IMPROPER INTEGRALS AND ABSOLUTELY INTEGRABLE FUNCTIONS - Multiple Integrals - Advanced Calculus of Several Variables

Advanced Calculus of Several Variables (1973)

Part IV. Multiple Integrals

Chapter 6. IMPROPER INTEGRALS AND ABSOLUTELY INTEGRABLE FUNCTIONS

Thus far in this chapter on integration, we have confined our attention to bounded functions that have bounded support. This was required by our definition of the integral of a nonnegative function as the volume of its ordinate set (if contented)—if f: ImagenImage is to be integrable, then its ordinate set must be contented. But every contented set is a priori bounded, so f must be bounded and have bounded support.

However there are certain types of unbounded subsets of Imagen with which it is natural and useful to associate a notion of volume, and analogously functions which are either unbounded or have unbounded support, but for which it is nevertheless natural and useful to define an integral. The following two examples illustrate this.

Example 1Let f: [1, ∞) → Image be defined by f(x) = 1/x2. Let A denote the ordinate set of f, and An the part of A lying above the closed interval [1, n] (Fig. 4.39). Then

Image

Since Image and limn→∞ v(An) = 1, it would seem natural to say that

Image

despite the fact that f does not have bounded support. (Of course what we really mean is that our definitions of volume and/or the integral ought to be extended so as to make this true.)

Image

Figure 4.39

Example 2Let f be the unbounded function defined on (0, 1 ] by Image. Let A denote the ordinate set of f, and An the part of A lying above the closed interval [1/n, 1] (Fig. 4.40). Then

Image

Since Image and limn→∞ v(An) = 2, it would seem natural to say that

Image

These two examples indicate both the need for some extension (to unbounded situations) of our previous definitions, and a possible method of making this extension. Given a function f : UImage, with either f or U (or both) unbounded,

Image

Figure 4.40

we might choose a sequence Image of subsets of U which “fill up” U in some appropriate sense, on each of which f is integrable, and then define

Image

provided that this limit exists. Of course we would have to verify that ∫U f is thereby well-defined, that is, that its value is independent of the particular chosen sequence Image. Partly because of this problem of well-definition, we shall formulate our initial definition of ∫U f in a slightly different manner, and then return later to the interpretation of ∫Uf as a sequential limit (as above).

We deal first with the necessity for f : UImage to be integrable on enough subsets of U to “fill up” U. We say that f is locally integrable on U if and only if f is integrable on every compact (closed and bounded) contented subset of U. For instance the function f(x) = 1/x2 of Example 1 is not integrable on (1, ∞), but is locally integrable there. Similarly the function f(x) = 1/x½ of Example 2 is not integrable on (0, 1), but is locally integrable there.

It will be convenient to confine our attention to open domains of definition. So let f : UImage be a locally integrable function on the open set Image. We then say that f is absolutely integrable on U if and only if, given ε > 0, there exists a compact contented subset Bε of U such that

Image

for every compact contented set Image which contains Bε.

The student should verify that the sum of two absolutely integrable functions on U is absolutely integrable, as is a constant multiple of one, so the set of all absolutely integrable functions on U is a vector space. This elementary fact will be used in the proof of the following proposition, which provides the reason for our terminology.

Proposition 6.1If the function f : UImage is absolutely integrable, then so is its absolute value ImagefImage.

We will later prove the converse, so f is absolutely integrable if and only if ImagefImage is.

PROOFWriting f = f+f, it suffices by the above remark to show that f+ and f are both absolutely integrable, since ImagefImage = f+ + f. We consider f+. Given ε > 0, choose Bε as in the definition, such that Image implies

Image

if A is a compact contented subset of U. It will suffice to show that Image also implies that

Image

Given a compact contented set A with Image, define

Image

so that f(x) = f+(x) for Image, and

Image

because f+(x) = 0 if Image. Then

Image

as desired, because A+ is a compact contented set such that Image.

Image

The number I, which the following theorem associates with the absolutely integrable function f on the open set Image, will be called the improper (Riemann) integral of f on U. It will be temporarily denoted by IUf (until we have verified that IUf = ∫Uf in case f is integrable on U).

The following definition will simplify the interpretation of IUf as a sequential limit of integrals of f on compact contented subsets of U. The sequence Image of compact contented subsets of U is called an approximating sequencefor U if (a) Image for each Image, and (b) Image.

Theorem 6.2 Suppose f is absolutely integrable on the open set Image. Then there exists a number I = IUf with the property that, given ε > 0, there exists a compact contented set Cε such that

Image

for every compact contented subset A of U containing Cε. Furthermore

Image

for every approximating sequence Image for U.

