INFINITE PRODUCTS
AND INTEGRATION.
DRAFT COPY ONLY.
8/4/2005
Raimond A. Struble, PhD.

Professor Emeritus
Department of Mathematics
North Carolina State University at Raleigh
Raleigh, NC.

http://www.infiniteproduct.info/struitgr.htm
© 2005, Raimond A. Struble, PhD.

Send comments and correspondence to: raimondstruble@yahoo.com
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ABSTRACT.

A brisk treatment of infinite products, which lead to Lebesgue integration, is presented at an elementary (undergraduate) level in mathematics. This includes further clarification (beyond that of references [1], and [2]) of the geometric role played by rectangle-filling by isosceles triangles˚ as related to infinite products˚ and integration˚, including now double integrals˚. We present completely new interpretations of all (real-valued) infinite products, using a visual scheme of filling double rectangles by isosceles triangles. One rectangle captures the increasing factors, while the other rectangle captures the decreasing factors. Every possibility (convergent or not) is included, and the indeterminate cases are resolved whenever appropriate.

TABLE OF CONTENTS.


Abstract.
Table of Contents.
Introduction.
Chapter 1: Rectangle-filling and infinite products.
       Section 1A: Decreasing Products and the Filling-Lemma.
      Section 1B: Riemann-zeta Case.
Chapter 2: Increasing Products.
Chapter 3: Mixed Increasing and Decreasing Products.
Chapter 4: Remanding by Inverted Triangles.
Chapter 5: Infinite Products and Integration.
      Section 5A: Link between Decreasing Products and Integration.
      Section 5B: Riemann-Zeta Case.
Chapter 6: Products Depending upon a Parameter and Double Integrals.
      Section 6A: The Cosine Example.
      Section 6B: The Sine Example.
      Section 6C: Double Integrals.
      Section 6D: Riemann-zeta Example.
Chapter 7: Decreasing Products and Integration with Remanded Triangles.
Chapter 8: Double-rectangle and Mixed Products.
Chapter 9: Quotient-products and Double-rectangle-Filling.
Chapter 10: Integration with Positive and Negative Tent-functions.
Chapter 11: The Reverse Question: Mikusiński's Theorem.
Chapter 12: Acknowledgments.
Chapter 13: References.

INTRODUCTION.

References [1], and [2], furnish the principal source of introduction to the present work, although this treatment is intended, more or less, as an independent development. The topics are separated into many chapters, for easy passage through the material. Hopefully, the journey is clear and rewarding to those interested in infinite products. Very many concrete examples and applications of infinite products are presented in the previous work. Here, we are mainly concerned with some theoretical underpinnings and the completion of the linkage to Lebesgue integration, relying heavily upon Mikusiński's Theorem.

CHAPTER 1: RECTANGLE-FILLING
AND INFINITE PRODUCTS.


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SECTION 1A. DECREASING PRODUCTS AND THE FILLING-LEMMA.

In two recent endeavors [1, 2], a geometric selection process, intending to fill up a rectangular area ˚ with isosceles triangles ˚ has been crystallized. The easiest way to explain this process has been through pictures of a few initial steps. One commences by filling one-half the rectangle with a single triangle, and then through a succession of steps, selects symmetrically located collections of isosceles triangles to fill the remaining half-rectangle, as illustrated below:

31.

32.

33.

34.

28.
The process is to continue indefinitely in infinitely many steps, with the ultimate question remaining: do the triangles completely fill the rectangle? One can quantitatively characterize this process by considering a corresponding sequence, r1, r2, r3, ... rn, ..., of non-negative numbers depicting the fractions of the unfilled portions assailed by the triangles at each step. At step 1, two triangles fill an r1-fraction of the remaining area. This leaves an (1-r1)-fraction of the area yet-to-be-filled. At step 2, six triangles fill an r2-fraction of the remaining area. This leaves a (1-r1)(1-r2)-fraction of the area yet-to-be-filled. At step 3, eighteen triangles fill an r3-fraction of the remaining area. This leaves a (1-r1)(1-r2)(1-r3)-fraction of the area yet-to-be-filled. Clearly, if this process is continued indefinitely, then the infinite product:
(1.1)        P = (1-r1)(1-r2)(1-r3) ... (1-rn)....
gives the final fraction of the area remaining. Only if P=0, do the isosceles triangles˚ completely fill the rectangle. If P>0, then that number is the fraction of the one-half rectangle that has not been filled.

In [2], it is demonstrated that, in any event, the product P always satisfies the inequalities:
(1.2)        (1-R)S/R < P < e-S,
where S is the sum of the corresponding infinite series (finite or infinite):
(1.3)        S = r1 + r2 + r3 + ... + rn + ...,
and R = l.u.b. 1<n<∞ rn. Various ramifications of these inequalities are expounded upon in [2]. Here we simply declare what we call the Filling-Lemma:
FILLING-LEMMA: The sequence of isosceles triangles conforming to the fractions rn of unfilled portions of the rectangle, will completely fill the rectangle, if and only if the corresponding infinite series (1.3) diverges (S = ∞). Whenever S converges (S < ∞), the corresponding infinite product (1.1) is positive and P is the fraction of the one-half rectangle that is not filled.
The above illustration depicts the case rn=1/4 (all n), and so, indeed, such triangles do completely fill the rectangle. In many other interesting examples, the triangles do not completely fill the rectangle.

There are aspects of this special rectangle-filling process that should be noted. First of all, the process is characterized by a sequence of steps in which specific, non-overlapping collections of isosceles triangles are placed so as to be in contact with earlier triangles (and the rectangle). Except for the preliminary, large triangle, every individual isosceles triangle occupies an unfilled portion within a skewed triangle pointing downward, and making contact with the edges of the latter. This process can result in either a fat or slim isosceles triangle with reduced area, or a maximum-area triangle having one-half the dimensions of the skewed triangle:
FAT:

36.

MAX (1/4):

73.

SLIM:

38.


Because of this, any infinite product (1.1), reflecting this particular geometric filling-process, is limited by the condition rn < 1/4 for all n. This is not a very severe limitation, much as it appears, since if rn > 1/4 for infinitely many n, then the series S diverges, and the immediate conclusion is that P=0. (The omission of finitely many factors only changes the numerical value of an infinite product, which can easily be accounted for.) Whenever rn is appreciably less than 1/4, then the corresponding isosceles triangles must either be very slim or very fat ones, and the pictures illusrating the filling-steps are severely altered. For example, with the fractions rn = 1/n (n > 1/4), such as from the harmonic series˚, the slim triangles˚ rapidly degenerate into essentially vertical-line segments:

355.
Nonetheless, in this case the triangles do completely fill the rectangle, since the harmonic series diverges.

SECTION 1B. RIEMANN-ZETA CASE.

In [1, 2], we have pointed out a more dramatic (and interesting) case, wherein the fractions are the Riemann-zeta-numbers, rn = 1/ps, where p is the nth prime, and s is the (real) Riemann variable, greater than 1. Four of the many, many possibilities using slim and fat triangles in various ways are illustrated below, when s=2.

356.

357.

390.

391.
Regardless of how the triangles are chosen, however, they do not completely fill the rectangle, since the corresponding infinite series:
(1.4)        S = 1/2s + 1/3s + 1/5s + 1/7s + ...
converges for all s > 1. Therefore, by the Filling-Lemma, a fraction:
(1.5)        P = (1 - 1/2s)(1 - 1/3s)(1 - 1/5s)(1 - 1/7s) ...
of the one-half rectangle remains at the end. This familiar infinite product is equal to the reciprocal of the Riemann-zeta-function:
(1.6)        ζ(s) = 1 + 1/2s + 1/3s + 1/4s + 1/5s + 1/6s + 1/7s + ....        (all integers)
by Euler's well-known product formula. The fraction of the one-half rectangle that is actually filled, therefore, is given by:
(1.7)        F = 1 - P = 1 - 1/ζ(s).
A graph of this particular situation for s > 0 is rather interesting:

407.
This is not just an artificial piecing together of the F-values for the two ranges, 0 < s <1 and 1 < s, but rather, is an actual plot of the convergent infinite series:
(1.8)        F = 1/2s + (1 - 1/2s)1/3s + (1 - 1/2s)(1 - 1/3s)1/5s + (1 - 1/2s)(1 - 1/3s)(1 - 1/5s)1/7s + ...
for 0 < s. This is but a special case of the general formula:
(1.9)        F = 1 - P = r1 + (1 - r1)r2 + (1 - r1)(1 - r2)r3 + (1 - r1)(1 - r2)(1 - r3)r4 + ...,
which holds for any P in (1.1), and always converges (This formula can be obtained simply by expanding the factors in (1.1)). In the Riemann-zeta case, (1.8) converges to the correct constant value 1 for 0 < s <1, and these are the successful filling-cases, while (1.8) converges to 1 - 1/ζ(s) for s > 1, the unsuccessful filling-cases. Many more examples of rectangle-filling, and non-filling, by isosceles triangles are included in [2] (a few we shall revisit here), and for which the geometric interpretation is given by an infinite product, (1.1). These products have been labeled, decreasing products.

