Asymptotic behavior of a Fourier/Laplace transform
Happy new year! Here's a late answer.
It will be shown that, when $h(v)=\exp\left(-\alpha\psi\left(\frac14+\frac{iv}{2}\right)\right)$,
$$\widehat{h}(x)=(2\alpha)^{1/4}\sqrt{\pi}e^{\alpha\gamma}\cdot\frac{e^{-x/2+2\sqrt{2\alpha}\sqrt{x}}}{x^{3/4}}\left(1+O\left(\frac1{\sqrt x}\right)\right)$$ where $\gamma$ is the Euler-Mascheroni constant.
On the other hand, it is trivial that when $h(v)=\exp\left(-\alpha\psi\left(\frac14-\frac{iv}{2}\right)\right)$, $\widehat{h}(x)=0$ for $x>0$.
By the substitution $u=\frac14+\frac{iv}{2}$,
$$\begin{align} \widehat{h}(x) &:=\int_{\mathbb R}\exp\left(-\alpha\psi\left(\frac14+\frac{iv}{2}\right)\right)e^{ixv}dv \\ &=-2ie^{-x/2}\int_{\frac14+i\mathbb R}\underbrace{e^{-\alpha\psi(u)}e^{2xu}}_{:=f(u)}du \\ H(x)&:=\frac i2e^{x/2}\cdot\widehat{h}(x)=\int_{\frac14+i\mathbb R}e^{-\alpha\psi(u)}e^{2xu}du \\ \end{align} $$
By residue theorem and considering the exponential decay of $f(z)$, it can be shown that $$H(x)-\int_{-\frac12+i\mathbb R}f(u)du=2\pi i\operatorname*{Res}_{z=0}f(z)$$ $$H(x)=2\pi i\operatorname*{Res}_{z=0}f(z)+\underbrace{\int_{-\frac12+i\mathbb R}f(u)du}_{:=J_0}$$
Lemma 1: $J_0=O(e^{-x})$.
Proof:
$$\begin{align} \left|\int_{-\frac12+i\mathbb R}f(u)du\right| &=\left|\int_{\mathbb R}e^{-\alpha\psi(-1/2+iu)}e^{-x+2xiu}du\right| \\ &\le\int_{\mathbb R}\left|e^{-\alpha\psi(-1/2+iu)}e^{-x+2xiu}\right|du \\ &=e^{-x}\int_{\mathbb R}\left|e^{-\alpha\psi(-1/2+iu)}\right|du \\ &=e^{-x}\int_{\mathbb R}\left|e^{-\alpha\psi(3/2-iu)-\alpha\pi i\tanh(\pi u)}\right|du \quad (1)\\ &=e^{-x}\int_{\mathbb R}\left|e^{-\alpha\psi(3/2-iu)}\right|du \\ &=Ce^{-x} \qquad (2) \end{align} $$
$(1)$: By the reflection formula $\psi(1-x)-\psi(x)=\pi\cot(\pi x)$.
$(2)$: The last integral can be considered as a constant $C$ because it converges (as $e^{-\alpha\psi(3/2-iu)}\approx u^{-\alpha}$ for large $|u|$ and $\alpha>1$) and is independent of $x$.
Now, let's focus on the residue at $0$. Trivially, $$2\pi i\operatorname*{Res}_{z=0}f(z)=\oint_{|z|=R}f(z)dz \qquad R<1$$
The trick here is to take $R=\sqrt{\frac{\alpha}{2x}}$ (I will explain how to come up with this choice of contour on request).
Define $\phi(z)=\psi(z)+\frac1z+\gamma$. We have $\phi(z)=O(|z|)$ as $z\to 0$.
$$\begin{align} 2\pi i\operatorname*{Res}_{z=0}f(z) &=\oint_{|z|=R}f(z)dz \\ &=\oint_{|z|=R}\exp\left(-\alpha\left(-\frac1z-\gamma+\phi(z)\right)+2xz\right)dz \\ &=e^{\alpha\gamma}\oint_{|z|=R}\exp\left(\frac{\alpha}{z}+2xz\right)\left(e^{-\alpha\phi(z)}-1+1\right)dz \\ &=e^{\alpha\gamma}\underbrace{\oint_{|z|=R}\exp\left(\frac{\alpha}{z}+2xz\right)dz}_{:=J_1} \\ &+e^{\alpha\gamma}\underbrace{\oint_{|z|=R}\exp\left(\frac{\alpha}{z}+2xz\right)\left(e^{-\alpha\phi(z)}-1\right)dz}_{:=J_2} \\ \end{align} $$
Lemma 2: $$J_1=i\cdot\frac{\alpha^{1/4}\sqrt{\pi}}{2^{3/4}}\cdot\frac{e^{2\sqrt{2\alpha}\sqrt{x}}}{x^{3/4}}\left(1+O\left(\frac1{\sqrt x}\right)\right)$$
$$\begin{align} J_1 &=\int^\pi_{-\pi}\exp\left(\frac{\alpha}{R}e^{-i\theta}+2xRe^{i\theta}\right)iRe^{i\theta}d\theta \\ &=i\sqrt{\frac{\alpha}{2x}}\int^\pi_{-\pi}\exp\left(\alpha\sqrt{\frac{2x}{\alpha}}e^{-i\theta}+2x\sqrt{\frac{\alpha}{2x}}e^{i\theta}\right)e^{i\theta}d\theta \\ &=i\sqrt{\frac{\alpha}{2x}}\int^\pi_{-\pi}\exp\left(2\sqrt{2\alpha x}\cos\theta\right)e^{i\theta}d\theta \\ &=2i\sqrt{\frac{\alpha}{2x}}\int^\pi_{0}\cos\theta \, e^{2\sqrt{2\alpha x}\cos\theta} d\theta \qquad (1)\\ &=2i\sqrt{\frac{\alpha}{2x}}\cdot\pi I_1\left(2\sqrt{2\alpha x}\right) \qquad (2)\\ &=2\pi i\sqrt{\frac{\alpha}{2x}}\cdot \frac{e^{2\sqrt{2\alpha}\sqrt{x}}}{\sqrt{2\pi}\sqrt{2\sqrt{2\alpha x}}}\left(1+O\left(\frac1{\sqrt x}\right)\right) \qquad (3) \\ &=i\cdot\frac{\alpha^{1/4}\sqrt{\pi}}{2^{3/4}}\cdot\frac{e^{2\sqrt{2\alpha}\sqrt{x}}}{x^{3/4}}\left(1+O\left(\frac1{\sqrt x}\right)\right) \\ \end{align} $$
$(1)$: The imaginary part cancels out due to oddness, and the extra factor of $2$ is due to the evenness of the real part.
