MATH 2300-005 QUIZ 3
Name:
Due Tuesday, September 13th at the beginning of class. Please use additional paper as
necessary to submit CLEAR and COMPLETE solutions.
Z ∞
dx
1. We know that
diverges but increasing the exponent of x in the denominator
x
1
by any amount produces a convergent improper integral. Show that the family of
functions {x(ln x)p : p > 0} is “between” x and the family {xp : p > 1} in the following
sense:
x(ln x)p
= 0, for any q > 1, p > 0.
lim
x→∞
xq
(If you find this too confusing, you may do only the cases p = 1, 2, 3.)
As an example, for p = 2 we have
(ln x)2 L0 H
2 ln x L0 H
2
lim
lim
= 0.
x→∞ xq−1 = x→∞ (q − 1)xq−1 = x→∞ (q − 1)2 xq−1
lim
In general, you can use L’Hopital’s rule n times, where n is smallest integer greater than
p:
(ln x)p
p(p − 1) . . . (p − n + 1)(ln x)p−n
=
·
·
·
=
lim
= 0,
x→∞ xq−1
x→∞
(q − 1)n xq−1
lim
since the exponent p − n of ln x is less than or equal to zero.
Z ∞
dx
converge/diverge? Find the value of the
2. For which values of p does
x(ln x)p
e
improper integral when it is convergent.
If p 6= 1 we have
Z ∞
e
which is
1
p−1
dx
= lim
x(ln x)p t→∞
Z
t
dx
= lim
p
t→∞
e x(ln x)
1−p
1
(ln t)
−
= lim
t→∞ 1 − p
1−p
Z
1
ln t
du
u1−p ln t
=
lim
t→∞ 1 − p 1
up
if p > 1 and ∞ if p > 1. For p = 1 we get lim ln(ln t) = ∞ and the integral
t→∞
diverges as well. In summary
Z
∞
e
dx
=
x(ln x)p
∞
p≤1
.
p>1
1
p−1
Z
3. For what values of p does the improper integral
0
1
dx
converge?
xp
For p 6= 1 we have
Z
0
1
Z 1
dx
dx
x1−p 1
=
lim
=
lim
xp
t→0+ t xp
t→0+ 1 − p t
1
t1−p
=
− lim
1 − p t→0+ 1 − p
which is
1
1−p
if p < 1 and ∞ if p > 1. For p = 1 we get lim − ln t = ∞ and the integral
t→0+
diverges as well. In summary
Z
0
Z
1
dx
=
xp
∞
1
1−p
p≥1
.
p<1
∞
dx
converges by comparison. Second, find the value of the
+1
0
√ ).
improper integral (you should get 32π
3
4. First, show that
x3
1
1
For instance, we can compare 1+x
3 ≤ x3 on [1, ∞) so that
Z ∞
Z 1
Z ∞
Z 1
Z ∞
dx
dx
dx
dx
dx
=
+
≤
+
< ∞.
3
3
3
3
1+x
1+x
x3
0
0 1+x
1
0 1+x
1
As for the actual value, we use partial fractions:
1
A
Bx + C
1
1
2
=
, 1 = (A+B)x2 +(−A+B+C)x+(A+C), A = , B = − , C = .
3
2
1+x
x+1x −x+1
3
3
3
Hence the improper integral is
Z 1 t
1
−x + 2
lim
+
dx
t→∞ 3 0
x + 1 x2 − x + 1
Z
Z
1
1 t −2x + 1
1 t
dx
= lim ln |t + 1| +
dx
+
t→∞ 3
6 0 x2 − x + 1
2 0 x2 − x + 1
Z
dx
1
1
1 t
2
= lim ln |t + 1| − ln |t − t + 1| +
t→∞ 3
6
2 0 (x − 1/2)2 + 3/4
√
1
1
1
2t − 1
1
√
= lim ln |t + 1| − ln |t2 − t + 1| + √ arctan
− √ arctan(−1/ 3)
t→∞ 3
6
3
3
3
2π
π
π
1 t+1
= √ .
= √ + √ + lim ln √
2 3 6 3 t→∞ 3
t2 − t + 1 3 3
In the last line we are using the limit
lim √
t→∞
1 + 1/t
t+1
= lim p
= 1.
− t + 1 t→∞ 1 − 1/t + 1/t2
t2
5. Find the value of C for which the following improper integral converges and evaluate the
integral for this value of C:
Z ∞
1
C
√
−
dx.
x2 + 4 x + 2
0
The integral is (with x = 2 tan θ integrating the first summand)
Z arctan(t/2)
lim
sec θ dθ − C ln |t + 2| + C ln 2
t→∞ 0
= lim ln | sec(arctan(t/2)) + tan(arctan(t/2))| − C ln |t + 2| + C ln 2
t→∞
√
t2 + 4 + t 1p 2
= lim ln |
t + 4 + t/2| − C ln |t + 2| + C ln 2 = lim ln + C ln 2.
t→∞
t→∞
2(t + 2)C 2
√
2
t +4+t
Now we see that C = 1 is the only possibility, else 2(t+2)
C goes to 0 or ∞ as t → ∞ and
the logarithm will diverge. For C = 1, the value of the integral is ln 2.