Let $A$ be a constant, say larger than $1/2$.
I am interested in the integral $$I_s = \int_0^\infty (A+\varepsilon+\nu^2)^{-s} e^{ix\nu} d\nu$$ where $s$ is a complex variable, ultimately with fixed real part ($\Re(s) = 2+\delta$).
Option 1, moving contour
By moving the integration line until $\Im \nu = A$, for which everything in holomorphic, we are led to consider the integral $$I_s = e^{-Ax}\int_0^\infty (\varepsilon + 2iAt + t^2)^{-s} e^{ixt} dt$$ and I do want to see this exponential decay. However, this integral does not converge absolutely (the integrand absolute value is about $t^{\Re(s)}$).
Question 1. Is this integral convergent? How does it depend on $s$ and on $x$?
Option 2, obtaining an explicit formula
The original integral is a Basset integral for the $K$-Bessel function (given e.g. here) which states that it looks like $$K_{s}(zx) \frac{x^{s}}{z^s\Gamma(s)}$$ where $z^2 = A+\varepsilon$
Question 2. Knowing that $\Re(s) = 2 + \delta$ for a small $\delta>0$, how does this depend upon $s$ (when its imaginary part grows)? How does it depend upon $x$?
It looks like it blows up vertically in $s$, since the $\Gamma$ function does by Stirling formula.
The integral converges as long as $\operatorname{Re}s>0$. Let us be a bit more precise about the relation between $I_s$ and $K_s$. We have that, with $z=(A+\varepsilon)^{1/2}$, $$\operatorname{Re} I_s= \int_0^\infty \frac{\cos(x t)}{\left(z^2+t^2\right)^s}\,dt=K_{s-1/2}(xz)\frac{\sqrt{\pi}}{\Gamma(s)}\left(\frac{x}{2z}\right)^{s-1/2}$$ provided that $\operatorname{Re}s>0$.
If we can fix $s$ and let $x\to\infty$, we use the known asymptotics of the Bessel function to discover that $$\operatorname{Re} I_s\sim \left(\frac{\pi}{2xz}\right)^{1/2}e^{-zx}\frac{\sqrt{\pi}}{\Gamma(s)}\left(\frac{x}{2z}\right)^{s-1/2}=(2z)^{-s}\frac{\pi}{\Gamma(s)}x^{s-1}e^{-zx}\quad \text{ as } x\to\infty.$$ Now, if we fix $x>0$, and let $s\to\infty$, we find that $$\operatorname{Re} I_s\sim \left(\frac{\pi}{2s}\right)^{1/2}\frac{\sqrt{\pi}}{\Gamma(s)}\left(\frac{exz}{2s-1}\right)^{-s+1/2}\sim \frac{1}{2}\sqrt{\frac{\pi}{s}}\left(\frac{2}{xze}\right)^{s-1/2}.$$ Note also that by the unitarity of the Fourier transform, for real $s>1/4$, \begin{aligned} \|I_s\|_{L^2(\mathbb{R})}^2&=\int_0^\infty \frac{dt}{(t^2+z^2)^{2s}} \\ &=z^{-4s+1}\int_0^\infty \frac{dt}{(1+t^2)^{2s}} \\ &=\frac{1}{2}z^{-4s+1}\int_0^\infty \frac{t^{-1/2}}{(1+t)^{2s}}\,dt \\ &=\frac{1}{4}z^{-4s+1}B(1/2, 2s-1/2) \\ &=\frac{\sqrt{\pi}}{4}z^{-4s+1}\frac{\Gamma(2s-1/2)}{\Gamma(2s)} \\ &\sim\frac{\sqrt{\pi}}{4}z^{-4s+1} (2s)^{-1/2} \end{aligned} as $s\to\infty$. Since $$\|I_s\|_{L^\infty(\mathbb{R})}\le \int_0^\infty (z^2+t^2)^{-s}\,dt<\infty$$ for $s>1/2$, by interpolation we obtain that $I_s\in L^p(\mathbb{R})$ for any $2\le p\le\infty$ provided that $s>1/2$.