2
$\begingroup$

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.

  • 1
    You got the substitution wrong; the exponents of $z$ and $u$ have the same sign.2012-12-22

2 Answers 2

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 $\frac{1}{2} \sqrt{\frac{\pi}{i x}}$.

  • 0
    2 tries answering, 2 times the page seized up. I'll get back to you later.2012-12-23
2

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 $x \rightarrow \infty$.

  • 0
    Thanks Zarrax! I am aware of the Fresnel Integral, but I was not looking for an answer using contour integration.2012-12-23