1

$$f(z)=\frac{e^{-iz}}{z}$$

What is the value of:

$$\frac{1}{\pi i }\int_{C} f(z) dz$$

if ${C}$ is the arc of the semicircle with radius $R \to \infty$ ,going counterclockwise from point $(R,0)$ to $(-R,0)$.

At first I used the residue theorem and got the correct answer according to the solutions, which is $2$. However i quickly noticed that the curve is a semicircle (angle $0 \to \pi$ instead of $0 \to 2\pi$) so I got confused. I've tried some other things in hope of discovering the way the exercise is supposed to be solved but I honestly just get the feeling it might be something very simple I'm overlooking. As far as I know, Residue Theorem can only be applied when the curve is simple and closed, also the fact that the singularity is at $z = 0$ doesn't help very much.

emacs drives me nuts
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FRIENDS
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  • Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. – Community Jun 29 '22 at 15:36
  • Try to use ML inequality. – Riemann Jun 29 '22 at 15:53
  • The curve is simple and closed, and we can handle the residue at $0$ with the semi-residue lemma. But since the exponent is $-iz$ rather than $iz$ (unless that $-$ sign was a misprint), the arc's behaviour is divergent. – J.G. Jun 30 '22 at 13:08
  • @J.G. Thank you for answering. The exponent being -iz is not a misprint, big part of why I'm having trouble is because of that. When you talk about the semi-residue lemma does that refer to the formula piiRes(f, 0) ? I had thought about this but I think the answer would be 1 if I'm not wrong, which is not the correct answer according to the solutions. – FRIENDS Jun 30 '22 at 13:23
  • I'm referring to this. – J.G. Jun 30 '22 at 13:24
  • @J.G. That is what I meant. – FRIENDS Jun 30 '22 at 13:25
  • Consider $\int_{|z| = R} e^{-i z}/z , dz$ and $\lim_{R \to \infty} \int_{|z| = R \land -\pi < \arg z < 0} e^{-i z}/z , dz$ and apply Jordan's lemma. – Maxim Jun 30 '22 at 15:57
  • @Maxim I appreciate the suggestion maxim, however since $e^{-i az}/z = e^{-i z}/z$ where a = -1 < 0 I don't think I can use the Jordan's lemma. According to Wikipedia, $a$ must be a positive parameter. – FRIENDS Jun 30 '22 at 16:11

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