I have to integrate the function $$f(z) = \frac{1}{1+z^2} e^{-2\pi i z \xi}$$ on the upper part of the circumference with center $0$ and radius $R$ in the complex plane. $\xi$ is a real number. I tried to use polar coordinates but it doesn't help. Can you give me a hint?
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For this integral, you might consider using the so-called ML inequality, which states that if $ |f(z)| \leq M$ and the length of the contour $\gamma$ is less than $L$, then $$ \left|\int_\gamma f(z) dz\right| \leq ML.$$
Back to the question, can you try to compute the bound of $f$ on the semi-circle on the upper plane? Try to substitute $z = Re^{i\theta}$ directly and simplify the exponent.
Ken Hung
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$|f(z)| \leq 1 $ and the length of $\gamma \leq \pi R$. – Nicola Apr 12 '20 at 18:55
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You might want to use a 'tighter' bound, $ |f(z)| \leq 1$ is not that useful. [Hint: you may just leave the fraction as it would be, cause the $z^2 = R^2e^{2i\theta}$ in the denominator is useful in eliminating the $R$] – Ken Hung Apr 12 '20 at 18:57
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The upper bound $M$ should not depend on $\theta$, right? – Nicola Apr 12 '20 at 19:06
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Yeah. You should eliminate the dependence of upper bound on any variables except $\theta$. – Ken Hung Apr 12 '20 at 19:07
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So $|f(z)| \leq \frac{1}{1+R^2}$. – Nicola Apr 12 '20 at 19:17
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And $\frac{\pi R}{1+R^2}$ tends to $0$ as $R$ tends to infinity. Thus the integral is $0$. Right? – Nicola Apr 12 '20 at 19:53
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I don't think that $e^{-2 \pi i \xi z}$ is always bounded... – Nicola Apr 12 '20 at 22:12