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Math Help - integrating 1/x with lower limit of 0

  1. #1
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    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?
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  2. #2
    Moo
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    Hello,
    Quote Originally Posted by da kid View Post
    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
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  3. #3
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    Quote Originally Posted by Moo View Post
    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
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  4. #4
    MHF Contributor chisigma's Avatar
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    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
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  5. #5
    MHF Contributor matheagle's Avatar
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    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.
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  6. #6
    MHF Contributor chisigma's Avatar
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    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
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  7. #7
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    Quote Originally Posted by matheagle View Post
    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?
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