FUNCTIONS DEFINED BY IMPROPER INTEGRALS

FUNCTIONS DEFINED BY

IMPROPER INTEGRALS

William F. Trench

Professor Emeritus

Department of Mathematics

Trinity University

San Antonio, Texas, USA

[email protected]

©Copyright May 2012, William F. Trench This is a supplement to the author’s

Introduction to Real Analysis

1

Foreword

This is a revised version of Section 7.5 of myAdvanced Calculus(Harper & Row, 1978). It is a supplement to my textbookIntroduction to Real Analysis, which is refer-enced several times here. You should review Section 3.4 (Improper Integrals) of that book before reading this document.

2

Introduction

In Section 7.2 (pp. 462–484) we considered functions of the form

F .y/_D

Z b

a

f .x; y/ dx; cyd:

We saw that iff is continuous onŒa; bŒc; d , thenF is continuous onŒc; d  (Exer-cise 7.2.3, p. 481) and that we can reverse the order of integration in

Z d

c

F .y/ dy _D

Z d

c Z b

a

f .x; y/ dx

!

dy

to evaluate it as

Z d

c

F .y/ dy _D

Z b

a Z d

c

f .x; y/ dy

!

dx

(Corollary 7.2.3, p. 466).

Here is another important property ofF.

Theorem 1 Iff andfyare continuous onŒa; bŒc; d ;then

F .y/_D

Z b

a

f .x; y/ dx; cyd; (1)

is continuously differentiable onŒc; d andF0_{.y/can be obtained by differentiating(1)}

under the integral sign with respect toy_Ithat is,

F0.y/_D

Z b

a

fy.x; y/ dx; cyd: (2)

HereF0.a/ andfy.x; a/are derivatives from the right andF0.b/andfy.x; b/are

derivatives from the left:

Proof Ifyandy_Cy are inŒc; d andy_¤0, then

F .y_Cy/ F .y/

y D

Z b

a

f .x; y_Cy/ f .x; y/

y dx: (3)

From the mean value theorem (Theorem 2.3.11, p. 83), ifx₂Œa; bandy,y_Cy ₂ Œc; d , there is ay.x/betweenyandy_Cysuch that

From this and (3),

ˇ ˇ ˇ ˇ ˇ

F .y_Cy/ F .y/ y

Z b

a

fy.x; y/ dx ˇ ˇ ˇ ˇ ˇ

Z b

a j

fy.x; y.x// fy.x; y/jdx: (4)

Now suppose > 0. Sincefyis uniformly continuous on the compact setŒa; bŒc; d 

(Corollary 5.2.14, p. 314) andy.x/is betweeny andy_Cy, there is aı > 0such that ifj_j< ıthen

jfy.x; y/ fy.x; y.x//j< ; .x; y/2Œa; bŒc; d :

This and (4) imply that

ˇ ˇ ˇ ˇ ˇ

F .y_Cy F .y// y

Z b

a

fy.x; y/ dx ˇ ˇ ˇ ˇ ˇ

< .b a/

ify andy_Cy are inŒc; d and0 <_jy_j< ı. This implies (2). Since the integral in (2) is continuous onŒc; d (Exercise 7.2.3, p. 481, withf replaced by fy),F0 is

continuous onŒc; d .

Example 1 Since

f .x; y/_Dcosxy and fy.x; y/D xsinxy

are continuous for all.x; y/, Theorem1implies that if

F .y/_D

Z

0

cosxy dx; ₁< y <₁; (5)

then

F0.y/_D

Z

0

xsinxy dx; ₁< y <₁: (6)

(In applying Theorem1for a specific value ofy, we takeR_DŒ0; Œ ; , where >_jy_j.) This provides a convenient way to evaluate the integral in (6): integrating the right side of (5) with respect toxyields

F .y/_D sinxy y

ˇ ˇ ˇ ˇ

xD0D

siny

y ; y¤0:

Differentiating this and using (6) yields

Z

0

xsinxy dx_D siny y2

cosy

y ; y¤0:

To verify this, use integration by parts.

We will study the continuity, differentiability, and integrability of

F .y/_D

Z b

a

whereS is an interval or a union of intervals, andF is a convergent improper integral for eachy ₂S. If the domain off isŒa; b/S where 1 < a < b ₁, we say thatF ispointwise convergent onSor simplyconvergent onS, and write

Z b

a

f .x; y/ dx_D lim

r!b Z r

a

f .x; y/ dx (7)

if, for eachy ₂ Sand every > 0, there is anr _D r0.y/(which also depends on)

such that

ˇ ˇ ˇ ˇ

F .y/

Z r

a

f .x; y/ dx

ˇ ˇ ˇ ˇD

ˇ ˇ ˇ ˇ ˇ

Z b

r

f .x; y/ dx

ˇ ˇ ˇ ˇ ˇ

< ; r0.y/y < b: (8)

If the domain off is.a; bSwhere ₁ a < b <₁, we replace (7) by

Z b

a

f .x; y/ dx_D lim

r!aC

Z b

r

f .x; y/ dx

and (8) by

ˇ ˇ ˇ ˇ ˇ

F .y/

Z b

r

f .x; y/ dx

ˇ ˇ ˇ ˇ ˇ

D

ˇ ˇ ˇ ˇ

Z r

a

f .x; y/ dx

ˇ ˇ ˇ ˇ

< ; a < r r0.y/:

In general, pointwise convergence ofF for ally ₂ S does not imply thatF is continuous or integrable onŒc; d , and the additional assumptions thatfyis continuous

andR_abfy.x; y/ dxconverges do not imply (2).

Example 2 The function

f .x; y/_Dye jyjx is continuous onŒ0;₁/. ₁;₁/and

F .y/_D

Z 1 0

f .x; y/ dx_D

Z 1 0

ye jyjxdx converges for ally, with

F .y/_D

8 ˆ <

ˆ :

1 y < 0; 0 y_D0; 1 y > 0_I

therefore,F is discontinuous aty_D0.

Example 3 The function

f .x; y/_Dy3e y2x

is continuous onŒ0;₁/. ₁;₁/. Let

F .y/_D

Z 1 0

f .x; y/ dx_D

Z 1 0

Then

F0.y/_D1; ₁< y <₁: However,

Z 1 0

@ @y.y

3

e y2x/ dx_D

Z 1 0

.3y2 2y4x/e y2xdx_D

(

1; y_¤0; 0; y_D0;

so

F0.y/_¤

Z 1 0

@f .x; y/

@y dx if yD0:

3

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 thatf is locally integrable on an intervalI if it is integrable on every finite closed subinterval ofI.

