Re: Integration of 0

On Wed, 10 Aug 2005 11:30:17 GMT, "n3wb1e" <n3wb1e@xxxxxxxxxxxxxx>
wrote:

>
>"agentmath" <Agent.Math@xxxxxxxxx> ha scritto nel messaggio
>news:1123617224.977190.237230@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
>> In indefinite integraion,
>>
>> (*) integral 0 dx = constant.
>>
>> But by the property of integration,
>>
>> (**) integral 0 dx = integral (0 product 1) dx = 0 product (integral 1
>> dx) = 0.
>>
>> Why are (*) and (**) different?
>>
>
>
>Even though the totality of books you may read says that, the homogeneity
>property for indefinite integral DOES NOT HOLD for 0.

That's not really the answer. The homogeneity property does hold
if you look at things right - the actual problem is something
else.

There _is_ a problem with what you see in books, which has
already been pointed out. One way to put it is to say it's
an abuse of the equals sign, since the thing on the right
is not just one thing. Another way to look at it:

If we want to talk about _the_ indefinite integral of a
function then the integral of a function cannot be a function,
it has to be an equivalence class of functions. Given a
function f, let's say [f] is the class of all functions
of the form g = f + c for some constant c. Say V is the
space of all these equivalence classes [f].

Now the integral of a function should really be regarded
as an element of V. (Why? If we say int x dx = x^2/2 + c
we seem to be saying that x^2/2 = x^2/2 + 1, which is
wrong. But if we say that int x dx = [x^2/2] there's
no problem, because [x^2/2] _is_ precisely equal to
[x^2/2 + 1].)

And V is a vector space in a natural/standard way
(standard because V is just the quotient space of
the space of all functions modulo the subspace of
constant functions); if f and g are functions and
a is a scalar we define

(i) [f] + [g] = [f+g]
(ii) a[f] = [af].

(You need to check that these are "well-defined";
for (ii) that amounts to showing that if [f] = [g]
then [af] = [ag], which is easy to see. Note for
example that we _cannot_ define the product of
two elements of V by

(iii) [f][g] = [fg];

why not?)

And now the problem goes away - homogeneity
_does_ hold, even for the integral of 0 times
the scalar 0. Because 0[0] = [0]; both sides
are exactly the class of all constant functions.


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

David C. Ullrich
.