Solutions 6
Integral Convergence
1.
We have
converge.
Zx
1
Z∞
cos t dt = sin x − sin 1 . Since lim sin x does not exist,
x→ ∞
cos x dx does not
1
2. Since all the integrands are positive functions we can use the integral comparison test.
Z∞
x3
1
x3
1
(i) For x ≥ 1, we have 4
≥ 4
=
≥ 0 . Since
dx diverges,
3
4
4
x + 2x + 7
x + 2x + 7x
10x
x
1
Z∞
x3
dx diverges by the comparison test for integrals.
x4 + 2x3 + 7
1
tan−1 x
π
(ii) For x ≥ 1 we have 0 ≤
≤ 2 . Since
2
x
2x
by the comparison test for integrals.
(iii)
Z∞
π
dx converges,
2x2
1
Z∞
tan−1 x
dx converges
x2
1
x2
1
x2
≥ 3 = ≥ 0 . Since
For x ≥ 2 we have 3
x −1
x
x
Z∞
1
dx diverges,
x
2
Z∞
2
x2
dx diverges
x3 − 1
by the comparison test for integrals.
(iv) For x ≥ 1, we have 0 ≤ exp(−x4 ) ≤ exp(−x) as exp(−x) is a decreasing function. Since
Z∞
Z∞
−x
exp(−x4 )dx converges by the comparison test for integrals.
e dx converges,
1
1
3. In this question we use the integral absolute convergence test with the integral comparison
test.
Z∞
Z∞
cos x 1
cos x
1
(i) For x ≥ 1 we have 3 ≤ 3 . Since
dx converges,
dx converges by the
3
x
x
x
x3
1
1
absolute convergence and comparison tests for integrals.
Z∞
Z∞
sin x 1
sin x
1
(ii) For x ≥ 1 we have 4/3 ≤ 4/3 . Since
dx
converges,
dx converges by
x
x
x4/3
x4/3
1
1
the absolute convergence and comparison tests for integrals.
Z∞
Z∞
cos x 1
cos x
1
≤
(iii)
For x ≥ 1 we have . Since
dx converges,
dx
100
100
100
1+x
x
x
1 + x100
1
converges by the absolute convergence and comparison tests for integrals.
4. (i) As t−(α+1) sin t ≤ t−(α+1) for all t ≥ 1 and
test and absolute convergence test for integrals
Z∞
1
Z∞
1
t−(α+1) dt converges, by the comparison
1
t−(α+1) sin t dt converges.
(ii)
Integrating by parts, I(x) =
Zx
t−α cos t dt = x−α sin x − sin 1 + α
1
Since α > 0 ,
lim x−α sin x = 0 and
x→ ∞
Z∞
Zx
t−(α+1) sin t dt .
1
t−(α+1) sin t dt converges by part (i) lim I(x) exists,
x→ ∞
1
so
Z∞
x−α cos x dx converges.
1
Zx
Using the substitution u = tp , J(x) =
(iii)
cos(tp ) dt = p−1
and α =
p−1
. Since α > 0, by part (ii)
p
u−α cos u du where y = xp
1
1
Z∞
Zy
u−α cos u du converges. Thus lim J(x) exists,
x→ ∞
1
so
Z∞
cos(xp ) dx converges.
1
5. The first step in each of the following is to determine where the integrand is unbounded.
e
ex
≤√
. Consider
(i) The problem occurs at 1. For 0 ≤ x < 1, we have 0 ≤ √
2
1−x
1 − x2
Z1
0
1
√
dx =
1 − x2
Z1
0
1
p
u(2 − u)
du
p
R1 √
√
via the substitution u = 1 − x. Now 1/ u(2 − u) ≤ 1/ u for u ∈ [0, 1], and 0 1/ u du
Z1
1
√
converges. Therefore
dx converges by the comparison test for integrals, and hence
1 − x2
0
Z1
ex
dx converges too, by the comparison test for integrals.
1 − x2
0
Z2π
cos x 1
1
√ dx
(ii) The problem occurs at 0. For 0 < x ≤ 2π, we have √ ≤ √ . Since
x
x
x
√
0
converges,
Z2π
cos x
√ dx converges by the absolute convergence and comparison tests for integrals.
x
0
6.
As 0 ≤ f (x) ≤ 3x−1/2 for all x ∈ (0, 1] and
by the integral comparison test.
As 0 ≤ x−1 ≤ (f (x))2 for all x ∈ (0, 1] and
integral comparison test.
Z1
0
Z1
3x−1/2 dx converges
0
x−1 dx diverges
Z1
f (x) dx convereges
0
Z1
0
(f (x))2 dx diverges by the
1
is decreasing with
x(log x)3
∞
X
1
converges if and
f (x) ≥ 0 for all x ≥ 1. Hence by the integral test for series
n(log n)3
n=3
x
Z∞
Z∞
Zx
1
1
1
−1
only if
so
dt converges. Now
dt =
dt converges.
3
3
2
t(log t)
t(log t)
2(log t) 3
t(log t)3
7.
(i)
The function f : [3, ∞] → R defined by f (x) =
3
Hence
∞
X
n=3
3
3
1
converges.
n(log n)3
1
is decreasing with
x log x log(log x)
f (x) ≥ 0 for all x ≥ 1. Hence by the integral test for series
Z∞
∞
X
1
1
converges if and only if
dt converges. Now
n
log
n
log(log
n)
t
log
t
log(log
t)
n=3
(ii)
Zx
The function f : [3, ∞] → R defined by f (x) =
3
1
dt = [log(log(log t))]x3 so
t log t log(log t)
3
3
∞
X
n=3
Z∞
1
diverges.
n log n log(log n)
1
dt diverges. Hence
t log t log(log t)