Theorem 2 (CauchyCriterion for Convergence of an Improper Integral I) Suppose

gis locally integrable onŒa; b/and denote

G.r /_D

Z r

a

g.x/ dx; ar < b:

Then the improper integralR_abg.x/ dx converges if and only if;for each > 0;there is anr02Œa; b/such that

jG.r / G.r1/j< ; r0r; r1< b: (9)

Proof For necessity, supposeR_abg.x/ dx _D L. By definition, this means that for each > 0there is anr02Œa; b/such that

jG.r / L_j<

2 and jG.r1/ Lj<

2; r0r; r1< b:

Therefore

jG.r / G.r1/j D j.G.r / L/ .G.r1/ L/j

jG.r / L_{j C j}G.r1/ Lj< ; r0r; r1< b:

For sufficiency, (9) implies that

jG.r /_{j D j}G.r1/C.G.r / G.r1//j<jG.r1/j C jG.r / G.r1/j jG.r1/j C;

r0r r1< b. SinceGis also bounded on the compact setŒa; r0(Theorem 5.2.11,

p. 313),Gis bounded onŒa; b/. Therefore the monotonic functions

G.r /_Dsup˚

G.r1/ ˇ

ˇrr1< b and G.r /Dinf˚G.r1/ ˇ

are well defined onŒa; b/, and

lim

r!b G.r /DL and r!limb G.r /DL

both exist and are finite (Theorem 2.1.11, p. 47). From (9),

jG.r / G.r1/j D j.G.r / G.r0// .G.r1/ G.r0//j

jG.r / G.r0/j C jG.r1/ G.r0/j< 2;

so

G.r / G.r /2; r0r; r1< b:

Sinceis an arbitrary positive number, this implies that

lim

r!b .G.r / G.r //D0;

soL_DL. LetL_DL_DL. Since

G.r /G.r /G.r /;

it follows that limr!b G.r /DL.

We leave the proof of the following theorem to you (Exercise2).

Theorem 3 (Cauchy Criterion for Convergence of an Improper Integral II) Suppose

gis locally integrable on.a; band denote

G.r /_D

Z b

r

g.x/ dx; ar < b:

Then the improper integralRb

a g.x/ dx converges if and only if;for each > 0;there

is anr02.a; bsuch that

jG.r / G.r1/j< ; a < r; r1r0:

To see why we associate Theorems2and3with Cauchy, compare them with The-orem 4.3.5 (p. 204)

4

Uniform convergence of improper integrals

Henceforth we deal with functionsf _Df .x; y/with domainsI S, whereS is an interval or a union of intervals andI is of one of the following forms:

Œa; b/with ₁< a < b ₁;

.a; bwith ₁ a < b <₁;

In all cases it is to be understood thatf is locally integrable with respect tox onI. When we say that the improper integralR_abf .x; y/ dxhas a stated property “on S” we mean that it has the property for everyy ₂S.

Definition 1 If the improper integral

Z b

a

f .x; y/ dx_D lim

r!b Z r

a

f .x; y/ dx (10)

converges onS;it is said to converge uniformly.or be uniformly convergent/onS if;

for each > 0;there is anr02Œa; b/such that

ˇ ˇ ˇ ˇ ˇ

Z b

a

f .x; y/ dx

Z r

a

f .x; y/ dx

ˇ ˇ ˇ ˇ ˇ

< ; y ₂S; r0r < b;

or;equivalently;

ˇ ˇ ˇ ˇ ˇ

Z b

r

f .x; y/ dx

ˇ ˇ ˇ ˇ ˇ

< ; y ₂S; r0r < b: (11)

The crucial difference between pointwise and uniform convergence is thatr0.y/in

(8) may depend upon the particular value ofy, while ther0in (11) does not: one choice

must work for ally ₂S. Thus, uniform convergence implies pointwise convergence, but pointwise convergence does not imply uniform convergence.

Theorem 4 .Cauchy Criterion for Uniform Convergence I/The improper integral in(10)converges uniformly onS if and only if;for each > 0;there is anr02Œa; b/

such that

ˇ ˇ ˇ ˇ

Z r1

r

f .x; y/ dx

ˇ ˇ ˇ ˇ

< ; y₂S; r0r; r1< b: (12)

Proof SupposeR_abf .x; y/ dx converges uniformly onS and > 0. From Defini-tion1, there is anr02Œa; b/such that

ˇ ˇ ˇ ˇ ˇ

Z b

r

f .x; y/ dx

ˇ ˇ ˇ ˇ ˇ

< 2 and

ˇ ˇ ˇ ˇ ˇ

Z b

r1

f .x; y/ dx

ˇ ˇ ˇ ˇ ˇ

<

2; y 2S; r0r; r1< b: (13)

Since

Z r1

r

f .x; y/ dx_D

Z b

r

f .x; y/ dx

Z b

r1

f .x; y/ dx;

(13) and the triangle inequality imply (12). For the converse, denote

F .y/ _D

Z r

a

Since (12) implies that

jF .r; y/ F .r1; y/j< ; y2S; r0r; r1< b; (14)

Theorem2withG.r /_DF .r; y/(yfixed but arbitrary inS) implies thatRb

a f .x; y/ dx

converges pointwise fory ₂ S. Therefore, if > 0then, for eachy ₂S, there is an r0.y/2Œa; b/such that

ˇ ˇ ˇ ˇ ˇ

Z b

r

f .x; y/ dx

ˇ ˇ ˇ ˇ ˇ

< ; y ₂S; r0.y/r < b: (15)

For eachy₂S, chooser1.y/maxŒr0.y/; r0. (Recall (14)). Then

Z b

r

f .x; y/ dx_D

Z r1.y/

r

f .x; y/ dx_C

Z b

r1.y/

f .x; y/ dx;

so (12), (15), and the triangle inequality imply that

ˇ ˇ ˇ ˇ ˇ

Z b

r

f .x; y/ dx

ˇ ˇ ˇ ˇ ˇ

< 2; y₂S; r0 r < b:

In practice, we don’t explicitly exhibitr0 for each given . It suffices to obtain

estimates that clearly imply its existence.

