.1.1

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f (x, y) = cos xy\quad and\quadfy (x, y) = −x sin xy

(1.1.11)

are continuous for all (x, y), Theorem [theorem:1] implies that if π

F (y) = ∫

cos xy dx,

−∞ < y < ∞,

(1.1.12)

0

then π ′

F (y) = − ∫

x sin xy dx,

−∞ < y < ∞.

(1.1.13)

0

(In applying Theorem [theorem:1] for a specific value of y , we take R = [0, π] × [−ρ, ρ], where ρ > |y|.) This provides a convenient way to evaluate the integral in [eq:6]: integrating the right side of [eq:5] with respect to x yields π

sin xy ∣ ∣ ∣ y

F (y) =

sin πy =

,

y ≠ 0.

(1.1.14)

y

x=0

Differentiating this and using [eq:6] yields π

∫

sin πy x sin xy dx = y

0

2

π cos πy −

,

y ≠ 0.

(1.1.15)

y

To verify this, use integration by parts. We will study the continuity, differentiability, and integrability of b

F (y) = ∫

f (x, y) dx,

y ∈ S,

(1.1.16)

a

where

S

is an interval or a union of intervals, and F is a convergent improper integral for each y ∈ S . If the domain of where −∞ < a < b ≤ ∞ , we say that F is pointwise convergent on S or simply convergent on S , and write

f

is

[a, b) × S

b

∫

r

f (x, y) dx = lim ∫ r→b−

a

if, for each y ∈ S and every ϵ > 0 , there is an r = r

0 (y)

∣ ∣F (y) − ∫ ∣ a

r

f (x, y) dx

(1.1.17)

a

(which also depends on ϵ) such that

∣ ∣ f (x, y) dx ∣ = ∣∫ ∣ ∣ r

b

∣ f (x, y) dx ∣ < ϵ,

r0 (y) ≤ y < b.

(1.1.18)

∣

If the domain of f is (a, b] × S where −∞ ≤ a < b < ∞ , we replace [eq:7] by b

∫

b

f (x, y) dx = lim ∫ r→a+

a

f (x, y) dx

(1.1.19)

r

and [eq:8] by b

∣ ∣F (y) − ∫ ∣

r

∣ ∣ f (x, y) dx ∣ = ∣∫ ∣ a ∣

r

∣ f (x, y) dx ∣ < ϵ, ∣

a < r ≤ r0 (y).

(1.1.20)

In general, pointwise convergence of F for all y ∈ S does not imply that F is continuous or integrable on [c, d], and the additional assumptions that f is continuous and ∫ f (x, y) dx converges do not imply [eq:2]. b

y

a

y

[example:2] The function f (x, y) = ye

−|y|x

(1.1.21)

is continuous on [0, ∞) × (−∞, ∞) and ∞

F (y) = ∫

∞

f (x, y) dx = ∫

0

ye

−|y|x

dx

(1.1.22)

0

converges for all y , with

1.1.2

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⎧ −1 F (y) = ⎨ ⎩

y < 0,

0

y = 0,

1

y > 0;

(1.1.23)

therefore, F is discontinuous at y = 0 . [example:3] The function 3

f (x, y) = y e

−y

2

x

(1.1.24)

is continuous on [0, ∞) × (−∞, ∞) . Let ∞

F (y) = ∫

∞ 3

f (x, y) dx = ∫

0

y e

−y

2

x

dx = y,

−∞ < y < ∞.

(1.1.25)

0

Then ′

F (y) = 1,

−∞ < y < ∞.

(1.1.26)

However, ∞

∂

∫ 0

∞ 3

(y e

−y

2

x

) dx = ∫

∂y

(3 y

2

4

− 2 y x)e

0

−y

2

x

dx = {

1,

y ≠ 0,

0,

y = 0,

(1.1.27)

so ∞ ′

∂f (x, y)

F (y) ≠ ∫ 0

dx\quad if\quady = 0.

(1.1.28)

∂y

This page titled 1.1: Introduction to Improper Functions is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by William F. Trench via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.

1.1.3

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1.2: Preparation We begin with two useful convergence criteria for improper integrals that do not involve a parameter. Consistent with the definition on p. 152, we say that f is locally integrable on an interval I if it is integrable on every finite closed subinterval of I . [theorem:2] Suppose g is locally integrable on [a, b) and denote r

G(r) = ∫

g(x) dx,

a ≤ r < b.

(1.2.1)

a

Then the improper integral ∫

b

a

g(x) dx

converges if and only if, for each ϵ > 0, there is an r

0

|G(r) − G(r1 )| < ϵ,

For necessity, suppose ∫

b

a

g(x) dx = L

∈ [a, b)

r0 ≤ r, r1 < b.

(1.2.2)

. By definition, this means that for each ϵ > 0 there is an r

0

ϵ |G(r) − L| < 2

such that

∈ [a, b)

such that

ϵ \quad and\quad|G(r1 ) − L|
0, there is an r

0

1.2.1

∈ (a, b]

such that

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|G(r) − G(r1 )| < ϵ,

a < r, r1 ≤ r0 .

(1.2.11)

To see why we associate Theorems [theorem:2] and [theorem:3] with Cauchy, compare them with Theorem 4.3.5 (p. 204) This page titled 1.2: Preparation is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by William F. Trench via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.

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1.3: Uniform convergence of improper integrals Henceforth we deal with functions f of the following forms:

with domains I × S , where S is an interval or a union of intervals and I is of one

= f (x, y)

with −∞ < a < b ≤ ∞ ; with −∞ ≤ a < b < ∞ ; (a, b) with −∞ ≤ a ≤ b ≤ ∞ . [a, b) (a, b]

In all cases it is to be understood that ∫

b

a

f (x, y) dx

f

is locally integrable with respect to

x

on I . When we say that the improper integral

has a stated property “on S” we mean that it has the property for every y ∈ S .

[definition:1] If the improper integral b

∫

r

f (x, y) dx = lim ∫ r→b−

a

f (x, y) dx

(1.3.1)

a

converges on S, it is said to converge uniformly (or be uniformly convergent) on S if, for each ϵ > 0, there is an that b

∣ ∣∫ ∣

r

f (x, y) dx − ∫

a

r0 ∈ [a, b)

such

∣ f (x, y) dx ∣ < ϵ,

y ∈ S,

r0 ≤ r < b,

(1.3.2)

∣

a

or, equivalently, b

∣ ∣∫ ∣

∣ f (x, y) dx ∣ < ϵ,

y ∈ S,

r0 ≤ r < b.

(1.3.3)

∣

r

The crucial difference between pointwise and uniform convergence is that r (y) in [eq:8] may depend upon the particular value of y , while the r in [eq:11] does not: one choice must work for all y ∈ S . Thus, uniform convergence implies pointwise convergence, but pointwise convergence does not imply uniform convergence. 0

0

(Cauchy Criterion for Uniform Convergence I) [theorem:4] The improper integral in [eq:10] converges uniformly on S if and only if, for each ϵ > 0, there is an r ∈ [a, b) such that 0

∣ ∣∫ ∣ r

Suppose ∫

b

a

f (x, y) dx

r1

∣ f (x, y) dx ∣ < ϵ, ∣

y ∈ S,

r0 ≤ r, r1 < b.

(1.3.4)

converges uniformly on S and ϵ > 0 . From Definition [definition:1], there is an r

0

b

∣ ∣∫ ∣

∣

\, and\, ∣∫ 2

∣

r

b

∣

ϵ

f (x, y) dx ∣
0 ; however, it does not converge uniformly on any neighborhood of y = 0 , since, for any r > 0 , e > if |y| is sufficiently small. 0

1

−r|y|

2

[definition:2] If the improper integral b

∫

b

f (x, y) dx = lim ∫ r→a+

a

f (x, y) dx

(1.3.14)

r

converges on S, it is said to converge uniformly (or be uniformly convergent) on S if, for each ϵ > 0, there is an that b

∣ ∣∫ ∣

b

f (x, y) dx − ∫

a

r0 ∈ (a, b]

such

∣ f (x, y) dx ∣ < ϵ,

y ∈ S,

a < r ≤ r0 ,

(1.3.15)

∣

r

or, equivalently, ∣ ∣∫ ∣ a

r

∣ f (x, y) dx ∣ < ϵ, ∣

y ∈ S,

a < r ≤ r0 .

(1.3.16)

We leave proof of the following theorem to you (Exercise [exer:3]). (Cauchy Criterion for Uniform Convergence II) [theorem:5] The improper integral b

∫

b

f (x, y) dx = lim ∫ r→a+

a

converges uniformly on S if and only if, for each ϵ > 0, there is an r

0

r

∣ ∣∫ ∣

f (x, y) dx

(1.3.17)

r

∈ (a, b]

such that

∣ f (x, y) dx ∣ < ϵ,

y ∈ S,

a < r, r1 ≤ r0 .

(1.3.18)

∣

r1

We need one more definition, as follows. [definition:3] Let f all ∫

b

c

= f (x, y)

be defined on

(a, b) × S,

and let c be an arbitrary point in (a, b). Then f (x, y) dx both converge uniformly on S. y ∈ S

where ∫

b

a

−∞ ≤ a < b ≤ ∞.

f (x, y) dx

1.3.2

Suppose

f

is locally integrable on

is said to converge uniformly on

S

if

∫

c

a

(a, b)

f (x, y) dx

for

and

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c

We leave it to you (Exercise [exer:4]) to show that this definition is independent of c ; that is, if ∫ f (x, y) dx and both converge uniformly on S for some c ∈ (a, b) , then they both converge uniformly on S for every c ∈ (a, b) . a

∫

b

c

f (x, y) dx

b

We also leave it you (Exercise [exer:5]) to show that if f is bounded on [a, b] × [c, d] and ∫ f (x, y) dx exists as a proper integral for each y ∈ [c, d], then it converges uniformly on [c, d] according to all three Definitions [definition:1]–[definition:3]. a

[example:5] Consider the improper integral ∞ −1/2

F (y) = ∫

x

e

−xy

dx,

(1.3.19)

0

which diverges if y ≤ 0 (verify). Definition [definition:3] applies if y > 0 , so we consider the improper integrals 1

∞ −1/2

F1 (y) = ∫

x

e

−xy

−1/2

dx\quad and\quadF2 (y) = ∫

0

x

e

−xy

dx

(1.3.20)

dx,

(1.3.21)

1

separately. Moreover, we could just as well define c

∞ −1/2

F1 (y) = ∫

x

e

−xy

−1/2

dx\quad and\quadF2 (y) = ∫

0

x

e

−xy

c

where c is any positive number. Definition [definition:2] applies to F . If 0 < r 1

and y ≥ 0 , then

0

−ry

, since the change of variable

(0, ρ)

u = xy

yields r1

∫

r1 y −1/2

x

e

−xy

dx = y

−1/2

−1/2

∫

r

u

e

−u

du,

(1.3.24)

ry

which, for any fixed r > 0 , can be made arbitrarily large by taking y sufficiently small and r = 1/y . Therefore we conclude that F (y) converges uniformly on [ρ, ∞) if ρ > 0. Note that that the constant c in [eq:16] plays no role in this argument. [example:6] Suppose we take ∞

sin u

∫

π du =

u

0

(1.3.25) 2

as given (Exercise [exer:31](b)). Substituting u = xy with y > 0 yields ∞

sin xy

∫ 0

What about uniform convergence? Since apply here. If 0 < r < r and y > 0 , then

π dx =

x

(sin xy)/x

,

y > 0.

(1.3.26)

2

is continuous at

x =0

, Definition [definition:1] and Theorem [theorem:4]

1

r1

sin xy

∫ r

x

( y

r1

r1

1 dx = −

cos xy ∣ ∣ ∣ x

r

+∫ r

cos xy 2

x

∣ dx) , \, so\quad ∣∫ ∣ r

r1

sin xy x

∣ 3 dx ∣ < . ∣ ry

(1.3.27)

Therefore [eq:18] converges uniformly on [ρ, ∞) if ρ > 0 . On the other hand, from [eq:17], there is a δ > 0 such that

1.3.3

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∞

sin u

∫ u0

π du >

u

, 4

0 ≤ u0 < δ.

(1.3.28)

This and [eq:18] imply that ∞

r

∞

sin xy

∫

sin u

dx = ∫ x

yr

π du >

u

(1.3.29) 4

for any r > 0 if 0 < y < δ/r . Hence, [eq:18] does not converge uniformly on any interval (0, ρ] with ρ > 0 . This page titled 1.3: Uniform convergence of improper integrals is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by William F. Trench via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.

