MA244 Analysis III
Solutions. Sheet 3.
NB. THESE ARE SKELETON SOLUTIONS ONLY, USE THEM WISELY!
0.1
Further exercises in integration
1. (Q.1) Let
1
Z
xp−1 (1 − x)q−1 dx = L1 + L2 ,
0
where
1/2
Z
xp−1 (1 − x)q−1 dx,
L1 = lim
1 ↓0
1
Z
2
xp−1 (1 − x)q−1 dx.
L2 = lim
2 ↓0
1/2
The improper integral exists iff L1 , L2 exists. Consider L1 . The function
(1 − x)q−1 is bounded on [0, 1/2]. Let B > 0 be the bound. The improper
integral
Z 1/2
(1/2)p
xp
,
lim
=
B
Bxp−1 dx = lim B |1/2
1 ↓0 1 ↓0
p 1
p
1
exists (here we used that p > 0). Therefore, L1 exists by the comparison
test. Similarly, using the fact that xp is bounded on [1/2, 1] we can establish
the existence of L2 . Therefore, the Euler integral of the first kind exists by
definition.
2. (Q.2) If p, q ∈ N, the Euler integral exists in the usual sense. The calculation
goes as follows:
Z 1 p 0
Z
Z 1
(q − 1) 1 p
x
q−1
p−1
q−1
(1 − x)
=
x (1 − x)q−2 dx
x (1 − x) dx =
p
p
0
0
0
(q − 1)(q − 2)
=
p(p + 1)
Z
1
x
p+1
(1−x)
q−3
0
=
(q − 1)(q − 2) . . . 1
dx = . . . =
p(p + 1) . . . (p + q − 2)
Z
1
xp+q−2 dx
0
(q − 1)!(p − 1)!
.
(p + q − 1)!
3. (Q.3) In,m exists in the usual sense. Consider the following change of variables:
y = sin2 (x). Notice that the inverse function x(y) is not differentiable at
y = 0, 1. Therefore, the ‘clean’ computation goes as follows:
Z
Z 1−2
1 1 (m−1)/2
m
n
0
y
(1−y)(n−1)/2 dy
In,m = lim
sin (x(y)) cos (x(y))x (y)dy =
1 ↓0,2 ↓0 2
0
1
1
= B((m + 1)/2, (n + 1)/2).
2
2
When resolving y = sin (x) with respect to sin(x) we used sin(x) ≥ 0, cos(x) ≥
0 for x ∈ [0, π/2].
4. (Q.4) Let us represent Γ(p) = L1 + L2 , where
Z
L1 = lim
↓0
1
x
Z
p−1 −x
R
xp−1 e−x dx.
e dx, L2 = lim
R↑0
1
L1 converges by the
test which uses xp−1 e−x ≤ xp−1 for x ∈ (0, 1]
R 1 comparison
and the fact that 0 xp−1 dx converges for any p > 0. L2 also converges by the
comparison test: notice that the function xp−1 e−x/2 is bounded on [1, ∞):
−xc /2
xp−1 e−x/2 ≤ e−1/2 (p ≤ 3/2), xp−1 e−x/2 ≤ xp−1
(p ≤ 3/2),
c e
where xc = 2(p − 1). Let us call this bound B(p). We found that
xp−1 e−x ≤ B(p)e−x/2 , x ∈ [1, ∞).
R∞
As the integral 0 e−x/2 dx converges, L2 exists by the comparison test. Therefore, the Euler integral of the second kind exists by definition.
5. (Q.5) For p > 0,
Z
xp
xp
1
1
1 ∞ p 0 −x
(x ) e = lim e−x |1l + lim e−x |u1 + Γ(p + 1) = Γ(p + 1)
Γ(p) =
l↓0 p
u↑∞ p
p 0
p
p
The result Γ(n + 1) = n! follows from the above by induction started from
Γ(1) = 1.
