# Integral of 1/x

by kasse
Tags: 1 or x, integral
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 P: 463 Is the integral of 1/x ln x og log x?
 P: 1,017 The integral is lnx.
 P: 529 Proof: y=ln(x) d/dx ln(x)= dy/dx x=e^y dx/dy=x so dy/dx=1/x
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,559 Integral of 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 loga(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).
P: 1,520
 Quote by christianjb 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.
P: 529
 Quote by Werg22 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.
 HW Helper P: 3,348 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 $$\int_1^e \frac{1}{x} dx = 1$$
HW Helper
P: 3,348
 Quote by christianjb 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.
P: 529
 Quote by 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 $$\int_1^e \frac{1}{x} dx = 1$$

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.
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,559 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.
HW Helper
P: 3,348
 Quote by HallsofIvy ...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?
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,559 Yes, I meant $$\lim_{x\rightarrow 0}\frac{a^x-1}{x}$$ Thanks

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