Given $f\in H^1(\mathbb{R}^3)$ and $t>0$, consider the following integral:
$$I\_f(t):=\int\_{\mathbb{R}^3}\int\_0^{+\infty}e^{-s+i\frac{(s+|x|)^2}{t}}f(x)dsdx$$
I need to show that $I\_f(t)$ is finite, and to find how fast it grows as $t\rightarrow +\infty$.


If $f$ decays fast enough at infinity so that $f\in L^1(\mathbb{R}^3)$, we easily get:


$$|I(f)|\lesssim\int\_{\mathbb{R}^3}|f(x)|dx<+\infty$$
so in this case $I\_f(t)$ is finite, uniformily in $t$.


If $f$ doesn't dacays fast enough, we can't simply pass to absolute value. We must take into account the oscillatory term $e^{i(s+|x|)^2/t}$, which should produce some polynomial growth in $t$. How to proceed in this case?


Thank you for any suggestions.



Convergence of an oscillatory integral
Here is a more detailed version of my sketch in the comments above.


First of all, let $\rho$ be a compactly supported smooth function (cutoff function) such that $\rho=1$ on $B\_1$. Write $f=\rho f+(1-\rho)f$. Since $\rho f$ is compactly supported, it is $L^1$, and your old argument applies to it, so it remains to bound that for $(1-\rho)f$, whose support is disjoint from $B\_1$, which we simply denote as $f$ henceforth.


Then we do the $s$ integral first. Integration by parts once gives


$$ \int\_0^\infty e^{-s}e^{i(s+|x|)^2/t}ds=\frac{te^{i|x|^2/t}}{2i|x|}-\int\_0^\infty t\frac{d}{ds}(\frac{e^{-s}}{2i(s+|x|)})e^{i(s+|x|)^2/t}ds. $$


Integrate by parts a second time gives


$$ \int\_0^\infty e^{-s}e^{i(s+|x|)^2/t}ds
=\frac{te^{i|x|^2/t}}{2i|x|}+\frac{t}{2i|x|}(\frac{t}{2i|x|}+\frac{t}{2i|x|^2})e^{i|x|^2/t}
-\int\_0^\infty t^2\frac{d}{ds}(\frac{1}{2i(s+|x|)}\frac{d}{ds}(\frac{e^{-s}}{2i(s+|x|)}))e^{i(s+|x|)^2/t}ds. $$


Write $I(t)=I\_1(t)+I\_2(t)+I\_3(t)$, corresponding to the three terms above.


We need oscillation in $x$ to bound $I\_1$; we don't need it for $I\_2$ and $I\_3$.


$$I\_1(t)=\frac{t}{2i}\int \frac{e^{i|x|^2/t}}{|x|}f(x)dx=-\frac{t}{2i}\int e^{i|x|^2/t}\vec x\cdot\nabla(\frac{f(x)}{|x|(3+2i|x|^2/t)})$$


is bounded by a constant (depending on $t$) times


$$\int \frac{|\nabla f(x)|}{|x|^2}+\frac{|f(x)|}{|x|^3}\le(\int |\nabla f|^2)^{1/2}(\int\_{|x|\ge1} |x|^{-4})^{1/2}+(\int |f|^2)^{1/2}(\int\_{|x|\ge1} |x|^{-6})^{1/2}\le C.$$


For $I\_2$ we have


$$ |I\_2(t)|\le t^2\int |x|^{-2}f(x)\le t^2(\int |f|^2)^{1/2}(\int\_{|x|\ge1} |x|^{-4})^{1/2}.$$


For $I\_3$ we note that both $e^{-s}$ and $1/(s+|x|)$ are decreasing and convex in $s$, so is their product. Hence $\frac{d}{ds}\frac{e^{-s}}{s+|x|}\le0$ and is increasing. Since $1/(s+|x|)>0$ is decreasing, their product is negative increasing. Hence


$$ \int\_0^\infty |\frac{d}{ds}(\frac{1}{2i(s+|x|)}\frac{d}{ds}(\frac{e^{-s}}{2i(s+|x|)}))|ds=\frac{t}{2|x|}(\frac{t}{2|x|}+\frac{t}{2|x|^2}) $$


and the bound for $I\_3$ follows the same way as $I\_2$.


P.S. Going through the above estimates with an explicit dependence on $t$ shows that the bound growth like $t^2$, though I think the actually growth may well be slower.


P.P.S. A place to study oscillatory integrals systematically is Chapter 6 of Stein, *Harmonic Analysis*.