6. (Q.6)
Z
Γ(p)Γ(q) =
∞
p−1 −u
u
e
du
0
Z
∞
Z
∞
=4
v
Z
q−1 −v
x
2p−1 −x2
Z
e
Z
2p−1
sin
y 2q−1 e−y
2
0
π/2
=4
0
∞
e
0
2p−1 2q−1 −x2 −y 2
y
∞
e dv = 4
0
dxdyx
0
∞
Z
2q−1
(φ) cos
0
Z
∞
(φ)
r2(p+q)−1 e−r
0
= B(p, q)Γ(p + q).
The second inequality in the above is the change of variables (u, v) → (x2 , y 2 );
the third inequality is a formal replacement of the repeated integral with a
double integral; the fourth is a change to polar coordinates; the fifth equality
uses an appropriate generalization of Question 3 and the change of variables
r2 = x.
7. (Q.7) If x = 0 and α > 0, the Bessel integral diverges:
Z
∞
R
Z
cosh(αt)dt := lim
R↑∞
0
If x > 0 and α > 0,
Z ∞
Z
− x2 et αt
Kα (x) ≤
e
e dt =
0
1
∞
0
1
cosh(αt)dt ≥ lim
2 R↑∞
− x2 y α−1
e
y
Z
dy ≤
∞
Z
R
eαt dt = ∞.
0
x
e− 2 y y α−1 dy = (x/2)−α Γ(α).
0
The right hand side of the above chain is finite as the gamma integral is finite
for α > 0, see Question 4. We conclude that Kα (x) converges for x > 0, α > 0.
2
Rx
8. (Q.8) Let F (x) = 0 f . As f is continuous, F is differentiable and F 0 = f
(FTC1). By FTC2,
Z u(x)
f = F (u(x)) − F (d(x)).
d(x)
Applying the chain rule,
d
dx
Z
u(x)
f = F 0 (u(x))u0 (x) − F 0 (d(x))d0 (x) = f (u(x))u0 (x) − f (d(x))d0 (x).
d(x)
9. (Q.9) Suggested change of variables: y = π/2 − x.
π/2
Z
Z
0
f (cos(π/2 − y))dy =
f (cos(x))dx = −
f (sin(y))dy,
0
π/2
0
π/2
Z
where we used cos(π/2 − y) = sin(y)
10. (Q.10) (i) Upper bound:
Z 200π
Z 200π
Z 200π
cos(x)
sin0 (x)
sin(x) 200π
sin(x)
dx =
dx =
|100π +
dx
x
x
x
x2
100π
100π
100π
Z 200π
Z 200π
sin(x)
1
1
1
1
dx ≤
dx =
−
<
.
=
2
2
x
100π 200π
100π
100π x
100π
(ii) Lower bound: using integration by parts as in part (i),
Z
200π
100π
=
cos(x)
dx =
x
Z
49
X
p=0
=
49 Z
X
p=0
200π
100π
100π+(2p+1)π
100π+2pπ
100π+(2p+1)π
sin(x)
100π+2pπ
Z
99
X
sin(x)
dx
=
x2
k=0
sin(x)
dx +
x2
1
1
−
2
x
(x + π)2
Z
Z
100π+(k+1)π
100π+kπ
100π+(2p+2)π
100π+(2p+1)π
dx =
49 Z
X
p=0
sin(x)
dx
x2
sin(x)
dx
x2
!
100π+(2p+1)π
100π+2pπ
π sin(x)
(2x+π)dx > 0,
+ π)2
x2 (x
as each of the integrals under the summation sign is positive (each of the
corresponding integrands is a non-negative continuous function which is not
identically equal to zero).
0.2
Uniform convergence
11. (Q.11) Uniform contunuity of f : A → R means that for any > 0 there is
δ > 0 such that for any x, x0 ∈ A: |x − x0 | < δ, |f (x) − f (x0 )| < . To prove that
f is continuous at any x0 ∈ A, let us just set x0 = x0 in the above inequalities,
which will then read as a standard definition of continuity at x0 .
