Re: Divergence Thm, 1/r^2 flux, and the origin


Then you recall wrong.

See, it is a subtle problem. In my
case, I'm dealing with Div(F) = 0 for all x != 0, and you can't get
much more integrable than that!

Let f(x) = 0 for x < 0, f(x) = 1 for x >= 0. Then f'(x) for all
x <> 0, in particular f is differentiable almost everywhere and
f' is integrable.

And so according to you it follows that

f(1) - f(-1) = int_{-1}^1 f'(t) dt = 0.

Jeez.

--Jeff

************************

David C. Ullrich


Thank you! A very nice and easy counter-example. I was remembering
wrong...it is the other half of the fundamental thm of calculus that
is so tolerant of discontinuities.

--Jeff

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