4.10. IMPROPER INTEGRALS
4.10
Improper Integrals
4.10.1
Introduction
237
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In Calculus II, students de…ned the integral a f (x) dx over a …nite interval [a; b].
The function f was assumed to be continuous, or at least bounded, otherwise
the integral was not guaranteed to exist. Under these conditions, assuming an
Rb
antiderivative of f could be found, a f (x) dx always existed, and was a number.
In this section, we investigate what happens when these conditions are not met.
De…nition 337 (Improper Integral) An integral is an improper integral
if either the interval of integration is not …nite (improper integral of type
1) or if the function to integrate is not continuous (not bounded that is becomes
in…nite) in the interval of integration (improper integral of type 2).
Example 338
Z
1
e
x
dx is an improper integral of type 1 since the upper limit
0
of integration is in…nite.
Example 339
tinuous at 0.
Z
1
0
1
dx
is an improper integral of type 2 because
is not conx
x
R 1 dx
is an improper integral of type 1 since the upper limit
0 x
1
1
of integration is in…nite. It is also an improper integral of type 2 because
x 1
is not continuous at 1 and 1 is in the interval of integration.
Example 340
Z
2
dx
1
is an improper integral of type 2 because 2
is
2
x
1
x
1
2
not continuous at 1 and 1.
Z
Example 342
tan xdx is an improper integral of type 2 because tan x is
Example 341
0
not continuous at
2
.
We now look how to handle each type of improper integral.
4.10.2
Improper Integrals of Type 1
These are easy to identify. Simply look at the interval of integration. If either
the lower limit of integration, the upper limit of integration or both are not
…nite, it will be an improper integral of type 1.
De…nition 343 (improper integral of type 1) Improper integrals of type 1
are evaluated as follows:
238
CHAPTER 4. INTEGRALS
1. If
Z
t
f (x) dx exists for all t
a, then we de…ne
a
Z
1
f (x) dx = lim
Z
t
f (x) dx
t!1 a
a
Z
1
provided the limit exists as a …nite number. In this case,
f (x) dx is
Z 1 a
said to be convergent (or to converge). Otherwise,
f (x) dx is said
a
to be divergent (or to diverge).
Z b
f (x) dx exists for all t b, then we de…ne
2. If
t
Z
b
f (x) dx = lim
Z
b
t! 1 t
1
f (x) dx
Z b
provided the limit exists as a …nite number. In this case,
f (x) dx is
1
Z b
said to be convergent (or to converge). Otherwise,
f (x) dx is said
1
to be divergent (or to diverge).
Z 1
Z b
3. If both
f (x) dx and
f (x) dx converge, then we de…ne
1
a
Z
1
f (x) dx =
1
Z
a
f (x) dx +
1
Z
1
f (x) dx
a
The integrals on the right are evaluated as shown in 1: and 2.
4.10.3
Improper Integrals of Type 2
These are more di¢ cult to identify. One needs to look at the interval of integration, and determine if the integrand is continuous or not in that interval.
Things to look for are fractions for which the denominator becomes 0 in the interval of integration. Keep in mind that some functions do not contain fractions
explicitly like tan x, sec x.
De…nition 344 (improper integral of type 2) Improper integrals of type
2 are evaluated as follows:
1. if f is continuous on [a; b) and not continuous at b then we de…ne
Z
a
b
f (x) dx = lim
t!b
Z
a
t
f (x) dx
4.10. IMPROPER INTEGRALS
239
Z b
provided the limit exists as a …nite number. In this case, f (x) dx is said
a
Z b
to be convergent (or to converge). Otherwise,
f (x) dx is said to be
a
divergent (or to diverge).
2. if f is continuous on (a; b] and not continuous at a then we de…ne
Z
a
b
f (x) dx = lim+
t!a
Z
b
f (x) dx
t
Z b
provided the limit exists as a …nite number. In this case, f (x) dx is said
a
Z b
to be convergent (or to converge). Otherwise,
f (x) dx is said to be
a
divergent (or to diverge).