CHAPTER 2: INCREASING PRODUCTS.


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Also of interest are increasing products, such as:
(2.1)        P = (1+r1)(1+r2)(1+r3)...,
where the non-negative numbers, rn, are not, as yet, restricted. A complete treatment of these products as a separate topic is given in [2], but here we will recast these products in terms of decreasing products. The simple trick is to consider reciprocal products:
(2.2)        Pˇ = 1/(1+r1)(1+r2)(1+r3)...,
and replace each reciprocal, 1/(1+rn), by its equal, (1-řn), where:
             řn = rn/(1+rn)
Using inequalities (1.2) for:
(2.3)        Pˇ = (1 - ř1)(1 - ř2)(1 - ř3)...
one obtains:
(2.4)        (1 - Ř)Š/Ř < Pˇ < e-Š
where:
(2.5)        Š = r1/(1 + r1) + r2/(1 + r2) + r3/(1 + r3) + ...,
and Ř = l.u.b.k rk/(1 + rk). Thus when one reconsiders the increasing product, (2.1) itself, we obtain:
(2.6)        eŠ < P < (1 - Ř)-Š/Ř = [(1 + Ř/(1 - Ř))]Š/Ř = (1 + rk)Š/Ř.
The last expression holds whenever the l.u.b. above is an actual maximum, rk/(1 + rk). Moreover, a geometric interpretation of the filling of a rectangle by isosceles triangles follows from (2.3), whenever řn < 1/4 holds for all n. This outcome is guaranteed by the restriction rn < 1/4 in (2.1). Again, the latter is not a very severe restriction, since if rn > 1/4 for infinitely many n, then P = ∞ anyway. In this case, P = ∞ is the circumstance Pˇ = 0, so that P in (2.1) is finite if and only if the series (2.5) converges, and the rectangle is not completely filled by the triangles. The inequalities (2.6) provide useful information as to the numerical value of P, directly in terms of the rn-values. The right-hand member can be conveniently replaced by eS, where S = r1 + r2 + r3 ... (see (2.5) of [2]). We should emphasize that all the discussion and examples of Chapter 1: Rectangle filling and infinite products apply directly to these increasing products, requiring only that one take into account the fact that P = 1/Pˇ.

NEW CHAPTER 3: MIXED INCREASING AND DECREASING PRODUCTS.


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The trick of employing reciprocals is particularly useful for analyzing mixed increasing and decreasing infinite products:
(3.1)        P = (1 - r1)(1 + r2)(1 - r3) (1 + r4) ... (1 - r2n+1)(1 + r2n+2) ....
This formulation is completely general, since any of the rn-values can be zero, and the corresponding factors just disappear from the product. Replacing (1 + r2n+2) by its equal, 1/(1 - ř2n+2), with ř2n+2 = r2n+2/(1 + r2n+2), we obtain the equivalent quotient-product:
(3.2)        P = PQ = [(1 - r1)/(1 - ř2)] [(1 - r3)/(1 - ř4)] [(1 - r5)/(1 - ř6)] ....
which will be analyzed in detail in Chapter 8: Double-rectangle and Mixed Products. and in Chapter 9: Quotient-products and Double-rectangle-Filling. Here we make the simple observation that whenever the two infinite series:
(3.3)        r1 + r3 + r5 + ....
and
             ř2 + ř4 + ř6 + .....
converge, this quotient-product can be replaced (upon rearrangement of factors) by:
(3.4)        P = PQ = (1 - r1)(1 - r3)(1 - r5) ... / (1 - ř2)(1 - ř4)(1 - ř6)...,
which exhibits two decreasing products. Each of these can be depicted by rectangular-filling schemes (unsuccessful, of course), and whose numerical quotient yields the valid product number. The discussions and examples of Chapter 1: Rectangle-filling and infinite products apply directly to each of these.

When the series (3.3) may not converge, a possible treatment of (3.1), referred to as "turning a mixed increasing-decreasing product into a decreasing product" in [2], is available whenever the pairwise products:
             (1 - r2n+1)(1 - r2n+2) = (1 - r'2n+1)
result in non-negative r'2n+1 = r2n+1 - r2n+2 + r2n+1r2n+2 values. In such cases, (3.1) becomes
(3.5)        P = P' = (1 - r'1)(1 - r'3)(1 - r'5) ...,
and the inequalities
(3.6)        (1 - R')S'/R') < P = P' < e-S'/R'
hold, where S' = r'1 + r'3 + r'5 + ... and R' = l.u.b. r'2n+1. In the form (3.5), the mixed product (3.1) is subject to the filling lemma and to all the discussions and examples of Chapter 1: Rectangle filling and infinite products, and the inequalities (3.6) furnish useful information as to the numerical value of P.

Should one presuppose the opposite effect, where the pairwise products:
             (1 - r2n+1)(1 + r2n+2) = 1 + (1 - r"2n+2)
result in a non-negative r"2n+2 = r2n+2 - r"2n+1 - r2n+1r2n+2 values, then (3.1) becomes:
(3.7)        P = P" = (1 + r"2)(1 + r"4)(1 + r"6) ....
In [2], it is shown that P" satisfies the inequalities:
(3.8)        (1 + R")S"/R") < P = P" < eS"/R",
where S" = r"2 + r"4 + r"6 + ..., and R" = l.u.b. r"2n+1. Hence in this circumstance, P = P" is finite if and only if S" < ∞.

Finally, one can replace (1- r2n+1) with its equal, 1/(1 + ř2n+1) in (3.1),
where ř2n+1 = r2n+1/(1 + r2n+1), and obtain another intriguing equivalent quotient form:
(3.9)        P = PQ = [(1 + r2)/(1 + ř1)] [(1 + r4)/(1 + ř3)] [(1 + r6)/(1 + ř5)] ...,
to complete the landscape (so to speak). Again, we observe that whenever the two series
(3.10)        ř1 + ř2 + ř3 + ...
and
             r2 + r4 + r6 + ...
converge, this quotient-product (3.9) can be replaced (upon rearrangement of factors) by:
              P = PQ = (1 + r2)(1 + r4)(1 + r6) ... / (1 + ř1)(1 + ř3)(1 + ř5) ...
Each of these increasing products converges, and their numerical quotient yields the valid product number. (The convergences in (3.3) and (3.10) follow from those of the primary series r1 + r3 + r5 + ... and r2 + r4 + r6 + ...).

CHAPTER 4: REMANDING BY INVERTED TRIANGLES.


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Pursuing the rectangle-filling process still further, we now reconsider, and extend the "remanding" concept illustrated in [2]. This concept is best explained by again illustrating a few beginning steps. Following step 1, after having placed two isosceles triangles according to the fraction r1, step 2 becomes a remanding step, consisting of returning to the rectangle a portion r2 of the two r1-triangles. This remanding step is depicted by two inverted isosceles triangles.

393.
Then the portion of the 1/2-rectangle yet-to-be-filled becomes:
             (1 - r1) + r1r2,
since r1 is the fraction un-remanded at step 1, and r2 is the fraction of the latter that is actually-remanded. This quantity, r1r2, is returned to the rectangle, so the plus sign is appropriate. Following step 3, after having placed eight isosceles triangles (two in the r2 remanded, inverted triangles) according to the fraction r3, an r4-portion of these eight r3-triangles is remanded to the rectangle. Again, the remanding step is depicted by eight inverted isosceles triangles.

