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Improper integral

  1. Oct 17, 2006 #1
    hi, why is ln|x|, from -1 to 1, converging?

    is "0" the bad point, and must i break up the integral from 1 to $, where $ = 0, and from $ to -1... so i have xlnx-x as my derivative... and i get -2?

  2. jcsd
  3. Oct 18, 2006 #2


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    Since this has nothing to do with "differential equations" I am moving it to "calculus".

    I can't tell you "why is ln|x|, from -1 to 1, converging" or even the question I think you intended: "why is the integral of ln|x|, from -1 to 1, converging", because it isn't! The definition of an improper integral of f(x), from -1 to 1 where there is a singularity at x= 0, is
    [tex]\lim_{h\rightarrow 0}\int_{-1}^h f(t)dt+ \lim_{k\rightarrow 0}\int_k^1 f(t)dt[/tex]
    Notice that the two limits are taken separately. Yes, an anti-derivative of ln x is x ln x- x (for x> 0). An anti-derivative of ln(-x), for x< 0 is x ln(-x)+ x. Therefore
    [tex]\int_{-1}^1 ln|x|dx= \lim_{h\rightarrow 0^+}(-hln(h)-h)+\lim_{k\rightarrow 0^+}(kln(k)-k)[/tex]
    and neither of those limits exist.
    What you found, by taking both limits with "h" so that they cancelled out is the "Cauchy Principal Value" which has some occaisional uses but I don't think gives any good information here.
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