Let $f\in H^1(\mathbb{R}^3)$. Define, for $M>0$, $$I(M)=\int_{B(0,M)}e^{i|y|^2}f(y)dy$$ where $B(0,M)$ is the ball centered in the origin and of radius $M$ in $\mathbb{R}^3$.
Is it true that $|I(M)|$ is uniformly bounded for $M>0$? In that case, can i bound $\sup_{M>0}|I(M)|$ in term of the $H^1$ norm of $f$?
Some progress:
We can write $$I(M)=\int_{B(0,M)}e^{i|y|^2}\int_{\mathbb{R}^3}e^{i(y,\xi)}\widehat{f(\xi)}d\xi dy=\int_{\mathbb{R}^3}\widehat{f(\xi)}\int_{B(0,M)}e^{i|y|^2+i(y,\xi)}dyd\xi$$ Denote $$g_M(\xi)=\int_{B(0,M)}e^{i|y|^2+i(y,\xi)}dy$$ Using Cauchy-Schwartz we get $$|I(M)|=\left|\int_{\mathbb{R}^3}\widehat{f(\xi)}g_M(\xi)d\xi\,\right|=\left|\int_{\mathbb{R}^3}<\xi>\widehat{f(\xi)}\frac{g_M(\xi)}{<\xi>}d\xi\,\right|\leq\Vert f\Vert_{H^1}\sqrt{\int_{\mathbb{R}^3}\frac{|g_M(\xi)|^2}{1+|\xi|^2}d\xi}$$
So the thesis would follow if one prove that $g_M(\xi)$ has some decay in $\xi$ (a bit more than $1/\sqrt{|\xi|}$), unifomly on $M$.
EDIT: Using wolfram alpha i rewrite $g_M(\xi)$ in terms of (complex) error functions, and using the asymptotics for these objects i get indeed $|g_M(\xi)|\lesssim 1/|\xi|$ for large $\xi$ and uniformly in $M$.
Is this bound actually true? In this case, there is a proof that doesn't use the asymptotics of error functions?
Thank you for any suggestions
We can assume that $\xi = L e_3$, and then (spherical coordinates) $$ g_M(\xi) = C_1 \int_0^M r^2e^{ir^2}\, dr \int_{-1}^1 dw\, e^{irLw} = \frac{C_2}{L} \int_0^M re^{ir^2}\sin rL\, dr . $$ To show that $|g_M|\lesssim 1/L$, with a constant that is independent of $M>0$, it now suffices to show that $I=\int_0^M re^{i(r^2\pm rL)}\, dr = O(1)$, uniformly in $M$. An integration by parts gives that (for the $+$ sign) $$ I = -i \int_0^M \frac{r}{2r + L} \frac{d}{dr} e^{i(r^2+rL)}\, dr = O(1) + iL\int_0^M \frac{e^{i(r^2+rL)}}{(2r+L)^2}\, dr = O(1) , $$ as desired.