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?
$\begingroup$
$\endgroup$
1
-
4$\begingroup$ Mathematica calculates the integral as $$ \frac{2 \pi e^{\frac{2 \pi ^3}{27 x^2}} K_{\frac{1}{3}}\left(\frac{2 \pi ^3}{27 x^2}\right)}{3 \sqrt{3} \left| x\right| }$$ which is positve because the modified Bessel function $K_\nu(z)$ is. $\endgroup$– Karl FabianCommented Aug 24, 2020 at 12:59
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
5
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 $[(2k+1)\pi,(2k+2)\pi]$.
answered Aug 24, 2020 at 13:19
-
3$\begingroup$ alternatively, we may use $\sin \tau d\tau=d(1-\cos \tau)$ that yields $\int_0^\infty f(\tau)\sin \tau d\tau=-\int_0^\infty f'(\tau)(1-\cos \tau) d\tau\ge 0$ $\endgroup$ Commented Aug 24, 2020 at 13:25
-
$\begingroup$ so a double integration by parts, even better $\endgroup$ Commented Aug 24, 2020 at 13:33
-
$\begingroup$ yes, and may be done without change of variables (here this is a matter of taste, but when the change of variables is not so explicit, it looks more suitable), I posted the computation as an alternative answer $\endgroup$ Commented Aug 24, 2020 at 13:44
-
$\begingroup$ Very nice, thanks for both answers. $\endgroup$– BazinCommented Aug 24, 2020 at 14:00
-
$\begingroup$ Similarly $\int_0^\infty f(t) \cos t\, dt \ge 0$ if $f \ge0$ is decreasing and convex. $\endgroup$ Commented Aug 24, 2020 at 15:59
$\begingroup$
$\endgroup$
$$ \frac{F(x)}2=\int_0^\infty e^{-\pi t^2}\cos(x t^3) dt=\int_0^\infty \frac{e^{-\pi t^2}}{3xt^2}d\sin(x t^3)=-\int_0^\infty \left(\frac{e^{-\pi t^2}}{3xt^2}\right)'\sin(x t^3)dt\\=\int_0^\infty \frac1{3xt^2}\left(\frac{e^{-\pi t^2}}{3xt^2}\right)'d\left(1-\cos(x t^3)\right)=-\int_0^\infty \left(\frac1{3xt^2}\left(\frac{e^{-\pi t^2}}{3xt^2}\right)'\right)'\left(1-\cos(x t^3)\right)dt\\= \int_{0}^\infty\frac{2e^{-\pi t^2}(2\pi^2t^4+5\pi t^2+5)}{9x^2t^6}\left(1-\cos(x t^3)\right)dt\geqslant 0. $$
answered Aug 24, 2020 at 13:41