This question is one I came up with after considering the rules of integration.

We know that the integration of a function from a to b such that a = b is 0: INT{n:n} f(x) = 0

BUT, is this always true? By this I mean, would this be true in a function where f(n) is undefined?

For example:

Does INT{1:1} 1/(1-x)^2 = 0? I will work this out. I just want someone to tell me if my work is correct.

Since f(1) in this case is undefined, I'll use limits:

lim{n->1} INT{n:n} 1/(1-x)^2

lim{n->1} [1/(1-x)] from {n:n}

lim{n->1} 1/(1-n) - 1/(1-n) ... which is 0/0, so using L'Hopital's Rule:

lim{n->1} 0/(-n) = 0

As we see, this does equal 0, so the rule seems to work. However, if I try a slightly different equation, such that f(n) is undefined and lim{x->n} f(x) is undefined, does the rule still work?

Does INT{1:1} 1/(1-x) = 0?

lim{n->1} INT{n:n} 1/(1-x)

lim{n->1} -ln(1-x) from {n:n}

lim{n->1} -ln(1-n) + ln(1-n) ... which is 'negative' undefined plus undefined.

lim{n->1} ln[(1-n)/(1-n)]

lim{n->1} ln(1) = 0 ... but can I do this???

Since

lim{n->1} ln(1-n) is not defined

Because

lim{n->1-} ln(1-n) does not exist

Is it legal to say

lim{n->1} ln[(1-n)/(1-n)] = lim{n->1} ln(1) ???