Understanding the Relationship between Integration and Natural Logarithms

[morrobay]

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

 

[Integral]

ln x = \int_1^x \frac {dx} {x} 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 1 = \int_1^e \frac {dx} {x}

 

[christianjb]

int[1/x]=ln(x); ln(1)=0 by definition.

It follows that d/dx ln(x)=d/dx(int[1/x])=1/x.

See: http://en.wikipedia.org/wiki/Natural_logarithm

 

[christianjb]

ln x = \int_1^x \frac {dx} {x} 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 1 = \int_1^e \frac {dx} {x}

You beat me by seconds!

 

[HallsofIvy]

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

One can, however, also first define e^x by defining a^x in general, and showing that the derivative of a^x is of the form C_a a^x for some constant C_a. (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 C_e= 1 so that the derivative of de^x/dx= e^x.

If you have defined e^x 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 e^x.

morrobay, if that is the way you have defined ln(x), then saying y= ln(x) is the same as saying x= e^y. Differentiating x with respect to y, dx/dy= e^y so
dy/dx= 1/e^y. (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= e^y, dy/dx= 1/e^y= 1/x. y= ln(x) so that says d(ln(x))/dx= 1/x.

 

[dodo]

I understand ... that the derivative of ln x = 1/x

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.

 

[morrobay]

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

 

[Gib Z]

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.

 

[HallsofIvy]

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.

 

[mathwonk]

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.

 

[ice109]

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.

 

[d_leet]

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.

This is your first post in this thread and you basically only do it to insult others? What exactly was the point in that?

 

[Gib Z]

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.