Why $\lim\limits_{R\to\infty} \int_{C_R}\frac{e^{iz}}{z}dz=0$? Why does $$\lim_{R\to\infty} \int_{C_R}\frac{e^{iz}}{z}dz=0$$ where $C_R = \{Re^{it}: 0\le t\le \pi\}$?
 A: $$
\begin{eqnarray}
\lim_{R\to\infty} \int_{C_R}\frac{e^{iz}}{z}dz &=& \lim_{R\to\infty} \int_0^\pi d\left(R e^{i t}\right) \frac{\exp\left(iR e^{i t}\right)}{R e^{i t}} \\
&=& i \lim_{R\to\infty}  \int_0^\pi dt \ R e^{i t} \frac{\exp\left(iR e^{i t}\right)}{R e^{i t}} \\
&=& i \lim_{R\to\infty} \int_0^\pi dt \ \exp\left[iR \left(\cos t + i \sin t\right)\right] \\
&=& i \lim_{R\to\infty} \int_0^\pi dt \ e^{i R \cos t} e^{- R \sin t} \\
&=& 0
\end{eqnarray}
$$
since $\sin t \ge 0$ for $0 \le t \le \pi$ and $\left|e^{i R \cos t}\right| = 1$.
A: Since $\left|e^{iz}\right|=e^{-y}$ and $\left|\frac{\mathrm{d}z}{z}\right|=\mathrm{d}t$, we have
$$
\begin{align}
\left|\lim_{R\to\infty}\int_{C_R}\frac{e^{iz}}{z}\,\mathrm{d}z\,\right|
&\le\lim_{R\to\infty}\int_0^\pi e^{-R\sin(t)}\,\mathrm{d}t\\[6pt]
&=0
\end{align}
$$
by dominated convergence.
In fact,
$$
\begin{align}
\int_0^\pi e^{-R\sin(t)}\,\mathrm{d}t
&=2\int_0^{\pi/2} e^{-R\sin(t)}\,\mathrm{d}t\\
&\le2\int_0^{\pi/2} e^{-2Rt/\pi}\,\mathrm{d}t\\
&=\frac\pi R\int_0^R e^{-u}\,\mathrm{d}u\\
&\le\frac\pi R
\end{align}
$$
A: Put $z:=x+iy\,\,,\,\,x,y\in\Bbb R\,$ , so if $\,Re^{it}=z=x+iy\,\,,\,R\to\infty\Longrightarrow x^2+y^2\to\infty$ and $\,0\leq t\leq \pi\Longrightarrow y\geq 0\,$:
$$\left|\oint_{C_R}\frac{e^{iz}}{z}dz\right|\xrightarrow [R\to\infty\Longrightarrow y\to\infty]{}0$$
Added: The above follows from Jordan's Lemma since
$$\left|\frac{1}{Re^{it}}\right|=\frac{1}{R}\xrightarrow [R\to\infty]{}0$$
