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.