Bounds on $f(k ;a,b) =\frac{ \int_0^\infty \cos(a x) e^{-x^k} \, dx}{ \int_0^\infty \cos(b x) e^{-x^k}\, dx}$ Suppose we define a function 
\begin{align}
f(k ;a,b) =\frac{ \int_0^\infty \cos(a x) e^{-x^k} \,dx}{ \int_0^\infty \cos(b x) e^{-x^k} \,dx}
\end{align}
can we show that 
\begin{align}
|f(k ;a,b)| \le 1
\end{align}
for $ 0<k \le 2$  and $a\ge b$? 
This question was motivated by the discussion here. 
Note that for $k=1$ and $k=2$ this can be done, since
\begin{align}
 \int_0^\infty \cos(a x) e^{-x^1} \,dx=\frac{1}{1+a^2}\\
 \int_0^\infty \cos(a x) e^{-x^2} \,dx=\frac{\sqrt{\pi}}{2}e^{-a^2/4}\\
\end{align}
So, we have that 
\begin{align}
f(1;a,b)&=\frac{1+b^2}{1+a^2} \\
f(2;a,b)&=e^{ \frac{b^2-a^2}{4}}
\end{align}
In which case, we have that the conjectured bound is true. 
Edit: The bounty was posted specifically to address this question and a question raised by Jack D'Aurizio in the comments.
The question is:
Let 
\begin{align}
g_k(z)=\int_0^\infty \cos(zx) e^{-x^k} dx
\end{align}
What is the largest value of $k$ such that $g_k(z)$ is non-negative and decreasing for $z\in \mathbb{R}^{+}$?
 A: We have to prove that the Fourier cosine transform of $e^{-x^k}$ is non-negative and decreasing on $\mathbb{R}^+$.
We have:
$$ \int_{0}^{+\infty}\cos(\xi x)e^{-x}\,dx = \frac{1}{1+\xi^2},\qquad \int_{0}^{+\infty}\cos(\xi x)e^{-x^2}\,dx = \frac{\sqrt{\pi}}{2}\,e^{-\xi^2/4} \tag{1} $$
hence the claim is expected to hold by some kind of interpolation argument. Let $f_k(x)=e^{-x^k}$ and
$$ g_k(\xi)=\int_{0}^{+\infty}\cos(\xi x)\,e^{-x^k}\,dx.\tag{2} $$
We have:
$$ g_k'(\xi) = -\int_{0}^{+\infty}\sin(\xi x) x e^{-x^k}\,dx =-\frac{1}{\xi}\int_{0}^{+\infty}\cos(\xi x)\left(e^{-x^k}-kx^k e^{-x^k}\right)\,dx\tag{3}$$
hence:
$$ g_k(\xi)+\xi g_k'(\xi) = \frac{k}{\xi}\int_{0}^{+\infty}\cos(\xi x) x^k e^{-x^k}\,dx \tag{4} $$
and as soon as we prove that the RHS of $(4)$ cannot be too large (for instance, through $w e^{-w}\in\left[0,\frac{1}{e}\right]$ for any $w\in\mathbb{R}^+$) we have that $g_k(\xi)$ is close to $g_1(\xi)$ or $g_2(\xi)$ by a Gronwall-like inequality.
As already shown in your other question, $g_k(\xi)$ does not preserve its non-negativity and monotonicity for every $k$. That is not surprising, since $e^{-x^k}$, regarded as a distribution over $\mathbb{R}^+$, weakly converges to $c_k\cdot\mathbb{1}_{(0,1)}(x)$ as $k\to +\infty$, and the Fourier transform of $\mathbb{1}_{(0,1)}(x)$ is extremely regular (as any Fourier transform of a compact-supported function, by the Paley-Wiener theorem) but has plenty of real zeroes. Another interesting (but probably very hard) question arises:

What is
  $$ \sup\left\{k\in\mathbb{R}^+: g_k(\xi)\text{ is non-negative and decreasing over } \mathbb{R}^+\right\}$$
  ?

Numerical simulations seem to suggest that such a supremum is exactly $\color{red}{2}$.
I strongly suspect that the Fejer-Riesz theorem is involved.
A: Yes, the Fourier transform of $\exp(-|x|^k)$ is positive and decreasing
for all $k$ such that $0 < k \leq 2$.
This follows from the known case of $k=2$ (Gaussians) via an argument of
B.F.Logan cited in my 1991 paper with Odlyzko and Rush:

Noam D. Elkies, Andrew M. Odlyzko, and Jason A. Rush: On the packing densities of superballs and other bodies, Invent. Math. 105 (1991), 613-639.

See Lemma 5 on page 626 (with $k=\sigma$; in that paper
we needed only positivity, not that the Fourier transform is decreasing).
The key is that $\exp(-t^{k/2})$ is a "totally monotone" function of $t>0$
(its $n$-th derivative has sign $(-1)^n$ for all $t>0$),
and decays to zero as $t \to \infty$, whence it is
a nonnegative mixture of decreasing exponentials $e^{-ct}$ $(c>0)$ by
Bernstein's theorem.
Taking $t=x^2$ we deduce that $\exp(-|x|^k)$ is a nonnegative mixture of
Gaussians $\exp(-cx^2)$ $(c>0)$.  Since the Fourier transform of $\exp(-cx^2)$
is positive and decreasing for all $c>0$, the same is true of the Fourier
transform of $\exp(-|x|^k)$, QED.
