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Integral of 1/x 
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#1
Mar2007, 11:46 PM

P: 463

Is the integral of 1/x
ln x og log x? 


#2
Mar2107, 01:15 AM

P: 1,017

The integral is lnx.



#3
Mar2107, 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 


#4
Mar2107, 05:35 AM

Math
Emeritus
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Thanks
PF Gold
P: 39,338

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 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). 


#5
Mar2107, 04:41 PM

P: 1,520




#6
Mar2107, 04:54 PM

P: 529

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. 


#7
Mar2207, 03:59 AM

HW Helper
P: 3,352

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] 


#8
Mar2207, 04:05 AM

HW Helper
P: 3,352

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. 


#9
Mar2207, 04:08 AM

P: 529

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. 


#10
Mar2207, 01:39 PM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,338

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. 


#11
Mar2307, 03:06 AM

HW Helper
P: 3,352

= 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? 


#12
Mar2307, 05:49 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,338

Yes, I meant
[tex]\lim_{x\rightarrow 0}\frac{a^x1}{x}[/tex] Thanks 


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