Example 4 For the improper integral of Example2,

ˇ ˇ ˇ ˇ

Z 1 r

f .x; y/ dx

ˇ ˇ ˇ ˇD

Z 1 r j

y_je jyjx _De rjyj; y _¤0: Ifjy_j , then

ˇ ˇ ˇ ˇ

Z 1 r

f .x; y/ dx

ˇ ˇ ˇ ˇ

e r;

soR1

0 f .x; y/ dx converges uniformly on . 1; [Œ;1/if > 0; however, it

does not converge uniformly on any neighborhood ofy _D 0, since, for any r > 0, e rjyj> 1₂if_jy_jis sufficiently small.

Definition 2 If the improper integral

Z b

a

f .x; y/ dx_D lim

r!aC

Z b

r

f .x; y/ dx

converges onS;it is said to converge uniformly.or be uniformly convergent/onS if;

for each > 0;there is anr02.a; bsuch that ˇ

ˇ ˇ ˇ ˇ

Z b

a

f .x; y/ dx

Z b

r

f .x; y/ dx

ˇ ˇ ˇ ˇ ˇ

< ; y ₂S; a < rr0;

or;equivalently;

ˇ ˇ ˇ ˇ

Z r

a

f .x; y/ dx

ˇ ˇ ˇ ˇ

We leave proof of the following theorem to you (Exercise3).

Theorem 5 .Cauchy Criterion for Uniform Convergence II/The improper integral

Z b

a

f .x; y/ dx_D lim

r!aC

Z b

r

f .x; y/ dx

converges uniformly onSif and only if;for each > 0;there is anr0 2 .a; bsuch

that

ˇ ˇ ˇ ˇ

Z r

r1

f .x; y/ dx

ˇ ˇ ˇ ˇ

< ; y₂S; a < r; r1r0:

We need one more definition, as follows.

Definition 3 Letf _D f .x; y/be defined on.a; b/S;where ₁ a < b ₁:

Supposef is locally integrable on.a; b/for ally ₂S and letcbe an arbitrary point in.a; b/:ThenR_abf .x; y/ dxis said to converge uniformly onSifR_acf .x; y/ dxand

Rb

c f .x; y/ dxboth converge uniformly onS:

We leave it to you (Exercise 4) to show that this definition is independent ofc; that is, ifR_acf .x; y/ dx andR_cbf .x; y/ dx both converge uniformly onS for some c₂.a; b/, then they both converge uniformly onSfor everyc₂.a; b/.

We also leave it you (Exercise5) to show that iff is bounded onŒa; bŒc; d  andRb

a f .x; y/ dx exists as a proper integral for each y 2 Œc; d , then it converges

uniformly onŒc; d according to all three Definitions1–3.

Example 5 Consider the improper integral

F .y/_D

Z 1 0

x 1=2e xydx;

which diverges ify 0(verify). Definition3applies if y > 0, so we consider the improper integrals

F1.y/D Z 1

0

x 1=2e xydx and F2.y/D Z 1

1

x 1=2e xydx

separately. Moreover, we could just as well define

F1.y/D Z c

0

x 1=2e xydx and F2.y/D Z 1

c

x 1=2e xydx; (16)

wherecis any positive number.

Definition2applies toF1. If0 < r1< randy0, then

ˇ ˇ ˇ ˇ

Z r1

r

x 1=2e xydx

ˇ ˇ ˇ ˇ

<

Z r

r1

x 1=2dx < 2r1=2;

Definition1applies toF2. Since

ˇ ˇ ˇ ˇ

Z r1

r

x 1=2e xydx

ˇ ˇ ˇ ˇ

< r 1=2

Z 1 r

e xydx_D e

ry

yr1=2;

F2.y/ converges uniformly onŒ;1/if > 0. It does not converge uniformly on

.0; /, since the change of variableu_Dxyyields

Z r1

r

x 1=2e xydx_Dy 1=2

Z r1y

ry

u 1=2e udu;

which, for any fixedr > 0, can be made arbitrarily large by takingysufficiently small andr _D 1=y. Therefore we conclude that F .y/ converges uniformly onŒ;₁/if > 0:

Note that that the constantcin (16) plays no role in this argument.

Example 6 Suppose we take

Z 1 0

sinu u duD

2 (17)

as given (Exercise31(b)). Substitutingu_Dxywithy > 0yields

Z 1 0

sinxy x dxD

2; y > 0: (18)

What about uniform convergence? Since.sinxy/=x is continuous atx _D 0, Defini-tion1and Theorem4apply here. If0 < r < r1andy > 0, then

Z r1

r

sinxy x dxD

1 y

cosxy x

ˇ ˇ ˇ ˇ r1

r C Z r1

r

cosxy x2 dx

; so

ˇ ˇ ˇ ˇ

Z r1

r

sinxy x dx

ˇ ˇ ˇ ˇ

< 3 ry:

Therefore (18) converges uniformly onŒ;₁/if > 0. On the other hand, from (17), there is aı > 0such that

Z 1 u0

sinu u du >

4; 0u0< ı:

This and (18) imply that

Z 1 r

sinxy x dxD

Z 1 yr

sinu u du >

4

5

Absolutely Uniformly Convergent Improper Integrals

Definition 4 .Absolute Uniform Convergence I/The improper integral

Z b

a

f .x; y/ dx_D lim

r!b Z r

a

f .x; y/ dx

is said to converge absolutely uniformly onS if the improper integral

Z b

a j

f .x; y/_jdx_D lim

r!b Z r

a j

f .x; y/_jdx

converges uniformly onS; that is, if, for each > 0, there is anr02Œa; b/such that

ˇ ˇ ˇ ˇ ˇ

Z b

a j

f .x; y/_jdx

Z r

a j

f .x; y/_jdx

ˇ ˇ ˇ ˇ ˇ

< ; y₂S; r0 < r < b:

To see that this definition makes sense, recall that iff is locally integrable onŒa; b/ for allyinS, then so is_jf_j(Theorem 3.4.9, p. 161). Theorem4withf replaced by

jf_jimplies thatR_abf .x; y/ dxconverges absolutely uniformly onS if and only if, for each > 0, there is anr02Œa; b/such that

Z r1

r j

f .x; y/_jdx < ; y₂S; r0r < r1< b:

Since _ˇ

ˇ ˇ ˇ

Z r1

r

f .x; y/ dx

ˇ ˇ ˇ ˇ

Z r1

r j

f .x; y/_jdx;

Theorem4implies that ifR_abf .x; y/ dxconverges absolutely uniformly onS then it converges uniformly onS.

