32.15
Evaluate
Hence,
32.7
For positive p, show that
converges.
IMPROPER INTEGRALS
261
By Problem 32.6,
For
Hence,
converges. Now let us consider
By the reduction formula of Problem 28.42,
Hence, the question eventually reduces to the case of
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
(Note that we used L'Hopital's rule to show
Hence,
So,
sufficiently large x. Thus, for some x 0 , if x ^ x a , (In x)
p < x,
1 /(In x)
p > 1 Ix.
Hence, the integral must be divergent for arbitrary
32.9
32.10
Show that
show that
is divergent for p < 1.
For x > e, (In x)
p < In x,
Evaluate
32.11
32.12
Evaluate
and, therefore, l/(ln xY s 1/ln x. Now apply Problems 32.8 and 32.9.
But,
Hence,
Then
Hence,
Let
32.13
Evaluate
cos x dx.
By Problem 28.9,
Hence,
since
and
32.14
Evaluate J 0 " e~
x dx.
If
f(x) dx = +<*> and gW s/(*) for all A: >; x 0 .
g(x) dx is divergent.
g(x)dx =
g(x) dx + g(x) dx > g(x) dx + f(x)dx->+*.
e~" cos AC dx = \e "'(sin x — cos x).
e * cos x dx = lim [ | e *(sin A: —
cos x) = lim |[e "(cosy-sine;)-(-!)]= i,
P<1.
P<1.