394.
Then the portion of the 1/2-rectangle yet-to-be-filled becomes the product:
              [(1 - r1) + r1r2] [(1 - r3) + r3r4].
When this complicated remanding procedure is continued indefinitely, the ultimate fractional part of the 1/2-rectangle remaining is given by the infinite product:
(4.1)        PR = [(1 - r1) + r1r2] [(1 - r3) + r3r4] ... [(1 - r2n+1) + r2n+1r2n+2] ...
= [(1 - r1(1 - r2)] [(1 - r3(1 - r4)] ... [(1 - r2n+1(1 - r2n+2)] ...,
where the even-numbered fractions depicted the remanding steps, following the filling, odd-numbered steps. In the procedure illustrated above, the r2n+2-values are bounded by 1/4, since the inverted triangles are placed within the right side-up ones. As if this remanding process isn't complicated enough already, we now ask if it is possible to find some way to increase the value of the remanding fractions r2n+2 in (4.1). The answer is yes, and it can even be illustrated pictorially. For if at the remanding step 2, several inverted isosceles triangles are employed to exhaust more of the area of the two r1-triangles (the Filling-lemma guarantees the extreme possibility of remanding it all) then r2 will be increased accordingly:

395.
This modification can be continued indefinitely, and thus, without being specific, one imagines an extension of (4.1) to cover all situations where the even-numbered fractions r2n+2 are restricted only by the inequality:
(4.2)        r2n+2< 1


With this relaxed inequality, it is possible to convert many mixed increasing and decreasing products:
(3.1)        P = (1 - r1)(1 + r2)(1 - r3) (1 + r4) ... (1 - r2n+1)(1 + r2n+2) ...
into visible form of remanding products (4.1). For if the inequality
(4.3)        r2n+2 < r2n+1 / (1 - r2n+1)
holds, then the fractions
(4.4)        r2n+2* = (1 - r2n+1) r2n+2 / r2n+1
satisfy (4.2), and
             [(1 - r2n+1(1 - r2n+2*)] = (1 - r2n+1)(1 + r2n+2).
Therefore,
(4.5)        P = PR = [1 - r1(1 - r2*)] [1 - r3(1 - r4*)] ... [1 - r2n+1(1 - r2n+2*)] ...
The inequality (4.3), (same ones leading to (3.5)) allows for the r2n+2 values greater than r2n+1 values. In fact, whenever r2n+1 < 1/4 holds, then r2n+2 < 1/3 will suffice, which is not very restrictive. However, when the opposite inequality:
(4.6)        r2n+1 / (1 - r2n+1) < r2n+2
holds, then the mixed products (3.1) can be expressed as an increasing product:
(3.7)        P = P" = (1 + r"2)(1 + r"4)(1 + r"6) ....
where r2n+2" = r2n+2(1 - r2n+1) - r2n+1 > 0. This (in turn) leads to a reciprocal product (2.2) for evaluation and analysis as a decreasing product. It appears that a majority of mixed products, therefore, can be displayed as a decreasing product, using either the remanding process or (2.2). This is particularly important for resolving indeterminacies, when both of the infinite products (1 - r1)(1 - r3)(1 - r5)... and (1 + r2)(1 + r4)(1 + r6)... diverge. (See Chapter 9: Quotient-products and Double-rectangle-Filling.)

It is of interest to interpret (4.1) as a decreasing infinite product, for which the rate of decrease is simply retarded by the internal (1 - r2n+2) factors. Of course, these can now retard the decrease sufficiently to produce a positive value of P, where the odd-numbered factors (1 - r2n+1) by themselves, would not. The geometric picture of rectangle-filling (and remanding) by isosceles triangles can be very, very complicated, but the meaning of the numerical results in (4.1) are straightforward enough.

Finally, we remark that the remanding procedure can be instituted at the very beginning. For a portion of the area of the preliminary big triangle (step 0) can be remanded by inverted triangles as illustrated.

404.
These remanded, inverted triangles then participate in the subsequent filling-procedure. We choose not to pursue the relative arithmetical details, such an 0-step entails, but will continue to work with the 1/2-rectangle, as initially envisioned.

CHAPTER 5: INFINITE PRODUCTS
AND INTEGRATION.


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SECTION 5A. LINK BETWEEN DECREASING PRODUCTS AND INTEGRATION.

An interesting link between decreasing infinite products (1.1) and integration was exploited in [1]. (Actually, initiating 20 years earlier in a similar special colloquium given, 17 April 1984, at N.C.S.U., entitled, "Can one do research using calculus and pictures?") For this purpose, using the geometrical picture, one envisions a process of lowering all filling isosceles triangles to the base of the rectangle, and subsequently interpreting their images as being defined by simple functions of a variable, x, along the base. Because these are triangular in shape and their graphs resemble tents, we have chosen to call them tent-functions:

396.
The combined graph of these infinitely many tents is, in general, very complicated indeed, most especially when coming from very slim triangles˚ and very fat triangles˚ (as those displayed in Chapter 1. Rectangle-filling and infinite products). But if we imagine that these functions become linearly-ordered in some manner or other:
(5.1)        t1(x), t2(x), t3(x), ..., tn(x), ...,
then their integrals over the base of the rectangle add up to:
(5.2)        F = ∫t1(x)dx + ∫t2(x)dx + ∫t3(x)dx + ... + ∫tn(x)dx + ...,
which is just the sum of the areas of the filling isosceles triangles, where F = 1 - P. Now the pointwise sums of the tent-functions in (5.1)
(5.3)        L(x) = t1(x) + t2(x) + t3(x) + ... + tn(x) + ....
define a Lebesgue-integrable-function of x, of perhaps a severe nature:

397.
Nonetheless, the infinite decreasing product:
(1.1)        P = (1-r1)(1-r2)(1-r3) ... (1-rn)....
is given directly in terms of the integral of any such function by:
(5.4)        P = 1 - F = 1 - ∫ L(x)dx.
Of course, there are many, many such Lebesgue-integrable-functions, L(x), (5.3) satisfying (5.4) for any given product, P. But so long as the restriction, rn < 1/4 is maintained for all n, the geometric picture stemming from any rectangle-filling process by isosceles trianges conforming to the fraction, rn, can be used to display some of the more interesting ones, which are closely connected to the infinite product. Since increasing products (2.1) can be expressed in terms of decreasing products, the analogous formula for these becomes:
(5.5)        P = 1/(1 - ∫ Lˇ(x)dx),
where Lˇ(x) is one of the integrable-functions obtained by lowering the filling-triangles associated with the reciprocal product in (2.3).

SECTION 5B. RIEMANN-ZETA CASE.

It is instructive to examine this situation in the particular case of the rectangle-filling (or not) by isosceles triangles conforming to the Riemann-zeta-numbers. Since the product, P, in (1.5) is given by:
(5.6)        P(s) = 1/ζ(s)
(at least for s > 1), and F(s) = 1 - P(s) yields the fraction of filling for any of the various choices, (5.4) yields, more explicitly:
(5.7)        F(s) = 1 - 1/ζ(s) = ∫ L(s,x)dx,
for s > 1. However, as noted previously, F(s)=1 for 0 <s < 1, and can be expressed as a convergent infinite series (1.8) throughout the entire range s > 0. The myriad of integrable-functions L(s,x) are best visualized by reference to the illustrations of Chapter 1: Rectangle-filling and infinite products for s > 0. For 0 <s < 1, not only is F(s)=1, but for any possible choice of collections of tent-functions (stemming from the successful filling-triangles), the corresponding pointwise sum:
(5.8)        L(s,x) = t1(s,x) + t2(s,x) + t3(s,x) + ... + tn(s,x) + ....
will produce the identity L(s,x) = 1 along the associated rectangle. This result, of course, is because the corresponding triangles completely fill up the rectangle. The interesting functions, L(s,x), are those for s > 1, which one can only guess at.

CHAPTER 6: PRODUCTS DEPENDING UPON A PARAMETER
AND DOUBLE INTEGRALS.


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In this Riemann-zeta-number example, the infinite products involved depend upon a parameter, which is typical of many examples of such products. In [2], we have examined quite a number of these, and for which the letter x was used as the parameter. Here we would rather retain the letter x to symbolize the extent of a variable along the base of various rectangles considered, as above for L(s,x) in (5.8). So we will make something of a heretical choice, and use s as a parameter throughout.