12. (Q.12) (i) f (x) = x2 is a polynomial, hence continuous on R (Analysis II). On
the other hand, |f (x + δ/2) − f (x − δ/2)| = 2|x||δ| → ∞ for x → ∞. Therefore,
|f (x + δ/2) − f (x − δ/2)| can be made as large as we like by varying x, no
matter how small δ is. Therefore, f is not uniformly continuous. (ii) Take for
example f (x) = 1/(1 + x2 ). Then for δ > 0,
|f (x+δ/2)−f (x−δ/2)| = δ
2|x|
δ
2|x|
≤δ
≤√ ,
2
2
2
(1 + (x + δ/2) )(1 + (x − δ/2) )
(1 + 2x )
2
where the last bound comes from a direct analysis of the function f (R) =
2R/(1 + 2R2 ) for R ≥ 0. Therefore, f is uniformly continuous with δ = √2 .
13. (Q.13) Due
R ∞ to (i) all integrals over finite intervals exist. Fix any R>∞0. As the
integral a g < ∞ (condition (ii)), there exists b > a such that b g < /3.
As fn → f uniformly on [a, b], there exists N ∈ N such that for any n > N
and any x ∈ [a, b], |f (x) − fn (x)| < /(3(b − a)). Condition (ii) implies that
|f (x)| ≤ g(x), x ∈ [a, ∞). Therefore, for any n > N ,
Z ∞
Z ∞
Z b
Z ∞
Z ∞
g < .
|fn − f | ≤ /3 + 2
|fn − f | +
fn | ≤
f−
|
a
Therefore, limn→∞
R∞
a
R∞
fn =
b
b
a
a
a
f by definition.
14. (Q.14) Fix x > 1. Let fn (t) = tx−1 (1 − t/n)n I[0,n] (t), n ∈ N, f (t) = g(t) =
tx−1 e−t . Here I[0,n] is the indicator function of the interval [0, n]: I[0,n] (t) = 1
if t ∈ [0, n] and zero otherwise. Notice that all functions introduced
R n above as
continuous,R hence regulated on [0, R] for any R > 0. Moreover, 0 tx−1 (1 −
∞
t/n)n dt = 0 fn and
fn (t) = tx−1 en log(1−t/n) I[0,n] (t) ≤ tx−1 e−t = g(t), t ≥ 0,
where we used the inequality log(1
R ∞ + x) ≤ x, x > 0. Therefore g dominates
the sequence (fn ). Notice that 0 g converges, this is just Euler’s integral of
the second kind, see Question 4. In order to apply the dominated convergence
theorem it remains to prove that fn → f uniformly on [0, R]. Accoridng to
Taylor’s theorem,
log(1 − y) = −y −
1
1
y 2 , y > 0,
2 (1 − ξ)2
where ξ ∈ (0, y). Therefore,
2
||fn − f ||∞ ≤ R
x−1
||e
t+n log(1−t/n)
2
−R
2n
≤ Rx−1 |e
1
(1−R/n)2
− 1||∞ = R
x−1
||e
t
− 2n
1
(1−ξ)2
− 1||∞
− 1| → 0, n → ∞.
Therefore, fn → f uniformly on [0, R]. So we can use the dominated convergence theorem to claim that
Z n
Z ∞
Z ∞
Z ∞
x−1
n
t (1 − t/n) dt =
fn →
f=
tx−1 e−t dt = Γ(x), n → ∞.
0
0
0
0
15. P
(Q.15) (i) For x ∈ [1 + δ, ∞), 1/k x ≤ (1/k)(1+δ) , k = 1, 2, . . .. As the series
∞
1+δ
converges, the zeta-function series converges uniformly on [1 +
k=1 1/k
δ, ∞) by the Weierstrass M -test. (ii) No, the zeta-function series does not
converge uniformly on (1, ∞). Really,
Z ∞
n
∞
∞
X
X
X
n1−x
x
x
x
x
|
1/k −
1/k | =
1/k ≥
1/y dy =
.
x−1
n
k=1
k=1
k=n+1
Choosing x = 1 + 1/n, we find that
|
n
X
x
1/k −
k=1
∞
X
1/k x | ≥ n1+1/n ≥ n,
k=1
which can be made as large as we like by increasing n. This contradicts the
definition of uniform convergence (if unsure, write down the negation of the
definition).
