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Let $\gamma_r(t) = 1+re^{it}$ for $0\leq t \leq \pi/4$. Calculate $\lim\limits_{r\rightarrow 0+} \int_{\gamma_r} \frac{\cos (\pi z)}{\log z}$ where $\log z$ is the principle branch of the logarithm.

I made a few calculations. $\frac{\cos(\pi z)}{\log z}$ is analytic on $\mathbb{C}$ except a simple pole at $z=1$ and the residue at $z=1$ is $-1$.

I tried to enlarge $\gamma_r(t)$ to get a closed path which has $z=1$ in its interior. However, it is hard to calculate the integral on other parts of the path.

TH Wang
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1 Answers1

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I must again refer to @SangchulLee ’s super useful lemma:

Lemma. Suppose $f(z)$ is holomorphic near $z = z_0$. Fix $\theta_0 \in (0, 2\pi)$. If $\gamma_{\epsilon}$ denotes a counter-clockwise oriented arc of angle $\theta_0$ on the circle of radius $\epsilon$ centered at $z_0$, then $$ \lim_{\epsilon\to0} \int_{\gamma_{\epsilon}} \frac{f(\zeta)}{\zeta-z_0}\;d\zeta=i\theta_0 f(z_0).$$

Thus, your integral equals $$\frac\pi4 i(-1)=-\frac{\pi i}4$$

Szeto
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