One of the simplest examples from [2] is the increasing product:
(6.1)        P0(s) = (1 + 1/s)(1 + 1/s2)(1 + 1/s3) ... (s > 4 for rectangle-filling)
which, in turn, after inversion (see (2.1)) leads to a decreasing one:
(6.2)        P0ˇ(s) = (1 - 1/(1+s))(1 - 1/(1 + s2)) (1 - 1/(1 + s3)) ...,
If a rectangle is filled (actually unsuccessfully) by a selection of collections of isosceles triangles conforming to the fractions rn = 1/(1 + sn), then the pointwise sum of the corresponding tent-functions, t1(s,x), t2(s,x), t3(s,x), ..., yields an integrable-function:
(6.3)        Lˇ(s,x) = t1(s,x) + t2(s,x) + t3(s,x) + ...
for which:
              P0(s) = 1/(1 - ∫ Lˇ(s,x)dx)
holds for s > 4. So:
(6.4)        ∫ Lˇ(s,x) = 1 - 1/P0(s).


SECTION 6A. THE COSINE EXAMPLE.

The following is a much more interesting example from [2], stemming from the well-known cosine infinite product:
(6.5)        cos π s = (1 - 4s2)(1 - 4s2/32) (1 - 4s2/52) ... (1 - 4s2/(2n-1)2) ...,
and we limit s to the interval 0 < s < 1/2, where it remains a decreasing product (s < 1/4 is needed for rectangle-filling). If a rectangle is filled (unsuccessfully) by a selection of collections of isosceles triangles conforming to the fractions rn = 4s2 / (2n - 1)2, then the pointwise sum of the resulting tent-functions, t1(s,x), t2(s,x), t3(s,x)..., yield an integrable-function:
              L(s,x) = t1(s,x) + t2(s,x) + t3(s,x) + ...,
for which:
(6.6)        cos π s = 1 - ∫ L(s,x) dx
holds. The graph of cos πs gives no indication of which L(s,x) function of x might be involved for each s.

SECTION 6B. THE SINE EXAMPLE.

Another well-known example stems from the sine infinite product (quotient form):
(6.7)        sin πs/πs = (1 - s2)(1 - s2/22)(1 - s2/32)(1 - s2/42) ...
which for 0 < s < 1 remains a decreasing product. (For rectangle-filling, s < 1/2). Again, for each such s, we envision an integrable-function:
             L(s,x) = t1(s,x) + t2(s,x) + t3(s,x) + ...,
resulting from some selection of collections of isosceles triangles conforming to the fractions rn = s2/n2 of the filling-triangles, and for which:
(6.8)        sin πs/πs = 1 - ∫ L(s,x) dx
holds.

SECTION 6C. DOUBLE INTEGRALS.

We are particularly concerned with these latter two examples, since they facilitate the extension to double integrals. Indeed, there is very little mystery as to what the integrals with respect to the paramater s give. From (6.6), one obtains (using the not uncommon symbolic confusion about variables and integration):
             ∫ cos π s ds = sin π s / π = s - ∫ ∫ L(s,x) dx ds,
or by (6.5):
(6.9)        ∫0s (1 - 4s2)(1 - 4s2/32) (1 - 4s2/52) ... ds = s - ∫0s ∫ L(s,x) dx ds,
in infinite product form explicitly for the definite integral. From (6.7), one obtains directly the infinite product form for the definite integral:
(6.10)        ∫0s (1 - s2) (1 - s2/22) (1 - s2/32) ... ds = s - ∫0s ∫ L(s,x) dx ds,
where we do not have an antiderivative in closed form to use. For s=1/4 and s=1/2, one can write, respectively:
(6.11)        ∫01/4 cos π s ds = 1/4 - ∫01/4 ∫ L(s,x) dx ds = 1/21/2 π.
and:
(6.12)        ∫01/2 sin π s / π s ds = 1/2 - ∫01/2 ∫ L(s,x) dx ds = ? (numerical?)
The point to be made with these two concrete examples is that in the general case, (5.4) often becomes of special interest when the product factors involve and auxiliary parameter s, and we can write:
(6.13)        P(s) = 1 - F(s) = 1 - ∫ L(s,x) dx.
Therefore, the definite (and even indefinite) integrals of such a product P(s), simple or not, might be contemplated and readily identified. In such situations, these can be throught to involve double-integrals of two-variable-functions, L(s,x). The construction (definition) of the two-variable-function involves a selection, for each fixed s, of collections of rectangle-filling-triangles conforming to the fractions rn(s) of the product:
(6.14)        P(s) = (1 - r1(s)) (1 - r2(s)) (1 - r3(s)) ... (with rn(s) < 1/4)
and lowering these to the base of the rectangle, to form a sequence, t1(s,x), t2(s,x), t3(s,x), ..., of tent-functions (of x). Then the pointwise sum:
(6.15)        L(s,x) = t1(s,x) + t2(s,x) + t3(s,x) + ...
is what ends up on the right-hand side in (6.13). A single integration then of the left-hand member leads to the formula:
(6.16)        ∫ P(s) ds = s - ∫ ∫ L(s,x) dx ds.
for the double-integral on the right. In many examples, the left-hand-member is readily evaluated, or at least easily interpreted from a graph of P(s), as in the above cases. It is of passing interest to observe that if the above treatment of the sine example is modified so as to employ the (non-quotient form) infinite product:
             sin πs = πs (1 - s2) (1 - s2/22) (1 - s2/32) ...,
then the selections of filling triangles need not change at all, since the factor, πs merely changes the scale of any associated rectangle (See Chapter 11: The Reverse Question. Mikusiński's Theorem). We would then have explicitly,
             0∫1/2 sin πs = 1/2 - 0∫1/2 L(x,s) dx ds = 1/π,
in lieu of (6.12).
Chapter 4: Remanding by Inverted Triangles.

SECTION 6D. RIEMANN-ZETA EXAMPLE.

Below we attempt to illustrate this overall phenomenon with the Riemann-zeta example.

398.
At each s-station, the function-values of L(s,x) (along the rectangle) are obtained from the summation in (6.15), where the tent-functions, tn(s,x), have resulted from the lowering of the filling-isosceles-triangles corresponding to the zeta-numbers, 1/ps. The degree of success of the filling-process is given by the final fraction, F(s), displayed along the right-hand edge of the rectangle at s. This display, when viewed along the entire s-axis, is the same as in the graph following (1.7), and at each s-station, for s > 1, gives the area F(s) under the very ragged L(s,x) curve (explicitly: the number 1 - P(s) = 1 - 1/ζ(s)). The composite then of all these L(s,x) values defines the two-variable-function in the (s,x)-plane and for which F(s) = ∫ L(s,x)dx holds at each s-station. (For 0 < s < 1, L(s,x)=1). Then by Fubini's Theorem, the double-integral:
              ∫0s ∫ L(s,x) dx ds = ∫0s F(s) ds
is equal to the repeated integral, now represented as:
(6.17)        ∫0s F(s)ds = ∫0s (1 - P(s))ds = ∫01 ds - ∫1s 1/ζ(s) ds
             = 1 - ∫1s 1/ζ(s) ds,
for s > 1. Formulated as in (6.16), this expression becomes simply:
(6.18)        ∫0s P(s)ds = ∫1s 1/ζ(s) ds,
where:
(6.19)        P(s) = (1 - 1/2s)(1 - 1/3s)(1 - 1/5s) ... = 1/ζ(s),
for s > 1. Because of the latter, (6.18) is readily "verifiable", using the fundamental theorem of calculus! So the above ((6.17) to (6.19)) appears to be nothing more than the "manipulation" of symbols. However, embedded in it all is the interpretation of (6.17) as a double-integral:
(6.20)        ∫0s F(s)ds = ∫0s ∫ L(s,x) dx ds.
Also, the "manipulation" in this case owes much of its success to the Euler product formula, where the infinite product is comfortably identified. So a graph of (6.17) versus s is readily obtained, without having to work with an infinite product:

408.
It is a very nice, smooth curve, very unlike that of L(s,x) versus x at any s-station, or possibly, L(s,x) versus s at any x-station, for s > 1. The integral, ∫0s L(s,x) ds versus s, might be a somewhat smoother function, but essentially unattainable, unless one uses, say, slim triangles˚ throughout, or perhaps, fat triangles˚ throughout. This question requires further study.

