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Integral of 1/x 
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#1
Feb1011, 05:30 AM

P: 67

G'day all,
I am trying to gain a deeper understanding of the integral of 1/x. I understand that ln(a) is the area under graph 1/x from x=1 to x=a, where a>0, this is a definite integral. What I am trying to wrap my head around is the integral of 1/x being lnx, with the absolute value of x causing me the most confusion. How does this relate to the above definition? What does this indefinite integral mean? I have seen it explained that for y = ln(x) where x<0, by chain rule dy/dx = 1/(x) = 1/x, thus the integral of 1/x is lnx, but it is here that I am losing all intuition of the concept. I understand the process of using the chain rule and how this solution is arrived at, but I'm left wondering what the indefinite integral of 1/x really is as a mentally tangible concept. 


#2
Feb1011, 06:00 AM

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PF Gold
P: 1,783

Note that the function 1/x is defined for both x>0 and x<0. ln(x) has domain only x>0.
So solve the two parts in pieces. What function had derivative 1/x when x< 0? answer ln(x). What function has derivative 1/x when x > 0? answer ln(x). Define a(x) = x if x>0 and x if x < 0 and you have the general answer ln(a(x)). This 'a' is a commonly defined function with a standard notion a(x) = x. Remember the definition of "the" indefinite integral of a function. It is the set of antiderivatives i.e. the solution set of y'=f(x). So [tex] \int x^{1}dx = lnx + c[/tex]. One other point. If it bothers you that we must throw in that absolute value, think about how ln(x) is defined. We define it in terms of solution to an exponential equation. We could have defined the notation to be the antiderivative of 1/x and absorb the x into the notation: Ln(x) = ln(x), and then derive that Ln(x) is the "antiexponential" for positive x, (and is also an even function since 1/x is an odd function). It is a matter of historic convention that Napier considered solutions to exponential equations and we invented notation for that first. 


#3
Feb1011, 07:16 AM

P: 67

Thanks for the speedy reply.
If, for arguments sake, the very definition of ln(x) was the area under 1/x, could this be defined over R\{0} or would we be stuck with the conventional R+? Could the asymptote be crossed in calculating the area? The area under 1/x from x=1 to x=1 would be zero (if it can be defined), as the two areas on either side of the asmptote would be equal and of opposite signs, but infinite in size. This corrosponds to ln(1) = 0. Could a version of ln(x) be defined as the area from x=1 to x=a, where a is R\{0} to include values of x<0? To me this definition of ln(x) = ln(x), fits very well with it's associated derivitive 1/x. I can see that 1/x is the derivitive of ln(X) for all R\{0}, but is my above reasoning another way of looking at the same mathematical phenomena that is 1/x? Is this explanation of why the integral of 1/x is indeed ln(x) valid, or are my asymptote crossing antics blasphemy? Thanks again. 


#4
Feb1011, 08:59 AM

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P: 11,955

Integral of 1/x
The natural logarithm defined on (0,infty) can be extended on the complex plane, so that any complex number except 0 can serve as a logarithm's argument.
This extension is tackled in all books on complex analysis, I presume. 


#5
Feb1011, 09:02 AM

Math
Emeritus
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Thanks
PF Gold
P: 39,691

(This is in response to SpruceMoose's last post. bigubau snuck in ahead of me.)
What you are talking about is the "Cauchy Principle Value" of an improper integral. If f(x) is not continuous at x= c, and a< c< b, then the standard definiton of the improper integral is [tex]\int_a^b f(x)dx= \lim_{\alpha\to c^}\int_a^\alpha f(x)dx+ \lim_{\lim_{\beta\to c^}\int_\beta^b f(x)dx[/tex] while the Cauchy Principal Value is [tex]C.P.V.\int_a^b f(x)dx= \lim_{\gamma\to 0^+}\int_a^{c\gamma} f(x)dx+ \int_{c+ \gamma}^b f(x)dx[/tex] If the "standard" improper integral exists, then so does the Cauchy Principal Value and they are equal but the Cauchy Principal Value may exist when the standard integral does not. In particular, for f(x)= 1/x, a= 1, b= 1, the Cauchy principal value is [tex]C.P.V.\int_{1}^1 \frac{1}{x}dx= \lim_{\gamma\to 0+}\int_{1}^{\gamma} \frac{1}{x}dx+ \int_\gamma^1 \frac{1}{x}dx[/tex] [tex]= \lim_{\gamma\to 0^+}\left(ln(1) ln\gamma\right)+ \left(ln(1) ln(\gamma)\right)[/tex] which equals 0 because everything on the left and right cancels. If we had use "[itex]\alpha[/itex]" and "[itex]\beta[/itex]" as in the standard definition, that would not happen and we would have [tex]\lim_{\alpha\to 0} ln(\alpha)+ \lim_{\beta\to 0} ln(\beta)[/tex] and now neither limit exists. 


#6
Feb1011, 10:14 AM

P: 67

Thanks Bigubau and Ivy.
Does this mean that the C.P.V. of 1/x taken from x=1 to x=a, where a is R\{0} = ln(a)? 


#7
Feb1011, 04:46 PM

Engineering
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P: 7,295

A more elementary way of thinking about this, if you restrict yourself to real variables, is to see (by looking at the graph of y = 1/x) that your "integral" definition of ln(x) only makes obvious sense for x > 0.
From that definition, is it clear that [tex]\int_a^b \frac 1 x dx = \log b  \log a[/tex] when a and b are both positive However it is equally clear from the graph (or by substituting u = x) what the value of definite integral [tex]\int_a^b \frac 1 x dx[/tex] is when both a and b are both negative. That explains where the x comes from. Personally I think the x notation is confusing for beginners. It's probably easier to think about the "positive" and "negative" cases separately, till you are confident about what is going on. 


#8
Feb1011, 08:15 PM

P: 67

Thanks AlephZero.
I never thought of it that way and that makes sense if you never cross the assymptote. But my question still remains. Does the C.P.V. of 1/x taken from x=1 to x=a, where a is R\{0} = ln(a), more specifically where a<0? 


#9
Feb1111, 09:11 AM

Math
Emeritus
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Thanks
PF Gold
P: 39,691

Yes, that is correct. And as for a> 0, that is the same as the integral.



#10
Feb1111, 01:20 PM

P: 67

Thank you very much.
Your help is greatly appreciated. 


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