- im·prop·er (from
Merriam-Webster Online Dictionary)- Pronunciation:
- \(ˌ)im-ˈprä-pər\
- Function:
adjective- Etymology:
- Middle English, from Middle French
impropre,from Latinimproprius,fromin-+propriusproper:not proper: as (a:)not in accord with fact, truth, or right procedure:incorrect <improperinference> (b:)not regularly or normally formed or not properly so called (c:)not suited to the circumstances, design, or end <impropermedicine> (d:)not in accord with propriety, modesty, good manners, or good taste <improperlanguage>synonymssee 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."

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