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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

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

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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.2012-12-13
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    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.2012-12-13
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    $y=\mathrm{Im}(z)$ 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.2012-12-13
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    It's definitely *not* correct: $$\max_{z\in C_R} \left| \frac{e^{i(x+iy)}}{z} \right| = \frac1R.$$2012-12-13
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    Will you please read Theorem 2 in http://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.2012-12-13
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    @mrf, I think you're right as that equality is for the max and $\,y\geq 0\,$. Yet Jordan's Lemma ends the problem and I edited my answer. I still can't see what exactly was Rob's problem with "taking out "y"", though...2012-12-13
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    @DonAntonio: if you are not taking part of the integral out, what is $y$ in terms of $R$?2012-12-13
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    But @robjohn, can you see I was trying to estimate *the module* (absolute value) of the integral?! This is a very common thing when trying to do real integrals by contour methods, but in this case, as mrf noted, it doesn't work. Yet I still cannot see what "taking out of the integral" you're talking about. Did you take a peek at the link I wrote to you in my past comment?2012-12-13
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    Now that you cite Jordan's Lemma, your answer is good. In fact, Jordan's Lemma is a quantification of the convergence that I get by Dominated Convergence.2012-12-14
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    @robjohn, I still can't understand your problem with that $\,y\,$ thing. What I tried to do wasn't the real thing, but the estimation theorem allows that, just as in plain, basic Riemann integrals we have $$\left|\int_a^bf(x)\,dx\right|\leq(b-a)\max_{x\in [a,b]}|f(x)|$$2012-12-14
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    @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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    I think you meant dominated convergence theorem.2012-12-13
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    @TCL: Indeed. Thanks.2012-12-13