PROOFTo start with, we choose a (fixed) approximating sequence Image for U (whose existence is provided by Exercise 6.16). Given η > 0, the Heine-Borel theorem (see the Appendix) implies that Image if k is sufficiently large. Here Bη/2 is the compact contented subset of U (provided by the definition) such that Image implies that

Image

Consequently, if k and l are sufficiently large, it follows that

Image

Thus Image is a Cauchy sequence of numbers, and therefore has a limit I (see the Appendix).

Now let Cε = Bε/3. Given a compact contented subset A of U which contains Cε, choose a fixed k sufficiently large that Image and Image. Then

Image

as desired. This completes the proof of the first assertion.

Now let Image be an arbitrary approximating sequence for U. Given δ > 0, choose K sufficiently large that Image, for all Image. Then what we have just proved gives Image for all Image, so it follows that

Image

We can now show that IUf and ∫Uf are equal when both are defined. Note that, if f is integrable on the open set Image, then it follows from Exercise 3.1 that f is locally integrable on U (why?).

Theorem 6.3If Image is open and f : UImage is integrable, then f is absolutely integrable, and

Image

PROOFLet M be an upper bound for the values of Imagef(x)Image on U. Given ε > 0, let Bε be a compact contented subset of U such that v(U − Bε) < ε/M. If A is a compact contented subset of U which contains Bε, then

Image

so it follows that f is absolutely integrable on U.

If in addition A contains the set Cε of Theorem 6.2, then

Image

This being true for all ε > 0, it follows that IUf = ∫Uf.

Image

In order to proceed to compute improper integrals by taking limits over approximating sequences as in Theorem 6.2, we need an effective test for absolute integrability. For nonnegative functions, this is easy to provide. We say that the improper integral ∫Uf is bounded (whether or not it exists, that is, whether or not f is absolutely integrable on U) if and only if there exists M > 0 such that

Image

for every compact contented subset A of U.

Theorem 6.4Suppose that the nonnegative function f is locally integrable on the open set U. Then f is absolutely integrable on U if and only if ∫Uf is bounded, in which case ∫Uf is the least upper bound of the values ∫Af, for all compact contented sets Image.

PROOFSuppose first that (1) holds, and denote by I the least upper bound of {∫Af} for all compact contented Image. Given ε > 0, I − ε is not an upper bound for the numbers {∫Af}, so there exists a compact contented set Imagesuch that

Image

If A is any compact contented set such that Image then, since Image, we have

Image

so it follows that both

Image

The first inequality implies that f is absolutely integrable, and then the second implies that ∫Uf = I as desired (using Theorem 6.2).

Conversely, if f is absolutely integrable on U, then it is easily seen that Image for every compact contented subset U of A, so we can take M = ∫Uf in (1).

Image

Corollary 6.5Suppose the nonnegative function f is locally integrable on the open set U, and let Image be an approximating sequence for U. Then f is absolutely integrable with

Image

provided that this limit exists (and is finite).

PROOFThe fact that the monotone increasing sequence of numbers Image is bounded (because it has a finite limit) implies easily that ∫Uf is bounded, so f is absolutely integrable by Theorem 6.4. Hence (2) follows from Theorem 6.2.

Image

As an application, we will now generalize see Examples 1 and 2 to Imagen. Let A denote the solid annular region Image (Fig. 4.41), and let f be a spherically symmetric continuous function on A. That is,

Image

for some function g, where T : ImagenImagen is the n-dimensional spherical coordinates mapping of Exercise 5.19. By part (a) of that exercise, and the change of variables theorem, we have

Image

Image

Figure 4.41

where

Image

Example 3Let U = ImagenB1n and f(x) = 1/ImagexImagep where p > n. Writing Ak = BknB1n, Corollary 6.5 and (3) above give

Image

because kn−p → 0 since n − p < 0.

Example 4Let U denote the interior of the unit ball with the origin deleted, and f(x) = 1/ImagexImagep with p < n. So now the problem is that f, rather than U, is unbounded. Writing Image, Corollary 6.5 and (3) give

Image

because p < n.

For nonnegative locally integrable functions, Corollary 6.5 plays in practice the role of a “working definition.” One need not worry in advance about whether the improper integral ∫Uf actually exists (that is, whether f is absolutely integrable on U). Simply choose a convenient approximating sequence Image for U, and compute Image. If this limit is finite, then ∫Uf does exist, and its value is the obtained limit.

In the case of an arbitrary (not necessarily nonnegative) function, one must know in advance that f is absolutely integrable on U. Once this has been established, Theorem 6.2. enables us to proceed as above—choose an approximating sequence Image for U, and then compute Image.

The simplest way to show that a given locally integrable function f : UImage is absolutely integrable is to compare it with a function g : UImage which is already known to be absolutely integrable, using the comparison test stated below. First we need the converse of Proposition 6.1.

