Riemann Sums
2.5 Improper Integrals
In this section, we will define the integral of a function over a set of real numbers, where the set of real numbers need not be a closed, bounded, interval; we also define integrals when the value of the integrand is unbounded, even if the interval of integration is itself bounded. This really requires a new definition, for Theorem 2.3.6 tells us that the Riemann integral of an unbounded function does not exist.
The actual calculation of our new type of integral will involve calculating our usual integrals on intervals of the form [a, b], where now either a or b varies, and we will then take a limit as a or b approaches some “problematic” value. Of course, the calculation of the integrals on the intervals [a, b] can/will still use the Fundamental Theorem of Calculus, and, hence, we may apply all of our techniques from earlier sections to find anti-derivatives.
As we wish to be able to discuss integrals over intervals [a, b] and over intervals (a, b] (and over other sets), the notationRb
a does not suffice to distinguish between the types of intervals that we care about. Thus, we adopt some new notation; if E is a subset of the real numbers, we will write
Z
E
f (x) dx
for the integral of f over the set E. Of course, right now, this has no meaning for us, unless E is a closed interval [a, b], in which caseRb
a f (x) dx =R
[a,b]f (x) dx. Our goal in this section is to defineR
Ef (x) dx for sets E that need not be closed intervals.
Before we really start looking at new types of integrals, it will be helpful to have a new piece of terminology.
Definition 2.5.1. If f is a real function, with domain D, an extension of f is a function f whose domain is larger (or equal to) D, and such that f and ˆˆ f agree at all points of D, i.e., for all x in D, ˆf (x) = f (x).
In other words, an extension ˆf , of f , is a function whose domain includes the domain, D, of f , and such that the restriction of ˆf to D is equal to f .
Example 2.5.2. For example, the function ˆf , with domain [0, 1] given by extension of f to [0, 1], since
lim
x→0+
sin x x = 1.
Now, let’s start our discussion of more general notions of integration by looking at the easiest case, one where we already know the answer: the case of a function which is Riemann integrable.
Example 2.5.3. Consider f (x) = x2on the interval [0, 1]. Certainly, f is Riemann integrable on [0, 1], since f is continuous on [0, 1]. Thus, we know whatR
[0,1]x2dx means; it meansR1 0 x2dx, which, by the Fundamental Theorem of Calculus, is equal to (x3/3)
1 0= 1/3.
What if we omit the 0 from the interval of integration, i.e., what should R
(0,1]x2dx mean?
Theorem 2.3.14 tells us that, for Riemann integrals on closed intervals, altering the function at a finite number of points does not change the integrability of the function or the value of the integral. Thus, intuitively, it seems reasonable that omitting a single point of integration, like 0, should not affect the integral. Therefore, in this case, it seems reasonable to make the definition that
While this seems reasonable, it should, in a way, seem like “cheating”; we wanted to integrate f (x) = x2on the interval (0, 1], and yet we used information about f on the larger interval [0, 1], i.e., we used that there was a continuous extension of f from the interval (0, 1] to the interval [0, 1]. Could we have definedR
(0,1]x2dx without using the extension to [0, 1]? Yes.
For all a such that 0 < a ≤ 1, f (x) = x2 is Riemann integrable on the interval [a, 1], and
Now, we can take the limit as a approached 0 from the right to obtain our previous answer.
That is, we could have definedR
(0,1]x2dx by
The point is that, in this example, we obtain the same value for R
(0,1]x2dx, regardless of whether we use the fact that x2extends to [0, 1] or whether we instead use the limits of integrals, as our lower-limit of integration approaches 0.
Example 2.5.4. Let’s consider a more-complicated example. Let f be the function, with domain (0, 1], given by
f (x) = sin x x .
How many choices do we have for reasonable ways to define the integralR
(0,1]f (x) dx? At least two.
First, we can define an extension ˆf to the closed interval [0, 1], and then define
Z
Of course, we either need to pick a particular extension, or show that the value ofR
[0,1]f (x) dxˆ is always the same, regardless of what extension we select. We could, in fact, take ˆf to be the continuous extension of f given in Example 2.5.2; then, ˆf would certainly be Riemann integrable by Theorem 2.3.8.