16. (Q.16) We will prove that the sequence of partial sums is uniformly Cauchy.
The sum of the n-th and the (n + 1)-st terms of the series is
√
√
(1 − x2 )n + 1
n n n+1−x n
√
= (−1)n xn √ √
fn (x)+fn+1 (x) = (−1) x √ √
√ .
n n+1
n n + 1( n + 1 + x n)
For any x ∈ [0, 1], the absolute value of the above can is bounded by
xn (1 − x2 )n + 1
√
.
n(n + 1)
On the interval [0, 1], xn (1 − x2 ) is maximised at xc =
p
n/(1 + n), which gives
2
||fn (x) + fn+1 (x)||∞ ≤ √
.
n(n + 1)
Therefore,
||
m
X
fn (x)||∞ ≤
k=n
∞
X
√
k=n
2
.
k(k + 1)
The right hand
of the above inequality vanishes in the limit n → ∞, as
P side
√ 2
the series
k k(k+1) converges. Therefore, the sequence of partial sums is
uniformly Cauchy and the series in question converges uniformly on [0, 1].
P
k−1 x2k−1
17. (Q.17) f (x) = ∞
. Let
k=1 (−1)
(2k−1)
fn (x) =
n
X
n
(−1)k−1
k=1
X
x2k−1
, fn0 (x) =
(−1)k−1 x2k−2 .
(2k − 1)
k=1
Note that (fn ) and (fn0 ) converge uniformly by the M -test with Mk = max(|a|, b)2k−1
and Mk = max(|a|, b)2k−2 correspondingly. Therefore, the termwise differentiation is justified and
f 0 (x) = lim fn0 (x) =
n→∞
1
.
1 + x2
Notice that f (0) = 0 and f 0 is continuous on [a, b]. By FTC2,
Z x
f (x) =
f 0 (t)dt = arctan(x).
0
P∞
18. (Q.18) The series f (x) = k=0 (k + 1)xk converges uniformly on [a, b] by the
Weierstrass M -test with Mk = (k + 1)max(−a, b)k . Therefore, f is continuous
and the series can be integrated termwise:
Z x
∞ Z x
∞
X
X
1
k
f (t)dt =
(k + 1)t dt =
xk+1 = 1 +
.
1−x
0
k=0 0
k=0
BY FTC1,
d
f (x) =
dx
Z
x
f (t)dt =
0
1
.
(1 − x)2
P
k−1
19. (Q.19)The series f (x) = ∞
converges uniformly on [a, b] by the
k=1 k(k +1)x
Weierstrass M -test with Mk = k(k +1)max(−a, b)k . Therefore, f is continuous
and the series can be integrated termwise:
Z
x
f (t)dt =
0
∞ Z
X
k=1
x
0
k−1
k(k + 1)t
∞
X
dt =
(k + 1)xk = −1 +
k=1
1
,
(1 − x)2
where the last equality uses the result of Question 18. BY FTC1,
Z x
2
d
f (t)dt =
.
f (x) =
dx 0
(1 − x)3
20. (Q.20)
∞
X
n=1
√
2n−1
Z 1/
∞
∞
√ X
√ X
1
(−1)n−1
(−1)n−1
n−1
√
= 3
= 3
(−1)
(2n − 1)3n−1
(2n − 1)
3
0
n=1
n=1
√ Z
= 3
0
√
1/ 3
3
x2n−2 dx
√
√
√ π
1
dx
=
3
arctan(1/
3)
=
3· .
1 + x2
6
The
of the order of summation and integration is justified, as
P∞ interchange
n
(−y)
converges
uniformly in [0, 1/3].
n=0
November the 9th, 2015
Sergey Nazarenko and Oleg Zaboronski.