1.3.4

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1.4: Absolutely Uniformly Convergent Improper Integrals (Absolute Uniform Convergence I) [definition:4] The improper integral b

r

∫

f (x, y) dx = lim ∫ r→b−

a

f (x, y) dx

(1.4.1)

|f (x, y)| dx

(1.4.2)

a

is said to converge absolutely uniformly on S if the improper integral b

∫

r

|f (x, y)| dx = lim ∫ r→b−

a

converges uniformly on S ; that is, if, for each ϵ > 0 , there is an r b

∣∫ ∣

r

|f (x, y)| dx − ∫

a

such that

∈ [a, b)

0

∣

a

∣ |f (x, y)| dx ∣ < ϵ,

y ∈ S,

r0 < r < b.

(1.4.3)

∣

a

To see that this definition makes sense, recall that if f is locally integrable on [a, b) for all y in S , then so is |f | (Theorem 3.4.9, p. 161). Theorem [theorem:4] with f replaced by for each ϵ > 0 , there is an r ∈ [a, b) such that

|f |

implies that

∫

b

a

f (x, y) dx

converges absolutely uniformly on

S

if and only if,

0

r1

∫

|f (x, y)| dx < ϵ,

y ∈ S,

r0 ≤ r < r1 < b.

(1.4.4)

r

Since ∣ ∣∫ ∣ r

Theorem [theorem:4] implies that if ∫

b

a

r1

∣ f (x, y) dx ∣ ≤ ∫ ∣ r

r1

|f (x, y)| dx,

(1.4.5)

converges absolutely uniformly on S then it converges uniformly on S .

f (x, y) dx

[theorem:6] ( Weierstrass’s Test for Absolute Uniform Convergence I) Suppose ∫ M (x) dx < ∞, and

M = M (x)

is nonnegative on

[a, b),

b

a

|f (x, y)| ≤ M (x),

Then ∫

b

a

f (x, y) dx

Denote ∫

b

a

y ∈ S,

a ≤ x < b.

(1.4.6)

converges absolutely uniformly on S.

M (x) dx = L < ∞

. By definition, for each ϵ > 0 there is an r

0

∈ [a, b)

such that

r

L−ϵ < ∫

M (x) dx ≤ L,

r0 < r < b.

(1.4.7)

a

Therefore, if r

0

< r ≤ r1 ,

then r1

0 ≤∫

r1

M (x) dx = ( ∫

r

r

M (x) dx − L) − ( ∫

a

M (x) dx − L) < ϵ

(1.4.8)

a ≤ r0 < r < r1 < b.

(1.4.9)

a

This and [eq:19] imply that r1

∫ r

r1

|f (x, y)| dx ≤ ∫

M (x) dx < ϵ,

y ∈ S,

r

Now Theorem [theorem:4] implies the stated conclusion. [example:7] Suppose g = g(x, y) is locally integrable on [0, ∞) for all y ∈ S and, for some a such that

0

|g(x, y)| ≤ K e

p0 x

,

y ∈ S,

x ≥ a0 .

≥0

, there are constants K and p

0

(1.4.10)

If p > p and r ≥ a , then 0

0

1.4.1

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∞

∫

∞

e

−px

|g(x, y)| dx =

∫

r

e

−(p−p0 )x

e

−p0 x

|g(x, y)| dx

r ∞

≤

K∫

e

−(p−p0 )x

Ke

∞

e

0

−px

g(x, y) dx

0

, p − p0

r

so ∫

−(p−p )r

dx =

converges absolutely on S . For example, since α

|x

sin xy| < e

p0 x

α

\quad and \quad| x ∞

cos xy| < e

p0 x

(1.4.11) ∞

for x sufficiently large if p > 0 , Theorem [theorem:4] implies that ∫ e x sin xy dx and ∫ e x cos xy dx converge absolutely uniformly on (−∞, ∞) if p > 0 and α  ≥  0 . As a matter of fact, ∫ e x sin xy dx converges absolutely on (−∞, ∞) if p > 0 and α > −1 . (Why?) 0

−px

α

0

−px

α

0

∞

−px

α

0

(Absolute Uniform Convergence II) [definition:5] The improper integral b

∫

b

f (x, y) dx = lim ∫ r→a+

a

f (x, y) dx

(1.4.12)

|f (x, y)| dx

(1.4.13)

r

is said to converge absolutely uniformly on S if the improper integral b

∫

b

|f (x, y)| dx = lim ∫ r→a+

a

converges uniformly on S ; that is, if, for each ϵ > 0 , there is an r

0

b

∣ ∣∫ ∣

∈ (a, b]

b

|f (x, y)| dx − ∫

a

r

such that

∣ |f (x, y)| dx ∣ < ϵ,

y ∈ S,

a < r < r0 ≤ b.

(1.4.14)

∣

r

We leave it to you (Exercise [exer:7]) to prove the following theorem. [theorem:7] (Weierstrass’s Test for Absolute Uniform Convergence II) Suppose ∫ M (x) dx < ∞, and

M = M (x)

is nonnegative on

(a, b],

b

a

|f (x, y)| ≤ M (x),

Then ∫

b

a

f (x, y) dx

y ∈ S,

x ∈ (a, b].

(1.4.15)

converges absolutely uniformly on S .

[example:8] If g = g(x, y) is locally integrable on (0, 1] for all y ∈ S and −β

|g(x, y)| ≤ Ax

,

0 < x ≤ x0 ,

(1.4.16)

for each y ∈ S , then 1 α

∫

x

g(x, y) dx

(1.4.17)

0

converges absolutely uniformly on S if α > β − 1 . To see this, note that if 0 < r < r

1

r

∫

r α

x

r1

|g(x, y)| dx ≤ A ∫

α−β+1

α−β

x

Ax

dx =

r1

r

≤ x0

∣ ∣ < α − β + 1 ∣r1

, then α−β+1

Ar

.

(1.4.18)

α −β +1

Applying this with β = 0 shows that 1 α

F (y) = ∫

x

cos xy dx

(1.4.19)

sin xy dx

(1.4.20)

0

converges absolutely uniformly on (−∞, ∞) if α > −1 and 1 α

G(y) = ∫

x

0

converges absolutely uniformly on (−∞, ∞) if α > −2 . By recalling Theorem 4.4.15 (p. 246), you can see why we associate Theorems [theorem:6] and [theorem:7] with Weierstrass.

1.4.2

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This page titled 1.4: Absolutely Uniformly Convergent Improper Integrals is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by William F. Trench via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.

1.4.3

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1.5: Dirichlet’s Tests Weierstrass’s test is useful and important, but it has a basic shortcoming: it applies only to absolutely uniformly convergent improper integrals. The next theorem applies in some cases where ∫ f (x, y) dx converges uniformly on S , but ∫ |f (x, y)| dx does not. b

b

a

[theorem:8] (Dirichlet’s Test for Uniform Convergence I) If g,

gx ,

a

and h are continuous on [a, b) × S, then

b

∫

g(x, y)h(x, y) dx

(1.5.1)

a

converges uniformly on S if the following conditions are satisfied: lim { sup |g(x, y)|} = 0; x→b−

(1.5.2)

y∈S

There is a constant M such that ∣ sup ∣∫ y∈S ∣ a

∫

b

a

| gx (x, y)| dx

x

∣ h(u, y) du ∣ < M , ∣

a ≤ x < b;

(1.5.3)

converges uniformly on S.

If x

H (x, y) = ∫

h(u, y) du,

(1.5.4)

a

then integration by parts yields r1

∫

r1

g(x, y)h(x, y) dx =

∫

r

g(x, y)Hx (x, y) dx

r

=

g(r1 , y)H (r1 , y) − g(r, y)H (r, y) r1

−∫

gx (x, y)H (x, y) dx.

r

Since assumption (b) and [eq:20] imply that |H (x, y)| ≤ M , (x, y) ∈ (a, b] × S , Eqn. [eq:21] implies that ∣ ∣∫ ∣ r

on [r, r

1]

×S

r1

∣ g(x, y)h(x, y) dx ∣ < M (2 sup |g(x, y)| + ∫ ∣ x≥r r

r1

| gx (x, y)| dx)

(1.5.5)

.

Now suppose ϵ > 0 . From assumption (a), there is an r ∈ [a, b) such that |g(x, y)| < ϵ on S if r and Theorem [theorem:6], there is an s ∈ [a, b) such that 0

0

≤x 0.

(1.5.8)

x +y

0

The obvious inequality 1 ∣ cos xy ∣ ∣ ∣ ≤ ∣ x +y ∣ x +y

(1.5.9)

is useless here, since ∞

dx

∫ 0

= ∞.

(1.5.10)

x +y

However, integration by parts yields r1

cos xy

∫ r

∣ ∣ y(x + y) ∣r

x +y sin r1 y

=

r1

r1

sin xy

dx =

2

r1

sin xy

+∫ y(r + y)

r

dx

y(x + y)

r

sin ry −

y(r1 + y)

sin xy

+∫

2

dx.

y(x + y)

Therefore, if 0 < r < r , then 1

∣ ∣∫ ∣

r1

r

cos xy x +y

∣ 1 2 dx ∣ < ( +∫ ∣ y r+y r

∞

1

3 2

) ≤

(x + y)

3 2

y(r + y)

≤ ρ(r + ρ)

if y ≥ ρ > 0 . Now Theorem [theorem:4] implies that I (y) converges uniformly on [ρ, ∞) if ρ > 0 . We leave the proof of the following theorem to you (Exercise [exer:10]). [theorem:9] (Dirichlet’s Test for Uniform Convergence II) If g,

gx ,

and h are continuous on (a, b] × S, then

b

∫

g(x, y)h(x, y) dx

(1.5.11)

a

converges uniformly on S if the following conditions are satisfied: lim { sup |g(x, y)|} = 0; x→a+

y∈S

There is a constant M such that b

∣ sup ∣∫ y∈S

b

∫

a

| gx (x, y)| dx

∣

x

∣ h(u, y) du ∣ ≤ M ,

a < x ≤ b;

(1.5.12)

∣

converges uniformly on S .

By recalling Theorems 3.4.10 (p. 163), 4.3.20 (p. 217), and 4.4.16 (p. 248), you can see why we associate Theorems [theorem:8] and [theorem:9] with Dirichlet. This page titled 1.5: Dirichlet’s Tests is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by William F. Trench via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.

1.5.2

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1.6: Consequences of Uniform Convergence [theorem:10] If f

= f (x, y)

is continuous on either [a, b) × [c, d] or (a, b] × [c, d] and b

F (y) = ∫

f (x, y) dx

(1.6.1)

a

converges uniformly on [c, d], then F is continuous on [c, d]. Moreover, d

∫

b

(∫

c

b

f (x, y) dx) dy = ∫

a

d

(∫

a

f (x, y) dy) dx.

(1.6.2)

c

We will assume that f is continuous on (a, b] × [c, d] . You can consider the other case (Exercise [exer:14]). We will first show that F in [eq:23] is continuous on [c, d]. Since F converges uniformly on (specifically, [eq:11]) implies that if ϵ > 0 , there is an r ∈ [a, b) such that b

∣ ∣∫ ∣

Therefore, if c ≤ y, y

0

≤ d]

, Definition [definition:1]

[c, d]

∣ f (x, y) dx ∣ < ϵ,

c ≤ y ≤ d.

(1.6.3)

∣

r

, then b

∣ |F (y) − F (y0 )| =

∣∫ ∣ r

∣ ∣∫ ∣

≤

a

b

f (x, y) dx − ∫

a

∣ f (x, y0 ) dx ∣ ∣

a

∣

∣ [f (x, y) − f (x, y0 )] dx ∣ + ∣∫ ∣ ∣

b

∣ f (x, y) dx ∣ ∣

r

b

∣ + ∣∫ ∣

∣ f (x, y0 ) dx ∣ ,

r

∣

so r

|F (y) − F (y0 )| ≤ ∫

|f (x, y) − f (x, y0 )| dx + 2ϵ.

(1.6.4)

a

Since f is uniformly continuous on the compact set [a, r] × [c, d] (Corollary 5.2.14, p. 314), there is a δ > 0 such that |f (x, y) − f (x, y0 )| < ϵ

if (x, y) and (x, y

0)

are in [a, r] × [c, d] and |y − y

0|

(1.6.5)

. This and [eq:25] imply that

0,

(1.6.14)

y1

it follows that ∞

e

−xy1

−e

−xy2

∫

dx = log x

0

y2

,

y2 ≥ y1 > 0.

y1

(1.6.15)

[example:11] From Example [example:6], ∞

sin xy

∫

π dx =

x

0

,

y > 0,

(1.6.16)

2

and the convergence is uniform on [ρ, ∞) if ρ > 0 . Therefore, Theorem [theorem:10] implies that if 0 < y

1

y

π 2

∞

2

(y2 − y1 ) =

∫

∞

sin xy

(∫

y1

dx) dy = ∫ x

0

y

2

y1 ∞

=

∫

dy) dx x

cos x y1 − cos x y2 x2

0

, then

sin xy

(∫

0

< y2

dx.

The last integral converges uniformly on (−∞, ∞) (Exercise 10(h)), and is therefore continuous with respect to y on (−∞, ∞), by Theorem [theorem:10]; in particular, we can let y → 0+ in [eq:27] and replace y by y to obtain 1

1

∞

∫ 0

2

1 − cos xy

πy dx =

2

,

y ≥ 0.

