How was the integral of 1/x defined?

[gsingh2011]

I looked up the derivation for the fact that the integral of 1/x from 1 to infinity is ln(x), but I couldn't find anything. If you have one I would love to see it.

A lot of people were saying stuff like "It's a definition. You can't derive it". Is that correct? If so how could anyone know that the integral of 1/x from 1 to infinity is ln(x)? It seems like someone must have had a derivation to know that...

 

[Kevin_Axion]

I know you wanted the integral but here is a video describing the derivative of $$ln(x)$$ is equal to $$1/x$$, which might be useful.

 

[l'Hôpital]

Well, an equally good question is, what is ln x?

We CAN define it as the integral of 1/x, though perhaps you might be more familiar with it being the inverse of e^x.

In which case, consider the following:
$$e^{\ln x} = x \ \ Thus... \ \ \frac{d(e^{\ln x})}{dx} = 1$$

By the chain rule...

$$e^{\ln x} \frac{d(\ln x)}{dx} = 1$$

But we know e^ln x = x so...

$$\frac{d(\ln x)}{dx} = \frac{1}{x}$$

 

[gsingh2011]

Well, an equally good question is, what is ln x?

We CAN define it as the integral of 1/x, though perhaps you might be more familiar with it being the inverse of e^x.

In which case, consider the following:
$$e^{\ln x} = x \ \ Thus... \ \ \frac{d(e^{\ln x})}{dx} = 1$$

By the chain rule...

$$e^{\ln x} \frac{d(\ln x)}{dx} = 1$$

But we know e^ln x = x so...

$$\frac{d(\ln x)}{dx} = \frac{1}{x}$$

That makes sense, but isn't that a derivation and not a definition? And in response to the question what is ln(x), you could define a quantity called natural log of x as the integral of 1/x but how would you know that the integral would follow a logarithmic relationship, let alone a relationship with a base e.

Also, kevin_axion, i did not see any video.

 

[HallsofIvy]

You can, in fact start by defining
$$ln(x)= \int_1^x \frac{1}{t}dt$$
From that, because 1/x is continuous for all positive x but not defined for x= 0, we can immediately conclude:
1) ln(x) is defined for all positive x
2) it is continuous and differentiable for all positive x.
3) its derivative is 1/x which is positive for all positive x
4) so ln(x) is an increasing function.
5) $ln(1)= \int_1^1 dt/t= 0$
6) so ln(x)< 0 for 0<x< 1 and ln(x)> 0 for 1< x.

7) If x is any positive number, then so is 1/x and, by this definition
$$ln(1/x)= \int_1^{1/x} \frac{1}{t}dt$$
Let u= xt so that t= u/x and dt= du/x. 1/t= x/u so (1/t)dt= (x/u)(du/x)= du/u.
When t= 1, u= x and when t= 1/x, u= 1 so
$$ln(1/x)= \int_x^1 \frac{1}{u}du= -\int_1^x\frac{1}{u}du= -ln(x)$$

8) If x and y are both positive the so is xy and, by this definition,
$$ln(xy)= \int_1^{xy} \frac{1}{t}dt$$

Let u= t/x so that t= xu and dt= xdu. Then dt/t= (xdu)/xu= du/u. When t= 1, u= 1/x and when t= xy, u= y so
$$ln(xy)= \int_{1/x}^y \frac{1}{u}du= \int_{1/x}^1 \frac{1}{u}du+ \int_1^y \frac{1}{u}du$$
$$= -\int_1^{1/x} \frac{1}{u}du+ \int_1^y \frac{1}{u}du= -ln(1/x)+ ln(y)$$
$$= -(-ln(x))+ ln(y)= ln(x)+ ln(y)$$

9) If x is a positive real number and y is any real number, then $x^y$ is a positive real number and, by this definition,
$$ln(x^y)= \int_1^{x^y} \frac{1}{t}dt$$

If y is not 0, let $u= t^{1/y}$ so that $t= u^y$ and $dt= y u^{y-1}du$ and $dt/t= yu^{y-1}du/u^y= y du/u$. When t= 1 u= 1 and when $t= x^y$ u= x so
[tex]ln(y^x)= y\int_1^x \frac{1}{u}du= y ln(x)[/itex].

If y= 0, then $x^y= x^0= 1$ so $ln(x^y)= ln(1)= 0= 0 ln(x)$ so that "$ln(y^x)= y ln(x)$ for any real y.

