Circumventing Jordan's Lemma

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?

• $K$ isn't a constant it depends on $R$! – azarel Sep 18 '13 at 14:48
• 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 $C_R$. – Daniel Fischer Sep 18 '13 at 14:55
• @DanielFischer: Oh my! I am an idiot... So my solution actually works with your modification, right? – Prism Sep 18 '13 at 15:01
• I agree with the part after the "...". Not sure about the part before. – Daniel Fischer Sep 18 '13 at 15:02
• @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 $\sup\limits_{z \in C_R} \left\lvert e^{iz}\right\rvert=\sup\limits_{x^2+y^2=R^2, y\ge 0}\left\lvert e^{-y+ix}\right\rvert=\sup\limits_{y\in(0, R)}\left\lvert e^{-y}\right\rvert = 1$ – Prism Sep 18 '13 at 15:09

$$z=Re^{it}=R\cos t+iR\sin t\;,\;\;0< t< \pi\implies \left|\;e^{iz}\;\right|=e^{-R\sin t}$$

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

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

• 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 $0$. – Daniel Fischer Sep 18 '13 at 15:01
• I don't need $\;t=0\,,\,\pi\;$ as these two points on the semicircle $\,C_R\;$ are covered by the real interval $\;[-R,R]\;$ ... – DonAntonio Sep 18 '13 at 15:03
• 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. – Daniel Fischer Sep 18 '13 at 15:05
• 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. ... – DonAntonio Sep 18 '13 at 15:10