# Asymptotics of fast decaying oscillating integrals

Let $g(t):\mathbb{R}\rightarrow \mathbb{C}$ be a smooth even function such that for each positive integer $n$ the function $g^{(n)}(t)$ tends to 0 when $t$ goes to infinity and $\int_0^\infty |g^{(n)}(t)|dt<\infty$. How to study the asymptotics of $I(A)=\int_0^\infty \cos(At) g(t)dt$ for large positive $A$? Integrating by parts several times we see that it is $o(A^{-n})$ for all $n$ (here it is important that $g$ is even), but what next?

Examples: if $g(t)=1/(t^2+c^2)$, $c>0$ we have $I(A)=\frac{\pi}{2c}e^{-cA}$, for $g(t)=e^{-t^2}$ we have $I(A)=\frac{\sqrt{\pi}}2 e^{-A^2/4}$, for $g(t)=(t^2+1)^{-1/2}$ we have $I(A)=K_0(A)\sim \sqrt{\frac{\pi}{2A}}e^{-A}$.

I may assume that $g$ has analytic continuation to some neighborhood of a real line and singularities of fractional power types at some complex points, like in $(t^2+1)^{-1/2}$ example.

• How representative are the examples you gave? Taking the analytic continuation you can evaluate the Fourier transforms along the line $x - i \epsilon$ and get exponential decay. But this of course require your analytic continuation to be to a strip of a fixed width, which I don't know if is what you meant. – Willie Wong Oct 31 '16 at 15:47
• And conversely, if the FT has exponential decay, then $g$ is holomorphic on a strip of that width. – Christian Remling Oct 31 '16 at 16:47