5.3 Improper Integrals Involving Rational and …md90/PH3066/Asmar 5.3.pdfSection 5.3 Improper...

Section 5.3 Improper Integrals Involving Rational and Exponential Functions 299

11.! 2!

0

d!

1 + a cos !=

2"!1 " a2

, 0 < |a| < 1.

12.! 2!

0

sin2 !

a + b cos !d! =

2"

b2

"a "

#a2 " b2

$, 0 < |b| < a.

13.! 2!

0

d!

a + b cos2 !=

2"!

a!

a + b, 0 < a.

14.! 2!

0

d!

a + b sin2 !=

2"!

a!

a + b, 0 < a.

15.! 2!

0

d!

a cos ! + b sin ! + c=

2"!c2 " a2 " b2

, a2 + b2 < c2.

16.! 2!

0

d!

a cos2 ! + b sin2 ! + c=

2"#(a + c)(b + c)

, 0 < c < a, c < b.

5.3 Improper Integrals Involving Rational and Exponential Functions

In this section we present a useful technique to evaluate improper integralsinvolving rational and exponential functions. Let a and b be arbitrary realnumbers. Consider the integrals

(1)! b

!"f(x) dx and

! "

af(x) dx,

where in each case f is continuous in the interval of integration and atits finite endpoint. These are called improper integrals, because the in-terval of integration is infinite. The integral

" "a f(x) dx is convergent if

limb#"" ba f(x) dx exists as a finite number. Similarly,

" b!" f(x) dx is con-

vergent if lima#!"" ba f(x) dx exists as a finite number.

Now let f(x) be continuous on the real line. The integral" "!" f(x) dx

is also improper since the interval of integration is infinite, but here it isinfinite in both the positive and negative direction. Such an integral is saidto be convergent if both

" "0 f(x) dx and

" 0!" f(x) dx are convergent. In

this case, we set (Figure 1)

300 Chapter 5 Residue Theory

Figure 1 Splitting an im-proper integral over the line.

(2)! "

!"f(x) dx = lim

a#!"

! 0

af(x) dx + lim

b#"

! b

0f(x) dx.

We define the Cauchy principal value of the integral" "!" f(x) dx to be

(Figure 2)

(3) P.V.

! "

!"f(x) dx = lim

a#"

! a

!af(x) dx, if the limit exists.

Figure 2 The Cauchy princi-pal value of the integral.

The Cauchy principal value of an integral may exist even though the integralitself is not convergent. For example,

" a!a x dx = 0 for all a, which implies

that P.V." "!" x dx = 0, but the integral itself is clearly not convergent

since" "0 x dx = !. However, whenever

" "!" f(x) dx is convergent, then

P.V." "!" f(x) dx exists, and the two integrals will be the same. This is

because lima#"" 0!a f(x) dx and lima#"

" a0 f(x) dx both exist, and so

P.V.

! "

!"f(x) dx = lim

a#"

! a

!af(x) dx

= lima#"

#! 0

!af(x) dx +

! a

0f(x) dx

$

= lima#"

! 0

!af(x) dx + lim

a#"

! a

0f(x) dx

=! "

!"f(x) dx.

Because of this fact, we can compute a convergent integral over the realline by computing its principal value, which can often be obtained by use ofcomplex methods and the residue theorem.

The following test of convergence for improper integrals is similar to teststhat we have proved for infinite series. We omit the proof.


PROPOSITION 1INEQUALITIES FOR

IMPROPERINTEGRALS

Let" BA f(x) dx represent an improper integral as in (1), where A = "! or

B = !.(i) If

" BA |f(x)| dx is convergent, then

" BA f(x) dx is convergent and we have%%%

" BA f(x) dx

%%% #" BA |f(x)| dx.

(ii) If |f(x)| # g(x) for all A < x < B and" BA g(x) dx is convergent, then

" BA f(x) dx is convergent and we have

%%%" BA f(x) dx

%%% #" BA g(x) dx.

You are encouraged to use Proposition 1 or any other test from calculus(such as the limit comparison test) to show that an improper integral is con-vergent. Once the convergence is determined, we may compute the integralvia its principal value, as we will illustrate in the examples.

EXAMPLE 1 Improper integrals and residues: the main ideasEvaluate

I =! !

"!

x2

x4 + 1dx.

