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Calc II proof for integral of 1/x

  1. Aug 31, 2011 #1
    1. The problem statement, all variables and given/known data
    My professor gave us the problem: "Explain why not all antiderivatives of [itex]\frac{dx}{x}[/itex] are in the form of [itex]ln|x|[/itex]"


    2. Relevant equations
    [itex]\int \frac{dx}{x} = {ln|x|, x\neq 0}[/itex]


    3. The attempt at a solution
    I honestly don't know where to even start on this one, from all of my background I have been taught that [itex]\int \frac{dx}{x} = {ln|x|, x\neq 0}[/itex]. I have some clue that it has something to do with what happens when [itex] x = 0 [/itex] because of the fourm at this link (http://forums.xkcd.com/viewtopic.php?f=17&t=68042) but the poster that proposed [itex]\int \frac{dx}{x}[/itex] may not equal [itex]ln|x|[/itex] never explained why. Any help would be greatly appreciated.
     
  2. jcsd
  3. Aug 31, 2011 #2

    ehild

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    Is only a single anti-derivative in general?

    ehild
     
  4. Aug 31, 2011 #3
    Correct (from what I gathered). He just wants any antiderivative that isn't [itex]ln|x|[/itex] or [itex]ln(x)[/itex] and why that is an antiderivative of [itex] \frac {1}{x}[/itex]
     
  5. Aug 31, 2011 #4
    I think the last post on your xkcd link does explain it pretty thoroughly, though the tex formatting doesn't seem to be coming through. The last poster mentions that the general anti-derivative is of the form

    [itex]f(x) = \left\{\begin{array}x \ln|x| + C_1 & x > 0 \\ \ln|x| + C_2 & x < 0\end{array}\right.[/itex]

    and the reasoning is exactly as you have suspected - the discontinuity at x=0.
     
  6. Aug 31, 2011 #5
    But, that piecewise function is still in the form of [itex]ln|x|[/itex]. What my professor was looking for was an equation [itex]f(x)[/itex] not in that form that still satisfied [itex]F'(x) = f(x)[/itex]
     
  7. Aug 31, 2011 #6
    Hmm... I think some specifics of the question are being lost here. What does "in the form of" mean? Does your professor mean that it must be an anti-derivative which is not equal to [itex]\ln|x|+C[/itex], where C is a constant? If so, then if [itex]C_1 \ne C_2[/itex], the piecewise function defined above is an answer to the question, it is not equal to [itex]\ln|x|+C[/itex] for any constant C.

    I would go further to say that a function such as
    [itex]f(x) = \left\{\begin{array}x \ln|x| + C & x \ne 0 \\ 42 & x = 0\end{array}\right.[/itex]
    where C is a constant would also be an answer to the question. It's derivative is [itex]\frac{1}{x} \hspace{10mm} \forall x \ne 0[/itex], which is precisely what the expression [itex]\int\frac{1}{x}dx = f(x)[/itex] is saying.
     
    Last edited: Aug 31, 2011
  8. Aug 31, 2011 #7
    What I took him to mean was there would be no [itex]ln[/itex] in the formula at all.
     
  9. Aug 31, 2011 #8
    Well, it is easy to show (using the mean value theorem) that any anti-derivative on a continuous interval differs from any other anti-derivative on that interval by a constant. I think, perhaps, that your professor wanted you to play with that idea, and really prove that statement using the mean value theorem. This should yield the result that any anti-derivative will incorporate ln|x| in some way.
     
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