The sign of the tail of Fourier transform of a positive function/ characteristic function I am interested in  a specific density (positive function) and would like to prove that the tail of its characteristic function (Fourier transform) is positive ($>0$). Here is the density $f(x)=c_\alpha \exp(-|x|^{\alpha})(|x|^\alpha\log(|x|)+1)$, $\alpha \in(1,2)$, $x\in R$, $c_\alpha>0$ is a normalizing constant. I can show that the tail of the characteristic function  is $O(\log(\omega)/\omega^{\alpha+1})$. I just cannot find any theorems for the sign of the limit.  From the numerical analysis I see it is positive.
 A: This is an extended version of my comment (and now heavily edited). Your function $f$ is
$$
 f(x) = c_\alpha e^{-|x|^\alpha} (1 + |x|^\alpha \log |x|) = c_\alpha (1 + |x|^\alpha \log |x| - |x|^\alpha) + O(x^2)
$$
as $x \to 0$. Similarly,
$$
 f'(x) \operatorname{sign} x = c_\alpha (\alpha |x|^{\alpha - 1} \log |x| - (\alpha - 1) |x|^{\alpha - 1}) + O(|x|) ,
$$
and
$$
 f''(x) = c_\alpha (\alpha (\alpha - 1) |x|^{\alpha - 2} \log |x| - (\alpha (\alpha - 1) - (2 \alpha + 1)) |x|^{\alpha - 1}) + O(1) .
$$
Define
$$
 G(z) = \frac{1}{(1 + z^2)^{(\alpha + 1)/2}}
$$
and
$$
 H(z) = \frac{\log(1 + |z|^2)}{(1 + |z|^2)^{(\alpha + 1)/2}} = -\frac{d}{d\alpha} \frac{2}{(1 + |z|^2)^{(\alpha + 1)/2}} \, .
$$
Then the inverse Fourier transforms of $G(z)$ and $H(z)$ are
$$
 g(x) = \frac{1}{2^{\alpha/2} \sqrt{\pi} \, \Gamma(\tfrac{\alpha + 1}{2})} \, |x|^{\alpha/2} K_{\alpha/2}(|x|)
$$
and (by the dominated convergence theorem)
$$
 h(x) = -\frac{d}{d\alpha} \biggl( \frac{1}{2^{\alpha/2} \sqrt{\pi} \, \Gamma(\tfrac{\alpha + 1}{2})} \, |x|^{\alpha/2} K_{\alpha/2}(|x|) \biggr) .
$$
Note that
$$
 g(x) = \frac{A_\alpha}{2} - \frac{B_\alpha}{2} \, |x|^\alpha + O(x^2) ,
$$
where
$$
 A_\alpha = \frac{\Gamma(\tfrac{\alpha}{2})}{\sqrt{\pi} \, \Gamma(\tfrac{1 + \alpha}{2})} , \qquad B_\alpha = \frac{-2^{-\alpha} \Gamma(-\tfrac{\alpha}{2})}{\sqrt{\pi} \, \Gamma(\tfrac{1 + \alpha}{2})} .
$$
Similarly,
$$
 g'(x) = -\frac{B_\alpha}{2} \, \alpha |x|^{\alpha - 1} + O(|x|) ,
$$
and
$$
 g''(x) = -\frac{B_\alpha}{2} \, \alpha (\alpha - 1) |x|^{\alpha - 2} + O(1) .
