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Why is this proof wrong

  1. May 13, 2009 #1
    Another one of those 0 = 1 proofs.

    If you dislike them, please navigate away to cats doing things with captions. Otherwise stay tuned.

    Tell me what's wrong (obviously I know what's wrong but I just want to put it out there for people to mull over).

    Note, you need to know basic calculus.

    [tex]

    \int \frac{1}{x} dx = \int \frac{1}{x} dx
    [/tex]

    [tex]
    u = \frac{1}{x}
    [/tex]

    [tex]
    dv = dx
    [/tex]

    [tex]
    du = \frac{-1}{x^2} dx
    [/tex]

    [tex]
    v = x
    [/tex]

    [tex]
    \int \frac{1}{x} dx = u * v - \int v du
    [/tex]

    [tex]
    \int \frac{1}{x} dx = \frac{1}{x} * x - \int x * \frac{-1}{x^2} dx
    [/tex]

    [tex]
    \int \frac{1}{x} dx = 1 - \int \frac{-1}{x} dx
    [/tex]

    [tex]
    \int \frac{1}{x} dx = 1 + \int \frac{1}{x} dx
    [/tex]

    [tex]
    0 = 1
    [/tex]
     
    Last edited: May 13, 2009
  2. jcsd
  3. May 13, 2009 #2

    Pengwuino

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    Gold Member

    Right off the bat, [tex]u = \frac{{dv}}{x} = dx[/tex] makes no sense. The second line makes no sense either.... am i completely missing something?
     
  4. May 13, 2009 #3
    Sorry those are supposed to be separate. I will edit that in.

    This is what it should look like

    [tex]
    u = \frac{1}{x}
    [/tex]

    [tex]
    dv = dx
    [/tex]

    I have also added an extra step just to show that I am going to be using integration by parts to do the "proof"
     
  5. May 13, 2009 #4

    Pengwuino

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    Gold Member

    [tex]\int \frac{1}{x} dx = 1 - \int \frac{-1}{x} dx[/tex] isn't valid. Integration by part goes like:

    [tex]\int\limits_a^b {udv} = [uv]_a^b - \int\limits_a^b {vdu} [/tex]

    The term you think is 1 is actually 0.
     
  6. May 14, 2009 #5
    We are dealing with antiderivatives. So, given F(x) and G(x) that are antiderivatives of 1/x, it is true that F(x) = G(x) + C, for some constant C. For example, the second to last line you have can be written as

    ln(x) + C = 1 + ln(x) + D for some constants C, and D. We do not then conclude that 0 = 1.

    The integral of 1/x dx is a FAMILY of functions that all differ by a constant.
     
  7. May 14, 2009 #6

    neu

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    I only got to the end of the second sentence
     
  8. May 14, 2009 #7

    D H

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    Staff Emeritus
    Science Advisor

    Russell has the right answer. To be strictly correct, the rule for integration by parts is better written as

    [tex]\int u dv = u*v - \int v du + C[/tex]

    We drop the arbitrary constant because it is implied by the very use of indefinite integrals, or antiderivatives. The inverse of the derivative is not unique.
     
  9. May 14, 2009 #8
    Russell and D_H are correct, the constant term is missing. Gj guys :)
     
  10. May 16, 2009 #9

    gel

    User Avatar

    and Pengwuino too. You either put in constants of integration, or use definite integrals. Either way resolves the error.
     
  11. May 16, 2009 #10
    Right, but since I defined the problem in the start as indefinite integrals, I was looking for the answer that was related to indefinite integrals. Anyways. It's just a simple problem
     
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