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Principal value and integral of 1/z

  1. Mar 6, 2010 #1
    I am slightly confused about the definition of principle value. If we have an integral
    [tex]\int 1/z,[/tex]
    where the integration from [tex]-\infty[/tex] to [tex]\infty[/tex] is implied, then by Cauchy integral theorem we know that the principle value
    [tex]P \int 1/z=i\pi.[/tex]

    However, I would like to write down this principle value explicitly. My best shot is
    [tex] \lim_{\epsilon\rightarrow0}\lim_{R\rightarrow\infty}\int_{-R}^{-\epsilon}1/z+\int_{\epsilon}^{R}1/z.[/tex]

    Assuming that this is correct (is it?) I can (can I?) calculate the integrals first and take limits afterwards. I get

    [tex]\lim_{\epsilon\rightarrow0}\lim_{R\rightarrow\infty} \ln\left(-\frac{\epsilon}{\epsilon}\right) + \ln\left(-\frac{R}{R}\right)=2\ln(-1)=0.[/tex]

    Can you tell me what am I doing wrong?
     
  2. jcsd
  3. Mar 6, 2010 #2
    You need to think about how to define the function ln(z). There is no unique definition, precisely because of the path dependence of the integral of 1/z.
     
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