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Let $C_R$ be the semi-circle of radius $R$ in the upper half plane, centered at the origin (oriented counter-clockwise). I would like to prove that $$ \lim_{R\to\infty} \int_{C_R} \frac{e^{iz}}{z} dz = 0 $$ I am aware that the above limit is a corollary of Jordan's Lemma. Is there a more direct, elegant way of showing that the limit is $0$?

I have one idea using integration by parts. We have $$ \int_{C_R} \frac{e^{iz}}{z}dz = \frac{e^{iz}}{iz}\left.\right|_{-R}^{R}+\int_{C_R}\frac{e^{iz}}{iz^{2}} dz $$ Using basic estimates we get that $$ \left|\int_{C_R} \frac{e^{iz}}{z}dz \right| \le \frac{1}{R}+\frac{1}{R}+\frac{K\pi}{R}\to 0 $$ as $R\to\infty$. Here $K$ is some constant. I used: $$ \left|\int_{C_R}\frac{e^{iz}}{iz^{2}} dz\right| \le \underbrace{(\pi R)}_{\textrm{length}}\cdot\left|\sup_{y\in(-R, R)}e^{-y}\right|\frac{1}{R^2}\le \frac{K\pi}{R} $$ So my questions are:

1) Is my solution actually correct?

2) Is there a possibly better approach for this problem?

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  • $\begingroup$ $K$ isn't a constant it depends on $R$! $\endgroup$ – azarel Sep 18 '13 at 14:48
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    $\begingroup$ You don't have $\sup\limits_{y \in (-R,R)} e^{-y}$, you have $\sup\limits_{z \in C_R} \left\lvert e^{iz}\right\rvert$, and that is $1$, since $\operatorname{Im} z \geqslant 0$ on $C_R$. $\endgroup$ – Daniel Fischer Sep 18 '13 at 14:55
  • $\begingroup$ @DanielFischer: Oh my! I am an idiot... So my solution actually works with your modification, right? $\endgroup$ – Prism Sep 18 '13 at 15:01
  • $\begingroup$ I agree with the part after the "...". Not sure about the part before. $\endgroup$ – Daniel Fischer Sep 18 '13 at 15:02
  • $\begingroup$ @DanielFischer: Haha! Thanks Daniel. Okay, since I like $x, y$ notation to keep track of what's happening, I guess this could also be written as $\sup\limits_{z \in C_R} \left\lvert e^{iz}\right\rvert=\sup\limits_{x^2+y^2=R^2, y\ge 0}\left\lvert e^{-y+ix}\right\rvert=\sup\limits_{y\in(0, R)}\left\lvert e^{-y}\right\rvert = 1$ $\endgroup$ – Prism Sep 18 '13 at 15:09
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$$z=Re^{it}=R\cos t+iR\sin t\;,\;\;0< t< \pi\implies \left|\;e^{iz}\;\right|=e^{-R\sin t}$$

Note that since $\,\sin t> 0\;$ , we get

$$\lim_{R\to\infty}e^{-R\sin t}=0\;,\;\;\forall\,t\in (0,\pi)$$

so applying the estimation lemma directly we get

$$\left|\;\int\limits_{C_R}\frac{e^{iz}}zdz\;\right|\leq\max_{0\leq t\leq \pi}\frac{e^{-R\sin t}}RR\pi\xrightarrow[R\to\infty]{}0$$

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  • $\begingroup$ For $t \in\{0,\pi\}$, you have the exponential factor with modulus $1$, so the estimation by $\sup \lvert f(z)\rvert \cdot \operatorname{length}\gamma$ doesn't converge to $0$. $\endgroup$ – Daniel Fischer Sep 18 '13 at 15:01
  • $\begingroup$ I don't need $\;t=0\,,\,\pi\;$ as these two points on the semicircle $\,C_R\;$ are covered by the real interval $\;[-R,R]\;$ ... $\endgroup$ – DonAntonio Sep 18 '13 at 15:03
  • $\begingroup$ Still, $\sup \left\lvert \frac{e^{iz}}{z}\right\rvert = \frac{1}{R}$. You can show that the integral converges to $0$ in many ways, but it's not as easy as that. $\endgroup$ – Daniel Fischer Sep 18 '13 at 15:05
  • $\begingroup$ I think now I see your point: my thought was going into $\;f(Re^{it})\;$ and etc. but this is in fact almost the same as proving, and using, J.L. ... $\endgroup$ – DonAntonio Sep 18 '13 at 15:10

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