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Integral of 1/x=ln(x), or

  1. Dec 20, 2014 #1
    so i was messing around and found that
    is always true.
    taking the integrand as x-1
    at first we have
    but the limit is 2 sided(as m approaches the positive side of zero i get a positive value, while as m approaches the negative side of zero, i get a negative value.)

    So i decided to find the average of both sides of the limit:


    i graphed this with m=a really small (positive) number and found that it converges almost exactly to ln(x)

    anyway, i just thought it was pretty cool how the power property always holds so i figured id share it. most of the sites i see just say its ln(x), period.
    Last edited: Dec 20, 2014
  2. jcsd
  3. Dec 20, 2014 #2


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    No, it's not true when n = -1.
    What does this have to do with the integral above? Furthermore, this limit does not exist.
  4. Dec 20, 2014 #3
    in a way it is, i mean if its possible to find the integral with the standard power rule using limits, and that limit just happens to converge to ln(x) then its gotta be true, right?
    its the limit form of x0/0
    Ive heard people say that if the function is 2-sided the limit doesnt exist. But here the limit is clearly the average of both sides. I dont know if this is considered "standard" but obviously it has to mean something.'

    its really pretty odd that wikipedia doesnt offer
    00%7D%5Cfrac%7B1%7D%7B2%7D%28%5Cfrac%7Bx%5E%7Bm%7D%7D%7Bm%7D-%5Cfrac%7Bx%5E%7B-m%7D%7D%7Bm%7D%29.gif as an alternative definition for ln(x). it seems it would be pretty fundamental
    Last edited: Dec 20, 2014
  5. Dec 20, 2014 #4


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    I'm sorry, but nothing you post will (or even can) support this claim. Whenever you write a fraction, say ##\frac{a}{b}##, the notation explicitly requires ##b\neq0##. This is the requirement of the notation. This means that from the right hand side of the equation, ##n\neq -1##, always. So the claim that "the equation is always true" is by default false.

    Again by definition. A limit is said "to exist" if (a) it is independent of the path and (b) the value is an element of the co-domain.

    First, where's your proof?
    Second, I certainly don't consider that fundamental.
  6. Dec 20, 2014 #5
    note that i had to use rather large values of m (for m=.1 the difference is almost imperceptable to the eye)
    so id have to say that not only does it converge, but it converges quite nicely as well.
    and i derived it by evaluating a limit that (supposedly) doesnt exist.

    Attached Files:

  7. Dec 20, 2014 #6


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    DivergentSpectrum, that's not a proof. It's plausible, but that's not a proof. "almost imperceptable to the eye" is nowhere near good enough. A famous example is the integral
    ##\int_0^\infty \cos(2x) \prod_{n=1}^{\infty} \cos\left(\frac{x}{n}\right) \, dx##
    This was believed to be equal to ##\frac{\pi}{8}##, but it turns out that it differs at the 42nd decimal place.
  8. Dec 20, 2014 #7


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    This definition of a function seems too contrived and restrictive to be a good intuitive definition. The terms being averaged have to be exactly matched, and limits taken. Why would anyone be interested in that strange limit, or the function it defines? I might say it is an interesting property of ln(x), but not a good definition. On the other hand, defining ln(x) as the inverse function of ex seems much more natural to me.
  9. Dec 20, 2014 #8
    Well i kinda already did prove it.
    But i guess given that i did evaluate a limit that "does not exist" by averaging the right and left limit(which to me seems totally rational) im guessing theres a hole in the math somewhere.
    thats like saying e^x is 2.718282... times itself x times is better than saying its the function for which f(x)=d f(x)/dx
  10. Dec 20, 2014 #9


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    Both defining e(x) as the function that equals it's derivative and defining ln(x) as the integral of 1/x are well motivated if I care about slopes of functions and integrals of basic functions. It's the strange limit that I see no motivation for..
  11. Dec 20, 2014 #10


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    What? Where?

    The only time I've seen that technique used is in the context of Fourier series. It's very non-standard.

    If you mean "the stuff in the original post", then nothing in that post flows from one logic step to the next.
  12. Dec 21, 2014 #11
    let me try this again:
    The objective here is to generalize the power rule for all numbers.

    So, lets for a second assume the power rule is true for x-1

    then a careless calculation would give x0/0=nonsense.

    But maybe we could calculate the LIMIT


    here m is very small.
    Also notice that if m was positive:
    we would have a positive when x is positive; and a negative when x is negative

    but if m was negative:
    we would have a positive when x is negative; or a negative when x is positive.
    So, you could say:
    gif.latex?%5Clim_%7Bm%5Crightarrow%200%5E%7B+%7D%7D%20%5Cfrac%7Bx%5E%7Bm%7D%7D%7Bm%7D.gif gif.gif gif.gif
    limit DNE and give up.

