
Re: Indefinite integral puzzle
Posted:
Dec 31, 2010 3:53 AM


On Thu, 30 Dec 2010, Craig Feinstein wrote: > On Dec 30, 10:32 am, Pfss...@aol.com wrote: >> On Wed, 29 Dec 2010 10:35:22 0800 (PST), Craig Feinstein
>>> Let I(f(x))dx be the indefinite integral of a function f(x). Then what >>> is: >>> I(I(I(I(...)dx)dx)dx)dx ? >> >> well, first of all, what does this mean??? >> Give a precise definition!!! > > I(I(I(I(...)dx)dx)dx)dx means that we want the (indefinite) integral > of the integral of the integral of the integral ..... ad infinitum.
That's not a precise definition. It's not even a definition. It's the same intuitive description of your notion, as was depicted with your notation.
Let f(x) = 0. Then integral 0 dx = c1 integral^2 0 dx^2 = c1.x + c2 ... integral^j 0 dx^j is any polynomial in R[x] of degree at most j.
Thus in the limit one may opine that integral^oo 0 dx^j is any polynomial p in R[[x]], ie all formal infinite series of the form sum(j=0,oo) aj.x^j.
See my other post where I indicate how the concept is ludicrous by sketching out how integral^oo f(x) dx^oo = integral^oo g(x) dx^oo for all f,g with Taylor series.
You can't even define an operator I:R^R > R^R and take the limit lim(n>oo) I^n(f) which you could do if I(f) = integral(0,x) f(x) dx.
You have to define the operator I:P(R^R) > P(R^R) and then create a topology for P(R^R) so that lim(n>oo) I^n({f}) is definable.
Though there are several topologies for R^R, which you may want to review and pick on the most likely to use, there are none for P(R^R) that I know, which are in use.
In other words, you've got a heap of work to do to give a precise definition of integral^oo f(x) dx^oo for indefinite integration. In addition, my intuitive exploration given in my other post, indicates that the result of integral^oo f(x) dx^oo, for a wide range of functions, lacks significance.
For example integral^oo 0 dx^oo = R[[x]] is every function with a Taylor series and a lot of other infinite series in powers of x that have a variety of convergent and divergent properties.
That's because you've a infinite number of constants of integration. Happily, it's merely a countable infinite.
My claim is integral^oo f(x) dx^oo = R[[x]] for all f with a Taylor series.
By default, the functions are taken to have domain R. R[[x]] however, does include partial functions, ie functions with domains other than R.