However, we don’t have to use continuous extension of f . Since (sin x)/x does possess a continuous extension to the interval [0, 1], it follows that (sin x)/x is bounded on (0, 1]. Thus, Theorem 2.3.8 and Theorem 2.3.14 imply that it doesn’t matter how we define ˆf at x = 0; no matter what value we chose for ˆf (0), ˆf will be bounded and, at least, piecewise-continuous – hence, Riemann integrable – and changing the value at one point will not affect the value of the integral. Therefore, we could define
Z
for any extension of f to a function ˆf on [0, 1].
However, we could take the second approach, as we did in Example 2.5.3; without using an extension at all, we could define
Z if the limit does exist, how do we know that that limit is equal to what we’d get by extending (sin x)/x to [0, 1] and then taking the Riemann integral of our extension over [0, 1]?
Actually, the answers to these questions are easy, given our earlier results. Suppose that ˆf is an extension of f to [0, 1]. Then, as we discussed above, ˆf is Riemann integrable on [0, 1]. By Remark 2.4.4, the function on [0, 1] that sends a toR1
a f (x) dx is continuous; in particular,ˆ Z 1
a f (x) dx exists and equals what we would obtain by calculating the Riemann integral of any extension of f to [0, 1]. Therefore, it seems reasonable to define the integral of f over the half-open interval (0, 1] by
Z
What we have seen in the previous two examples is that, in those cases, it made sense to defineR
a f (x) dx is a Riemann integral. However, in those cases, we could also have defined R
(0,1]f (x) dx by extending f to the closed interval [0, 1] and then using the Riemann integral of
the extended function. Now we will look at an example where the approach via extensions does not work, but the limit idea still yields a meaningful result.
Example 2.5.5. Consider the function
f (x) = 1
√x = x−1/2
on the half-open interval (0, 1]. As x approaches 0 from the right, 1/√
x approaches ∞, and so f is unbounded on (0, 1]. Hence, any extension of f to the closed interval [0, 1] will also be unbounded and, therefore, will not be Riemann integrable (Theorem 2.3.6).
On the other hand, using the Fundamental Theorem, and the Power Rule for Integration, we find
lim
a→0+
Z 1 a
x−1/2dx = lim
a→0+
x1/2 1/2
1 a
= lim
a→0+(2 − 2√
a) = 2.
This means that the area under the graph of y = 1/√
x and over the interval [a, 1] approaches 2 as a → 0+. We say, simply, that the area under the graph of y = 1/√
x and over the interval (0, 1] equals 2.
-0.5 0a 0.5 1 1.5
Figure 2.33: Area under the graph of y = 1/√
x over [a, 1].
What we see in this example is that defining the integral over a half-open interval (a, b]
in terms of limits of integrals over closed intervals gives us a well-defined number, while the Riemann integral of an extension to the closed interval [a, b] does not exist.
Before making a definition, we wish to look at one more example.
Example 2.5.6. Suppose that f (x) = e−x and we wish to integrate f over the interval [0, ∞).
Note that −e−x is an anti-derivative of e−x.
In this example, the function f itself is bounded on the given interval; e−xis between 0 and 1 for x in [0, ∞). On the other hand, the interval that we want to integrate over is unbounded, since it “goes out to infinity”. How should we define
Z
[0,∞)
e−xdx?
Here, there is no closed interval to which we can possibly extend f . However, the limit approach still yields an answer:
lim
b→∞
Z b 0
e−xdx = lim
b→∞
−e−x
b 0
= lim
b→∞ − e−b− (−e0)
= 0 + 1 = 1.
This means that the area under the graph of y = e−xand over the interval [0, b] approaches 1 as b → ∞. We say, simply, that the area under the graph of y = e−x and over the interval [0, ∞) equals 1.
0 5
1
b
Figure 2.34: Area under the graph of y = e−xover [0, b].
In light of the above discussion and examples, we make the following definition:
Definition 2.5.7. Suppose that f is defined on the half-open interval [a, b), where b may be
∞, and suppose that, for all c such that a ≤ c < b, f is Riemann integrable on the closed interval [a, c].
provided that the limit exists, in which case we say that f is integrable on [a, b), or that the integral Rb
af (x) dx converges. Otherwise, we say that Rb
af (x) dx diverges.
Similarly, suppose that f is defined on the half-open interval (a, b], where a may be −∞, and suppose that, for all c such that a < c ≤ b, f is Riemann integrable on the closed interval [c, b].
provided that the limit exists, in which case we say that f is integrable on (a, b], or that the integral Rb
af (x) dx converges. Otherwise, we say that Rb
af (x) dx diverges.
