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Logarithmic Integral

  1. Apr 11, 2010 #1
    I was trying to evaluate this integral,

    [tex]\int\frac{dx}{\ln x}[/tex]

    I substituted [tex]x=e^{i\theta}[/tex] and I get,


    which is,

    [tex]\int\frac{\cos \theta}{\theta}+i\frac{\sin \theta}{\theta} \ d\theta[/tex]

    [tex]\int\frac{\cos \theta}{\theta} \ d\theta+i\int\frac{\sin \theta}{\theta} \ d\theta[/tex]

    [tex]Ci(\theta)+i \ Si(\theta)[/tex]

    [tex]Ci(\theta)[/tex] and [tex]Si(\theta)[/tex] are the cosine and sine integrals, respectively.


    [tex]\int\frac{dx}{\ln x}=Ci(-i\ln x)+i \ Si(-i\ln x)[/tex]

    I was just asking if anybody has seen the logarithmic integral( [tex]li(x)[/tex] ) expressed this way.
  2. jcsd
  3. Apr 11, 2010 #2


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    Science Advisor

    Interesting. The Mathematica functions site (http://functions.wolfram.com/GammaBetaErf/LogIntegral/27/01/0003/) lists the following identity:

    [tex] \text{li}(z)=\text{Ci}(i \log (z))-i\text{ Si}(i \log (z))-\log (i
    \log (z))+\frac{1}{2} \left(\log (\log (z))-\log
    \left(\frac{1}{\log (z)}\right)\right) [/tex]

    This contains your two terms but has some added terms. I'm not sure why.
  4. Apr 11, 2010 #3
    Pay attention to limits of integration.

    Your first integral is from 0 to x, but your substitution means that your second integral is now a contour integral over from [itex]-i \ln{x}[/itex] to [itex]i\infty[/itex]. So, you can't use cosine and sine integrals already. An added difficulty is that 1/ln(x) has a singularity at x=1, and, consequently, [itex]e^{i\theta}/{\theta}[/itex] has a singularity at 0. You need to think how to do the contour integral so that the result matches the Cauchy principal value of your original integral.
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