(1.6.17)

2

x

The next theorem is analogous to Theorem 4.4.20 (p. 252). [theorem:11] Let f and f be continuous on either [a, b) × [c, d] or (a, b] × [c, d]. Suppose that the improper integral y

b

F (y) = ∫

f (x, y) dx

(1.6.18)

fy (x, y) dx

(1.6.19)

a

converges for some y

0

∈ [c, d]

and b

G(y) = ∫ a

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converges uniformly on [c, d]. Then F converges uniformly on [c, d] and is given explicitly by y

F (y) = F (y0 ) + ∫

G(t) dt,

c ≤ y ≤ d.

(1.6.20)

y

0

Moreover, F is continuously differentiable on [c, d]; specifically, ′

F (y) = G(y),

where F

′

(c)

and f

y (x,

c)

are derivatives from the right, and F

′

(d)

c ≤ y ≤ d,

and f

y (x,

(1.6.21)

are derivatives from the left.

d)

We will assume that f and f are continuous on [a, b) × [c, d] . You can consider the other case (Exercise [exer:15]). y

Let r

Fr (y) = ∫

f (x, y) dx,

a ≤ r < b,

c ≤ y ≤ d.

(1.6.22)

a

Since f and f are continuous on [a, r] × [c, d] , Theorem [theorem:1] implies that y

r ′

Fr (y) = ∫

fy (x, y) dx,

c ≤ y ≤ d.

(1.6.23)

a

Then y

Fr (y) =

Fr (y0 ) + ∫

r

(∫

y

fy (x, t) dx) dt

a

0

y

=

F (y0 ) + ∫

G(t) dt

y

0

y

+(Fr (y0 ) − F (y0 )) − ∫

b

(∫

y0

fy (x, t) dx) dt,

c ≤ y ≤ d.

r

Therefore, y

∣ ∣Fr (y) − F (y0 ) − ∫ ∣

∣ G(t) dt∣ ≤

| Fr (y0 ) − F (y0 )|

∣

y

0

y

∣ + ∣∫ ∣

Now suppose ϵ > 0 . Since we have assumed that lim

r→b−

Fr (y0 ) = F (y0 )

| Fr (y0 ) − F (y0 )| < ϵ,

1

b

∣∫ ∣

y0

∣ fy (x, t) dx ∣ dt. ∣

r

exists, there is an r in (a, b) such that 0

r0 < r < b.

Since we have assumed that G(y) converges for y ∈ [c, d], there is an r ∣

b

∫

∈ [a, b)

(1.6.24)

such that

∣ fy (x, t) dx ∣ < ϵ,

t ∈ [c, d],

r1 ≤ r < b.

(1.6.25)

∣

r

Therefore, [eq:29] yields y

∣ ∣Fr (y) − F (y0 ) − ∫ ∣

if max(r

0,

r1 ) ≤ r < b

∣ G(t) dt∣ < ϵ(1 + |y − y0 |) ≤ ϵ(1 + d − c)

(1.6.26)

∣

y0

and t ∈ [c, d] . Therefore F (y) converges uniformly on [c, d] and y

F (y) = F (y0 ) + ∫

G(t) dt,

c ≤ y ≤ d.

(1.6.27)

y0

Since G is continuous on [c, d] by Theorem [theorem:10], [eq:28] follows from differentiating this (Theorem 3.3.11, p. 141). [example:12] Let

1.6.3

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∞ 2

I (y) = ∫

e

−yx

dx,

y > 0.

(1.6.28)

0

Since r

e

−yx

r√y

1

2

∫

dx =

2

∫ √y

0

e

−t

dt,

(1.6.29)

0

it follows that ∞

1

2

I (y) =

∫ √y

and the convergence is uniform on then

[ρ, ∞)

−t

dt,

(1.6.30)

0

if ρ > 0 (Exercise [exer:8](i)). To evaluate the last integral, denote ρ

2

J

e

ρ

ρ

2

(ρ) = ( ∫

e

du) ( ∫

0

e

−v

2

dv) = ∫

0

ρ

0

2

e

−t

dt

;

ρ

2

−u

J(ρ) = ∫

∫

0

e

2

−( u +v )

du dv.

(1.6.31)

0

Transforming to polar coordinates r = r cos θ , v = r sin θ yields − −−−−−−− − π/2

J

2

π(1 − e

2

(ρ) = ∫

∫

0

2

√ π(1 − e−ρ )

2

ρ

re

−r

−ρ

)

dr dθ =

, \quad so\quadJ(ρ) =

.

4

0

(1.6.32)

2

Therefore − √π

∞ 2

∫

e

−t

dt = lim J(ρ) =

1

2

2

ρ→∞

0

∞

\quad and\quad ∫

e

−yx

dx = 2

0

− − π , y

√

y > 0.

(1.6.33)

Differentiating this n times with respect to y yields − 1 ⋅ 3 ⋯ (2n − 1)√π

∞

∫

2n

x

2

e

−yx

dx =

y > 0,

n

n = 1, 2, 3, … ,

where Theorem [theorem:11] justifies the differentiation for every ρ > 0 (Exercise [exer:8](i)).

n

, since all these integrals converge uniformly on b

Some advice for applying this theorem: Be sure to check first that F (y ) = ∫ f (x, y so, differentiate ∫ f (x, y) dx formally to obtain ∫ f (x, y) dx. Then F (y) = ∫ f this improper integral converges uniformly. 0

b

a

(1.6.34)

2 y n+1/2

0

b

a

′

y

0)

a

b

a

dx

[ρ, ∞)

if

converges for at least one value of y . If if y is in some interval on which

y (x, y) dx

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1.6.4

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1.7: Applications to Laplace transforms The Laplace transform of a function f locally integrable on [0, ∞) is ∞

F (s) = ∫

e

−sx

f (x) dx

(1.7.1)

0

for all s such that integral converges. Laplace transforms are widely applied in mathematics, particularly in solving differential equations. We leave it to you to prove the following theorem (Exercise [exer:26]). [theorem:12] Suppose f is locally integrable on [0, ∞) and |f (x)| ≤ M e F converges uniformly on [ s , ∞) if s > s . 1

1

∞

0

1

for sufficiently large x. Then the Laplace transform of

0

[theorem:13] If f is continuous on [0, ∞) and H (x) = ∫ converges uniformly on [s , ∞) if s > s . 1

s0 x

e

−s0 u

f (u) du

is bounded on

[0, ∞),

then the Laplace transform of

f

0

If 0 ≤ r ≤ r , 1

r1

∫

r1

e

−sx

f (x) dx = ∫

r

r1

e

−(s−s0 )x

e

−s0 x

f (x) dt = ∫

r

e

−(s−s0 )t

′

H (x) dt.

(1.7.2)

r

Integration by parts yields r1

∫

r1

r1

e

−sx

f (x) dt = e

−(s−s0 )x

r

∣ H (x)∣ ∣r

+ (s − s0 ) ∫

e

−(s−s0 )x

H (x) dx.

(1.7.3)

r

Therefore, if |H (x)| ≤ M , then ∣ ∣∫ ∣

r1

e

−sx

r

∣ f (x) dx ∣ ≤ ∣

∣ −(s−s ) r −(s−s0 )r 0 1 M ∣e +e + (s − s0 ) ∫ ∣

r1

e

−(s−s0 )x

r

≤

3M e

−(s−s0 )r

Now Theorem [theorem:4] implies that F (s) converges uniformly on [s

1,

≤ 3M e

−( s1 −s0 )r

,

∣ dx ∣ ∣

s ≥ s1 .

.

∞)

The following theorem draws a considerably stonger conclusion from the same assumptions. [theorem:14] If f is continuous on [0, ∞) and x

H (x) = ∫

e

−s0 u

f (u) du

(1.7.4)

0

is bounded on [0, ∞), then the Laplace transform of f is infinitely differentiable on (s

0,

with

∞),

∞

F

(n)

n

(s) = (−1 )

∫

e

−sx

n

x f (x) dx;

(1.7.5)

0

that is, the n -th derivative of the Laplace transform of f (x) is the Laplace transform of (−1)

n

.

n

x f (x)

First we will show that the integrals ∞

In (s) = ∫

e

−sx

n

x f (x) dx,

n = 0, 1, 2, …

(1.7.6)

0

all converge uniformly on [s

1,

∞)

if s

1

> s0

. If 0 < r < r , then 1

r1

∫ r

r1

e

−sx

n

x f (x) dx = ∫

r1

e

−(s−s0 )x

e

−s0 x

n

x f (x) dx = ∫

r

e

−(s−s0 )x

n

′

x H (x) dx.

(1.7.7)

r

Integrating by parts yields

1.7.1

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r1

∫

e

−sx

n

n

x f (x) dx =

r e

−(s−s0 ) r1

n

H (r) − r e

1

−(s−s0 )r

H (r)

r r1

′

−∫

H (x)(e

−(s−s0 )x

n

x ) dx,

r

where indicates differentiation with respect to x. Therefore, if |H (x)| ≤ M ′

∣ ∣∫ ∣

r1

e

r

Therefore, since e

−(s−s0 )r

−sx

≤∞

∣ n −(s−s0 )r n −(s−s0 )r n x f (x) dx ∣ ≤ M ( e r +e r +∫ ∣

∞

|(e

−(s−s0 )x

n

′

)x ) | dx) .

(1.7.8)

r

n

r

decreases monotonically on (n, ∞) if s > s (check!), 0

r1

∣ ∣∫ ∣

e

−sx

r

∣ n −(s−s0 )r n x f (x) dx ∣ < 3M e r , ∣

n < r < r1 ,

so Theorem [theorem:4] implies that I (s) converges uniformly [s , ∞) if s F = −F , and an easy induction proof yields [eq:30] (Exercise [exer:25]). n

n+1

on [0, ∞), then

1

1

(1.7.9)

. Now Theorem [theorem:11] implies that

> s0

′ n

[example:13] Here we apply Theorem [theorem:12] with f (x) = cos ax (a ≠ 0 ) and s

0

x

∫

=0

. Since

sin ax cos au du =

(1.7.10) a

0

is bounded on (0, ∞), Theorem [theorem:12] implies that ∞

F (s) = ∫

e

−sx

cos ax dx

(1.7.11)

0

converges and ∞

F

(n)

n

(s) = (−1 )

∫

e

−sx

n

x

cos ax dx,

s > 0.

(1.7.12)

0

(Note that this is also true if a = 0 .) Elementary integration yields s F (s) =

2

2

s

.

(1.7.13)

+a

Hence, from [eq:31], ∞

∫ 0

e

−sx

n

x

n

cos ax = (−1 )

d

n

n

ds

s 2

s

2

,

n = 0, 1, … .

(1.7.14)

+a

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1.7.2

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1.8: Improper Functions (Exercises) [exer:1] Suppose g and h are differentiable on [a, b], with a ≤ g(y) ≤ b\quad and\quada ≤ h(y) ≤ b,

c ≤ y ≤ d.

(1.8.1)

Let f and f be continuous on [a, b] × [c, d] . Derive Liebniz’s rule: y

h(y)

d ∫ dy

f (x, y) dx =

′

′

f (h(y), y)h (y) − f (g(y), y)g (y)

g(y) h(y)

+∫

fy (x, y) dx.

g(y)

(Hint: Define H (y, u, v) = ∫

v

u

f (x, y) dx

and use the chain rule.)

[exer:2] Adapt the proof of Theorem [theorem:2] to prove Theorem [theorem:3]. [exer:3] Adapt the proof of Theorem [theorem:4] to prove Theorem [theorem:5]. c

[exer:4] Show that Definition [definition:3] is independent of c ; that is, if ∫ f (x, y) dx and ∫ on S for some c ∈ (a, b) , then they both converge uniformly on S and every c ∈ (a, b) . a

b

c

f (x, y) dx

both converge uniformly

[exer:5] ()0pt0pt14pt 8pt22pt0pt b

Show that if f is bounded on [a, b] × [c, d] and ∫ f (x, y) dx exists as a proper integral for each uniformly on [c, d] according to all of Definition [definition:1]–[definition:3]. a

y ∈ [c, d]

, then it converges

Give an example to show that the boundedness of f is essential in (a). [exer:6] Working directly from Definition [definition:1], discuss uniform convergence of the following integrals: ∞

∞

dx

(a) ∫

(b) ∫

dx

2

2

1 +y x

0

−xy

e

2

x

dx

0

∞

∞

(c) ∫

2n

x

2

−yx

e

(d) ∫

dx

0

sin x y

2

dx

0

∞

(e) ∫

∞

(3 y

2

−y

− 2xy)e

2

x

(f) ∫

dx

0

2

2

−xy

(2xy − y x )e

dx

0

[exer:7] Adapt the proof of Theorem [theorem:6] to prove Theorem [theorem:7]. [exer:8] Use Weierstrass’s test to show that the integral converges uniformly on S : ()0pt0pt14pt 8pt22pt0pt ∞

∫

e

−xy

sin x dx

, S = [ρ, ∞) ,ρ > 0

0 ∞

sin x

∫

,

dx S = [c, d]

y

,1 0

1 ∞

∫

e

xy

2

e

,

−x

,

dx S = [−ρ, ρ] ρ > 0

−∞ ∞

cos xy − cos ax

∫

dx

2

, S = (−∞, ∞)

x

0 ∞

∫

2n

x

2

e

−yx

,

,

dx S = [ρ, ∞) ρ > 0

, n = 0 , 1, 2,…

0

[exer:9] ()0pt0pt14pt 8pt22pt0pt Show that ∞ y−1

Γ(y) = ∫

x

e

−x

dx

(1.8.2)

y ≥ 0,

(1.8.3)

0

converges if y > 0 , and uniformly on [c, d] if 0 < c < d < ∞ . Use integration by parts to show that Γ(y + 1) Γ(y) =

, y

and then show by induction that Γ(y + n) Γ(y) =

,

y > 0,

n = 1, 2, 3, … .