    But heres the idea why not find the average of the two limits.

    we need to put the limits together, so we replace

    Now that the limits are the same, we plug it back into the average equation:


    = gif.gif
    which is my original formula.
    Last edited: Dec 21, 2014
  13. Dec 21, 2014 #12


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    Well the obvious problem is that the formula
    ##\lim_{x\to a} \left( f(x)+g(x)\right) = \lim_{x\to a} f(x) + \lim_{x\to a} g(x)##
    is only allowed if ##\lim_{x\to a} f(x)## and ##\lim_{x\to a} g(x)## individually exist, and in this case they don't.
  14. Dec 21, 2014 #13
    Honestly ive seen less rigorous proofs in textbooks. but ok, suppose i "got the right answer the wrong way"
    what is this formula 7D%7D%5Cfrac%7B1%7D%7B2%7D%28%5Cfrac%7Bx%5E%7Bm%7D%7D%7Bm%7D-%5Cfrac%7Bx%5E%7B-m%7D%7D%7Bm%7D%29.gif =ln(x) called? how would it be "properly" derived?
    added another graph for the skeptics. notice the size of the tick marks.

    whats this fourier series method called? im interested

    Attached Files:

    • 2.jpg
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  15. Dec 21, 2014 #14


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    I don't think it has a name.

    No idea. My initial thought was L'Hopital, but I couldn't get it to work.

    If ##f(x)## is a periodic function then under relatively weak conditions there exists constants ##a_n## and ##b_n## such that
    ##f(x) = \sum_{n=\infty}^\infty a_n \sin(2\pi nx/L) + b_n \cos(2\pi nx/L)##
    where L is the period.
    Importantly if ##f(x_0)## is a jump discontinuity then the Fourier series converges to ##\frac{f(x_0+) +f(x_0 -)}{2}## which is similar expression to what you have.
  16. Dec 21, 2014 #15
    Hi DivergentSpectrum !
    Your idea about
    with n tending to -1 (or m=n+1 tending to 0) is good in essence. But you should not use an indefinite integral.
    The value of the integral is not xm/m , but is (xm-x0m)/m where x0 is the lower bound.
    For m tending to 0 it tends to (1-1)/0 = 0/0 at first sight. But a more correct expression is :
    (em ln(x) - em ln(x0))/m = (1+m ln(x)+O(m2)... -1-m ln(x0)-O(m2))/m = ln(x)-ln(x0)+O(m)
    which limit is well defined = ln(x)-ln(x0)
    The definition of the natural logarithm is stated with x0=1 so that ln(1)=0.
    So, the definite integral tends to ln(x) as you claimed it.
  17. Dec 21, 2014 #16
    you can also work backwards to find the derivative:
    d 7D%7D%5Cfrac%7B1%7D%7B2%7D%28%5Cfrac%7Bx%5E%7Bm%7D%7D%7Bm%7D-%5Cfrac%7Bx%5E%7B-m%7D%7D%7Bm%7D%29.gif /dx=
    lim (xm-1+x-m-1)/2


    therefore ln(x)= 7D%7D%5Cfrac%7B1%7D%7B2%7D%28%5Cfrac%7Bx%5E%7Bm%7D%7D%7Bm%7D-%5Cfrac%7Bx%5E%7B-m%7D%7D%7Bm%7D%29.gif =∫x-1 dx

    I think if you were into numerically calculating ln(x), the limit definition would be immensely better than the taylor series.

    I really doubt i discovered something new, but its still a really cool result.

    edit: in my first proof i said
    i think i meant to say if x is negative we would get some complex number, doesnt really affect anything though.
    Last edited: Dec 21, 2014
  18. Dec 21, 2014 #17


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    if you want a general rule just write

    $$\int \! x^a\dfrac{\mathrm{d}x}{x}=C+\lim_{t\rightarrow a} \dfrac{x^t-1}{t}$$
  19. Dec 21, 2014 #18


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    How would you calculate ##x^{m}## faster than a Taylor series? A Taylor series only needs to worry about ##x^k## for integer k. You are asking for 1/m th roots.

    Edit: this is really a moot question: calculators don't use the Taylor series of Log anyway.
  20. Dec 22, 2014 #19
    im not sure entirely how floating point arithmetic works, but it converges everywhere as opposed to a series which is only good in the general area.
    Regardless, its a lot "prettier" than a taylor series lol.
    This definitely qualifies as a mathematical beauty type of thing. Personally it doesnt surprise me the method of averaging limits works for terms of sin and cos as well, considering how theyre so closely related to exp/ln. I guess you could almost say it "had to be true" that ln(x) is some kind of limiting case of the power rule. but im not going to go as far to say ln(x)=x0/0, lest i get stoned to death or burnt at the stake or something :P (maybe thats why i got a bad reception at first).

    Maxima recognizes the limit as ln(x), so im guessing mathematica does as well. What i dont get is why i cant find a single web page that states this formula
  21. Dec 22, 2014 #20


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    $$e^x=\lim_{h\rightarrow 0}(1+h\, x)^{1/h}\\
    \log(x)=\lim_{h\rightarrow 0}\dfrac{x^h-1}{h}$$

    Are dual limits. Both are well known. I will say a quick web search finds the first limit more often then the second. That is an interesting curiosity.
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