CHAPTER 32
Improper Integrals
32.1 Determine whether the area in the first quadrant under the curve y = l/x, for *£!, is finite.
This is equivalent to determining whether the improper integral J* (1 Ix) dx is convergent. J* (1 Ix) dx =
Thus, the integral diverges and the area is infinite.
32.2 Determine whether J" (1 Ix2) dx converges.
Thus, the integral converges.
32.3 For what values of p is J" (1 /x)p dx convergent?
By Problem 32.1, we know that the integral is divergent when p = 1.
The last limit is l/(p-l) if p>l, and+=° if p<l.Thus, the integral converges if and only if p > 1.
32.4 For p>l, is dx convergent?
p
First we evaluate J [(In x)/xp] dx by integration by parts. Let u = lnx, dv = (l/* ) dx, du = (\lx)dx.
Thus, Hence,
I In the last step, we used L'Hopital's rule to evaluate
Thus, the integral converges for all p > 1.
32.5 For is convergent?
for Hence, by Problem 32.3. Hence, is
divergent for p :£ 1.
32.6 Evaluate £ xe~'dx.
By integration by parts, we find J xe * dx = -e *(x + 1) Hence, J
[In the last step, we used L'Hopital's rule to evaluate
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, IMPROPER INTEGRALS 261
32.7 For positive p, show that converges.
By Problem 32.6, For Hence,
converges. Now let us consider By the reduction formula of Problem 28.42,
Hence,
(Note that we used L'Hopital's rule to show
Hence, the question eventually reduces to the case of P<1. Thus, we have
convergence for all positive p.
32.8 Is convergent when p a 1?
By successive applications of L'Hopital's rule, we see that Km (In x)p/x = 0. Hence, (In x)"lx < 1 for
p p
sufficiently large x. Thus, for some x0, if x ^ xa, (In x) < x, 1 /(In x) > 1 Ix. So,
Hence, the integral must be divergent for arbitrary P<1.
32.9 If f(x) dx = +<*> and gW s/(*) for all A: >; x0. show that g(x) dx is divergent.
g(x)dx = g(x) dx + g(x) dx > g(x) dx + f(x)dx->+*.
32.10 Show that is divergent for p < 1.
For x > e, (In x)p < In x, and, therefore, l/(ln xY s 1/ln x. Now apply Problems 32.8 and 32.9.
32.11 Evaluate
But, Hence,
32.12 Evaluate
Let Then Hence,
32.13 Evaluate cos x dx.
By Problem 28.9, e~" cos AC dx = \e "'(sin x — cos x). Hence, e * cos x dx = lim [ | e *(sin A: —
cos x) = lim |[e "(cosy-sine;)-(-!)]= i, since and
32.14 Evaluate J0" e~x dx.
32.15 Evaluate
Hence,
Improper Integrals
32.1 Determine whether the area in the first quadrant under the curve y = l/x, for *£!, is finite.
This is equivalent to determining whether the improper integral J* (1 Ix) dx is convergent. J* (1 Ix) dx =
Thus, the integral diverges and the area is infinite.
32.2 Determine whether J" (1 Ix2) dx converges.
Thus, the integral converges.
32.3 For what values of p is J" (1 /x)p dx convergent?
By Problem 32.1, we know that the integral is divergent when p = 1.
The last limit is l/(p-l) if p>l, and+=° if p<l.Thus, the integral converges if and only if p > 1.
32.4 For p>l, is dx convergent?
p
First we evaluate J [(In x)/xp] dx by integration by parts. Let u = lnx, dv = (l/* ) dx, du = (\lx)dx.
Thus, Hence,
I In the last step, we used L'Hopital's rule to evaluate
Thus, the integral converges for all p > 1.
32.5 For is convergent?
for Hence, by Problem 32.3. Hence, is
divergent for p :£ 1.
32.6 Evaluate £ xe~'dx.
By integration by parts, we find J xe * dx = -e *(x + 1) Hence, J
[In the last step, we used L'Hopital's rule to evaluate
260
, IMPROPER INTEGRALS 261
32.7 For positive p, show that converges.
By Problem 32.6, For Hence,
converges. Now let us consider By the reduction formula of Problem 28.42,
Hence,
(Note that we used L'Hopital's rule to show
Hence, the question eventually reduces to the case of P<1. Thus, we have
convergence for all positive p.
32.8 Is convergent when p a 1?
By successive applications of L'Hopital's rule, we see that Km (In x)p/x = 0. Hence, (In x)"lx < 1 for
p p
sufficiently large x. Thus, for some x0, if x ^ xa, (In x) < x, 1 /(In x) > 1 Ix. So,
Hence, the integral must be divergent for arbitrary P<1.
32.9 If f(x) dx = +<*> and gW s/(*) for all A: >; x0. show that g(x) dx is divergent.
g(x)dx = g(x) dx + g(x) dx > g(x) dx + f(x)dx->+*.
32.10 Show that is divergent for p < 1.
For x > e, (In x)p < In x, and, therefore, l/(ln xY s 1/ln x. Now apply Problems 32.8 and 32.9.
32.11 Evaluate
But, Hence,
32.12 Evaluate
Let Then Hence,
32.13 Evaluate cos x dx.
By Problem 28.9, e~" cos AC dx = \e "'(sin x — cos x). Hence, e * cos x dx = lim [ | e *(sin A: —
cos x) = lim |[e "(cosy-sine;)-(-!)]= i, since and
32.14 Evaluate J0" e~x dx.
32.15 Evaluate
Hence,