(1.8.4)

y(y + 1) ⋯ (y + n − 1)

How can this be used to define Γ(y) in a natural way for all y ≠ 0 , −1, −2, …? (This function is called the gamma function.) Show that Γ(n + 1) = n! if n is a positive integer. Show that ∞

∫

e

−st

α

t

−α−1

dt = s

Γ(α + 1),

α > −1,

s > 0.

(1.8.5)

0

[exer:10] Show that Theorem [theorem:8] remains valid with assumption (c) replaced by the assumption that monotonic with respect to x for all y ∈ S .

| gx (x, y)|

is

[exer:11] Adapt the proof of Theorem [theorem:8] to prove Theorem [theorem:9]. [exer:12] Use Dirichlet’s test to show that the following integrals converge uniformly on S = [ρ, ∞) if ρ > 0 : ∞

(a) ∫

x

∞

0

(b) ∫

dx

y

1

(c) ∫

∞

sin xy

x +y

dx

∞

cosxy 2

(d) ∫

dx

[exer:13] Suppose g,

sin xy dx

1

gx

sin xy log x

2

1 + xy

and h are continuous on [a, b) × S, and denote H (x, y) = ∫

x

a

h(u, y) du, a ≤ x < b.

Suppose also that

b

lim { sup |g(x, y)H (x, y)|} = 0\quad and \quad ∫ x→b−

y∈S

gx (x, y)H (x, y) dx

(1.8.6)

a

converges uniformly on S. Show that ∫

b

a

g(x, y)h(x, y) dx

converges uniformly on S .

[exer:14] Prove Theorem [theorem:10] for the case where f

= f (x, y)

is continuous on (a, b] × [c, d] .

[exer:15] Prove Theorem [theorem:11] for the case where f

= f (x, y)

is continuous on (a, b] × [c, d] .

1.8.2

https://math.libretexts.org/@go/page/17329

[exer:16] Show that ∞

C (y) = ∫

∞

f (x) cos xy dx\quad and\quadS(y) = ∫

−∞

f (x) sin xy dx

(1.8.7)

−∞

are continuous on (−∞, ∞) if ∞

∫

|f (x)| dx < ∞.

(1.8.8)

−∞

[exer:17] Suppose f is continuously differentiable on [a, ∞), lim

x→∞

f (x) = 0

, and

∞ ′

∫

| f (x)| dx < ∞.

(1.8.9)

a

Show that the functions ∞

C (y) = ∫

∞

f (x) cos xy dx\quad and\quadS(y) = ∫

a

f (x) sin xy dx

(1.8.10)

a

are continuous for all y ≠ 0 . Give an example showing that they need not be continuous at y = 0 . [exer:18] Evaluate F (y) and use Theorem [theorem:11] to evaluate I : ()0pt0pt14pt 8pt22pt0pt ∞

∞

dx

F (y) = ∫

, y ≠ 0; I

1 + y 2 x2

0

dx

, y > −1 ; I

bx dx

∞ y

x

−1

ax − tan x

0

∞

F (y) = ∫

−1

tan

=∫

a

x

b

−x

=∫

0

dx log x

0

, a, b > 0

, a , b > −1

∞

F (y) = ∫

e

−xy

cos x dx

,y >0

0 ∞

e

−ax

−e

−bx

I =∫

cos x dx x

0

, a, b > 0

∞

F (y) = ∫

e

−xy

sin x dx

,y >0

0 ∞

e

−ax

−e

−bx

I =∫

sin x dx x

0

, a, b > 0

∞

F (y) = ∫

∞

e

−x

sin xy dx

;

I =∫

0

e

−x

1 − cos ax dx x

0 ∞

F (y) = ∫

∞

e

−x

cos xy dx

;

I =∫

0

e

−x

sin ax

0

dx x

[exer:19] Use Theorem [theorem:11] to evaluate: ()0pt0pt14pt 8pt22pt0pt 1

∫

n

y

(log x ) x

,

,

dx y > −1 n = 0

, 1, 2,….

0 ∞

dx

∫

2

(x

0

+ y )n+1

dx

, y > 0 , n = 0 , 1, 2, ….

∞ 2n+1

∫

x

2

e

−yx

dx

, y > 0 , n = 0 , 1, 2,….

0 ∞

∫

xy

x

,

dx 0 < y < 1

.

0

[exer:20]

1.8.3

https://math.libretexts.org/@go/page/17329

()0pt0pt14pt 8pt22pt0pt Use Theorem [theorem:11] and integration by parts to show that ∞ 2

F (y) = ∫

e

−x

cos 2xy dx

(1.8.11)

0

satisfies F

′

+ 2yF = 0.

(1.8.12)

Use part (a) to show that − √π F (y) =

e

−y

2

.

(1.8.13)

2

[exer:21] Show that ∞

y 2

∫

e

−x

sin 2xy dx = e

−y

2

2

∫

0

e

u

du.

(1.8.14)

0

(Hint: See Exercise [exer:20].) [exer:22] State a condition implying that ∞

C (y) = ∫

∞

f (x) cos xy dx\quad and\quadS(y) = ∫

a

f (x) sin xy dx

(1.8.15)

a

are n times differentiable on for all y ≠ 0 . (Your condition should imply the hypotheses of Exercise [exer:16].) [exer:23] Suppose f is continuously differentiable on [a, ∞), ∞

∫

k

′

|(x f (x)) | dx < ∞,

0 ≤ k ≤ n,

(1.8.16)

a

and lim

n

x→∞

x f (x) = 0

. Show that if ∞

C (y) = ∫

∞

f (x) cos xy dx\quad and\quadS(y) = ∫

a

f (x) sin xy dx,

(1.8.17)

a

then ∞

C

(k)

∞ k

(y) = ∫

x f (x) cos xy dx\quad and\quadS

(k)

(y) = ∫

a

0 ≤k ≤n

k

x f (x) sin xy dx,

(1.8.18)

a

.

[exer:24] Differentiating ∞

F (y) = ∫

y cos

dx

(1.8.19)

x

1

under the integral sign yields ∞

1

−∫

y sin

1

x

dx,

(1.8.20)

x

which converges uniformly on any finite interval. (Why?) Does this imply that F is differentiable for all y ? [exer:25] Show that Theorem [theorem:11] and induction imply Eq. [eq:30]. [exer:26] Prove Theorem [theorem:12]. ∞

[exer:27] Show that if F (s) = ∫ e f (x) dx converges for difference between this and Theorem [theorem:13]?) −sx

0

[exer:28] Prove: If f is continuous on [0, ∞) and ∫

∞

0

e

−s0 x

s = s0

f (x) dx

, then it converges uniformly on

. (What’s the

[ s0 , ∞)

converges, then

1.8.4

https://math.libretexts.org/@go/page/17329

∞

lim

∫

s→s0 +

∞

e

−sx

f (x) dx = ∫

0

e

−s0 x

f (x) dx.

(1.8.21)

0

(Hint: See the proof of Theorem 4.5.12, p. 273.) [exer:29] Under the assumptions of Exercise [exer:28], show that ∞

lim

∫

s→s0 +

∞

e

−sx

f (x) dx = ∫

r

e

−s0 x

f (x) dx,

r > 0.

(1.8.22)

r

[exer:30] Suppose f is continuous on [0, ∞) and ∞

F (s) = ∫

e

−sx

f (x) dx

(1.8.23)

0

converges for s = s . Show that lim 0

s→∞

F (s) = 0

. (Hint: Integrate by parts.)

[exer:31] ()0pt0pt14pt 8pt22pt0pt Starting from the result of Exercise [exer:18](d), let b → ∞ and invoke Exercise [exer:30] to evaluate ∞

∫

e

−ax

sin x dx,

a > 0.

(1.8.24)

x

0

Use (a) and Exercise [exer:28] to show that ∞

sin x

π

∫

dx =

.

(1.8.25)

0 ≤ x ≤ ∞.

(1.8.26)

x

0

2

[exer:32] ()0pt0pt14pt 8pt22pt0pt Suppose f is continuously differentiable on [0, ∞) and |f (x)| ≤ M e

s0 x

,

Show that ∞

G(s) = ∫

e

−sx

′

f (x) dx

(1.8.27)

0

converges uniformly on [s

1,

∞)

if s

1

> s0

. (Hint: Integrate by parts.)

Show from part (a) that ∞

G(s) = ∫

e

−sx

2

xe

x

2

sin e

x

dx

(1.8.28)

0

converges uniformly on [ρ, ∞) if ρ > 0 . (Notice that this does not follow from Theorem [theorem:6] or [theorem:8].) [exer:33] Suppose f is continuous on [0, ∞), f (x) lim

(1.8.29) x

x→0+

exists, and ∞

F (s) = ∫

e

−sx

f (x) dx

(1.8.30)

0

converges for s = s . Show that 0

∞

∫ s0

∞

F (u) du = ∫ 0

1.8.5

e

−s0 x

f (x) dx.

(1.8.31)

x

https://math.libretexts.org/@go/page/17329

Answers to selected exercises If f (x, y) = 1/y for y ≠ 0 and f (x, 0) = 1, then ∫

b

a

f (x, y) dx

does not converge uniformly on [0, d] for any d > 0 .

, (d), and (e) converge uniformly on (−∞, ρ] ∪ [ρ, ∞) if ρ > 0 ; (b), (c), and (f) converge uniformly on [ρ, ∞) if ρ > 0 . ∞

Let C (y) = ∫ π

x

=

y y

2

a

;I

2

y2 + 1 1

(f) F (y) = y

2

n

2

b

2

a

. Then C (0) = ∞ and S(0) = 0 , while S(y) = π/2 if y ≠ 0 .

;I

x a+1 = log b +1

+1

2

+1

;I

= tan

;I

=

;I

= tan

−1

−1

b − tan

a

+1 y

(e) F (y) =

y +1

1 =

1 y

F (y) = b

+1

(d) F (y) =

1

log 2

sin xy dx

1

π

;I

2|y|

F (y) =

and S(y) = ∫

dx

1

F (y) =

∞

cos xy

1

2

log(1 + a ) 2

−1

a

+1

−n−1

(−1 ) n!(y + 1 )

−2n−1

π2

(

2n −n−1/2 )y n

n!

(c)

2y

−2

n+1

(log y)

1

(d)

2

(log x)

∞ n

∫

| x f (x)| dx < ∞

−∞

No; the integral defining F diverges for all y . π

−1

− tan

a

2

Beginning of manual v

1. If H (y, u, v) = ∫

f (x, y) dx

then

u

Hu (y, u, v) = −f (u, y),

Hv (y, u, v) = f (v, y),

(1.8.32)

v

and, by Theorem 1, H

y (u,

. If

v, y) = ∫

fy (x, y) dx

u h(y)

F (y) = H (y, g(y), h(y)) = ∫

f (x, y) dx,

(1.8.33)

g(y)

then ′

F (y) =

′

′

Hv (y, g(y), h(y))h (y) + Hu (y, g(y), h(y))g (y) + Hy (y, g(y), h(y)) h(y)

=

′

′

f (h(y), y)h (y) − f (g(y), y)g (y) + ∫

fy (x, y) dx.

g(y)

2. Theorem 3 (Cauchy Criterion for Convergence of an Improper Integral II) Suppose subinterval of (a, b] and denote

g

is integrable on every finite closed

b

G(r) = ∫

g(x) dx,

a < r ≤ b.

(1.8.34)

r

Then the improper integral ∫

b

a

g(x) dx

converges if and only if, for each ϵ > 0, there is an r

0

|G(r) − G(r1 )| ≤ ϵ,

1.8.6

a < r, r1 ≤ r0 .