Theorem 6 . Weierstrass’s Test for Absolute Uniform Convergence I/ Suppose

M _DM.x/is nonnegative onŒa; b/;Rb

a M.x/ dx <1;and

jf .x; y/_j M.x/; y₂S; ax < b: (19)

ThenR_abf .x; y/ dxconverges absolutely uniformly onS:

Proof DenoteRb

a M.x/ dx D L < 1. By definition, for each > 0there is an

r02Œa; b/such that

L <

Z r

a

M.x/ dx L; r0< r < b:

Therefore, ifr0< rr1;then

0

Z r1

r

M.x/ dx_D

Z r1

a

M.x/ dx L

Z r

a

M.x/ dx L

This and (19) imply that

Z r1

r j

f .x; y/_jdx

Z r1

r

M.x/ dx < ; y₂S; ar0< r < r1< b:

Now Theorem4implies the stated conclusion.

Example 7 Supposeg_Dg.x; y/is locally integrable onŒ0;₁/for ally ₂Sand, for somea00, there are constantsKandp0such that

jg.x; y/_j Kep0x_{; y}

2S; xa0:

Ifp > p0andr a0, then

Z 1 r

e px_jg.x; y/_jdx _D

Z 1 r

e .p p0/x e p0x

jg.x; y/_jdx

K

Z 1 r

e .p p0/x

dx_D Ke

.p p0/r

p p0

;

soR₀1e px_{g.x; y/ dxconverges absolutely onS. For example, since}

jx˛sinxy_j< ep0x

and jx˛cosxy_j< ep0x

for x sufficiently large if p0 > 0, Theorem 4 implies that R01e pxx˛sinxy dx

andR1

0 e pxx˛cosxy dxconverge absolutely uniformly on. 1;1/ifp > 0and

˛ 0. As a matter of fact,R1

0 e pxx˛sinxy dxconverges absolutely on. 1;1/

ifp > 0and˛ > 1. (Why?)

Definition 5 .Absolute Uniform Convergence II/The improper integral

Z b

a

f .x; y/ dx_D lim

r!aC

Z b

r

f .x; y/ dx

is said to converge absolutely uniformly onS if the improper integral

Z b

a j

f .x; y/_jdx_D lim

r!aC

Z b

r j

f .x; y/_jdx

converges uniformly onS; that is, if, for each > 0, there is anr02.a; bsuch that

ˇ ˇ ˇ ˇ ˇ

Z b

a j

f .x; y/_jdx

Z b

r j

f .x; y/_jdx

ˇ ˇ ˇ ˇ ˇ

< ; y ₂S; a < r < r0b:

We leave it to you (Exercise7) to prove the following theorem.

Theorem 7 .Weierstrass’s Test for Absolute Uniform Convergence II/ Suppose

M _DM.x/is nonnegative on.a; b;Rb

a M.x/ dx <1;and

jf .x; y/_j M.x/; y ₂S; x₂.a; b:

ThenRb

Example 8 Ifg_Dg.x; y/is locally integrable on.0; 1for ally₂Sand

jg.x; y/_j Ax ˇ; 0 < xx0;

for eachy₂S, then

Z 1

0

x˛g.x; y/ dx

converges absolutely uniformly onS if˛ > ˇ 1. To see this, note that if0 < r < r1x0, then

Z r

r1

x˛_jg.x; y/_jdx A

Z r

r1

x˛ ˇdx_D Ax

˛ ˇC1

˛ ˇ_C1

ˇ ˇ ˇ ˇ r

r1 < Ar

˛ ˇC1

˛ ˇ_C1:

Applying this withˇ_D0shows that

F .y/ _D

Z 1

0

x˛cosxy dx

converges absolutely uniformly on. ₁;₁/if˛ > 1and

G.y/_D

Z 1

0

x˛sinxy dx

converges absolutely uniformly on. ₁;₁/if˛ > 2.

By recalling Theorem 4.4.15 (p. 246), you can see why we associate Theorems6 and7with Weierstrass.

6

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 whereRb

a f .x; y/ dx converges uniformly on S, but Rb

a jf .x; y/jdx

does not.

Theorem 8 .Dirichlet’s Test for Uniform Convergence I/Ifg; gx;andhare

con-tinuous onŒa; b/S;then

Z b

a

g.x; y/h.x; y/ dx

converges uniformly onSif the following conditions are satisfied_W

(a) lim

x!b (

sup

y2Sj

g.x; y/_j

)

D0_I

(b) There is a constantM such that

sup

y2S ˇ ˇ ˇ ˇ

Z x

a

h.u; y/ du

ˇ ˇ ˇ ˇ

(c) R_ab_jgx.x; y/jdxconverges uniformly onS:

Proof If

H.x; y/_D

Z x

a

h.u; y/ du; (20)

then integration by parts yields

Z r1

r

g.x; y/h.x; y/ dx _D

Z r1

r

g.x; y/Hx.x; y/ dx

D g.r1; y/H.r1; y/ g.r; y/H.r; y/ (21) Z r1

r

gx.x; y/H.x; y/ dx:

Since assumption(b)and (20) imply thatjH.x; y/_j M; .x; y/₂.a; bS, Eqn. (21) implies that

ˇ ˇ ˇ ˇ

Z r1

r

g.x; y/h.x; y/ dx

ˇ ˇ ˇ ˇ

< M

2sup

xrj

g.x; y/_{j C}

Z r1

r j

gx.x; y/jdx

(22)

onŒr; r1S.

Now suppose > 0. From assumption(a), there is an r0 2 Œa; b/ such that

jg.x; y/_j < onS ifr0 x < b. From assumption(c) and Theorem6, there is

ans02Œa; b/such that

Z r1

r j

gx.x; y/jdx < ; y 2S; s0< r < r1< b:

Therefore (22) implies that

ˇ ˇ ˇ ˇ

Z r1

r

g.x; y/h.x; y/

ˇ ˇ ˇ ˇ

< 3M; y₂S; max.r0; s0/ < r < r1< b:

Now Theorem4implies the stated conclusion.

The statement of this theorem is complicated, but applying it isn’t; just look for a factorizationf _Dgh, wherehhas a bounded antderivative onŒa; b/andgis “small” near b. Then integrate by parts and hope that something nice happens. A similar comment applies to Theorem 9, which follows.

