Estimate for an Airy integral Let me define for $x\in\mathbb R$,
$
F(x)=\int_{\mathbb R} e^{-π t^2}\cos(x t^3) dt.
$
I claim that $F(x)>0$ for all $x\in\mathbb R$.
Well, it is obvious for $x=0$ since $F(0)=1$ and also for $x$ near $0$ by continuity of $F$.
I guess that for $x$ large, a van der Corput method or a version of the stationary phase method should give the result. How can I prove this at "finite distance", in a situation where no asymptotic method could help?
 A: The most elementary way is to exploit the periodicity and compare positive and negative contributions. Changing  variable in the integral, with $\tau=xt^3$, and integrating by parts, you will find an analogous statement for
$$\int_0^\infty f(\tau)\sin(\tau)d\tau>0,$$
for some  positive and decreasing function (also depending on $x$) $f$, which makes the claim evident since any positive contribute on $[2k\pi,(2k+1)\pi]$ is larger in absolute value than the successive negative contribution on $[(2k+1)\pi,(2k+2)\pi]$.
A: $$
\frac{F(x)}2=\int_0^\infty e^{-\pi t^2}\cos(x t^3) dt=\int_0^\infty \frac{e^{-\pi t^2}}{3xt^2}d\sin(x t^3)=-\int_0^\infty \left(\frac{e^{-\pi t^2}}{3xt^2}\right)'\sin(x t^3)dt\\=\int_0^\infty \frac1{3xt^2}\left(\frac{e^{-\pi t^2}}{3xt^2}\right)'d\left(1-\cos(x t^3)\right)=-\int_0^\infty \left(\frac1{3xt^2}\left(\frac{e^{-\pi t^2}}{3xt^2}\right)'\right)'\left(1-\cos(x t^3)\right)dt\\=
\int_{0}^\infty\frac{2e^{-\pi t^2}(2\pi^2t^4+5\pi t^2+5)}{9x^2t^6}\left(1-\cos(x t^3)\right)dt\geqslant 0.
$$