∈ (a, b]

such that (A)

https://math.libretexts.org/@go/page/17329

For necessity, suppose ∫

b

a

g(x) dx = L

. By definition, this means that for each ϵ > 0 there is an r

0

ϵ |G(r) − L| < 2

∈ (a, b]

such that

ϵ \quad and\quad|G(r1 ) − L|
0, there is an r

0

r

∣ ∣∫ ∣

Suppose ∫

b

a

f (x, y) dx

(1.8.41)

such that

∈ (a, b]

∣ f (x, y) dx ∣ < ϵ,

r1

f (x, y) dx

r

y ∈ S,

a < r, r1 ≤ r0 .

(A)

∣

converges uniformly on S and ϵ > 0 . From Definition 2, there is an r

∣ ∣∫ ∣ a

0

r

∣ ϵ ∣ f (x, y) dx ∣ < \quad and\quad ∣∫ ∣ 2 ∣ a

r1

∈ (a, b]

such that

∣ ϵ f (x, y) dx ∣ < , y ∈ S, a < r, r1 ≤ r0 . ∣ 2

(B)

Since r

∫ r1

b

f (x, y) dx = ∫

b

f (x, y) dx − ∫

r1

f (x, y) dx

(1.8.42)

r

(B) and the triangle inequality imply (A). For the converse, denote b

F (y) = ∫

f (x, y) dx.

(1.8.43)

r

1.8.7

https://math.libretexts.org/@go/page/17329

Since (A) implies that |F (r, y) − F (r1 , y)| ≤ ϵ,

y ∈ S,

Theorem 2 with G(r) = F (r, y) (y fixed but arbitrary in S ) implies that if ϵ > 0 then, for each y ∈ S , there is an r (y) ∈ (a, b] such that

a < r, r1 ≤ r0 ,

∫

b

f (x, y) dx

a

(1.8.44)

converges pointwise for y ∈ S . Therefore,

0

r

∣ ∣∫ ∣ a

For each y ∈ S , choose r

1 (y)

∣ f (x, y) dx ∣ ≤ ϵ, ∣

≤ min[ r0 (y), r0 ]

y ∈ S,

a < r ≤ r0 (y).

. Then

r

r1 (y)

∫

(C)

r

f (x, y) dx = ∫

a

f (x, y) dx + ∫

a

f (x, y) dx,

(1.8.45)

r1 (y)

so (A), (C), and the triangle inequality imply that ∣ ∣∫ ∣ a

r

∣ f (x, y) dx ∣ ≤ 2ϵ, ∣

y ∈ S,

a < r ≤ r0

b

(1.8.46)

c

b

4. From Definition 3, ∫ f (x, y) dx converges uniformly on S if and only if ∫ f (x, y) dx and ∫ f (x, y) dx both converge uniformly on S , where c ∈ (a, b) . From Theorems 4 and Theorem 5, this is true if and only if, for any ϵ > 0 there are points r and s in (a, b) such that a

a

c

0

0

r1

∣ ∣∫ ∣ r

∣ f (x, y) dx ∣ ≤ ϵ, ∣

y ∈ S,

r0 ≤ r, r1 < b

(1.8.47)

a < s, s1 < s0 .

(1.8.48)

and s

∣ ∣∫ ∣

∣ f (x, y) dx ∣ ≤ ϵ,

y ∈ S,

∣

s1

These conditions are independent of c . 5. (a) If |f (x, y)| ≤ M on [a, b] × [c, d] then r2

∣ ∣∫ ∣

∣ f (x, y) dx ∣ ≤ M | r2 − r1 |

(1.8.49)

∣

r1

so the Cauchy convergence theorems imply the conclusion. Define f

= f (x, y)

on [0, 1] × [0, 1] by ⎧ 1 f (x, y) = ⎨ y ⎩ 1

if\quad0 < y ≤ 1, (1.8.50) if\quady = 0.

Then

r1

r2 − r1

⎧

r2

∫

f (x, y) dx = ⎨ ⎩

if\quad0 < y ≤ 1,

y

(1.8.51)

r2 − r1

if\quady = 0.

Therefore f does not satisfy the requirements of Cauchy’s convergence theorems. 6. In all parts I (y) denotes the given integral. I (0) = ∞ ∞

1 ∫ |y|

r

. If

y ≠0

du 2

0

and

ϵ>0

, choose

r

so that

∫ r

du 2

1 +u

< ρϵ

. Then

if |y| ≥ ρ, so I (y) converges uniformly on (−∞, ρ] ⋃[ρ, ∞) if ρ > 0 .

1 +u

1.8.8

https://math.libretexts.org/@go/page/17329

if

I (y) = ∞

y ≤0

. If

let

y >0

I (y) = y

∞

∫

e

−u

2

u

3

du < ρ ϵ

y

r

∫

3

e

−u

2

u

. If

du

and

ρ >0

ϵ>0

, choose

so that

r

0

∞

1

. Then

∞

1

; then

u = xy

∫

3

e

−u

2

u

if y ≥ ρ , so I (y) converges uniformly on [ρ, ∞) if ρ > 0 .

du < ϵ

r ∞

I (y) = ∞

if y ≤ 0 . If

let

y >0

u = xy

1/2

; then

∞

−n−1/2

u

e

−u

n+1/2

du < ϵρ

. Then y

−n−1/2

u

e

−u

du

. If

ρ >0

and

ϵ>0

, we can choose

so that

r

0 2n

∫

r

2n

∫

∞ 2n

∫

I (y) = y

u

e

−u

du < ϵ

if y ≥ ρ , so I (y) converges uniformly on S = [ρ, ∞) if ρ > 0 .

r

Since

I (−y) = −I (y)

, it suffices to assume that

162), this integral converges conditionally. If

. If

y >0

and

ρ >0

ϵ>0

∞

1

then

2

u = yx

I (y) =

sin u du

∫ 2 √y

, we can choose

r

− √u

0

so that

∞

∣

sin u du ∣ − − ∣ < 2ϵ√ρ − ∣ √u

, so

∣∫ ∣

. From Example 3.4.14 (p.

r

I (y)

converges uniformly on (−∞, −ρ] ⋃[ρ, ∞) if ρ > 0. ∞

If

then

2

u =y x

I (y) = 3 ∫

e

−u

y3

0 ρ

∫

∞

2 du −

∞

∫

ue

−u

du

. If

ρ >0

, we can choose

so that

r

0

3∫

e

−u

ϵ du < 2

r

and

3

ue

ρ ϵ

−u

du < 2

r

. Then ∞

∣ ∣3 ∫ ∣ r

e

−u

∞

3 du − y

∫

3

ue

−u

r

∣ du ∣ < ϵ\quad if\quad|y| ≥ ρ, ∣

(1.8.52)

so I (y) converges uniformly on (−∞, ρ] ⋃[ρ, ∞) if ρ > 0 . I (y) = −∞

if

y ≤0

. If

y >0

, let

; then

u = xy

∞ 2

|2u − u | e

−u

du < ϵρ

2

∫ y

∫

∞

1 I (y) =

(2u − u )e

−u

du

. If

ρ >0

and

ϵ>0

, we can choose

r

so that

0

, so I (y) converges uniformly on S = [ρ, ∞) if ρ > 0 .

r

7. Theorem 7 (Weierstrass’s Test for Absolute Uniform Convergence II) Suppose some b ∈ (a, b],

f = f (x, y)

is locally integrable

(a, b]

and, for

0

|f (x, y)| ≤ M (x), y ∈ S, x ∈ (a, b0 ],

(A)

where b0

∫

M (x) dx < ∞.

(1.8.53)

a

Then ∫

b

a

f (x, y) dx

Denote ∫

b0

a

converges absolutely uniformly on S.

M (x) dx = L < ∞

. By definition, for each ϵ > 0 there is an r

0

∈ (a, b0 ]

such that

b0

L−ϵ ≤ ∫

M (x) dx ≤ L,

a < r ≤ r0 .

(1.8.54)

r

Therefore, if a < r

1

< r ≤ r0

, then r

b0

0 ≤∫

b0

M (x) dx = ( ∫

r1

M (x) dx − L) − ( ∫

r1

M (x) dx − L) < ϵ.

(1.8.55)

r

This and (A) imply that r

∫

r

|f (x, y)| dx ≤ ∫

r1

M (x) dx < ϵ, y ∈ S, a < r1 ≤ r0 ≤ b.

(1.8.56)

r1

Now Theorem 5 implies the stated conclusion. |e

−xy

sin x| ≤ e

−ρx

if y ≥ ρ and ∫

∞

ρ

e

−ρx

dx < ∞

.

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∞

sin x dx

∫

, where

= I1 (y) + I2 (y)

y

x

0

1

∞

sin x dx

I1 (y) = ∫

y

x

0

sin x dx

\quad and\quadI2 (y) = ∫

(1.8.57)

y

x

1

are both improper integrals. Since 3

5

x sin x = x − (

7

x −

9

x

x

) −(

3!

−

5!

7!

) + ⋯ < x,

0 ≤ x ≤ 1.

(1.8.58)

9!

If 0 ≤ 1 and y ≤ d ≤ 2 , then 1

∣ sin x ∣ 1−y 1−d ∣ ∣ ≤x ≤x \:so\: ∫ ∣ xy ∣ 0

so I

1 (y)

x

1 dx =

(1.8.59) 2 −d

0

∞ −c

≤x

y

1

−c

\quad and\quad ∫

x

x

dx =

\quad if \quad

(1.8.60)

c −1

1

converges absolutely uniformly on S .

If x ≥ 1 then e e

−1+d

dx < ∫

converges absolutely uniformly on S . Since c > 1 , | sin x|

I2 (y)

1 −1+y

x

−px

xy

e y

∣ sin xy ∣ −px ∣ ∣ ≤e ∣ ∣ x

e

−px

dx < ∞

, since p > 0 .

1

b

1

≤

(1 − x)

∞

for all y and ∫

, if 0 ≤ x < 1 and y ≤ b , and ∫

b

(1 − x)

−b

(1 − x )

dx < ∞

if b < 1 .

0

∣

cos xy

If |y| ≥ ρ > 0 then ∣

∞

∣

1 ∣ ≤ ∣ 1 + ρ2 x2

∣ 1 + x2 y 2

dx

for all x, and ∫

1 + ρ2 x2

0

0 then e

−x/y

≤e

and ∫

−x/ρ

e

−x/ρ

dx < ∞

.

0 ∞

If |y| ≤ ρ then e

xy

2

e

−x

≤e

xρ

2

e

−x

and ∫

e

xρ

2

e

−x

dx < ∞

.

−∞ ∞

2 dx

If |x| ≥ 1 then | cos xy − cos ax| ≤ 2 and ∫

2

0 then e

2

−yx

2

≤e

−ρx

and ∫

2n

x

2

e

−ρx

dx < ∞

.

0 1

9. (a) If 0 < x < 1 then |x

y−1

e

−x

c−1

| 0 , ∫

1

0

y−1

x

e

−x

dx

converges uniformly on

0

if

[c, ∞)

c >0

, by Theorem 7. If

x >1

then

y−1

|x

e

−x

∞ d−1

| ≤x

e

−x

if

y ≤d

. Therefore, since

∞

∫ 1

[c, d]

e

−x

dx

converges uniformly on (−∞, d] for every d , by Theorem 6. Hence, ∫

d−1

x

e

−x

dx < ∞

,

1

∞ y−1

x

∫

y−1

x

e

−x

dx

converges uniformly on

0

if c > 0 .

If y > 0 then ∞

Γ(y) = ∫

y

y−1

x

e

−x

x e

−x

dx = y

0

∞

∣ ∣ ∣

∞

1 +

0

∫ y

y

x e

0

−x

Γ(y + 1) dx =

.

(A)

y

Therefore Γ(y + n) Γ(y) =

(B) y(y + 1) ⋯ (y + n − 1)

is true when n = 1 . Now suppose it is true for given positive integer n , and replace y by y + n in (A):

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Γ(y + n + 1) Γ(y + n) =

.

(1.8.61)

y +n

Substituting this into (B) yields Γ(y + n + 1) Γ(y) =

,

(1.8.62)

y(y + 1) ⋯ (y + n)

which completes the induction. If −n < y < −n + 1 with n ≤ 1 then 0 < y + n < 1 and we can compute Γ(y + n) from the definition in part (a): ∞ y+n−1

Γ(y + n) = ∫

x

e

−x

dx.

(1.8.63)

0

Then we can define Γ(y) by (B). The assertion is true if n = 1 , since ∞

Γ(2) = ∫

∞

∞

xe

−x

dx = −x e

−x ∣

∣ ∣

+∫

0

0

e

−x

dx = 1.

