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Let $\gamma_R : [0, \pi/ 4] \ni \theta \mapsto R e^{i \theta}$. I want to show that $$ \lim_{R \rightarrow \infty} \int_{\gamma_R} e^{i z^2} dz = 0 $$ for $R > 1$. In order to use Jordan's lemma I want to show that $$ \lim_{R \rightarrow \infty} \max_{z \in \gamma_R} | e^{iz^2-iz} | = 0 $$ How can I do that ?

Further thoughts: On $\gamma_R$ we have \begin{align} e^{iz(z-1)} = \exp(iRe^{i\theta}(Re^{i\theta}-1)) = \exp(iR^2e^{2i\theta}-iRe^{i\theta}) \\ = \exp(iR^2(\cos 2\theta +i \sin 2\theta)-iRe^{i\theta}) \\ \leq \exp(-R^2 \sin 2\theta) \end{align} where we take the absolute value in the second line.

Can I just ignore the case $\theta = 0 ?$

  • Corrected it. My intentions was the limit as $R \rightarrow \infty$. –  May 09 '13 at 15:22
  • Since $|e^u|=e^{\mathrm{Re}(u)}$ you can study the real part of $iz^2-iz$ when $z=Re^{i\theta}$ with $\theta\in[0,\pi/4]$ and find its maximum. It's not too hard. Maybe there's a shorter way? – Philippe Malot May 09 '13 at 15:28
  • By the way, are you sure you're really using Jordan's lemma ? I think it gives the upper bound $\mathrm{length}(\gamma_R)\max_{z\in\gamma_R} |e^{iz^2}|$ – Philippe Malot May 09 '13 at 15:34
  • I am not sure if I can ignore $\theta = 0$. Otherwise the claim would follow since for $\theta \in (0,2\pi]$ we have that $\lim_{R \rightarrow \infty} \exp(-R^2 \sin 2 \theta) = 0$ for $R > 1$. Probably I can do this because leaving out a countable number of points from an integral doesn't change it –  May 09 '13 at 15:35
  • I use this http://en.wikipedia.org/wiki/Jordan's_lemma –  May 09 '13 at 15:36
  • I see, I was talking about the so-called "estimation lemma". – Philippe Malot May 09 '13 at 15:38
  • Hint : use $\sin(x)\geq \frac{2}{\pi}x$ for all $x\in[0,\frac \pi 2]$. I wouldn't use the Jordan Lemma, it's useless. – Philippe Malot May 09 '13 at 15:41

1 Answers1

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Use the relation $\sin{2 \theta} \ge \frac{4 \theta}{\pi}$ when $\theta \in [0,\pi/4]$. Then

$$\int_{C_R} dz\, e^{i z^2} = i R \int_0^{\pi/4} d\theta \, e^{i \theta} e^{-R^2 \sin{2 \theta}} e^{i R^2 \cos{2 \theta}}$$

and

$$\left | \int_{C_R} dz\, e^{i z^2}\right | \le R \int_0^{\pi/4} d\theta \, e^{-R^2 \sin{2 \theta}} \le R \int_0^{\pi/4} d\theta \, e^{-4 R^2 \theta/\pi} \le \frac{\pi}{4 R}$$

as $R \to \infty$.

Ron Gordon
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  • Voilà, it works really well. +1 – Philippe Malot May 09 '13 at 15:43
  • Yeah. But the strange thing is, this is an exam question and before I was asked to prove the Jordan lemma in order to use it for this question. Indeed I have used $\sin \theta \geq 2 \theta /\pi$ on $[0,\pi /2]$ for the proof of the lemma. –  May 09 '13 at 15:44