Naturally, ifRb
af (x) dx converges, we defineRa
b f (x) dx = −Rb
af (x) dx.
Note that we do not have a notational conflict; if f is, in fact, Riemann integrable on [a, b], then, by Theorem 2.4.3 and Remark 2.4.4, the Riemann integralRb
af (x) dx equals both of the one-sided limits of integrals given above.
The way that we usually conclude that f is Riemann integrable on all of the closed intervals [a, c] (or [c, b]) contained in [a, b) (or (a, b]) is that f is continuous on the entire half-open interval.
Assuming this is the case, the only way that the integralRb
af (x) dx can possibly fail to converge is for either the interval of integration to be unbounded, or for the function f to be unbounded on the interval of integration. We give these two types of integrals which involve unbounded activity a name:
Definition 2.5.8. An integral Rb
a f (x) dx, in which a or b is ±∞, or such that f is un-bounded on the interval (a, b) is called an improper integral.
The integrals in Example 2.5.5 and Example 2.5.6 are improper integrals, and yet the inte-grals converge. The importance of improper inteinte-grals is that, for continuous functions, they’re the only types of integrals which might diverge.
While the integrals in Definition 2.5.7 are the basic new types of integrals that we are defining in this section, we are also interested in more-complicated integrals, ones which break up into a finite number of pieces which are of the types found in Definition 2.5.7 .
Example 2.5.9. Consider the integral
Z 8
−1
1 x2/3dx.
The point x = 0 is in the interval [−1, 8], the interval over which we’re supposed to integrate.
However, as x approaches 0 from the left or right, the integrand goes to ∞. This is a type of improper integral. The question is: how should we define what such an integral means?
-2 -1 0 1 2 3 4 5 6 7 8 9
Figure 2.35: The y = x−2/3 becomes unbounded from either side of x = 0.
The answer is: we want Theorem 2.3.16, on splitting up integrals, to remain true. This means that we want it to be true that
Z 8
−1
1
x2/3dx = Z 0
−1
1
x2/3dx + Z 8
0
1 x2/3dx,
where each of the summands on the right is an integral of the type we defined in Definition 2.5.7.
We now calculate by taking limits:
lim
There can also be multiple “problem points”.
Example 2.5.10. Consider the integral
Z ∞ 1
1
(x − 2)(x − 4)dx.
After possibly removing a finite number of points (problem points, where unboundedness comes into play), we want to split the interval [1, ∞) into a finite number of closed or half-open intervals on which the integrand 1/[(x − 2)(x − 4)] is continuous; in this splitting, we allow a pair of closed or half-open intervals to intersect each other in, at most, one point. We then add together the resulting integrals, provided all of them exist; otherwise, we say that the original integral diverges.
Thus, we start with the interval [1, ∞). We remove the two points x = 2 and x = 4, where 1/[(x − 2)(x − 4)] is undefined. For now, this gives us intervals [1, 2), (2, 4), and (4, ∞). The intervals (2, 4) and (4, ∞) are not closed or half-open. Why do we care? We do not want to have to deal with some sort of simultaneous limits at the two endpoints of the intervals.
Therefore, we further split the intervals (2, 4), and (4, ∞). Where do we split them? At any point in-between the endpoints. It is a theorem, which we incorporated into the statement of Definition 2.5.11, that is doesn’t matter where the splitting occurs. So, we split the interval (2, 4) into (2, 2.5] and [2.5, 4) (pairs of half-open intervals may intersect at a point), and we split (4, ∞) into (4, 7] and [7, ∞).
We end up with five half-open intervals: I1= [1, 2), I2 = (2, 2.5], I3 = [2.5, 4), I4 = (4, 7], and I5 = [7, ∞), whose union is equal to the original interval [1, ∞), minus a finite number of points, and pairs of the half-open intervals intersect each other in, at most, one point.
Now, we define
Z ∞ 1
1
(x − 2)(x − 4)dx to equal the sum
Z provided that all of these integrals exist, in which case we say thatR∞ 1
1
(x−2)(x−4)dx converges.
If one (or more) of the five separate integrals, appearing in the summation, diverges, then we say thatR∞
1 1
(x−2)(x−4)dx diverges.
