Re: integral zeta

On Aug 16, 7:40 am, David C. Ullrich <ullr...@xxxxxxxxxxxxxxxx> wrote:
Well, integral(x)
could be x^2 + 1, in which case it has no zeroes.
Or it could be x^2 - 4, which has two zeroes. Etc.

By "x^2 + 1" and "x^2 - 4" I assume you mean
"x^2/2 + 1" and "x^2/2 - 4," respectively. The OP
is already confused about the definition of
primitive, and now you just confused him even more
by writing the _wrong_ primitive for integral(x).

In the old thread given in Mr. Revilla's link, I
notice that Prof. Ullrich himself points out
how the integral of a function can be thought of
as a equivalence class of functions. All of this
raises an interesting question. Is there a
canonical way to choose, without resorting to
the Axiom of Choice, one particular primitive
for each integrable function.

Notice that if we resticted ourselves to the
set of polynomial functions, then one may
consider the canonical primitive of such a
polynomial function f(x) to be F(x) such that
F'(x) = f(x) and F(0) = 0. This would agree with
the OP's intuitions.

What if f is not a polynomial? Actually, we can
use the above definition as long as f is
actually defined and integrable in a
neighborhood of zero. But as for a way to
choose a canonical primitive for the set of all
functions that are integrable on _any_ open
set appears to be impossible in ZF.

Thus, by this definition, the canonical
primitive for Prof. Ullrich's example, namely
2 sin(t) cos(t) is sin^2 (t), since it's the
only one that vanishes as t = 0.

Notice that ironically, in the 2005 post in
Mr. Revilla's thread, someone referred to
the OP's question as stupid, and Prof.
Ullrich was quick to defend the question as
not being stupid at all. Compare this to
his attitude in this thread.

.

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