Solution To highlight the main ideas, we present the solution in basic steps.Step 1: Show that the improper integral is convergent. Because the integrandis continuous on the real line, it is enough to show that the integral outside afinite interval, say ["1, 1], is convergent. For |x| # 1, we have x2

x4+1 $ x2

x4 = 1x2 ,

and since% !1

1x2 dx is convergent, it follows by Proposition 1 that

% !1

x2

x4+1 dx and% "1"!

x2

x4+1 dx are convergent. Thus% !"!

x2

x4+1 dx is convergent, and so

(4)! !

"!

x2

x4 + 1dx = lim

R#!

! R

"R

x2

x4 + 1dx.

Step 2: Set up the contour integral. We will replace x by z and consider the functionf(z) = z2

z4+1 . The general guideline is to integrate this function over a contour thatconsists partly of the interval ["R, R], so as to recapture the integral

% R"R

x2

x4+1 dxas part of the contour integral on #R. Choosing the appropriate contour is notobvious in general. For a rational function, such as x2

x4+1 , where the denominatordoes not vanish on the x-axis and the degree of the denominator is two more thanthe degree of the numerator, a closed semi-circle #R as in Figure 3 will work. Since#R consists of the interval ["R, R] followed by the semi-circle $R, using the additiveproperty of path integrals (Proposition 2(iii), Section 3.2), we write

(5) I"R =!

"R

f(z) dz =!

["R, R]f(z) dz +

!

#R

f(z) dz = IR + I#R .

Figure 3 The path and polesfor the contour integral in Ex-ample 1.

For z = x in ["R, R], we have f(z) = f(x) = x2

x4+1 and dz = dx, and so IR =% R"R

x2

x4+1 dx, which according to (4) converges to the desired integral as R % &.So, in order to compute the desired integral, we must get a handle on the otherquantities, I"R and I#R , in (5). Our strategy is as follows. In Step 3, we computeI"R by the residue theorem; and in Step 4, we show that limR#! I#R = 0. This


will give us the necessary ingredients to complete the solution in Step 5.Step 3: Compute I"R by the residue theorem. For R > 1, the function f(z) = z2

z4+1

has two poles inside #R. These are the roots of z4 + 1 = 0 in the upper half-plane.To solve z4 = "1, we write "1 = ei!; then using the formula for the nth roots fromSection 1.3, we find the four roots

z1 =1 + i!

2, z2 =

"1 + i!2

, z3 ="1 " i!

2, z4 =

1 " i!2

(see Figure 3). Since these are simple roots, we have simple poles at z1 and z2 andthe residues there are (Proposition 1(ii), Section 5.1)

Res (z1) =z2

ddz (z4 + 1)

&&&&&z=z1

=z2

4z3

&&&&z=z1

=14z

&&&&z=z1

=!

24(1 + i)

=1 " i

4!

2,

and similarly

Res (z2) =14z

&&&&z=z2

="1 " i

4!

2.

So, by the residue theorem, for all R > 1

(6) I"R =!

"R

z2

z4 + 1dz = 2"i

"Res (z1) + Res (z2)

$= 2"i

"2i

4!

2=

"!2.

Step 4: Show that limR#! I#R = 0. For z on $R, we have |z| = R, and so&&&&

z2

z4 + 1

&&&& $R2

R4 " 1= MR.

Appealing to the integral inequality (Theorem 2, Section 3.2), we have

|I#R | =&&&&!

#R

z2

z4 + 1dz

&&&& $ l($R)MR = "RR2

R4 " 1=

"

R " 1/R3% 0, as R % &.

Step 5: Compute the improper integral. Using (5) and (6), we obtain

"!2

= IR + I#R .

Let R % &, then IR %% !"!

x2

x4+1 dx and I#R % 0, and so! !

"!

x2

x4 + 1dx =

"!2. !

Figure 4 An alternative pathfor the integral in Example 1.

In Example 1, we could have used the contour ! $R in the lower half-plane

in Figure 4. In this case, it is easiest to take the orientation of ! $R to be

negative in order to coincide with the orientation of the interval ["R, R].The five steps that we used in the solution of Example 1 can be used to

prove a the following geberal result.


PROPOSITION 2INTEGRALS OF

RATIONALFUNCTIONS

Let f(x) = p(x)q(x) be a rational function with degree q(x) $ 2 + degree p(x),

and let "R denote an arc of the circle centered at 0 with radius R > 0. Then

(7) limR#"

%%%%

!

!R

p(z)q(z)

dz

%%%% = 0.

Moreover, if q(x) has no real roots and z1, z2, . . . , zN denote all the polesof p(z)

q(z) in the upper half-plane, then

(8)! "

!"

p(x)q(x)

dx = 2#iN&

j=1

Res'p(z)q(z)

, zj(.