Example 9 Let

I.y/_D

Z 1 0

cosxy

x_Cy dx; y > 0:

The obvious inequality

ˇ ˇ ˇ ˇ

cosxy x_Cy

ˇ ˇ ˇ ˇ

1 x_Cy

is useless here, since

Z 1 0

dx

However, integration by parts yields

Z r1

r

cosxy x_Cy dx D

sinxy y.x_Cy/

ˇ ˇ ˇ ˇ r1

r

C

Z r1

r

sinxy y.x_Cy/2dx

D sinr1y

y.r1Cy/

sinry y.r _Cy/C

Z r1

r

sinxy y.x_Cy/2 dx:

Therefore, if0 < r < r1, then

ˇ ˇ ˇ ˇ

Z r1

r

cosxy x_Cy dx

ˇ ˇ ˇ ˇ

< 1 y

₂

r_Cy C

Z 1 r

1 .x_Cy/2

3

y.r_Cy/2

3 .r_C/

ify > 0. Now Theorem 4implies thatI.y/converges uniformly onŒ;₁/if > 0.

We leave the proof of the following theorem to you (Exercise10).

Theorem 9 .Dirichlet’s Test for Uniform Convergence II/Ifg; gx;andhare

con-tinuous on.a; bS;then

Z b

a

g.x; y/h.x; y/ dx

converges uniformly onSif the following conditions are satisfied_W

(a) lim

x!aC

(

sup

y2Sj

g.x; y/_j

)

D0_I

(b) There is a constantM such that

sup

y2S ˇ ˇ ˇ ˇ ˇ

Z b

x

h.u; y/ du

ˇ ˇ ˇ ˇ ˇ

M; a < xb_I

(c) Rb

a jgx.x; y/jdxconverges uniformly onS.

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 Theorems8and9with Dirichlet.

7

Consequences of uniform convergence

Theorem 10 Iff _D f .x; y/is continuous on eitherŒa; b/Œc; d or.a; bŒc; d 

and

F .y/ _D

Z b

a

f .x; y/ dx (23)

converges uniformly onŒc; d ;thenF is continuous onŒc; d :Moreover;

Z d

c Z b

a

f .x; y/ dx

!

dy_D

Z b

a Z d

c

f .x; y/ dy

!

Proof We will assume thatf is continuous on.a; bŒc; d . You can consider the other case (Exercise14).

We will first show thatF in (23) is continuous onŒc; d . SinceF converges uni-formly onŒc; d , Definition1 (specifically, (11)) implies that if > 0, there is an r₂Œa; b/such that

ˇ ˇ ˇ ˇ ˇ

Z b

r

f .x; y/ dx

ˇ ˇ ˇ ˇ ˇ

< ; cyd:

Therefore, ifcy; y0d , then

jF .y/ F .y0/j D ˇ ˇ ˇ ˇ ˇ

Z b

a

f .x; y/ dx

Z b

a

f .x; y0/ dx ˇ ˇ ˇ ˇ ˇ

ˇ ˇ ˇ ˇ

Z r

a

Œf .x; y/ f .x; y0/ dx ˇ ˇ ˇ ˇC

ˇ ˇ ˇ ˇ ˇ

Z b

r

f .x; y/ dx

ˇ ˇ ˇ ˇ ˇ

C

ˇ ˇ ˇ ˇ ˇ

Z b

r

f .x; y0/ dx ˇ ˇ ˇ ˇ ˇ

;

so

jF .y/ F .y0/j Z r

a j

f .x; y/ f .x; y0/jdxC2: (25)

Sincef is uniformly continuous on the compact setŒa; r Œc; d (Corollary 5.2.14, p. 314), there is aı > 0such that

jf .x; y/ f .x; y0/j<

if.x; y/and.x; y0/are inŒa; r Œc; d andjy y0j< ı. This and (25) imply that

jF .y/ F .y0/j< .r a/C2 < .b aC2/

ifyandy0are inŒc; d andjy y0j< ı. ThereforeF is continuous onŒc; d , so the

integral on left side of (24) exists. Denote

I _D

Z d

c Z b

a

f .x; y/ dx

!

dy: (26)

We will show that the improper integral on the right side of (24) converges toI. To this end, denote

I.r /_D

Z r

a Z d

c

f .x; y/ dy

!

dx:

Since we can reverse the order of integration of the continuous functionf over the rectangleŒa; r Œc; d (Corollary 7.2.2, p. 466),

I.r /_D

Z d

c Z r

a

f .x; y/ dx

From this and (26),

I I.r /_D

Z d

c Z b

r

f .x; y/ dx

!

dy:

Now suppose > 0. SinceRb

a f .x; y/ dx converges uniformly onŒc; d , there is an

r02.a; bsuch that

ˇ ˇ ˇ ˇ ˇ Z b r

f .x; y/ dx

ˇ ˇ ˇ ˇ ˇ

< ; r0< r < b;

so_jI I.r /_j< .d c/ifr0< r < b. Hence,

lim

r!b Z r

a Z d

c

f .x; y/ dy

! dx _D Z d c Z b a

f .x; y/ dx

!

dy;

which completes the proof of (24).

Example 10 It is straightforward to verify that

Z 1 0

e xydx_D 1

y; y > 0;

and the convergence is uniform onŒ;₁/if > 0. Therefore Theorem10implies that if0 < y1< y2, then

Z y2

y1 dy

y D

Z y2

y1

Z 1 0

e xydx

dy_D

Z 1 0

Z y2

y1

e xydy

dy

D

Z 1 0

e xy1 _e xy2

x dx:

Since

Z y2

y1 dy

y Dlog y2

y1

; y2y1> 0;

it follows that

Z 1 0

e xy1 _e xy2

x dxDlog y2

y1

; y2y1> 0:

Example 11 From Example6,

Z 1 0

sinxy x dxD

2; y > 0;

and the convergence is uniform onŒ;₁/if > 0. Therefore, Theorem10implies that if0 < y1< y2, then

2.y2 y1/ D

Z y2

y1 Z 1 0 sinxy x dx dy_D Z 1 0

Z y2

y1 sinxy x dy dx D Z 1 0

cosxy1 cosxy2

The last integral converges uniformly on. ₁;₁/(Exercise 10(h)), and is therefore continuous with respect toy1on. 1;1/, by Theorem10; in particular, we can let

y1!0Cin (27) and replacey2byyto obtain

Z 1 0

1 cosxy x2 dxD

y

2 ; y0:

The next theorem is analogous to Theorem 4.4.20 (p. 252).

