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Questionable justifications for the absolute value in ln(|x|) in the integral of 1/x?

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? Serious answers only, please. Thanks.

Update:

ah, please do not just repeat one of the insufficient justifications. I know what the domain of ln is, but this does not make it a good answer. A better idea is to examine the idea of antiderivative, where derivative of 1/x = ln x if x is positive, and the derivative of 1/x = ln(-x) if x is negative, and then look at the definition of the absolute value. Someone go with this, not telling me that the domain of ln is x>0.

1 Answer

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  • 10 years ago

    It's more a matter of domain of logarithms. Think about what logarithms do - they return the exponent on the power.

    log 0? What does that mean? What exponent do you put on a base to make the power = 0?

    Similarly, ln(-1) : What exponent can you put on e to make the power = -1?

    Likewise, looking at the derivative of ln x = 1/x, x=0 is undefined.

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