Corollary 6.6Let f : UImage be locally integrable. If ImagefImage is absolutely integrable on U, then so is f.

PROOFWrite f = f+f, so ImagefImage = f+ + f. Since ImagefImage is absolutely integrable, ∫UImagefImage is bounded (by Theorem 6.4). It follows immediately that ∫Uf+ and ∫Uf are bounded. Hence Theorem 6.4 implies that f+ and f are both absolutely integrable on U, so f = f+f is also.

Image

The import of Corollary 6.6 is that, in practice, we need only to test nonnegative functions for absolute integrability.

Corollary 6.7(Comparison Test) Suppose that f and g are locally integrable on U with Image. If g is absolutely integrable on U, then so is f.

PROOFSince g is absolutely integrable, ∫Ug is bounded (by Theorem 6.4). But then the fact that Image implies immediately that ∫Uf is also bounded, so f is also absolutely integrable.

Image

Example 5 The functions

Image

are all absolutely integrable on U = (1, ∞), by comparison with the absolutely integrable function g(x) = 1/x2 of Examples 1 and 3. For the latter, we note that

Image

so Image sin x/x2Image is absolutely integrable by Corollary 6.7. But then it follows from Corollary 6.6 that (sin x)/x2 itself is absolutely integrable on (1, ∞).

Similarly the functions

Image

are absolutely integrable on U = (0, 1), by comparison with the function g(x) = 1/x1/2 of Examples 2 and 4.

Next we want to define a different type of improper integral for the case of functions of a single variable. Suppose that f is locally integrable on the open interval (a, b), where possibly a = − ∞ or b = + ∞ or both. We want to define the improper integral denoted by

Image

to be contrasted with the improper integral ∫(a, b)f(x) dx which is defined if f is absolutely integrable on (a, b). It will turn out that Image exists in some cases where f is not absolutely integrable on (a, b), but that Image if both exist.

The new improper integral Image is obtained by restricting our attention to a single special approximating sequence for (a, b), instead of requiring that the same limit be obtained for all possible choices of the approximating sequence (as, by Theorem 6.2., is the case if f is absolutely integrable). Separating the four possible cases, we define

Image

(where a and b are finite), provided that the appropriate limit exists (and is finite) and say that the integral converges; otherwise it diverges.

Example 6The integral Image converges, because

Image

However f(x) = (1 + x)/(1 + x2) is not absolutely integrable on (− ∞, ∞) = Image. If it were, we would have to obtain the same limit π for any approximating sequence Image for Image. But, taking An = [−n, 2n], we obtain

Image

Example 7Consider Image. Since limx→0[(sin x)/x] = 1, sin x/x is continuous on [0, 1]. So we need only consider the convergence of

Image

But (cos x)/x2 is absolutely convergent on (1, ∞), by comparison with 1/x2.

Therefore Image converges; its actual value is π/2 (see Exercise 6.14). However the function f(x) = (sin x)/x is not absolutely integrable on (0, ∞) (see Exercise 6.15).

The phenomenon illustrated by Examples 6 and 7, of Image converging despite the fact that f is not absolutely convergent on (a, b), does not occur when f is nonnegative.

Theorem 6.8Suppose f : (a, b) → Image is locally integrable with Image. Then f is absolutely integrable if and only if Image converges, in which case

Image

PROOFThis follows immediately from Theorem 6.2. and Corollary 6.5.

Image

Example 8We want to compute Image, which converges because Image, while

Image

To obtain the value, we must resort to a standard subterfuge.

Consider f : Image2Image defined by Image. Since

Image

when x2 + y2 is sufficiently large, it follows from Example 3 and the comparison test that Image is absolutely integrable on Image2. If Dn denotes the disk of radius n in Image2, then

Image

On the other hand, if Sn denotes the square with edgelength 2n centered at the origin, then

Image

Comparing the two results, we see that

Image

Since Image is an even function, it follows that

Image

Example 9The important gamma function Γ : (0, ∞) → Image is defined by

Image

We must show that this improper integral converges for all x > 0. Now

Image

If Image, then f(t) = tx−1 et is continuous on [0, 1]. If Image, then Image, so the first integral on the right converges, because g(t) = tx−1 is absolutely convergent on (0, 1) by Example 4, since 1 − x < 1.

To consider the second integral, note first that

Image

so

Image

for t sufficiently large. Since 1/t2 is absolutely convergent on (1, ∞), it follows that Image converges.

The fundamental property of the gamma function is that Γ(x + 1) = xΓ(x) if x > 0:

Image

Since Γ(1) = 1, it follows easily that

Image

if n is a positive integer (Exercise 6.1).

