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Integrating 1/x

  1. Nov 20, 2004 #1
    Yeah, this questions may be a little elementary for some, but I don't seem to have any sources which would be able to tell me how do i integrate a function 1/x. Any help would be great.

    : )
  2. jcsd
  3. Nov 20, 2004 #2


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    I assume that you know that an antiderivative is ln(x), but you don't know how to get there?
  4. Nov 20, 2004 #3
    Go to www.google.com Type in "int(1/x)" and click "I'm Feeling lucky"

  5. Nov 20, 2004 #4
    whoa crap, ln (x) didn't cross my mind. Thanks alot........
  6. Nov 20, 2004 #5
    ln(x) will never "cross your mind." You have to either memorize it or know how to derive it. :wink:
  7. Nov 20, 2004 #6
    [tex]\int (1/x) dx=\int dy = y+constant = lnx+constant[/tex]
  8. Nov 20, 2004 #7


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    [tex] \int_1^x t^{-1 + \epsilon} dt [/tex]

    For [tex]\epsilon[/tex] greater than 0. Obviously this would be:

    [tex] \frac{x^{\epsilon} - 1}{\epsilon} [/tex]

    And you can see by applying l'hopitals that this goes to ln(x) as [tex]\epsilon[/tex] goes to 0.

    Now if you want to get more technical, and not assume anything, you can see that the derivative of a^x is:

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

    Now if the limit exists, which you can see it clearly does by looking at the graph, this means that the derivative is some function of a times a^x. You can easily check that for a less than one, this function of a is negative, its equal to 0 at one, and its greater than 0 for a greater than one, and continues getting bigger and bigger. For some a, it will be exactly equal to one, and we'll call this e. So the derivative of e^x is just e^x. The inverse of exponentiation is logarithm, and ln(x) is defined as the log base e. Then the derivative of a^x is the derivative of e^(ln(a)*x) which you can see from the chain rule is ln(a)*e^(ln(a)*x), or just ln(a)*a^x. So the function of a was just ln(a). This can be used in the limit above:

    [tex] \lim_{\epsilon \rightarrow 0} \frac{x^{\epsilon} - 1}{\epsilon} = \lim_{\epsilon \rightarrow 0} \frac{ln(x) \cdot x^{\epsilon}}{1} = ln(x) [/tex]

    Actually, you don't even have to use l'hopitals at this point, because if you look carefully you'll see this limit and the funtion of a are indentical, just with a replaced by x and h replaced by [tex]\epsilon[/tex].
    Last edited: Nov 20, 2004
  9. Nov 20, 2004 #8


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    Actually, you can take this even farther and use it to find the value of e:

    [tex] \ln(x) = \lim_{\epsilon \rightarrow 0} \frac{x^{\epsilon} - 1}{\epsilon} [/tex]

    Now noting that e^x is the inverse of ln(x):

    [tex] y = e^x [/tex]

    [tex] x = \ln(y) = \lim_{\epsilon \rightarrow 0} \frac{y^{\epsilon} - 1}{\epsilon} [/tex]

    rearranging (and noting that these expressions are only valid in the limit as [tex]\epsilon[/tex] goes to 0):

    [tex] \epsilon \cdot x = y^\epsilon - 1 [/tex]

    [tex] y^\epsilon = 1 + x \cdot \epsilon [/tex]

    [tex] y = (1 + \epsilon \cdot x)^{1/\epsilon} [/tex]

    which leaves:

    [tex] e^x =\lim_{\epsilon \rightarrow 0} (1 + \epsilon \cdot x)^{1/\epsilon}[/tex]

    or, plugging in 1 for x, and replacing [tex]\epsilon[/tex] with 1/n:

    [tex] e = \lim_{n \rightarrow \infty} (1 + 1/n)^n[/tex]

    Which is the standard way of defining e. I'm sorry if this is much more than you wanted. I never really thought about this stuff, just accepting it at face value, and your question prompted me to try to derive it myself. I was suprised how nicely it all works out.
  10. Nov 20, 2004 #9
    Don't forget that in calculus they use log[x] as the natural log, instead of ln[x].
  11. Nov 20, 2004 #10


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    Can you explain that...?
  12. Nov 20, 2004 #11
    Hey, thanks for showing how the integral for 1/y was obtained, that was helpful....

    : )
  13. Nov 20, 2004 #12
    Pretty much every calculus book uses log[x] instead of ln[x] to represent the natural log with base e. Remeber that in high school they told you that if you had log[x] without a base it was understood to be 10? Well, that's not ture in calculus. Log[x] has a base e. Is that better?
  14. Nov 20, 2004 #13


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    I have never seen log(x) used as the natural log
  15. Nov 20, 2004 #14
    I see it used all the time. Even Mathematica uses it like that.
  16. Nov 20, 2004 #15
    Yes, but only mathematicians.

    I don't go near engineers, because they smell of booze and modernism, but I'm told that they use log to mean log to the base 10.
  17. Nov 21, 2004 #16


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    It's just a notational convention, not a 'new' definition for the notation log(x) because it's calculus. In my book on statistical mechanics they also use log for ln.
    The author writes:
    My (university) calculus book uses ln, for log base e. Simply to seperate it from the other bases.
  18. Nov 21, 2004 #17
    My statistical mechanics book (Kittel) also uses log to mean ln. It has something to do with clarity. ln can be confused easily as a product of l and n or something...
  19. Nov 21, 2004 #18


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    Base 10 log is used to measure gain in power amplification, used to measure sound levels, used to measure earthquakes, used to measure power of radio signals and gain of antennas among other things.

    Even more importantly,
    What do they mean slide rules are only a historical curiosity!! I still use mine!


    And what's all this talk of smelling like booze Geez, I've gotta find a stronger aftershave
    Last edited: Nov 21, 2004
  20. Nov 21, 2004 #19

    matt grime

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    But as log to any base only differs by a multiplicative constant, it is merely a change of unit away from being log in any other base.

    log_{10} is useful since it is, give or take a small number, the number of digits in the original number. but it's still only factor of 2 and a bit away from log proper....

    as we were told:

    log will always mean log base e from now on. After all, why on earth would you want to use anything else?
  21. Nov 21, 2004 #20
    why would log mean log_e ?? thats what we have ln for !! Everything I've learned says that ln is log_e and log is log_10.
    As for the integration part you just have to remember that any equation with 1/x or x^-1 will have a ln in it's answer.
    I alway remember by doing this problem.
    what is the antiderivative of 1/chair ?
    I'll let you figure it out.
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