Re: Fallacious (!) proof that 1 = 0

if f is ANY antiderivative of 1/u we will still have f(1) - f(1) = 0.

f(1) is not a single value; that was the point. f(1) is really a whole
set of values:
{ 0 + k * 2 * pi * i : k is an integer}. When you subtract this set
from itself, you don't get just the number 0.

I don't really follow you here:

u(0) = exp(i.0) = 1
u(2pi) = exp(i.2pi) = 1

So if f is an antiderivative of 1/u we have

f(u(0)) = f(exp(i.0)) = f(1)
f(u(2pi)) = f(exp(i.2pi)) = f(1)

So, the value of the integral at x= i.2pi minus the value of the
integral at x = 0 is f(1)-f(1) = 0.

Unless you're suggesting that we use a different antiderivative at
x = 0 and at x = i.2pi?

I am at an advantage over you: I know what the mistake is because
I'm the one who put it in! The REAL problem is that 1/u doesn't HAVE
an antiderivative on the whole of the unit circle in the complex
plane, even though one can be defined on a small disk centred on any
point on it.

For real functions f, we can always assume that the integral from
0 to pi of [f(sin(x)]cos(x) will be zero because it is equal to the
integral from sin(0) to sin(pi) of f, and sin(0) = sin(pi) = 0.

For complex functions, though, we CAN'T assume that the integral from
0 to 2pi of [f(exp(ix))].iexp(ix) will be zero, even though
exp(i.0) = exp(i.2pi) = 1.

The reason for the difference is interesting, but it's late.

I discuss this issue is my post 'An interpretation of Cauchy's integral

theorem' on 12th Jan.

.

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