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I'm trying to evaluate $$ \int_0^\infty \mathrm{d}x\ e^{-ix^2}. $$ I tried to integrate on the following contour $\Gamma_R$: the frontier of a circular sector, centered at the origin, of angle $\pi / 4$, used counterclockwise.

Therefore by Cauchy's theorem: $$ \oint _{\Gamma_R}\mathrm{d}z\ e^{-z^2} = 0. $$ Splitting the contour into its three natural curves we get: $$ \int_0 ^R \mathrm{d}x\ e^{-x^2}+\int_0 ^{\pi/4} iRe^{i\phi } \mathrm{d}\phi \ e^{-R^2e^{i2\phi}} + \int_R^0 \mathrm{d}re^{i\pi/4} e^{-ir^2}=0 $$

$$ e^{i\pi/4}\int_0 ^R \mathrm{d}r\ e^{-ir^2} = \Delta(R) + \int_0^R \mathrm{d}x e^{-x^2}, $$ where $\Delta(R)$ denotes the integral in $\mathrm{d}\phi$.

Now, \emph{if} $\lim_{R\to\infty}\Delta(R)=0$, we get precisely the desired result: $$ \int_0 ^\infty \mathrm{d}x\ e^{-ix^2} = e^{-i\pi/4} \int_0 ^\infty \mathrm{d}r e^{-r^2} = e^{-i\pi/4} \frac{\sqrt\pi}{2}. $$ My problem is showing the fact that $\Delta$ is negligible for very large $R$; I wanted to use some form of the Great Circle Lemma but I encounter a difficulty as $\phi\to\pi/4$:

$$ \Big| \int_0 ^{\pi/4} iRe^{i\phi } \mathrm{d}\phi \ e^{-R^2e^{i2\phi}} \Big| \le \int_0 ^{\pi/4} R \mathrm{d}\phi \Big| e^{-R^2e^{i2\phi}} \Big| = \int_0 ^{\pi/4} R \mathrm{d}\phi e^{-R^2\cos2\phi} $$ but now? Should I take a limiting procedure of some sort?

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    $\begingroup$ Use the fact that if $u\in[0,\frac\pi2]$, then $\cos u\geq 1-\frac2\pi\, u$. $\endgroup$ – Etienne Feb 27 '14 at 20:18
  • $\begingroup$ @Etienne: it looks like even then the exponent vanishes in $\pi/4$ and I get a dependence on $R$!! $\endgroup$ – Brightsun Feb 27 '14 at 21:11
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    $\begingroup$ I think it's OK: just compute $\int_0^{\frac\pi4} e^{-R^2(1-\frac4\pi \phi)}d\phi$. $\endgroup$ – Etienne Feb 27 '14 at 21:22
  • $\begingroup$ Why then? Sorry I can't see your point... $\endgroup$ – Brightsun Feb 27 '14 at 21:24
  • $\begingroup$ I GOT IT perhaps $\endgroup$ – Brightsun Feb 27 '14 at 21:31
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Thanks to Etienne, I got an idea that just might work: let's set $2\phi = \varphi$, so our $\Delta$ becomes overestimated by: $$ \int_0^{\pi/2}\mathrm{d}\varphi \frac{R}{2}e^{-R^2\sin\varphi} \le \int_0^{\pi/2}\mathrm{d}\varphi \frac{R}{2}e^{-2R^2\varphi/\pi} $$ where we used the inequality $\sin\varphi\ge 2\varphi/\pi$ for $\varphi\in[0,\pi/2]$. Thus we get by directly calculating the last one: $$ \ldots \le \frac{\pi}{4R}\left(-e^{-R^2}+1\right) \to0 $$

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    $\begingroup$ I think it should be $1-e^{-R^2}$. $\endgroup$ – Etienne Feb 27 '14 at 22:55
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$\exp{(-ix^2)}=\cos{(x^2)}-i\,\sin{(x^2)}$

And then more than enough information is given here: Intégrale de Fresnel (Wikipedia, french). Scroll down for a version of your proof.

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