What is the integral of log(z) over the unit circle?I tried three times, but in the end I am concluding that it equals infinity, after parametrizing, making a substitution and integrating directly (since the Residue Theorem is not applicable, because the unit circle encloses an infinite amount of non-isolated singularities.)Any ideas are welcome.Thanks,

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Mylee Mcintyre · Accepted Answer

If   log  ⁡  z is interpreted as principal value      L    o    g    z  :=  log  ⁡      |    z      |    +  i      A    r    g    z     ,where       A    r    g   denotes the polar angle in the interval      ]      −    π    ,  π  [   , then the integral in question is well defined, and comes out to   −  2  π  i. (This is the case   α  :=  −  π in the following computations).But in reality the logarithm   log  ⁡  z of a   z  ∈                    C              ∗   is, as we all know, not a complex number, but only an equivalence class modulo   2  π  i. Of course it could be that due to miraculous cancellations the integral in question has a unique value nevertheless. For this to be the case we should expect that for any   α  ∈            R       and any choice of the branch of the   log along  γ  :    t  ↦  z  (  t  )  :=      e          i      t          (  α  ≤  t  ≤  α  +  2  π  )we obtain the same value of the integral. This boils down to computing      ∫    α          α      +      2      π        (  i  t  +  2  k  π  i  )    i      e          i      t          d  t  =  −      ∫    α          α      +      2      π        t        e          i      t          d  t  =  2  π  i        e          i      α           .During the computation several things have cancelled, but the factor       e          i      α       remains. This shows that the integral in question cannot be assigned a definite value without making some arbitrary choices.

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