3. If f is not continuous at c where a < c < b and both
Z b
f (x) dx converge then we de…ne
Z
c
f (x) dx and
a
c
Z
b
f (x) dx =
a
Z
a
c
f (x) dx +
Z
b
f (x) dx
c
The integrals on the right are evaluated as shown in 1: and 2:
We now look at some examples.
4.10.4
Examples
Evaluating an improper integral is really two problems. It is an integral
problem and a limit problem. It is best to do them separately.
When breaking down an improper integral to evaluate it, make sure that
each integral is improper at only one place, that place should be either
the lower limit of integration, or the upper limit of integration.
Example 345
Z
1
1
dx
x2
This is an improper integral of type 1. We evaluate it by …nding lim
t!1
Z t
Z 1
dx
1
1
dx
First,
= 1
and lim 1
= 1 hence
= 1.
2
2
t!1
t
t
1 x
1 x
Z
1
t
dx
.
x2
240
Example 346
CHAPTER 4. INTEGRALS
Z
1
1
dx
x
This is an improper integral of type 1. We evaluate it by …nding lim
t!1
Z t
Z 1
dx
dx
First,
= ln t and lim ln t = 1 hence
diverges.
t!1
x
1 x
1
Z
1
t
dx
x
R1
dx
1 + x2
This is an improper integral of type 1. Since both limits of integration are
in…nite, we break it into two integrals.
Z 1
Z 0
Z 1
dx
dx
dx
=
+
2
2
1 + x2
1 1+x
1 1+x
0
Example 347
1
R0
R 1 dx
1
dx
Note that since the function
is even, we have 1
= 0
;
1 + x2
1 + x2
1 + x2
R 1 dx
.
we only need to do 0
1 + x2
Z 1
Z t
dx
dx
= lim
2
t!1
1
+
x
1
+
x2
0
0
and
Z
t
0
dx
= tan
1 + x2
= tan
= tan
1
x
1
t
1
t
t
0
tan
1
0
Thus
Z
1
0
dx
= lim tan
t!1
1 + x2
=
1
t
2
It follows that
Z
Example 348
R
2
0
1
1
dx
= +
1 + x2
2
2
=
sec xdx
This is an improper integral of type 2 because sec x is not continuous at . We
2
Z t
evaluate it by …nding lim
sec xdx.
t! 2
0
4.10. IMPROPER INTEGRALS
241
First,
Z
t
t
sec xdx = ln jsec x + tan xjj0
0
= ln jsec t + tan tj
Z 2
and lim (ln jsec t + tan tj) = 1 hence
sec xdx diverges.
t! 2
0
Example 349
R
0
sec2 xdx
This is an improper integral of type 2, sec2 x is not continuous at
Z
sec2 xdx =
0
First, we evaluate
R
Z
2
sec2 xdx +
0
2
0
Z
2
. Thus,
sec2 xdx
2
2
sec xdx.
Z 2
Z t
sec2 xdx = lim
sec2 xdx
t! 2
0
Z
0
t
sec2 xdx = tan t
tan 0
0
= tan t
Therefore,
Z
0
It follows that
R
2
sec2 xdx = lim (tan t)
t! 2
=1
2
0
sec2 xdx diverges, therefore,
R
0
sec2 xdx also diverges.
Remark 350 If we had failed to see that the above integral is improper, and
had evaluated it using the Fundamental Theorem of Calculus, we would have
obtained a completely di¤ erent (and wrong) answer.
Z
sec2 xdx = tan
tan 0
0
= 0 (this is not correct)
Z
1
dx
2
1x
This integral is improper for several reasons. First, the interval of integration
is not …nite. The integrand is also not continuous at 0. To evaluate it, we break
it so that each integral is improper
placeZbeing one
of the
Z 1at only Zone1place, Zthat
Z 1
0
1
dx
dx
dx
dx
dx
limits of integration, as follows:
=
+
+
+
.
