DOUBLE PRECISION FUNCTION OMEGA(XI4) - Fermionic overlap

Calculation of the fermionic overlap, ${\rm OMEGA}=\omega$, as function of the conformal $I\overline{I}$-distance, ${\rm XI4}=\xi$.

The fermionic overlap has been computed in Ref. [1],

$$\omega(\xi)= \frac{6 B(\frac{3}{2},\frac{5}{2})}{z^{3/2}}\... ...e{6ex} z \equiv \frac{1}{2}\left(\xi +\sqrt{\xi^2-4}\right).$$ (1)

In the present routine the following simple, but accurate approximation for $\omega$ is used:

$$\omega (\xi )\approx \frac{4}{(\xi +1/2)^{3/2}}.$$ (2)