## Proof for intergral dx/x=lnx?

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### Proof for intergral dx/x=lnx?

All in the title, can someone explain to me the proof that

intergrating dx/x gives ln(x)?

Also, what would this make things like dx/(x^2) intergrate to?

intergrating dx/x gives ln(x)?

Also, what would this make things like dx/(x^2) intergrate to?

- jestingrabbit
- Factoids are just Datas that haven't grown up yet
**Posts:**5965**Joined:**Tue Nov 28, 2006 9:50 pm UTC**Location:**Sydney

### Re: Proof for intergral dx/x=lnx?

Ratio wrote:All in the title, can someone explain to me the proof that

intergrating dx/x gives ln(x)?

Also, what would this make things like dx/(x^2) integrate to?

To answer your first question we really have to know how you're defining ln. Sometimes that integral is given as the definition, for instance. You might want to look up the inverse function theorem if you're defining it as the inverse of exp.

As for your second question, the answer is a lot simpler than you think. hint:

**Spoiler:**

ameretrifle wrote:Magic space feudalism is therefore a viable idea.

### Re: Proof for intergral dx/x=lnx?

I apologise, I was brought up with ln always meaning log to the base e. I forgot that not everyone had that definition.

- gmalivuk
- GNU Terry Pratchett
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### Re: Proof for intergral dx/x=lnx?

Yes, but then how are you defining e. The point is, one way of defining e is for it to be the base of the logarithm formed by integrating 1/x. (Or, more simply, e is such that the derivative of log base e is 1/x.)

### Re: Proof for intergral dx/x=lnx?

First some change of variable magic:

[math]\int_a^{ab} \frac{1}{x}dx = \int_1^b \frac{a}{at} dt = \int_1^b \frac{1}{t} dt[/math]

Now apply this magic to the function of interest [imath]f(a) = \int_1^{a}\frac{1}{x} dx[/imath]:

[math]f(ab) = \int_1^{ab}\frac{1}{x} dx = \int_1^{a}\frac{1}{x} dx + \int_a^{ab}\frac{1}{x} dx = \int_1^{a}\frac{1}{x} dx+\int_1^{b}\frac{1}{t} dt = f(a)+f(b)[/math]

So this function has the property [imath]f(ab)=f(a)+f(b)[/imath] which means that it is some logarithm. Now "just" evaluate it at any particular point to show that it is the right one.

Also [math]\int\frac{dx}{x^2} = \int x^{-2} dx = \frac{x^{(-2+1)}}{(-2+1)}[/math] just like any other power of x. Only if the exponent is exactly -1 does anything weird happen. Think about it: You know the power rule for derivatives: [imath]\frac{d}{dx} x^n = nx^{n-1}[/imath] for any n other than 0. Since this doesn't work for [imath]x^0=1[/imath] there is no power of x that has 1/x a its derivative. But for any other power of x, there is some other power with that as its derivative.

[math]\int_a^{ab} \frac{1}{x}dx = \int_1^b \frac{a}{at} dt = \int_1^b \frac{1}{t} dt[/math]

Now apply this magic to the function of interest [imath]f(a) = \int_1^{a}\frac{1}{x} dx[/imath]:

[math]f(ab) = \int_1^{ab}\frac{1}{x} dx = \int_1^{a}\frac{1}{x} dx + \int_a^{ab}\frac{1}{x} dx = \int_1^{a}\frac{1}{x} dx+\int_1^{b}\frac{1}{t} dt = f(a)+f(b)[/math]

So this function has the property [imath]f(ab)=f(a)+f(b)[/imath] which means that it is some logarithm. Now "just" evaluate it at any particular point to show that it is the right one.

Also [math]\int\frac{dx}{x^2} = \int x^{-2} dx = \frac{x^{(-2+1)}}{(-2+1)}[/math] just like any other power of x. Only if the exponent is exactly -1 does anything weird happen. Think about it: You know the power rule for derivatives: [imath]\frac{d}{dx} x^n = nx^{n-1}[/imath] for any n other than 0. Since this doesn't work for [imath]x^0=1[/imath] there is no power of x that has 1/x a its derivative. But for any other power of x, there is some other power with that as its derivative.

