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Integral of 1/x

  1. Mar 20, 2007 #1
    Is the integral of 1/x

    ln x og log x?
     
  2. jcsd
  3. Mar 21, 2007 #2
    The integral is lnx.
     
  4. Mar 21, 2007 #3
    Proof:
    y=ln(x)
    d/dx ln(x)= dy/dx
    x=e^y
    dx/dy=x
    so dy/dx=1/x
     
  5. Mar 21, 2007 #4

    HallsofIvy

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    Possible cause of the confusion: some texts, especially advanced texts, use "log" to mean the natural logarithm (ln) rather than the common logarithm (base 10). Common logarithms are seldom used now.

    In general, since loga(x)= ln(x)/ln(a),
    [tex]\frac{d log_a(x)}{dx}= \frac{1}{x ln(a)}[/tex]

    Of course, ln(e)= 1 so that reduces to the 1/x for the derivative of ln(x).
     
    Last edited: Mar 21, 2007
  6. Mar 21, 2007 #5
    That is not a proof at all. The natural number e is found with the definition of ln, the integral of 1/x. There is no "proof", it's a definition to start with. The only thing that needs to be proven is that ln behaves as a logarithmic function.
     
  7. Mar 21, 2007 #6
    Whatever dude.
    It's a proof given the defn. d/dx e^x=e^x and defining ln to be the inverse operation to e^.

    I agree that you could define ln(x) by its derivative. A check on Wikipedia shows both definitions.

    If you define ln(x) from its derivative- it remains to be shown that e^ln(x)=x. The proof is essentially the same as I gave- working the other way.
     
  8. Mar 22, 2007 #7

    Gib Z

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    I remember mathwonk saying things are much easier and more rigourous if the ln function is defined as

    [tex]\int_1^x \frac{1}{t} dt = \ln x[/tex]. From this definition we can define its base, and we Euler decided to call it e. So e can be defined to be the number which fulfills the condition that

    [tex]\int_1^e \frac{1}{x} dx = 1[/tex]
     
  9. Mar 22, 2007 #8

    Gib Z

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    That is eqivalent to the definition that e is the unique number that fulfills [tex]\lim_{h\to 0} \frac{e^h -1}{h} = 1 [/tex].
    Definitons of e and Ln are not hard to make. The real challenge is to prove alternate definitons are in fact eqivalent. Wergs why makes things much easier.
     
  10. Mar 22, 2007 #9

    That's fine. You have to define ln(x) somehow. I prefer to define it as the inverse operation to e^x, and then derive the integral of 1/x from there.

    So, in fact, my full defn. would start from showing that e^x satisfies certain properties and working upwards- whereas you're starting from the other end and working backwords.
     
  11. Mar 22, 2007 #10

    HallsofIvy

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    Either way is fine- and off the point, which was really the distinction between ln x and log x!

    The reason mathwonk says that defining ln x in terms of the integral is easier, and I agree, is that it avoids having to prove that
    [tex]\lim_{x\rightarrow 0}\frac{a^x- a}{x}[/tex]
    exists.
     
  12. Mar 23, 2007 #11

    Gib Z

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    [tex]\lim_{h\to 0} \frac{e^{x+h} - e^x}{h}
    = e^x \lim_{h\to 0} \frac{e^h -1}{h}[/tex]

    Did you mean 1 or instead of a, or am i missing something?
     
  13. Mar 23, 2007 #12

    HallsofIvy

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    Yes, I meant
    [tex]\lim_{x\rightarrow 0}\frac{a^x-1}{x}[/tex]
    Thanks
     
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