(1.8.64)

0

If Γ(n + 1) = n! for some n ≥ 1 , then ∞

Γ(n + 2) =

∞

∞

n+1

∫

x

e

−x

dx = −e

−x

n+1 ∣

x

0

=

∣ ∣0

+ (n + 1) ∫

n+1

x

e

−x

dx

0

(n + 1)Γ(n + 1) = (n + 1)n! = (n + 1)!,

which completes the induction proof. The change of variable x = st yields ∞

∫

e

−st

α

t

∞

1 dt =

α+1

s

0

α

∫

x

e

−x

1 dx =

α+1

Γ(α + 1),

(1.8.65)

s

0

from the definition of the Gamma function. Since |g

x (x,

y)|

is monotonic with respect to x, r1

∫

| gx (x, y)| dx = |g(r1 , y) − g(r, y)|,

a ≤ r < r1 < b.

(A)

r

From Assumption (a) of Theorem 8, if ϵ > 0 there is an r

0

∈ [a, b)

|g(s, y)| ≤ ϵ,

such that

y ∈ S,

r0 ≤ s < b.

(1.8.66)

Therefore, (A) implies that r1

∫

| gx (x, y) dx ≤ 2ϵ,

y ∈ S,

r0 ≤ r ≤ r1 < b.

(1.8.67)

r

Now Theorem 4 implies that ∫

b

a

) If g,

gx ,

|g(x, y)| dx

converges uniformly on S , which is assumption (c) of Theorem 8.

and h are continuous on (a, b] × S then b

∫

g(x, y)h(x, y) dx

(1.8.68)

a

converges uniformly on S if the following conditions are satisfied: ()0pt0pt14pt 8pt22pt0pt lim { sup |g(x, y)|} = 0; x→a+

y∈S

There is a constant M such that

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b

∣ sup ∣∫ ∣

y∈S

∫

b

a

| gx (x, y)| dx

∣ h(u, y) du ∣ ≤ M ,

a < x ≤ b;

(1.8.69)

∣

x

converges uniformly on S .

If b

H (x, y) = ∫

h(u, y) du

(A)

x

then integration by parts yields r

r

∫

g(x, y)h(x, y) dx =

−∫

r1

g(x, y)Hx (x, y) dx

r1

=

−g(r, y)H (r, y) + g(r1 , y)H (r1 , y) r

+∫

gx (x, y)H (x, y) dx.

r1

Therefore, since assumption (b) and (A) imply that |H (x, y)| ≤ M , (x, y) ∈ (a, b] × S , r

∣ ∣

r

∣

∣∫

g(x, y)h(x, y) dx ∣ ≤ M (2 ∣

r1

sup

|g(x, y)| + ∫

a 0 . From assumption (a), there is an From assumption (c) and Theorem 5, there is an s ∈ (a, b] such that

r0 ∈ [a, b)

1

such that

|g(x, y)| < ϵ

on

S

if

a < x ≤ r0 ≤ b

.

0

r

∫

| gx (x, y)| dx < ϵ,

y ∈ S,

a < r1 < r ≤ s0 .

(1.8.70)

y ∈ S,

a < r1 < r min(r0 , s0 )

(1.8.71)

r1

Therefore (B) implies that r

∣ ∣∫ ∣

∣ g(x, y)h(x, y)∣ < 3M ϵ, ∣

r1

Now Theorem 5 implies the stated conclusion. ∞

sin xy

Denote F (y) = ∫

y

x

1

and, with 1 ≤ r < r , 1

r1

dx =

y

−

x

r

r1

r1

sin xy

F (r, r1 , y) = ∫

cos xy ∣ ∣ y ∣ yx

r

cos ry =

y

−

cos r1 y y

yr

cos xy

−∫

y+1

r1

cos xy

−∫

yr

r

1

dx

x

r

y+1

dx.

x

Therefore r1

2 |F (r, r1 , y)| ≤

+∫

y

yr

−y−1

x

3 dx
0.

(1.8.72)

yr

r

Now Theorem 4 implies that F (y) converges uniformly on [ρ, ∞) if ρ > 0 . ∞

(b) Denote F (y) = ∫ 2

sin xy dx log x

and, with 2 ≤ r < r , 1

r1

r

r1

sin xy

F (r, r1 , y) = ∫

dx = − log x

cos xy ∣ ∣ y log x ∣

r

r1

1 −

cos xy

∫ y

r

2

dx.

(1.8.73)

x(log x)

Therefore |F (r, r1 , y)| ≤

1∣ 2 ∣ +∫ y ∣ log r r

r1

∣ dx 3 ∣ ≤ . 2 x(log x) ∣ y log r

(1.8.74)

Now Theorem 4 implies that F (y) converges uniformly on [ρ, ∞) if ρ > 0 .

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https://math.libretexts.org/@go/page/17329

∞

(c) Denote F (y) = ∫

cos xy x +y

0

, and, with 0 < r < r , 1

2

r1

F (r, r1 , y) = ∫ r

cos xy x + y2

r1

r

1 dx =

( y

sin xy ∣ 1 ∣ +∫ x + y 2 ∣r

sin xy (x + y 2 )2

r

dx) ,

(1.8.75)

so 3 |F (r, r1 , y)| ≤

.

2

(1.8.76)

y(r + y )

Now Theorem 4 implies that F (y) converges uniformly on [ρ, ∞) if ρ > 0 . ∞

(d) Denote F (y) = ∫

sin xy 1 + xy

0

, and, with 0 < r < r , 1

r1

sin xy

F (r, r1 , y) = ∫ r

r1

r1

cos xy

∣ ∣ y(1 + xy) ∣r

dx = − 1 + xy

−∫ r

cos xy 2

2

dx,

(1.8.77)

y (1 + xy )

so 3 |F (r, r1 , y)| ≤

.

(1.8.78)

y(1 + ry)

Now Theorem 4 implies that F (y) converges uniformly on [ρ, ∞) if ρ > 0 . 13. Integration by parts yields r1

∫

r1

g(x, y)h(x, y) dx =

∫

r

g(x, y)Hx (x, y) dx

r

=

g(r1 , y)H (r1 , y) − g(r, y)H (r, y) r1

−∫

gx (x, y)H (x, y) dx,

r

so ∣ ∣∫ ∣ r

r1

∣ ∣ g(x, y)h(x, y) dx ∣ ≤ 2 sup {{ sup |g(x, y)H (x, y)|}} + ∣∫ ∣ ∣ r x≥r y∈S

Now suppose ϵ ≥ 0 . From our first assmption, there is an s

0

∈ [a, b)

Since ∫

b

a

gx (x, y)H (x, y) dx

∣ gx (x, y)H (x, y) dx ∣ . ∣

s0 ≤ r < b.

(1.8.80)

y∈S

converges uniformly on S , Theorem 4 implies that there is an r

0

∣ ∣∫ ∣ r

r1

(1.8.79)

such that

sup {{ sup |g(x, y)H (x, y)|}} < ϵ, x≥r

r1

∣ gx (x, y)H (x, y) dx ∣ ≤ ϵ, ∣

y ∈ S,

∈ [a, b)

such that

r0 ≤ r < r1 < b.

(1.8.81)

max(r0 , s0 ) ≤ r < r1 < b.

(1.8.82)

Therefore, ∣ ∣∫ ∣

r

Now Theorem 4 implies that ∫

b

a

14. Theorem 10 If f

= f (x, y)

r1

∣ g(x, y)h(x, y) dx ∣ ≤ 2ϵ, ∣

gx (x, y)h(x, y) dx

y ∈ S,

converges uniformly on S .

is continuous on (a, b] × [c, d] and b

F (y) = ∫

f (x, y) dx

(A)

a

converges uniformly on [c, d], then F is continuous on [c, d]. Moreover,

1.8.13

https://math.libretexts.org/@go/page/17329

d

b

∫

b

(∫

c

d

f (x, y) dx) dy = ∫

a

(∫

a

f (x, y) dy) dx.

(B)

c

We will first show that F in (A) is continuous on [c, d]. Since F converges uniformly on [c, d], Definition 1 implies that if there is an r ∈ (a, b] such that ∣ ∣∫ ∣ a

r

∣ f (x, y) dx ∣ ≤ ϵ, ∣

c ≤ y ≤ d.

ϵ>0

,

(1.8.83)

Therefore, if y and y are in [c, d], then 0

b

∣ |F (y) − F (y0 )| =

∣∫ ∣ b

∣ ≤

∣∫ ∣

r

b

f (x, y) dx − ∫

a

∣ f (x, y0 ) dx ∣ ∣

a

∣

∣ [f (x, y) − f (x, y0 )] dx ∣ + ∣∫ ∣ a ∣ ∣ + ∣∫ ∣ a

r

r

∣ f (x, y) dx ∣ ∣

∣ f (x, y0 ) dx ∣ ∣

so b

|F (y) − F (y0 )| ≤ ∫

|f (x, y) − f (x, y0 )| dx + 2ϵ.

(C)

r

Since f is uniformly continuous on the compact set [r, b] × [c, d] (Corollary 5.2.14, p. 314), there is a δ > 0 such that |f (x, y) − f (x, y0 )| < ϵ

if (x, y) and (x, y

0)

are in [r, b] × [c, d] and |y − y

0|

(1.8.84)

. This and (C) imply that

0 . Since we have assumed that lim

r→a+

Fr (y0 ) = F (y0 )

| Fr (y0 ) − F (y0 )| < ϵ,

1

b

∣∫ ∣

y0

r

∫

∣ fy (x, t) dx ∣ dt.

(B)

∣

a

exists, there is an r in (a, b) such that 0

a < r < r0 .

Since we have assumed that G(y) converges for y ∈ [c, d], there is an r ∣

y

∣

G(t) dt∣ ≤ | Fr (y0 ) − F (y0 )| + ∣∫

∈ (a, b]

(1.8.96)

such that

∣ fy (x, t) dx ∣ < ϵ,

t ∈ [c, d],

a < r ≤ r1 .

(1.8.97)

∣

a

Therefore, (B) yields y

∣ ∣Fr (y) − F (y0 ) − ∫ ∣

if a < r < min(r

0,

r1 )

∣ G(t) dt∣ < ϵ(1 + |y − y0 |) ≤ ϵ(1 + d − c)

(1.8.98)

∣

y

0

and t ∈ [c, d] . Therefore F (y) converges uniformly on [c, d] and y

F (y) = F (y0 ) + ∫

G(t) dt,

c ≤ y ≤ d.

(1.8.99)

y0

1.8.15

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Since G is continuous on [c, d] by Theorem 10, (A) follows from differentiating this (Theorem 3.3.11, p. 141). 16. Since ∞

|f (x) cos xy| ≤ |f (x)|,

|f (x) sin xy| ≤ |f (x)|, \quad and\quad ∫

|f (x)| dx < ∞,

(1.8.100)

−∞ ∞

∞

−∞

−∞

Theorems 6 and 7 imply that ∫ f (x) cos xy dx and ∫ implies that C (y) and S(y) are continuous on (−∞, ∞).

converge uniformly on

f (x) sin xy dx

(−∞, ∞)

, so Theorem 10

If y ≠ 0 , integrating by parts yields ∞

C (y) =

sin xy ∣ ∣ ∣ y

f (x)

′

∫ y

a

sin ay =

∞

1 −

f (x) sin xy dx

a ∞

1

−f (a)

−

′

∫

y

y

f (x) sin xy dx

a

and ∞

S(y) =

cos xy ∣ ∣ ∣a y

−f (x)

+ y

f (x) cos xy dx

a ∞

1

f (a)

′

∫ y

cos ay =

∞

1 +

′

∫ y

f (x) cos xy dx,

a ∞

since lim f (x) = 0 . From Exercise [exer:17] with f replaced by f , ∫ f (x) cos xy dx and continuous on (−∞, ∞). Therefore C (y) and S(y) are continuous on (−∞, 0) ∪ (0, ∞) . ′

x→∞

′

1

∫

∞

1

′

f (x) cos xy dx

are

To see that C and S are not necessarily continuous at y = 0 , let a = 1 and f (x) = 1/x, so ∞

x→∞

∞

dx

′

lim f (x) = 0\quad and\quad ∫

| f (x)| = ∫

1

= 1.

2

(1.8.101)

x

1

Then r

r→∞

r

cos xy

C (y) = lim ∫

sin xy

dx\quad and\quadS(y) = lim ∫ x

1

r→∞

dx,

y ≠ 0.

(1.8.102)

x

1

If y > 0 make the change of variable u = xy to see that ry

∞

cos u

C (y) = lim ∫ r→∞

cos u

du = ∫ u

y

du

(1.8.103)

u

y

and ry

r→∞

∞

sin u

S(y) = lim ∫

sin u

du. S(y) = ∫ u

y

du.

