FREE CALCULUS BOOK

The Calculus Integral [Beta0.2],
B S Thomson,
ClassicalRealAnalysis.com (2009).

Download a free PDF file from http://www.youpublish.com/files/23072.


Well "calculus book" doesn't quite describe it. It is an account of integration theory on the real line that starts in the initial chapters with the "calculus integral" (i.e., the original integral of Newton) and carries the development as far as the integrals of Lebesgue and Henstock-Kurzweil.

If you are, however, teaching a calculus course and have ever considered dropping the Riemann integral and its ugly step-sister the improper Riemann integral from the syllabus, then you might want to use the first three chapters as a basis for the teaching of the basic integral of the calculus.

A trade paperback version will be available shortly.

TOP TEN REASONS FOR DUMPING THE RIEMANN INTEGRAL #5

It clears up the mystery of the popcorn function.


Suppose a function is zero at every irrational point. What is its integral? Well, certainly, the points left out are insignificant. Not merely a set of measure zero but even countable. Such a set plays no role in determining the integral so your function can have any values on the rationals, or even remain undefined on the rationals. The simple answer is that your function is integrable on every interval with a zero integral.

Oh, what? Sorry? You have only learned the Riemann integral. Alas, the answer now is entirely different. Now the function must be defined at these missing rational points and the answer depends on how you define them. If the resulting function is integrable then certainly the value of the integral is zero, but it may or may not be integrable. Let xn be a listing of all the rationals and let your function be defined to be f(xn)=cn and with f(x)=0 at x irrational. What is a necessary and sufficient condition for f to be integrable [i.e., integrable in the dumb Riemann sense]? That's a tough question, but one that is not particularly important.


In 1875, K. J. Thomae discovered the now-famous example of a function of this kind that is continuous at all the irrationals and discontinuous at the rationals. This function has many names: the modified Dirichlet function, Thomae function, Riemann function, raindrop function, ruler function, and popcorn function.



His example is a nice curiosity in the study of continuous functions. But it is usually presented to students of integration theory as example of a seriously discontinuous function that is integrable. The student gets the impression that it is important to have continuity, that discontinuities must be controlled, that without proper configuration of the values of a function the integral is badly affected, and that integration theory has its mysteries. That's good teaching?

If we drop the Riemann integral then the popcorn function would not be mentioned in the context of integration theory and can return to its proper place in the study of continuity.

Is there a function continuous at every rational and discontinuous at every irrational?
Is there a function discontinuous at every rational and continuous at every irrational?

The answer to the first question is "no" and the answer to the second question is "popcorn."

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August 18, 2008 gale.holland@latimes.com

The annual college textbook rush starts this month, a time of reckoning for many students who will struggle to cover eye-popping costs of $128, $156, even $198 a volume.

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Trade Paperback of BBT now available!

[BBT] Real Analysis,
Second Edition (2008)
Andrew M. Bruckner, Judith B. Bruckner and Brian S. Thomson
ISBN:1434844129
EAN-13:9781434844125

This is the second edition of our graduate level real analysis textbook formerly published by Prentice Hall (Pearson) in 1997.

In addition to "nearly" free PDF files of this text that have been available for a while now, we are now offering a trade paperback version. This will soon be listed on Amazon, and is already listed on CreateSpace. For details on ordering or for download instructions visit our website.

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TOP TEN REASONS FOR DUMPING THE RIEMANN INTEGRAL #6

#6. It's all that improper stuff.

im·prop·er (from Merriam-Webster Online Dictionary)

Pronunciation:

\(ˌ)im-ˈprä-pər\

Function:

adjective

Etymology:

Middle English, from Middle French impropre, from Latin improprius, from in- + proprius proper

: not proper: as (a:) not in accord with fact, truth, or right procedure : incorrect <improper inference> (b:) not regularly or normally formed or not properly so called (c:) not suited to the circumstances, design, or end <improper medicine> (d:) not in accord with propriety, modesty, good manners, or good taste <improper language>

synonyms see indecorous .

How is an integral improper? Well, if your main point of reference is the unfortunate Riemann integral then, in any situation in which you need to perform an integration of a non-Riemann integrable function, that can only be considered "improper". Thus calculus students are drilled on the need to interpret all "proper" integrals as Riemann integrals. If a function is unbounded then it cannot have an integral but it might have an "improper" integral.

In all advanced mathematics the class of integrable functions includes an abundance of unbounded functions. It is a feature of the theory. There is nothing improper about such integrals because the definition of the integral includes them. All of the calculus drill on handling unbounded functions is dropped for advanced classes. But, in most cases, we still teach beginning students using the terminology and the methods of the nineteenth century.


The Riemann integral should be dumped if only to get rid of all this improper nonsense.

It even has its particularly stupid aspects. Suppose we wish to integrate f(x)=x^(-1/2) on the interval [0,1]. The fundamental theorem of the calculus says search for a suitable antiderivative, and F(x) = 2 x^(1/2) comes to mind fairly soon. So the integral has the value F(1)-F(0)= 2.

"But", howls the calculus professor, "You didn't notice that the function f(x) here is unbounded, therefore it has no integral properly speaking. You should have integrated on [t,1] for all t between 0 and 1 and then taken the limit as t tends to zero on the right. Then you must assert that the function does not have a proper integral but does have an improper integral equal to ...well yes, the same value 2 as before, but this is the correct procedure that must be followed."