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I wish to evaluate the integral $I= \int_{0}^{\infty} e^{ix^{2}}\, dx $ Consider the contour $I_{R}=\oint_{C_{(R)}} e^{iz^{2}}\, dz$ where $C_{(R)}$ is the closed circular sector in the upper half plane with boundary points 0, R, and $Re^{\frac{i \pi}{4}}$. Show that $I_{R} = 0$ and that $\lim_{R \to{+}\infty}{\oint_{C_{1(R)}} e^{iz^{2}}\, dz}$, where $C_{1(R)}$ is the line integral along the circular sector from R to $Re^{\frac{i\pi}{4}}.$

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I don't see any effort on your part. –  Joel Cohen Apr 27 '12 at 1:37
What have you tried so far? –  Brett Frankel Apr 27 '12 at 1:38
Your continued refusal to demonstrate any thought about the problems you are posing, despite repeated requests to do so, borders on abusive to the community. This is a valid reason for suspension. Please explain what it is that you've tried or thought about so far. –  Zev Chonoles Apr 27 '12 at 1:39
Would you like a cup of coffee while you wait for the solution to pop out? –  Domagoj Pandža Apr 27 '12 at 1:39

1 Answer 1

up vote 3 down vote accepted

Note that $$\operatorname{Im} [\int_{0}^{\infty} e^{ix^{2}}\, dx] = \operatorname{Re} [\int_{0}^{\infty} e^{ix^{2}}\, dx] = \int_{0}^{\infty} \sin(x^2)\, dx=\int_{0}^{\infty} \cos(x^2)\, dx=\frac{1}{2}\sqrt{\frac{\pi}{2}}$$

So we can clearly find the value of the integral this way. However, we may also use substitutions to achieve this:

We want this integral to resemble the Gaussian integral, whose value we know ($\int_{0}^{\infty}e^{-x^2}=\sqrt{\pi}/2$).

Letting $ix^2=-u^2 \Rightarrow u^2=-ix^2 \Rightarrow u=e^{-i\pi /4}x \Rightarrow x=e^{i\pi /4}u$. Thus $dx = e^{i \pi/4} du$.

$$I=\int_{0}^{\infty} e^{ix^{2}}\, dx=e^{i\pi/4}\int_{0}^{\infty} e^{-u^2}\, du=\frac{1}{2}e^{i\pi/4}\sqrt{\pi}=\frac{(1+i)}{2}\sqrt{\frac{\pi}{2}}$$

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That gets the right answer, but there's a bit of analysis needed to justify that change of variables, since, so to speak, the "$\infty$" that $x$ goes to is not the same as the one that $u$ goes to. –  Robert Israel Apr 27 '12 at 2:10
@RobertIsrael You are correct, I should have noted this. –  Argon Apr 27 '12 at 20:05

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