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Say we are given an oscillatory integral of the form

$\Psi(x)=\int_{-\infty}^\infty e^{i\psi(x,t)} a(t)dt$.

where $a(t)$ is a sufficiently nice function. When, for instance, $|\psi(x,t)_t| \gg x^\alpha$ uniformly in $t$ for some $\alpha>0$, one can prove that $\Psi(x)$ is rapidly decreasing in $x\to \infty$.

My question here is to determine if we can find a constant $c>0$ and $C>0$ such that $\Psi(x)\ll e^{-cx^C}$.

For example, if we put

$f(x)=\int_{-\infty}^\infty \frac{\cos xt dt}{\sqrt{1+t^2}}$

then it is known through the asymptotic expansion of the modified Bessel function of the second kind that

$f(x) \ll x^{-1/2}e^{-x}.$

Other example will be the following integral:

$K_{ix}(1)=\int_{-\infty}^\infty e^{-\cosh t}\cos xt dt$

which is $\ll x^{-1/2}e^{-\frac{\pi x}{2}}$.

However, if I were to use the integration by parts, as far as I know, I can only prove that these functions are rapidly decaying in $x$.

To summarize: I would like to learn if there is any general method which can go beyond polynomial-decay of the given oscillatory integral.

Thank you.

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Contour integration often does the trick. – Noam D. Elkies Jan 25 '12 at 22:51
The method of steepest descent may be what you are looking for... – Tom Dickens Jan 26 '12 at 3:47
Noam and Tom: Thank you for the comments. I used those two examples since in these representations, I thought it is not clear how to apply contour integration or the method of steepest descent, unless I change the integrals. If I were to use Basset's integral for the $K$-bessel function, the method of the steepest descent can be applied and give the exact asymptotic for the second example. – Junehyuk Jung Jan 26 '12 at 21:45
Point was that one can not always find such nice integral representation. So I was wondering if there's any trick that can be applied without modifying the integration too much. – Junehyuk Jung Jan 26 '12 at 21:48

Such a statement won't hold in generality. Imagine you choose $\psi(x,t) = x^{\alpha + \epsilon}t$. Then $\Psi(x) = \hat{a}(x^{\alpha + \epsilon})$. Since the Fourier transform is a bijection from the space of Schwartz functions to itself, you can select a Schwartz function $a(t)$ whose Fourier transform does not decay faster than any ${\displaystyle e^{-cx^{c}}}$, and then $\hat{a}(x^{\alpha + \epsilon})$ will also not have this quasi-exponential decay.

share|cite|improve this answer
I think I would have to clarify my question. What I want to know is if one can determine such $C$ and $c$ exist given the integral, not to find those constants in every cases. Anyway thank you for the answer! – Junehyuk Jung Jan 26 '12 at 21:34

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