(1.8.104)

u

y

∞

Therefore

limy→0+ C (y) = ∞

, so

C

is not continuous at

y =0

. Since

and

S(0) = 0

limy→0+ S(y) = ∫

not continuous at y = 0 .

sin u du ≠ 0

0

u

,

S

is

The integral diverges if y = 0 . If y ≠ 0 substitute u = |y|x to obtain ∞

dx

F (y) = ∫ 0

∞

1

2

1 +x y

2

=

du

∫ |y|

2

1 +u

0

∞

1

−1

=

tan |y|

∣ u∣ ∣

π =

,

(A)

2|y|

0

so F (y) converges for all y ≠ 0 . To test for uniform convergence, suppose |y| > 0 and 0 < r < r . Then 1

r1

∫ r

if

. If

|y| ≥ ρ

ϵ>0

there is an

α >0

dx 2

1 +x y

r1 |y|

1 2

such that

= |y|

∞

1 ∫ ρ

du

∫

α

2

1 +u

r|y|

∞

1
−1 . Now Theorem 11 implies that 1

a

x

log x

∞

e

−yx

a

dx = ∫

0

F (y) = ∫

1

b

−x

I =∫

y cos x dx = y

0

2

0

a y

dx ∫

x

1

dy = ∫

b

b

a y

dy ∫

x

dy

a+1

dx = ∫

0

= log y +1

b

.

(1.8.112)

b +1

. Since

+1 ∞

∣ ∣∫ ∣

e

−yx

r

∞

∣ cos x dx ∣ ≤ ∫ ∣

e

e

−xy

−yr

dx =

,

(1.8.113)

y

r

Theorem 4 (or Theorem 6) implies that F (y) converges uniformly on [ρ, ∞) if ρ > 0 . Therefore, Theorem implies that if a , b > 0 then ∞

I =

e

−ax

∫

0

e

−yx

1 sin x dx = y

2

b

e

−yx

b

cos x dx ∫

0

∞

dy ∫

a ∞

∞

x

b

=

−bx

cos x dx = ∫

0

F (y) = ∫

−e

∫

y

cos x dx = ∫

0

y

a

2

e

−yx

dy

a

1 dy =

+1

2

2

+1

2

+1

b log

a

.

. Since

+1 ∣ ∣∫ ∣ r

∞

e

−yx

∣ sin x dx ∣ ≤ ∫ ∣ r

∞

1.8.17

e

−yx

e

−yr

dx =

,

(1.8.114)

y

https://math.libretexts.org/@go/page/17329

Theorem 4 (or Theorem 6) implies that F (y) converges uniformly on every [ρ, ∞) if ρ > 0 . Therefore, if a , b > 0 then Theorem 11 implies that ∞

I =

e

−ax

−e

sin x dx = ∫ x

0 b

=

∫

∞

e

−x

y

0

2

−yx

y

a

−1

dy = tan

2

e

−yx

dy

a

1

sin x dx = ∫

0

y sin xy dx =

b

e

b

sin x dx ∫

0

∞

dy ∫

a

(e) F (y) = ∫

∞

−bx

∫

−1

b − tan

a.

+1

. Since

+1 ∣ ∣∫ ∣ r

∞

e

∞

∣ sin xy dx ∣ ≤ ∫ ∣ r

−x

e

−x

dx = e

−r

,

(1.8.115)

Theorem 4 (or Theorem 6) implies that F (y) converges uniformly on [ρ, ∞) if ρ > 0 . Therefore Theorem 11 implies that ∞

I =

∫

e

−x

a

x

0 ∞

F (y) = ∫

e

−x

1 cos xy dx = y

0

2

a

dy ∫

e

−x

e

−x

dx ∫

0

∞

∫

a

dx = ∫

0

=

∞

1 − cos ax

y

sin xy dx = ∫

0

1 dy =

y2 + 1

0

sin xy dy

0 2

log(1 + a ). 2

. Since

+1 ∣ ∣∫ ∣ r

∞

e

−x

∣ cos xy dx ∣ ≤ ∫ ∣ r

∞

e

−x

dx = e

−r

,

(1.8.116)

Theorem 4 (or Theorem 6) implies that F (y) converges uniformly on [ρ, ∞) if ρ > 0 . Therefore Theorem 11 implies that ∞

I =

∫

∞

e

−x

0 a

=

∫

e

−x

dx ∫

0

∞

dy ∫

0

a

sin ax dx = ∫ a

e

−x

1

cos xy dx = ∫

0

0

cos xy dy

0

y

2

−1

dy = tan

a.

+1

19. (a) We start with 1 y

F (y) = ∫

x

1 dx =

y > −1.

(A)

y +1

0

Formally differentiating this yields 1

F

(n)

n

n

(y) = ∫

y

(log x ) x

(−1 ) n! dx =

n+1

,

y > −1.

(B)

(y + 1)

0

To justify this we will show by induction that the improper integrals 1

In (y) = ∫

n

y

(log x ) x

dx,

n = 0, 1, 2, …

(1.8.117)

0

converge uniformly on [ρ, ∞) if ρ > −1 . We begin with n = 0 :. r

y+1

0

y+1

[ρ, ∞)

if ρ > −1 . Now suppose that

0

so I (y) = F (y) converges uniformly on by parts yields

y+1

r r ∣ ∣ = ≤ , y + 1 ∣r1 y +1 ρ+1

x

x

r

dx =

y

∫

y+1

r

∫

n+1

(log x )

y

x

r

−1 < ρ ≤ y.

In (y)

n+1

(log r)

converges uniformly on y+1

−r

1

dx =

(1.8.118)

. Integrating

[ρ, ∞)

n+1

(log r1 )

y +1

r1 r

n+1 −

∫ y +1

n

y

(log x ) x

dx,

−1 < y < ∞.

r1

1.8.18

https://math.libretexts.org/@go/page/17329

Letting r

1

→ 0

yields r

y+1

n+1

∫

r

y

(log x )

x

n+1

(log r)

r

n+1

dx =

− y +1

0

n

∫ y +1

y

(log x ) x

dx.

(C)

0

Since the integral on the right converges, it follows that the integral on the left converges; in fact 1 n+1

∫

(log x )

1

n+1

y

x

dx = −

n

∫ y +1

0

y

(log x ) x

dx.

(1.8.119)

0

We must still show that the integral on the left converges uniformly on [ρ, ∞) if ρ > −1 . To this end, note from (C) that ∣ ∣∫ ∣ 0

r n+1

(log x )

if y ≥ ρ , Now suppose ϵ > 0 . Since

∣ rρ+1 (log r)n+1 ∣ ∣ n+1 ∣ dx ∣ ≤ ∣ ∣+ ∣∫ ∣ ρ+1 ρ+1 ∣ 0 ∣ ∣

y

x

ρ+1

n+1

lim r

(log r)

=0

, there is an r

n

y

(log x ) x

∣ dx ∣ ∣

(D)

such that

∈ (0, 1)

1

r

r→0+

ρ+1

∣r

n+1

(log r)

∣

r

n

y

(log x ) x

0

dx

ϵ

∣ ≤ ρ+1

∣

Since ∫

∣

2

∣

\quad if \quad0 < r < r1 .

is uniformly convergent (by our induction assumption), there is r

∈ (0, 1)

2

n+1 ∣ ∣∫ ρ+1 ∣ 0

r n

y

(log x ) x

ϵ ∣ dx ∣ ≤ , ∣ 2

(1.8.120)

such that

y ≥ ρ,

(1.8.121)

Now (D) implies that r

∣ ∣∫ ∣ 0

n+1

(log x )

∣ dx ∣ ϵ, ∣

y

x

y ∈ [ρ, ∞),

0 < r < min(r1 , r2 ).

(1.8.122)

This, Theorem 11, and an easy induction argument imply (B). (b) Substituting x = u √y yields ∞

dx

F (y) = ∫

2

x

0

∞

1 =

du

∫ √y

+y

2

u

0

π =

,

y > 0.

(A)

2 √y

+1

Formally differentiating this yields yields ∞

dx

∫ 0

2

π =

n+1

(x

1 ⋅ 3 ⋯ (2n − 1)y

−n−1/2

2n + 1

+ y)

(2n)!

π =

y

2n+1

π =

2n+1

(

−n−1/2

n!

2

2n −n−1/2 )y ,

y > 0.

n

2

Theorem 11 implies that the formal differentiation is legitimate, since, if y ≥ 0 and r > 0 , then ∞

2

(x

r ∞

dx

hence, the improper integrals ∫

2

(x

0

∞

dx

∫

n+1

+ y)

−2n−1

−2n−2

≤∫

x

r dx =

;

(1.8.123)

(2n − 1)

r

, n = 0 , 1, 2, … converge uniformly on [0, ∞).

n+1

+ y)

∞

Denote I

n (y)

=∫

2n+1

x

2

e

−yx

dx

. Then

0 ∞

xe

−yx

∞

1

2

I0 (y) = ∫

= 2

0

2x e

−yx

dx = −

2

e 2y

0

∞

1

2

∫

−yx

∣ ∣ ∣

0

1 =

.

(1.8.124)

2y

Since ∞

∫ r

∞ 2n+1

x

2

e

−yx

dx ≤ ∫

2n+1

x

2

e

−ρx

dx\quad if\quad0 < ρ ≤ r,

(1.8.125)

r

1.8.19

https://math.libretexts.org/@go/page/17329

if n ≥ 0 , we can differentiate I formally with respect to y ∈ (0, ∞) to obtain n

(n)

n

In (y) = (−1 ) I

0

n! = 2y

n+1

.

(1.8.126)

(d) Denote ∞

I (y) =

∞

∫

y

x

dx = ∫

0

e

∞

1

x log y

dx =

∫ log y

0

y

x

∞

∣ ∣ log y ∣0

=

(log y)y

x

dx

0

1 =−

0 < y < 1. log y

∞

Formally differentiating this yields

′

I (y) = ∫

x−1

. There are two improper integrals here:

dx

J1 (y) = ∫

0

∞

J2 (y) = ∫

1

xy

xy

x−1

dx

xy

x−1

dx

and

0

. If r < 1 then

1 r

∫

xy

x−1

1 (y)

∫ y

0

Therefore J

r

1 dx =

converges uniformly on [ρ

xy

x

y

2

r

dx =

2

(1.8.127)

∞

1

x−1

xρ

0 < ρ1 ≤ y ≤ 1.

2ρ1

then

r > 1 and ρ

xy

2

r x dx =

1]

∞

∫

∫

0

1,

r

1 dx ≤

x

∫ ρ2

r

xρ

2

dx,

(1.8.128)

r

Since ∞

1 lim x→∞

x

∫ ρ2

xρ

dx = 0

2

(1.8.129)

r ∞

Theorem 7 implies that uniformly on [ρ

1,

J2 (y)

converges uniformly on

. Therefore, if

[0, ρ2 ]

then

0 < ρ1 < ρ2 < 1

∫

xy

x−1

converges

0

. Now Theorem 11 implies that

ρ2 ]

∞ ′

I (y) = ∫

xy

x−1

dx,

0 < y < 1.

(A)

0

However, since I (y) = −

1 log y

, we know that I

′

∞

1 (y) =

2

y(log y)

. This and (A) imply that ∫

xy

x

1 dx =

2

(log x)

0

.

∞

20. Here F (y) = ∫

2

e

−x

, so Theorem 11 implies that

cos 2xy dx

0 ∞ 2

′

F (y) = −2 ∫

xe

−x

sin 2xy dx,

(A)

0

since the integral on the right converges uniformly on (−∞, ∞), by Theorem 6. Integration by parts yields ∞

1 F (y) =

= 2y 1

=

2

(e

−x

2y

2

∫

e ∞

2

+2 ∫

0

−x

sin 2xy dx)

0

1

2

∫ y

xe

∞

1 =

(2y cos 2xy) dx

0

∞

∣ sin 2xy dx ∣ ∣

−x

xe

−x

sin 2xy dx = −

′

F (y). 2y

0

′

From this and (A),

− √π

∞ 2

F (0) = ∫ 0

e

−x

′

F (y) + 2yF (y) = 0

dx = 2

, so

F (y) = −2y F (y) − √π 2

(Example 12), F (y) =

e 2

−y

,

log F (y) = −y

2

+ log F (0)

, and

F (y) = F (0)e

−y

2

. Since

.

1.8.20

https://math.libretexts.org/@go/page/17329

∞ 2

Here F (y) = ∫

e

−x

sin 2xy dx

, so Theorem 11 implies that

0 ∞ 2

′

F (y) = 2 ∫

xe

−x

cos 2xy dx,

(A)

0

since the integral on the right converges uniformly on (−∞, ∞). Integrating this by parts yields ∞

∞

2

′

F (y) =

−e

−x

∣ cos 2xy ∣ ∣

2

− 2y ∫

0

e

−x

=

so

′

F (y) + 2yF (y) = 1

,

e

2

y

′

F (y) + 2 e

sin 2xy dx

0

1 − 2yF (y),

y

2

yF (y) = e

y

2

,

and

(e

y

′

2

F (y)) = e

y

2

.

Therefore,

since

F (0) = 0

,

y

F (y) = e

−y

2

2

∫

e

u

du

.

