I am trying to find the inverse Laplace Transform of $$F(\theta) = \frac{\theta ^{\alpha -\frac{1}{2}}\,{\mathrm{e}}^{\theta /2}\,\Gamma \left(\alpha \right)\,{\mathrm{K}}_{\alpha -\frac{1}{2}}\left(\frac{\theta }{2}\right)}{\sqrt{\pi }\,\Gamma \left(2\,\alpha -1\right)}$$ for $\alpha \in (0, 1/2)$ wehere $K$ means the modified Bessel function of the second kind.
Does anyone has any idea how I could proceed?
It should be the case, that the function of which $F$ is a Laplace transform is a probability density function.
All the best and thx in advance!
If you make the change of variable $\theta=2z$ and consider the function $$f(z)=(2z)^{\alpha-1/2}\mathrm e^zK_{\alpha-1/2}(z)$$ Then, my version of Mathematica finds $$\mathcal{L}^{-1}(f)(t)\\=\frac{-\mathrm i~\mathrm e^{-\mathrm i\pi\alpha}}{4\pi^{3/2}}~(t(2+t))^{-\alpha}\cdot\\\Bigg[4^{\alpha}\mathrm e^{2\mathrm i\pi\alpha}\pi\Gamma(\alpha)-(t(2+t))\Gamma(-\alpha)\Gamma(2\alpha )\sin(2\pi\alpha)~\bigg(-1+{}_2F_1\left(1,-\alpha;~1-2\alpha;~\frac{-2}{t}\right)\bigg)\Bigg]$$ So good luck proving that....