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
    $\begingroup$ try $\psi=1$, and you'll convince yourself this is hopeless. $\endgroup$ – Carlo Beenakker Jan 21 '15 at 7:33
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.)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.