Homework Help: What is Integral of 1/x

The integral is lnx.

 

Proof:
y=ln(x)
d/dx ln(x)= dy/dx
x=e^y
dx/dy=x
so dy/dx=1/x

 

Possible cause of the confusion: some texts, especially advanced texts, use "log" to mean the natural logarithm (ln) rather than the common logarithm (base 10). Common logarithms are seldom used now.

In general, since log_a(x)= ln(x)/ln(a),
[tex]\frac{d log_a(x)}{dx}= \frac{1}{x ln(a)}[/tex]

Of course, ln(e)= 1 so that reduces to the 1/x for the derivative of ln(x).

 

That is not a proof at all. The natural number e is found with the definition of ln, the integral of 1/x. There is no "proof", it's a definition to start with. The only thing that needs to be proven is that ln behaves as a logarithmic function.

 

Whatever dude.
It's a proof given the defn. d/dx e^x=e^x and defining ln to be the inverse operation to e^.

I agree that you could define ln(x) by its derivative. A check on Wikipedia shows both definitions.

If you define ln(x) from its derivative- it remains to be shown that e^ln(x)=x. The proof is essentially the same as I gave- working the other way.

 

I remember mathwonk saying things are much easier and more rigourous if the ln function is defined as

[tex]\int_1^x \frac{1}{t} dt = \ln x[/tex]. From this definition we can define its base, and we Euler decided to call it e. So e can be defined to be the number which fulfills the condition that

[tex]\int_1^e \frac{1}{x} dx = 1[/tex]

 

That is eqivalent to the definition that e is the unique number that fulfills [tex]\lim_{h\to 0} \frac{e^h -1}{h} = 1 [/tex].
Definitons of e and Ln are not hard to make. The real challenge is to prove alternate definitons are in fact eqivalent. Wergs why makes things much easier.

 

That's fine. You have to define ln(x) somehow. I prefer to define it as the inverse operation to e^x, and then derive the integral of 1/x from there.

So, in fact, my full defn. would start from showing that e^x satisfies certain properties and working upwards- whereas you're starting from the other end and working backwords.

 

Either way is fine- and off the point, which was really the distinction between ln x and log x!

The reason mathwonk says that defining ln x in terms of the integral is easier, and I agree, is that it avoids having to prove that
[tex]\lim_{x\rightarrow 0}\frac{a^x- a}{x}[/tex]
exists.

 

[tex]\lim_{h\to 0} \frac{e^{x+h} - e^x}{h}
= e^x \lim_{h\to 0} \frac{e^h -1}{h}[/tex]

Did you mean 1 or instead of a, or am i missing something?

 

Yes, I meant
[tex]\lim_{x\rightarrow 0}\frac{a^x-1}{x}[/tex]
Thanks