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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.
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
- 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.