CHAPTER 7: DECREASING PRODUCTS AND INTEGRATION
WITH REMANDED TRIANGLES.


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As a most important development, we now take note of the interesting link between decreasing products and integrals, in the presence of the remanded triangles as described in Chapter 4: Remanding by Inverted Triangles. Just how does one interpret the molding of these inverted isosceles triangles into tent-functions? The simplest arithmetical answer is to imagine that the inverted triangles are re-inverted into right-side-up triangles, and then lowered to the base of the rectangle, yielding negative-valued tent-functions in (5.1). Then their contributions to the sums in (5.2) and (5.3) result in the correct (arithmetical) integral, F, and the function, L(x). Because of the rectangle-filling picture, these series are absolutely convergent (and so conform to the ADVANCED THEOREM in [1], concerning Lebesgue integrals). A more satisfactory geometric view of the molding process involves lowering the inverted triangles, as they are, to the base of the rectangle, so that they become negative-valued tent-functions, with graphs protruding below the x-axis (rectangle base):

400.
This picture shows why the inverted triangle tent-functions enter the above sums as negative numbers, while the right-side-up triangles enter as positive numbers. Consequently, for this geometric interpretation, one should display a double-rectangle picture (above and below), with the usual conventions of positive and negative values above and below the x-axis, when it concerns tent-functions:

401.
Such a picture not only accommodates the above inverted triangles, but more general filling-schemes (beyond the remanding ones), where other positive and negative tent-functions can be included. This is how one arrives at negative and positive terms in the sums (5.2) and (5.3). For such schemes, the double-rectangle picture can often be used to insure absolute convergence of the series of these positive and negative tent-functions. Another important observation: The remanded (upside-down) triangles can be transported as collections of filling, non-overlapping, isosceles triangles into the lower rectangle. (See the following chapter.)

CHAPTER 8: DOUBLE-RECTANGLE
AND MIXED PRODUCTS.


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On the other hand, using this double-rectangle picture for mixed infinite products:
(3.1)        P = (1 - r1)(1 + r2)(1 - r3)(1 + r4) ....
necessitates a slight re-examination. So long as rn < 1/4, the odd-numbered sequence of fractions, r2n+1, correspond to the ordinary filling of the upper-rectangle, while the even-numbered sequence of fractions, r2n+2, reflect (not correspond to) the filling of the lower-rectangle. Both rectangles are now (temporarily) viewed as having positive areas. The actual triangles to be selected for the filling of the lower-rectangles are to correspond to the reciprocal fractions, ř2n = r2n/(1+r2n), where (1 + r2n) = 1/(1 - ř2n holds, and where (3.1) assumes the quotient form (3.2).

402.

In general, we can envision here separate filling-choices between the upper and lower ones, with outcomes determined by the separate infinite series:
(8.1)        S- = r1 + r3 + r5 + ...
and:
(8.2)        S+ = r2 + r4 + r6 + ...
(For an extensive mathematical treatment of these mixed products using (8.1) and (8.2), see [2]).

In particular, the convergence (or not) of these reflect the complete filling (or not) of the rectangles. The product (3.1) itself may end up as any number from 0 to ∞, or it may not converge at all. Briefly, its value is bounded by the inequalities (see (3.3) in [2]):
(8.3)        (1 + R+)S+/R+ (1 - R-)S-/R- < P < eS+-S-,
where R+ = l.u.b.k r2k+2 and R- = l.u.b.k r2k+1. These also allow for P=0 and P=∞, where S- = ∞, S+ < ∞, and S- < ∞, S+ = ∞, but not in cases where S- = ∞, S+ = ∞, but where (3.1) may converge and yet not absolutely. It is instructive to contemplate such indeterminacy with the double-rectangle picture where, of course, there is a special order to the above and below filling-processes, which must be maintained for convergence. When both S+ and S- are finite, the order of filling is immaterial. (See [2] for another geometric scheme depicting an alternative filling-process: of an isosceles triangle by other isosceles triangles.) Under the restrictions rn < 1/4, this double-rectangle filling-picture with the odd r2n+1 fractions above and the even ř2n+2 fractions below, correspond to the mixed product (3.1) in the quotient form:
(3.2)        P = PQ = ( [(1 - r1)/(1 - ř2)] [(1 - r3)/(1 - ř4)] [(1 - r5)/(1 - ř6)] ....
In the situation with S- < ∞ and S+ < ∞, the mixed product (3.1) is also given by P = P+P-, where:
             P+ = (1 + r2)(1 + r4)(1 + r6)...
and:
             P- = (1 - r1)(1 - r3)(1 - r5)...
Now by (2.3):
             P+ = 1 /(1 - ř2)(1 - ř4) (1 - ř6) ... = 1 / Pˇ.
Therefore, P = PQ = P+ / Pˇ. With this expression, (8.3) can be re-expressed in the more revealing form:
(8.4)        (1 - R-)S-/R- eŠ < P < (1 - Ř)-Š/Ř e-S-,
where the series, Š = ř2 + ř4 + ř6 + ... corresponds directly to the filling of the lower rectangle, with Ř = l.u.b.k ř2k, and the series S- = r1 + r3 + r5 + ... corresponds to the filling of the upper-rectangle, with R- = l.u.b.k r2k+1.

It is, perhaps, appropriate to illustrate the idea of double-rectangle-fillings, as exemplified by mixed increasing and decreasing products, by a concrete example. To this end, we will examine the case of the Euler-type infinite product, where, however, we alternate the signs of the fractions involving the s-power of the primes:
(8.5)        PA(s) = (1 - 1/2s)(1 + 1/3s)(1 - 1/5s)(1 + 1/7s)(1 - 1/11s)(1 + 1/13s)...
for s > 1. If we recast this product into the standard quotient form, (3.2):
(8.6)        PA(s) = PAQ(s) = [(1 - 1/2s) / (1 - 1/(1+3s))] [(1 - 1/5s) / (1 - 1/(1+7s))] [(1 - 1/11s) / (1 - 1/(1+13s))] ...,
then R- = 1/2s, Ř(s) = 1/(1 + 3s), and S-(s) = 1/2s + 1/5s + 1/11s + ..., Š = 1/3s + 1/7s + 1/13s + .... So the other exponents in (8.4) become:
             S- / R- = 1 + (2/5)s + (2/11)s + ...
and:
             Š / Ř = (1 + 3s) [ 1/3s + 1/7s + 1/13s + ... ]
and (8.4) itself becomes (explicitly with...):
(8.7)        (1 - 1/2s)[1 + (2/5)s + (2/11)s + ...] e[1/3s + 1/7s + 1/13s + ...] < PAQ(s) =
             < (1 - 1 /(1 + 3s)-(1+3s)[1/3s+1/7s+1/13s+...] e-[1/2s + 1/5s + 1/11s + ...].
When only the dominant sums and factors are retained here as s ⇒ ∞, one finds that PAQ(s) is asympototic to e-1/2s, as would be expected from (8.5). Of course, the easiest way to obtain the actual value of the product for any s is to use (8.5) or (8.6) directly. In this special case, one can also obtain alternate inequalities using (3.3):
(8.8)        (1 + 1/(2s - 1))[-1 + 2s/[1 + 3s -(2/5)s+2s(1 / (1 + 1/7s))-(2/11)s +2s(1 / (1 + 1/13s)) + ...]
             < PA(s) < e[same exponent]
These last inequalities are valid over the extended domain, 0 < s, where (8.5) converges conditionally.

Turning now to the idea of double-rectangle-filling, according to the quotient form (8.6), the following picture illustrates the situation when s = 2, using slim triangles. The triangles actually shown (above and below) produce the (partial) quotient:
             (1 - 1/(22)) (1 - 1/(52)) (1 - 1/(112)) / (1 - 1/(1 + 32)) (1 - 1/(1 + 72)) (1 - 1/(1 + 132)) = 0.7140495 / 0.8768117 = 0.8143704
(a number within 0.07% of the value of the infinite product (8.6) for s=2).

