MA244 Analysis III
Solutions. Sheet 3.
Questions for credit: 6 (4 points), 8 (5 points), 12 (8 points) and 15 (8 points)
0.1
Further exercises in integration
R∞
R∞
R∞
R1
2
2
2
2
2
e−x dx = 0 e−x dx + 1 e−x dx. Therefore, 0 e−x dx and 1 e−x dx
are either both convergent or both divergent. The exponential function is
2
2
increasing
x ≥ 1. Therefore, e−x ≤ e−x for x ≥ 1.
R ∞ −xand −x ≤ −x
R R for
But 1 e := limR→∞ 1 e−x dx = limR→∞ (e−1 − e−R ) = e−1 , convergent.
Therefore, the probability integral converges by the comparison principle.
R 1/2
R1
2. B(p, q) = 0 xp−1 (1 − x)q−1 dx + 1/2 xp−1 (1 − x)q−1 dx. The Euler integral converges iff each of the integrals in the right hand side converges. For
R 1/2
x ∈ (0, 1/2], xp−1 (1 − x)q−1 ≤ xp−1 max(1, (1/2)q−1 ). The integral 0 xp−1 dx
R 1/2
converges for p > 0 (lecture material). Therefore, 0 xp−1 (1 − x)q−1 dx converges by the comparison principle. Similarly, for x ∈ [1/2, 1), xp−1 (1−x)q−1 ≤
R1
R 1/2
(1 − x)q−1 max(1, (1/2)p−1 ). The integral 1/2 (1 − x)q−1 dx = 0 xq−1 dx conR1
verges for q > 0 (lecture material). Therefore, 1/2 xp−1 (1 − x)q−1 dx converges
by the comparison principle. We conclude that the Euler integral of the first
kind converges for p > 0, q > 0.
R∞
R1
R∞
3. 0 xp−1 e−x dx = 0 xp−1 e−x dx + 1 xp−1 e−x dx. The Euler integral converges
iff each integral in Rthe right hand side converges. For x ∈ (0, 1], xp−1 e−x ≤
1
xp−1 . The integral 0 xp−1 dx converges for p > 1 (lecture material). Therefore,
R 1 p−1 −x
x e dx converges by the comparison principle. For x ∈ [1, ∞), xp−1 e−x ≤
0
x|p−1| e−x = (x|p−1| e−x/2 )e−x/2 ≤ supx∈[0,∞) (x|p−1| e−x/2 )e−x/2 ≤ Cp e−x/2 , where
Cp > 0 is a p-dependent constant. (The global maximum of g(x) = x|p−1| e−x/2
is achieved at xc = 2|p − 1|. The valueR at the global maximum is Cp =
∞ −x/2
−|p−1|
(2|p − 1|)|p−1|
dx converges (question 1).
R ∞e p−1 −x> 0.) The integral 1 e
Therefore, 1 x e dx converges by the comparison principle. We conclude
that the Euler integral of the second kind converges for p > 0.
1.
R∞
0
4. Let {p0 = a < p1 < . . . < pN = b} be a finite set of points such that F
is continuously differentiable on each of [pk−1 , pk ], k = 1, 2, . . . N . Such a set
exists
the definition of piece-wise continuous differentiability. By FTC2,
R pk by
0
F
=
F (pk ) − F (pk−1 ). Therefore,
pk−1
Z
b
0
F =
a
N Z
X
k=1
pk
pk−1
0
F =
N
X
(F (pk ) − F (pk−1 )) = F (b) − F (a).
k=1
Notice the crucial role the continuity of F plays in the proof.
5. Let k ∈ Z. sign(sin(x)) = 1 for x ∈ (2kπ, (2k + 1)π) and sign(sin(x)) = −1
for x ∈ ((2k + 1)π, (2k + 2)π). Let F : F (x) = x − 2kπ for x ∈ [2kπ, (2k + 1)π],
F (x) = π−(x−(2k+1)π) for x ∈ [(2k+1)π, (2k+2)π]. (A ’saw tooth’ function.)
