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

1. ## 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.

2. ## Re: 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.
$\displaystyle \frac{d}{dx}\left[\ln{(x)}\right] = \frac{1}{x}$, 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 $\displaystyle |x|$?

3. ## Re: 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.
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

$\chi$ $\sigma$

4. ## 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.