414.
The filling triangles rapidly morph into vertical line segments. We note that since the geometric series expansions for the reciprocals of the increasing factors of (8.5):
             1/(1 + 1/ps) = 1 - 1/ps + 1/p2s - 1/p3s + ...
have alternating signs, an Euler formula for a "modified" zeta-type function has minus signs scattered about:
(8.9)        ζm(s) = 1 + 1/2s - 1/3s + 1/4s + 1/5s - 1/6s - 1/7s + 1/8s + 1/9s + 1/10s + 1/11s - 1/12s - 1/13s + ....
Here, the sign rule is complicated, but straightforward: every other power of every other prime factor of each integer n contributes a minus sign. Whatever -- our modified Euler product (8.5) is equal to the reciprocal, 1/ζm(s) of this "modified" zeta-type function. (Recall the standard (common) proof of Euler's product formula, using geometric expansions, say, in Refs. [8] and [19]). The sum of the product terms in (8.9) for s=2 is approximately ζm(2)= 1.23, which yields PA = 1/ζm(2) = 0.813, as compared to the partial product, 0.814, above.

CHAPTER 9: QUOTIENT-PRODUCTS
AND DOUBLE-RECTANGLE-FILLING.


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In Chapter 4: Remanding by Inverted Triangles, we derived the infinite product:
(4.1)        PR = [(1 - r1) + r1r2] [(1 - r3) + r3r4] ... [(1 - r2n+1) + r2n+1r2n+2] ...
= [(1 - r1(1 - r2)] [(1 - r3(1 - r4)] ... [(1 - r2n+1(1 - r2n+2)] ...,
depicting rectangle-filling in the presence of remanding triangles. In particular, it was shown that by the increased employment of inverted triangles, the interior factors, (1 - r2n+2) can be accentuated sufficiently to change the nature of the product, PR from a zero-value to a positive value. To pursue this possibility further, we first recast (4.1) in the usual form of a decreasing product:
(9.1)        PR = P' = (1 - r1')(1 - r3') ... (1 - r2n+1') ...,
where r2n+1' = r2n+1(1 - r2n+2). This form emphasizes the fact that PR = P' can be visualized by a simple rectangle-filling process using isosceles triangles, without resorting to the remanding process using inverted triangles. Here, any filling-triangles simply conform to the fractions r2n+1'. Focusing upon the rather special situation in which PR = P' > 0 (the r2n+1' triangles do not completely fill the rectangle), but where the (unremanded) product:
(9.2)        PN = (1 - r1)(1 - r3) ... (1 - r2n+1) ...
is zero (any choice of r2n+1 triangles do completely fill a rectangle), we now recast (4.1) into the interesting quotient form (as in (3.2)):
(9.3)        PR = P' = PQ = [(1 - r1) / (1 - ř2)] [(1 - r3) / (1 - ř4)] ... [(1 - r2n+1) / (1 - ř2n+2)] ...
where ř2n+2 = (1 - r2n+1)/ [1 - r2n+1(1 - r2n+2)] (in this case). Here, the quotient form displays a numerator product (9.2) which vanishes, and this implies that the denominator product:
(9.4)        PD = (1 - ř2)(1 - ř4) ... (1 - ř2n+2) ...
must also vanish, since the quotient, PQ = PR = P' is greater than zero. Thus if one envisions the filling by triangles of an upper rectangle according to the product PN in (9.2), and a lower rectangle according to the product PD in (9.4), then the quotient form (9.3) degenerates into the indeterminate form, 0/0. And yet, either the remanding picture (4.1) or the decreasing picture (9.1) resolves this indeterminacy, and yields the correct limiting value for the quotient. Of course, this demonstrates that the numerator and denominator factors in the quotient (9.3) can not be rearranged, if the correct value of the infinite product is to be obtained. For this example, both upper- and lower-rectangles are completely filled by the associated triangles, and such will be the case for any indeterminate situation encountered for mixed increasing and decreasing products (3.1), whenever S+ = ∞ and S- = ∞. (Recall the related discussion in Chapter 3: Mixed Increasing and Decreasing Products and Chapter 4: Remanding by Inverted Triangles.)

It is noteworthy that the rather bizarre series (8.9) is conditionally convergent for 0 < s < 1, and (because of the modified Euler product formula) resolves the 0/0 indeterminacy of the quotient form (8.6) of our modified Euler-type product, for this range of s-values.

It is clear that some indeterminacies in the quotient form (9.3) may be resolved by direct recourse to an equivalent decreasing product:
(9.5)        P = PQ = P" = (1 - r1")(1 - r3") ... (1 - r2n+1") ...
where:
(9.6)        r2n+1" = (r2n+1 - ř2n+1) / (1 - ř2n+2),
provided only that r2n+1" > 0 holds for all n sufficiently large. Should the opposite inequality entail, then (9.5) becomes an (eventually) increasing product, which again may resolve any indeterminacies. However, since many mixed products can be reformulated as decreasing products, many indeterminate cases can be resolved, when appropriate.

In all other situations where at least one of S+ or S- are finite, the double-rectangle-filling picture is readily applicable, and depicts the process vividly, whether the limit is positive, zero, or infinite.

Finally, we note that since every pair of factors, (1 - r2n+1)(1 + r2n+2) can be replaced by a single factor, (1 - r2n+1') or (1 + r2n+2") (or 1), a mixed infinite product (3.1) can be successively condensed by iteration of this process. If the condensed product ultimately becomes either increasing or decreasing, then one has rectangle-filling available to furnish the outcome. If, on the other hand, should the product be truly mixed, then this condensing process allows for the detection of the necessary reduction of its oscillations, if convergence is to occur at all, and there is any resolution of the indeterminacy possible. In the non-convergence cases, this condensation process may still be helpful in detecting such behavior.

CHAPTER 10: INTEGRATION WITH
POSITIVE AND NEGATIVE TENT-FUNCTIONS.


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Now, the link back to the integration concept using the sum:
(5.2)        F = ∫t1(x)dx + ∫t2(x)dx + ∫t3(x)dx + ... + ∫tn(x)dx + ...,
with:
(5.4)        1 - F = P = (1 - r1)(1 + r2)(1 - r3)(1 + r4) ...
requires some special considerations when the double-rectangle is employed. The tent-functions, tn(x), represented in the lower rectangular are obtained by raising the inverted triangles (which correspond to the reciprocal fractions, ř2n+2) to the x-axis, and must then be given negative values, as was done above for the remanded triangles. In this instance, the reason for the negative values lies in the fact that these triangles correspond to the reciprocal product:
             Pˇ = (1 - ř2)(1 - ř4)(1 - ř6) ...,
in the denominator of the quotient product PQ = P+ / Pˇ, where:
             PQ = P = P+ × P- = P+/Pˇ
(at least when not both S+ and S- are infinite). So, for purposes of showing the evaluation of the mixed product (3.1), using filling-triangles in the double-rectangles, the lower one is to be viewed as positive. On the other hand, for purposes of obtaining the appropriate tent-functions for (5.2), the lower triangle is viewed as negative. This integrable-function L(x) is then given by an absolutely convergent series:
(5.3)        L(x) = t1(x) + t2(x) + t3(x) + ...
of positive and negative numbers. These, of course, stem from certain collections of filling-triangles for the upper and lower rectangles, corresponding to fractions r2n+1 above and ř2n+2 below. Both collections of filling-triangles become tent-functions by movement to the x-axis. Those triangles pointing upward become positive ones, while those triangles pointing downward become negative ones. The numerical sums in (5.2) and (5.3) are then the correct ones, which correlate with the correct mixed product in (5.4).

CHAPTER 11: THE REVERSE QUESTION:

MIKUSIŃSKI'S THEOREM.


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The "reverse" question that we ask is simple: given a Lebesgue-integrable-function L(x), can one obtain an infinite product leading to rectangle-filling by isosceles triangles whose related tent-function series converges to L(x) (a.e.)? This is certainly an exceptional question to ask, as throughout the extensive history of Lebesgue integration, such questions are rarely (if ever) even considered. When given an integrable-function, one does not wander about looking for some kind of limit-mechanism to represent it. As done here, the brevy of convergence theorems (bounded convergence theorem, Lebesgue convergence theorem, etc.) are concerned with the results of limiting processes, leading to integrable-functions, without concern for (or interest in) the reverse question. So one should not expect to create a whole new, rigorous theory, attempting to answer the "reverse" question, but we can make some interesting observations leaning in that direction; with the assistance of the late Prof. J. Mikusiński, one obtains the complete answer in our case.

