Thread: absolute values for integrals equal to log?

1. Re: absolute values for integrals equal to log?

Originally Posted by fan96
What would be the proper method of dealing with a definite integral of $1/x$ that involves both the positive and negative sides, taking into account the discontinuity at $x=0$?

For example, Wolfram Alpha tells me that

$\int_{-1}^1 \frac 1x \, dx$

does not converge (and I could see why this might be the case), even though one might expect it to be equal to zero.
$\noindent To deal with such an integral, we have to split it up at the discontinuity and separately consider limits over the continuous parts. That is, we must consider$

$\lim\limits_{a\to 0^{-}}\int_{-1}^a \frac{1}{x}\,dx +\lim\limits_{b\to 0^{+}}\int_{b}^{1}\frac{1}{x}\, dx.$

$\noindent The value of the integral is defined to be the value of the sum of these limits, if this exists. However, if we evaluate those limits, we get -\infty +\infty, which is indeterminate. Hence that integral is undefined.$

$\noindent It is possible though to attach a \textit{Cauchy Principal Value} to the integral. The Cauchy Principal Value of this integral is 0. See the examples at:$

https://en.wikipedia.org/wiki/Cauchy_principal_value.

2. Re: absolute values for integrals equal to log?

Originally Posted by peter ringout
It has nothing to do with either sides. It is a simple fact that the integral of 1/x IS
ln|x|+C.

THIS IS NOT COMPLICATED!!!!

Do we at least agree that the integral of 1/(x^2) IS -1/x??
Technically, the anti-derivative of 1/x^2 should be
-1/x+C1 for x>0
-1/x+C2 for x<0
where C1 and C2 are (possibly different) constants.

It definitely matters which side you are looking at.
Suppose f'(x)=1/x^2 and f(1)=1. You cannot determine f(-1).

3. Re: absolute values for integrals equal to log?

Originally Posted by peter ringout
f'=1/x^2 so f=-1/x+C. f(1)=1 imples C=2 and hence f=-1/x+2. So f(-1)=-1+2=1.
Consider two antiderivatives below.
f(x)=-1/x+2 for x>0
f(x)=-1/x for x<0
f'(x)=1/x^2 for all non-zero real values of x
f(1)=1, f(-1)=1

f(x)=-1/x+2 for x>0
f(x)=-1/x+2 for x<0
f'(x)=1/x^2 for all non-zero real values of x
f(1)=1, f(-1)=3

Both satisfy the requirement. Therefore, f(-1) cannot be uniquely determined.

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