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Math Help - An interesting integration

  1. #1
    Senior Member ecMathGeek's Avatar
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    An interesting integration

    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) ???
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  2. #2
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    Quote Originally Posted by ecMathGeek View Post
    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?
    We define,
    INT(a,b)f(x) dx = Limit of Riemann Integral (if it exists).
    For function f defined on closed interval [a,b].

    Note, under this strict definition, we require that f to be defined on [a,b] (and to be integratable). Hence a<b.

    What about b>a?
    We define

    INT(a,b)f(x) dx=-INT(b,a)f(x)dx.
    Again, this IS NOT a theorem.
    This is a definition.

    And if a=b we define INT(a,b)f(x)dx=0.
    For f defined at "a".
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  3. #3
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    We define,
    INT(a,b)f(x) dx = Limit of Riemann Integral (if it exists).
    For function f defined on closed interval [a,b].

    Note, under this strict definition, we require that f to be defined on [a,b] (and to be integratable). Hence a<b.

    What about b>a?
    We define

    INT(a,b)f(x) dx=-INT(b,a)f(x)dx.
    Again, this IS NOT a theorem.
    This is a definition.

    And if a=b we define INT(a,b)f(x)dx=0.
    For f defined at "a".
    Thank you for responding, but I already know ALL of those definitions. All you needed to point out was this:

    Quote Originally Posted by ThePerfectHacker View Post
    We define,
    INT(a,b)f(x) dx = Limit of Riemann Integral (if it exists).
    For function f defined on closed interval [a,b].
    I merely forgot that f(x) MUST be defined on the interval [a,b] for the integration to be defined. Everything else I did was irrelevant.
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