Looking for sufficient conditions for positive Fourier transforms I am looking for some sufficient conditions for an even, continuous, nonnegative, non-increasing, non-convex function to be non-negative definite.  In other words
$$
\int_0^\infty f(x)\cos(x\omega) \, dx\ge 0, \quad   \omega \in \mathbb{R}.
$$
The function $f(x)=\exp(-x^{\alpha})[x^{\alpha} \log(x) + \delta/\alpha ]$, $\alpha\in(1,2)$, $\delta\ge 2$.  Note that for $\alpha\in(0,1)$, $f(x)$ is convex.
I have tried complex detour numerically (as in Tuck, E. O.: On positivity of Fourier transforms, Bull. Austral. Math. Soc., 74 (2006) 133–
138) – did not work out. It seems the imaginary part always increases around $0$.
I have tried Polya type conditions () - no luck either.
For $\alpha\in(0,1)$ everything works out as the function is convex.
Thank you in advance for any hints or references!
 A: This can be very complicated, so I will only give a reference which contains many highly non-trivial results of this sort:
MR0428382 J. V. Linnik and  I. V. Ostrovskiĭ,  Decomposition of random variables and vectors. Translated from the Russian. Translations of Mathematical Monographs, Vol. 48. American Mathematical Society, Providence, R. I., 1977.
Look especially in the chapter which is called Necessary conditions for $I_0$,
I think it is Chapter IV.
A: How about the following reference T. Gneiting, Criteria of Po´lya type for radial positive definite functions, Proc. Am. Math. Soc. 129 (2001), 2309–2318. I have plotted the results for 6th derivative all numeric but it seems working,
A: please show me how to prove for $|\omega|>20$
With great pleasure. We shall just show that $F(y)=\int_0^\infty e^{-x^a}x^a\log x\cos(yx)\,dx>0$ for large enough $y>0$. Note that $\cos(yx)=\Re e^{iyx}$, so we have the real part of a contour integral from $0$ to $+\infty$ of $e^{-z^a}z^a\log z e^{iyz}\,dz$. The integrand oscillates like crazy on the line, so we would like to move the contour up to get less oscillation. Ideally we would like to have a curve $\Gamma$ parameterized by $z(t)$ so that $\Re \left[e^{-z^a}z^a\log z e^{iyz}z'(t) \right] \ge 0$ everywhere on $\Gamma$. If we can do it, it would be the end of the story. However, for the integral in question it is, clearly, impossible (without the extra $1$ or $2$, the Fourier transform has zero integral, so it cannot be positive everywhere), so we'll settle for less: the integral over the "head part" of the contour will have relatively large positive real part and the tail will be small.
We will use the same curve $\Gamma$ that is normally used to prove that the Fourier transform of $e^{-|x|^a}$ is non-negative, namely, the curve $z(x)=x+i\xi(x)$, $x>0$, where $\xi(x)$ satisfies $(x+i\xi)^a=g+iyx$, $\xi,g\in\mathbb R$. This will make $e^{-z^a}e^{iyz}$ real positive on $\Gamma$. Note that $\xi$ is well-defined and continuous in $x$ for $1\le a<2$ and $\xi(x)\asymp x^{2-a}$ at $+\infty$, so this change of the contour is legitimate.
On $\Gamma$, we have $e^{-z^a}e^{iyz}=e^{-H}$ where $H=g+y\xi$. Differentiating the identity defining $\Gamma$, we get $az^{a-1}(1+i\xi_x)=g_x+iy$, so
$$
H_x=g_x+y\xi_x=\Re [(g_x+iy)(1-i\xi_x)]=\Re[az^{a-1}|1+\xi_x|^2]>0 
$$
for $z$ in the first quadrant, so $H$ is strictly increasing.
We now want to evaluate
$$
\Re\int_{\Gamma} e^{-H}z^a\log z\,dz=\int_{\Gamma} \Re\left[\frac{z^{a+1}}{a+1}\left(\log z-\frac 1{a+1}\right)\right]d(-e^{-H})\,.
$$
To do it, we will switch to polar coordinates $z=Re^{i\theta}$. Notice that the equation for $\Gamma$ becomes $R^a\sin(a\theta)=yR\cos\theta$. Since $\theta\mapsto \frac{\sin(a\theta)}{\cos\theta}$ increases from $0$ to $+\infty$ as $\theta$ runs from $0$ to $\pi/2$, the curve $\Gamma$ intersects every circumference centered at $0$ only once, so the radius $R$ is a legitimate parameter on $\Gamma$.