Notice that F is continuous, piece-wise continuously differentiable
and F 0 (x) =
R πn
sign(sin(x))
for x 6= πk. By the theorem of Question 4, 0 sign(sin(x))dx =
R πn 0
F = F (πn) − F (0) = F (πn), which is zero for even n and π for odd n.
0
Ra
R0
Ra
R0
Ra
t=−x
6. In each case −a f (x)dx = −a f (x)dx+ 0 f (x)dx = − a f (−t)dt+ 0 f (x)dx =
Ra
(f (x) + f (−x))dx. Therefore:
0
Ra
Ra
(a) If f (x) = f (−x), −a f = 2 0 f .
Ra
Ra
(b) If f (x) = −f (−x), −a f = 0 0 · dx = 0.
Both statements are true.
R π/2
π/2 R π/2
7. In = 0 sin0 (x) cosn−1 (x)dx = sin(x) cosn−1 (x)|x=0 − 0 sin(x)(cosn−1 )0 (x)dx =
R π/2
R π/2
(n − 1) 0 sin2 (x) cosn−2 (x)dx = (n − 1) 0 (1 − cos2 (x)) cosn−2 (x)dx =
(n − 1)(In−2 − In ). So,
n−1
In−2 .
In =
n
By explicit calculation, I0 = π/2, I1 = 1.
If n is even,
n−1
(n − 1)(n − 3)
(n − 1)(n − 3) . . . 3 · 1
In−2 =
In−4 = . . . =
I0
n
n(n − 2)
n(n − 2) . . . 4 · 2
(n − 1)(n − 3) . . . 3 · 1 π
.
=
n(n − 2) . . . 4 · 2 2
In =
If n is odd,
n−1
(n − 1)(n − 3)
(n − 1)(n − 3) . . . 2 · 4
In−2 =
In−4 = . . . =
I1
n
n(n − 2)
n(n − 2) . . . 3 · 1
(n − 1)(n − 3) . . . 4 · 2
=
n(n − 2) . . . 3 · 1
In =
So,
9·7·5·3
π
63π
63π
· = 9 =
,
10 · 8 · 6 · 4 · 2 2
2
512
8·6·4·2
27
128
I9 =
=
=
.
9·7·5·3
315
315
I10 =
8. The theoretical underpinning for this question is Theorem 22 of the course.
Let xk = k/n, k = 1, . . . , n − 1.
(a)
n−1
n−1
X
1X
k
lim
= lim
xk
n→∞
n2 n→∞ n k=1
k=1
Z
T hm.22
=
0
1
1
xdx = .
2
(b)
lim
n→∞
n
X
k=1
n
1
1X 1
= lim
n + k n→∞ n k=1 1 + xk
T hm.22
Z
1
=
0
1
dx = log(1+x)|10 = log(2).
1+x
(c)
n
n
X
kp
1X p
xk
lim
= lim
p+1
n→∞
n→∞ n
n
k=1
k=1
T hm.22
Z
=
0
1
xp dx =
1
.
p+1
9. The integral is positive: f is continuous on [0, 2π], so the integral exists in the
usual sense.
Z 2π
Z π
Z 2π
Z π
f=
f+
f=
(f (x) + f (x + π))dx,
0
0
π
0
f (x) + f (x + π) = sin(x)/x + sin(x + π)/(xR + π) = sin(x)(1/x − 1/(x + π)) =
π
> 0 for x ∈ (0, π). Therefore, f > 0 by integral bounds.
