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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.2012-12-22

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