$(2)$: $I_1$ is the first order modified Bessel function of the first kind.
$(3)$: Due to the well-known asymptotic expansion $I_1(z)=\frac{e^z}{\sqrt{2\pi z}}\left(1+O\left(\frac1z\right)\right)$ for $z\to\infty$.
Lemma 3: $$J_2=O\left(\frac{e^{2\sqrt{2\alpha}\sqrt{x}}}{x^{5/4}}\right)$$
Proof:
$$\begin{align} |J_2| &=\left|\oint_{|z|=R}\exp\left(\frac{\alpha}{z}+2xz\right)\left(e^{-\alpha\phi(z)}-1\right)dz\right| \\ &=\left|\int^\pi_{-\pi}e^{2\sqrt{2\alpha}\sqrt{x}\cos\theta}\left(\exp\left(-\alpha\phi(Re^{i\theta})\right)-1\right)iRe^{i\theta}d\theta\right| \\ &\le R\int^\pi_{-\pi}e^{2\sqrt{2\alpha}\sqrt{x}\cos\theta}\left|\exp\left(-\alpha\phi(Re^{i\theta})\right)-1\right|d\theta \\ &\le R\int^\pi_{-\pi}e^{2\sqrt{2\alpha}\sqrt{x}\cos\theta}C|Re^{i\theta}|d\theta \qquad (1)\\ &=CR^2\int^\pi_{-\pi}e^{2\sqrt{2\alpha}\sqrt{x}\cos\theta}d\theta \\ &=CR^2\cdot 2\pi I_0(2\sqrt{2\alpha}\sqrt{x}) \\ &=C\cdot\frac{\alpha}{2x}\cdot 2\pi \cdot \frac{e^{2\sqrt{2\alpha}\sqrt{x}}}{\sqrt{2\pi}\sqrt{2\sqrt{2\alpha x}}}\left(1+O\left(\frac1{\sqrt x}\right)\right) \qquad (2)\\ &=O\left(\frac{e^{2\sqrt{2\alpha}\sqrt{x}}}{x^{5/4}}\right) \end{align} $$
$(1)$: As $\phi(z)=O(|z|)$, $\exp\left(-\alpha\phi(z)\right)-1=\exp(O(|z|))-1=1+O(|z|)-1=O(|z|)$.
$(2)$: Due to the well-known asymptotic expansion $I_0(z)=\frac{e^z}{\sqrt{2\pi z}}\left(1+O\left(\frac1z\right)\right)$ for $z\to\infty$.
Therefore, $$2\pi i\operatorname*{Res}_{z=0}f(z)=i\cdot\frac{\alpha^{1/4}\sqrt{\pi}e^{\alpha\gamma}}{2^{3/4}}\cdot\frac{e^{2\sqrt{2\alpha}\sqrt{x}}}{x^{3/4}}\left(1+O\left(\frac1{\sqrt x}\right)\right)+O\left(\frac{e^{2\sqrt{2\alpha}\sqrt{x}}}{x^{5/4}}\right)$$ $$\implies 2\pi i\operatorname*{Res}_{z=0}f(z)=i\cdot\frac{\alpha^{1/4}\sqrt{\pi}e^{\alpha\gamma}}{2^{3/4}}\cdot\frac{e^{2\sqrt{2\alpha}\sqrt{x}}}{x^{3/4}}\left(1+O\left(\frac1{\sqrt x}\right)\right)$$
In conclusion, $$H(x)=i\cdot\frac{\alpha^{1/4}\sqrt{\pi}e^{\alpha\gamma}}{2^{3/4}}\cdot\frac{e^{2\sqrt{2\alpha}\sqrt{x}}}{x^{3/4}}\left(1+O\left(\frac1{\sqrt x}\right)\right)+O(e^{-x})$$
$$\implies H(x)=i\cdot\frac{\alpha^{1/4}\sqrt{\pi}e^{\alpha\gamma}}{2^{3/4}}\cdot\frac{e^{2\sqrt{2\alpha}\sqrt{x}}}{x^{3/4}}\left(1+O\left(\frac1{\sqrt x}\right)\right)$$
As we defined $H(x)=\frac i2e^{x/2}\cdot\widehat{h}(x)$, it can be concluded, eventually, $$\widehat{h}(x)=(2\alpha)^{1/4}\sqrt{\pi}e^{\alpha\gamma}\cdot\frac{e^{-x/2+2\sqrt{2\alpha}\sqrt{x}}}{x^{3/4}}\left(1+O\left(\frac1{\sqrt x}\right)\right)$$