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Improper integral of 1/x

  1. Aug 11, 2004 #1
    OK, so I'm trying to work out this:
    [tex] \int^{\infty}_a \frac{\dx}{x} [/tex]
    Where [tex] a [/tex] is a positive constant. Can you evaluate this analytically? I'm thinking the limit must exist, but [tex] \ln \left( \infty \right) = \infty [/tex] , or at least tends to it in the limit. So can someone tell me the deal?

    p.s. There's a dx ontop of that fraction, which has mysteriously disappeared into the abyss.
    Last edited: Aug 11, 2004
  2. jcsd
  3. Aug 11, 2004 #2
    [tex]\int_a^\infty\frac{dx}{x}=\lim_{b\rightarrow\infty}\int_a^b\frac{dx}{x}=\lim_{b\rightarrow\infty}[\ln b-\ln a]=\infty[/tex]

    So the limit does not exist.
  4. Aug 11, 2004 #3
    This seems very strange to me, seeing as
    [tex] \lim_{b \rightarrow \infty } \frac{1}{b} = 0 [/tex] . The matter arose while I was trying to find an electric potential in electrostatics by integrating the electric field from infinity to the point in question. The field was proportional to the reciprocal of the distance.
  5. Aug 11, 2004 #4
    Just because the function you are integrating goes to zero, it doesn't mean that the integral will be defined. Integrals like [tex]\int_a^\infty\frac{dx}{x^2}[/tex] are defined, but functions like 1/x go to zero too "slowly" for the integral to converge.
  6. Aug 12, 2004 #5


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    Yeah, it's odd that the the integral of [itex]1/x[/itex] diverges, even if it diverges very very slowly. It's the same for the summation:

    [tex]\lim_{n \rightarrow \infty} \sum_{k=1}^{n} \frac{1}{k} = \infty[/tex]
  7. Aug 12, 2004 #6
    Two important points. The electric field decreases with the square of the distance. That's why the integral converges. As far as 1/x goes, just keep in mind that while the curve might approach the axis as x->inf it doesn't mean that the the area under the curve does.

  8. Aug 12, 2004 #7


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    Notice that in the series 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 +....

    That the first term is 1/2, the sum of next two terms is > 1/2, the sum of th3 next four is > 1/2, and the sum of the next eight terms is > 1/2, etc.....,

    so the sum grows very slowly to as large as you like. I.e. at each stage, it always takes "twice as long" to get larger by another 1/2 as it did before, but it eventually does so.
  9. Aug 13, 2004 #8
    OK thanks, I think the trouble with my electrostatics lay in the fact that I was using an approximation to an infinite charge distribution, in which case (i.e. an idealized one) the potential may well be infinite. How ill-behaved of 1/x to keep eeking out a living.
  10. Aug 18, 2004 #9
    The logarithmic functions are increasing functions if the base is larger than 1. That's why this integral is improper.
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