Theorem 11 Letf andfy be continuous on eitherŒa; b/Œc; d or.a; bŒc; d :

Suppose that the improper integral

F .y/ _D

Z b

a

f .x; y/ dx

converges for somey02Œc; d and

G.y/_D

Z b

a

fy.x; y/ dx

converges uniformly onŒc; d :ThenF converges uniformly onŒc; d and is given ex-plicitly by

F .y/ _DF .y0/C Z y

y0

G.t / dt; cy d:

Moreover,F is continuously differentiable onŒc; d ; specifically,

F0.y/_DG.y/; cyd; (28)

whereF0.c/andfy.x; c/are derivatives from the right, andF0.d /andfy.x; d /are

derivatives from the left:

Proof We will assume thatf andfy are continuous on Œa; b/Œc; d . You can

consider the other case (Exercise15). Let

Fr.y/D Z r

a

f .x; y/ dx; ar < b; cyd:

Sincef andfyare continuous onŒa; r Œc; d , Theorem1implies that

F_r0.y/_D

Z r

a

fy.x; y/ dx; cyd:

Then

Fr.y/ D Fr.y0/C Z y

y0

Z r

a

fy.x; t / dx

dt

D F .y0/C Z y

y0

G.t / dt

C.Fr.y0/ F .y0// Z y

y0

Z b

r

fy.x; t / dx !

Therefore,

ˇ ˇ ˇ ˇ

Fr.y/ F .y0/ Z y

y0

G.t / dt

ˇ ˇ ˇ ˇ j

Fr.y0/ F .y0/j

C ˇ ˇ ˇ ˇ ˇ Z y y0 Z b r

fy.x; t / dx ˇ ˇ ˇ ˇ ˇ dt: (29)

Now suppose > 0. Since we have assumed that limr!b Fr.y0/D F .y0/exists,

there is anr0in.a; b/such that

jFr.y0/ F .y0/j< ; r0< r < b:

Since we have assumed thatG.y/converges fory ₂Œc; d , there is anr12Œa; b/such

that _ˇ ˇ ˇ ˇ ˇ Z b r

fy.x; t / dx ˇ ˇ ˇ ˇ ˇ

< ; t ₂Œc; d ; r1r < b:

Therefore, (29) yields

ˇ ˇ ˇ ˇ

Fr.y/ F .y0/ Z y

y0

G.t / dt

ˇ ˇ ˇ ˇ

< .1_{C j}y y0j/.1Cd c/

if max.r0; r1/r < bandt 2 Œc; d . ThereforeF .y/converges uniformly onŒc; d 

and

F .y/ _DF .y0/C Z y

y0

G.t / dt; cy d:

SinceG is continuous onŒc; d by Theorem10, (28) follows from differentiating this (Theorem 3.3.11, p. 141).

Example 12 Let

I.y/_D

Z 1 0

e yx2dx; y > 0:

Since

Z r

0

e yx2dx_{D p}1 y

Z rpy 0

e t2dt;

it follows that

I.y/_{D p}1 y

Z 1 0

e t2dt;

and the convergence is uniform onŒ;₁/if > 0(Exercise8(i)). To evaluate the last integral, denoteJ./_DR

0 e t2

dt; then

J2./_D

Z

0

e u2du

Z

0

e v2dv

D Z 0 Z 0

e .u2Cv2/du dv: Transforming to polar coordinatesr_Drcos,v_Drsinyields

J2./_D

Z =2

0 Z

0

r e r2d r d_D .1 e

2 /

4 ; so J./D

q

.1 e 2 /

Therefore

Z 1 0

e t2dt_D lim

!1J./D

p

2 and

Z 1 0

e yx2dx _D 1 2

r

y; y > 0:

Differentiating thisntimes with respect toyyields

Z 1 0

x2ne yx2dx_D 13 .2n 1/

p

2n_ynC1=2 y > 0; nD1; 2; 3; : : : ;

where Theorem 11 justifies the differentiation for every n, since all these integrals converge uniformly onŒ;₁/if > 0(Exercise8(i)).

Some advice for applying this theorem: Be sure to check first that F .y0/ D Rb

a f .x; y0/ dxconverges for at least one value ofy. If so, differentiateR b

a f .x; y/ dx

formally to obtainRb

a fy.x; y/ dx. ThenF0.y/ D Rb

a fy.x; y/ dx if y is in some

interval on which this improper integral converges uniformly.

8

Applications to Laplace transforms

TheLaplacetransformof a functionf locally integrable onŒ0;₁/is

F .s/_D

Z 1 0

e sxf .x/ dx

for allssuch that integral converges. Laplace transforms are widely applied in mathe-matics, particularly in solving differential equations.

We leave it to you to prove the following theorem (Exercise26).

Theorem 12 Supposef is locally integrable onŒ0;₁/and_jf .x/_j Mes0x_for

suf-ficiently largex. Then the Laplace transform ofF converges uniformly onŒs1;1/if

s1> s0.

Theorem 13 Iff is continuous onŒ0;₁/andH.x/_DR1 0 e

s0u_{f .u/ duis bounded}

onŒ0;₁/;then the Laplace transform off converges uniformly onŒs1;1/ifs1> s0:

Proof If0rr1, Z r1

r

e sxf .x/ dx _D

Z r1

r

e .s s0/x e s0x

f .x/ dt _D

Z r1

r

e .s s0/t

H0.x/ dt: Integration by parts yields

Z r1

r

e sxf .x/ dt _De .s s0/x_H.x/

ˇ ˇ ˇ ˇ r1

r

C.s s0/ Z r1

r

Therefore, if_jH.x/_j M, then

ˇ ˇ ˇ ˇ

Z r1

r

e sxf .x/ dx

ˇ ˇ ˇ ˇ

M

ˇ ˇ ˇ ˇ

e .s s0/r1

Ce .s s0/r

C.s s0/ Z r1

r

e .s s0/x dx

ˇ ˇ ˇ ˇ

3Me .s s0/r

3Me .s1 s0/r

; ss1:

Now Theorem4implies thatF .s/converges uniformly onŒs1;1/.

The following theorem draws a considerably stonger conclusion from the same assumptions.

Theorem 14 Iff is continuous onŒ0;₁/and

H.x/_D

Z x

0

e s0u

f .u/ du

is bounded onŒ0;₁/;then the Laplace transform off is infinitely differentiable on

.s0;1/;with

F.n/.s/_D. 1/n

Z 1 0

e sxxnf .x/ dx_I (30)

that is, then-th derivative of the Laplace transform off .x/is the Laplace transform of. 1/nxnf .x/.

