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Math Help - Strange Integrals

  1. #1
    Forum Admin topsquark's Avatar
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    Strange Integrals

    What is the most general antiderivative of \frac{1}{x}? I'll be you are expecting that:

    \int \frac{dx}{x} = ln|x| + C.

    BUT

    Note that \frac{d}{dx}ln ( -|x| ) = \frac{1}{x}.

    Now, I'll be the first to admit that ln ( -|x| ) has no real values. But it does manage to solve the differential equation y^{\prime} = \frac{1}{x}.

    So what, if anything, does this all mean? And what does it mean to have a function with no (real) domain, but that has a real valued derivative? There's something I'm just not picturing right here.

    -Dan
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    Super Member Rebesques's Avatar
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    Oh Tospq, you are so affected by Physics, that you look for a meaning behind everything in the Formalistic Playground of Maths!
    Last edited by Rebesques; August 31st 2007 at 11:46 PM. Reason: sleepless :(
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Rebesques View Post
    Oh Tospq, you are so affected by Physics, that you look for a meaning behind everything in the Formalistic Playground of Maths!
    But of course! I wouldn't have it any other way.

    -Dan
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    Quote Originally Posted by topsquark View Post
    What is the most general antiderivative of \frac{1}{x}? I'll be you are expecting that:

    \int \frac{dx}{x} = ln|x| + C.

    BUT

    Note that \frac{d}{dx}ln ( -|x| ) = \frac{1}{x}.

    Now, I'll be the first to admit that ln ( -|x| ) has no real values. But it does manage to solve the differential equation y^{\prime} = \frac{1}{x}.

    So what, if anything, does this all mean? And what does it mean to have a function with no (real) domain, but that has a real valued derivative? There's something I'm just not picturing right here.

    -Dan
    I am not sure what you are doing.

    \ln |x| is the most general anti-derivative. You need to prove that f: \mathbb{R} - \{ 0 \}\mapsto \mathbb{R} has this an an anti-derivative which it does.

    What you wrote it not a function.
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    I am not sure what you are doing.

    \ln |x| is the most general anti-derivative. You need to prove that f: \mathbb{R} - \{ 0 \}\mapsto \mathbb{R} has this an an anti-derivative which it does.

    What you wrote it not a function.
    I am aware of all this. Especially the last part. What I am interested in is why is the ln(-|x^2|) solution even there at all?

    -Dan
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    Quote Originally Posted by topsquark View Post
    I am aware of all this. Especially the last part. What I am interested in is why is the ln(-|x^2|) solution even there at all?

    -Dan
    How is it a solution?
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    How is it a solution?
    Whoops! Made an oopsie. I meant, of course, that y = ln(-|x|) is a solution.

    Then
    y^{\prime} = \frac{1}{-|x|} \cdot -sign(x)

    When x is negative this means that y^{\prime} = \frac{1}{-(-x)} \cdot -(-1) = \frac{1}{x}

    When x is positive this means that y^{\prime} = \frac{1}{-x} \cdot -(1) = \frac{1}{x}

    (Hah! I did it right after all! )

    -Dan
    Last edited by topsquark; September 1st 2007 at 08:05 PM. Reason: I goofed. Yay!
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    Quote Originally Posted by topsquark View Post
    Whoops! Made an oopsie. I meant, of course, that y = ln(-|x|) is a solution.

    Then
    y^{\prime} = \frac{1}{-|x|} \cdot -sign(x)

    When x is negative this means that y^{\prime} = \frac{1}{-(-x)} \cdot -(-1) = \frac{1}{x}

    When x is positive this means that y^{\prime} = \frac{1}{-x} \cdot -(1) = \frac{1}{x}

    (Hah! I did it right after all! )

    -Dan
    The chain rule for f(g(x)) works only for functions so that g is differenciable and f is differenciable on the range of g. Now the range of g over here is (-\infty,0) certainly f is not differenciable on this interval.


    Another explanation is that you just cannot use the chain rule here. The "function" \ln (-|x|) is simply now a function. It is like taking the derivative of \sqrt{-(1+x^2)}.
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    The chain rule for f(g(x)) works only for functions so that g is differenciable and f is differenciable on the range of g. Now the range of g over here is (-\infty,0) certainly f is not differenciable on this interval.


    Another explanation is that you just cannot use the chain rule here. The "function" \ln (-|x|) is simply now a function. It is like taking the derivative of \sqrt{-(1+x^2)}.
    Hmmm....

    Never thought about it that way. (sigh) I suppose that is the more sensible approach. Though not nearly as intriguing...

    All right, problem solved to my satisfaction.

    -Dan
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