We also have $\Re [z^a\bar z]=\Re[(x-i\xi)(g+iyx)]=x(g+y\xi)=xH$, so
$$
H=R^a\frac{\cos((a-1)\theta)}{\cos\theta}
$$
Now come a couple of observations. The first one is that
$$
\sin(a\theta)=\sin((a-1)\theta)\cos\theta+\cos((a-1)\theta)\sin\theta
$$
and
$$
-\sin((a-1)\theta)\cos\theta+\cos((a-1)\theta)\sin\theta=\sin((2-a)\theta)>0
$$
for $\theta\in[0.\frac{\pi}2]$. Hence
$$
R^a\cos((a-1)\theta)\sin\theta\le R^a\sin(a\theta)=yR\cos\theta\le 2R^a\cos((a-1)\theta)\sin\theta\,.
$$
Juxtaposing this with the polar formula for $H$, we see that
$H\sin\theta\le yR\le 2H\sin\theta$ on $\Gamma$.
The second observation is that $\sin(a\theta)\le 1$, so for $R\in[0,1]$, we have $\cos\theta\le\frac 1y$, i.e., $\theta\in[\theta_0,\frac\pi 2]$ where $\theta_0=\arccos\frac 1y$ is quite close to $\frac\pi 2$ for large $y$.
Now we are ready to look at the real part of $\frac{z^{a+1}}{a+1}\left(\log z-\frac 1{a+1}\right)$ when $R\le 1$. It is
$$
\frac{R^{a+1}}{a+1}\left[\cos((a+1)\theta)\left(\log R-\frac 1{a+1}\right)-\theta\sin((a+1)\theta)\right]\,.
$$
Note that $(a+1)\theta\in [2\theta_0,\frac{3\pi}2)$, so as soon as $\theta_0\ge\pi/4$ ($y>\sqrt 2$), the cosine is negative. Thus we can ignore $\log R<0$, which leaves us with the expression
$$
-\cos((a+1)\theta)\frac 1{a+1}-\theta\sin((a+1)\theta)
$$
on $[\theta_0,\frac {\pi}2]$. Taking the derivative, we see that it is increasing in $\theta$, so $\theta=\theta_0$ is the worst case. Again, it can easily be verified that this expression is greater than, say, $c(a)=\frac 1{2(a+1)}$ if $\theta_0$ is close enough to $\frac\pi 2$. Taking into account the bound $H\le yR/\sin\theta_0=:H_1R$ on $[0,1]$ and noting that for $R=1$, we have $H=\Re[z^a]+y\xi\ge -1+y\sin\theta_0=:H_0$, we immediately conclude that
$$
\int_{\Gamma:R<1}\dots d(-e^{-H})\ge c(a)\int_0^{H_0/H_1}\frac {R^{a+1}}{a+1}H_1e^{-H_1R}\,dR
$$
For large $y$, we have $H_0\approx H_1\approx y$ (see the accurate estimates above), so the whole integral is about
$$
\frac{c(a)y}{a+1}\int_0^1 R^{a+1}e^{-yR}\,dR\approx \frac{c(a)}{a+1}y^{-a-1}\int_0^\infty r^{a+1}e^{-r}\,dr
$$
i.e., we have just power decay in $y$ here.
On the other hand, since $R\le 2H/y=R(H)$ all the way through, we have
$$
\left|\int_{\Gamma: R>1}\dots d(-e^{-H})\right|\le\frac 1{a+1}\int_{H_0}^\infty R(H)^{a+1}\sqrt{\left(\log R(H)+\frac 1{a+1}\right)^2+\frac{\pi^2}4} \,e^{-H}\,dH\,,
$$
which decays exponentially in $y$, so the whole integral is positive for large $y$. I leave the accurate estimates of the minimal $y$ for which it works to you. $y>20$ is certainly enough, but I won't be suprised if one can push it down to $y>10$.
I'm still struggling with small $y$... :-( Meanwhile, feel free to ask questions if something is unclear.
