Practice Problems: Improper Integrals
Written by Victoria Kala
December 6, 2014
Solutions to the practice problems posted on November 30.
For each of the following problems:
(a) Explain why the integrals are improper.
(b) Decide if the integral is convergent or divergent. If it is convergent, find which value it
converges to.
Z ∞
1
1. √4
dx
0 1+x
Solution:
(a) Improper because it is an infinite integral (called a Type I).
(b) Rewrite:
Z ∞ Z t Z t t
1 1 4
√ dx = lim √ dx = lim (1 + x)−1/4 dx = lim (1 + x)3/4
0
4
1+x t→∞ 0
4
1+x t→∞ 0 t→∞ 3 0
4 4
lim (1 + t)3/4 − = ∞
t→∞ 3 3
So the integral diverges.
Z 2
1
2. 2
dx
−2 x
Solution: This question was on my subject GRE.
(a) Improper because the function x12 is discontinuous at x = 0 (called a Type II).
(b) There are two ways to do this problem, so I will post both solutions.
One way: Split up the integral at x = 0:
Z 2 Z 0 Z 2 Z t Z 2
1 1 1 1 1
2
dx = 2
dx + 2
dx = lim 2
dx + lim 2
dx
−2 x −2 x 0 x t→0 −
−2 x s→0 +
s x
−1 t −1 2 −1 1 1 1
= lim− + lim+ = lim− − − + lim+
t→0 x −2 s→0 x s t→0 t 2 2 s→0 s
Both of the limits diverge so the integral diverges.
Another way: x12 is an even function, so it is symmetric about x = 0:
Z 2 Z 2 Z 2
1 1 1 −1 2 1
2
dx = 2 2
dx = lim+ 2 2
dx = lim+ 2 = −1 + 2 lim+ = ∞
−2 x 0 x t→0 t x t→0 x t t→0 t
So the integral diverges.
1
, Z 0
3. 2r dr
−∞
Solution:
(a) Improper because it is an infinite integral (called a Type I).
(b) Rewrite:
0 0
2r 2t
Z Z
0 1 1 1
r r
2 dr = lim 2 dr = lim = − lim = −0 =
−∞ t→−∞ t t→−∞ ln 2 t ln 2 t→−∞ ln 2 ln 2 ln 2
Convergent!
Z ∞
4. (y 3 − 3y 2 ) dy
−∞
Solution:
(a) Improper because it is an infinite integral (called a Type I).
(b) Need to split it up, try about y = 0:
Z ∞ Z 0 Z ∞
(y 3 − 3y 2 ) dy = (y 3 − 3y 2 ) dy + (y 3 − 3y 2 ) dy
−∞ −∞ 0
0 s
y4 y4
Z Z
0 s
= lim (y 3 −3y 2 ) dy+ lim − y3
(y 3 −3y 2 ) dy = lim + lim − y3
t→−∞ t s→∞ 0 t→−∞ 4 t s→∞ 4 0
4 4
t s
= − lim − t3 + lim − s3
t→−∞ 4 s→∞ 4
Both of these limits diverge, so the integral diverges.
Z ∞
5. cos πt dt
−∞
Solution:
(a) Improper because it is an infinite integral (called a Type I).
(b) Need to split it up, try about t = 0:
Z ∞ Z 0 Z ∞ Z 0 Z r
cos πt dt = cos πt dt + cos πt dt = lim cos πt dt + lim cos πt dt
−∞ −∞ 0 s→−∞ s r→∞ 0
1 0 1 r 1 1
= lim sin πt + lim sin πt = − lim sin πs + lim sin πr
s→−∞ π s r→∞ π 0 s→−∞ π r→∞ π
Both of these limits diverge, so the integral diverges.
2
Written by Victoria Kala
December 6, 2014
Solutions to the practice problems posted on November 30.
For each of the following problems:
(a) Explain why the integrals are improper.
(b) Decide if the integral is convergent or divergent. If it is convergent, find which value it
converges to.
Z ∞
1
1. √4
dx
0 1+x
Solution:
(a) Improper because it is an infinite integral (called a Type I).
(b) Rewrite:
Z ∞ Z t Z t t
1 1 4
√ dx = lim √ dx = lim (1 + x)−1/4 dx = lim (1 + x)3/4
0
4
1+x t→∞ 0
4
1+x t→∞ 0 t→∞ 3 0
4 4
lim (1 + t)3/4 − = ∞
t→∞ 3 3
So the integral diverges.
Z 2
1
2. 2
dx
−2 x
Solution: This question was on my subject GRE.
(a) Improper because the function x12 is discontinuous at x = 0 (called a Type II).
(b) There are two ways to do this problem, so I will post both solutions.
One way: Split up the integral at x = 0:
Z 2 Z 0 Z 2 Z t Z 2
1 1 1 1 1
2
dx = 2
dx + 2
dx = lim 2
dx + lim 2
dx
−2 x −2 x 0 x t→0 −
−2 x s→0 +
s x
−1 t −1 2 −1 1 1 1
= lim− + lim+ = lim− − − + lim+
t→0 x −2 s→0 x s t→0 t 2 2 s→0 s
Both of the limits diverge so the integral diverges.
Another way: x12 is an even function, so it is symmetric about x = 0:
Z 2 Z 2 Z 2
1 1 1 −1 2 1
2
dx = 2 2
dx = lim+ 2 2
dx = lim+ 2 = −1 + 2 lim+ = ∞
−2 x 0 x t→0 t x t→0 x t t→0 t
So the integral diverges.
1
, Z 0
3. 2r dr
−∞
Solution:
(a) Improper because it is an infinite integral (called a Type I).
(b) Rewrite:
0 0
2r 2t
Z Z
0 1 1 1
r r
2 dr = lim 2 dr = lim = − lim = −0 =
−∞ t→−∞ t t→−∞ ln 2 t ln 2 t→−∞ ln 2 ln 2 ln 2
Convergent!
Z ∞
4. (y 3 − 3y 2 ) dy
−∞
Solution:
(a) Improper because it is an infinite integral (called a Type I).
(b) Need to split it up, try about y = 0:
Z ∞ Z 0 Z ∞
(y 3 − 3y 2 ) dy = (y 3 − 3y 2 ) dy + (y 3 − 3y 2 ) dy
−∞ −∞ 0
0 s
y4 y4
Z Z
0 s
= lim (y 3 −3y 2 ) dy+ lim − y3
(y 3 −3y 2 ) dy = lim + lim − y3
t→−∞ t s→∞ 0 t→−∞ 4 t s→∞ 4 0
4 4
t s
= − lim − t3 + lim − s3
t→−∞ 4 s→∞ 4
Both of these limits diverge, so the integral diverges.
Z ∞
5. cos πt dt
−∞
Solution:
(a) Improper because it is an infinite integral (called a Type I).
(b) Need to split it up, try about t = 0:
Z ∞ Z 0 Z ∞ Z 0 Z r
cos πt dt = cos πt dt + cos πt dt = lim cos πt dt + lim cos πt dt
−∞ −∞ 0 s→−∞ s r→∞ 0
1 0 1 r 1 1
= lim sin πt + lim sin πt = − lim sin πs + lim sin πr
s→−∞ π s r→∞ π 0 s→−∞ π r→∞ π
Both of these limits diverge, so the integral diverges.
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