Here is a straightforward application.

EXAMPLE 2 Improper integral of a rational function


Evaluate ! !

"!

1(x2 + 1)(x2 + 4)

dx.

Solution The integrand satisfies the two conditions of Proposition 2: degree q(x) =4 # 2 + degree p(x) = 2, and the denominator q(x) = (x2 + 1)(x2 + 4) has no realroots. So we may apply (8). We have

(z2 + 1)(z2 + 4) = (z + i)(z " i)(z + 2i)(z " 2i),

and hence 1(z2+1)(z2+4) has simple poles at i and 2i in the upper half-plane (Fig-

ure 5). The residues there are

Res (i) = limz#i

(z " i)1

(z " i)(z + i)(z2 + 4)=

1(2i)("1 + 4)

= " i

6

andRes (2i) = lim

z#2i(z " 2i)

1(z2 + 1)(z " 2i)(z + 2i)

=i

12.

According to (8),! !

"!

1(x2 + 1)(x2 + 4)

dx = 2"i"" i

6+

i

12$

="

6. !

Integrals Involving Exponential FunctionsThe success of the method of contour integration depends crucially on thechoice of contours. In the previous examples, we used expanding semi-circular contours. To evaluate integrals involving exponential functions, wewill use rectangular contours. We will consider integrals of the form

! "

!"

eax

ebx + cdx, where a < b, and c > 0.


Since c > 0 and eax > 0, the denominator does not equal zero. Also thecondition a < b guarantees that the integral is convergent (Exercise 28).

EXAMPLE 3 Integral involving exponential functionsLet % be a real number > 1. Establish the identity

(9)! !

0

ex

e$x + 1dx =

"

% sin !$

.

Solution Step 1: As noted before, since % > 1, the integral converges. Theintegrand leads us to consider the function f(z) = ez

e!z+1 , whose poles are the rootsof e$z + 1 = 0. Since the exponential function is 2"i-periodic, then

e$z = "1 = ei! ' %z = i"+2k"i ' zk = (2k+1)"

%i, k = 0, ±1, ±2, . . . .

Thus f(z) = ez

e!z+1 has infinitely many poles at zk, all lying on the imaginary axis(Figure 6).


Step 2: Selecting the contour of integration. Our contour should expand in thex-direction in order to cover the entire x-axis. To avoid including infinitely manypoles on the y-axis, we will not expand the contour in the upper half-plane, as we didwith the semi-circles in the previous examples. Instead, we will use a rectangularcontour #R consisting of the paths #1, #2, #3, and #4, as in Figure 6, and let Ij

denote the path integral of f(z) over #j (j = 1, . . . , 4). As R % &, #R will expandin the horizontal direction, but the length of the vertical sides remains fixed at 2!

$ .To understand the reason for our choice of the vertical length, let us consider I1

and I3. On #1, we have z = x, dz = dx,

(10) I1 =! R

"R

ex

e$x + 1dx,

and so I1 converges to the desired integral I as R % &. On #3, we have z = x+i 2!$ ,

dz = dx, and using the 2"i-periodicity of the exponential function, we get

(11) I3 =! "R

R

ex+i 2"!

e$(x+i 2"! ) + 1

dx = "e2"! i

! R

"R

ex

e$x + 1dx = "e

2"! iI1.

This last equality explains the choice of the vertical sides: They are chosen so thatthe integral on the returning horizontal side #3 is equal to a constant multiple ofthe integral on #1. From here the solution is straightforward.Step 3: Applying the residue theorem. From Step 2, we have only one pole of f(z)inside #R at z0 = !

$ i. Since e$z + 1 has a simple root, this is a simple pole. UsingProposition 1(ii), Section 5.1, we find

Res" ez

e$z + 1,

"

%i$

=e

"! i

%e$ "! i =

e"! i

%ei!= "

e"! i

%,

and so by the residue theorem

(12) I1 + I2 + I3 + I4 =!

"R

ez

e$z + 1dz = 2"i Res

" ez

e$z + 1,

"

%i$

= "2"ie

"! i

%.


Step 4: Show that the integrals on the vertical sides tend to 0 as R % &. For I2,z = R + iy (0 $ y $ 2!

$ ) on #2, hence |e$z| = |e$Rei$y| = e$R,

|e$z + 1| # |e$z|" 1 = e$R " 1 (1

|e$z + 1| $1

e$R " 1,

and so for z on #2

|f(z)| =&&&&

ez

e$z + 1

&&&& $&&eR+iy

&&e$R " 1

=eR

e$R " 1=

1e($"1)R " e"R

.

