Inverting a Characteristic Function for half-cubic Student's T entailing a Modified Bessel of 2nd kind The Characteristic function of the Student's T with $\alpha$ degrees of freedom, 
$C(t)=\frac{2^{1-\frac{\alpha }{2}} \alpha ^{\alpha /4} \left| t\right| ^{\alpha /2}
   K_{\frac{\alpha }{2}}\left(\sqrt{\alpha } \left| t\right| \right)}{\Gamma
   \left(\frac{\alpha }{2}\right)}$ entails a modified Bessel function of the second kind
$K_{\alpha/2}\left(\sqrt{\alpha } \left| t\right| \right)$. To invert the Fourier to get the probability density of the $n$-summed variable when $\alpha$ is not an integer poses problem as the equation below seems integrable otherwise. Of particular interest is the distribution for $\alpha= 3/2$ ("halfcubic"). With $n$ an integer ( $n >2$):
  $$f_n(x)= \left(\frac{3^{3/8}}{\sqrt[8]{2} \,\Gamma \left(\frac{3}{4}\right)}\right)^n \int_{-\infty }^{\infty } e^{-i\, t x}  \left| t\right| ^{\frac{3 n}{4}} K_{\frac{3}{4}}\left(\sqrt{\frac{3}{2}} \left| t\right| \right)^n \, dt$$
   I tried all manner of expansions and reexpressions of the Bessel into other functions (Hypergeometric, Gamma) to no avail. One good news is that $n=2$ works on Mathematica because the Wolfram library has the square of a Bessel function. It would be great to get the solution for at least $n=3$.
 A: Well, according to Mathematica
FourierTransform[
 Abs[t]^(9/4)*BesselK[3/4, (\[Sqrt](3/2)*Abs[t])^3], t, w],
which translates to
$$\mathcal{F}_t\left[\left| t\right| ^{9/4} K_{3/4}\left(\left(\sqrt{\frac{3}{2}} \left| t\right| \right)^3\right)\right](w)$$
is equal to
$$\frac{4\ 2^{3/8} \pi  \, _1F_4\left(\frac{11}{12};\frac{1}{3},\frac{1}{2},\frac{2}{3},\frac{5}{6};-\frac{w^6}{39366}\right)}{9\ 3^{5/8} \Gamma \left(\frac{5}{12}\right)}+\frac{2\ 2^{13/24} \sqrt{\pi } w^4 \Gamma \left(\frac{19}{12}\right) \, _1F_4\left(\frac{19}{12};\frac{7}{6},\frac{4}{3},\frac{3}{2},\frac{5}{3};-\frac{w^6}{39366}\right)}{729\ 3^{5/8} \Gamma \left(\frac{7}{6}\right)}-\frac{w^2 \Gamma \left(\frac{1}{4}\right) \, _1F_4\left(\frac{5}{4};\frac{2}{3},\frac{5}{6},\frac{7}{6},\frac{4}{3};-\frac{w^6}{39366}\right)}{27 \sqrt[8]{2} 3^{5/8}}.$$
The Fourier transform in Mathematica is
$$\dfrac{1}{\sqrt{2 \pi }}\int _{-\infty }^{\infty } d t\, f(t)\, e^{i t \omega }.$$
To find that result by hand is... hard, I guess.
