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

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

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## 3 Answers

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

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

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$y$ does go to $\infty$ on some of the circle, but not on all of the circle. It is not really valid to pull $y$ out of the integral like that. – robjohn♦ Dec 13 '12 at 5:22
Of course it is: By Cauchy's Estimate Formula, $$\left|\int_{C_R}\frac{e^{iz}}{z}dz\right|\leq\max_{z\in C_R}\left|\frac{e^{i(x+iy)}}{z}\right|\cdot l(C_R)\leq \frac{e^{-y}}{R}\cdot R\pi$$ Perhaps there's a mistake somewhere else, but the fact of "pulling out y", if this is what you meant, is correct. – DonAntonio Dec 13 '12 at 11:55
$y=\mathrm{Im}(z)$ and $z$ is the variable of integration, so the scope of $y$ is inside the integral. If $z=Re^{it}$, then $y=R\sin(t)$ inside the integral, but outside the integral only $R$ is known. – robjohn♦ Dec 13 '12 at 12:21
It's definitely not correct: $$\max_{z\in C_R} \left| \frac{e^{i(x+iy)}}{z} \right| = \frac1R.$$ – mrf Dec 13 '12 at 13:09
Will you please read Theorem 2 in tinyurl.com/9wtzao3 ? I'm sure you know this but for some reason you seem to be forgetting it. With this, and since $\,|e^{i(x+iy)}|=|e^{-y}e^{ix}|=e^{-y}\,$, we get the formula above. It is not that I'm taking the integration variable (or part of it) "out" of the integral, of course. I'm just estimating, or bounding above, the module of the integral using that theorem of Cauchy, since this is all is needed to show the integral's limit is zero when $\,R\to\infty\,$ . I hope this clears up things unless something else is bothering you. – DonAntonio Dec 13 '12 at 13:15
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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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I think you meant dominated convergence theorem. – TCL Dec 13 '12 at 3:19
@TCL: Indeed. Thanks. – robjohn♦ Dec 13 '12 at 3:28
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