Asymptotic behavior of a Fourier/Laplace transform I see many results concerning the asymptotics of Fourier transforms. These link in particular the regularity/continuation properties of the function to the polynomial/exponential decay of its Fourier transform. However, these results often hold only in the real variable. I am interested in the Fourier transform "along the imaginary axis" instead.
Let us be more precise. I am interested by the digamma function $\psi = \frac{\Gamma'}{\Gamma}$, and in the function
$$h(\nu) = \exp\left(-\alpha \psi \left( \frac14 \pm \frac{i\nu}{2} \right)\right),$$
where $\alpha$ is a fixed parameter, say $\alpha > 1$. I am interested in the asymptotic behavior of the Fourier transform of $h$ at $+\infty$. More precisely,
$$\widehat{h}(x) = \int_{\mathbb{R}} h(\nu) e^{ix\nu} d\nu.$$
How to get asymptoptics when $x \to +\infty$ in this situation? I have no feeling about what determines it: size? variations? only asymptotics of $h$?
I had many trials, not convincing. Typically, just changing variables, I can get an expression of the shape
$$e^{-\frac{x}{2}} \int_{i\mathbb{R}} e^{-\alpha \psi(u)} e^{2xu} du$$
which looks more like a Laplace (?) transform than a Fourier transform. I was motivated by the fact that I am expecting for other reasons an exponential decay as above, so that I am hoping for a polynomial behavior in $x$ for the remaining integral. However, is the growth/decay estimate of this last integral easier to understand than the original one?
So my question could be synthzised into

Do we have $\int_{i\mathbb{R}} e^{-\alpha \psi(u)} e^{xu} du \ll x^A$ for a certain $A$?

 A: I'll provide a sketch answer, to illustrate a general process. It's all about poles! Admittedly I acted as though the function was the transform of something un smooth, like a counting function. Perhaps
$$h(z) := \psi\left(\frac 14 + \frac i2 z\right).$$
Contour integral approach
First consider the poles of $h(z)$, which can be deduced from those for the original Digamma function (in turn corresponding just to poles of $\Gamma$). These are simple poles at
$$z_n := (2n + 1/2)i,\qquad n \in \{0,1,2,3,\ldots\},$$
each with residue $2/i = -2i$ (since those for the digamma function each have residue $1$, which we have "scaled" by a factor of $i/2$).
Fix $S$ such that $ \mathrm{Im}(z_N) < S < \mathrm{Im}z_{N+1}$ for some $N$, and let $T>0$. Consider the following contour integral over the rectangle:
(My bad! The labels on the $x_n$ should start from $0$ in the picture.)

Fix $x>0$ for now.
For $f(z) := h(z) e^{ixz}$, Cauchy's Residue theorem gives that
$$\int_{A_T} f(z) = \int_{B_T} f(z) + \left(\int_{C_T} f(z) + \int_{D_T} f(z)\right) + 2\pi i\sum_{n=0}^N \mathrm{res}(f,z_n)$$
This simplifies in the limit with the following claim, which should follow from a naive uniform estimate on $|h|$ on $D_T$ and $C_T$. Since I haven't proven it, I'll label it as an assumption:

Assumption: As $T → ∞$,
$$ \int_{C_T} f(z)\ dz + \int_{D_T} f(z)\ dz \to 0.$$

Also noting that
$$\mathrm{res}(f,z_n) = -2i e^{ixz_n} = -2i e^{-\frac{4n+1}2x},$$
in the limit we would have an asymptotic expansion, with the resonances given precisely by the residues:
$$\hat h(x) = \lim_{T→ ∞}\int_{A_T} f(z)\ dz= 4\pi \sum_{n=0}^N e^{-x(4n+1)/2} + \int_{\mathbb R + iS} h(z) e^{ixz}\ dz.$$
To finish off, one would have to show that the last integral decays at a faster rate than the other terms (in terms of $x$), which I shall not do. Does Paley–Wiener still apply?
Full asymptotic expansion?
Perhaps simpler is to ignore the unrigourousness above and investigate the natural conjecture which results from taking $ N → ∞ $:
$$\hat h(x) = 4\pi \sum_{n=0}^∞ e^{-x(4n+1)/2} = 4\pi \frac{ e^{3 x/2}}{e^{2 x}-1}.$$
This conjecture would also come from some formal-series argument; it should be enough to show that the inverse transform of this gives the original $h$. I'm not convinced that it is true, but it's worth a pop.
