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In regards to the above proof, I'm a little confused as to how the last conclusion was made --

How does the fact that

$$\int_{-\infty}^{\infty}e^{-x^2}dx = \sqrt{\pi}$$

to conclude that:

$$\int_0^{\infty}\cos{x^2} + i\sin{x^2}dx = \int_0^{\infty}e^{ix^2}dx = \frac{\sqrt{2\pi}}{4} + i\frac{\sqrt{2\pi}}{4}?$$

r123454321
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  • They're using the $\sqrt{\pi}$ equality to finish the computation of the integral on the line above. – Josh Keneda Sep 15 '14 at 01:22
  • $\int_{\infty}^{\infty}$ should be $\int_{\color{red}{-}\infty}^{\infty}$ – mike Sep 15 '14 at 01:23
  • So $ie^{i\pi / 4}\int_R^0 e^{-u^2}(-1/R)du = (-1/R)ie^{i\pi / 4} \int_R^0 \sqrt{\pi} du = \pi i e^{i\pi / 4}$, if I've done that correctly. How are the $\frac{\sqrt{2\pi}}{4}$ terms derived? – r123454321 Sep 15 '14 at 01:47
  • Be careful: they're not saying that $e^{-u^2} = \sqrt{\pi}$. They're saying that the integral of $e^{-u^2}$ over the real line is $\sqrt{\pi}$. I've posted a full answer below. – Josh Keneda Sep 15 '14 at 03:20
  • possible duplicate of Some way to integrate $\sin(x^2)$? – leo Jul 23 '15 at 16:11
  • Can anyone explain why we have the inequalities $|\int_0^{\pi/4}e^{i(Re^{i\theta})^2} i * Re^{i\theta}d\Theta| \leq \int_0^{\pi/4}Re^{-R^2 sin(\theta)}d\Theta$? How can we derive it?Thanks for your help! – user110320 Jul 31 '18 at 20:57
  • i think it should say $$|\int_0^{\pi/4}e^{i (Re^{i\theta})^2}iRe^{i\theta} d\theta| \leq \int_0^{\pi/4} Re^{-R^2\sin(2\theta)}d\theta$$ But in any case this follows from Euler's Formula $e^{2i\theta} = \cos2\theta + i \sin 2\theta$ – Math_Day Jan 25 '24 at 05:18

1 Answers1

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There's a typo in their parametrization of $\gamma_3$. They have $dz = (-iRe^{i \frac{\pi}{4}}) dt$ instead of $dz = (-Re^{i \frac{\pi}{4}}) dt$. With this correction, their line after showing that the integral over $\gamma_2$ drops out should read: $$\lim_{R\rightarrow \infty} \int_0^R e^{iz^2} dz = \lim_{R\rightarrow \infty} R e^{i \pi/4} \int_0^1 e^{i(Re^{i \pi/4}(1-t))^2} dt. \quad\quad\quad (*)$$

Now, if we do their $u$-substitution, the right hand side becomes \begin{align*}\lim_{R \rightarrow \infty} R e^{i\pi/4} \int_R^0 e^{-u^2} \frac{-1}{R} du &= \lim_{R\rightarrow \infty} R e^{i\pi/4} \int_0^R e^{-u^2} \frac{1}{R} du\\ &= \lim_{R \rightarrow \infty} e^{i \pi/4} \int_0^R e^{-u^2} du\\ &= e^{i\pi/4} \int_0^\infty e^{-u^2} du\\&=e^{i\pi/4}\frac{\sqrt{\pi}}{2},\end{align*}

where the last equality follows from the fact that $e^{-u^2}$ is even and so $$\int_0^\infty e^{-u^2} du = \frac{1}{2}\int_{-\infty}^\infty e^{-u^2} du = \frac{1}{2} \sqrt{\pi}.$$

Combining this result with $(*)$, we have $$\int_0^\infty e^{iz^2} dz = e^{i\pi/4}\frac{\sqrt{\pi}}{2} = \frac{\sqrt{2\pi}}{4}+ i \frac{\sqrt{2\pi}}{4}.$$

Josh Keneda
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  • Not sure if you'll see this, but a brief followup -- how does this proof show that $\int_0^{\infty}\sin{x^2}dx = \frac{\sqrt{2\pi}}{4}$? Doesn't it only show that $\int_0^{\infty}\cos{x^2}dx = \frac{\sqrt{2\pi}}{4}$? (i.e. $\cos{x^2}$ is exactly the real part of $e^{iz^2}$?) – r123454321 Sep 16 '14 at 07:12
  • Similarly, $\sin{x^2}$ is exactly the imaginary part of $e^{iz^2}$ on the real line, so taking imaginary parts gives us the $\sin{x^2}$ integral. – Josh Keneda Sep 16 '14 at 07:21