Part II: The Answers
590
Answers
901–1,001
990.
divergent, compare to 1
1 x
dx
∞
∫
The given improper integral is
x
x
dx
+
−
∞
∫
2
1
4
1
. Notice that on the interval [1, ∞], you have
0 ≤ x
4
– 1 ≤ x
4
, so it follows that 0
1
4
4
≤
− ≤
x
x and that
1
1
1
4
4
x
x
−
≥
. Also,
because x + 2 > x, you get x
x
x
x
x
x
x
+
−
≥
=
=
2
1
1
4
4
2
. Therefore, it follows that
x
x
dx
x
x
dx
x
x
dx
x
dx
+
−
≥
=
=
≥
∞
∞
∞
∞
∫
∫
∫
∫
2
1
1
0
4
1
4
1
2
1
1
. An integral of the form
1
1 x
dx
p
∞
∫
converges if and only if p > 1, so 1
1 x
dx
∞
∫
is divergent (you can also show this
directly by using a limit to write the integral and evaluating). Therefore, because
x
x
dx
x
dx
+
−
≥
≥
∞
∞
∫
∫
2
1
1
0
4
1
1
, you can conclude that
x
x
dx
+
−
∞
∫
2
1
4
1
is divergent.
991.
convergent, compare to π
2
1
5
1 x
dx
∞
∫
The given improper integral is tan
−
∞
∫
1
5
1
x
x
dx. Notice that 0
2
1
≤
≤
−
tan x π on the interval
[1, ∞), so 0
2
1
5
5
≤
≤
−
tan x
x
x
π
. Note that you could actually bound tan
−1
x below by π
4
,
but zero also works and makes the inequalities cleaner.
It follows that 0
2
2
1
1
5
1
5
1
5
1
≤
≤
=
−
∞
∞
∞
∫
∫
∫
tan x
x
dx
x
x
dx
π
π
. An integral of the form
1
1 x
dx
p
∞
∫
converges if and only if p > 1; because
1
5
1 x
dx
∞
∫
is convergent, π
2
1
5
1 x
dx
∞
∫
is also convergent
(you can also show this directly by using a limit to write the integral and evaluating).
Because 0
2
2
1
1
5
1
5
1
5
1
≤
≤
=
−
∞
∞
∞
∫
∫
∫
tan x
x
dx
x
x
dx
π
π
, you can conclude that tan
−
∞
∫
1
5
1
x
x
dx is also
convergent.
992.
divergent, compare to 1
1 x
dx
∞
∫
The given improper integral is
x
x
dx
2
6
2
1
−
∞
∫
. Notice that on the interval [2, ∞), you have
x
x
x
x
x
x
x
2
6
2
6
2
3
1
1 0
−
≥
=
= ≥ . Therefore, you know that
x
x
dx
x
dx
2
6
2
2
1
1
−
≥
∞
∞
∫
∫
. The integral
1
1 x
dx
p
∞
∫
diverges if and only if p ≤ 1, so 1
2 x
dx
∞
∫
also diverges (note that the lower
limit of 2 instead of 1 does not affect the divergence). Therefore, because
x
x
dx
x
dx
2
6
2
2
1
1
−
≥
∞
∞
∫
∫
, you can conclude that
x
x
dx
2
6
2
1
−
∞
∫
also diverges.
590
Answers
901–1,001
990.
divergent, compare to 1
1 x
dx
∞
∫
The given improper integral is
x
x
dx
+
−
∞
∫
2
1
4
1
. Notice that on the interval [1, ∞], you have
0 ≤ x
4
– 1 ≤ x
4
, so it follows that 0
1
4
4
≤
− ≤
x
x and that
1
1
1
4
4
x
x
−
≥
. Also,
because x + 2 > x, you get x
x
x
x
x
x
x
+
−
≥
=
=
2
1
1
4
4
2
. Therefore, it follows that
x
x
dx
x
x
dx
x
x
dx
x
dx
+
−
≥
=
=
≥
∞
∞
∞
∞
∫
∫
∫
∫
2
1
1
0
4
1
4
1
2
1
1
. An integral of the form
1
1 x
dx
p
∞
∫
converges if and only if p > 1, so 1
1 x
dx
∞
∫
is divergent (you can also show this
directly by using a limit to write the integral and evaluating). Therefore, because
x
x
dx
x
dx
+
−
≥
≥
∞
∞
∫
∫
2
1
1
0
4
1
1
, you can conclude that
x
x
dx
+
−
∞
∫
2
1
4
1
is divergent.
991.
convergent, compare to π
2
1
5
1 x
dx
∞
∫
The given improper integral is tan
−
∞
∫
1
5
1
x
x
dx. Notice that 0
2
1
≤
≤
−
tan x π on the interval
[1, ∞), so 0
2
1
5
5
≤
≤
−
tan x
x
x
π
. Note that you could actually bound tan
−1
x below by π
4
,
but zero also works and makes the inequalities cleaner.
It follows that 0
2
2
1
1
5
1
5
1
5
1
≤
≤
=
−
∞
∞
∞
∫
∫
∫
tan x
x
dx
x
x
dx
π
π
. An integral of the form
1
1 x
dx
p
∞
∫
converges if and only if p > 1; because
1
5
1 x
dx
∞
∫
is convergent, π
2
1
5
1 x
dx
∞
∫
is also convergent
(you can also show this directly by using a limit to write the integral and evaluating).
Because 0
2
2
1
1
5
1
5
1
5
1
≤
≤
=
−
∞
∞
∞
∫
∫
∫
tan x
x
dx
x
x
dx
π
π
, you can conclude that tan
−
∞
∫
1
5
1
x
x
dx is also
convergent.
992.
divergent, compare to 1
1 x
dx
∞
∫
The given improper integral is
x
x
dx
2
6
2
1
−
∞
∫
. Notice that on the interval [2, ∞), you have
x
x
x
x
x
x
x
2
6
2
6
2
3
1
1 0
−
≥
=
= ≥ . Therefore, you know that
x
x
dx
x
dx
2
6
2
2
1
1
−
≥
∞
∞
∫
∫
. The integral
1
1 x
dx
p
∞
∫
diverges if and only if p ≤ 1, so 1
2 x
dx
∞
∫
also diverges (note that the lower
limit of 2 instead of 1 does not affect the divergence). Therefore, because
x
x
dx
x
dx
2
6
2
2
1
1
−
≥
∞
∞
∫
∫
, you can conclude that
x
x
dx
2
6
2
1
−
∞
∫
also diverges.