sin(x) x(x+π)
R1 √
R1√
R
√
2 dx − 1 1 · dx =
2 −
(
1
+
x
1
+
x
1 + 0)dx > 0, as the inte0
0
0
grand is continuous, non-negative,
and
positive
R1√
R 1 at at least one point of
2
the integration interval. So, 0 1 + x dx > 0 1 · dx
R1
R1
R1
(b) 0 x2 sin2 (x)dx − 0 x sin2 (x)dx = 0 x(x − 1) sin2 (x)dx < 0, as the integrand is continuous, non-positive,
negative
R 1 2 and
R 1 at at2 least one point of
2
the integration interval. So, 0 x sin (x)dx < 0 x sin (x)dx
R2
R2
R2 2
2
(c) 1 ex dx− 1 ex dx = 1 ex (ex −x −1)dx > 0, as the integrand is continuous,
2
non-negative (x2 − x ≥ 0 for x ≥ 1, so ex −x
at at least
R 2 ≥x21), andR 2positive
x
one point of the integration interval. So, 1 e dx > 1 e dx
10. (a)
0.2
Uniform convergence.
11. The integrand is continuous (hence uniformly continuous) on D = [0, 10] ×
[0, 10]. More over, the partial derivative of the integrand with respect to x is
also continuous on D. Therefore, the integral is differentiable with Rrespect to
2
b
the limits of integration (FTC1) and the c-derivative of the integral a e−cy dy
is equal to the integral of the c-derivative of the integrand. Therefore, applying
the chain rule formula from the hint we find
Z 10
2
0
−x3
f (x) = −e
−
y 2 e−xy dy
x
R ∞ −αx −βx
R ∞ −αx
0
12. Let F (α) = 0 e −e
dx.
Then
F
(α)
=
−e
= − α1 . Therefore,
x
0
F (α) = − log(α) + C(β), where C(β is a constant of integration. It can be
found by noticing that F (β) = 0, which gives C(β) = log(β). Therefore,
F (α) = log(β/α).
R∞
13. (a) F (p) = 0 e−pt dt = p1
R∞
1
(b) F (p) = 0 e−pt e−αt dt = p+α
R∞
R∞
R∞
1
(c) F (p) = 0 e−pt cos(βt)dt = 2i1 ( 0 e−pt eiβt dt + 0 e−pt e−iβt dt) = 21 ( p−iβ
+
p
1
) = p2 +β 2
p+iβ
P
P∞ xn
1
14. (a) For any x ∈ [−1, 1], |xn /n2 | ≤ 1/n2 . As ∞
n=1 n2 < ∞,
n=1 n2 converges
uniformly on [−1, 1] by the M-test.
P
P∞ sin(nx)
1
(b) For any x ∈ R, sin(nx)/2n ≤ 1/2n . But ∞
n=1 2n < ∞. So,
n=1
2n
converges on R uniformly by the M-test.
15. In each of the three cases below, both the power series S(x) and the series for
S 0 (x) converge uniformly on [a, b]. Therefore, the term-wise differentiation is
justified. Integrating S 0 to obtain S is done using F T C2.
P∞ k−1
P
1
xk
0
= 1−x
. Therefore, S(x) =
(a) Let S(x) = ∞
k=1 x
k=1 k . Then S (x) =
−log(1 − x) + C, where C is an integration constant. As S(0) = 0, C = 0.
P
xk
Therefore, ∞
k=1 k = −log(1 − x).
P
P
1
k−1 xk
k−1
(b) Let S(x) = ∞
. Then S 0 (x) = ∞
= 1+x
. Therek=1 (−1)
k=1 (−x)
k
fore, S(x) = log(1 + x) + C, where C is an integration constant. As
P
xk
S(0) = 0, C = 0. Therefore, ∞
k=1 k = log(1 + x).
P
P∞ 2k−2
x2k−1
1
0
(c) Let S(x) = ∞
= 1−x
2 . Therefore,
k=1 2k−1 . Then S (x) =
k=1 x
1+x
1
S(x) = 2 log 1−x + C, where C is an integration constant. As S(0) = 0,
P
xk
1
1+x
C = 0. Therefore, ∞
k=1 k = − 2 log 1−x .
November the 3rd, 2014
Daniel Ueltschi and Oleg Zaboronski.