derivates and anti-derivatives of ln x and 1/x (respectively)

Posted on 2011-05-10
Last Modified: 2012-05-11
So I know derivative of ln(x) = 1/x, but the anti-derivative of 1/x = ln(|x|), why the absolute value? It is possible to have negative values for x in ln, so why isn't it just x?
Question by:Zenoture
    LVL 84

    Expert Comment

    the values of ln for negative x are not the values of the anti-derivative of 1/x for negative x
    LVL 18

    Accepted Solution

    F(x) = ln(|x|)

    means that for x<0

    F(x) = ln(-x)

    F'(x) = (-1) / (-x) = 1/x

    so the anti-derivative is shown to be valid

    but also of course

    d/dx(ln(x)) = 1/x


    in the real numbers ln(x) is not defined for x<0, but in complex numbers

    -x = x e^(pi i)

    since e^(pi i) = -1

    so for example

    ln(-5) = ln(5 e^(pi i)) = ln(5) + pi i

    and in general

    ln(-x) = ln(x) + pi i

    therefore the difference between ln(-x) and ln(x) is a constant which can be absorbed into a constant of integration, and ultimately would not affect a definite integral


    integral ln(x) is found by considering ln(x) = 1 ln(x) , then using integration by parts to give

    x ln(x) - x
    LVL 5

    Expert Comment

    Hi Zenoture.

    Im sure you will find the following link interesting.
    LVL 26

    Expert Comment

    deighton gave the key to the answer to your question when he said
     "in the real numbers ln(x) is not defined for x<0"
    He then went on to give a nice illustration of what the situation is if one considers complex numbers.

    Author Closing Comment

    Sorry, forgot to award points.

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