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Let $$f(z)=\frac{e^{\pi iz}}{z^2-2z+2}$$ and $\gamma_R$ is the closed contour made up by the semi-circular contour $\sigma_1$ given by, $\sigma_1(t)=Re^{it}$, and the straight line $\gamma_2$ from $-R$ to $R$ (so the contour of $\gamma_R$ appears to be a semi circle).

Prove that $$\lim_{R \rightarrow \infty} \int_{\sigma_1}f(z) dz=0.$$

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1 Answer 1

Note that $|e^{i\pi z}| = |e^{i\pi(x+iy)}| = e^{-i\pi y} \le 1$ in the upper half-space ($y>0$). Hence

$$\left| \int_{\sigma_1} f(z)\,dz \right| \le \pi R \cdot \frac{C}{R^2} \to 0 $$

as $R \to \infty$.

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