1
$\begingroup$

Consider the following oscillatory integral with two parameters: $$I(a,b)=\int_{\mathbb{R}}e^{i(ax^2+bx)}\psi(x)\,dx$$ where $\psi$ is smooth and supported in $\{x:|x|\in[1/2,2]\}$.

Can we write $I(a,b)$ as $f(a)g(b)$ for some functions $f$ and $g$? i.e. we want to separate $a$ and $b$. I guess we have to find an asymptotic formula for $I(a,b)$ first but I do not know how to deal with the two parameters oscillatory integral (If there were only one parameter $\lambda$, then I can possibly use the Van der corput Lemma).

$\endgroup$
1
  • 1
    $\begingroup$ try $\psi=1$, and you'll convince yourself this is hopeless. $\endgroup$ Commented Jan 21, 2015 at 7:33

1 Answer 1

1
$\begingroup$

If the critical point of the phase is in the domain of $\psi$, use $|a|$ in place of $\lambda$. Then (if $a>0$) you have a phase $\Phi(x)=x^2+bx/a$, and $|\Phi''(x)|\geq 2$, so you get decay of order $|a|^{-1/2}$.

See p. 328 in Stein and Shakarchi vol. 4 for the estimate.

Alternatively, you can complete the square and do a (linear) change of variables. The result is the same, $a$ takes the place of $\lambda$ and $b$ has no effect.

If the critical point is not in the domain of $\psi$, then it shouldn't be hard to beat the first case. Try $\lambda=a$ or $\lambda=b$; one will be larger and give you a better estimate. (Or do a change of variables as above and take $\lambda=a$, since the result will beat the decay you get when you have a stationary point.)

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .