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

3 Answers 3

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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

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

  • 0
    @DonAntonio: Your statement that $ \left|\int_{C_R}\frac{e^{iz}}{z}\,\mathrm{d}z\right|\le\frac{\pi}{e^{y}} $ makes no sense. The only variable whose scope extends outside the integral on the left is $R$. Inside the integral, $y=\mathrm{Im}(z)\in[0,R]$, but that varies over the contour.2012-12-14
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$ \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 $\left|e^{i R \cos t}\right| = 1$.

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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} $

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    @TCL: Indeed. Thanks.2012-12-13