Math 414
Professor Lieberman
April 28, 2003
HOMEWORK #12 SOLUTIONS
Section 6.4
5. (c) Integrate by parts (with u = sin(ln x) and dv = dx) to obtain
Z
Z
Z
cos(ln x)
dx = x sin(ln x) − cos(ln x) dx.
sin(ln x) dx = x sin(ln x) − x
x
Another integration by parts (with u = cos(ln x) and dv = dx) then gives
Z
Z
Z
− sin(ln x)
cos(ln x) dx = x cos(ln x) − x
dx = x sin(ln x) + sin(ln x) dx.
x
Therefore
Z
Z
sin(ln x) dx = x sin(ln x) − x cos(ln x) −
sin(ln x) dx,
so
Z
1
sin(ln x) dx = [x sin(ln x) − x cos(ln x)].
2
(i) The integrand (x−2 ) is undefined (and unbounded) at x = 0, so the integral does not
exist.
7. (c) F 0 (x) = 2x arctan(x4 ) − 2 arctan(4x2 ).
8. We have
x
Z
F (x) − F (t) =
f (s) ds
t
for a ≤ x < t ≤ b, and there are numbers m and M such that m ≤ f (s) ≤ M for s ∈ [a, b].
It follows from Theorem 6.3.2 that
Z x
Z x
Z x
m(t − x) =
m ds ≤
f (s) ds ≤
m ds ≤ M (t − x),
t
t
t
and therefore
|F (x) − F (t)| ≤ max{|m|, |M |}|x − t|,
so F is Lipschitzian.
10. (d) If we rewrite
n
−3/2
√
1
k=
n
r
k
,
n
then Exercise 6.2.10 gives
lim n−3/2
n→∞
n √
X
Z
k=
0
k=1
1
1√
2
x dx = .
3
2
(f) Now we have
n
1
1
=
,
n2 + k 2
n 1 + (k/n)2
Z 1
n
1 π
X
n
1
=
dx
=
arctan
x
lim
= .
2
2
2
n→∞
n +k
4
0
0 1+x
so
k=1
22. To make the formulas easier to read, define
Z x
F (x) =
f (t) dt.
0
Now we integrate by parts with u = [f (x)]2 and dv = f (x) dx (so v = F (x)) to obtain
Z 1
1 Z 1
3
2
[f (x)] dx = [f (x)] F (x) −
2f (x)f 0 (x)F (x) dx
0
0
0
Z 1
= [f (1)]2 F (1) −
2f (x)f 0 (x)F (x) dx
0
Z 1
2
= [f (1)] F (1) −
2f (x)f 0 (x)[F (1) − F (x)] dx.
0
f0
Now we use the hypothesis 0 <
to conclude that f ≥ 0 (because f is increasing and
f (0) = 0). It also follows that F ≥ 0 by similar reasoning. Therefore f (x)f 0 (x)[F (1) −
F (x)] ≤ f (x)[F (1) − F (x)] because f 0 ≤ 1. From our previous calculation, we conclude
that
Z 1
Z 1
Z 1
3
[f (x)] dx ≤
2f (x)[F (1) − F (x)] dx =
2f (x)F (1) − 2f (x)F (x) dx
0
0
0
Z
1
2
2
= 2F (x)F (1) − [F (x)] = F (1) =
0
Section 6.5
3. (a) p > 1 because
Z
2
1
f (x) dx
0
b
b
1
−p+1 dx
=
(1
−
p)x
= p − 1 − (p − 1)b1−p ,
p
1
1 x
and this expression has a limit as b → ∞ if p > 1.
(b) p < 1 because
Z b
1
dx = p − 1 − (p − 1)b1−p ,
p
x
1
and this expression has a limit as b → 0 if p < 1.
R1
7. (a) The Cauchy principal value for −1 (1/x) dx is
Z −ε
Z 1
1
1
lim
dx +
dx = lim (ln ε − ln ε) = 0,
ε→0+
ε→0+
−1 x
ε x
but
Z
0
1
1
dx = lim
x
ε→0+
Z
ε
1
1
dx = lim − ln ε = ∞,
x
ε→0+
.