### Re: Proof for intergral dx/x=lnx?

start with d/dx (e^x) = e^x, and e^x is the inverse of ln(x)

e^ln(x) = x

differentiate both sides with respect to x

[e^ln(x)][d/dx(ln(x))] = 1

x[d/dx(ln(x))] = 1

d/dx(ln(x)) = 1/x

fundamental theorem of calculus => ln(x) = integral (1/x)

e^ln(x) = x

differentiate both sides with respect to x

[e^ln(x)][d/dx(ln(x))] = 1

x[d/dx(ln(x))] = 1

d/dx(ln(x)) = 1/x

fundamental theorem of calculus => ln(x) = integral (1/x)

- gmalivuk
- GNU Terry Pratchett
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### Re: Proof for intergral dx/x=lnx?

yeyui wrote:You know the power rule for derivatives: [imath]\frac{d}{dx} x^n = nx^{n-1}[/imath] for any n other than 0.

Actually, the derivative rule does work perfectly fine for n=0, too, since [imath]\frac{d}{dx} x^0 = 0x^{-1}=0[/imath], as we'd expect.

It just doesn't tell us anything useful about 1/x, because the zero there annihilates any information that could otherwise normally be gleaned from the derivative.

### Re: Proof for intergral dx/x=lnx?

I'd like to thank everyone for their responses, I understand it now.

### Re: Proof for intergral dx/x=lnx?

Can someone show with the graphs of: y = 1/x and y= ln(x) that integral of 1/x = ln(x). I have seen math proofs for this as well as d/dx ln(x) = 1/x

And graphs proving d/dx ln(x) = 1/x. But besides a Riemman summation is there a graphic way to show area under curve of 1/x = ln(x) ?

And graphs proving d/dx ln(x) = 1/x. But besides a Riemman summation is there a graphic way to show area under curve of 1/x = ln(x) ?

### Re: Proof for intergral dx/x=lnx?

I have a question. Does anyone have a proof that f(xy) = f(x) + f(y), f(e) = 1, defines a unique function for positive, real x and y, provided that it truly does? In that case, I would think that to be the most efficient definition of ln.

http://aselliedraws.tumblr.com/ - surreal sketches and characters.

- jestingrabbit
- Factoids are just Datas that haven't grown up yet
**Posts:**5965**Joined:**Tue Nov 28, 2006 9:50 pm UTC**Location:**Sydney

### Re: Proof for intergral dx/x=lnx?

Eastwinn wrote:I have a question. Does anyone have a proof that f(xy) = f(x) + f(y), f(e) = 1, defines a unique function for positive, real x and y, provided that it truly does? In that case, I would think that to be the most efficient definition of ln.

It doesn't. You need continuity (or convexity or some other regularity condition) to get uniqueness. Otherwise, that fixes its values on things of the form [imath]e^{p/q}[/imath], where p and q are positive integers, but you can do whatever you like on the rationals.

Edit: made a product a power (damn logarithms...).

ameretrifle wrote:Magic space feudalism is therefore a viable idea.

- NathanielJ
**Posts:**882**Joined:**Sun Jan 13, 2008 9:04 pm UTC

### Re: Proof for intergral dx/x=lnx?

Eastwinn wrote: Does anyone have a proof that f(xy) = f(x) + f(y), f(e) = 1, defines a unique function for positive, real x and y, provided that it truly does?

To perhaps expand a bit on what jesting said, a change of variables transforms the equation that you gave into Cauchy's functional equation, which isn't quite as nice over the real numbers as you might initially expect.

### Re: Proof for intergral dx/x=lnx?

Ah, continuity, I knew I was forgetting something. I'll try my hand at proving that that's enough

http://aselliedraws.tumblr.com/ - surreal sketches and characters.

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