$\delta^{(1/2)}_\mu(x)=\frac{1}{2\pi}\int_{-\infty}^{\infty}|k|^{1/2}e^{ikx}e^{-\mu|k|}dk$ which provides the half-Dirac delta distribution in the limiting case of $\mu\to0$.
I know the solution of $\delta^{(1/2)}_\mu=\frac{1}{2\pi}\int_{-\infty}^{\infty}|k|^{1/2}e^{ikx}e^{-\mu|k|}dk$ is $\sqrt{\frac{1}{4\pi}}(x^2+\mu^2)^{-3/4}\cos(\frac{3}{2}\tan^{-1}(\frac{x}{\mu}))$, but I would appreciate knowing how the evaluation is accomplished.
I can get to the point of $\delta^{(1/2)}_\mu=2\int_0^{\infty}\cos(kx)e^{-\mu k}k^{1/2}dk$, but beyond here I have no idea.
EDIT: Obvious thing that I missed on the first go-about is that this is related to the Laplace transform of $k^{1/2}\cos(k)$. Evaluating this seems "straight-forward" albeit incredibly nontrivial.
As I said you need some complex analysis for showing that
$$\int_0^\infty t^{a-1} e^{-zt}dt = \Gamma(a) z^{-a}, \qquad \text{Re}(z) > 0,\ \text{Re}(a) > 0$$ The proof is that for $\text{Re}(z) > 0$ the LHS and the RHS are analytic in $z$, and for $z \in (0,\infty)$ they are equal (change of variable $u = zt$), hence by analytic continuation (or by the identity theorem, or by the Cauchy integral theorem) they are equal for every $\text{Re}(z) > 0$.
Then you have (for $x \in \mathbb{R}$, $\mu > 0$) $$2\pi \delta^{(1/2)}_\mu(x) = \int_{-\infty}^\infty |k|^{1/2} e^{-\mu |k|} e^{i k x}dk = \int_0^\infty k^{1/2} (e^{-(\mu+ix) k}+e^{-(\mu-ix) k})dk$$ $$ = \Gamma(3/2) ((\mu+ix)^{-3/2}+(\mu-ix)^{-3/2}) = \Gamma(3/2)\, 2\,\text{Re}((\mu+ix)^{-3/2})$$ $$ = 2 \Gamma(3/2) \, |\mu+ix|^{-3/2} \cos(\text{arg}((\mu+ix)^{-3/2})) = \sqrt{\pi}\, (\mu^2+x^2)^{-3/4} \cos(\frac{3}{2}\arctan(x/\mu))$$
There is a plot of $\delta_\mu^{(1/2)}$ for $\mu = 1,0.7,0.5$ :