A first step is to consider a modified filling-process by rectangles, very much like that by triangles. The "twist" here is to begin by selecting a sufficiently elementary example, and then to proceed with the construction of the desired infinite product. The procedures used will carry over to the more general case, almost by analogy.

We consider a smooth, continuous, positive function L(x), and proceed to fill the area under its graph, using non-overlapping rectangles, as illustrated.

410.
This process is reminiscent of the elementary calculus definition of a Riemann integral, but where the filling-rectangles are chosen successively (somewhat arbitrarily, yet judiciously) to fill the area under the curve y = L(x) from a to b. Then ∫ab L(x) dx is just the sum of the areas of the rectangles. If we choose to express these rectangles:
(11.1)        b1(x), b2(x), b3(x), ..., bn(x), ...,
as simple (two) step-functions, (when lowered to the x-axis), then:
(11.2)        ∫ab L(x) dx = ∫ab b1(x) dx + ∫ab b2(x) dx + ∫ab b3(x) dx + ... + ∫ab bn(x) dx + ...
while:
(11.3)        L(x) = b1(x) + b2(x) + b3(x) + ... + bn(x) + ...
for a < x < b. This last equation is called a brick-expansion of the function L(x), and the bn(x) are called bricks (suggested by their graphs, of course).

For each of the original rectangles (before the lowering takes place) we assume one has selected a completely-filling collection of isosceles triangles. Then a reordering of these infinitely many collections of isosceles triangles presents us with a sequence of non-overlapping triangles, completely filling the area under the graph of L(x). When these are lowered to the x-axis, we obtain a sequence:
(11.4)        t1(x), t2(x), t3(x), ..., tn(x), ...
of tent-functions satisfying:
(11.5)        ∫ab L(x) dx = ∫ab t1(x) dx + ∫ab t2(x) dx + ∫ab t3(x) dx + ... + ∫ab tn(x) dx + ...
and:
(11.6)        L(x) = t1(x) + t2(x) + t3(x) + ... + tn(x) + ...
for a < x < b. Here we have moved from fractions to areas, which will require some further reconciliations. In this intermediate step, each of the rectangle-fillings can be described by a decreasing infinite product:
              P = (1 - r1)(1 - r2)(1 - r3) ....
The selection of the filling-triangles is arbitrary, except for the requirement that P=0, i.e, S = r1 + r2 + r3 + ... = ∞. However, each of the triangles must be scaled up or down by a scale-factor representing the actual area of the corresponding rectangle. When this is done, (11.5) and (11.6) hold, as claimed.

For the first triangle, its area is ∫ab t1(x) dx, constituting the fraction:
ρ1 = ∫ab t1(x) dx / ∫ab L(x) dx
of the total area under y = L(x). With this triangle removed, the second triangle has area ∫ab t2(x) dx, constituting the fraction:
             ρ2 = ∫ab t2(x) dx / [ ∫ab L(x) dx - ∫ab t1(x) dx]
of the remaining area. With the first and second triangles removed, the third triangle has area ∫ab t3(x) dx, constituting the fraction:
             ρ3 = ∫ab t3(x) dx / [ ∫ab L(x) dx - ∫ab t1(x) dx - ∫ab t2(x) dx]
of the remaining area. Continuing in this manner, where:
(11.7)        ρn = ∫ab tn(x) dx / [ ∫ab L(x) dx - ∫ab t1(x) dx - ∫ab t2(x) dx - ... - ∫ab tn-1(x) dx],
we obtain a sequence of fractions:
(11.8)        ρ1, ρ2, ρ3, ..., ρn, ...,
depicting the exhaustion of the area under y = L(x) by non-overlapping isosceles triangles, for which the corresponding infinite product:
(11.9)        P = (1 - ρ1)(1 - ρ2)(1 - ρ3) ... (1 - ρn) ...
is zero. The end-product itself seems rather unexciting, except for the picture generated by partial products displaying the progress of the filling of the area under the curve y = L(x). Should we replace the integral value ∫ab L(x) dx in the prescription (11.7) by a larger one, say B, equal to the area of a super-rectangle which encompasses the curve y = L(x), then (11.9) will converge to the value 1 - ∫ab L(x) dx/B > 0, and we can regard the triangles as filling (unsuccessfully) the super-rectangle, as envisioned from the beginning of Chapter 1: Rectangle-filling and infinite products, albeit one triangle at-a-time! Of course, one could lump together non-overlapping collections of the filling-triangles, and obtain alternative fractions, resulting in the same limit for the corresponding infinite product. This is how the original, very special selection process of Section 1A: Decreasing Products and the Filling-Lemma comes about in this context. The "super-rectangle" in these cases touches the graph of y = L(x) infinitely often, of course, and is merely the original rectangle.

409.
We observe that the quantity, B - ∫ab L(x) dx = BP, is just the area above the graph, y = L(x), and that B(1-P) = BF = ∫ab L(x) dx is the area under the graph, y = L(x). For this elementary case, we have produced an infinite product leading to rectangle-filling by isosceles triangles, whose related tent-function series converges to L(x). We will be able to do the same for more general integrable L(x).

It turns out that the very first step used in that elementary example is available for any Lebesgue-integrable-function L(x). Without going into all the finer details, we state (loosely) the pertinent result needed, which is due to Jan Mikusiński [11, 12]:

MIKUSIŃSKI EXPANSION THEOREM.

If L(x) is a Lebesgue-integrable-function (on the entire real line, if desired), then there exists a sequence of brick-functions (possibly positive and/or negative):
(11.10)        b1(x), b2(x), b3(x), ...
such that:
(11.11)        L(x) = b1(x) + b2(x) + b3(x) + ... (a.e.)
and for which:
(11.12)        ∫ab L(x) dx = ∫ab b1(x) + ∫ab b2(x) + ∫ab b3(x) + ...
holds. These two last series also converge absolutely. Because of this convergence, the positive and negative terms can be treated separately, with (11.11) and (11.12) substituting, in turn, for (11.3) and (11.2).

For each of these, the second step of replacing the brick-expansion (11.3) by the tent-expansion (11.6), is again done with completely-filling collections of isosceles triangles, which result in a single sequence of triangles, not necessarily non-overlapping as before. However, for these, (11.5) still holds for the corresponding sequence of tent-functions, (11.4). Using the definition (11.7), with ∫ab L(x) dx again replaced by a larger value B, results in a convergent infinite product (11.9). If the brick-functions are non-overlapping, then, as for the elementary example, the filling-triangles are also, and the modified limit in (11.9) will become 1 - ∫ab L(x) dx / B. This is no real limitation, since (by further refining the sequence) one can require the bricks in the expansion (11.11) be non-overlapping. Thus all the requirements of the elementary example can be met for any non-negative integrable-function L(x), for the existence of filling-triangles whose related tent-function series converges to L(x) (a.e.). But what is filled in this general case is not so clear, since the brick-functions (and triangles) may not be bounded! Thus we do need to limit our L(x)-functions to be bounded-functions, so that a rectangle of sufficient size can be chosen so as to encompass the graph of y = L(x). Other than that, the answer to the reverse question for a positive function is in the affirmative.