In the remainder of this example, we will show that
Z ∞ 1
1
(x − 2)(x − 4)dx
diverges, by showing that the first improper integral in the summation above diverges, i.e., we will show that
We find an anti-derivative of (x−2)(x−4)1 via partial fractions, as in Section 1.3. So, we first determine constants A and B such that
1
(x − 2)(x − 4) = A
x − 2 + B x − 4,
for all x, other than x = 2 and x = 4. Clearing the denominators, by multiplying each side of the equality by the big denominator on the left, i.e., by (x − 2)(x − 4), we obtain
1 = A(x − 4) + B(x − 2),
which needs to hold for all x. Plugging in x = 2, we find that 1 = A · (−2) + B · 0, so that
A = −1/2. Plugging in x = 4, we find that 1 = A · 0 + B · 2, and so B = 1/2. Thus,
1
(x − 2)(x − 4) = −1/2
x − 2 + 1/2 x − 4,
and we want to calculate
lim
or show that it doesn’t exist. We need to find an anti-derivative of 1/(x − a), where a is 2 or 4.
We accomplish this by making the substitution u = x − a, so that du = dx, and
Z 1
x − adx = Z 1
udu = ln |u| + C = ln |x − a| + C.
Thus, Formula 2.3 becomes
lim
As b is approaching 2 from the left, we know that all of the x’s that we are considering are less than 2 and, hence, |x − 2| = 2 − x and |x − 4| = 4 − x. Now, evaluating at x = b and subtracting
(x−2)(x−4)dx diverges, and so does the integral over the bigger interval [1, ∞).
We should mention that, just becauseR2 1
1
(x−2)(x−4)dx diverges to ∞, that does not imply that R∞
1 1
(x−2)(x−4)dx also diverges to ∞; other parts of the summand that we split the integral into may (and do) diverge to −∞. Thus, we simply say that R∞
1 1
(x−2)(x−4)dx diverges, without trying to specify the manner in which it diverges.
We summarize our discussion and example above into a definition. In this definition, we will split a set into a union of closed or half-open intervals (as we did above); the point is that each of the intervals that we split things into has a “problem”, an unboundedness issue, at, at most, one endpoint. It is important that the definition gives you the same result, regardless of how you pick the half-open intervals. The proof of this is in Theorem 2.A.11.
Definition 2.5.11. Let E be a subset of the real numbers which is the union of a finite number of intervals. Let f be a real function, which is defined and continuous on the set E except, perhaps, at a finite set of points P . Let E − P denote the set of points in E which are not in the set P .
Then, E − P is a union of a finite number of closed or half-open intervals I1, I2, . . . , In on which f is defined and continuous. Given any such decomposition of E − P into intervals, we define the integral of f on E by
Z
E
f (x) dx = Z
I1
f (x) dx + Z
I2
f (x) dx + . . . + Z
In
f (x) dx,
provided that each integral in the summation on the right converges; in this case, we say that R
Ef (x) dx converges or that f is integrable on E. If any one of the integrals in the summation above diverges, then we say that R
Ef (x) dx diverges.
These definitions of converges, diverges, and the value of the integral are independent of the choice of the intervals I1, I2, . . . , In, as long as the given conditions are satisfied.
If E is an interval [a, b], [a, b), (a, b], or (a, b), then we may also use other notation; we set
Z b a
f (x) dx = Z
E
f (x) dx, and Z a
b
f (x) dx = − Z
E
f (x) dx.
Note that, with our terminology, we may say that a function f is integrable on a set E, even if f is not defined at a finite number of the points in E. For instance, in Example 2.5.9, we would say that 1/x2/3is integrable on (−1, 8), even though 1/x2/3is not defined at x = 0. Be aware that other books might use different terminology, and might not say that f is integrable on a set on which f is not defined.
Just as we have linearity for Riemann integrals, Theorem 2.3.19, we also have linearity for our more general integrals, provided the individual integrals converge. The proof is essentially identical, except that, to deal with the improper integrals, you must use that limits are “linear”, i.e., you can pull constants out of limits and split up sums.
Theorem 2.5.12. (Linearity of Improper Integrals) Let E be a subset of the real numbers which is the union of a finite number of intervals. Suppose that f and g are integrable on E, and that a and b are any real numbers. Then, af + bg is integrable on E
and Z
E
af (x) + bg(x) dx = a Z
E
f (x) dx + b Z
E
g(x) dx.
