Fourier transform of $f_a(x)= a^{-2}\exp(-|x|^a)$, $a \in (0,2)$, is decreasing in $a$ Can one show that Fourier transform of 
$$ f_a(x) = a^{-2} \exp(-|x|^a), \qquad a \in (0,2)$$
is decreasing in $a$?
I have a solution for $a \in (0,1]$ which cannot be used for $a\in (1,2)$.
 A: Too long for a comment.
We have
$
\widehat{f_a}(\xi)=\int_{\mathbb R}  a^{-2} e^{-\vert x\vert ^a-ix \xi} dx=
2\int_0^{+\infty}  a^{-2} e^{-x^a} \cos(x\vert\xi\vert)dx, 
$
so that
$$
J_a(\xi)=-\frac{a^3}2\partial_a\widehat{f_a}(\xi)=
\int_0^{+\infty} \bigl( 2  + x^a \ln (x^a)\bigr)e^{-x^a}\cos(x\vert\xi\vert)dx,\quad\text{and}
$$
$$
aJ_a(\xi)=\int_0^{+\infty} t^{\frac 1a-1}\bigl( 2  + t \ln t\bigr)e^{-t}\cos(t^{1/a}\vert\xi\vert)dt.
$$
We note that
$
\frac d{dt}(t\ln t)=\ln t+1
$
which is positive iff $t>1/e$ so that
$
2+t\ln t\ge 2-\frac{1}{e}>0,
$
proving that $J_a(0)>0$.
Remark. Positive values of $\xi$ remain to be checked. Note that for $\chi_0$ smooth equal to 1 near 0 and vanishing outside of $[0,1]$, we have
$$
t^{\frac 1a-1}\bigl( 2  + t \ln t\bigr)e^{-t}=t^{\frac 1a-1}\bigl( 2  + t \ln t\bigr)e^{-t}\chi_0(t) +\psi(t),
$$
where $\psi$ belongs to the Schwartz space, thus as well as its cosine transform. Somehow the main part of the integral is located near $t=0$ and
$$
\int_0^{+\infty}  \cos(2π s\vert\xi\vert)  ds 2πa=π a\delta_0(\xi),
$$
which is indeed positive for $a>0$.
