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Ln x

  1. Feb 12, 2008 #1


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    Can someone explain how the integration of 1/x = ln x
    I understand the usual way results in division by zero and that the derivative of ln x = 1/x
    So my question is, how is the quantity of area under the curve from 1 to x relate to the
    natural log of x
  2. jcsd
  3. Feb 13, 2008 #2


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    [tex] ln x = \int_1^x \frac {dx} {x} [/tex] is the definition of lnx. What more do you need?

    So the area below the inverse curve between 1 and x IS lnx by definition.

    A key relationship is [tex] 1 = \int_1^e \frac {dx} {x} [/tex]
  4. Feb 13, 2008 #3
  5. Feb 13, 2008 #4
    You beat me by seconds!
  6. Feb 13, 2008 #5


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    That is one definition of ln(x). (And, in fact, my preference. Although it would be better to write it as
    [tex]ln(x)= \int_1^x \frac{1}{t} dt[/tex].
    Of course, we immediately have, by the "Fundamental Theorem of Calculus, that d ln(x)/dx= 1/x.)

    One can, however, also first define ex by defining ax in general, and showing that the derivative of ax is of the form Ca ax for some constant Ca. (Constant in that it does not depend on x. It does depend on a, of course.) Then you define "e" to be the number such that Ce= 1 so that the derivative of dex/dx= ex.

    If you have defined ex in such a way, it is easy to show that it is "one-to-one" and so has an inverse and then define ln(x) to be the inverse function to ex.

    morrobay, if that is the way you have defined ln(x), then saying y= ln(x) is the same as saying x= ey. Differentiating x with respect to y, dx/dy= ey so
    dy/dx= 1/ey. (That is not just "inverting" the fraction! It requires the chain rule to show that dy/dx= 1/(dx/dy) but it is true.)

    But, since x= ey, dy/dx= 1/ey= 1/x. y= ln(x) so that says d(ln(x))/dx= 1/x.
    Last edited by a moderator: Feb 13, 2008
  7. Feb 13, 2008 #6
    From here, the indefinite integral of 1/x is any function F(x) of the family ln x + c, with c an arbitrary constant; and the definite integral from 1 to x is the difference F(x) - F(1), for any F in the family above. The lower limit 1 is chosen because F(1) = (ln 1) + c = c, so that F(x) - F(1) is simply ln x.

    Perhaps a grahpic illustration of how the derivative of ln x is 1/x could help. See the picture below (sorry for the quick and cheap graph, the scales are not right, but it illustrates the point.)

    At the left is the graph of the exponential function. At the right, the natural log, which is the same graph but with the axis flipped (x and y interchanged).

    At the left we show the tangent to the curve at point P. Since the derivative of the exponential is the exponential itself, the tangent of the angle marked 'theta' is exp(x); and since the opposite side is also exp(x) (the y coordinate of P), it follows that the adjacent side (over the x axis) is always 1, no matter which point P you choose.

    When you move that triangle to the right graph, along with the flipping of the whole figure, the dotted line now represents the x coordinate of P. Now the slope at P (the tangent of the angle marked 'alpha') is 1/x.

    Attached Files:

    • ln.png
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    Last edited: Feb 13, 2008
  8. Feb 14, 2008 #7


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    Thanks for the Courant/NYU level information .
    Regarding the graphic explanation of (d/dx) ln x = 1/x
    Is there an equivalent graphic explanation of the integration of 1/x = ln x
    Consider the area under curve of y = 1/x and x = 44, ln 44 = 3.78418. From 1 to 44.
    Can this area be defined in a quantity of unit areas, mabey from a summation of a series.
    Such that the total number of these unit areas is equivalent in number to 3.78418,
    that is the ln x
  9. Feb 14, 2008 #8

    Gib Z

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    Graphic explanations :( Well, we can show you geometric results and interpretations...

    The area is defined in terms of the summation of a series. Look up the Riemann definition of the integral.
  10. Feb 14, 2008 #9


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    What, exactly, are you asking?

    As far as the "indefinite integral" or "anti-derivative" is concerned, the integral is defined as the opposite of the derivative. If you know that the derivative of ln x is 1/x, then it follows immediately that the anti-derivative (integral) of 1/x is ln x.

    The fact that the definite integral, say from a to b, of 1/x is ln(b)- ln(a) follows from the fundamental theorem of calculus.
  11. Feb 14, 2008 #10


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    if it walks and talks like a duck it is a duck.

    Log functions are the only continuous non constant functions defined on R+, that take 1 to 0, and change multiplication into addition.

    If you use the fact that the derivative of that integral is 1/x, and the base point is 1, you can prove that integral does these things too, so it walks and talks like a log function, hence it must be one.

    i.e. any (differentiable) function that takes 0 to 1, and has derivative 1/x, takes multiplication to addition.
  12. Feb 14, 2008 #11
    thanks to everyone except mathwonk and gib z for their semi-lucid explanations. as for you two i hope being cryptic and confusing makes you both extremely smug and happy because it does nothing for anyone else's edification.
  13. Feb 14, 2008 #12
    This is your first post in this thread and you basically only do it to insult others? What exactly was the point in that?
  14. Feb 15, 2008 #13

    Gib Z

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    My post was in direct response to the post just before mine, in particular, lines 3 and 5. I thought I was being quite clear cut actually, obvious you don't.

    Mathwonks post was far more informative than mine, basically saying if a function has the same, and distinguishing properties as another one, then both functions are equal.
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