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Math Help - Infinite areas cancel out?

  1. #1
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    Infinite areas cancel out?

    My math teacher and I got in a debate the other day, and i couldn't find any information anywhere so i figured i'd post it here

    We were working on calculating infinite converging areas and we got to the following problem

    \begin{array}{cc}-1\\1\end{array}\int\frac{1}{x}dx

    Since i'm always looking for ways to avoid doing extra work, i said it would be possible to simply cancel out the areas from -1 to 0 and from 0 to 1 based on symetry and end up with zero

    however, when you actually work out the integral you get \infty - \infty which is undefined

    As far as i can tell, the reason that is undefined is because the two infinities could be coming from anywhere and it's impossible to make an assumption, but in this case, when we know the two infinite areas are exactly equal (except for the minus sign) why can't we say that is equal to zero?
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  2. #2
    ynj
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    Because for the positive part, say the "area" between [0,1], there area is actuall infinite, that is to say, does not exist. So you can not add the area between [-1,0] and [0,1] and say it is 0, since there is no "area" between [0,1].
    Or you can explain like this:
    This improper integral that have a spot 0 is defined iff \int_0^1 f(x) \ dx,\int_{-1}^0 f(x) \ dxare both defined.
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  3. #3
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    Quote Originally Posted by mrmidget81 View Post
    My math teacher and I got in a debate the other day, and i couldn't find any information anywhere so i figured i'd post it here

    We were working on calculating infinite converging areas and we got to the following problem

    \begin{array}{cc}-1\\1\end{array}\int\frac{1}{x}dx

    Since i'm always looking for ways to avoid doing extra work, i said it would be possible to simply cancel out the areas from -1 to 0 and from 0 to 1 based on symetry and end up with zero

    however, when you actually work out the integral you get \infty - \infty which is undefined

    As far as i can tell, the reason that is undefined is because the two infinities could be coming from anywhere and it's impossible to make an assumption, but in this case, when we know the two infinite areas are exactly equal (except for the minus sign) why can't we say that is equal to zero?
    To add a little more from the previous post:

    \int_{-1}^{1} \frac{1}{x} \, dx is an improper integral. It is undefined and not necessarily well defined.

    It can be taken to mean \lim_{a \rightarrow 0} \left( \int_{-1}^{a} \frac{1}{x} \, dx + \int_{a}^{1} \frac{1}{x} \, dx \right) = 0, which is the Cauchy Principle Value.

    But it can also be taken to mean \lim_{a \rightarrow 0} \left( \int_{-1}^{2a} \frac{1}{x} \, dx + \int_{a}^{1} \frac{1}{x} \, dx \right) = \ln 2 .... etc.

    The previous poster has given the conditions under which an improper integral of this type will be convergent.
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  4. #4
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    Thanks for the quick replies

    and sorry to ask a follow up question but how is it then you can calculate two different areas for the same integral?
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    Quote Originally Posted by mrmidget81 View Post
    Thanks for the quick replies

    and sorry to ask a follow up question but how is it then you can calculate two different areas for the same integral?
    When things don't exist, contradictions and inconsistencies are often present ....
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  6. #6
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    but can you really say it doesn't exist just because you can't define it?
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  7. #7
    ynj
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    Quote Originally Posted by mrmidget81 View Post
    but can you really say it doesn't exist just because you can't define it?
    The area is defined by integral...
    So when integral does not exists, area does not exists..
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