2
2
2
2
2
1x
1x
1x
0 x
1 x
We then evaluate each improper integral . The reader will verify that it diverges.
Example 351
242
CHAPTER 4. INTEGRALS
Several important results about improper integrals are worth remembering,
they will be used with in…nite series. The proof of these results is left as an
exercise.
Z 1
dx
converges if p > 1, it diverges otherwise.
Theorem 352
p
1 x
Proof. This is an improper integral of typeZ1 since the upper limit of integration
t
dx
is in…nite. Thus, we need to evaluate lim
. When p = 1, we have already
t!1 1 xp
seen that the integral diverges. Let us assume that p 6= 1. First, we evaluate the
integral.
Z
1
t
dx
xp
=
Z
t
x
p
dx
1
1 p t
=
=
x
1
t
1
p
1
1 p
p
1
1
p
The sign of 1 p is important. When 1 p > 0 that is when p < 1, t1 p is in the
t1 p
1
numerator. Therefore, lim
= 1 thus the integral diverges.
t!1 1
p 1 p
When 1 p < 0 that is when p > 1, t1 p is really
Z 1 in the denominator so that
1
1
dx
t1 p
=
and therefore
converges. In conclusion,
lim
p
t!1 1
p 1 p
p 1
1 x
we have looked at the following cases:
Case 1: p = 1. In this case, the integral diverges.
Case 2: p < 1. In this case, the integral diverges.
Case 3: p > 1. In this case, the integral converges.
Z
1
dx
converges if p < 1, it diverges otherwise.
p
0 x
Proof. See problems.
Theorem 353
4.10.5
Comparison Theorems for Improper Integrals
Sometimes an improper integral is too di¢ cult to evaluate. One technique is to
compare it with a known integral. The theorem below shows us how to do this.
Theorem 354 (Direct Comparison Test) Suppose that f and g are two continuous functions for x a such that 0 g (x) f (x). Then, the following is
true:
4.10. IMPROPER INTEGRALS
243
Z1
Z1
1. If f (x) dx converges then g (x) dx also converges.
a
a
Z1
Z1
2. If g (x) dx diverges, then f (x) dx also diverges.
a
a
The theorem is not too di¢ cult to understand if we think about the integral
in terms of areas. Since both functions are positive, the integrals simply represent the are of the region below their graph. Let Af be the area of the region
below the graph of f . Use a similar notation for Ag . If 0 g (x) f (x), then
Ag
Af . Part 1 of the theorem is simply saying that if Af is …nite, so is Ag ;
this should be obvious from the inequality. Part 2 says that if Ag is in…nite, so
is Af .
Remark 355 When using the comparison theorem, the following inequalities
are useful:
p
x2 x
x 1
and
ln x
ex
x
Z1
Example 356 Study the convergence of e
x2
dx
1
We cannot evaluate the integral directly, e
We note that
x
x2
1 () x2
()
x
() e
x
2
x2
does not have an antiderivative.
x
x
e
Now,
Z1
e
x
dx = lim
t!1
1
Zt
x
e
dx
1
= lim e
1
t!1
1
e
t
=e
Z1
and therefore converges. It follows that e
1
theorem.
x2
dx converges by the comparison
244
CHAPTER 4. INTEGRALS
R1
dx
1 + x2
Note that we did this problem above and found that this integral converges to .
2
If we only need to know whether it converges or not, we can use a comparison
R 1 dx
1
1
theorem instead. We notice that 0 < 2
< 2 when x 1 and since 1
x +1
x
x2
R 1 dx
converges (see above), by the direct comparison test, we conclude that 1
1 + x2
also converges.
Example 357 Study the convergence of
1
Remark 358 The technique we used in the previous example would not have
R 1 dx
1
1
worked with the integral 1
because 2 1 > 2 , so the direct comx
x2 21
x
2
parison test does not allow us to conclude. However, the next comparison test
would work in such cases.
