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.

Cherish Rivera

Cherish Rivera

Answered question

2022-11-24

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,

Answer & Explanation

Mylee Mcintyre

Mylee Mcintyre

Beginner2022-11-25Added 11 answers

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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