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Antiderivative of 1/x: ln(x) or ln(|x|)?

  1. Jan 7, 2017 #1
    1. The problem statement, all variables and given/known data
    Calculate the integral:
    ## \int_{a}^{b} \frac{1}{x} dx ##

    2. Relevant equations
    -

    3. The attempt at a solution
    In highschool we learned that:
    ## \int_{a}^{b} \frac{1}{x} dx = ln(|x|) + C ##
    because the logarithm of a negative number is undefined.

    However, in my current maths course about complex numbers we leared to take the logarithms of negative numbers as well, using: ##ln(-x) = ln(x) + \pi i ##. With this extension, can we say:
    ## \int_{a}^{b} \frac{1}{x} dx = ln(x) + C ##
    or is this still wrong?

    Thank you very much!
     
  2. jcsd
  3. Jan 7, 2017 #2

    andrewkirk

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    If ##a=b## the integral exists, trivially, and is equal to zero.
    If ##a<0<b## the definite integral does not exist because the explosion at 0 makes the function not integrable over the interval ##[a,b]##.
    If ##a<b=0## or ##0=a<b## is 0, the proper integral does not exist because ##\log 0## and ##\log|0|## are undefined. The improper integral doesn't exist either, since ##\lim_{x\to 0}\log x## and ##\lim_{x\to 0}\log |x|## do not exist.

    If ##0<a<b## then the integral exists and both formulas give the same result, since on that range of integration ##|x|=x##.

    If ##a<b<0##, again the integral exists and gives the same result from both formulas, because the constant ##\pi i## cancels out.

    PS:
    For one-dimensional integration the integration constant ##C## should not be used in a definite integral. It is used in an indefinite integral, but always cancels out when one makes a definite integral (ie inserts lower and upper integration limits).
     
  4. Jan 7, 2017 #3

    Mark44

    Staff: Mentor

    @Alettix, please post problems involving integrals and other calculus topics in the Calculus & Beyond section, not the Precalc section where you posted this thread. I have moved your thread.
     
  5. Jan 8, 2017 #4

    pasmith

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    If [itex]a < b < 0[/itex] then [tex]
    \int_{a}^{b} \frac1x\,dx = \left[ \ln x \right]_{a}^{b} = \left[ \ln|x| + i\pi \right]_{a}^{b} = \ln|b| - \ln|a|,[/tex] consistent with the first rule you were taught. Integrals where [itex]a < 0 < b[/itex] are not defined due to the singularity at the origin.
     
  6. Feb 8, 2017 #5

    I like Serena

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    If we're talking about complex numbers, we have ##\ln b - \ln a##.
    If we're talking about real numbers, we need to consider the undefined point we have at 0, so that ##a## and ##b## both have to be either positive, or negative.
    Otherwise we have to get to more "tricks" to deal with the discontinuity.
    Either way, ##\ln |x| + C## is a poor man's anti-derivative since the integration constant can be different on either side of ##0##.
     
  7. Feb 8, 2017 #6

    Ray Vickson

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    I hope nobody in high school told you that formula, because you have ##a## and ##b## on one side but no ##a## or ##b## on the other side of the equation. That cannot be right.

    Anyway, if ##0 < a < b##. then ##|x| = x## for any ##x## between ##a## and ##b##, so it makes no difference whether you use ##\ln |x|## or ##\ln x##. If ##a < b < 0## the integral ##\int_a^b dx/x## computes the (negative) area from ##x = a## to ##x = b## below the ##x##-axis and above the curve ##y = 1/x## (which lies below the axis). If you put ##t = -x## the integral becomes ##\int_{-a}^{-b} (-dt)/(-t)),## which equals ##\int_{|a|}^{|b|} dt/t = \ln |b| - \ln |a| < 0## because ##|b| < |a|##. That is as it should be.
     
    Last edited: Feb 8, 2017
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