Consequently, by the ML-inequality for path integrals,

|I2| =&&&&!

"2

ez

e$z + 1dz

&&&& $l(#2)

e($"1)R " e"R=

2!$

e($"1)R " e"R% 0, as R % &.

The proof that I4 % 0 as R % & is done similarly; we omit the details.Step 5: Compute the desired integral (9). Using (10), (11), and (12), we find that

I1

"1 " e

2"! i

$+ I2 + I4 = "2"i

e"! i

%.

Letting R % & and using the result of Step 4, we get

"1 " e

2"! i

$ ! !

"!

ex

e$x + 1dx = "2"i

e"! i

%,

and after simplifying! !

"!

ex

e$x + 1dx =

2"i

%"e

"! i " e"

"! i

$ ="

% sin !$

,

which implies (9). !There are interesting integrals of rational functions that are not com-

putable using semi-circular contours as in Example 1, such as

(13)! "

0

dx

x3 + 1.

This integral can be reduced to an integral invovling expoential functions,by using the substitution x = et. We outline this useful technique in thefollowing example.

EXAMPLE 4 The substitution x = et

Establish the identity

(14)! !

0

1x$ + 1

dx ="

% sin !$

, % any real number > 1.

Solution Step 1: Show that the integral converges. The integrand is continuous,so it is enough to show that the integral converges on [1,&). We have 1

x!+1 $ 1x! ,


and the integral is convergent since% !1

1x! dx < &.

Step 2: Apply the substitution x = et. Let x = et, dx = et dt, and note that as xvaries from 0 to &, t varies from "& to &, and so

I =! !

0

1x$ + 1

dx =! !

"!

et

e$t + 1dt =

! !

"!

ex

e$x + 1dx,

where, for convenience, in the last integral we have used x as a variable of integrationinstead of t. Identity (14) follows now from Example 3. !

Figure 7 Splitting an im-proper integral.

The tricky part in Example 3 is choosing the contour. Let us clarify thispart with one more example. We will compute the integral

" "0

ln xx4+1 dx. The

integral is improper because the interval is infinite and because the integrandtends to "! as x % 0. To define the convergence of such integrals, we followthe general procedure of taking all one-sided limits one at a time. Thus(Figure 7)

! "

0

ln x

x4 + 1dx = lim

"%0

! 1

"

ln x

x4 + 1dx + lim

b#"

! b

1

ln x

x4 + 1dx.

It is not di!cult to show that both limits exist and hence that the integralconverges. We will use the substitution x = et to solve the problem. If youlike, you could instead check the convergence after changing variables.

EXAMPLE 5 An integral involving lnxDerive

(15)! !

0

lnx

x4 + 1dx = " "2

8!

2.

Solution Let x = et, lnx = t, dx = etdt. This transforms the integral into! !

"!

t

e4t + 1et dt =

! !

"!

xex

e4x + 1dx,


where we have renamed the variable x, just for convenience. In evaluating thisintegral, we will integrate f(z) = zez

e4z+1 over the rectangular contour #R in Figure 8,and let Ij denote the integral of f(z) on #j . Here again, we chose the vertical sidesof #R so that on the returning path #3 the denominator equals to e4x + 1. As wewill see momentarily, this will enable us to relate I3 to I1. Let us now computeI"R =

%"R

f(z) dz. As you can check, f(z) has one (simple) pole at z = i!4 inside

#R. By Proposition 1(ii), Section 5.1, the residue there is

Res" zez

e4z + 1, i

"

4$

=i!4 ei "

4

4ei!= " i"(1 + i)

16!

2.

So by the residue theorem

(16) I"R = 2"i

'" i"(1 + i)

16!

2

(=

"2(1 + i)8!

2= I1 + I2 + I3 + I4.


Moving to each Ij (j = 1, . . . , 4), we have

I1 =!

"1

zez

e4z + 1dz =

! R

"R

xex

e4x + 1dx.

On #3, z = x + i!2 , dz = dx, so using e

i"2 = i, we get

I3 =!

"3

zez

e4z + 1dz =

! "R

R

(x + i!2 )exei "

2

e4x + 1dx

= "i

! R

"R

xex

e4x + 1dx +

"

2

! R

"R

ex

e4x + 1dx

= "iI1 +"

2

! R

"R

ex

e4x + 1dx.

