Evaluate . y ⫺⬁ 2 dx

EXAMPLE 3 Evaluate . y ⫺⬁ 2 dx

1⫹x SOLUTION It’s convenient to choose a苷 0 in Definition 1(c):

1⫹x 2 y ⫺⬁ 1⫹x 2 y 0 1⫹x 2

We must now evaluate the integrals on the right side separately:

2 dx 苷 lim y 0 2 lim tan tl⬁ x 0 1⫹x tl⬁ 1⫹x ]

苷 lim 共tan ⫺1 t⫺ tan ⫺1 0 兲 苷 lim tan ⫺1

SECTION 7.8 IMPROPER INTEGRALS

y 0 ⫺⬁ 1⫹x 2 dx 苷 lim t l t 2 ⫺⬁ 苷 y t lim ⫺⬁ tan 1⫹x x l ] t

l ⫺⬁ 共tan

Since both of these integrals are convergent, the given integral is convergent and

1 y= y 1+≈ ⬁

y ⫺⬁ 2 dx 苷

area =π

1⫹x

Since 1 0 2 x 兾共1 ⫹ x 兲⬎0 , the given improper integral can be interpreted as the area of the infinite region that lies under the curve y苷 1 兾共1 ⫹ x 2 兲 and above the -axis (see x

FIGURE 6

Figure 6).

EXAMPLE 4 For what values of is the integral p

1 x p dx convergent?

SOLUTION We know from Example 1 that if p苷 1 , then the integral is divergent, so let’s

assume that p苷 1 . Then

t ⬁ y 1 x x dx l

1 dx 苷 lim

⫺p

l ⫺p ⫹ 1 册 x苷 1

1⫺p 冋 t 册

1 1 苷 lim t ⬁ p⫺ 1 ⫺1

If p⬎ 1 , then p⫺ 1⬎0 , so as

, t p⫺ tl⬁ 1 l⬁ and . 1 p⫺ 兾t 1 l0 Therefore

and so the integral converges. But if p⬍ 1 , then p⫺ 1⬍0 and so

1 苷 t p⫺ 1⫺p l⬁

as t l ⬁

and the integral diverges.

We summarize the result of Example 4 for future reference:

1 dx x is convergent if p ⬎ 1 and divergent if p 艋 1. p

TYPE 2: DISCONTINUOUS INTEGRANDS

Suppose that is a positive continuous function defined on a finite interval f 关a, b兲 but has

a vertical asymptote at . Let be the unbounded region under the graph of and above b S f the -axis between and . (For Type 1 integrals, the regions extended indefinitely in a x a b

CHAPTER 7 TECHNIQUES OF INTEGRATION

horizontal direction. Here the region is infinite in a vertical direction.) The area of the part

of between and (the shaded region in Figure 7) is S a t

y=ƒ

x=b

If it happens that ⫺ A approaches a definite number A as tlb , then we say that the

area of the region is S A and we write

FIGURE 7

We use this equation to define an improper integral of Type 2 even when is not a posi- f

tive function, no matter what type of discontinuity has at . f b

3 DEFINITION OF AN IMPROPER INTEGRAL OF TYPE 2

(a) If is continuous on f and is discontinuous at , then b

N Parts (b) and (c) of Definition 3 are illustrated

in Figures 8 and 9 for the case where t f b

f and has vertical asymptotes at and , f a c y a t l b ⫺ y a

respectively.

if this limit exists (as a finite number).

(b) If is continuous on f and is discontinuous at , then a

a f ⫹ y tla t f

if this limit exists (as a finite number).

The improper integral b x

a f is called convergent if the corresponding limit

0 at

exists and divergent if the limit does not exist.

(c) If has a discontinuity at , where c f c a⬍c⬍b , and both

x a f and

FIGURE 8

x c f are convergent, then we define

EXAMPLE 5 Find . y 2 dx

sx ⫺ 2

SOLUTION We note first that the given integral is improper because f sx ⫺ 2 has the vertical asymptote x苷 2 . Since the infinite discontinuity occurs at the left end-

FIGURE 9

point of

, we use part (b) of Definition 3:

2 t lim l 2 ⫹ y sx ⫺ 2 y t sx ⫺ 2

2 sx ⫺ 2 ] t

苷 lim

2 ⫹ 2 ( s3 st ⫺ 2 )

area =2œ„3

苷2 s3

Thus the given improper integral is convergent and, since the integrand is positive, we

FIGURE 10

can interpret the value of the integral as the area of the shaded region in Figure 10.

SECTION 7.8 IMPROPER INTEGRALS

V ␲ EXAMPLE 6 Determine whether y

0 sec x dx converges or diverges. SOLUTION Note that the given integral is improper because lim

⫺ sec x 苷 ⬁ . Using part (a) of Definition 3 and Formula 14 from the Table of Integrals, we have

y 0 sec x dx t l lim ⫺ ln sec x ⫹ tan x 0

0 sec x dx 苷 t l lim ⫺

苷 t lim

because ⫺ sec t l ⬁ and tan t l ⬁ as . Thus the given improper integral is divergent.

3 dx

EXAMPLE 7 Evaluate if y 0 possible.

x⫺ 1

SOLUTION Observe that the line x苷 1 is a vertical asymptote of the integrand. Since it occurs in the middle of the interval

, we must use part (c) of Definition 3 with c苷 1 :

3 y dx

3 dx

1 dx

x⫺ 1 0 x⫺ 1 y 1

x⫺ 1

where t y

1 dx

苷 t dx

0 t x⫺ lim 1 l 1 ⫺ y 苷 0 t lim 1 ⫺ ln x⫺ 1 x⫺ 0 1 l

苷 lim t

l 1 ⫺ ( ln t⫺ 1 ⫺ ln ⫺1 )

苷 lim t ⫺ ln

because 1 1⫺tl0 as tl1 . Thus x

0 dx

is divergent. This implies that

is divergent. [We do not need to evaluate x 1 dx

0 dx

| WARNING If we had not noticed the asymptote x苷 1 in Example 7 and had instead confused the integral with an ordinary integral, then we might have made the following erroneous calculation:

3 dx

0 苷 ln x⫺ 1 0 苷 x⫺ ln 2 ⫺ ln 1 苷 ln 2 1 This is wrong because the integral is improper and must be calculated in terms of limits.

From now on, whenever you meet the symbol b x

a f you must decide, by looking at the function f on

, whether it is an ordinary definite integral or an improper integral.