Theorem 359 (Limit Comparison Test) Let f and g be two positive and
Z1
f (x)
continuous functions on [a; 1). If 0 < lim
< 1 then f (x) dx and
x!1 g (x)
a
Z1
g (x) dx behave the same way, that is they both converge or both diverge.
a
Remark 360 The theorem does not say anything about what these integrals
converge to. In particular, it does not say that if they converge they converge to
the same value.
Remark 361 Of course, one di¢ culty of this theorem is to …nd an integral to
compare to.
Example 362 Study the convergence of
R1
1
dx
x2
1
2
.
We need to …nd an integral to compare it to. We see that
1
for large x. We compute
x2
lim
x!1
1
x2
1
x2
=
x2
x!1
x2
=
1
lim
1
x2
1
2
is similar to
1
2
1
2
Hence, the two integrals behave the same way. Since
R 1 dx
rem 352 it follows that 1
also converges.
x2 21
R 1 dx
converges by theo1 x2
4.10. IMPROPER INTEGRALS
4.10.6
245
Things to know
Be able to tell if an integral is improper or not and what type it is.
Be able to tell if an improper integral converges or diverges. If it converges, be able to …nd what it converges to.
Be able to write an improper integral as a limit of de…nite integral(s).
4.10.7
Problems
1. Determine why each integral below is improper. Then, determine if they
are convergent or divergent. When they converge, evaluate them.
(a)
R1
1
dx
2
(3x + 1)
R 1
dx
p
(b)
1
2 x
R1
(c) 2 sin tdt
R1
2
(d)
xe x dx
1
R1
(e) 0 xe 5x dx
R 1 ln x
dx
(f) 1
x
R 1 ln x
(g) 1
dx
x2
R1
x2
dx
(h)
1 9 + x6
R1 3
(i) 0 5 dx
x
R 14 dx
p
(j)
2 4
x+2
R1
ex
dx
(k)
1 ex
1
2. Use one of the comparison theorems to determine if
or diverges.
3. Use one of the comparison theorems to determine if
or diverges.
4. Use one of the comparison theorems to determine if
or diverges.
R 1 cos2 x
converges
1 1 + x2
R1
1
R1
1
dx
converges
+1
x3
p
dx
converges
x+1
246
CHAPTER 4. INTEGRALS
5. Find the values of a for which
6. Prove theorem 353.
7. Find the values of p for which
4.10.8
R1
0
R1
e
eax dx converges and diverges.
dx
p converges and diverges.
x (ln x)
Answers
1. Determine why each integral below is improper. Then, determine if they
are convergent or divergent. When they converge, evaluate them.
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
(j)
(k)
R1
dx
1
2
=
1
12
(3x + 1)
dx
p
diverges
2 x
R1
sin tdt diverges
2
R1
2
xe x dx = 0
1
R1
1
xe 5x dx =
0
25
R 1 ln x
dx diverges
1
x
R 1 ln x
dx = 1
1
x2
R1
1
x2
dx =
1 9 + x6
9
R1 3
dx diverges
0 x5
R 14 dx
32
p
=
2 4
3
x+2
R1
ex
dx diverges
1 ex
1
R
1
1
2. Use the comparison theorem to determine if
verges.
It converges.
R 1 cos2 x
converges or di1 1 + x2
3. Use one of the comparison theorems to determine if
or diverges.
It converges.
4. Use one of the comparison theorems to determine if
or diverges.
It diverges.
R1
1
R1
1
dx
converges
x3 + 1
p
dx
converges
x+1
4.10. IMPROPER INTEGRALS
247
R1
5. Find the values of a for which 0 eax dx converges and diverges.
It diverges if a 0 and converges otherwise.
6. Prove theorem 353.
It diverges if p
R1
dx
p converges and diverges.
x (ln x)
1 and converges otherwise.
7. Find the values of p for which
e