To show that I2 and I4 tend to 0 as R % &, we proceed as in Step 5 of Example 3.For z = R + iy (0 $ y $ !

2 ) on #2, we have |z| $ R + y $ R + !2 , and so, as in

Example 3,

|f(z)| =&&&&

zez

e4z + 1

&&&& = |z|&&&&

ez

e4z + 1

&&&& $ (R +"

2)

eR

e4R " 1=

R + !2

e3R " e"R.

Consequently, by the ML-inequality for path integrals

|I2| =&&&&!

"2

zez

e4z + 1dz

&&&& $l(#2)(R + !

2 )e3R " e"R

=!2 (R + !

2 )e3R " e"R

% 0, as R % &.

The proof that I4 % 0 as R % & is done similarly; we omit the details. Substitutingour finding into (16) and taking the limit as R % &, we get

"2(1 + i)8!

2= lim

R#!

)I1 + I2 " iI1 +

"

2

! R

"R

ex

e4x + 1dx + I4

*

= (1 " i)! !

"!

xex

e4x + 1dx +

"

2

! !

"!

ex

e4x + 1dx.

Taking imaginary parts of both sides, we obtain our answer%!"!

xex

e4x+1 dx = " !2

8$

2.

If we take real parts of both sides we get the value of the integral in (14) thatcorresponds to % = 4. !

The substitution x = et is also useful even when we do not use complexmethods to evaluate the integral in t. For example,

! "

0

ln x

x2 + 1dx =

! "

!"

tet

e2t + 1dt = 0

since the integral is convergent and the integrand tet

e2t+1 = tet+e!t is an odd

function of t.

Exercises 5.3


In Exercises 1–8, evaluate the given improper integral by the method of Example 1.

1.! !

"!

dx

x4 + 1=

"!2

2.! !

"!

dx

x4 + x2 + 1=

"!3

3.! !

"!

dx

(x2 + 1)(x4 + 1)=

"

24.

! !

"!

dx

(x " i)(x + 3i)=

"

2

5.! !

"!

dx

(x2 + 1)3=

3"

86.

! !

"!

dx

(x4 + 1)2=

3"

4!

2

7.! !

"!

dx

(4x2 + 1)(x " i)=

"

3i 8.

! !

"!

dx

(x + i)(x " 3i)=

"

2In Exercises 9–12, evaluate the given improper integral by the method of Example 3.

9.! !

"!

ex

e3x + 1dx =

2"

3!

310.

! !

"!

13ex + e"x

dx ="

2!

3

11.! !

"!

xex

e2x + 1dx = 0 12.

! !

"!

eax

ebx + 1dx =

"

b sin !ab

(0 < a < b)

In Exercises 13–20, evaluate the given improper integral by the method of Example 5.In some cases, the integral follows from more general formulas that we derivedearlier. You may use these formulas to verify your answer.

13.! !

0

dx

x3 + 1=

2"

3!

314.

! !

"!

dx

(1 + x2)n+1=

(2n)!22n(n!)2

"

15.! !

0

x

x5 + 1dx =

"

5 sin"

2!5

$ 16.! !

0

x

x$ + 1dx =

"

% sin"

2!$

$ (% > 2)

17.! !

0

!x

x3 + 1dx =

"

318.

! !

0

x$

(x + 1)2dx =

"%

sin "%("1 < % < 1)

19.! !

0

x2ex

e2x + 1dx =

"3

1620.

! !

0

x ln(2x)x3 + 1

dx =2"2

27+

2" ln 23!

3

21.! !

0

ln(2x)x2 + 4

dx ="

2ln2 22.

! !

0

ln(ax)x2 + b2

dx ="

2bln(ab) (a, b > 0)

23.! !

0

"ln x

$2

x2 + 1dx =

"3

824.

! !

0

(lnx)2

x3 + 1dx =

10 "3

81!

3

Figure 9 The path and polesfor the contour integral in Ex-ercise 21.

25. Use the contour #R in Figure 9 to evaluate! !

0

1x3 + 1

dx.

[Hint: I"R = I1 + I2 + I3; I3 = "ei 2"3 I1; and I2 % 0 as R % &.]

26. Follow the steps in Example 1 to prove Proposition 2.

27. A property of the gamma function. (a) Show that for 0 < % < 1,! !

0

x$"1

1 + xdx =

"

sin "%.

5.3 Improper Integrals Involving Rational and …md90/PH3066/Asmar 5.3.pdfSection 5.3 Improper...

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