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Math Help - integral of 1/x is ln|x| +C; why the absolute value?

  1. #1
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    integral of 1/x is ln|x| +C; why the absolute value?

    I have seen two justifications for the absolute value sign in ln|x| as the antiderivative of 1/xm, but neither one seems sufficient. The first one is quite lame, that ln can only deal with a non-zero positive domain (as long as we are sticking to the real numbers). But this would not rule out a definition such as (as example only)
    ln(x) if x is positive
    -ln(|x|) if x is negative.
    Or something like this. I am not proposing this as a definition; only showing how the justification above is insufficient.
    The next justification I have seen is that the area under the curve 1/x over an interval (a,b), with a<b<0, will be the same as the area under the curve 1/x over the interval (|b|,|a|), so we take the absolute value. However, the integral is not the same thing as the area; if we are looking at the negative side of the x-axis, we get negative "signed areas" from the integral, not the areas.

    So, why the absolute value? Thanks.
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  2. #2
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    Re: integral of 1/x is ln|x| +C; why the absolute value?

    Quote Originally Posted by nomadreid View Post
    I have seen two justifications for the absolute value sign in ln|x| as the antiderivative of 1/xm, but neither one seems sufficient. The first one is quite lame, that ln can only deal with a non-zero positive domain (as long as we are sticking to the real numbers). But this would not rule out a definition such as (as example only)
    ln(x) if x is positive
    -ln(|x|) if x is negative.
    Or something like this. I am not proposing this as a definition; only showing how the justification above is insufficient.
    The next justification I have seen is that the area under the curve 1/x over an interval (a,b), with a<b<0, will be the same as the area under the curve 1/x over the interval (|b|,|a|), so we take the absolute value. However, the integral is not the same thing as the area; if we are looking at the negative side of the x-axis, we get negative "signed areas" from the integral, not the areas.

    So, why the absolute value? Thanks.
    \displaystyle \frac{d}{dx}\left[\ln{(x)}\right] = \frac{1}{x}, where \displaystyle x > 0, and \displaystyle \frac{d}{dx}\left[\ln{(-x)}\right] = \frac{1}{x}, where \displaystyle x < 0.

    So that means \displaystyle \int{\frac{1}{x}\,dx} = \begin{cases}\ln{(x)}\textrm{ if }x > 0 \\ \ln{(-x)} \textrm{ if }x < 0\end{cases}.

    Now what is the definition of \displaystyle |x|?
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  3. #3
    MHF Contributor chisigma's Avatar
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    Re: integral of 1/x is ln|x| +C; why the absolute value?

    Quote Originally Posted by nomadreid View Post
    I have seen two justifications for the absolute value sign in ln|x| as the antiderivative of 1/xm, but neither one seems sufficient. The first one is quite lame, that ln can only deal with a non-zero positive domain (as long as we are sticking to the real numbers). But this would not rule out a definition such as (as example only)
    ln(x) if x is positive
    -ln(|x|) if x is negative.
    Or something like this. I am not proposing this as a definition; only showing how the justification above is insufficient.
    The next justification I have seen is that the area under the curve 1/x over an interval (a,b), with a<b<0, will be the same as the area under the curve 1/x over the interval (|b|,|a|), so we take the absolute value. However, the integral is not the same thing as the area; if we are looking at the negative side of the x-axis, we get negative "signed areas" from the integral, not the areas.

    So, why the absolute value? Thanks.
    In my opinion for any x \ne 0 is...

    \frac{d}{dx} \ln x = \frac{1}{x} (1)

    ... and the 'absolute value' isn't necessary. If You consider that...

    \ln (-x)= \ln x + i\ \pi (2)

    ... the difference between them is only the constant i\ \pi the derivative of which is 0. For the same reason is...

    \int_{a}^{b} \frac{dx}{x} = \ln b - \ln a (3)

    ... if 0 \ni [a,b]. What You have to keep in mind however is that that is my own opinion and I am not sure that in a test exam You can sustain that without negative consequences...

    Kind regards

    \chi \sigma
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    Re: integral of 1/x is ln|x| +C; why the absolute value?

    Thanks to both of you, Prove It and chisigma. Both excellent answers.
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