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?

  • $\begingroup$ 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, 2017 at 22:01
  • $\begingroup$ Thank you, I've searched but didn't find. $\endgroup$ Jun 14, 2017 at 22:18
  • 1
    $\begingroup$ 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, 2017 at 22:22


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