0 ∞

22. Theorems 6 and 11 imply that S and C are n times continuously differentiable on (−∞, ∞) if ∫

.

n

| x f (x)| dx < ∞

−∞

23. We will show first that ∞

∞ k

Ck (y) = ∫

k

x f (x) cos xy dx\: and\: Sk (y) = ∫

a

x f (x) sin xy dx, 0 ≤ k ≤ n,

converge uniformly on U = (−∞, −ρ] ∪ [ρ, ∞) if ρ > 0 . Note that if lim 2 ,…n . If 0 ≤ k ≤ n , then ρ

n

x f (x) = 0

x→∞

r1

∣ k [x f (x) sin xy ∣ ∣ y r

x f (x) cos xy dx =

r

, then lim

x→∞

k

x f (x) = 0

, k = 0 , 1,

r1

r1

1

k

∫

(1.8.130)

a

k

−∫

′

(x f (x)) sin xy dx] .

(A)

r

Henceforth k is fixed. Our assumptions imply that if ρ > 0 and ϵ > 0 then there is an r

0

∈ [a, ∞)

such that

∞ k

∫

′

k

|(x f (x)) | dx < ρϵ\quad and \quad| x f (x)| < ρϵ,

x ≥ r0 .

(1.8.131)

r0

Therefore (A) implies that ∣ ∣∫ ∣ r

r1

∣ k x f (x) cos xy dx ∣ < 3ϵ, ∣

r0 ≤ r < r1 , y ∈ (−∞, −ρ] ∪ [ρ, ∞).

Now Theorem 4 implies that C , C ,…, C converge uniformly on (−∞, −ρ] ∪ [ρ, ∞). Since every Theorem 11 now implies that that if y ≠ 0 then 0

1

k

(1.8.132)

y ≠0

is in such an interval,

∞

C

(k)

k

(y) = ∫

x f (x) sin xy dx,

0 ≤ k ≤ n.

(1.8.133)

a

A similar argument applies to S , S ,…S ′

r1

Let I (y; r, r

1)

1

=∫

y sin

r

x

dx x

(n)

.

. Assume for the moment that y ≥ 0 . Substituting u = y/x yields y/r

I (y; r, r1 ) = ∫ y/r1

Therefore, since

∣ sin u ∣ ∣ ∣ ≤1 ∣ ∣ u

for all u,

u (

y/r

y ) sin u (−

y

2

u

,

|I (y; r, r1 )| ≤ y/r 1 ≤ r ≤ r1

,

. Therefore, Theorem 4 implies that ∫ 1

r

Now denote

Fr (y) = ∫ 1

Since F

r (−y)

= Fr (y)

dx x

1

y sin

x

dx x y

y cos

. substituting

u = y/x

yields

Fr (y) = y ∫ y/r

, it follows that lim

r→∞

Fr (y) = ∞

du.

, we can write

I (−y; r, r1 ) = −I (y; r, r1 )

converges uniformly on every finite interval.

cos u 2

(1.8.134)

u

y/r1

. In fact, since ∞

|I (y; r, r1 ) ≤ |y|/r 1 ≤ r ≤ r1

sin u

) du = ∫

du

, so

limr→∞ Fr (y) = ∞

for all

y ≥0

.

u

for all y , so the answer to the question is “no.”

1.8.21

https://math.libretexts.org/@go/page/17329

Let P be the induction assumption n

∞

F

(n)

n

(s) = (−1 )

∫

e

−sx

n

x f (x) dx,

s > s0 ,

(1.8.135)

0

which is true by the definition of F for n = 0 . If P is true, then Theorems 11 and 13 imply that n

F

(n+1)

∞

d

n

(s) =

(−1 )

∫ ds

∞

e

−sx

n

n

x f (x) dx = (−1 )

d

∫

0

(e

−sx

n

x f (x)) dx

ds

0

∞ n+1

=

(−1 )

∫

e

−sx

n+1

x

f (x) dx,

0

so P implies P n

n+1

, which completes the induction proof.

x

Let G(x) = ∫

e

−s0 t

f (t) dt

. If s > s then 0

0 ∞

∞

F (s) = ∫

e

−sx

f (x) dx = ∫

0

∞

e

−(s−s0 )x

′

G (x) dx = (s − s0 ) ∫

0

e

−(s−s0 )x

G(x) dx

(A)

0

∞

(integration by parts). Since (s − s

0) ∫

e

−(s−s0 )x

dx = 1

, (A) implies that

0 ∞

F (s) − F (s0 ) = (s − s0 ) ∫

e

−(s−s0 )x

(G(x) − F (s0 )) dx.

(B)

0 ∞

Now suppose ϵ > 0 . Since F (s

0)

=∫

e

−s0 t

f (t) dt = lim G(x)

from (B), then

, there is an

such that

r

t→∞

0

|G(x) − F (s0 )| < ϵ

r

|F (s) − F (s0 )| ≤

(s − s0 ) ∫

if

x ≥r

; hence,

∞

e

−(s−s0 )x

|G(x) − F (s0 )| + ϵ(s − s0 ) ∫

0

e

−(s−s0 )x

dx

r r


s0

then |e

−sx

f (x)| = | e

−(s−s0 )x

e

s0 x

f (x)| ≤ M e

−(s−s0 )x

≤ Me

−( s1 −s0 )x

.

(1.8.136)

Since ∞

∫

Me

−( s1 −s0 )x

M dx =

< ∞,

(1.8.137)

s1 − s0

0

Theorem 6 implies the stated conclusion. In Theorem 13 we assumed only that ∫

x

0

e

−s0 u

f (u) du

is bounded; here we are assuming that ∫

∞

0

e

−s0 u

f (u) du

is convergent.

Let ∞

G(x) = ∫

e

−s0 t

f (t) dt\quad and\quadH (x) = sup {|G(t)| ∣ ∣ t ≥ x} .

(1.8.138)

x

Then |G(x)| ≤ H (x)\quad and \quad lim G(x) = lim H (x) = 0, x→∞

since ∫

∞

0

e

−s0 x

f (x) dx

converges. Since f is continuous on [0, ∞), G (x) = −e ′

1.8.22

(A)

x→∞

−s0 x

f (x)

. Integration by parts yields

https://math.libretexts.org/@go/page/17329

∞

∫

∞

e

−sx

f (x) dx =

∫

r

∞

e

−(s−s0 )x

(e

−s0 x

f (x)) dx = − ∫

r

e

−(s−s0 )x

′

G (x) dx

0 ∞

∞

=

−e

−(s−s0 )x

∣ G(x)∣ ∣

+ (s − s0 ) ∫

r

e

−(s−s0 )x

G(x) dx

r ∞

=

e

−(s−s0 )r

G(r) + (s − s0 ) ∫

e

−(s−s0 )x

G(x) dx,

s ≥ s0 .

r

Therefore ∞

∣ ∣∫ ∣ r

e

−sx

∞

∣ f (x) dx ∣ ≤ ∣

|G(r)| e

−(s−s0 )r

+ H (r)(s − s0 ) ∫

e

dx

r

=

(G(r) + H (r))e

so (A) implies that F (s) converges uniformly on [s

0,

−(s−s0 )r

≤ 2H (r)e

−(s−s0 )

,

s ≥ s0 ,

.

∞)

∞

From Theorem 13, F (s) = ∫

−(s−s0 )x

x

e

−sx

converges for all s > s . Denote G(x) = ∫

f (x) dx

e

0

0

−s0 t

f (t) dt

, x ≥ 0 . Then

0 ∞

F (s) =

∫

∞

e

−(s−s0 )x

e

−s0 x

f (x) dx = ∫

0

e

−(s−s0 )x

′

G (x) dx

0 ∞

=

(s − s0 ) ∫

e

−(s−s0 )x

G(x) dx

0 ∞

(integration by parts). Since (s − s

0)

∫

e

−(s−s0 )x

dx = 1

,

0 ∞

F (s) − F (s0 ) = ∫

e

−(s−s0 )x

(G(x) − F (s0 )) dx

(1.8.139)

0

If ϵ > 0 there is an R such that |G(x) − F (s

0 )|

s then 0

R

|F (s) − F (s0 )|
0 , there is an r > 0 such that

0 r

∫

e

−s0 x

|f (x)| dx < ϵ.

(A)

0

If s > s , then 0

1.8.23

https://math.libretexts.org/@go/page/17329

∞

∞

∫

e

−sx

f (x) dx =

∫

r

e

−(s−s0 )x

′

G (x) dx

r ∞

∞

=

e

−(s−s0 )x

∣ G(x)∣ ∣

+ (s − s0 ) ∫

r

G(x)e

−(s−s0 )x

dx

r ∞

=

−e

−(s−s0 )r

G(r) + (s − s0 ) ∫

G(x)e

−(s−s0 )x

dx.

r

Therefore, since |G(x)| ≤ M , ∞

∣ ∣∫ ∣ r

e

−sx

∞

∣ f (x) dx ∣ ≤ ∣

Me

−(s−s0 )r

+ M (s − s0 ) ∫

e

−(s−s0 )x

dx

r ∞

=

M (e

−(s−s0 )r

−e

−(s−s0 )x ∣

∣ ∣

) = 2M e

−(s−s0 )r

.

r

This and (A) imply that ∣ ∣∫ ∣ 0

∞

e

∣ −(s−s0 )r f (x) dx ∣ ≤ ϵ + 2M e . ∣

−sx

(1.8.141)

Therefore, ∣ lim sup ∣∫ ∣ 0 s→∞

∞

e

−sx

∣ f (x) dx ∣ ≤ ϵ. ∣

(1.8.142)

f (x) dx = 0,

(1.8.143)

Since ϵ is arbitrary, this implies that ∞

lim sup ∫ s→∞

∞

e

From Exercise 18(d), ∫

−ax

−e

e

−sx

0

−bx −1

−1

sin x dx = tan

b − tan

From Exercise 30, letting b → ∞ yields

a.

x

0 ∞

∫

e

−ax

sin x x

0

∞

π dx =

−1

− tan

sin x

a, \quad so \bf{(b)}\quad ∫

2

π dx =

x

0

.

(1.8.144)

2

Integrating by parts yields r

r

∫

e

−sx

′

f (x) dt = e

−sr

f (r) − f (0) + ∫

0

Suppose s ≥ s [ s , ∞). Since

. Since |f (x)| ≤ M e

> s0

1

se

−sx

f (x) dx.

(A)

0

s0 x

,e

−sr

|f (r)| ≤ M e

−( s1 −s0 )r

. Therefore e

−sr

f (r) = 0

converges uniformly to zero on

1

∣ ∣∫ ∣

∞

se

−sx

r

∞

∣ f (x) dx ∣ ≤ ∣

M |s| ∫

e

M |s|e

−(s−s0 )x

s − s0

r

M (s − s0 + | s0 |)e

≤

−( s1 −s0 )r

dx ≤

−( s1 −s0 )r

≤ M (1 +

s − s0

| s0 |

)e

−( s1 −s0 )r

∞

it follows that

∫

uniformly on [s

1,

0

=0

−sx

f (x) dx

converges to zero uniformly on

. Since this implies that

[ s1 , ∞)

, (A) implies that G(s) converges uniformly on [s

∞)

(b) In this case let with s

r

se

r

,

s1 − s0

1,

2

′

f (x) = x e

x

2

sin e

x

, so

1 f (x) = −

2

cos e 2

x

. Since

∫

se

−sx

f (x) dx

converges

0

.

∞)

2

| cos e

x

| ≤1

for all x, the hypotheses stated in (a) hold

. Therefore G(s) converges uniformly on [ρ, ∞) if ρ > 0 . ∞

33. We will first show that

∫

e

−s0 x

dx x

0

|G(x)| ≤ M

x

f (x)

converges. Denote

G(x) = ∫

e

−s0 t

f (t) dt

. Since

F (s0 )

is convergent; say

0

, 0 ≤ x < ∞ . If 0 < r < r then 1

r1

∫ r

e

−s0 x

r1

f (x)

′

G (x)

dx = ∫ x

r

G(r) dx =

x

r

1.8.24

r1

G(r1 ) −

−∫ r1

r

G(x) 2

dx.

(1.8.145)

x

https://math.libretexts.org/@go/page/17329

Therefore r1

∣ ∣∫ ∣

e

uniformly on [s

0,

H (s) = ∫

e

f (x) x

r

∞

so Theorem 2 implies that

−s0 x

f (x)

−st

dx x

0

∣ 3M dx ∣ ≤ , ρ ∣

converge when

ρ < r < r1

s = s0

(1.8.146)

. Therefore Exercise 27 implies that it converges

, Therefore Theorem 10 implies that

∞)

s

∫

s

F (u) du =

s0

∫

∞

(∫

s0

∞

e

−ux

f (x) dx) du = ∫

0

s

(∫

0

e

−ux

du) f (x) dx

s0 ∞

=

∫ 0 ∞

From Exercise 30 (with f (x) replaced by f (x)/x),

lim ∫ s→∞

0

e

−sx

f (x) dx = 0

(e

−s0 x

−e

−sx

f (x) )

dx x

, which implies the stated conclusion.

x

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Index D Dirichlet’s Tests 1.5: Dirichlet’s Tests

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