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Integral of 1 over x

  1. Jan 9, 2015 #1
    As we know, [itex]\int \frac {1}{x} \, dx = ln (x) + C[/itex]
    But, what it would be, if [itex]x[/itex] is negative?
    I calculated this way,
    [itex]\int \frac {1}{x} \, dx = - \int \frac {1}{-x} \, dx = \int \frac {1}{-x} \, d(-x) = ln (-x) + C[/itex] [because, [itex]-x[/itex] has a positive value.]
     
  2. jcsd
  3. Jan 9, 2015 #2
    The variable ##x## can be either positive or negative because the fact is, [tex]\int \frac{1}{x} dx = \ln({|x|}) + C[/tex]
     
  4. Jan 9, 2015 #3

    dextercioby

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    The function 1/x is (maximally) defined at all points in R minus 0, the natural logarithm only at points >0. It turns out that ## \ln -x ## is not a valid expression in the reals, because it conflicts with the known ##\ln ab = \ln a + \ln b## by being forced to consider ##\ln -1##.
     
  5. Jan 10, 2015 #4

    pwsnafu

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    ##ln(-x)## is a valid expression for negative x.

    ##\ln(ab) = \ln(a) + \ln(b)## is a property which is valid for positive a and b, it is not a definition of the logarithm. The property being unsatisfied for negatives has nothing to do with whether ##\ln(-x)## is defined or not.
     
    Last edited: Jan 10, 2015
  6. Jan 16, 2015 #5

    Svein

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    To maximize the confusion: ln(-x) = ln((-1)*x) = ln(-1) + ln(x) = iπ + ln(x)
     
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