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I need to calculate this $\int_{-\infty}^\infty \frac{sin(z)}{z}dz$ using the residue of $\frac{sin(z)}{z}$.

Then I write $\int_{-\infty}^\infty \frac{sin(z)}{z}dz$ = $\lim_{R\to\infty}Im(\int_{-R}^R \frac{e^{iz}}{z}dz+\int_{CR}\frac{e^{iz}}{z}dz)=2\pi i Res(\frac{e^{iz}}{z},0)$ where CR is the the half circunference of radius R>0 over the plane.

Then I need to show that $\lim_{R\to\infty}|\int_{CR}\frac{e^{iz}}{z}dz|=0$

Can someone help me?

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marked as duplicate by Community Jun 14 '17 at 22:23

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Note $e^{iz}/z$ has a pole at $z=0$ so you will need another semicircle at the origin to avoid it. Also this is one of the most commonly asked questions on this site, so you can find your answer with a search: math.stackexchange.com/questions/1739621/… $\endgroup$ – Grant B. Jun 14 '17 at 22:01
  • $\begingroup$ Thank you, I've searched but didn't find. $\endgroup$ – Ettore Moura Jun 14 '17 at 22:18
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    $\begingroup$ See Jordan's lemma. Alternatively, you may find it easier to use a rectangle than a semicircle, since the integrals over the line segments involved are easier to estimate. $\endgroup$ – Chappers Jun 14 '17 at 22:22