Is the Integral of 1/x ln x or log x?

[kasse]

Is the integral of 1/x

ln x og log x?

 

[chaoseverlasting]

The integral is lnx.

 

[christianjb]

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

 

[HallsofIvy]

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

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

 

[Werg22]

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

 

[christianjb]

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.

 

[Gib Z]

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

$$\int_1^x \frac{1}{t} dt = \ln x$$. 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

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

 

[Gib Z]

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^x.

That is eqivalent to the definition that e is the unique number that fulfills $$\lim_{h\to 0} \frac{e^h -1}{h} = 1$$.
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.

 

[christianjb]

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

$$\int_1^x \frac{1}{t} dt = \ln x$$. 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

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

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.

 

[HallsofIvy]

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
$$\lim_{x\rightarrow 0}\frac{a^x- a}{x}$$
exists.

 

[Gib Z]

...it avoids having to prove that
$$\lim_{x\rightarrow 0}\frac{a^x- a}{x}$$
exists.

$$\lim_{h\to 0} \frac{e^{x+h} - e^x}{h} = e^x \lim_{h\to 0} \frac{e^h -1}{h}$$

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

 

[HallsofIvy]

Yes, I meant
$$\lim_{x\rightarrow 0}\frac{a^x-1}{x}$$
Thanks