Finally, for a general, bounded L(x), with arbitrary values, suppose that one has selected completely-filling, non-overlapping triangles in a double-rectangle, so that the positive part:
(11.13)        L+(x) = t1(x) + t3(x) + t5(x) + ... (all > 0)
and the negative part:
(11.14)        L-(x) = t2(x) + t4(x) + t6(x) + ... (all < 0)
satisfy L(x) = L+(x) + L-(x). Then let B be the area of a rectangle of sufficient size to encompass the graphs of y = L+(x) and y = L-(x) (above and below):

413.
Now define the (positive and negative) fractions:
(11.15)        ρn = ∫ab tn(x) dx / [ B - ∫ab t1(x) dx - ∫ab t2(x) dx - ... - ∫ab tn-1(x) dx ]
for n = 1, 2, 3, .... Having insured that:
             ∫ab L+(x) dx = ∫ab t1(x) dx + ∫ab t3(x) dx + ∫ab t5(x) dx + ...,
             ∫ab L-(x) dx = ∫ab t2(x) dx + ∫ab t4(x) dx + ∫ab t6(x) dx + ...,
and
             ∫ab L(x) dx = ∫ab L+(x) dx + ∫ab L-(x) dx
hold, it follows that the mixed increasing and decreasing product:
(11.16)        P = (1 - ρ1)(1 - ρ2)(1 - ρ3) (1 - ρ4) ... (1 - ρ2n+1)(1 - ρ2n+2) ...
converges to the fraction, (1 - ∫ab L(x) dx/B) = 1 - F as a final affirmative response to the "inverse" question. Note that:
(11.17)        ∫ab L(x) dx = BF
             = B[ρ1 + (1 - ρ1)ρ2 + (1 - ρ1)(1 - ρ2)ρ3 + ... + (1 - ρ1)(1 - ρ2)...(1 - ρn-1) ρn + ... ]
as in (1.9) (which may be positive or negative). Here we have violated one of our notational conventions, namely, allowing the even-numbered fractions, ρ2n to be negative numbers. One could replace these increasing factors by their equals, 1/(1-ρˇ2n), where ρˇ2n = -ρ2n / (1 - ρ2n) (> 0), and obtain the equivalent quotient-form:
(11.18)        P = PQ = [(1 - ρ1)/(1 - ρˇ2)] [(1 - ρ3)/(1 - ρˇ4)] ... [(1 - ρ2n+1)/(1 - ρˇ2n+2)] ...,
and thereby recapture our notational conventions, as well as an appropriate form for our double-rectangle picture (recall Chapter 9: Quotient-products and Double-rectangle-Filling). In any case, we have demosntrated how one constructs an infinite product depicting rectangle-filling by isosceles triangles, whose related tent-function series converges to a given Lebesgue-integrable-function L(x) (a.e.), provided only that L+(x) and L-(x) are bounded. This demonstration seems to complete the simple story of infinite products and integration.
CONCLUDING COMMENTS. It comes as a surprise to the writer that infinite products and Lebesgue integration can be merged so effortlessly using simple pictures that depict a visual process of filling up a rectangle by non-overlapping triangles. The writer stumbled upon this picture in a desire to replace Mikusiński's Theorem with one using triangles (more suitable for purposes of Fourier transforms), in lieu of bricks, but subsequently dropped the topic (and all interest in mathematics, for that matter), upon retirement in 1987. It is more than just a coincidence that Professor Mikusiński passed away earlier that same year.

A critical reader of this work might (very well) observe that triangles can be dispensed with completely, and replaced by non-overlapping collections of rectangles in analogous filling schemes of a fundamental rectangle. In such a situation, Mikusiński's theorem applies directly, while the filling lemma also applies directly. So if rn is the fraction of the unfilled portion at each step, then the infinite product, P = (1 - r1)(1 - r2) ... (1 - rn) ... again produces the ultimate unfilled portion. Thereupon, the lowering of the filling rectangles may be viewed as producing a brick expansion of a Lebesgue integrable function, whose integral is F = 1-P. Also, the double-rectangle filling interpretations carry over without change. Though less picturesque, perhaps, the main advantage of sticking to bricks throughout is that the treatment carries over to higher (real) dimensions, visually. Thus, Lebesgue-integrable functions on k-dimensional rectangles have reflections in infinite products and rectangular fillings.

Other critical readers may (rightly) ask: what about infinite products in the complex plane? The simplest treatment of an infinite product:
              P = Z1 • Z2 • Z3 • • • •
here is to express each complex number in its polar form:
              Zn = (1 ± rn) eiθn,
where 1 ± rn = |Zn| and, say, |θn| < π. Then:
              P = (1 ± r1)(1 ± r2) ... (1 ± rn) ... ei (θ1 + θ2 + ...),
where the real infinite product:
              |P| = (1 ± r1)(1 ± r2) ... (1 ± rn) ...
is then subject to the complete real-analysis given in this work (including double rectangle filling and the link to integration). Complex convergence, of course, requires further that the infinite series:
              θ1 + θ2 + ... + θn + ...
converges, mod 2π. This treatment contrasts with the customary one based upon the symbolism, Zn = 1 + Un, where (using the logarithm function), P, is shown to converge if and only if the series U1 + U2 + U3 + ..., converges in the complex plane.

It is revealing to note that instead of considering the infinite product example:
              cos π z = (1 - 4z2)(1 - 4z2/32)(1 - 4z2/52)(1 - 4z2/72) ...
in the complex variable, z = x + iy, one can express this quantity as two infinite products in the real variable, x:
             cos π z = (1 - 4x2)(1 - 4x2/32)(1 - 4x2/52) ... ]cosh πy
             + i[πx(1 - x2)(1 - x2/22)(1 - x2/32) ... ] sinh πy.
Such a procedure, of course, is not exactly the same thing as above. In fact, with 1 ± rn = |1 - 4z2 / n2| and θn = arg(1 - 4z2 / n2), then the "exact" procedure becomes:
              cos π z = Z1 • Z3 • Z3 • • •,
as the infinite product to employ here for this quantity.

However, one can give "exact" interpretations of each of the two infinite products above, by admitting an extra complex factor, Z0 to account for the extra πx and/or y dependent factor in each. In the first case, with
              Z0 = cosh π y = (1 + r0) eiθ0, θ0 = 0,
and for odd n
              |1 ± rn| = |1 - 4x2/n2|, θn = 0, π
(depending upon the sign of (1 - 4x2/n2), which can be negative for finitely many n), we obtain the "exact" procedure for:
              Re cos π z = (1 + r0)(1 ± r1) (1 ± r3) ... (1 ± rn) ... ei (θ0 + θ1 + θ3 + ...),
Of course, x = Re z and y = Im z are to be used throughout this formulation. In the second case, with Z0 = πx sinh πy = (1 ± r0) ei θ0, θ0 = 0, π, and |1 ± rn| = |1 - x2/n2|, θn = 0, π, we obtain the "exact" procedure for
              Im cos π z = (1 ± r0)(1 ± r1) (1 ± r2) ... (1 ± rn) ... ei (θ0 + θ1 + θ2 + ...),
with x = Re z and y = Im z. These two can then be combined to give:
              cos π z = Re cos π z + i Im cos π z
itself, in an alternative (computationally simpler) version, using the complex variable, z, throughout. This alternative procedure, using f(z) = Re f(z) + i Im f(z), might possibly be the more advantageous method of employing the real-value theory of infinite products developed in this work. For if a two-variable, real-valued, function, φ(x,y) is known to have a convergent infinite product expansion, then there is a way (ref [26], not necessarily simple) of determining whether or not φ(x,y) is the real part of an analytic function. Moreover, there is a way (ref [26], again not necessarily simple) of obtaining its real conjugate complement, ψ(x,y) which might then be expected to possess an analogous infinite product expansion. These can be formulated, as in the above example, and combined to yield:
              f(z) = φ(x,y) + i ψ(x,y)
with x = Re z and y = Im z. Of course, sin π x furnishes another immediate example of this alternative procedure, as well as ez, upon expanding cos y and sin y as infinite products. Notably, in this last instance, we do NOT analytically extend an infinite product of ex "by replacing the real variable, x with the complex variable z", as is customary.

CHAPTER 12: REFERENCES.


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CHAPTER 13: ACKNOWLEDGEMENTS.


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I acknowledge the assistance of G. William Moore, MD, PhD, in reviewing and formatting the manuscript. Dr. Moore has participated actively in the organization of much of the material, and as prime editor of it all. His assistance has been of utmost value, without which the paper would never have been written. I owe him much thanks for his expert assistance, but all errors are of my own doing.

In addition, I acknowledge the inspiration supplied by my wife of fifty-nine years, Marilyn Struble, who rekindled in me a dormant enthusiasm in mathematics, after a sixteen-year hiatus. She had the insight to give me a copy of John Derbyshire's stimulating book, Prime Obsession [9].


Last updated: 8/4/2005, by Raimond A. Struble, PhD.