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

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.

• 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. – ireallydonknow Jan 13 '14 at 3:04
• Hint: consider the parametrization of the circle $z=Re^{it}$ with $t=0..2\pi$, and note that $dz=Rie^{it}\,dt$ then. – Berci Jan 13 '14 at 3:05
• Note that, $|ze^{iz}|= Re^{-R\sin t},\quad 0<t<\pi$. – Mhenni Benghorbal Jan 13 '14 at 3:08
• I can think of the estimation $\sin(\theta) \geq \frac{2}{\pi} \theta$ which is only valid for $0\leq \theta \leq \frac{\pi}{2}$ – Zaid Alyafeai Jan 13 '14 at 3:11

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.
There's even no need for parametrization. Your $f$ has an antiderivative, $g(z)=-ie^{iz}$, valid on the whole plane. Hence $$\int_{C_R} f(z)\,dz = g(-R)-g(R) = -2\sin R.$$