$$
The derivative of the Bessel function $K_\nu$ with respect to the parameter $\nu$ can in fact be expressed in terms of other special functions, but the formula is quite complicated and I will not copy it here and only refer to the paper Higher derivatives of the Bessel functions with respect to the order by Yu.A. Brychkov, DOI:10.1080/10652469.2016.1164156. An integral expression for this derivative follows directly from the integral expression for $K_\nu(x)$; see formula 10.32.9 in DLMF, for example. What is important here is that the series expansion of $K_\nu$ near $0$ can be differentiated with respect to $\nu$ term-by-term (and either the explicit expression or the integral formula, both mentioned above, provide a way to make this claim rigorous; I did not verify this carefully, though). It follows that
$$
 h(x) = -A_\alpha' + B_\alpha |x|^\alpha \log |x| - B_\alpha' |x|^\alpha + O(x^2 \log |x|) ,
$$
where primes stand for derivatives with respect to $\alpha$. Similarly,
$$
 h'(x) \operatorname{sign} x = \alpha B_\alpha |x|^{\alpha - 1} \log |x| - (\alpha B_\alpha' - B_\alpha) |x|^{\alpha - 1} + O(|x| \log |x|) ,
$$
and
$$
 h''(x) = \alpha (\alpha - 1) B_\alpha |x|^{\alpha - 2} \log |x| - (\alpha (\alpha - 1) B_\alpha' - (2 \alpha - 1) B_\alpha) |x|^{\alpha - 1} + O(\log |x|) .
$$
Consider an auxiliary function
$$
 \phi(x) = f(x) - \frac{c_\alpha}{B_\alpha} h(x) - \frac{2 c_\alpha (B_\alpha - B_\alpha')}{B_\alpha^2} g(x) .
$$
The constants are chosen in such a way that
$$\begin{gathered}
 \phi(x) = C_\alpha + O(x^2 \log |x|) , \\
 \phi'(x) = O(|x| \log |x|) , \\
 \phi''(x) = O(\log |x|) .
\end{gathered}$$
In order to estimate the Fourier transform of $\phi$, decompose $\phi$ into $\phi_1(x) = \phi(x) u(x)$ and $\phi_2(x) = \phi(x) (1 - u(x))$, where $u$ is smooth, $u = 1$ in $[-1, 1]$ and $u = 0$ outside $(-2, 2)$. Since $\phi_2$ is infinitely smooth and all derivatives of $\phi_2$ are integrable, the Fourier transform of $\phi_2$ decays faster than $|z|^{-p}$ for every $p > 0$, so we only need to handle $\phi_1$.
Observe that for an arbitrary $\varepsilon \in (0, 1)$, we have $|\phi_1''(x)| \leqslant a_\alpha \log (e + \tfrac{1}{|x|}) \leqslant b_{\alpha, \varepsilon} (1 + |x|^{-\varepsilon})$ for some constants $a_\alpha$ and $b_{\alpha, \varepsilon}$. Thus, $\phi_1'$ is Hölder continuous with exponent $\beta = 1 - \varepsilon$. As a consequence, the Fourier transform of $\phi_1'$ is $O(|z|^{-\beta})$ as $|z| \to \infty$. It follows that the Fourier transform of $\phi_1$ is $O(|z|^{-1 - \beta})$ for any $\beta \in (0, 1)$, and by choosing $\beta > \alpha - 1$ we find that, in particular, it is $o(|z|^{-\alpha})$. This implies that the Fourier transform of $\phi$ is $o(|z|^{-\alpha})$, and thus, finally
$$
 F(z) = \frac{c_\alpha}{B_\alpha} H(z) + \frac{2 c_\alpha (B_\alpha - B_\alpha')}{B_\alpha^2} G(z) + o(|z|^{-\alpha}) .
$$
This, of course, implies that $F(z) > 0$ when $|z|$ is large enough, as desired.

Note: Apparently it can be proved that the Fourier transform of $\phi$ is $O(|z|^{-3} \log |z|)$ as $|z| \to \infty$, and consequently the Fourier transform of $f(x)$ satisfies
$$
 F(z) = \frac{c_\alpha}{B_\alpha} H(z) + \frac{2 c_\alpha (B_\alpha - B_\alpha')}{B_\alpha^2} G(z) + O(|z|^{-3} \log |z|)
$$
as $|z| \to \infty$. The proof of the above bound, however, more effort: it involves monotonicity of $\phi''(x)$ for $x > 0$ small enough, not just Hölder continuity of $\phi'(x)$. The idea is the same as in the proof of $F(z) = O(|z|^{-1-\alpha} \log|z|)$ result mentioned in your post, I suppose. It is not really needed in this answer, and I am short of time, so I will skip the details if you do not mind.