Example 10The beta function is a function of two variables, defined on the open first quadrant Q of the xy-plane by

Image

If either x < 1 or y < 1 or both, this integral is improper, but can be shown to converge by methods similar to those used with the gamma function (Exercise 6.9).

Substituting t = sin2 θ in (6) (before taking the limit in the improper cases), we obtain the alternative form

Image

The beta function has an interesting relation with the gamma function. To see this, we first express the gamma function in the form

Image

by substituting t = u2 in (5). From this it follows that

Image

This last equality is easily verified by use of the approximating sequence Image for Q, where An = [1/n, n] × [1/n, n].

It follows from Corollary 6.5 that the integrand function in (9) is absolutely convergent, so we may use any other approximating sequence we like. Let us try Image, where Bn is the set of all those points with polar coordinates (r, θ) such that Image and Image (Fig. 4.42). Then

Image

using (7) and (8) and the obvious symmetry of the beta function. Hence

Image

Image

Figure 4.42

As a typical application of (7) and (10), we obtain

Image

Another typical example of the use of gamma functions to evaluate integrals is

Image

making the substitution u = x3 and using the fact that Image (Exercise 6.2).

Exercises

6.1Show that Γ(1) = 1, and then deduce from Γ(x + 1) = xΓ(x) that Γ(n + 1) = n!.

6.2Show that Image. Hint: Substitute t = u2 in the integral defining Image, and refer to Example 8. Then prove by induction on the positive integer n that

Image

6.3Recalling from the previous exercise section the formulas for the volume αn of the unit n-dimensional ball Image, deduce from the previous two exercises that

Image

for all Image.

6.4Apply Exercises 6.1 and 6.2, and formulas (7) and (10), to obtain a new derivation of

Image

6.5Show that Image by subtituting x = t1/2.

6.6Use the substitution t = xn to show that

Image

6.7 Use the substitution x = eu to show that

Image

6.8Show that

Image

6.9Prove that the integral defining the beta function converges.

6.10Show that the mass of a spherical ball of radius a, with density function d(x, y, z) = x2y2z2, is M = 4πa9/945. Hint: Calculate the mass of the first octant. Introduce spherical coordinates, and use formulas (7) and (10).

6.11Use ellipsoidal coordinates to show that the mass of the ellipsoidal ball Image, with density function d(x, y, z) = x2 + y2 + z2, is

Image

6.12By integration by parts show that

Image

Deduce from this recursion formula that

Image

and

Image

Apply Exercises 6.1 and 6.2 to conclude that

Image

for all integers Image.

6.13The purpose of this exercise is to give a final computation of the volume αn of the unit ball Image.

(a)Let T : ImagenImagen be the n-dimensional spherical coordinates mapping of Exercise 5.19. Note by inspection of the definition of T, without computing it explicitly, that the Jacobian of T is of the form

Image

for some function γ.

(b)Let f: ImagenImage be an absolutely integrable function such that g = f Image T is a function of ρ alone. Then show that

Image

for some constant Cn (independent of f). Setting f = g = 1 on Bn, a = 1, we see that αn = Cn/n, so it suffices to compute Cn.

(c)Taking limits in (*) as a → ∞, we obtain

Image

so it suffices to find an absolutely integrable function f for which we can evaluate both of these improper integrals. For this purpose take

Image

so Image. Conclude from Example 8 that Image, and then apply the previous exercise to compute Cn, and thereby αn.

6.14The purpose of the problem is to evaluate

Image

(a)First show that Image if x > 0.

(b)Then use integration by parts to show that

Image

(c)Hence

Image

provided that the interchange of limit operations can be justified.

6.15The object of this problem is to show that f(x) = (sin x)/x is not absolutely integrable on (0, ∞). Show first that

Image

Given any compact contented set Image, pick m such that Image, and then define

Image

for all n > m. Now conclude, from the fact that Image diverges, that

Image

Why does this imply that ∫(0, ∞) [(sin x)/x] dx does not exist?

6.16If U is an open subset of Imagen, show that there exists an increasing sequence Image of compact contented sets such that Image. Hint: Each point of U is contained in some closed ball which lies in U. Pick the sequence in such a way that Ak is the union of k closed balls.

6.17Let q(x) = xtAx be a positive definite quadratic form on Imagen. Then show that

Image

Outline:Let P be the orthogonal matrix provided by Exercise II.8.7, such that PtAP = P−1AP is the diagonal matrix whose diagonal elements are the (positive) eigenvalues λ1, . . . , λn of the symmetric matrix A. Use the linear mapping L : ImageynImagexn defined by x = Py to transform the above integral to

Image

Then apply the fact that Image by Example 8, and that fact that λ1λ2 · · · λn = det A by Lemma II.8.7.