Example 2.5.13. As we saw in Example 2.5.5 and Example 2.5.9 (with different upper-limits of integration), the integrals
Z 1 0
1
x1/2dx and
Z 1 0
1 x2/3dx
both converge; we leave it as an exercise for you to show that they converge to 2 and 3, respec-tively.
Therefore,
Z 1 0
5
x1/2 − 7 x2/3
dx converges, and equals 5 · 2 − 7 · 3 = −11.
It is sometimes possible to tell that an improper integral converges, without being able to determine what it converges to. This seemingly bizarre fact stems from a defining property of the real numbers: every non-empty set of real numbers, which has an upper bound, has a LEAST upper bound. See Definition 5.1.15 and Theorem 5.1.18. This least upper bound property is the main reason that we can sometimes tell that limits, as in improper integrals, exist without knowing the value.
We first give a theorem which says that there’s only one way for some types of improper integrals to diverge.
Theorem 2.5.14. Let I be an interval of the form [a, b) or (a, b], where we allow the intervals [a, ∞) or (−∞, b]. Suppose that, for all x in I, f (x) ≥ 0, and that, for all closed intervals [c, d] contained in I, f is Riemann integrable on [c, d].
Then, if there exists a real number M (an upper bound) such that, for all [c, d] contained in I,Rd
c f (x) dx ≤ M , thenRb
a f (x) dx converges, and what it converges to is less than, or equal to, M ; in other words, if there is an upper bound M on all of the Rd
c f (x) dx, then Rb
af (x) dx converges to the least such upper bound.
In particular, ifRb
a f (x) dx diverges, what it diverges to is ∞.
Proof. See Theorem 2.A.12.
Remark 2.5.15. The analogous statement for f (x) ≤ 0 is true, and is obtained by applying Theorem 2.5.14 to −f (x), which would be non-negative.
What you find for non-positive f is: if there exists a real number M (a lower-bound) such that, for all [c, d] contained in I, M ≤Rd
c f (x) dx, thenRb
af (x) dx converges to the greatest such lower bound. In particular, ifRb
af (x) dx diverges, what it diverges to is −∞.
Corollary 2.5.16. Let I be an interval of the form [a, b) or (a, b], where we allow the intervals [a, ∞) or (−∞, b]. Suppose that, for all x in I, 0 ≤ f (x) ≤ g(x), and that, for all closed intervals [c, d] contained in I, f and g are Riemann integrable on [c, d].
Then, ifRb
a g(x) dx converges, then so doesRb
af (x) dx, and what it converges to is some-thing less than, or equal to, Rb
ag(x) dx. This implies that, if Rb
a f (x) dx diverges, then so doesRb
a g(x) dx.
Proof. This is easy now. Since f ≤ g on I, Theorem 2.3.20 tells us that, for all [c, d] contained in I,
Z d c
f (x) dx ≤ Z d
c
g(x) dx.
IfRb
ag(x) dx converges, then Theorem 2.5.14 tells us that it converges to the least upper bound M on the integralsRd
c g(x) dx, but, by the inequality above, this M is an upper bound on all of the integralsRd
c f (x) dx. The corollary now follows by applying Theorem 2.5.14 again.
Example 2.5.17. Consider the integrals
Producing manageable anti-derivatives of these integrands is problematic/impossible. Nonethe-less, we can use Corollary 2.5.16 to determine quickly whether they converge or not.
First, note that, for all x in (0, 1],
Combining all of the above with Corollary 2.5.16, we conclude that Z 1
0
1 + sin x
√x dx converges (to something less than or equal to 4, but we don’t know what), while
Z ∞
In Exercises 1 through 17, evaluate the given integral if it converges; otherwise, show that the integral diverges.
1.
1 − cos v. Hint: use a half-angle identity to evaluate the integral.
9.
xn converge or diverge? If it converges, what does it converge to?
16. Let n = 1. Does Z 1
0
dx
xn converge or diverge? If it converges, what does it converge to?
17. Let n < 1. Does Z 1
0
dx
xn converge or diverge? If it converges, what does it converge to?
Use Corollary 2.5.16 to determine whether the integrals in Exercises 18 through 24 converge or diverge. If they converge, you need not calculate the integrals.
18.
25. Prove that the elliptic integral Z 1
0
dx
p(1 − x2)(1 − k2x2) converges. Assume k2< 1.
26. Formulate a similar comparison theorem for functions that grow without bound at some
26. Formulate a similar comparison theorem for functions that grow without bound at some