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Find the asymptotic behaviour as $f(x)=\int_{0}^{1}e^{ixz^2}dz$ as $x\rightarrow +\infty$.

Could anyone show me how to do this with either the method of stationary phase or integration by parts?

Here's what I've done for the second one:

Let $-iz^2x=u \implies z=i^{1/2}x^{-1/2}u^{-1/2}$, $dz=-{1\over 2}i^{1/2}x^{-1/2}u^{-3/2}$

Then $$f(x)=\int_{0}^{\infty} -{1 \over 2}e^{-u}i^{1/2}x^{-1/2}u^{-3/2}du =\\ =-{e^{i\pi/4}\over 2x^{1/2}} \int_{0}^{\infty}e^{-u}u^{-3/2}du$$

I don't know how to proceed from here since at the lower bound the integral is infinity.

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You got the substitution wrong; the exponents of $z$ and $u$ have the same sign. –  joriki Dec 22 '12 at 12:08
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2 Answers 2

Use stationary phase. The idea is that, as $x \rightarrow \infty$, contributions to the integral away from the stationary point (here, at $x=0$) vanish due to cancellations. Then the integral is, to first order,

$$\int_{0}^{\infty} dz \exp({i x z^2})$$

which takes the value $\frac{1}{2} \sqrt{\frac{\pi}{i x}}$.

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Could you show how you obtained the final answer, i.e. $1/2 \sqrt{\pi \over i x}$? –  adamG Dec 23 '12 at 9:55
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For now: use a contour in the complex plane for $\int_C dz \: \exp{(x z^2)}$, where $C$ goes from the origin to the point $(N,N)$, from $(N,N)$ along a circular arc of radius $\sqrt{2} N$ to the point $(\sqrt{2} N,0)$, and then from $(\sqrt{2} N,0)$ back to the origin. Because there are no singularities in the interior of $C$, the value of this integral is zero. The integral over the circular arc is zero as $N \rightarrow \infty$. That means the integral over the 45-degree line (the integral in my answer) is equal to the integral over the $x$-axis. –  Ron Gordon Dec 23 '12 at 10:26
    
2 tries answering, 2 times the page seized up. I'll get back to you later. –  Ron Gordon Dec 23 '12 at 10:26
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Letting $w = \sqrt{x}z$, your integral can be rewritten as $${1 \over \sqrt{x}}\int_0^{\sqrt{x}} e^{iw^2}\,dw$$ It can be shown using contour integration (look up "Fresnel Integral") that one has $$\int_0^{\infty} e^{iw^2}\,dw = \sqrt{\pi \over 8}\big(1 + i\big)$$ So asymptotically your integral behaves as $\sqrt{\pi \over 8x}\big(1 + i\big)$ as $x \rightarrow \infty$.

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Thanks Zarrax! I am aware of the Fresnel Integral, but I was not looking for an answer using contour integration. –  adamG Dec 23 '12 at 9:58
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