Proof First we will show that the integrals

In.s/D Z 1

0

e sxxnf .x/ dx; n_D0; 1; 2; : : :

all converge uniformly onŒs1;1/ifs1> s0. If0 < r < r1, then

Z r1

r

e sxxnf .x/ dx_D

Z r1

r

e .s s0/x e s0x

xnf .x/ dx_D

Z r1

r

e .s s0/x

xnH0.x/ dx: Integrating by parts yields

Z r1

r

e sxxnf .x/ dx _D r1ne .s s0/r1H.r / rne .s s0/rH.r /

Z r1

r

H.x/e .s s0/x_xn0 _dx;

where0indicates differentiation with respect tox. Therefore, if_jH.x/_j M ₁on Œ0;₁/, then

ˇ ˇ ˇ ˇ

Z r1

r

e sxxnf .x/ dx

ˇ ˇ ˇ ˇ

M

e .s s0/r

rn_Ce .s s0/r rn_C

Z 1 r j

.e .s s0/x

/xn/0_jdx

:

Therefore, sincee .s s0/r_rn

decreases monotonically on.n;₁/ifs > s0(check!),

ˇ ˇ ˇ ˇ

Z r1

r

e sxxnf .x/ dx

ˇ ˇ ˇ ˇ

< 3Me .s s0/r

rn; n < r < r1;

so Theorem4implies thatIn.s/converges uniformlyŒs1;1/ifs1 > s0. Now

The-orem11implies thatFnC1 D Fn0, and an easy induction proof yields (30)

Example 13 Here we apply Theorem12withf .x/_D cosax (a _¤ 0) ands0 D 0.

Since

Z x

0

cosau du_D sinax a

is bounded on.0;₁/, Theorem12implies that

F .s/_D

Z 1 0

e sxcosax dx

converges and

F.n/.s/_D. 1/n

Z 1 0

e sxxncosax dx; s > 0: (31)

(Note that this is also true ifa_D0.) Elementary integration yields

F .s/_D s s2_Ca2:

Hence, from (31),

Z 1 0

e sxxncosax_D. 1/n d

n

dsn

s

9

Exercises

1. Supposegandhare differentiable onŒa; b, with

ag.y/b and ah.y/b; cyd:

Letf andfybe continuous onŒa; bŒc; d . DeriveLiebniz’s rule:

d dy

Z h.y/

g.y/

f .x; y/ dx _D f .h.y/; y/h0.y/ f .g.y/; y/g0.y/

C

Z h.y/

g.y/

fy.x; y/ dx:

(Hint: DefineH.y; u; v/_DRv

u f .x; y/ dxand use the chain rule.)

2. Adapt the proof of Theorem2to prove Theorem3.

3. Adapt the proof of Theorem4to prove Theorem5.

4. Show that Definition3is independent ofc; that is, ifRc

a f .x; y/ dxand Rb

c f .x; y/ dx

both converge uniformly onSfor somec₂.a; b/, then they both converge uni-formly onSand everyc₂.a; b/.

5. (a) Show that iff is bounded onŒa; bŒc; d andRb

a f .x; y/ dxexists as a

proper integral for eachy ₂ Œc; d , then it converges uniformly onŒc; d  according to all of Definition1–3.

(b) Give an example to show that the boundedness off is essential in(a).

6. Working directly from Definition1, discuss uniform convergence of the follow-ing integrals:

(a)

Z 1 0

dx

1_Cy2_x2dx (b) Z 1

0

e xyx2dx

(c)

Z 1 0

x2ne yx2dx (d)

Z 1 0

sinxy2dx

(e)

Z 1 0

.3y2 2xy/e y2xdx (f)

Z 1 0

.2xy y2x2/e xydx

7. Adapt the proof of Theorem6to prove Theorem7.

8. Use Weierstrass’s test to show that the integral converges uniformly onS _W

(a)

Z 1 0

e xysinx dx, S_DŒ;₁/, > 0

(b)

Z 1 0

sinx

(c)

Z 1 1

e pxsinxy

x dx, p > 0, S D. 1;1/

(d)

Z 1

0

exy

.1 x/y dx, SD. 1; b/, b < 1

(e)

Z 1

1

cosxy

1_Cx2_y2dx, SD. 1; [Œ;1/, > 0.

(f)

Z 1 1

e x=ydx, S _DŒ;₁/, > 0

(g)

Z 1

1

exye x2dx, S _DŒ ; , > 0

(h)

Z 1 0

cosxy cosax

x2 dx, SD. 1;1/

(i)

Z 1 0

x2ne yx2dx, S_DŒ;₁/, > 0, n_D0,1,2,. . .

9. (a) Show that

.y/_D

Z 1 0

xy 1e xdx

converges ify > 0, and uniformly onŒc; d if0 < c < d <₁.

(b) Use integration by parts to show that

.y/_D .yC1/

y ; y 0;

and then show by induction that

.y/_D .yCn/

y.y_C1/ .y_Cn 1/; y > 0; nD1; 2; 3; : : : :

How can this be used to define .y/in a natural way for ally _¤ 0, 1, 2, . . . ? (This function is called thegamma function.)

(c) Show that .n_C1/_DnŠifnis a positive integer.

(d) Show that

Z 1 0

e stt˛dt_Ds ˛ 1 .˛_C1/; ˛ > 1; s > 0:

10. Show that Theorem8 remains valid with assumption(c) replaced by the as-sumption that_jgx.x; y/jis monotonic with respect toxfor ally2S.

11. Adapt the proof of Theorem8to prove Theorem9.

(a)

Z 1 1

sinxy

xy dx (b) Z 1

2

sinxy logx dx

(c)

Z 1 0

cosxy

x_Cy2 dx (d) Z 1

1

sinxy 1_Cxydx

13. Supposeg; gx and h are continuous on Œa; b/S; and denote H.x; y/ D Rx

a h.u; y/ du; ax < b:Suppose also that

lim

x!b (

sup

y2Sj

g.x; y/H.x; y/_j

)

D0 and

Z b

a

gx.x; y/H.x; y/ dx

converges uniformly onS:Show thatRb

a g.x; y/h.x; y/ dxconverges uniformly

onS.

14. Prove Theorem10for the case wheref _D f .x; y/is continuous on.a; b Œc; d .

15. Prove Theorem11for the case wheref _D f .x; y/is continuous on.a; b Œc; d .

16. Show that

C.y/_D

Z 1

1

f .x/cosxy dx and S.y/_D

Z 1

1

f .x/sinxy dx

are continuous on. ₁;₁/if

Z 1

1j

f .x/_jdx <₁:

17. Supposef is continuously differentiable onŒa;₁/, limx!1f .x/D0, and

Z 1 a j

f0.x/_jdx <₁: Show that the functions

C.y/_D

Z 1 a

f .x/cosxy dx and S.y/_D

Z 1 a

f .x/sinxy dx

are continuous for ally _¤ 0. Give an example showing that they need not be continuous aty_D0.

