Extending the Concept of the Integral

FIGURE 11.2 J* f (x) dx = J* f (*) dx + J" f (x) dx + jj f (*) dx.

f{x) in [a, c] by the continuous function f(x) that is equal to f(x) throughout [a, c) and has the value f] (c) = lim^c- f(x) at x — c. Then f (x) dx is well defined and it is reasonable to define J^ f(x)dx = Jac f(x) dx. By a similar trick, we define J^ f(x)dx and f(x) dx by considering continuous functions in the intervals [c, d] and [d,b], respectively, that are equal to / except at one or both of the end points. The only sensible definition now is

í f(x)dx= í f(x)dx+ Í f(x)dx+ [ f(x)dx Ja Ja Jc Jd

Then the interpretation of f(x)dx is simply the sum of the three areas in Fig. 11.2. This should make clear how f(x) dx can be defined for all functions f(x) that are piecewise continuous on [a, b].

Infinite Intervals of Integration

Suppose / is a function that is continuous for all x > a. Then f(x)dx is defined for each b > a. If the limit of this integral as b -»■ oo exists (and is finite), then we say that / is integrable over [a, oo), and define

The improper integral Ja°° f(x) dx is then said to converge. If the limit does not exist, the improper integral is said to diverge. If f(x) > 0 in [a, oo), we interpret the integral [11.6] as the area below the graph of / over the interval [a. oo). Analogously, we define

J-oo J a when / is continuous in (—00, b]. If this limit exists, the improper integral is said to converge. Otherwise, it diverges.

Example 11.7

The exponential distribution in statistics is defined by fix) = Xe"

Show that the area below the graph of / over [0, 00) is equal to 1. (See Fig. 11.3.)

Solution For b > 0, the area below the graph of / over [0, b] is equal to f

/ W** dx — lim I ke-'" dx = lim (-e~^ + l) = 1 JO b->ooJ o b—oo

Example 11.8

Show that

Then study the case a < 1. Solution For a^l and b > 1, f —dx= f x~adx =

J1 J1

FIGURE 11.3 Area A has an unbounded base, but the height decreases to 0 so rapidly that the total area is 1.

FIGURE 11.4 "A = 1/jf) dx = oc." 1 /x does not approach 0 fast enough, so the improper integral diverges.

For a > 1, one has bl~a = 1 /ba~} —► 0 as b oo. Hence, [1] follows from [2] by letting b ^ oo.

For ¿z = l, we have J^(l/x)dx = Inb - In 1 = ln£, which tends to oo as b tends to oo, so J]00(l/x)dx diverges. See Fig. 11.4.

For a < 1, the last expression in [2] tends to oo as b tends to oo. Hence, the integral diverges in this case.

If both limits of integration are infinite, the improper integral of a continuous function / on (—oo, oo) is defined by

If both integrals on the right-hand side converge, the improper integral f(x) dx is said to converge; otherwise, it diverges. Instead of using 0 as the point of subdivision, one could use an arbitrary fixed real number c. The value assigned to the integral will always be the same, provided that the integral does converge.

It is important to note that definition [11.8] requires both integrals on the right-hand side to converge. Note in particular that f"

b->oc J_b is not the definition of J^ f(x) dx. Problem 4 provides an example in which [*] exists, yet the integral in [11.8] diverges. So [*] is not an acceptable definition, whereas [11.8] is.

Example 11.9

For c > 0, examine the convergence of

Solution Let us begin with the indefinite integral f xe cx~ dx. Making the substitution u = —cx2, we have du = —2cx dx and so f If 1 1

/ xe-"" dx = / e" du = -—e" + C = -—e"«' + C J 2c J 2c 2c

According to [11.8], provided both integrals on the right side exist, one has

/oo cQ fOC

But now

/* _ , _ , 0 1 _ , 1 / xe cx dx = lim xe cx dx = lim--e cx = ——

In the same way, we see that the second integral in [*] is 1 /2c, so r°° 11

(This result is very important in statistics. See Problem 13.)

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