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Let $C_{R}$ be the upper half of the circle $|z|= R$.

Does $$ \lim_{R \to \infty} \int_{C_{R}} e^{iz} \ dz = 0 ?$$

I don't know if you can conclude from Jordan's lemma that it doesn't vanish. And the estimation lemma would appear to be inconclusive here.

But along the circle $|z|=R$, $ \displaystyle |e^{iz}| = e^{-R \sin t}$.

So as $R \to \infty$, the integrand decays exponentially.

Is that enough to conclude the integral vanishes?

And what about $ \displaystyle\lim_{R \to \infty} \int_{C_{R}} z e^{iz} \ dz$?

EDIT:

What's preventing both integrals from vanishing is the size of $|e^{iz}|$ near the endpoints of the contour.

If you were to integrate along only a portion of the contour that stays away from the endpoints, the estimation lemma would show that both integrals do vanish.

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  • $\begingroup$ en.wikipedia.org/wiki/Jordan's_lemma The function $g(z)=z$ does not have a bound that goes to $0$ as $R \to \infty$, so that condition is violated. $\endgroup$ – ireallydonknow Jan 13 '14 at 3:04
  • $\begingroup$ Hint: consider the parametrization of the circle $z=Re^{it}$ with $t=0..2\pi$, and note that $dz=Rie^{it}\,dt$ then. $\endgroup$ – Berci Jan 13 '14 at 3:05
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    $\begingroup$ Note that, $|ze^{iz}|= Re^{-R\sin t},\quad 0<t<\pi$. $\endgroup$ – Mhenni Benghorbal Jan 13 '14 at 3:08
  • $\begingroup$ I can think of the estimation $\sin(\theta) \geq \frac{2}{\pi} \theta $ which is only valid for $0\leq \theta \leq \frac{\pi}{2}$ $\endgroup$ – Zaid Alyafeai Jan 13 '14 at 3:11
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Let $z=Re^{i\theta}$. After substitution:

$\displaystyle \int_{C_R} e^{iz} dz = \int_0^{\pi} e^{iRe^{i\theta}}iRe^{i\theta} d\theta$

$= \displaystyle -i\int_0^{\pi} e^{iRe^{i\theta}}(-Re^{i\theta}) d\theta = -ie^{iRe^{i\theta}}\Big|_0^{\pi} = -i[e^{-iR} - e^{iR}] = -2\sin(R)$

So as $R\to\infty$, the limit does not exist.

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    $\begingroup$ Well, that was embarrassing. I was wasting my time trying to bound it when I could have just evaluated it directly. Thanks. $\endgroup$ – Random Variable Jan 13 '14 at 3:31
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There's even no need for parametrization. Your $f$ has an antiderivative, $g(z)=-ie^{iz}$, valid on the whole plane. Hence $$ \int_{C_R} f(z)\,dz = g(-R)-g(R) = -2\sin R. $$

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