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$?


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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

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.

  • 1
    $\begingroup$ 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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.