3
so 1/x is not improper Riemann integrable on [−1, 1].
R1
(b) The Cauchy principal value for −1 (1/x) dx is
Z −ε
Z 1
1
1
dx +
dx = lim (− ln ε − ln ε) = ∞.
lim
ε→0+
ε→0+
ε |x|
−1 |x|
(c) the Cauchy principal value is
Z 1−ε
Z 2
4
4
lim
dx +
dx = lim (4 ln ε − 4 ln 2 − [4 ln 1 − 4 ln ε]) = 4 ln 2.
+
x−1
ε→0
ε→0+
−1
1+ε x − 1
9. (i) First, notice that limx→∞ | ln x/(1+x1/2 )| = 0, so | ln x/(1+x2 )| ≤ x−3/2 for x sufficiently
large and therefore
Z ∞
ln x
dx
1 + x2
1
converges. Also,
Z 1
Z 1
ln x
− ln t −1
dx =
dt
2
1
+
x
1
+
(1/t)2 t2
k
1/k
(using the change of variables t = 1/x), so
Z 1
Z 1/k
ln x
ln x
dx = −
dx.
2
1 + x2
k 1+x
1
Therefore
Z 1
ln x
dx
1
+ x2
0
also converges and
Z ∞
Z 1
ln x
ln x
dx = −
dx.
2
2
1+x
1
0 1+x
It follows that
Z ∞
ln x
dx = 0.
1 + x2
0
19. If 0 < x < y, then integration by parts with u = 1/t and dv = sin t dx (so v = 1 − cos t)
gives
y Z
Z y
y
1 − cos t 1 − cos t
sin t
dt =
dt.
+
t
t
t2
x
x
x
Note that (1 − cos x)/x → 0 as x → 0+ , and (1 − cos y)/y → 0 as y → ∞. In addition,
|(1 − cos t)/t2 | ≤ 2/(1 + t2 ), so the integral
Z y
1 − cos t
dt
t2
x
converges as x → 0+ and y → ∞. Therefore
Z ∞
sin x
dx
x
0
converges and
Z ∞
Z ∞
sin x
1 − cos x
dx =
dx.
x
x2
0
0
4
To see that the integral is only conditionally convergent, we first observe that
Z (n+1)π
Z (n+1)π
| sin x|
1
2
dx ≥
| sin x| dx =
.
x
(n + 1)π nπ
(n + 1)π
nπ
It follows that
Z nπ
n
X
2
| sin x|
dx ≥
.
x
kπ
0
k=1
This sum diverges to infinity as n → ∞, so the integral diverges.
22. Since f is continuous on [0, ∞) and f (x) ≤ 1/x2 , the integral is convergent. Now we use
Exercise 19 to see that
Z ∞
Z ∞
Z ∞
sin x
1 − cos x
sin2 (x/2)
dx =
dx
=
2
dx
x
x2
x2
0
0
0
from the trig identity 1 − cos x = sin2 (x/2). Now the change of variables u = x/2 yields
Z ∞
Z ∞
Z
sin2 (x/2)
sin2 u
1 ∞ sin2 u
dx
=
2
du
=
dx.
x2
4u2
2 0
u2
0
0
Section 6.7
R∞
Rb
1. TRUE. 0 sin x dx diverges but 0 sin x dx is between 0 and 2 for any positive b.
8. TRUE. (This is similar to the argument for Dirichlet’s function.) First notice that L(P, f ) =
0 for any partition P . Now let q be an integer greater than 1 and set n = q 2 − q + 1. If P
is the partition {x0 , . . . , xn } with x1 = 1/q and xi = (q + i − 1)/q 2 for i = 2, . . . , n, then
the maximum of f is 1 on [x0 , x1 ] and also on at most 2q intervals [xi−1 , xi ] with i ≥ 2.
Therefore
1
1
3q + 1
U (P, f ) ≤ + 2q
= 2
.
q
n
q −q+1
Now we note that lim (3q + 1)/(q 2 − q + 1) = 0 to conclude that we can make U (P, f ) less
q→∞
than any given ε > 0 by taking q large enough.