# Thread: integrating 1/x with lower limit of 0

1. ## integrating 1/x with lower limit of 0

i know that the anitderivative of 1/x is ln(x) but ln(0) is undefined so how am i supposed to compute something like the integral of 1/x from 0 to 1 or from 0 to 2 or basically any positive real number as an upper limit?

2. Hello,
Originally Posted by da kid
i know that the anitderivative of 1/x is ln(x) but ln(0) is undefined so how am i supposed to compute something like the integral of 1/x from 0 to 1 or from 0 to 2 or basically any positive real number as an upper limit?
$\int_a^b \frac 1x ~dx=\ln(b)-\ln(a)$ (a and b positive)

$\int_0^b \frac 1x~dx$ actually means $\lim_{a \to 0^+} \int_a^b \frac 1x ~dx=\lim_{a \to 0^+} \ln(b)-\ln(a)$

but we know that $\lim_{a \to 0^+} \ln(a)=-\infty$

thus the integral is $+\infty$

3. Originally Posted by Moo
Hello,

$\int_a^b \frac 1x ~dx=\ln(b)-\ln(a)$ (a and b positive)

$\int_0^b \frac 1x~dx$ actually means $\lim_{a \to 0^+} \int_a^b \frac 1x ~dx=\lim_{a \to 0^+} \ln(b)-\ln(a)$

but we know that $\lim_{a \to 0^+} \ln(a)=-\infty$

thus the integral is $+\infty$
that's kind of what i thought. i mean it makes sense with the integral being the area under the curve and since 1/x is unbounded as x goes to 0 there should be infinite area under the curve. guess i just don't trust my intuition much heheh

4. Some year ago in a technical meeting I proposed the following discussion: if we define f(*) as ...

$f(x)= \int_{-1}^{x} \frac{dt}{t}$ (1)

... what is the range of the variable x for which f(*) is defined?... My opinion about the question was that, since is...

$\int \frac {dt}{t}= \ln |t| + c$

... the f(*) expressed as integral in (1) is...

$f(x)= \ln |x|$

... the diagram of which is...

For $x<0$ the defined integral (1) has clearly 'no problems'. Some problem exists however for $x>0$ because we have to cross the point $x=0$ in which the function $\ln |x|$ has a singularity. I proposed to 'overcame' the problem defining for $x>0$ f(*) as...

$f(x) = \lim_{a \rightarrow 0} (\int_{-1}^{-a}\frac{dt}{t} + \int_{a}^{x} \frac{dt}{t}) = \lim_{a \rightarrow 0} (- \ln |-1| + \ln |a| -\ln |a| + \ln |x|) = \ln |x|$ (2)

I remember that some collegues did accept my point of view and others didn't. More recently i discovered that the procedure you see in (2) was proposed by the illustrious French matematician Louis Augustin Cauchy more than a century ago...

http://en.wikipedia.org/wiki/Cauchy_principal_value

The discussion if accept or not the Cauchy definition of 'principal value' of a definite integral like (1) is in my opinion very far to be concluded...

Kind regards

$\chi$ $\sigma$

5. That is not the accepted limit.
You cannot force both bounds heading towards zero to go at the same rate.
Similarily $\int_{-\infty}^{\infty}xdx$ is not zero.
There is no reason these bounds should approach their infinities at the same rate, which would cancel the two triangular areas out.

6. The example in my post has the only purpose to 'illustrate' the problem. There is however another example of such type of integration which has great pratical significance: the 'Logarithmic Integral Function'...

http://mathworld.wolfram.com/LogarithmicIntegral.html

The 'story' of this function looks like a 'thriller'. Let's start with the indefinite integral...

$\int \frac{dt}{\ln t}$ (1)

How to solve it?... Firs step is the substitution $x=\ln t$ and the function we have to integrate becomes...

$\frac{e^{x}}{x}= \sum_{n=0}^{\infty} \frac {x^{n-1}}{n!}$ (2)

Second step is integration by series...

$\int \frac{e^{x}}{x}\cdot dx = \ln |x| + \sum_{n=1}^{\infty}\frac{x^{n}}{n\cdot n!}$ (3)

Third step is turning back to the variable t...

$\int \frac{dt}{\ln t}= \ln |\ln t| + \sum_{n=1}^{\infty} \frac{\ln^{n} t}{n\cdot n!} + c$ (4)

From 'historical' point of view the first definition of 'Logarithmic Integral Function' has been...

$li (x)= \int_{0}^{x} \frac{dx}{\ln t}$ (5)

Such definition has no problem if $x<1$... some minor problem seems to exist for $x>1$... what to do?... For 'solve' [in some sense however...] the problem, after long and terrible efforts, it has been assumed the existence of a constant $\mu$ [the so called 'Soldner constant'...] for which $li(\mu)=li(0)=0$ and the definition now is...

$li(x)= \int_{\mu}^{x}\frac {dt}{\ln t}$ (6)

After more long and terrible efforts the following numerical value for $\mu$ has been found...

$\mu= 1.45136923488\dots$ (7)

If we impose that value in (4) we obtain for the constant c...

$c= \gamma = .577215664901\dots$ (8)

... where $\gamma$ is the 'Euler's constant' ...

$\gamma= \lim_{n \rightarrow \infty} \sum_{k=1}^{n} \frac {1}{k} - \ln n$ (9)

At this point [finally!!!...] we can write...

$li(x)= \int_{\mu}^{x} \frac{dt}{\ln t} = \ln |\ln t| + \gamma + \sum_{n=1}^{\infty} \frac {\ln^{n} t}{n\cdot n!}$ (10)

Using (10) the following diagram of li(*) has been obtained...

It seems that, after one and half century, the 'suggestion' of Louis Augustin Cauchy is still very good!!! ...

Kind regards

$\chi$ $\sigma$

7. Originally Posted by matheagle
That is not the accepted limit.
You cannot force both bounds heading towards zero to go at the same rate.
Similarily $\int_{-\infty}^{\infty}xdx$ is not zero.
There is no reason these bounds should approach their infinities at the same rate, which would cancel the two triangular areas out.
i got my initial question answered obviously but this is making me think about things i haven't really thought hard about before. i like it. i don't know much about limits so this is making me curious. i know that when it comes to infinity (or infinities) that i should leave my intuitions behind but they've gotten me this far so i hate to abandon them outright. i'd rather try to make them adaptive to circumstance

why is it that there is no reason the bounds on this integral should approach their infinities at the same rate? i'm not trying to be presumptuous or think i know something everybody else doesn't, but i want to lay my intuitions out so some more knowledgeable people can critique them and advise me towards a new way of interpreting these things. that said, my intuition says something to the opposite--that there's no reason the bounds shouldn't approach their infinities at the same rate. after all, why can't i just tell them to? what is it that stops me from doing that?

one of my teachers told me that when i want to evaluate an integral it sometimes help to think of "starting" at the lower limit and "ending" at the upper limit, but of course this doesn't help when the lower limit is -infinity because it's not like i can stand at -infinity and go forward until i get to +infinity.

let's say that i try to salvage the idea and start at what i could imagine to be the "middle" which intuition tells me is the origin. now i split into two of me and the two of me start off in opposite directions each of me with the goal of finding infinity in one direction or the other. i'm just rehashing the question i posed a moment ago: what's stopping me from telling the bounds to approach their infinities at the same rate?