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Consider the following integral for $f \in C_c^{\infty}(\mathbb R^n)$, $x_0$ fixed (possibly zero), and $n \ge 3$

$$F(\lambda) = \int_{\mathbb R^n} e^{i\lambda \vert x-x_0 \vert^2} \frac{f(x)}{\vert x \vert^{n-2}} \ dx.$$ Does there exist a stationary phase type-formula describing the asymptotic of that integral?

This integral is finite as can be seen by switching to polar coordinates, observe that $\vert x \vert^{n-2}$ is also the Green's function of the Laplacian.

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  • $\begingroup$ For $x_0=0$ it serms that it goes to 0 faster than any power...is this what you guess? $\endgroup$
    – Giorgio Metafune
    Commented Aug 10, 2023 at 7:27
  • $\begingroup$ @GiorgioMetafune I actually don't have any intuition about the behaviour but would be curious to know how you see the behaviour for $x_0=0$? $\endgroup$
    – António Borges Santos
    Commented Aug 10, 2023 at 8:37
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    $\begingroup$ See the answer below (some details are missing). I was wrong before, since the final 1d integral is on the half line and not on the whole line, so the contributions at 0 give a power like decay. $\endgroup$
    – Giorgio Metafune
    Commented Aug 10, 2023 at 9:03

1 Answer 1

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Only the case $x_0=0$. Write $x=r\omega$ in spherical coordinates so that $$ F(\lambda)=\int_0^\infty e^{i\lambda r^2}r\, dr \int_S f(r\omega)\, d\omega $$ with $S$ being the unit sphere. Since $f \in C^\infty$, the function $u(r)=\int_S f(r\omega)\, d\omega$ belongs to $C^\infty ([0, \infty))$ and has vanishing odd derivatives at 0. Then $v(r)=u(\sqrt r)$ is smooth on $[0, \infty)$ and $$ F(\lambda)=\int_0^\infty e^{i \lambda r^2}ru(r)\, dr=\frac 12 \int_0^\infty e^{i\lambda s}v(s)\, ds. $$ Now the asymptotic behaviour depends on the value of $v$ and its derivatives at 0.

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