18. EvaluateF .y/and use Theorem11to evaluateI:

(a) F .y/ _D

Z 1 0

dx

1_Cy2_x2, y ¤ 0; I D Z 1

0

tan 1ax tan 1bx

x dx,

(b) F .y/ _D

Z 1 0

xydx,y > 1; I _D

Z 1 0

xa xb

logx dx, a,b > 1

(c) F .y/ _D

Z 1 0

e xycosx dx, y > 0

I _D

Z 1 0

e ax e bx

x cosx dx, a,b > 0

(d) F .y/ _D

Z 1 0

e xysinx dx, y > 0

I _D

Z 1 0

e ax _e bx

x sinx dx, a,b > 0

(e) F .y/ _D

Z 1 0

e xsinxy dx; I _D

Z 1 0

e x1 cosax

x dx

(f) F .y/ _D

Z 1 0

e xcosxy dx; I _D

Z 1 0

e xsinax x dx

19. Use Theorem11to evaluate:

(a)

Z 1

0

.logx/nxydx, y > 1, n_D0,1,2,. . . .

(b)

Z 1 0

dx

.x2_Cy/nC1 dx, y > 0, nD0,1,2, . . . .

(c)

Z 1 0

x2nC1e yx2dx, y > 0, n_D0,1,2,. . . .

(d)

Z 1 0

xyxdx, 0 < y < 1.

20. (a) Use Theorem11and integration by parts to show that

F .y/_D

Z 1 0

e x2cos2xy dx

satisfies

F0_C2yF _D0:

(b) Use part(a)to show that

F .y/_D

p

2 e

y2 :

21. Show that

Z 1 0

e x2sin2xy dx_De y2

Z y

0

eu2du:

(Hint: See Exercise 20.)

22. State a condition implying that

C.y/_D

Z 1 a

f .x/cosxy dx and S.y/_D

Z 1 a

aren times differentiable on for ally _¤ 0. (Your condition should imply the hypotheses of Exercise16.)

23. Supposef is continuously differentiable onŒa;₁/,

Z 1 a j

.xkf .x//0_jdx <₁; 0kn; and limx!1xnf .x/D0. Show that if

C.y/_D

Z 1 a

f .x/cosxy dx and S.y/_D

Z 1 a

f .x/sinxy dx;

then

C.k/.y/_D

Z 1 a

xkf .x/cosxy dx and S.k/.y/_D

Z 1 a

xkf .x/sinxy dx;

0kn.

24. Differentiating

F .y/_D

Z 1 1

cosy x dx under the integral sign yields

Z 1 1

1 xsin

y x dx;

which converges uniformly on any finite interval. (Why?) Does this imply that Fis differentiable for ally?

25. Show that Theorem11and induction imply Eq. (30).

26. Prove Theorem12.

27. Show that ifF .s/_D R1 0 e

sx_{f .x/ dxconverges fors}

Ds0, then it converges

uniformly onŒs0;1/. (What’s the difference between this and Theorem13?)

28. Prove: Iff is continuous onŒ0;₁/andR1

0 e s0xf .x/ dxconverges, then

lim

s!s0C

Z 1 0

e sxf .x/ dx _D

Z 1 0

e s0x

f .x/ dx:

(Hint: See the proof of Theorem 4.5.12, p. 273.)

29. Under the assumptions of Exercise28, show that

lim

s!s0C

Z 1 r

e sxf .x/ dx_D

Z 1 r

e s0x

30. Supposef is continuous onŒ0;₁/and

F .s/_D

Z 1 0

e sxf .x/ dx

converges fors_Ds0. Show that lims!1F .s/D0. (Hint: Integrate by parts.)

31. (a) Starting from the result of Exercise18(d), letb _{! 1}and invoke Exer-cise30to evaluate

Z 1 0

e axsinx

x dx; a > 0:

(b) Use(a)and Exercise28to show that

Z 1 0

sinx x dxD

2:

32. (a) Supposef is continuously differentiable onŒ0;₁/and

jf .x/_j Mes0x_{; 0}

x ₁:

Show that

G.s/_D

Z 1 0

e sxf0.x/ dx

converges uniformly onŒs1;1/ifs1> s0. (Hint: Integrate by parts.)

(b) Show from part(a)that

G.s/_D

Z 1 0

e sxxex2sinex2dx

converges uniformly onŒ;₁/if > 0. (Notice that this does not follow from Theorem6or8.)

33. Supposef is continuous onŒ0;₁/,

lim

x!0C

f .x/ x

exists, and

F .s/_D

Z 1 0

e sxf .x/ dx

converges fors_Ds0. Show that

Z 1 s0

F .u/ du_D

Z 1 0

10

Answers to selected exercises

5. (b)Iff .x; y/ _D 1=y fory _¤ 0andf .x; 0/ _D 1, thenRb

a f .x; y/ dx does not

converge uniformly onŒ0; d for anyd > 0.

6. (a),(d), and(e)converge uniformly on. ₁; _[Œ;₁/if > 0; (b),(c), and(f)

converge uniformly onŒ;₁/if > 0.

17. LetC.y/ _D

Z 1 1

cosxy

x dxandS.y/ D

Z 1 1

sinxy

x dx. ThenC.0/ D 1and S.0/_D0, whileS.y/_D=2ify_¤0.

18. (a)F .y/ _D

2_jy_j; I D 2 log

a

b (b)F .y/D 1

y_C1; I Dlog a_C1 b_C1

(c)F .y/ _D y

y2_C1; I D

1 2

b2_C1 a2_C1

(d)F .y/_D 1

y2_C1; I Dtan

1_{b tan} 1_a

(e)F .y/ _D y

y2_C1; I D

1

2log.1Ca

2_/

(f)F .y/_D 1

y2_C1; I Dtan 1_a

19. (a). 1/nnŠ.y_C1/ n 1 (b) 2 2n 1 2n n

!

y n 1=2

(c) nŠ

2ynC1 .logy/

2_(d) 1

.logx/2

22.

Z 1

1j

xnf .x/_jdx <₁

24.No; the integral definingF diverges for ally.

31. (a)

2 tan