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Integral of 1/x

by kasse
Tags: 1 or x, integral
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kasse
#1
Mar20-07, 11:46 PM
P: 463
Is the integral of 1/x

ln x og log x?
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chaoseverlasting
#2
Mar21-07, 01:15 AM
P: 1,017
The integral is lnx.
christianjb
#3
Mar21-07, 04:36 AM
P: 529
Proof:
y=ln(x)
d/dx ln(x)= dy/dx
x=e^y
dx/dy=x
so dy/dx=1/x

HallsofIvy
#4
Mar21-07, 05:35 AM
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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),
[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).
Werg22
#5
Mar21-07, 04:41 PM
P: 1,520
Quote Quote by christianjb View Post
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
#6
Mar21-07, 04:54 PM
P: 529
Quote Quote by Werg22 View Post
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
#7
Mar22-07, 03:59 AM
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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]
Gib Z
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Mar22-07, 04:05 AM
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Quote Quote by christianjb View Post
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 [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.
christianjb
#9
Mar22-07, 04:08 AM
P: 529
Quote Quote by Gib Z View Post
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'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
#10
Mar22-07, 01:39 PM
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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.
Gib Z
#11
Mar23-07, 03:06 AM
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Quote Quote by HallsofIvy View Post
...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?
HallsofIvy
#12
Mar23-07, 05:49 AM
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Yes, I meant
[tex]\lim_{x\rightarrow 0}\frac{a^x-1}{x}[/tex]
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