"Richard Hennessy" <rich.hennessy at verizon.net> writes:
> I meant the indefinite integral of the square wave function is
> continuous everywhere. Sorry. In general if a function involves step
> functions the integral in usually continuous, even though the original
> function is not.
>
> Rich
*sigh*
If f is a function from the real line to itself and you define a new
function F by
F(x) = integrate (f(y), y, a, x)
for any choice of a, then F is called the "indefinite integral" of f. A
theorem, which is sort of trivial once you've set up all the machinery
for Riemann integration, says that F is always continuous (unit steps or
no!)
Occasionally, people talk of "functions" like the Dirac delta function
?(x). This isn't, strictly speaking, a function R?R because it doesn't
evaluate to a number at x=0. Rather it is defined by
integrate (?(x), x, a, b) = 1 if a<0<b and 0 otherwise
An indefinite integral for ? looks like a unit step function (and is a
proper honest-to-god function). This isn't continuous.
If you're interested to know more about things like ? and more general
such objects, look up the theory of "distributions" (which is the more
general name for such objects) or "weak derivatives".
I hope this makes things a bit clearer,
Rupert
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