Why $\lim\limits_{R\to\infty} \int_{C_R}\frac{e^{iz}}{z}dz=0$?

Question

Why does $$\lim_{R\to\infty} \int_{C_R}\frac{e^{iz}}{z}dz=0$$ where $C_R = \{Re^{it}: 0\le t\le \pi\}$?

Answer 1

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

$$\left|\frac{1}{Re^{it}}\right|=\frac{1}{R}\xrightarrow [R\to\infty]{}0$$

Answer 2

$$ \begin{eqnarray} \lim_{R\to\infty} \int_{C_R}\frac{e^{iz}}{z}dz &=& \lim_{R\to\infty} \int_0^\pi d\left(R e^{i t}\right) \frac{\exp\left(iR e^{i t}\right)}{R e^{i t}} \\ &=& i \lim_{R\to\infty} \int_0^\pi dt \ R e^{i t} \frac{\exp\left(iR e^{i t}\right)}{R e^{i t}} \\ &=& i \lim_{R\to\infty} \int_0^\pi dt \ \exp\left[iR \left(\cos t + i \sin t\right)\right] \\ &=& i \lim_{R\to\infty} \int_0^\pi dt \ e^{i R \cos t} e^{- R \sin t} \\ &=& 0 \end{eqnarray} $$ since $\sin t \ge 0$ for $0 \le t \le \pi$ and $\left|e^{i R \cos t}\right| = 1$.

Answer 3

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, $$ \begin{align} \int_0^\pi e^{-R\sin(t)}\,\mathrm{d}t &=2\int_0^{\pi/2} e^{-R\sin(t)}\,\mathrm{d}t\\ &\le2\int_0^{\pi/2} e^{-2Rt/\pi}\,\mathrm{d}t\\ &=\frac\pi R\int_0^R e^{-u}\,\mathrm{d}u\\ &\le\frac\pi R \end{align} $$