Hi guys I am computing a complex integral and would just like some feed back if my work seems correct.

The question

Let γ\gamma be the circle |z|=1|z| = 1. Then compute ∫γLog(z)\int_\gamma Log(z). Where Log(z)Log(z) is the principal log.

My approach is direct I parametrize the curve and compute ∫baf(z(t))z′(t)dt\int_a ^b f(z(t))z'(t)dt.

Thus Let z(t)=eiθz(t)=e^{i\theta} where −π<θ≤π-\pi < \theta \leq \pi. As such we can compute z′(t)=ieiθz'(t)=ie^{i\theta}. ∫π−πLog(eiθ)ieiθdθ=−∫π−πθeiθdθ \int _{-\pi} ^{\pi} Log(e^{i \theta})ie^{i \theta}d\theta = -\int _{-\pi} ^{\pi} \theta e^{i\theta} d\theta Where we use integration by parts u=θ⇒du=dθu= \theta \Rightarrow du=d\theta and dv=eiθdθ⇒v=−ieiθdv= e^{i\theta} d\theta \Rightarrow v= -ie^{i \theta}. −∫π−πθeiθdθ=[iθeiθ−eiθ]π−π=πi+2-\int _{-\pi} ^{\pi} \theta e^{i\theta} d\theta=\large [i \theta e^{i\theta} -e^{i\theta} \large ]_{-\pi} ^{\pi} =\pi i +2 I would really appreciate it if people comment, also if people have a suggestions on how to calculate this in a better way I am curious. ================= What does it mean by principal Log? – ANUPAM BISHT 2 days ago principal value Log z is the logarithm whose imaginary part lies in the interval (−π,π](-\pi,\pi] – Kori 2 days ago note that log\log is discontinous at the unit circle, which means that your integration contour needs to be modified – tired 2 days ago But that would be true if I made it from 00 to2π2\pi. However shouldn't the one I set up avoid the discontinuity? – Kori 2 days ago Check your integration-by-parts computation -- I think you should be getting −2πi.-2 \pi i. – Mitchell Spector 2 days ago ================= =================