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I am trying to show that
$$ \lim_{R\to\infty}\int_{C_R} \frac{e^{iz}}{z} dz=0,\quad C_R(t)=Re^{it},\ \ t\in [0,\pi]. $$

I know we can do the following

$$\left| \int \frac{e^{iz}}{z} dz \right| \leq \frac{1}{R} \int \lvert e^{iz} \rvert \lvert dz \rvert$$

the integral is bounded and as $R$ goes to infinity it is zero. My question is what if I do the following

$$\left| \int_{0}^{\pi} \frac{e^{iR e^{it}}}{Re^{it}}\dot iRe^{it} dt \right|\leq \int_0^{\pi} |e^{-R\sin t} | dt$$

which is not zero as $R$ goes to infinity!

IrbidMath
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2 Answers2

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In fact, that integral does approach $0$ as $R\to\infty$. It is immediate from the dominated convergence theorem. However, an elementary estimate can be given as follows. For $t\in [0,\pi/2]$ note that $\sin t\ge \frac2{\pi}t$, and so $e^{-R\sin t} \le e^{-2Rt/\pi}$. Now you can do the integral estimate. (Oh, and note that $\int_0^\pi = 2\int_0^{\pi/2}$.)

By the way, your first approach is flawed. The integral is not bounded: It's the length of the semicircle, which is, of course, $\pi R$.

Ted Shifrin
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You can use Jordan's lemma:

Consider a complex-valued, continuous function $f$, defined on a semicircular contour $$ C_R = \{R e^{i \theta} \mid \theta \in [0, \pi]\} $$ of positive radius $R$ lying in the upper half-plane, centered at the origin. If the function $f$ is of the form $$ f(z) = e^{i a z} g(z) , \quad z \in C_R $$ with a positive parameter $a$, then Jordan's lemma states the following upper bound for the contour integral: $$ {\displaystyle \left|\int _{C_{R}}f(z)\,dz\right|\leq {\frac {\pi }{a}}M_{R}\quad {\text{where}}\quad M_{R}:=\max _{\theta \in [0,\pi ]}\left|g\left(Re^{i\theta }\right)\right|.} $$

Here you let $g(z)=1/z$ and $a=1$ and you have $M_R=\frac{1}{R}$.

The proof essentially uses the fact that ${\sin\theta}\le\frac{2\theta}{\pi}$ for all $\theta\in[0,\pi/2]$ as Ted mentions in his answer.