10) Since ln(x) is continuous and differentiable for all positive real numbers, we can apply the mean value theorem to any interval of positive real number and, in particular to the interval [1, 2].
$$\frac{ln(2)- ln(1)}{2- 1}= \frac{ln(2)- 0}{1}= ln(2)$$
and the mean value theorem says there must be some number, c, between 0 and 1 such that ln(2) is equal to the derivative at c: ln(2)= 1/c. Since c< 2, 1/c> 1/2 and it follows that ln(2)> 1/2. That is important because we can say that, for any positive real number, X, $ln(2^{2X})= 2X ln(2)> 2X(1/2)= X$. That is, given any positive real number, there exist x such that ln(x) is larger- ln(x) is NOT bounded above.

Since ln(x) is an increasing function, it follows that $\lim_{x\to \infty} ln(x)= \infty$. Also, since ln(1/x)= -ln(x), $\lim_{x\to 0}ln(x)= \lim_{y\to \infty} ln(1/y)= -\lim_{y\to\inty} ln(y)= -\infty$. That is, ln(x) is a one-to-one function mapping the set of all positive real numbers onto the set of all real numbers.

Therefore, ln(x) has an inverse function, which I will call "Exp(x)" from the set of all real numbers to the set of all positive real numbers. All of the properties of "Exp(x)", such as the fact that the derivative of Exp(x) is again Exp(x), that Exp(x+ y)= Exp(x)Exp(y), etc. can be derived from the corresponding ln laws above but the crucial point, answering your question, is this:

If y= Exp(x), then, by definition of "inverse" function, x= ln(y). If x is not 0, we can divide both sides of the equation by it: $1= (1/x)ln(y)= ln(y^{1/x})$. Now go back to the "exponential" form: $y^{1/x}= Exp(1)$ from which it follows that $y= (Exp(1))^x$. If x= 0, then y= Exp(0)= 1 because ln(1)= 0 and $y= 1= Exp(1)^0= Exp(1)^x$. That is, for all x, $Exp(x)= Exp(1)^x$ so that Exp(x), the inverse of ln(x), really is an exponential function. We can, of course, define e= Exp(1) to get that the inverse to ln(x) is $e^x$

 

[l'Hôpital]

That makes sense, but isn't that a derivation and not a definition? And in response to the question what is ln(x), you could define a quantity called natural log of x as the integral of 1/x but how would you know that the integral would follow a logarithmic relationship, let alone a relationship with a base e.

Also, kevin_axion, i did not see any video.

You can prove it.

Let

$$F(x) = \int_{1}^{x} \frac{1}{x} dx$$

You can easily show things like F(x) + F(y) = F(xy) . I'll let you play with that for a while but the point is, you can infact do it. With this definition, you can define 'e' to be the value of x such that F(x) = 1 and things like that.

EDIT: Halls beat me to the punch with a much better one haha.

 

[Char. Limit]

You can prove it by differentiating ln(x). Watch:

$$f(x) = ln(x)$$

$$f'(x) = lim_{h\rightarrow0} \frac{ln(x+h)-ln(x)}{h}$$

$$= lim_{h\rightarrow0} \frac{ln\left(\frac{x+h}{x}\right)}{h}$$

$$= lim_{h\rightarrow0} \frac{ln\left(1 + \frac{h}{x}\right)}{h}$$

$$= lim_{h\rightarrow0} \frac{1}{x} \frac{x}{h} ln\left(1+ \frac{h}{x}\right)$$

Now define t=x/h. Then...

$$= lim_{h\rightarrow0} \frac{1}{x} t ln\left(1+ \frac{1}{t}\right)$$

$$= \frac{1}{x} lim_{h\rightarrow0} ln\left(\left(1+\frac{1}{t}\right)^t\right)$$

$$= \frac{1}{x} ln\left( lim_{h\rightarrow0} \left(1+\frac{1}{t}\right)^t\right)$$

As h approaches 0, t approaches infinity...

$$= \frac{1}{x} ln\left( lim_{t\rightarrow\infty} \left(1+\frac{1}{t}\right)^t\right)$$

$$= \frac{1}{x} ln(e)$$

$$f'(x) = \frac{1}{x}$$

Q.E.D.

 

[Kevin_Axion]

Yea, sorry about that: http://www.khanacademy.org/video/proof--d-dx-ln-x----1-x?playlist=Calculus .

 

[morrobay]

I have seen some very good explanations with graphs showing that : d(ln x) / dx = 1/x
including this one by Dodo, post #6 https://www.physicsforums.com/showthread.php?t=215045
I would like to see a graphical explanation of : integral (1 to x) 1/t dt = ln x
(besides a Riemann summation )