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Consider an open circle $\mathcal{C}$ in the complex plane, which is centered on point $1$ of the radius $\varepsilon$ and doesn't intersect with $1-\varepsilon$ oriented in the anti-clockwise direction. Furthermore we define $f:\mathbb{C}\rightarrow\mathbb{C}$ $$f(z) = \cfrac{\log(z-1)}{z^2}\,x^z,$$ where $z\in\mathbb{C}$ and $x\in\mathbb{R}^+$.

Question: What is the exact value of $\int_\mathcal{C}f$ ?

Parameterization isn't helpful, because the integral does not tend to zero.

Daniel
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  • Related. A parametrization is not necessary, consider what the ML inequality gives for the integral over a small circle around $1$ and how it's different from taking a small circle around $0$. – Maxim Oct 20 '20 at 19:14
  • Is there a possibility that you could say how u would evaluate this contour? U don't have to be very rigorous, just show me please the "main steps". That would be very nice and helpful! – Daniel Oct 21 '20 at 12:25
  • The Wikipedia example from the linked question already has the details. You get $O(|\epsilon \ln \epsilon|)$ for the integral over a circle around $1$ and $O(|\epsilon^{-1} \ln \epsilon|)$ for the integral over a circle around $0$. So the first one is negligible, but we can't say anything about the second one. In fact, what you would get from this approach is an integral over $(-\infty, -\epsilon] \cup [\epsilon, 1]$, which doesn't have a finite limit as $\epsilon \to 0$, plus an integral over the part of the contour inside $|z| < \epsilon$, which doesn't have a finite limit either. – Maxim Oct 21 '20 at 13:19

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