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Thread: absolute values for integrals equal to log?

  1. #26
    Rambling Spirit
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    Re: absolute values for integrals equal to log?

    Quote Originally Posted by fan96 View Post
    What would be the proper method of dealing with a definite integral of that involves both the positive and negative sides, taking into account the discontinuity at ?

    For example, Wolfram Alpha tells me that



    does not converge (and I could see why this might be the case), even though one might expect it to be equal to zero.








    https://en.wikipedia.org/wiki/Cauchy_principal_value.
    fan96 and HeroWise like this.

  2. #27
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    Re: absolute values for integrals equal to log?

    Quote Originally Posted by peter ringout View Post
    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. #28
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    Re: absolute values for integrals equal to log?

    Quote Originally Posted by peter ringout View Post
    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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